id stringlengths 1 6 | url stringlengths 16 1.82k | content stringlengths 37 9.64M |
|---|---|---|
189200 | https://math.stackexchange.com/questions/361953/doubling-map-and-measure | dynamical systems - Doubling Map and Measure - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Doubling Map and Measure
Ask Question
Asked 12 years, 5 months ago
Modified12 years, 5 months ago
Viewed 414 times
This question shows research effort; it is useful and clear
2
Save this question.
Show activity on this post.
First off! This is a homework question, so I DO NOT want an answer to the question I'm writing, I really just want an explanation of the final bit (which I'll make clear).
So if we have T:[0,1)→[0,1),T x=2 x mod 1 T:[0,1)→[0,1),T x=2 x mod 1 which preserves Lebesgue measure m m, then we have ϕ:[0,1)→[0,1)ϕ:[0,1)→[0,1) defined as ϕ(x)=x 3 ϕ(x)=x 3. Find
lim n→∞1 n∑j=0 n−1 ϕ(T j x)lim n→∞1 n∑j=0 n−1 ϕ(T j x)
for m m almost every point x∈[0,1)x∈[0,1).
So essentially, my question is "what does it mean 'Find __ for m almost every point x'?" My question really is that I don't know what I'm computing? Does it mean compute m(∗∗∗)?m(∗∗∗)? If that's the case, I think I have a few ideas....
dynamical-systems
classical-mechanics
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
asked Apr 15, 2013 at 3:25
SquirtleSquirtle
6,983 39 39 silver badges 66 66 bronze badges
1
Fun question. I would start by using the fact that if x≡x+2−k(mod 1)x≡x+2−k(mod 1) for some natural number k k, then the sequences (T j x)(T j x) and (T j x′)(T j x′) share the infinite tail, but I really don't see where that leads? Also if x x is rational, then the sequence is eventually periodic. But I think that I'm approaching this limit from the atypical as opposed to a.e. side :-(Jyrki Lahtonen –Jyrki Lahtonen 2013-04-15 13:58:44 +00:00 Commented Apr 15, 2013 at 13:58
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
0
Save this answer.
Show activity on this post.
In measure theory, we say a property holds "almost everywhere" if the set of elements for which it doesn't hold has measure zero. Clearly, this definition depends on the measure, so we might say "m m-almost everywhere" if it's not clear from context.
In this case, what we're saying is, find an expression f(x)f(x) such that the equation
f(x)=lim n→∞1 n∑j=0 n−1 ϕ(T j x)(1)(1)f(x)=lim n→∞1 n∑j=0 n−1 ϕ(T j x)
holds m m-almost everywhere in [0,1)[0,1), i.e. m({x∈[0,1):(1) doesn't hold})=0 m({x∈[0,1):(1) doesn't hold})=0.
Of course such an f f will be non-unique. But that's probably okay – whoever is asking the question doesn't mind which you choose.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Apr 15, 2013 at 3:37
Ben MillwoodBen Millwood
14.6k 3 3 gold badges 35 35 silver badges 62 62 bronze badges
0
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
dynamical-systems
classical-mechanics
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
9Fixed, attracting points are Fatou points
3Ergodic theory question about the support of a measure.
2Finding periodic points of diffeomorphism of the circle
4Stationary points and their behavior of : x′=y−x 3,y′=−x+y 3 x′=y−x 3,y′=−x+y 3 via Linearisation
2Manipulating the 3 x mod 1 3 x mod 1 map for a discrete 1d system
2Showing the connection between the doubling map and Bernoulli shift map.
1The binary map is surjective.
Hot Network Questions
Overfilled my oil
Who is the target audience of Netanyahu's speech at the United Nations?
Where is the first repetition in the cumulative hierarchy up to elementary equivalence?
Is direct sum of finite spectra cancellative?
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
Origin of Australian slang exclamation "struth" meaning greatly surprised
What’s the usual way to apply for a Saudi business visa from the UAE?
Cannot build the font table of Miama via nfssfont.tex
Is it safe to route top layer traces under header pins, SMD IC?
Can you formalize the definition of infinitely divisible in FOL?
What meal can come next?
Sign mismatch in overlap integral matrix elements of contracted GTFs between my code and Gaussian16 results
I have a lot of PTO to take, which will make the deadline impossible
Discussing strategy reduces winning chances of everyone!
Passengers on a flight vote on the destination, "It's democracy!"
Why include unadjusted estimates in a study when reporting adjusted estimates?
Matthew 24:5 Many will come in my name!
Analog story - nuclear bombs used to neutralize global warming
Is it ok to place components "inside" the PCB
Exchange a file in a zip file quickly
Do we need the author's permission for reference
Xubuntu 24.04 - Libreoffice
PSTricks error regarding \pst@makenotverbbox
Childhood book with a girl obsessessed with homonyms who adopts a stray dog but gives it back to its owners
more hot questions
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
189201 | https://handwiki.org/wiki/Beatty_sequence | Anonymous
Not logged in
Create account
Log in
HandWiki
Beatty sequence
From HandWiki
Namespaces
Page
Discussion
More
More
Page actions
Read
View source
History
ZWI Export
Short description: Integers formed by rounding down the integer multiples of a positive irrational number
In mathematics, a Beatty sequence (or homogeneous Beatty sequence) is the sequence of integers found by taking the floor of the positive multiples of a positive irrational number. Beatty sequences are named after Samuel Beatty, who wrote about them in 1926.
Rayleigh's theorem, named after Lord Rayleigh, states that the complement of a Beatty sequence, consisting of the positive integers that are not in the sequence, is itself a Beatty sequence generated by a different irrational number.
Beatty sequences can also be used to generate Sturmian words.
Contents
1 Definition
2 Examples
3 History
4 Rayleigh theorem
4.1 First proof
4.2 Second proof
5 Properties
5.1 Relation with Sturmian sequences
6 Generalizations
7 References
8 Further reading
9 External links
Definition
Any irrational number [math]\displaystyle{ r }[/math] that is greater than one generates the Beatty sequence [math]\displaystyle{ \mathcal{B}_r = \bigl{ \lfloor r \rfloor, \lfloor 2r \rfloor, \lfloor 3r \rfloor,\ldots \bigr} }[/math] The two irrational numbers [math]\displaystyle{ r }[/math] and [math]\displaystyle{ s = r/(r-1) }[/math] naturally satisfy the equation [math]\displaystyle{ 1/r + 1/s = 1 }[/math]. The two Beatty sequences [math]\displaystyle{ \mathcal{B}_r }[/math] and [math]\displaystyle{ \mathcal{B}_s }[/math] that they generate form a pair of complementary Beatty sequences. Here, "complementary" means that every positive integer belongs to exactly one of these two sequences.
Examples
When [math]\displaystyle{ r }[/math] is the golden ratio [math]\displaystyle{ r=(1+\sqrt5)/2\approx 1.618 }[/math], the complementary Beatty sequence is generated by [math]\displaystyle{ s=r+1=(3+\sqrt5)/2\approx 2.618 }[/math]. In this case, the sequence [math]\displaystyle{ ( \lfloor nr \rfloor) }[/math], known as the lower Wythoff sequence, is
1, 3, 4, 6, 8, 9, 11, 12, 14, 16, 17, 19, 21, 22, 24, 25, 27, 29, ... (sequence A000201 in the OEIS),
and the complementary sequence [math]\displaystyle{ ( \lfloor ns \rfloor) }[/math], the upper Wythoff sequence, is
2, 5, 7, 10, 13, 15, 18, 20, 23, 26, 28, 31, 34, 36, 39, 41, 44, 47, ... (sequence A001950 in the OEIS).
These sequences define the optimal strategy for Wythoff's game, and are used in the definition of the Wythoff array.
As another example, for the square root of 2, [math]\displaystyle{ r=\sqrt2\approx 1.414 }[/math], [math]\displaystyle{ s=2+\sqrt2\approx 3.414 }[/math]. In this case, the sequences are
1, 2, 4, 5, 7, 8, 9, 11, 12, 14, 15, 16, 18, 19, 21, 22, 24, ... (sequence A001951 in the OEIS), and
3, 6, 10, 13, 17, 20, 23, 27, 30, 34, 37, 40, 44, 47, 51, 54, 58, ... (sequence A001952 in the OEIS).
For [math]\displaystyle{ r=\pi\approx 3.142 }[/math] and [math]\displaystyle{ s=\pi/(\pi-1)\approx 1.467 }[/math], the sequences are
3, 6, 9, 12, 15, 18, 21, 25, 28, 31, 34, 37, 40, 43, 47, 50, 53, ... (sequence A022844 in the OEIS), and
1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 24, 26, ... (sequence A054386 in the OEIS).
Any number in the first sequence is absent in the second, and vice versa.
History
Beatty sequences got their name from the problem posed in The American Mathematical Monthly by Samuel Beatty in 1926. It is probably one of the most often cited problems ever posed in the Monthly. However, even earlier, in 1894 such sequences were briefly mentioned by Lord Rayleigh in the second edition of his book The Theory of Sound.
Rayleigh theorem
Rayleigh's theorem (also known as Beatty's theorem) states that given an irrational number [math]\displaystyle{ r \gt 1 \,, }[/math] there exists [math]\displaystyle{ s \gt 1 }[/math] so that the Beatty sequences [math]\displaystyle{ \mathcal{B}_r }[/math] and [math]\displaystyle{ \mathcal{B}_s }[/math] partition the set of positive integers: each positive integer belongs to exactly one of the two sequences.
First proof
Given [math]\displaystyle{ r \gt 1 \,, }[/math] let [math]\displaystyle{ s = r/(r-1) }[/math]. We must show that every positive integer lies in one and only one of the two sequences [math]\displaystyle{ \mathcal{B}_r }[/math] and [math]\displaystyle{ \mathcal{B}_s }[/math]. We shall do so by considering the ordinal positions occupied by all the fractions [math]\displaystyle{ j/r }[/math] and [math]\displaystyle{ k/s }[/math] when they are jointly listed in nondecreasing order for positive integers j and k.
To see that no two of the numbers can occupy the same position (as a single number), suppose to the contrary that [math]\displaystyle{ j/r = k/s }[/math] for some j and k. Then [math]\displaystyle{ r/s }[/math] = [math]\displaystyle{ j/k }[/math], a rational number, but also, [math]\displaystyle{ r/s = r(1 - 1/r) = r - 1, }[/math] not a rational number. Therefore, no two of the numbers occupy the same position.
For any [math]\displaystyle{ j/r }[/math], there are [math]\displaystyle{ j }[/math] positive integers [math]\displaystyle{ i }[/math] such that [math]\displaystyle{ i/r \le j/r }[/math] and [math]\displaystyle{ \lfloor js/r \rfloor }[/math] positive integers [math]\displaystyle{ k }[/math] such that [math]\displaystyle{ k/s \le j/r }[/math], so that the position of [math]\displaystyle{ j/r }[/math] in the list is [math]\displaystyle{ j + \lfloor js/r \rfloor }[/math]. The equation [math]\displaystyle{ 1/r + 1/s = 1 }[/math] implies [math]\displaystyle{ j + \lfloor js/r \rfloor = j + \lfloor j(s - 1) \rfloor = \lfloor js \rfloor. }[/math]
Likewise, the position of [math]\displaystyle{ k/s }[/math] in the list is [math]\displaystyle{ \lfloor kr \rfloor }[/math].
Conclusion: every positive integer (that is, every position in the list) is of the form [math]\displaystyle{ \lfloor nr \rfloor }[/math] or of the form [math]\displaystyle{ \lfloor ns \rfloor }[/math], but not both. The converse statement is also true: if p and q are two real numbers such that every positive integer occurs precisely once in the above list, then p and q are irrational and the sum of their reciprocals is 1.
Second proof
Collisions: Suppose that, contrary to the theorem, there are integers j > 0 and k and m such that [math]\displaystyle{ j = \left\lfloor {k \cdot r} \right\rfloor = \left\lfloor {m \cdot s} \right\rfloor \,. }[/math] This is equivalent to the inequalities [math]\displaystyle{ j \le k \cdot r \lt j + 1 \text{ and } j \le m \cdot s \lt j + 1. }[/math]
For non-zero j, the irrationality of r and s is incompatible with equality, so [math]\displaystyle{ j \lt k \cdot r \lt j + 1 \text{ and } j \lt m \cdot s \lt j + 1, }[/math] which leads to [math]\displaystyle{ {j \over r} \lt k \lt {j + 1 \over r} \text{ and } {j \over s} \lt m \lt {j + 1 \over s}. }[/math]
Adding these together and using the hypothesis, we get [math]\displaystyle{ j \lt k + m \lt j + 1 }[/math] which is impossible (one cannot have an integer between two adjacent integers). Thus the supposition must be false.
Anti-collisions: Suppose that, contrary to the theorem, there are integers j > 0 and k and m such that [math]\displaystyle{ k \cdot r \lt j \text{ and } j + 1 \le (k + 1) \cdot r \text{ and } m \cdot s \lt j \text{ and } j + 1 \le (m + 1) \cdot s \,. }[/math]
Since j + 1 is non-zero and r and s are irrational, we can exclude equality, so [math]\displaystyle{ k \cdot r \lt j \text{ and } j + 1 \lt (k + 1) \cdot r \text{ and } m \cdot s \lt j \text{ and } j + 1 \lt (m + 1) \cdot s. }[/math]
Then we get [math]\displaystyle{ k \lt {j \over r} \text{ and } {j + 1 \over r} \lt k + 1 \text{ and } m \lt {j \over s} \text{ and } {j + 1 \over s} \lt m + 1 }[/math]
Adding corresponding inequalities, we get [math]\displaystyle{ k + m \lt j \text{ and } j + 1 \lt k + m + 2 }[/math] [math]\displaystyle{ k + m \lt j \lt k + m + 1 }[/math]
which is also impossible. Thus the supposition is false.
Properties
A number [math]\displaystyle{ m }[/math] belongs to the Beatty sequence [math]\displaystyle{ \mathcal{B}_r }[/math] if and only if [math]\displaystyle{ 1 - \frac{1}{r} \lt \left[ \frac{m}{r} \right]_1 }[/math] where [math]\displaystyle{ [x]_1 }[/math] denotes the fractional part of [math]\displaystyle{ x }[/math] i.e., [math]\displaystyle{ [x]_1 = x - \lfloor x \rfloor }[/math].
Proof: [math]\displaystyle{ m \in B_r }[/math] [math]\displaystyle{ \Leftrightarrow \exists n, m = \lfloor nr \rfloor }[/math] [math]\displaystyle{ \Leftrightarrow m \lt nr \lt m + 1 }[/math] [math]\displaystyle{ \Leftrightarrow \frac{m}{r} \lt n \lt \frac{m}{r} + \frac{1}{r} }[/math] [math]\displaystyle{ \Leftrightarrow n - \frac{1}{r} \lt \frac{m}{r} \lt n }[/math] [math]\displaystyle{ \Leftrightarrow 1 - \frac{1}{r} \lt \left[ \frac{m}{r} \right]_1 }[/math]
Furthermore, [math]\displaystyle{ m = \left\lfloor \left( \left\lfloor \frac{m}{r} \right\rfloor + 1 \right) r \right\rfloor }[/math].
Proof: [math]\displaystyle{ m = \left\lfloor \left( \left\lfloor \frac{m}{r} \right\rfloor + 1 \right) r \right\rfloor }[/math] [math]\displaystyle{ \Leftrightarrow m \lt \left( \left\lfloor \frac{m}{r} \right\rfloor + 1 \right) r \lt m + 1 }[/math] [math]\displaystyle{ \Leftrightarrow \frac{m}{r} \lt \left\lfloor \frac{m}{r} \right\rfloor + 1 \lt \frac{m + 1}{r} }[/math] [math]\displaystyle{ \Leftrightarrow \left\lfloor \frac{m}{r} \right\rfloor + 1 - \frac{1}{r} \lt \frac{m}{r} \lt \left\lfloor \frac{m}{r} \right\rfloor + 1 }[/math] [math]\displaystyle{ \Leftrightarrow 1 - \frac{1}{r} \lt \frac{m}{r} - \left\lfloor \frac{m}{r} \right\rfloor =\left[ \frac{m}{r} \right]_1 }[/math]
Relation with Sturmian sequences
The first difference [math]\displaystyle{ \lfloor (n+1)r\rfloor-\lfloor nr\rfloor }[/math] of the Beatty sequence associated with the irrational number [math]\displaystyle{ r }[/math] is a characteristic Sturmian word over the alphabet [math]\displaystyle{ {\lfloor r\rfloor,\lfloor r\rfloor+1} }[/math].
Generalizations
If slightly modified, the Rayleigh's theorem can be generalized to positive real numbers (not necessarily irrational) and negative integers as well: if positive real numbers [math]\displaystyle{ r }[/math] and [math]\displaystyle{ s }[/math] satisfy [math]\displaystyle{ 1/r + 1/s = 1 }[/math], the sequences [math]\displaystyle{ ( \lfloor mr \rfloor)_{m \in \mathbb{Z}} }[/math] and [math]\displaystyle{ ( \lceil ns \rceil -1)_{n \in \mathbb{Z}} }[/math] form a partition of integers. For example, the white and black keys of a piano keyboard are distributed as such sequences for [math]\displaystyle{ r = 12/7 }[/math] and [math]\displaystyle{ s = 12/5 }[/math].
The Lambek–Moser theorem generalizes the Rayleigh theorem and shows that more general pairs of sequences defined from an integer function and its inverse have the same property of partitioning the integers.
Uspensky's theorem states that, if [math]\displaystyle{ \alpha_1,\ldots,\alpha_n }[/math] are positive real numbers such that [math]\displaystyle{ (\lfloor k\alpha_i\rfloor)_{k,i\ge1} }[/math] contains all positive integers exactly once, then [math]\displaystyle{ n\le2. }[/math] That is, there is no equivalent of Rayleigh's theorem for three or more Beatty sequences.
References
↑ Beatty, Samuel (1926). "Problem 3173". American Mathematical Monthly 33 (3): 159. doi:10.2307/2300153.
↑ S. Beatty; A. Ostrowski; J. Hyslop; A. C. Aitken (1927). "Solutions to Problem 3173". American Mathematical Monthly 34 (3): 159–160. doi:10.2307/2298716.
↑ 3.0 3.1 John William Strutt, 3rd Baron Rayleigh (1894). The Theory of Sound. 1 (Second ed.). Macmillan. p. 123.
↑ J. V. Uspensky, On a problem arising out of the theory of a certain game, Amer. Math. Monthly 34 (1927), pp. 516–521.
↑ R. L. Graham, On a theorem of Uspensky, Amer. Math. Monthly 70 (1963), pp. 407–409.
Further reading
Holshouser, Arthur; Reiter, Harold (2001). "A generalization of Beatty's Theorem". Southwest Journal of Pure and Applied Mathematics 2: 24–29. Archived from the original on 2014-04-19.
Stolarsky, Kenneth (1976). "Beatty sequences, continued fractions, and certain shift operators". Canadian Mathematical Bulletin 19 (4): 473–482. doi:10.4153/CMB-1976-071-6. Includes many references.
External links
Weisstein, Eric W.. "Beatty Sequence".
Alexander Bogomolny, Beatty Sequences, Cut-the-knot
| |
| 0.00 (0 votes) Original source: sequence. Read more |
Retrieved from "
Categories:
Integer sequences
Theorems in number theory
Diophantine approximation
Articles containing proofs
Computing portal
Encyclopedia of Knowledge
Main page Data analysis Astronomy & Space Biology Computer concepts Chemistry Mathematics Physics Earth studies Unsolved problems History Philosophy Social studies Religion Medicine Engineering & Tech Software programs Finance & Business Biographies Organizations Companies Places Books Monographs Tutorials Manuals
Navigation
Navigation
Add a new article
Search in all topics
Search in namespaces
Search in categories
Search using prefix
Help
About HandWiki
FAQs
How to edit
Citation manager
Formatting articles
List of categories
Recent pages
Recent changes
Random page
Support & Donate
Translate
Wiki tools
Wiki tools
Special pages
Cite this page
Page tools
Page tools
User page tools
More
What links here
Related changes
Printable version
Permanent link
Page information
Page logs
Other projects
In other languages
Add links
Categories
Categories
Integer sequences
Theorems in number theory
Diophantine approximation
Articles containing proofs
This page was last edited on 6 February 2024, at 20:09.
Privacy policy
About HandWiki
Disclaimers |
189202 | https://youglish.com/pronounce/relinquish/english | Relinquish | 603 pronunciations of Relinquish in English
Toggle navigation
Login
Sign up
Daily Lessons
Submit
Get your widget
Donate!
for
English▼
• Arabic
• Chinese
• Dutch
• English
• French
• German
• Greek
• Hebrew
• Italian
• Japanese
• Korean
• Polish
• Portuguese
• Romanian
• Russian
• Spanish
• Swedish
• Thai
• Turkish
• Ukrainian
• Vietnamese
• Sign Languages
Say it!
All US UK AUS CAN IE SCO NZ
[x] All - [x] United States - [x] United Kingdom - [x] Australia - [x] Canada - [x] Ireland - [x] Scotland - [x] New Zealand
Close
How to pronounce relinquish in English (1 out of 603):
Speed:
arrow_drop_down
Normal
arrow_drop_up
vertical_align_top
emoji_objects
settings
▼
▲
↻
↻
U
×
that claimed to relinquish her throne, and they imprisoned her in the palace.
•••
[Feedback]
[Share]
[Save]
[Record]
[YouTube]
[G. TranslateB. TranslateDeepLReverso▼]
Google Translate
Bing Translate
DeepL
Reverso
Definition:
Click on any word below to get its definition:
that
claimed
to
relinquish
her
throne
and
they
imprisoned
her
in
the
palace
Discover more
Accent reduction software
Buy English grammar book
Buy English phonetics guide
Vocabulary building apps
Online English classes
Get English language assessment
Find English vocabulary builder
Foreign language courses
Purchase English dictionary subscription
Get professional language services
Nearby words:
Having trouble pronouncing 'relinquish' ? Learn how to pronounce one of the nearby words below:
relationship
related
relationships
religious
religion
release
relatively
relevant
released
relations
relative
relate
rely
relief
relation
relates
reliable
relax
religions
relatives
releasing
releases
relaxed
relating
relevance
relying
reluctant
reliability
relativity
relational
Discover more
Accent reduction software
Translation services
Word games
Foreign language courses
Language learning apps
Buy language learning app
Get English language assessment
Language learning platforms
Get professional language services
Online English classes
Phonetic:
When you begin to speak English, it's essential to get used to the common sounds of the language, and the best way to do this is to check out the phonetics. Below is the UK transcription for 'relinquish':
Modern IPA: rɪlɪ́ŋkwɪʃ
Traditional IPA: rɪˈlɪŋkwɪʃ
3 syllables: "ri" + "LIN" + "kwish"
Test your pronunciation on words that have sound similarities with 'relinquish':
relinquished
relinquishes
relinquishing
trilingual
reliquary
ruling class
bilingual
treblinka
helsinki
Discover more
English pronunciation course
English conversation partners
Daily pronunciation lessons
Language learning apps
Cultural immersion experiences
Contextual word usage
Language learning games
TOEFL IELTS preparation materials
High-quality headphones
English dictionary
Tips to improve your English pronunciation:
Here are a few tips that should help you perfect your pronunciation of 'relinquish':
Sound it Out: Break down the word 'relinquish' into its individual sounds "ri" + "lin" + "kwish". Say these sounds out loud, exaggerating them at first. Practice until you can consistently produce them clearly.
Self-Record & Review: Record yourself saying 'relinquish' in sentences. Listen back to identify areas for improvement.
YouTube Pronunciation Guides: Search YouTube for how to pronounce 'relinquish' in English.
Pick Your Accent: Mixing multiple accents can be confusing, so pick one accent (US or UK) and stick to it for smoother learning.
Here are a few tips to level up your english pronunciation:
Mimic the Experts: Immerse yourself in English by listening to audiobooks, podcasts, or movies with subtitles. Try shadowing—listen to a short sentence and repeat it immediately, mimicking the intonation and pronunciation.
Become Your Own Pronunciation Coach: Record yourself speaking English and listen back. Identify areas for improvement, focusing on clarity, word stress, and intonation.
Train Your Ear with Minimal Pairs: Practice minimal pairs (words that differ by only one sound, like ship vs. sheep) to improve your ability to distinguish between similar sounds.
Explore Online Resources: Websites & apps offer targeted pronunciation exercises. Explore YouTube channels dedicated to pronunciation, like Rachel's English and English with James for additional pronunciation practice and learning.
Discover more
Language learning platforms
Foreign language courses
Buy English pronunciation course
Accent reduction software
Speech therapy tools
Vocabulary building apps
English pronunciation apps
Purchase English practice materials
Purchase English audio lessons
Find online language tutor
YouGlish for:
Arabic
Chinese
Dutch
English
French
German
Greek
Hebrew
Italian
Japanese
Korean
Polish
Portuguese
Romanian
Russian
Spanish
Swedish
Thai
Turkish
Ukrainian
Vietnamese
Sign Languages
Choose your language:
English
Français
Español
Italiano
Português
Deutsch
العربية
HOME
ABOUT
CONTACT
PRIVACY & TERMS
SETTINGS
API
BROWSE
CONTRIBUTE
×Close
■Definitions■Synonyms■Usages■Translations
Translate to :
Close More
Dictionary not available Known issues Mother tongue required Content quota exceeded Subscription expired Subscription suspended Feature not available Login is required
A dictionary is not available for this language at this time. Buttons are not activated? Click here to see possible solutions. Login is required to perform this operation. Login here or sign up. Go to settings to set it up. This feature is currently not available, please try again later. Upgrade your subscription to increase your quota. Your subscription expired. Renew your subscription to get full access to your content. Your subscription is suspended. Contact us for more information.
CLOSE |
189203 | https://www.dlubal.com/es/soluciones/servicios-en-linea/wiki-de-analisis-de-estructuras/000171?srsltid=AfmBOor005tLesU3R5UMJlicTbDg7iLo9gnH9aAHqebkxK9gay_eF7E6 | Módulo de cortadura G – Definición y fórmula
Consent
Details
[#IABV2SETTINGS#]
About
This website uses cookies
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners who may combine it with other information that you’ve provided to them or that they’ve collected from your use of their services.
Consent Selection
Necessary
[x]
Preferences
[x]
Statistics
[x]
Marketing
[x]
Show details
Details
Necessary 17- [x] Necessary cookies help make a website usable by enabling basic functions like page navigation and access to secure areas of the website. The website cannot function properly without these cookies.
Cookiebot 1Learn more about this provider1.gifUsed to count the number of sessions to the website, necessary for optimizing CMP product delivery. Maximum Storage Duration: SessionType: Pixel Tracker
Dlubal 7Learn more about this providerthemeThis cookie is part of a bundle of cookies which serve the purpose of content delivery and presentation. The cookies keep the correct state of font, blog/picture sliders, color themes and other website settings.Maximum Storage Duration: PersistentType: HTML Local Storage ASP.NET_SessionId[x3]Preserves the visitor's session state across page requests.Maximum Storage Duration: SessionType: HTTP Cookie .AspNetCore.Antiforgery.#Helps prevent Cross-Site Request Forgery (CSRF) attacks.Maximum Storage Duration: SessionType: HTTP Cookie cookietestThis cookie is used to determine if the visitor has accepted the cookie consent box.Maximum Storage Duration: SessionType: HTTP Cookie PHPSESSIDPreserves user session state across page requests.Maximum Storage Duration: SessionType: HTTP Cookie
Google 4Learn more about this providerSome of the data collected by this provider is for the purposes of personalization and measuring advertising effectiveness.
rc::aThis cookie is used to distinguish between humans and bots. This is beneficial for the website, in order to make valid reports on the use of their website.Maximum Storage Duration: PersistentType: HTML Local Storage rc::bThis cookie is used to distinguish between humans and bots. Maximum Storage Duration: SessionType: HTML Local Storage rc::cThis cookie is used to distinguish between humans and bots. Maximum Storage Duration: SessionType: HTML Local Storage rc::fThis cookie is used to distinguish between humans and bots. Maximum Storage Duration: PersistentType: HTML Local Storage
apidocs.dlubal.com consent.cookiebot.com 4CookieConsent[x4]Stores the user's cookie consent state for the current domainMaximum Storage Duration: 1 yearType: HTTP Cookie
www.recaptcha.net 1_GRECAPTCHAThis cookie is used to distinguish between humans and bots. This is beneficial for the website, in order to make valid reports on the use of their website.Maximum Storage Duration: 180 daysType: HTTP Cookie
Preferences 0- [x] Preference cookies enable a website to remember information that changes the way the website behaves or looks, like your preferred language or the region that you are in.
We do not use cookies of this type.
Statistics 2- [x] Statistic cookies help website owners to understand how visitors interact with websites by collecting and reporting information anonymously.
Microsoft 2Learn more about this provider_clckCollects data on the user’s navigation and behavior on the website. This is used to compile statistical reports and heatmaps for the website owner.Maximum Storage Duration: SessionType: HTTP Cookie _clskRegisters statistical data on users' behaviour on the website. Used for internal analytics by the website operator. Maximum Storage Duration: SessionType: HTTP Cookie
Marketing 27- [x] Marketing cookies are used to track visitors across websites. The intention is to display ads that are relevant and engaging for the individual user and thereby more valuable for publishers and third party advertisers.
Google 5Learn more about this providerSome of the data collected by this provider is for the purposes of personalization and measuring advertising effectiveness.
_gaUsed to send data to Google Analytics about the visitor's device and behavior. Tracks the visitor across devices and marketing channels.Maximum Storage Duration: 2 yearsType: HTTP Cookie ga#Used to send data to Google Analytics about the visitor's device and behavior. Tracks the visitor across devices and marketing channels.Maximum Storage Duration: 2 yearsType: HTTP Cookie _gatUsed to send data to Google Analytics about the visitor's device and behavior. Tracks the visitor across devices and marketing channels.Maximum Storage Duration: 1 dayType: HTTP Cookie _gidUsed to send data to Google Analytics about the visitor's device and behavior. Tracks the visitor across devices and marketing channels.Maximum Storage Duration: 1 dayType: HTTP Cookie NIDRegisters a unique ID that identifies a returning user's device. The ID is used for targeted ads.Maximum Storage Duration: 6 monthsType: HTTP Cookie
YouTube 22Learn more about this provider#-#Used to track user’s interaction with embedded content.Maximum Storage Duration: SessionType: HTML Local Storage __Secure-ROLLOUT_TOKENUsed to track user’s interaction with embedded content.Maximum Storage Duration: 180 daysType: HTTP Cookie __Secure-YECStores the user's video player preferences using embedded YouTube videoMaximum Storage Duration: SessionType: HTTP Cookie __Secure-YNIDPendingMaximum Storage Duration: 180 daysType: HTTP Cookie iU5q-!O9@$Registers a unique ID to keep statistics of what videos from YouTube the user has seen.Maximum Storage Duration: SessionType: HTML Local Storage LAST_RESULT_ENTRY_KEYUsed to track user’s interaction with embedded content.Maximum Storage Duration: SessionType: HTTP Cookie LogsDatabaseV2:V#||LogsRequestsStoreUsed to track user’s interaction with embedded content.Maximum Storage Duration: PersistentType: IndexedDB remote_sidNecessary for the implementation and functionality of YouTube video-content on the website. Maximum Storage Duration: SessionType: HTTP Cookie ServiceWorkerLogsDatabase#SWHealthLogNecessary for the implementation and functionality of YouTube video-content on the website. Maximum Storage Duration: PersistentType: IndexedDB TESTCOOKIESENABLEDUsed to track user’s interaction with embedded content.Maximum Storage Duration: 1 dayType: HTTP Cookie VISITOR_INFO1_LIVETries to estimate the users' bandwidth on pages with integrated YouTube videos.Maximum Storage Duration: 180 daysType: HTTP Cookie youtube.comPendingMaximum Storage Duration: 1 yearType: HTTP Cookie YSCRegisters a unique ID to keep statistics of what videos from YouTube the user has seen.Maximum Storage Duration: SessionType: HTTP Cookie ytidb::LAST_RESULT_ENTRY_KEYUsed to track user’s interaction with embedded content.Maximum Storage Duration: PersistentType: HTML Local Storage YtIdbMeta#databasesUsed to track user’s interaction with embedded content.Maximum Storage Duration: PersistentType: IndexedDB yt-remote-cast-availableStores the user's video player preferences using embedded YouTube videoMaximum Storage Duration: SessionType: HTML Local Storage yt-remote-cast-installedStores the user's video player preferences using embedded YouTube videoMaximum Storage Duration: SessionType: HTML Local Storage yt-remote-connected-devicesStores the user's video player preferences using embedded YouTube videoMaximum Storage Duration: PersistentType: HTML Local Storage yt-remote-device-idStores the user's video player preferences using embedded YouTube videoMaximum Storage Duration: PersistentType: HTML Local Storage yt-remote-fast-check-periodStores the user's video player preferences using embedded YouTube videoMaximum Storage Duration: SessionType: HTML Local Storage yt-remote-session-appStores the user's video player preferences using embedded YouTube videoMaximum Storage Duration: SessionType: HTML Local Storage yt-remote-session-nameStores the user's video player preferences using embedded YouTube videoMaximum Storage Duration: SessionType: HTML Local Storage
Unclassified 15 Unclassified cookies are cookies that we are in the process of classifying, together with the providers of individual cookies.
Dlubal 13Learn more about this providermodePendingMaximum Storage Duration: PersistentType: HTML Local Storage com.grafana.faro.sessionPendingMaximum Storage Duration: SessionType: HTML Local Storage .DlubalWeb.SessionPendingMaximum Storage Duration: SessionType: HTTP Cookie iwPendingMaximum Storage Duration: 30 daysType: HTTP Cookie LoadZoneMaps_WebSessionPendingMaximum Storage Duration: 1 yearType: HTTP Cookie NG03092025PendingMaximum Storage Duration: 7 daysType: HTTP Cookie vbdiscussion_viewPendingMaximum Storage Duration: 1 yearType: HTTP Cookie vblastactivityPendingMaximum Storage Duration: 1 yearType: HTTP Cookie vblastvisitPendingMaximum Storage Duration: 1 yearType: HTTP Cookie vbnp_notices_displayedPendingMaximum Storage Duration: 1 yearType: HTTP Cookie vboptionsPendingMaximum Storage Duration: PersistentType: HTML Local Storage vbphrasePendingMaximum Storage Duration: PersistentType: HTML Local Storage vbsessionhashPendingMaximum Storage Duration: SessionType: HTTP Cookie
mia.dlubal.com unpkg.com 2com.grafana.faro.lastNavigationId[x2]PendingMaximum Storage Duration: SessionType: HTML Local Storage
Cross-domain consent[#BULK_CONSENT_DOMAINS_COUNT#] [#BULK_CONSENT_TITLE#]
List of domains your consent applies to: [#BULK_CONSENT_DOMAINS#]
Cookie declaration last updated on 9/3/25 by Cookiebot
[#IABV2_TITLE#]
[#IABV2_BODY_INTRO#]
[#IABV2_BODY_LEGITIMATE_INTEREST_INTRO#]
[#IABV2_BODY_PREFERENCE_INTRO#]
[#IABV2_LABEL_PURPOSES#]
[#IABV2_BODY_PURPOSES_INTRO#]
[#IABV2_BODY_PURPOSES#]
[#IABV2_LABEL_FEATURES#]
[#IABV2_BODY_FEATURES_INTRO#]
[#IABV2_BODY_FEATURES#]
[#IABV2_LABEL_PARTNERS#]
[#IABV2_BODY_PARTNERS_INTRO#]
[#IABV2_BODY_PARTNERS#]
About
Cookies are small text files that can be used by websites to make a user's experience more efficient.
The law states that we can store cookies on your device if they are strictly necessary for the operation of this site. For all other types of cookies we need your permission.
This site uses different types of cookies. Some cookies are placed by third party services that appear on our pages.
You can at any time change or withdraw your consent from the Cookie Declaration on our website.
Learn more about who we are, how you can contact us and how we process personal data in our Privacy Policy.
Please state your consent ID and date when you contact us regarding your consent.
[x] Do not sell or share my personal information
Use necessary cookies only Allow selection Customize Allow all cookies
Community
Regístrese en la extranet de Dlubal para obtener el máximo provecho del software y tener acceso exclusivo a sus datos personales.
Iniciar sesiónCrear una cuenta
0
Tienda en línea
Contacte al equipo de ventas
Español Español
Idioma
Deutsch
English
Français
Español
Português
Italiano
Česky
Polski
Pусский
中文(简体)
Soluciones
Sectores
Áreas de aplicación
Normas de cálculo de estructuras implementadas
Servicios en línea
Sectores
Básico
Estructuras de hormigón armado
Estructuras de hormigón pretensado
Estructuras de acero
Estructuras de madera
Estructuras de fábrica
Estructuras ligeras y de aluminio
Edificios
Estructuras industriales
Tuberías
Estructuras de puentes
Grúas y caminos de rodadura para grúas
Torres y mástiles
Estructuras de vidrio
Estructuras con membranas tensadas
Estructuras con cables y tensadas
Estructuras laminadas y tipo sándwich
Estructuras solares y sistemas de montaje
Especial
Estructuras temporales
Estructuras de andamios y estanterías
Estructuras costa afuera
Silos y tanques de almacenamiento
Instalaciones de energías renovables
Construcción naval y cuerpos flotantes
Estructuras de transportadoras
Estructuras de perforación
Piscinas y parques acuáticos
Estructuras de contenedores
Cimentaciones con pilotes perforados
Estructuras de escaleras
Centrales eléctricas
Ingeniería hidráulica de acero
Ingeniería mecánica
Estructuras infladas con aire
Enlaces rápidos | Soluciones
Herramienta de zonas geográficas
Simulación de viento y generación de cargas de viento en estructuras
Conexiones de acero
Software para análisis y diseño de estructuras de acero
Software para la búsqueda de forma y patrones de corte de estructuras de membrana
Edificios
Análisis dinámico y sísmico
Áreas de aplicación
Básico
Ingeniería estructural
Software de cálculo por elementos finitos (MEF)
Simulación de viento y generación de cargas de viento
Análisis de tensiones
Análisis no lineal
Análisis de estabilidad
Análisis no lineal de pandeo
Análisis de torsión de alabeo
Análisis dinámico y sísmico
Análisis dinámico no lineal
Análisis con el método del empuje incremental ("pushover")
Búsqueda de forma y patrones de corte
Uniones de acero
Planificación orientada al BIM
Especial
Cálculo de fábrica
Secciones conformadas en frío
Interacción suelo-estructura
Análisis de estructuras de fachadas
Cimentaciones e ingeniería de cimentaciones
Fases de construcción
Resistencia al fuego
Enlaces rápidos | Soluciones
Herramienta de zonas geográficas
Simulación de viento y generación de cargas de viento en estructuras
Conexiones de acero
Software para análisis y diseño de estructuras de acero
Software para la búsqueda de forma y patrones de corte de estructuras de membrana
Edificios
Análisis dinámico y sísmico
Normas de cálculo de estructuras implementadas
Eurocódigos (EC)
Normas alemanas (DIN)
Normas británicas (BS EN, BS)
Normas italianas (NTC)
Normas estadounidenses (AISC, ACI, AWC, ADM, ASCE 7 e IBC)
Normas canadienses (CSA)
Normas australianas (AS)
Normas suizas (SIA)
Normas chinas (GB, HK)
Normas de India (IS)
Normas mexicanas (RCDF, CFE Sismo 08)
Normas rusas (SP)
Normas sudafricanas (SANS)
Normas brasileñas (NBR)
Enlaces rápidos | Soluciones
Herramienta de zonas geográficas
Simulación de viento y generación de cargas de viento en estructuras
Conexiones de acero
Software para análisis y diseño de estructuras de acero
Software para la búsqueda de forma y patrones de corte de estructuras de membrana
Edificios
Análisis dinámico y sísmico
Servicios en línea
Mapas de cargas de nieve, velocidad del viento y cargas sísmicas
Cálculos en la nube
Wiki sobre análisis y dimensionamiento estructural
Propiedades de secciones de acero Calcular secciones de acero Valores de las secciones de
Enlaces rápidos | Soluciones
Herramienta de zonas geográficas
Simulación de viento y generación de cargas de viento en estructuras
Conexiones de acero
Software para análisis y diseño de estructuras de acero
Software para la búsqueda de forma y patrones de corte de estructuras de membrana
Edificios
Análisis dinámico y sísmico
Productos
Análisis por elementos finitos (AEF)
Análisis de estructuras de barras
Propiedades de secciones transversales
Simulación de viento
API de Dlubal
Productos antiguos
RFEM 6
RFEM 6 forma la base de la familia de programas modulares y se utiliza para la definición de estructuras, materiales y cargas para estructuras de placas, discos, láminas y barras, así como para elementos volumétricos y de contacto.
Complementos para RFEM 6
Análisis adicionalesAnálisis dinámicoSoluciones especialesCálculo y dimensionamientoUniones
Enlaces rápidos | Productos
Uniones de acero para RFEM 6
Programa independiente RWIND 3
Análisis con el método del empuje incremental ("pushover")
Modelo de edificio para RFEM 6
RSTAB 9
RSTAB 9 es un software potente de análisis y dimensionamiento en 3D de estructuras de vigas, pórticos o cerchas, que refleja el estado de la técnica actual y ayuda a los ingenieros y consultores de estructuras a cumplir con los requisitos de la ingeniería de estructuras moderna.
Complementos para RSTAB 9
Análisis adicionales para RSTAB 9Análisis dinámicoSoluciones especialesCálculo
Enlaces rápidos | Productos
Uniones de acero para RFEM 6
Programa independiente RWIND 3
Análisis con el método del empuje incremental ("pushover")
Modelo de edificio para RFEM 6
RSECTION 1
RSECTION apoya a los ingenieros estructurales determinando las propiedades de la sección para una amplia variedad de secciones transversales y permite un análisis de tensiones posterior.
Complemento para RSECTION
Secciones eficaces
Enlaces rápidos | Productos
Uniones de acero para RFEM 6
Programa independiente RWIND 3
Análisis con el método del empuje incremental ("pushover")
Modelo de edificio para RFEM 6
RWIND 3
RWIND 3 es un túnel de viento digital para la simulación de flujos de viento alrededor de geometrías de edificios arbitrarias y para el cálculo de las cargas de viento sobre sus superficies.
Enlaces rápidos | Productos
Uniones de acero para RFEM 6
Programa independiente RWIND 3
Análisis con el método del empuje incremental ("pushover")
Modelo de edificio para RFEM 6
API de Dlubal
El nuevo servicio API de Dlubal (gRPC) le ofrece una interfaz flexible para el software de estática basado en Python y C#, con acceso directo a toda la gama de productos de Dlubal. Benefíciese de una integración fluida y potente en su software Dlubal, ideal para la modelización paramétrica y tareas de optimización complejas.
Documentación de API
ÍndicePrimeros pasosAplicacionesObjetos del modeloSuscripciones y preciosEjemplos
Enlaces rápidos | Productos
Uniones de acero para RFEM 6
Programa independiente RWIND 3
Análisis con el método del empuje incremental ("pushover")
Modelo de edificio para RFEM 6
Productos antiguos
RFEM 5 | Software de análisis de estructuras por elementos finitos
RSTAB 8 | Software de análisis de estructuras de barras y cerchas
Módulos adicionales para RFEM 5
Programa para propiedades de secciones
Módulos adicionales para RSTAB 8
Programas independientes
Enlaces rápidos | Productos
Uniones de acero para RFEM 6
Programa independiente RWIND 3
Análisis con el método del empuje incremental ("pushover")
Modelo de edificio para RFEM 6
Apoyo
Soporte
Servicio
Asistente de soporte de IA
Ventas
Formación
Soporte
Preguntas frecuentes (FAQ)
Base de conocimientos
Características de los productos
Licencias
Formular una pregunta
Nuestro equipo de soporte
Enviar característica o idea del programa
Preguntas frecuentes sobre licencias y autorizaciones
Informar de algún problema o error en el programa
Actualizaciones del programa
Problemas del programa
Fórmulas | ¡Las matemáticas son divertidas!
Enlaces rápidos | Soporte y formación
Formular una pregunta
Cursos de formación en línea
Contactar con nuestro equipo de ventas
Vídeos
Servicio
Soporte técnico y servicio gratuitos
Extranet | Mi cuenta
Contrato de servicio
Herramienta de zonas geográficas para la determinación de cargas
Actualizaciones y nuevas versiones
Versiones anteriores de los programas
Enlaces rápidos | Soporte y formación
Formular una pregunta
Cursos de formación en línea
Contactar con nuestro equipo de ventas
Vídeos
Asistente de soporte de IA
Mia, su asistente de inteligencia artificial las 24 horas
Descubra su asistente personal de IA
Enlaces rápidos | Soporte y formación
Formular una pregunta
Cursos de formación en línea
Contactar con nuestro equipo de ventas
Vídeos
Ventas
Tienda en línea
Nuestro equipo de ventas
Contactar con nuestro equipo de ventas
Solicitar demostración de producto en línea
¿Por qué Dlubal Software?
Enlaces rápidos | Soporte y formación
Formular una pregunta
Cursos de formación en línea
Contactar con nuestro equipo de ventas
Vídeos
Formación
Primeros pasos con RFEM
Primeros pasos con RSTAB
Formación en línea
Formación en Dlubal
Cursos de formación individuales
Vídeos
Vídeos de aprendizaje en línea
Webinarios - Aprenda en línea
Cursos en línea
Enlaces rápidos | Soporte y formación
Formular una pregunta
Cursos de formación en línea
Contactar con nuestro equipo de ventas
Vídeos
Noticias
Novedades
Formación
Eventos
Novedades
Noticias actuales
Nuevas características de productos
Suscribirse al boletín de noticias
Nuevos programas
Blog de Dlubal
Enlaces rápidos | Noticias y eventos
Noticias actuales
Blog
Seminarios web
Cursos de formación en línea
Formación
Cursos de formación en línea
Formación individual
Enlaces rápidos | Noticias y eventos
Noticias actuales
Blog
Seminarios web
Cursos de formación en línea
Eventos
Resumen de eventos
Feria/Seminario web
Webinarios
Enlaces rápidos | Noticias y eventos
Noticias actuales
Blog
Seminarios web
Cursos de formación en línea
Recursos
Descargar versión completa
Zona gratis de Dlubal
Ejemplos
Documentos
Referencias
Versión completa de prueba gratis
¿Le gustaría probar las capacidades de los programas de Dlubal Software? ¡Tiene la oportunidad de hacerlo! La versión completa gratuita de 90 días le permite probar todos nuestros programas por completo.
Enlaces rápidos | Descargas e información
Descargar versión de prueba
Modelos para descargar
Manuales en línea
Proyectos de clientes
Área gratuita de Dlubal
En el área gratuita de Dlubal obtienes acceso a seminarios web, artículos y posibilidades de prueba del software, todo de manera gratuita y ordenado en un solo lugar.
Enlaces rápidos | Descargas e información
Descargar versión de prueba
Modelos para descargar
Manuales en línea
Proyectos de clientes
Ejemplos
Modelos de análisis estructural para descargar
Enviar modelo de análisis estructural
Ejemplos introductorios y tutoriales
Ejemplos de verificación
Vista general de imágenes
Enlaces rápidos | Descargas e información
Descargar versión de prueba
Modelos para descargar
Manuales en línea
Proyectos de clientes
Documentos
Manuales en línea
Manuales
Folletos, catálogos y certificados
Hojas informativas
Enlaces rápidos | Descargas e información
Descargar versión de prueba
Modelos para descargar
Manuales en línea
Proyectos de clientes
Referencias
Proyectos de clientes
¿Por qué enviar su proyecto?
¿Cómo presentar un proyecto de cliente?
Enviar un proyecto de cliente
Enlaces rápidos | Descargas e información
Descargar versión de prueba
Modelos para descargar
Manuales en línea
Proyectos de clientes
Formación
Aprendizaje en línea
Estudiantes y universidades
Plataforma de conocimientos
Infoentretenimiento
Aprendizaje en línea
RFEM 6 para principiantes
RFEM 6 para estudiantes
Programación con RFEM 6 y Python
RFEM 6 con Rhino y Grasshopper
RFEM 5 para principiantes
Modelado con RFEM 5
Lecciones de análisis de estructuras para estudiantes
Vídeo tutoriales cortos para los programas de Dlubal
Los mejores consejos y trucos en RFEM
Grabaciones de cursos en línea
Webinarios celebrados y grabados
Enlaces rápidos | Formación
Software de análisis estructural gratis para estudiantes
Software de análisis de estructuras gratuito para universidades técnicas
Tesis de graduación con Dlubal Software
Videos de formación en línea | Tutorial de RFEM 6 para principiantes
Nuestros clientes
Estudiantes y universidades
Software de análisis estructural gratuito para estudiantes
Solicitar o renovar licencia de estudiante gratuita
Solicitud de licencia para profesores gratuita
Enviar tesis de estudiante
¿Por qué enviar tu tesis?
Tesis de graduación con el software de análisis estructural de Dlubal
Software de análisis de estructuras gratuito para universidades técnicas
Solicitar paquete para universidades técnicas
Curso introductorio gratuito en su universidad
Solicitar fecha de un curso
Enlaces rápidos | Formación
Software de análisis estructural gratis para estudiantes
Software de análisis de estructuras gratuito para universidades técnicas
Tesis de graduación con Dlubal Software
Videos de formación en línea | Tutorial de RFEM 6 para principiantes
Nuestros clientes
Plataforma de conocimientos
Primeros pasos con RFEM
Vídeos
Manuales en línea
Wiki de ingeniería de estructuras
Base de datos de conocimientos
Preguntas frecuentes (FAQ)
Enlaces rápidos | Formación
Software de análisis estructural gratis para estudiantes
Software de análisis de estructuras gratuito para universidades técnicas
Tesis de graduación con Dlubal Software
Videos de formación en línea | Tutorial de RFEM 6 para principiantes
Nuestros clientes
Infoentretenimiento
Pódcast
Blog de Dlubal
Introducción al análisis y diseño de estructuras
Enlaces rápidos | Formación
Software de análisis estructural gratis para estudiantes
Software de análisis de estructuras gratuito para universidades técnicas
Tesis de graduación con Dlubal Software
Videos de formación en línea | Tutorial de RFEM 6 para principiantes
Nuestros clientes
Empresa
Acerca de la empresa
Contacto
Referencias
Nuestros clientes
Acerca de la empresa
Hechos y cifras
Filosofía de la empresa
¿Por qué Dlubal Software?
Comparación de productos
Política de calidad
Nuestro equipo
Enlaces rápidos | Empresa
Nuestro equipo
Oficinas de Dlubal Software
Distribuidores autorizados
Reseñas de los usuarios de Dlubal
Proyectos de clientes
Nuestros clientes
Contacto
Oficinas de Dlubal en todo el mundo
Distribuidores autorizados de Dlubal
Enlaces rápidos | Empresa
Nuestro equipo
Oficinas de Dlubal Software
Distribuidores autorizados
Reseñas de los usuarios de Dlubal
Proyectos de clientes
Nuestros clientes
Referencias
Opiniones de clientes
Proyectos de clientes
¿Por qué enviar su proyecto de cliente?
Ejemplos de verificación
Su reseña
Participación en proyectos de investigación
Enlaces rápidos | Empresa
Nuestro equipo
Oficinas de Dlubal Software
Distribuidores autorizados
Reseñas de los usuarios de Dlubal
Proyectos de clientes
Nuestros clientes
Nuestros clientes
Presentamos a nuestros clientes que realizan sus proyectos con el software de Dlubal. Descubra cómo nuestros clientes en todo el mundo implantan soluciones innovadoras en la construcción e ingeniería utilizando herramientas avanzadas para análisis estáticos y dinámicos de estructuras.
Enlaces rápidos | Empresa
Nuestro equipo
Oficinas de Dlubal Software
Distribuidores autorizados
Reseñas de los usuarios de Dlubal
Proyectos de clientes
Nuestros clientes
Carrera
Juntos logramos más
Equipos
Empleos
Noticias del equipo de Dlubal
Juntos logramos más
Carrera profesional en Dlubal
Enlaces rápidos | Carrera
Nuestro equipo
Oficinas de Dlubal Software
Equipos
Desarrollo de productos
Atención al cliente
Ventas
Marketing
Desarrollo de software
Administración
Enlaces rápidos | Carrera
Nuestro equipo
Oficinas de Dlubal Software
Empleos
Todas las ofertas de trabajo
Desarrollo de productos
Soporte al cliente
Ventas
Marketing
Desarrollo de software
Administración
Pasantes
Otros
Enlaces rápidos | Carrera
Nuestro equipo
Oficinas de Dlubal Software
Historias de equipo
¡Bienvenido a la sección de Historias del equipo! Aquí te traemos noticias, logros e ideas de nuestros equipos. Descubre en qué estamos trabajando, conoce a nuevos miembros del equipo y explora las innovaciones que tenemos preparadas.
Enlaces rápidos | Carrera
Nuestro equipo
Oficinas de Dlubal Software
Prueba gratuita de 90 días
Cuenta (Iniciar sesión)
Tienda en línea0 Cesta
Prueba gratuita de 90 días
Idioma
Español Español
Idioma
Deutsch
English
Français
Español
Português
Italiano
Česky
Polski
Pусский
中文(简体)
Community
Inicio
Soluciones
Servicios en línea
Wiki Esencial de Estructuras
Módulo de cortante
Wiki de ingeniería de estructuras
El wiki de análisis de estructuras de Dlubal explica los términos técnicos utilizados en el análisis y diseño de estructuras Los términos se proporcionan en orden alfabético y se utilizan normalmente en el software de Dlubal.
Buscar
¿Le ha resultado útil esta página?
Compartir
Copia
Imprimir
38605x
000171
02-07-2019
A(30)
Abertura
Acción
Acelerograma
Acoplamiento de giros
Administrador de bloques
Administrador de configuración
Administrador de proyectos
Ajuste de objetos
ajuste de resultados de superficie
Análisis de estructuras de fachadas
Análisis de segundo orden
Análisis dependiente del tiempo
Análisis modal
Apoyo de barra
Apoyo elástico en barra
Apoyo elástico en superficie
Apoyo en superficie
Apoyos en nudos
Armadura
Armadura básica
Armadura de cortante
Armadura longitudinal
Armadura mínima
Arrastrar y soltar
Arriostramiento
Articulación en barra
Articulación lineal
asistente para cargas
Asistente para cargas de nieve
Asistente para combinaciones
Á(1)
Árido
B(9)
Barra
Barra de resultados
Barra nula
barra representativa
Barra rígida
Biblioteca de materiales
Biblioteca de secciones
Biselado
Búsqueda de forma
C(57)
Capa de fondo
Carga circular libre
Carga climática
Carga de contorno de sólidos
Carga de impacto
Carga de masa
Carga de nudo
Carga en abertura
Carga en barra
Carga en conjunto de barras
Carga en conjunto de líneas
Carga en conjunto de sólidos
Carga lineal
carga lineal libre
Carga móvil
Carga poligonal libre
Carga puntual libre
Carga rectangular libre
Carga sísmica equivalente
Carga superficial
Cargar de viento
Cargas en barra a partir de una carga superficial
Caso de carga
Caso de carga
Caso de imperfección
Caso de masa
Caso de vibración natural
Celda
Cemento
Centro de Dlubal
Cercha
Chapa
Chapa de soporte
Chapa frontal
Coacción en nudo
Coacción lateral intermedia
Combinación de acciones
Combinación de carga
Combinación de masas
Combinación de resultados
Comentario
Complemento
Componente
Composición
comprobación del modelo
Comprobación plausible
Configuración del análisis estático
Conjunto de barras
conjunto de barras representativo
Conjunto de líneas
Conjunto de sólidos
conjunto de superficies
Construcción biónica
Contacto de superficies
Correas
Cubierta en forma de barril
Cubo de vistas
D(10)
Datos generales
Deformación
Deformación impuesta en línea
deformación impuesta en nudo
Diagrama de cálculo
Diagrama de tiempos
Diagrama de Voronoi
Dibujo de obra
Disparo de mástil
División de barras
E(23)
El informe
Elemento nulo
Elemento puntual
ELU – Estado límite último
enlace rígido
Ergebnisverläufe
Esfuerzo
Esfuerzo interno
Espectro de respuesta
Espesor
Esquema de combinaciones
Estabilidad de la estructura
Estado de tensión de la unidad
Estructura de pórtico
Estructura de soporte de placa
Estructura neumática
Estructura plegada
estructura superficial
Estructuras de barras
ETFE
Excentricidad
Excentricidad de barra
Excentricidad de la superficie
F(6)
Fase de servicio
Fases de construcción
FE-Netz
Fluencia
Fuentes
Funciones de teclado
G(6)
Generador de carga de viento
Generador de modelos
glTF
grado de libertad
Grupo de armaduras
Grupo de uniones
H(4)
Herramientas de modelado
HOAI
Hormigón
Hormigón armado
I(7)
Imperfección
Imperfección de conjunto de barras
Ingeniería de estructuras
Interfaz
Intersección
Isosuperficie
Iteración
L(12)
Lado de la superficie
Liberación de línea
Liberación de nudo
Liberación de superficie
Línea
Línea auxiliar
Linealidad
Lista de artículos
Lista de materiales
Longitud eficaz
Losa de piso plana
Losa elevada
M(15)
Masa modal
Medida
Membrana
Modelo
Modelo de análisis estructural
Modelo estructural
Modificación de rigidez de barra
Modificación de rigidez de superficie
modificación estructural
Modo de visualización
Modo propio
Módulo adicional
Módulo de cortante
Módulo de elasticidad
Momento de inercia de área
N(7)
Navegador de proyectos
Nervio
Nervio
Nota:
Nudo
Nudo de punzonamiento
Nudo libre
O(5)
Objeto auxiliar
Objeto visual
Opciones de selección
Optimización con enjambre de partículas (PSO)
Ortotropía
P(13)
Panel
Panel de control
Panel de cortante
Parametrización
Parte c/t
plano de la armadura
Plano de recorte
plano de trabajo
Preselección
Propuesta de armadura
Punto desconectado
Punto fijo
Punto intermedio de resultados de barra
R(25)
Reacción en el apoyo
Recuadro de recorte
refinamiento de la malla de EF
Refinamiento de la malla de superficie
refinamiento de malla de líneas
Refinamiento de malla de nudos
Refinamiento de malla de sólidos
Región media
Regla de combinación
Regularización
Rejilla
Rejilla de construcción
Rejilla de líneas
Relajación
Representación
Resistencia
Resistencia al alabeo
Resorte
Retracción
Retracción química
Rigidez
Rigidez de superficie
Rigidez definible en barra
Rigidización
Rigidizador
S(18)
Sección
Sección
Sección de resultados
Selección
Selección de objetos
Sistema de coordenadas
Situación de proyecto
Soldadura en ángulo
Sólido
Sólido de contacto
Stabimperfektion
Stablasten aus freier Linienlast
Stabnichtlinearität
Staborientierung
Supercombinación
Superficie
Superficie cuadrangular
Superficie de transmisión de cargas
T(7)
Techo de cúpula
Tensión equivalente
Tensión tangencial
Tipo de barra
Tipo de contacto de superficies
Torsión de alabeo
Trayectoria
U(2)
Unión con pasador
unión soldada en línea
V(6)
Valores de resultados en superficies
Viga
viga a flexión
Visibilidad
Vistas y visibilidades
Volumen de gas
M
Módulo de cortante
El módulo de corte G describe la deformación lineal elástica de un componente como resultado de una tensión cortante o fuerza de cizalladura. Esta constante de material se deriva del Módulo de elasticidad E y el coeficiente de contracción lateral (coeficiente de Poisson) ν:
G=E 2·1+ν
G Módulo de cortante
E Módulo de elasticidad
ν deformación transversal
Zusammenhang zwischen Schubmodul, E-Modul und Querdehnzahl
Enlaces
Herramientas útiles para la generación rápida de estructuras en RFEM
Análisis estructural lineal y no lineal en RSTAB
Módulo de corte
Productos recomendados para usted
RFEM 6 | Programa principal RFEM 6
RFEM 6
Programa principal
La nueva generación de software del MEF en 3D se utiliza para el análisis de estructuras de barras, superficies y sólidos.
Precio de la primera licencia 4.790,00 USD
RFEM 6 | Análisis adicional
Comportamiento no lineal del material para RFEM 6
Complemento
El complemento Comportamiento no lineal del material permite considerar las no linealidades del material en RFEM (por ejemplo, isótropo plástico, ortótropo plástico, daño isótropo).
Precio de la primera licencia 1.660,00 USD
RFEM 6 | Análisis adicional
Estabilidad de la estructura para RFEM 6
Complemento
El complemento Estabilidad de la estructura realiza el análisis de estabilidad de las estructuras.
Precio de la primera licencia 1.460,00 USD
RFEM 6 | Análisis adicional
Alabeo por torsión (7 GDL) para RFEM 6
Complemento
El complemento Alabeo por torsión (7 GDL) le permite considerar el alabeo de secciones como un grado de libertad adicional.
Precio de la primera licencia 1.660,00 USD
RFEM 6 | Soluciones especiales
Superficies multicapa (p. ej. laminadas, CLT) para RFEM 6
Complemento
El complemento Superficies multicapa permite al usuario definir estructuras con superficies multicapa. El cálculo se puede realizar con o sin acoplamiento a cortante.
Precio de la primera licencia 1.560,00 USD
RFEM 6 | Soluciones especiales
Optimización y coste / Estimación de emisiones de CO2
Complemento
El complemento de dos partes Optimización y estimación de coste / emisiones de CO2 encuentra los parámetros adecuados para los modelos y bloques parametrizados mediante la técnica de la inteligencia artificial (IA) de la optimización por enjambre de partículas (PSO) para el cumplimiento de los criterios de optimización comunes. Además, este complemento estima los costes del modelo o las emisiones de CO2 especificando los costes unitarios o las emisiones por definición de material para el modelo estructural.
Precio de la primera licencia 1.660,00 USD
RFEM 6 | Cálculo
Análisis tensión-deformación para RFEM 6
Complemento
El complemento Análisis tensión-deformación realiza un análisis de tensiones general calculando las tensiones existentes y comparándolas con las tensiones límite.
Precio de la primera licencia 1.360,00 USD
RFEM 6 | Cálculo
Cálculo de estructuras de hormigón para RFEM 6
Complemento
El complemento Cálculo de hormigón permite varias verificaciones según las normas internacionales. Es posible diseñar barras, superficies y pilares, así como realizar análisis de punzonamiento y deformaciones.
Precio de la primera licencia 2.970,00 USD
RFEM 6 | Cálculo
Cálculo de acero para RFEM 6
Complemento
El complemento de Cálculo de acero realiza las verificaciones de diseño de los estados límite último y de servicio de barras de acero según varias normas.
Precio de la primera licencia 2.970,00 USD
RFEM 6 | Cálculo
Cálculo de estructuras de madera para RFEM 6
Complemento
El complemento Cálculo de madera realiza las comprobaciones de cálculo de los estados límite últimos, de servicio y de resistencia al fuego de barras de madera según varias normas.
Precio de la primera licencia 2.170,00 USD
RFEM 6 | Cálculo
Cálculo de estructuras de fábrica para RFEM 6
Complemento
El complemento Cálculo de fábrica para RFEM permite el cálculo y dimensionamiento de estructuras de fábrica (mampostería) utilizando el método de los elementos finitos. Fue desarrollado como parte del proyecto de investigación titulado DDMaS – Digitalizing the Design of Masonry Structures. El modelo de material representa el comportamiento no lineal de la combinación de ladrillo y mortero en forma de un macro-modelado.
Precio de la primera licencia 1.860,00 USD
RFEM 6 | Cálculo
Cálculo de aluminio para RFEM 6
Complemento
El complemento de Cálculo de aluminio realiza las comprobaciones de diseño en el estado límite último y de servicio de elementos de aluminio según varias normas.
Precio de la primera licencia 1.970,00 USD
RFEM 6 | Uniones
Uniones de acero para RFEM 6
Complemento
El complemento Uniones de acero para RFEM le permite analizar conexiones de acero utilizando un modelo de elementos finitos. El modelo de elementos finitos se genera automáticamente en segundo plano y se puede controlar mediante la introducción simple y familiar de los componentes.
Precio de la primera licencia 2.670,00 USD
RSTAB 9 | Programa principal RSTAB 9
RSTAB 9
Programa principal
El moderno programa de análisis y cálculo estructural en 3D es adecuado para el análisis estructural y dinámico de estructuras de vigas, así como para el cálculo de hormigón, acero, madera y otros materiales.
Precio de la primera licencia 3.380,00 USD
RSTAB 9 | Análisis adicionales
Estabilidad de la estructura para RSTAB 9
Complemento
El complemento Estabilidad de la estructura realiza el análisis de estabilidad de las estructuras.
Precio de la primera licencia 1.460,00 USD
RSTAB 9 | Análisis adicionales
Alabeo por torsión (7º GL) para RSTAB 9
Complemento
El complemento Torsional Warping (7º GDL) permite considerar la deformación por torsión de la sección transversal como un grado de libertad adicional al calcular los miembros.
Precio de la primera licencia 1.660,00 USD
RSTAB 9 | Soluciones especiales
Optimización y estimación de costes/emisiones de CO2 para RSTAB 9
Complemento
El complemento de dos partes Optimización y estimación de coste/emisiones de CO2 encuentra los parámetros adecuados para los modelos y bloques parametrizados mediante la técnica de la inteligencia artificial (IA) de la optimización por enjambre de partículas (PSO) para el cumplimiento de los criterios de optimización comunes.
Precio de la primera licencia 1.660,00 USD
RSTAB 9 | Cálculo
Análisis tensión-deformación para RSTAB 9
Complemento
El complemento Análisis tensión-deformación realiza análisis generales de tensiones, calculando las tensiones existentes y comparándolas con las tensiones límite.
Precio de la primera licencia 1.260,00 USD
RSTAB 9 | Cálculo
Cálculo de hormigón para RSTAB 9
Complemento
El complemento Cálculo de hormigón permite varias verificaciones de barras y pilares según las normas internacionales.
Precio de la primera licencia 2.970,00 USD
RSTAB 9 | Cálculo
Cálculo de acero para RSTAB 9
Complemento
El complemento Cálculo de acero realiza las verificaciones del estado límite último y de servicio de barras de acero según varias normas.
Precio de la primera licencia 2.970,00 USD
RSTAB 9 | Cálculo
Cálculo de madera para RSTAB 9
Complemento
El complemento Cálculo de madera realiza las verificaciones de los estados límite últimos, de servicio y de resistencia al fuego de barras de madera según varias normas.
Precio de la primera licencia 2.170,00 USD
RSTAB 9 | Cálculo
Cálculo de aluminio para RSTAB 9
Complemento
El complemento Cálculo de aluminio realiza las verificaciones del estado límite último y de servicio de barras de aluminio según varias normas.
Precio de la primera licencia 1.970,00 USD
Mostrar familia de productos
Aplicar filtro
Mapas de cargas de nieve, velocidad del viento y cargas sísmicas
Versión completa de prueba gratis
Mia, su asistente de inteligencia artificial las 24 horas
RFEM 6
RSTAB 9
Manuales en línea
Base de conocimientos
Dlubal Software GmbH
Am Zellweg 2
93464 Tiefenbach
Alemania
Teléfono+34 960 13 67 00 (hablamos español)
Correo electrónicoinfo@dlubal.com
Boletín de noticias Dlubal
Unirse a Dlubal
Soluciones
Estructuras de hormigón armado
Estructuras de acero
Estructuras de madera
Uniones de acero
Productos
RFEM 6
RSTAB 9
RSECTION 1
RWIND 3
Apoyo
Preguntas frecuentes (FAQ)
Formular una pregunta
Herramienta de zonas geográficas para la determinación de cargas
Contactar con nuestro equipo de ventas
Noticias
Suscribirse al boletín de noticias
Noticias actuales
Resumen de eventos
Cursos de formación en línea
Recursos
Versión completa de prueba gratis
Enviar un proyecto de cliente
Proyectos de clientes
Manuales en línea
Formación
Solicitar o renovar licencia de estudiante gratuita
Software de análisis estructural gratuito para estudiantes
Software de análisis de estructuras gratuito para universidades técnicas
Tesis de graduación con el software de análisis estructural de Dlubal
Empresa
Acerca de la empresa
Oficinas de Dlubal en todo el mundo
¿Por qué Dlubal Software?
Carrera
Todas las ofertas de trabajo
Carrera profesional en Dlubal
Software de cálculo de estructuras y diseño
Español Español
Idioma
Deutsch
English
Français
Español
Português
Italiano
Česky
Polski
Pусский
中文(简体)
© 2001-2025 Dlubal Software GmbH | Todos los derechos reservados
Nota legal
Acerca de la empresa
Mapa del sitio |
189204 | https://spellingtesto9.quora.com/What-is-the-meaning-of-Opulence-What-are-the-similar-and-opposite-words-for-Opulence | What is the meaning of 'Opulence'?What are the similar and opposite words for “Opulence”? - Spelling test - Quora
Something went wrong. Wait a moment and try again.
Try again
Skip to content
Skip to search
Sign In
Spelling test
English word test
Follow
·
1.4K 1.4K
Vikas Sharma
4y ·
What is the meaning of "Opulence"?What are the similar and opposite words for “Opulence”?
Ans:-
Meaning:-
Opulence :-
(Noun)
Great wealth and luxuriousness
Synonyms off “Opulence”:-
1) Luxury
2) Riches
3) Fortune
4) Prosperity
5) Abundance
Antonyms of “Opulence”:-
1) Poverty
2) Restraint
3) Simplicity
4) Frugality
5) Miserliness
70 views ·
View upvotes
·
Upvote ·
9 1
About the Author
Vikas Sharma
721.3K content views 3K this month
Active in 12 Spaces
Joined April 2020
View more in Spelling test
About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press ·
© Quora, Inc. 2025 |
189205 | https://www.studocu.com/it/document/universita-degli-studi-di-cagliari/scienza-politica/relazone-1-serie-di-balmer/8002284 | Relazone 1 - Serie di balmer - SPETTRI ATOMICI - SERIE DI BALMER USAI ALBERTO 6060/ 27/10/ SOMMARIO - Studocu
Passa al documento
Insegnanti
Università
Scuola superiore
Esplora
Accedi
Benvenuto su Studocu Accedi alle risorse di studio
Accedi Registrati
Utente ospite Aggiungi la tua università o scuola
0 follower
0 Upload 0 upvotes
Novità
Home
La mia biblioteca
AI Notes
Ask AI
AI Quiz
Chat
Recenti
Non ci sono ancora elementi recenti.
La mia biblioteca
Corsi
Non hai ancora nessun corso.
Aggiungi Corsi
Libri
Non hai ancora nessun libro.
Studylists
Non hai ancora nessuna Studylist.
Crea una Studylist
Home
La mia biblioteca
Esplora
Esplora
Università
Scuole Superiori
Diplomi di scuola superiore
Risorse didattiche
Generatore di piani di lezione
Generatore di test
Generatore di quiz live
Ask AI
Relazone 1 - Serie di balmer
Corso
Scienza politica (SPS/04)
40 documenti
Università
Università degli Studi di Cagliari
Anno accademico:2019/2020
Caricato da:
alberto usai
Università degli Studi di Cagliari
0 follower
1 Upload1 upvotes
Segui
Consigliati per te
23 Schemi-lineamenti-di-diritto-pubblico-falcon 2 Scienza politica Altro 100% (4)
28 Dispense aggiuntive - Maurizio COTTA, Donatella DELLA PORTA, Leonardo MORLINO, Scienza politica, Bolo-gna,Scienza politica Appunti di lezione 100% (3)
15 Riassunto postdemocrazia colin crunch Scienza politica Riassunti 100% (2)
Commenti
accedi o registrati per pubblicare commenti.
Segnala il documento
Gli studenti hanno anche visualizzato
Capitolo 4 La dem. Tradita
storia stato sociale
Capitolo 11 Dimensione Sovranazionale Governace Multilivello
Capitolo 12 Crisi Della Politica
Piano Genitoriale Base - Istruzioni e Indicazioni Utili
Sistemi Maggioritari e Proporzionali: Analisi e Confronto
Altri documenti collegati
Domande Esame - Appunti prima parte
Ne bis in idem e omesso versamento iva
La Conferenza di Servizi
Diritto dell'Unione Europea
Diritto tribuario pdf
CAPITOLO 7: DEVIAZIONE E CRIMINE - ANALISI SOCIALE E TEORIE
Anteprima del testo
SPETTRI ATOMICI - SERIE DI BALMER
USAI ALBERTO 6060/
27/10/
SOMMARIO
E’ stato studiato lo spettro di emissione di una lampada all’idrogeno con il fine di dimostrare la quantizzazione dei livelli atomici negli atomi e, in particolare, che la separazione tra i livelli atomici scala come l’inverso del quadrato del numero quantico. E’ stato quindi determinato il valore della costante di Rydberg dai dati sperimentali ed è stato confrontato con il suo valore teorico. Sia i valori delle lunghezze d’onda di emissione ricavati dallo spettro che il valore della costante di Rydberg sono in accordo con i rispettivi valori teorici.
INTRODUZIONE
L’esistenza dello spettro di emissione è dovuta all’esistenza di livelli discreti di energia negli atomi. Un elettrone può passare da un livello di energia E 1 ad un livello superiore E 2 assorbendo un fotone. L’energia del fotone è legata alla sua frequenza dalla relazione di Planck:
(1) Efotone = E 1 - E 2 = hν = h
c λ
Per l’atomo di idrogeno vale la relazione :
(2) En = −( e 2 4 πε 0 )
2 m 2 n 2 ℏ 2
Il numero intero n è detto numero quantico, n = 1 corrisponde al livello di energia più basso, mentre n = 0 corrisponde allo stato in cui l’elettrone è completamente slegato dall’atomo. Se si considera il protone privo di massa la massa m che compare nella relazione (2) è la
massa dell’elettrone. Una migliore descrizione del sistema si ottiene però considerando la massa ridotta del sistema μ per cui vale la relazione: 1 μ
= 1
m
+ 1
M
, in cui M è la massa del
protone.
Mettendo insieme le due relazioni si ottengono le lunghezze d’onda corrispondenti a tutte le possibili transizioni tra livelli energetici :
(4)
1
λ
=( e
2 4 πε 0 )
2 μ 4 π 2 ℏ 3 c(
1
n 22
− 1
n 12 )
Nella relazione (4) n 1 ed n 2 rappresentano, rispettivamente, il numero quantico dello stato iniziale e finale.
In cui la quantità :
RH = ( e 2 4 πε 0 )
2 μ 4 π 2 ℏ 3 c
= 1,0968 × 10 − 7 m-
è detta costante di Rydberg.
In particolare transizioni da stati a energia maggiore allo stato finale n 2 = 2 producono lunghezze d’onda nella regione visibile dello spettro. Tali lunghezze d’onda sono date dalla formula di Balmer:
(4)
1
λ
=RH
(
1
22
−
1
n 2 )
con n = 3,4,5...
ESPERIMENTO
L’esperimento è stato realizzato utilizzando uno spettrometro a reticolo per disperdere la luce emessa da una lampada spettrale all’idrogeno. Lo spettrometro è connesso ad una camera CCD che misura l’intensità della luce in una determinata finestra spettrale.
Prima di procedere con la misurazione lo spettrometro è stato raffreddato con dell’azoto liquido in modo da diminuire il rumore termico che, soprattutto a basse frequenze, avrebbe reso difficile l’individuazione dei picchi.
In primo luogo è stata eseguita una calibrazione dello spettrometro attraverso la misura delle righe di emissione note di una lampada al mercurio. In questo modo è stato possibile eliminare eventuali errori sistematici. Prima di effettuare la misura dello spettro di H, il fascio di luce emesso dalla lampada è stato collimato con una lente.
Tutte le misure sono state eseguite in continuo. Il passo del reticolo è stato scelto piccolo (300 passi al mm) in modo tale da aumentare il numero di suddivisioni al mm e, di conseguenza, il potere risolutivo dello spettrometro.
Il tempo di integrazione è stato impostato su 7s.
L’apertura della fenditura è stata scelta di 0,2 mm. La lampada all’idrogeno è stata alimentata con 2,2 KV, tensione che non ha reso necessaria una correzione relativistica.
RISULTATI
Per calibrare lo spettrometro è stato misurato lo spettro di una lampada al mercurio e sono state confrontate le lunghezze d’onda di emissione note con quelle misurate. Il procedimento è illustrato più dettagliatamente nelle informazioni aggiuntive. I parametri ottenuti sono stati quindi utilizzati per correggere le misure relative all’idrogeno.
Lo spettro della lampada ad idrogeno è stato ottenuto eseguendo un'unica misurazione. Lo spettroscopio è stato impostato in modo tale da acquisire la misura di ciascuna finestra spettrale 20 volte. Questo ha permesso di ottenere una misura finale meno rumorosa.
A partire dallo spettro sono quindi state calcolatele lunghezze d’onda corrispondenti ai massimi.
λMISURATA(nm) λTEORICA(nm)
410,92 ± 0,73 410,
434,96 ± 0,75 434,
486,74 ± 0,81 486,
657,77 ± 0,99 656,
TAV 1. Nella tabella vengono riportate le lunghezza d’onda, teoriche e misurate (comprese di errore), corrispondenti alle righe di emissione dello spettro della lampada ad idrogeno.
L’errore sulle λ misurate è stato calcolato tenendo conto degli errori sui parametri
La scarsa precisione, in termini di cifre significative, di RH è principalmente dovuta al basso numero di punti sperimentali con cui è stata fatta la retta dei minimi quadrati.
INFORMAZION AGGIUNTIVE
La misura dello spettro della lampada al mercurio è stata eseguita in continuo (con 20 accumulazioni). Il tempo di integrazione è stato impostato su 7s.
E’ stata quindi fatta una calibrazione, supponendo una correlazione lineare tra λ misurate e λ note.
λTEOR=AλMIS+B
TAB 4. Nella tabella vengono riportati i valori dei coefficienti della retta dei minimi quadrati per la lampada al mercurio.
I parametri A e B sono stati utilizzati per calcolare l’errore sulle lunghezze d’onda di emissione della tabella 1.
ΔλTEOR= √(
∂λTEOR ∂A
ΔA)
2 +(
∂λTEOR ∂λMIS
ΔλMIS)
2 +(ΔB) 2
A B
1,0066±0,0013 -1,17±0,
Relazone 1 - Serie di balmer
Scarica
Scarica
Strumenti IA
Ask AI
Scelta multipla
Flashcards
Video quiz
Lezione audio
1 0
Salva
Relazone 1 - Serie di balmer
Corso: Scienza politica (SPS/04)
40 documenti
Università: Università degli Studi di Cagliari
Informazioni Più info
Scarica
Scarica
Strumenti IA
Ask AI
Scelta multipla
Flashcards
Video quiz
Lezione audio
1 0
Salva
SPETTRI ATOMICI - SERIE DI BALMER
USAI ALBERTO 6060/65180
27/10/2018
SOMMARIO
E’stato s tudiato lo spettro di emissione di una lampada all’idrogeno con il fine di dimostrare la
quantizzazione dei livelli atomici negli atomi e,in particolare,che la s eparazione tra i livelli
atomici scala come l’inverso del quadrato del numero quantico.E’stato quindi determinato il
valore della costante di Rydberg dai dati sperimentali ed è stato confrontato con il suo valore
teorico. Sia i valori delle lunghezze d’onda di emissione ricavati dallo spettro che il valore della
costante di Rydberg sono in accordo con i rispettivi valori te orici.
INTRODUZIONE
L’esistenza dello spettro di emissione è
dovuta all’esistenza di livelli discreti di
energia negli atomi. Un elettrone può passare
da un livello di energia E 1 ad un livello
superiore E 2 assorbendo un fotone.L’energia
del fotone è legata alla sua frequenza dalla
relazione di Planck:
(1) E fotone = E 1 - E 2 = hν =
h c
λ
Per l’atomo di idrogeno vale la relazi one :
(2) E n =
−
(
e 2
4 π ε 0
)
2 m
2 n 2 ℏ 2
Il numero intero n è detto numero quantico,
n=1 corrisponde al livel lo di energia più
basso,mentre n=0 corrisponde allo s tato in
cui l’elettrone è completamente slegato
dall’atomo.
Se si considera il protone privo di massa la
massa m che compare nella relazione(2)è la
massa dell’elettrone.Una migliore
descrizione de l siste ma si ot tiene però
considerando la massa ridotta del s istema µ
per cui vale la relazione:
1
μ=1
m+1
M
,in cui M è la massa del
protone.
Mettendo insieme le due relazioni si
ottengono le lunghezze d’onda
corrispondenti a tutte le possibili transizioni
tra livelli energetici :
(4)
1
λ=
(
e 2
4 π ε 0
)
2 µ
4 π 2 ℏ 3 c
(
1
n 2
2−1
n 1
2
)
Nella relazione(4)n 1 e d n 2 rappresent ano,
rispettivamente,i l numero quantico dell o
stato iniziale e finale.
In cui la quantità :
R H=
(
e 2
4 π ε 0
)
2 µ
4 π 2 ℏ 3 c
=1,0968
×10−7
m-1
è detta costante di Rydberg.
1
In particolare transizioni da stati a energia
maggiore allo stato finale n 2=2 producono
lunghezze d’onda nella regione visibile dello
spettro.Ta li lunghezze d’onda sono date
dalla formula di Balmer:
(4)
1
λ=R H
(
1
2 2−1
n 2
)
con n = 3,4,5…
ESPERIMENTO
L’esperimento è stato realizzato utilizzando
uno spettrometro a reticolo per disperdere la
luce emessa da una lampada spettrale
all’idrogeno.Lo spettrometro è connesso ad
una camera CCD che misura l’intensità della
luce in una determinata finestra spettral e.
Prima di procedere con la misurazione lo
spettrometro è stato raffreddato con
dell’azoto liquido in modo da diminuire il
rumore termico che,soprattutto a bass e
frequenze,avrebbe reso difficile
l’individuazione dei picchi.
In primo luogo è stata eseguita una
calibrazione dello spettromet ro attraverso la
misura delle righe di emissione note di una
lampada al mercurio.In questo modo è stato
possibile eliminare eventuali errori
sistematici.Prim a di effe ttuare la misura
dello spettro di H,il fascio di luce emesso
dalla lampada è stato collimato con una lente.
Tutte le misure sono stat e eseguit e in
continuo.Il pas so del reticolo è stato scelto
piccolo(300 passi al mm)in modo tale da
aumentare il numero di suddivisioni al mm e,
di conseguenza,il potere risoluti vo dello
spettrometro.
Il tempo di integrazione è s tato impostato su
7s.
L’a pertura della fenditura è stata scelta di
0,2 mm.La lampa da all’idrogeno è stata
alimentata con 2,2 KV,tensione che non ha
reso necessaria una correzione relati vistica.
RISULTATI
Per calibrare lo spettrometro è stato misurato
lo spettro di una lampada al mercurio e sono
state confrontate le lunghezze d’onda di
emissione note con quelle misurate.Il
procedimento è illustra to più
dettagliatamente nelle informazioni
aggiuntive.
I parametri ottenuti sono stati quindi
utilizzati per correggere le misure relative
all’idrogeno.
Lo spettro della lampada ad idrogeno è stato
ottenuto eseguendo un'unica misurazione.Lo
spettroscopio è stato impostato in modo tale
da acquisire la misura di ciascuna finestra
spettrale 20 volte.Questo ha permesso di
ottenere una misura finale meno rumorosa.
A partire dallo spettro sono quindi state
calcolatele lunghezze d’onda corrispondenti
ai massimi.
λ MISURATA(nm)λ TEORI CA(nm)
410,92 ± 0,73 410,47
434,96 ± 0,75 434,05
486,74 ± 0,81 486,13
657,77 ± 0,99 656,28
TAV 1.Nella tabella vengono riportate le
lunghezza d’onda, teoriche e misurate (comprese
di errore),corrispondenti alle righe di emissione
dello spettro della lampada ad idrogeno.
L’errore sulle λ misurate è s tato calcol ato
tenendo conto degli errori s ui parametri
2
ottenuti dalla calibrazione(una trattazione
più approfondita è riportata nelle
informazioni aggiuntive).
FIG 1.Nel seguente grafico viene rappresentato
lo spettro della lampada ad H misurato
(lunghezza d’onda in nm in funzione
dell’intensità in unità arbitrarie).
Per evidenziare che la separazione dei livelli
scala come l’inverso del numero quantico è
stata fatta una retta dei minimi quadrati,
riscrivendo la legge in questo modo:
1
λ n
=
(
1
n 2
)
R H+R H
4
Contemporaneamente è stata dimostrata
l’esattezza della costante di Rydberg,
confrontando il suo valore noto con il valore
del coefficiente angolare della retta.
R Hteor (nm-1)R Hsper (nm-1)
1096,7 × 10-7(1095,9 ±1,6)×10-7
TAB 2. Nella tabella viene riportato il valore
teorico e sperimentale(compreso di errore)
della costante di Rydberg.
Come errore su R H è stato considerato
l’errore del parametro A della retta dei
minimi quadrati.
FIG 2.Nel grafico è rappresentata la retta dei
minimi quadrati per le linee di assorbimento
descritte dalla serie di Balmer.I punti in blu
rappresentano l‘inverso delle lunghezze d’onda
di emissione(in nm-1); l’errore è compreso nelle
dimensioni del punto.
L’errore s ulle y è stato ca lcolato con la
formula standard della propagazione degli
errori.
CONCLUSIONI
I risultati ottenuti sono consistenti con le
previsioni teoriche.Nello specifico le
lunghezze d’onda di emissione dell’idrogeno
mostrate nella tabella 1 sono consistenti con i
loro rispettivi valori teorici.
La quantizzazione dell’energia nei livelli
atomici è inoltre dimostrata in maniera più
diretta dall’andamento lineare rappresentato
nella figura 2.Un ulteriore conferma è data
dal valore della costante di Rydberg ricavato,
che(come si evince dalla tabella 2)è in
ottimo accordo con il valore teorico.
3
Questo contenuto è troppo lungo per poterlo leggere sul telefono? Salvalo per leggerlo dopo sul computer.
Salva in una Studylist
La scarsa preci sione,in termini di cifre
significative,di R H è principalmente dovuta
al basso numero di punti sperimentali con cui
è stata fatta la retta dei mini mi quadrati.
INFORMAZION AGGIUNTIVE
La misura dello spettro del la lampada al
mercurio è stata eseguita in continuo(con 20
accumulazioni).Il tempo di integrazione è
stato impostato su 7s.
E’stata quindi fatta una calibrazione,
supponendo una correlazione lineare tra λ
misurate e λ note.
λ TEOR=A λ MIS+B
TAB 4.Nella tabella vengono riportati i valori
dei coefficienti della retta dei minimi quadrati
per la lampada al mercurio.
I parametri A e B sono stati utilizzati per
calcolare l’errore sull e lunghezze d’onda di
emissione della tabella 1.
Δ λ TEOR=
√
(
∂λ TEOR
∂A ΔA
)
2
+
(
∂λ TEOR
∂λ MIS
Δ λ MIS
)
2
+
(
ΔB
)
2
4
A B
1,0066±0,0013-1,17±0,54
1 su 4
Condividi
Scarica
Scarica
Più contenuti da:Scienza politica(SPS/04)
Più contenuti da:
Scienza politicaSPS/04Università degli Studi di Cagliari
40 documenti
Vai al corso
23 Schemi-lineamenti-di-diritto-pubblico-falcon 2 Scienza politica Altro 100% (4)
28 Dispense aggiuntive - Maurizio COTTA, Donatella DELLA PORTA, Leonardo MORLINO, Scienza politica, Bolo-gna,Scienza politica Appunti di lezione 100% (3)
15 Riassunto postdemocrazia colin crunch Scienza politica Riassunti 100% (2)
11 Capitolo 10 Politiche Pubbliche Scienza politica Appunti di lezione 100% (1)
Più contenuti da: Scienza politicaSPS/04Università degli Studi di Cagliari40 documenti Vai al corso
23 Schemi-lineamenti-di-diritto-pubblico-falcon 2 Scienza politica 100% (4)
28 Dispense aggiuntive - Maurizio COTTA, Donatella DELLA PORTA, Leonardo MORLINO, Scienza politica, Bolo-gna, Scienza politica 100% (3)
15 Riassunto postdemocrazia colin crunch Scienza politica 100% (2)
11 Capitolo 10 Politiche Pubbliche Scienza politica 100% (1)
98 Manuale di Scienza politica, Bologna, il Mulino, 2017 Scienza politica 75% (4)
8 Storia dell'Italia dal 1850 all'Unità Scienza politica Nessuno
Consigliati per te
23 Schemi-lineamenti-di-diritto-pubblico-falcon 2 Scienza politica Altro 100% (4)
28 Dispense aggiuntive - Maurizio COTTA, Donatella DELLA PORTA, Leonardo MORLINO, Scienza politica, Bolo-gna,Scienza politica Appunti di lezione 100% (3)
15 Riassunto postdemocrazia colin crunch Scienza politica Riassunti 100% (2)
23 Schemi-lineamenti-di-diritto-pubblico-falcon 2 Scienza politica 100% (4)
28 Dispense aggiuntive - Maurizio COTTA, Donatella DELLA PORTA, Leonardo MORLINO, Scienza politica, Bolo-gna, Scienza politica 100% (3)
15 Riassunto postdemocrazia colin crunch Scienza politica 100% (2)
Gli studenti hanno anche visualizzato
Capitolo 4 La dem. Tradita
storia stato sociale
Capitolo 11 Dimensione Sovranazionale Governace Multilivello
Capitolo 12 Crisi Della Politica
Piano Genitoriale Base - Istruzioni e Indicazioni Utili
Sistemi Maggioritari e Proporzionali: Analisi e Confronto
Altri documenti collegati
Domande Esame - Appunti prima parte
Ne bis in idem e omesso versamento iva
La Conferenza di Servizi
Diritto dell'Unione Europea
Diritto tribuario pdf
CAPITOLO 7: DEVIAZIONE E CRIMINE - ANALISI SOCIALE E TEORIE
Fatti aiutare a fare i compiti dall'IA, con l'app di Studocu Apri l'app
Italiano Italia
Azienda
Chi Siamo
Studocu Premium
Integrità accademica
Lavora con noi
Blog
Dutch Website
Strumenti di studio
Tutti gli strumenti
Ask AI
AI Notes
Generatore di Quiz IA
Notes to Quiz Videos
Notes to Audio
Generatore di infografiche
Contatti & Aiuto
F.A.Q.
Contattaci
Sala stampa
Aspetti Legali
Termini d'utilizzo
Informativa sulla privacy
Cookie Settings
Dichiarazione sui cookie
Copyright & DSA
Vedi le nostre recensioni su Trustpilot
Italiano Italia
Studocu non è affiliata a né sostenuta da alcun college, scuola o università.
Copyright © 2025 StudeerSnel B.V., Keizersgracht 424-sous, 1016 GC Amsterdam, KVK: 56829787, BTW: NL852321363B01
I cookie ti offrono un'esperienza personalizzata.
Non stiamo parlando del tipo croccante e gustoso. Questi cookie ci aiutano a mantenere sicuro il nostro sito web, offrirti un'esperienza migliore e mostrare annunci più pertinenti. Non li attiveremo a meno che tu non accetti. Vuoi saperne di più o modificare le tue preferenze?
Rifiuta tutti Accetta tutti i cookie
Impostazioni cookie |
189206 | https://teachy.ai/en/summaries/high-school/10th-grade/physics-en/summary-of-vectors-decomposition | Log In
Summary of Vectors: Decomposition
Lara from Teachy
SubjectPhysics
Physics
SourceTeachy Original
Teachy Original
TopicVectors: Decomposition
Vectors: Decomposition
Introduction
Relevance of the Topic
The concept of Vectors and their Decompositions is fundamental in Physics. They allow us to describe and understand physical quantities that have magnitude and direction, such as velocity, force, electric field, among others. Strengthening the knowledge of subjects like Calculus, Kinematics, and Newton's Laws. Mastering the idea of Vector Decomposition enhances the understanding of these quantities and directly impacts the resolution of complex problems in Physics.
Contextualization
Within the vast field of Natural Sciences, Physics is the mother discipline that studies the laws and properties of matter and energy. In the High School curriculum, the subject of Physics comes into play to deepen concepts of Basic Science and prepare students for more complex topics, such as the Study of Light, Thermodynamics, and Quantum Physics. In this scenario, the study of Vectors and their Decompositions holds a prominent place. It constitutes one of the pillars for the understanding of subsequent topics, therefore, it is a crucial step in the students' learning journey.
Theoretical Development
Components
Vector: The main characteristic of a vector is that it has both magnitude (i.e., size) and direction. Generally, a vector is represented by an arrow whose length is proportional to its magnitude and points in the direction of its effect. In Physics, vectors help describe motion, force, and other related phenomena.
Magnitude and Direction of a Vector: Magnitude refers to the quantity of something. For example, the magnitude of a car's velocity can be 100 km/h. Direction, on the other hand, is the path the vector takes to reach its destination. For example, the direction of velocity can be north or south.
Components of a Vector: Vectors can be subdivided into components. This decomposition is important to understand how vectors act in different directions. For example, a vector can be decomposed into its x and y components if it is acting in a two-dimensional Cartesian plane.
Key Terms
Vector Decomposition: The process of decomposing a vector involves separating the original vector into its x and y components or any other combination of components that are suitable for the problem at hand. This operation is crucial for solving physics problems involving vectors.
Cartesian Plane: A mathematical tool used to describe the location of points in a two-dimensional plane. It is formed by two lines, one horizontal (x-axis) and one vertical (y-axis), that intersect at a point called the origin. Each point in the plane is identified by an ordered pair (x, y), which represents the coordinates of the point relative to the origin.
Examples and Cases
Force Decomposition: Imagine a force F applied to an object at an angle of 30 degrees with the horizontal. To decompose this force, you can separate it into its x and y components using the 30-degree angle. The x component, called Fx, is the projection of F along the x-axis, and the y component, called Fy, is the projection of F along the y-axis.
Velocity Decomposition: Suppose an object is moving with a velocity v forming an angle of 45 degrees with the positive direction of the x-axis. To decompose this velocity, you can separate it into its x and y components using the 45-degree angle. The x component, called vx, is the projection of v along the x-axis, and the y component, called vy, is the projection of v along the y-axis.
Detailed Summary
Key Points
Vector and its Characteristics: The main characteristic of a vector is the combination of its magnitude (i.e., size) and direction. This idea is crucial to understand how vectors are used in Physics to describe natural phenomena.
Vector Decomposition: Vector Decomposition is the act of separating an original vector into its x and y components (in a two-dimensional plane) or into x, y, and z components (in a three-dimensional space). Mastering this technique is essential to solve physical problems involving vectors.
Components of a Vector: The components of a vector are the independent parts that, when summed, form the original vector. In a two-dimensional plane, a vector can have two components (x and y), while in a three-dimensional space, a vector can have three components (x, y, and z).
Conclusions
Vector Decomposition is an important tool in Physics that allows for easy manipulation of vectors in different directions. It helps us calculate and understand how vectors act in various physical contexts.
Vector Decomposition is a reversible process. That is, you can combine the individual components to obtain the original vector. This is useful to verify the work done when decomposing a vector.
Exercises
Force Decomposition: Given a force F of 500 N applied to an object at an angle of 60 degrees with the horizontal. Determine the Decomposition of F into its x and y components.
Velocity Decomposition: An object is moving with a velocity v of 50 m/s in a direction that forms an angle of 30 degrees with the positive direction of the x-axis. Determine the Decomposition of v into its x and y components.
Magnetic Field Decomposition: A magnetic field B has a magnitude of 2 T and is pointing in the direction of the x-axis. Perform the Decomposition of B in a two-dimensional situation considering angles with the positive direction of the x-axis of 0, 45, and 90 degrees.
Want access to more summaries?
On the Teachy platform, you can find a variety of resources on this topic to make your lesson more engaging! Games, slides, activities, videos, and much more!
People who viewed this summary also liked...
Summary
Waves: Equation | Active Summary
Lara from Teachy
-
Summary
Gravitation: Escape Velocity | Active Summary
Lara from Teachy
-
Summary
Modern Physics: Photons | Active Summary
Lara from Teachy
-
Summary
Gravitation: Gravitation Problems | Traditional Summary
Lara from Teachy
-
Join a community of teachers directly on WhatsApp
Connect with other teachers, receive and share materials, tips, training, and much more!
We reinvent teachers' lives with artificial intelligence
Audiences
TeachersStudentsSchools
ToolsSlidesQuestion BankLesson plansLessonsActivitiesSummariesBooks
2025 - All rights reserved
Terms of Use
Privacy Notice
|
Cookies Notice
| |
189207 | https://www.merriam-webster.com/sentences/frostbite | Examples of 'FROSTBITE' in a Sentence | Merriam-Webster
Chatbot
Chatbot
Games
Word of the Day
Grammar
Word Finder
Slang
NewNewsletters
Wordplay
Rhymes
Thesaurus
Join MWU
More
Games
Word of the Day
Grammar
Wordplay
Slang
Rhymes
Word Finder
Newsletters New
Thesaurus
Join MWU
Shop
Books
Merch
Log In
Username
My Words
Recents
Account
Log Out
Est. 1828
Example Sentencesfrostbite
noun
How to Use frostbite in a Sentence
frostbite
noun
Definition of frostbite
He wore gloves to prevent frostbite.
The most common sites for frostbitein dogs are their ears and the tips of their tails.
—Discover Magazine, 1 Feb. 2024
The more wind, the faster the body loses heat and the greater the chance of hypothermia or frostbite.
—NBC News, 16 Jan. 2024
Toss some in your gloves or boots in a pinch and avoid the dangers of frostbite.
— Collin Morgan, Car and Driver, 4 Jan. 2023
Fingers and toes are often the first to get cold — and among the biggest risks for frostbite.
—Washington Post, 6 Nov. 2020
Wet feet are the quickest way to blisters, frostbite, and even foot rot.
—Outdoor Life, 13 May 2020
The cub lost portions of both ears due to the frostbitefrom exposure, CPW said.
— David Williams, CNN, 15 May 2021
As was our car on the drive back — despite the probable frostbitein our hands (and paws).
— Travis M. Andrews, Washington Post, 12 Mar. 2024
In the 1960s, one such can caused frostbiteupon being touched.
— Desperation Town, ProPublica, 11 May 2020
Wear heavy gloves and boots to prevent cuts, bruises and frostbite.
— Julie Washington, cleveland, 28 Dec. 2020
He was found in the freezing cold, with what appears to be the beginnings of frostbiteon one paw.
— Jessie Yeung, CNN, 25 Nov. 2019
By then, Rousseau, too, had frostbiteacross his fingers.
— John Branch, New York Times, 1 Dec. 2023
Place a cloth between the ice pack and your skin, and apply the ice for no longer than 15 minutes to prevent frostbite.
— Nick Blackmer, Verywell Health, 29 June 2023
Be aware of the symptoms for frostbiteand hypothermia.
— Tamia Fowlkes, Journal Sentinel, 26 Feb. 2024
Ropes snap, skiffs break, fingers get exposed to the wind and frostbitequickly takes over.
— Ben Taub, The New Yorker, 28 Dec. 2021
The coldest wind chills can cause frostbitein a little as 10 minutes.
— Rob Shackelford, CNN, 30 Jan. 2023
To this day most Battle of the Bulge veterans suffer from the effects of frostbite.
— Bill Newcott, National Geographic, 13 Dec. 2019
Treatment for mild frostbiteincludes re-warming of the skin.
— Staff, Hartford Courant, 2 Feb. 2023
Symptoms of frostbiteinclude a loss of feeling and a white or pale appearance in fingers, toes, ear lobes, and the tip of the nose.
— Brittany Bowker, BostonGlobe.com, 3 Feb. 2023
That led to frostbite, said Tracy Woodard, a team leader for Intown Cares.
— Matt Kempner, ajc, 23 June 2023
The rescuers plowed her driveway and took Mr. White, who had severe frostbite, to the hospital.
— Sarah Maslin Nir, New York Times, 27 Dec. 2022
Still, the movie's DNA is on full display (arm-shattering frostbite!
— Wired Staff, WIRED, 20 July 2019
She was left with lingering effects of frostbitefor more than 50 years.
— Ed Shanahan, New York Times, 5 Oct. 2022
The final stage is severe frostbite, which results in numbness and skin that turns hard and black due to skin cells dying.
— Mary Kekatos, ABC News, 16 Jan. 2024
Lance had lost his mitts and was huddled close to a barrel stove, his fingers black with frostbitefrom cruel cold.
— Marc Lester, Anchorage Daily News, 1 Mar. 2023
There are fans whirring, windows open, and a space heater chug-chugging to keep my toes, still tucked inside winter boots, just north of frostbite.
— Maggie Bullock, Vogue, 1 June 2021
McBride survived the battle but ended up with frostbite.
—oregonlive, 9 Nov. 2020
Patches of mild frostbitewill have a dull waxy appearance.
— Tim MacWelch, Outdoor Life, 30 Dec. 2019
The early stage of frostbiteis frostnip, which doesn’t result in permanent damage to the skin.
— Chris Smith, BGR, 21 Feb. 2022
The weather service has placed widespread wind chill advisories and watches and has advised people to take precautions to avoid frostbiteand hypothermia.
—USA TODAY, 12 Jan. 2024
Some of these examples are programmatically compiled from various online sources to illustrate current usage of the word 'frostbite.' Any opinions expressed in the examples do not represent those of Merriam-Webster or its editors. Send us feedback about these examples.
Last Updated: 5 Sep 2024
More from Merriam-Webster
### Can you solve 4 words at once? Play Play ### Can you solve 4 words at once? Play Play
Word of the Day
obliterate
See Definitions and Examples »
Get Word of the Day daily email!
Merriam Webster
Learn a new word every day. Delivered to your inbox!
Help
About Us
Advertising Info
Contact Us
Privacy Policy
Terms of Use
Facebook
Twitter
YouTube
Instagram
© 2025 Merriam-Webster, Incorporated |
189208 | https://ws.engr.illinois.edu/sitemanager/getfile.asp?id=2853 | An initially uniform electric field, ˆ( ) oE r E z =
r r exists in between two infinitely large, parallel conducting plates, separated by a small vertical distance, d. The lower plate is at ground potential (0 volts) and the upper plate is held at a constant potential oV− . A hemi-spherical dielectric, with dielectric constant 1oK ε ε ≡ > and radius R d<< is placed on the lower plate with its center at the origin of coordinates of the parallel-plate capacitor as shown in the above figure. (a) State the boundary conditions for ( ) E r
r r and ( ) D r
r r at the surface of the hemi-spherical dielectric. (b) Calculate the potential inside ( ( ) in rΦ r ) and outside ( ( ) out rΦ r ) the hemi-spherical dielectric. Hint: You may use the following formula (without derivation):
where r and θ are spherical polar coordinates.
(c) Calculate the electric field inside ( ( ) in E r
r r ) and outside ( ( ) out E r
r r ) the hemi-spherical dielectric. (d) Make a detailed, careful sketch of the electric field lines and the equipotentials (lines of constant potential) between the parallel plates, and also inside and outside the hemi- spherical dielectric. Draw the electric field lines as solid lines and the equipotentials as dashed lines. (e) Calculate the bound charge density, bound σ on the upper surface of the hemi-spherical dielectric. Express yo ur answer in terms of 0 0 ,E V d K≡ and θ .
10
1( ) (cos )r a r b Pr θ
∞+=
Φ = +
∑ ll l lll
r
RKd
−Vo
0
yz
x
(0,0,0) |
189209 | https://mathworld.wolfram.com/MinimalSurfaceofRevolution.html | Minimal Surface of Revolution
Download Wolfram Notebook
Calculus of variations can be used to find the curve from a point to a point which, when revolved around the x-axis, yields a surface of smallest surface area (i.e., the minimal surface). This is equivalent to finding the minimal surface passing through two circular wire frames. The area element is
so the surface area is
and the quantity we are minimizing is
This equation has , so we can use the Beltrami identity
to obtain
which is called a catenary, and the surface generated by rotating it is called a catenoid. The two constants and are determined from the two implicit equations
| |
| (13) |
| (14) |
which cannot be solved analytically.
The general case is somewhat more complicated than this solution suggests. To see this, consider the minimal surface between two rings of equal radius . Without loss of generality, take the origin at the midpoint of the two rings. Then the two endpoints are located at and , and
| |
| (15) |
But , so
| |
| (16) |
Inverting each side
| |
| (17) |
so (as it must by symmetry, since we have chosen the origin between the two rings), and the equation of the minimal surface reduces to
| |
| (18) |
At the endpoints
| |
| (19) |
but for certain values of and , this equation has no solutions. The physical interpretation of this fact is that the surface breaks and forms circular disks in each ring to minimize area. Calculus of variations cannot be used to find such discontinuous solutions (known in this case as Goldschmidt solutions). The minimal surfaces for several choices of endpoints are shown above. The first two cases are catenoids, while the third case is a Goldschmidt solution.
To find the maximum value of at which catenary solutions can be obtained, let . Then (17) gives
| |
| (20) |
Now, denote the maximum value of as . Then it will be true that . Take of (20),
| |
| (21) |
Now set
| |
| (22) |
From (20),
| |
| (23) |
Take (23) (22),
| |
| (24) |
Defining ,
| |
| (25) |
This has solution . From (22), . Divide this by (25) to obtain , so the maximum possible value of is
| |
| (26) |
Therefore, only Goldschmidt ring solutions exist for .
The surface area of the minimal catenoid surface is given by
| |
| (27) |
but since
| |
| (28) |
| (29) |
| |
| (30) |
| (31) |
| (32) |
| (33) |
| (34) |
| (35) |
| (36) |
Some caution is needed in solving (◇) for . If we take and then (◇) becomes
| |
| (37) |
which has two solutions: ("deep"), and ("flat"). However, upon plugging these into (◇) with , we find and . So is not, in fact, a local minimum, and is the only true minimal solution.
The surface area of the catenoid solution equals that of the Goldschmidt solution when (◇) equals the area of two disks,
| |
| (38) |
| |
| (39) |
| |
| (40) |
Plugging in
| |
| (41) |
| |
| (42) |
Defining
| |
| (43) |
gives
| |
| (44) |
This has a solution . The value of for which
| |
| (45) |
is therefore
| |
| (46) |
For , the catenary solution has larger area than the two disks, so it exists only as a local minimum.
There also exist solutions with a disk (of radius ) between the rings supported by two catenoids of revolution. The area is larger than that for a simple catenoid, but it is a local minimum. The equation of the positive half of this curve is
| |
| (47) |
At ,
| |
| (48) |
At ,
| |
| (49) |
The area of the two catenoids is
| |
| (50) |
| (51) |
| (52) |
Now let , so
| |
| (53) |
| (54) |
| (55) |
| (56) |
| (57) |
The area of the central disk is
| |
| (58) |
so the total area is
| |
| (59) |
By Plateau's laws, the catenoids meet at an angle of , so
| |
| (60) |
| (61) |
| (62) |
and
| |
| (63) |
This means that
| |
| (64) |
| (65) |
| (66) |
so
| |
| (67) |
Now examine ,
| |
| (68) |
where . Finding the maximum ratio of gives
| |
| (69) |
| |
| (70) |
with as given above. The solution is , so the maximum value of for two catenoids with a central disk is .
If we are interested instead in finding the curve from a point to a point which, when revolved around the y-axis (as opposed to the x-axis), yields a surface of smallest surface area , we proceed as above. Note that the solution is physically equivalent to that for rotation about the x-axis, but takes on a different mathematical form. The area element is
| |
| (71) |
| |
| (72) |
and the quantity we are minimizing is
| |
| (73) |
Taking the derivatives gives
| |
| (74) |
| |
so the Euler-Lagrange differential equation becomes
Solving for then gives
which is the equation for a catenary. The surface area of the catenoid product by rotation is
| |
| |
| |
| |
| |
| |
Isenberg (1992, p. 80) discusses finding the minimal surface passing through two rings with axes offset from each other.
See also
Minimal Surface, Surface of Revolution
Explore with Wolfram|Alpha
More things to try:
binomial distribution n=40, p=0.32
factor x^12 - y^12
interval [-sqrt(5), 1+sqrt(5)]
References
Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 931-937, 1985.Goldstein, H. Classical Mechanics, 2nd ed. Reading, MA: Addison-Wesley, p. 42, 1980.Isenberg, C. The Science of Soap Films and Soap Bubbles. New York: Dover, pp. 79-80 and Appendix III, 1992.
Referenced on Wolfram|Alpha
Minimal Surface of Revolution
Cite this as:
Weisstein, Eric W. "Minimal Surface of Revolution." From MathWorld--A Wolfram Resource.
Subject classifications |
189210 | https://brainly.com/question/39582155 | [FREE] Lithium nitride has the formula \text{Li}_3\text{N}. What is the charge on the nitrogen ion? - brainly.com
Search
Learning Mode
Cancel
Log in / Join for free
Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions
Log in
Join for free
Tutoring Session
+54,5k
Smart guidance, rooted in what you’re studying
Get Guidance
Test Prep
+10,5k
Ace exams faster, with practice that adapts to you
Practice
Worksheets
+6,7k
Guided help for every grade, topic or textbook
Complete
See more
/
Chemistry
Textbook & Expert-Verified
Textbook & Expert-Verified
Lithium nitride has the formula Li 3N. What is the charge on the nitrogen ion?
1
See answer Explain with Learning Companion
NEW
Asked by hfuller6744 • 10/07/2023
0:01
/
0:15
Read More
Community
by Students
Brainly
by Experts
ChatGPT
by OpenAI
Gemini
Google AI
Community Answer
This answer helped 6233309 people
6M
0.0
0
Upload your school material for a more relevant answer
The ionic charge on the nitrogen ion in lithium nitride (Li3N) is -3, because nitrogen typically carries a -3 charge to reach a stable electron configuration.
Explanation
You're asking about the charge on the nitrogen ion in the compound Lithium Nitride, represented by the formula Li3N. To determine this, we consider the typical charges of the elements involved.
Lithium (Li) is an alkali metal which typically carries a +1 charge. Nitrogen (N), on the other hand, typically carries a -3 charge in order to achieve a stable electron configuration.
In the compound Li3N, we have three lithium ions, each carrying a +1 charge, which sums to +3. To balance this, the nitrogen ion must carry a -3 charge. Therefore, the charge on the nitrogen ion in Lithium Nitride (Li3N) is -3.
Learn more about Ionic Charge here:
brainly.com/question/38743515
SPJ11
Answered by SwethaJ •36.8K answers•6.2M people helped
Thanks 0
0.0
(0 votes)
Textbook &Expert-Verified⬈(opens in a new tab)
This answer helped 6233309 people
6M
0.0
0
The Basics of General, Organic, and Biological Chemistry - David W Ball
Basics of General, Organic, and Biological Chemistry - David W. Ball, John W. Hill, Rhonda J. Scott
Inorganic Chemistry
Upload your school material for a more relevant answer
In lithium nitride (Li3N), the charge on the nitrogen ion is -3, as nitrogen typically forms a -3 anion. This is necessary to balance the +3 charge contributed by three lithium ions. Thus, the nitrogen ion is represented as N³-.
Explanation
The
ionic charge on the nitrogen ion in lithium nitride, represented by the formula Li 3N, is -3. To understand this, we need to look at the typical charges of the elements involved.
Lithium (Li) is an alkali metal that typically has a charge of +1. This means each lithium ion contributes +1 to the overall charge of the compound.
Nitrogen (N) is a nonmetal found in group 15 of the periodic table. It typically forms an anion with a charge of -3. This is because nitrogen needs to gain three electrons to achieve the stable electron configuration of the nearest noble gas, neon.
In lithium nitride, there are three lithium ions contributing a total charge of +3:
3×+1=+3
To balance the +3 charge from the lithium ions, the nitrogen ion must have a charge of -3. Therefore, its charge can be represented as:
N 3−
This chemical balance results in a neutral compound, as the sum of the positive and negative charges equals zero:
+3+(−3)=0
In conclusion, the charge on the nitrogen ion in lithium nitride is indeed -3.
Examples & Evidence
For example, in compounds like aluminium nitride (AlN), aluminium has a charge of +3, and nitrogen again has a charge of -3, resulting in a neutral compound. Similarly, magnesium nitride (Mg3N2) involves magnesium with a charge of +2 and nitrogen still with -3, requiring two nitrogen ions for every three magnesium ions to maintain overall balance.
These conclusions are based on the standard oxidation states of lithium (+1), nitrogen (-3), and the rules governing the formation of ionic compounds.
Thanks 0
0.0
(0 votes)
Advertisement
hfuller6744 has a question! Can you help?
Add your answer See Expert-Verified Answer
### Free Chemistry solutions and answers
Community Answer 1 Nitrogen reacts with lithium to form the ionic compound lithium nitride, Li3N. (i) Write the equation for the reaction between lithium and nitrogen.
Community Answer lithium nitride consists of two ions chemically bonded together what are the charges of each ion
Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain?
Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized
Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __
Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta
Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction.
Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements
Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power.
Community Answer 5 Dimensional Analysis 1. I have 470 milligrams of table salt, which is the chemical compound NaCl. How many liters of NaCl solution can I make if I want the solution to be 0.90% NaCl? (9 grams of salt per 1000 grams of solution). The density of the NaCl solution is 1.0 g solution/mL solution.
New questions in Chemistry
Reactants undergo chemical reaction to form products. This chemical equation represents one such reaction. The coefficient for one of the reactants or products is incorrect. Which part of the chemical equation is incorrect? 2 C 4H 1010 O 2⋯8 C O 2+10 H 2O
Which of the samples most likely had the highest solubility? | Sample | Name | Chemical formula | Temperature of water (∘C) | :---: :---: | | 1 | Table sugar | C 12H 22O 11 | 80 | | 2 | Table sugar | C 12H 22O 11 | 45 | | 3 | Table salt | NaCl | 55 | | 4 | Table salt | NaCl | 63 | \bigcirc 1 \bigcirc 2 \bigcirc 3 \bigcirc 4
Which of the samples most likely had the highest solubility? | Sample | Name | Chemical formula | Temperature of water ( ∘C ) | :--- :--- | | 1 | Table sugar | C 12H 22O 11 | 80 | | 2 | Table sugar | C 12H 22O 11 | 45 | | 3 | Table salt | NaCl | 55 | | 4 | Table salt | NaCl | 63 | A. 1 B. 2 C. 3 D. 4
Consider the chemical equations shown here. 2 H 2(g)+O 2(g)→2 H 2O(g)Δ H 1=−483.6 k J+2=−241.8 k J/m o l 3 O 2(g)→2 O 3(g)Δ H 2=284.6 k J+2=142.3 k J/m o l What is the overall enthalpy of reaction for the equation shown below? 3 H 2(g)+O 3(g)→3 H 2O(g)□ kJ
Consider the chemical equations shown here. P 4(s)+3 O 2(g)→P 4O 6(s)Δ H 1=−1,640.1 k J P 4O 10(s)→P 4(s)+5 O 2(g)Δ H 2=2,940.1 k J What is the overall enthalpy of reaction for the equation shown below? Round the answer to the nearest whole number. P 4O 6(s)+2 O 2(g)→P 4O 10(s)
Previous questionNext question
Learn
Practice
Test
Open in Learning Companion
Company
Copyright Policy
Privacy Policy
Cookie Preferences
Insights: The Brainly Blog
Advertise with us
Careers
Homework Questions & Answers
Help
Terms of Use
Help Center
Safety Center
Responsible Disclosure Agreement
Connect with us
(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)
Brainly.com
Dismiss
Materials from your teacher, like lecture notes or study guides,
help Brainly adjust this answer to fit your needs.
Dismiss |
189211 | https://www.quora.com/How-can-you-find-the-angle-of-line-by-using-its-slope | Something went wrong. Wait a moment and try again.
Angles Between Lines
Plane Angle
Geometric Mathematics
Angles and Pairs of Angle...
Slope (Mathematics)
Lines and Angle
5
How can you find the angle of line by using its slope?
Robert Nichols
Author has 5K answers and 15.6M answer views
·
7y
Slope is rise/run. Rise is the change in y value, which will equal the length of the opposite side of the right triangle. Run is the change in x value, which is the length of the adjacent side. The trigonometric function Tangent is opposite/adjacent. Use the inverse tangent of the slope to find the angle.
The only catch is that some times inverse tangent will give you a negative angle. It is generally not acceptable to give the angle in a triangle as negative angle. If you get a negative angle, just use the absolute value of the angle.
Slope is rise/run. Rise is the change in y value, which will equal the length of the opposite side of the right triangle. Run is the change in x value, which is the length of the adjacent side. The trigonometric function Tangent is opposite/adjacent. Use the inverse tangent of the slope to find the angle.
The only catch is that some times inverse tangent will give you a negative angle. It is generally not acceptable to give the angle in a triangle as negative angle. If you get a negative angle, just use the absolute value of the angle.
Related questions
How do I determine the angle of a 1:12 slope?
How do I calculate the angle of a slope?
How do you find the slope of an angle?
How do I calculate the slope of a 30 degree angle?
How do I find the slope of a line if the angle between two lines and slope of other line is given ? See my question given in the link below .
Baijanath Tharu
Sometimes I am irrational like Pi.
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 92 answers and 234.3K answer views
·
9y
I assume you want to find the angle made by the line with the x-axis. In this context we can find the angle.
Slope of any line is given by the angle made by the line with the positve x-axis. Clearly we observe that if we know the slope of the line then we can find the angle.
Let the angle made by the line with the positive x-axis is θ and the slope is known then the angle θ is given by the formula:
I assume you want to find the angle made by the line with the x-axis. In this context we can find the angle.
Slope of any line is given by the angle made by the line with the positve x-axis. Clearly we observe that if we know the slope of the line then we can find the angle.
Let the angle made by the line with the positive x-axis is θ and the slope is known then the angle θ is given by the formula:
Stephen Zisk
Works at RedPoint Global Inc. (2016–present)
·
Author has 1.1K answers and 1.4M answer views
·
7y
Originally Answered: How do you find the angle based on the slope of a line?
·
The slope of a line is the rise over the run, which looks a lot like the tangent, so you can just take atan(slope) to get the angle. BUT, this only works for angles in
(
−
π
/
2
,
π
/
2
)
. That is, first and fourth quadrant. For vertical lines, the slope is undefined and the tangent of
\plusminus
π
/
2
is infinite. For angles in the second and third quadrants, the line “goes backward” and using the slope to get the angle will get you the complementary angle in the fourth and first quadrant, respectively.
L Viswanathan
Former Member of Technical Staff at Bell Telephone Labs (1977–1999)
·
Author has 3K answers and 1.4M answer views
·
7y
Originally Answered: How do you find the angle based on the slope of a line?
·
The slope of a line is the tangent of the angle the line makes with the horizontal axis (X-axis). The angle is therefore the arctangent of the slope number. If the slope is a non-negative number, the angle is between 0 and (pi/2) radians, else it is between (pi/2) radians and (pi) radians.
Promoted by The Penny Hoarder
Lisa Dawson
Finance Writer at The Penny Hoarder
·
Updated Sep 16
What's some brutally honest advice that everyone should know?
Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included.
And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did.
Cancel Your Car Insurance
You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix.
Don’t waste your time
Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included.
And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did.
Cancel Your Car Insurance
You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix.
Don’t waste your time browsing insurance sites for a better deal. A company calledInsurify shows you all your options at once — people who do this save up to $996 per year.
If you tell them a bit about yourself and your vehicle, they’ll send you personalized quotes so you can compare them and find the best one for you.
Tired of overpaying for car insurance? It takes just five minutes to compare your options with Insurify andsee how much you could save on car insurance.
Ask This Company to Get a Big Chunk of Your Debt Forgiven
A company calledNational Debt Relief could convince your lenders to simply get rid of a big chunk of what you owe. No bankruptcy, no loans — you don’t even need to have good credit.
If you owe at least $10,000 in unsecured debt (credit card debt, personal loans, medical bills, etc.), National Debt Relief’s experts will build you a monthly payment plan. As your payments add up, they negotiate with your creditors to reduce the amount you owe. You then pay off the rest in a lump sum.
On average, you could become debt-free within 24 to 48 months. It takes less than a minute to sign up and see how much debt you could get rid of.
Set Up Direct Deposit — Pocket $300
When you set up direct deposit withSoFi Checking and Savings (Member FDIC), they’ll put up to $300 straight into your account. No… really. Just a nice little bonus for making a smart switch.
Why switch? With SoFi, you can earn up to 3.80% APY on savings and 0.50% on checking, plus a 0.20% APY boost for your first 6 months when you set up direct deposit or keep $5K in your account. That’s up to 4.00% APY total. Way better than letting your balance chill at 0.40% APY.
There’s no fees. No gotchas.Make the move to SoFi and get paid to upgrade your finances.
You Can Become a Real Estate Investor for as Little as $10
Take a look at some of the world’s wealthiest people. What do they have in common? Many invest in large private real estate deals. And here’s the thing: There’s no reason you can’t, too — for as little as $10.
An investment called the Fundrise Flagship Fund lets you get started in the world of real estate by giving you access to a low-cost, diversified portfolio of private real estate. The best part? You don’t have to be the landlord. The Flagship Fund does all the heavy lifting.
With an initial investment as low as $10, your money will be invested in the Fund, which already owns more than $1 billion worth of real estate around the country, from apartment complexes to the thriving housing rental market to larger last-mile e-commerce logistics centers.
Want to invest more? Many investors choose to invest $1,000 or more. This is a Fund that can fit any type of investor’s needs. Once invested, you can track your performance from your phone and watch as properties are acquired, improved, and operated. As properties generate cash flow, you could earn money through quarterly dividend payments. And over time, you could earn money off the potential appreciation of the properties.
So if you want to get started in the world of real-estate investing, it takes just a few minutes tosign up and create an account with the Fundrise Flagship Fund.
This is a paid advertisement. Carefully consider the investment objectives, risks, charges and expenses of the Fundrise Real Estate Fund before investing. This and other information can be found in the Fund’s prospectus. Read them carefully before investing.
Cut Your Phone Bill to $15/Month
Want a full year of doomscrolling, streaming, and “you still there?” texts, without the bloated price tag? Right now, Mint Mobile is offering unlimited talk, text, and data for just $15/month when you sign up for a 12-month plan.
Not ready for a whole year-long thing? Mint’s 3-month plans (including unlimited) are also just $15/month, so you can test the waters commitment-free.
It’s BYOE (bring your own everything), which means you keep your phone, your number, and your dignity. Plus, you’ll get perks like free mobile hotspot, scam call screening, and coverage on the nation’s largest 5G network.
Snag Mint Mobile’s $15 unlimited deal before it’s gone.
Get Up to $50,000 From This Company
Need a little extra cash to pay off credit card debt, remodel your house or to buy a big purchase?
We found a company willing to help.
Here’s how it works: If your credit score is at least 620, AmONE can help you borrow up to $50,000 (no collateral needed) with fixed rates starting at 6.40% and terms from 6 to 144 months.
AmONE won’t make you stand in line or call a bank. And if you’re worried you won’t qualify, it’s free tocheck online. It takes just two minutes, and it could save you thousands of dollars.
Totally worth it.
Get Paid $225/Month While Watching Movie Previews
If we told you that you could get paid while watching videos on your computer, you’d probably laugh.
It’s too good to be true, right?
But we’re serious. By signing up for a free account with InboxDollars, you could add up to $225 a month to your pocket. They’ll send you short surveys every day, which you can fill out while you watch someone bake brownies or catch up on the latest Kardashian drama.
No, InboxDollars won’t replace your full-time job, but it’s something easy you can do while you’re already on the couch tonight, wasting time on your phone.
Unlike other sites, InboxDollars pays you in cash — no points or gift cards. It’s already paid its users more than $56 million.
Signing up takes about one minute, and you’ll immediately receive a $5 bonus to get you started.
Earn $1000/Month by Reviewing Games and Products You Love
Okay, real talk—everything is crazy expensive right now, and let’s be honest, we could all use a little extra cash. But who has time for a second job?
Here’s the good news. You’re already playing games on your phone to kill time, relax, or just zone out. So why not make some extra cash while you’re at it?
WithKashKick, you can actually get paid to play. No weird surveys, no endless ads, just real money for playing games you’d probably be playing anyway. Some people are even making over $1,000 a month just doing this!
Oh, and here’s a little pro tip: If you wanna cash out even faster, spending $2 on an in-app purchase to skip levels can help you hit your first $50+ payout way quicker.
Once you’ve got $10, you can cash out instantly through PayPal—no waiting around, just straight-up money in your account.
Seriously, you’re already playing—might as well make some money while you’re at it.Sign up for KashKick and start earning now!
Related questions
How do you go from the slope of a line to the angle from the x axis in degrees?
What is the formula for angle between 2 lines in slope form?
An angle between two lines is 45°, find the slope of L2, if the slope of L1 is 2/3?
A line makes an angle of 60° with the x-axis. What is the slope of the line? Explain how you know.
What angle is a 1 in 10 slope?
Drishti Sen
Student at Calcutta Airport English High School (2008–present)
·
Author has 74 answers and 126K answer views
·
6y
NOTHING TO WORRY !!!
!!! ITS VERY SIMPLE !!!
>>>>> You just need to know some basic formulae of straight lines. <<<<<
Slope of any straight line on the X-Y plane is given by tan θ , where θ is the angle which the straight line makes with the positive direction of x axis.
So , m = tan θ ( where m is the slope of that straight line which is inclined at an angle θ with the positive direction of x axis)
So, θ = tan^(-1)m
EXAMPLE , there is a straight line having a slope 1.
So, tan θ = 1
very clearly , θ = 45 degrees.
Madhav Polawar
Works at Student (AUC)
·
9y
You can easily do this just by taking inverse of tangent of the angle.
y=mx+c is the equation of line then m is slope of line.
y=((y2-y1)/(x2-x1))x+c is the two point form equation of line and (y2-y1)/(x2-x1) is the slope of line.
tan (angle)= slope of line.
angle=tan^(-1) (slope of line).
Sponsored by Stake
Stake: Online Casino games - Play & Win Online.
Play the best online casino games, slots & live casino games! Unlock VIP bonuses, bet with crypto & win.
Aleks Matveev
Studied at Massey - Vanier High School
·
7y
Originally Answered: How do you find the angle based on the slope of a line?
·
Find the components of the line (x,y)
And then use Tan ^-1(y/x)= Theta or angle
I hope you understand, and I go explain further.
-Aleks
STRUGGLES OF LIFE 2
Roger Bliss
Engineer / Programmer (1987–present)
·
Author has 21.4K answers and 8.4M answer views
·
Jan 5
Originally Answered: Can finding the slope of a line help determine the angle of the line?
·
yes, knowing the rise over run lets you use arctan() to get the angle - you might have to convert from radians depending on the arctan(() function and the metric for your angle…
Sponsored by Grammarly
Is your writing working as hard as your ideas?
Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step.
Daniel Doru Stanescu
Statistician at Eurostat (2010–present)
·
Author has 291 answers and 127.3K answer views
·
7y
Originally Answered: How do you find the angle based on the slope of a line?
·
The slope is the tangent (trigonometric function) of the angle. So if you have a scientific handheld computer or a trigonometric table you find the angle using the reverse function arctan.
Richard Polunsky
I've been interested in math since I was very young.
·
Author has 8.6K answers and 2M answer views
·
7y
Originally Answered: How do you find the angle based on the slope of a line?
·
Use the arctangent function.
arctan (slope ratio) = slope in degrees
where slope ratio = rise (y difference) of the line segment divided by run (x difference) of the line segment
Robert Nichols
Author has 5K answers and 15.6M answer views
·
Updated 6y
How do I determine the angle of a 1:12 slope?
How do I determine the angle of a 1:12 slope?
This is what we use inverse trigonometry for. If you know the lengths of both the opposite side in this case 1 (the rise), and the adjacent side 12 (the run), use tan−1(1/12) = approximately 4.76°.
Or if you prefer to measure in radians, the angle is 0.833333…
How do I determine the angle of a 1:12 slope?
This is what we use inverse trigonometry for. If you know the lengths of both the opposite side in this case 1 (the rise), and the adjacent side 12 (the run), use tan−1(1/12) = approximately 4.76°.
Or if you prefer to measure in radians, the angle is 0.833333…
Lucius Vorenus
You know the slope is: rise/run or distance traveled vertically divided by distance traveled horizontally . We can also write this as y/x. Make a right triangle with the slope and find the hypotenuse. Then use trigonometry to find the angle.
For example:
A line has slope 3/4.
arctan3/5= 30.96 degrees
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.8M answer views
·
7y
Related
How do I calculate the slope of a 30 degree angle?
You simply find
tan(30)
on your calculator or better still you use the following famous special triangle:
You simply find
tan(30)
on your calculator or better still you use the following famous special triangle:
Related questions
How do I determine the angle of a 1:12 slope?
How do I calculate the angle of a slope?
How do you find the slope of an angle?
How do I calculate the slope of a 30 degree angle?
How do I find the slope of a line if the angle between two lines and slope of other line is given ? See my question given in the link below .
How do you go from the slope of a line to the angle from the x axis in degrees?
What is the formula for angle between 2 lines in slope form?
An angle between two lines is 45°, find the slope of L2, if the slope of L1 is 2/3?
A line makes an angle of 60° with the x-axis. What is the slope of the line? Explain how you know.
What angle is a 1 in 10 slope?
A line makes an angle of 45 degrees with the x-axis. What is the slope of the line?
How do you find an angle from a slope and point on a graph?
How would I find the slope angle of a line y = 3x + 1?
What is meant by the slope of a straight line? Why is it used to find an angle of an equation?
What is the slope of a line with a given angle?
About
·
Careers
·
Privacy
·
Terms
·
Contact
·
Languages
·
Your Ad Choices
·
Press
·
© Quora, Inc. 2025 |
189212 | https://www.youtube.com/watch?v=4CbaVMxzLf0 | Ordering fractions using cross multiplication
Jim Meyer
335 subscribers
84 likes
Description
19375 views
Posted: 6 Oct 2013
Really we are using estimation and cross multiplication to figure out the order of fractions from least to greatest!
Transcript:
okay this is the third video for uh putting fractions in order from least to greatest and we've got three fractions here we've got to figure out which one is the least which one is the greatest and which one goes in the middle U the first step you remember always try and estimate find out which ones are greater than half or less than half and maybe even one will be greater than one whole but let's find out um let's see 2/3 2 out of three that's more than half I'm going to write a little M right there and four fths that's obviously more than half and two out of eight half of eight is four two is a lot less than that so what we can say is that just by looking at these we know that 28 is going to be the least okay so there's the least and I'm going to cross this out we don't need that anymore boom boom I'm left with 24s and four fths and I could try and find a common denominator but we're going to look at a new trick today called cross multiplying so uh what we're going to do is uh write 2/3 and 4 fs and by cross multiplying what I'm mean is I'm going to take this denominator the three and cross multiply we make a little uh up and diagonal motion 3 4 I going to write this 4 3 is 12 and now I'm going to take the five and I'm going to multiply the five the 2 2 five and that turns out to be 10 okay now you may have seen this before and you thought you were taught to think of this and this and see which one of those is the greater number right but I want to show you why that works really what you're doing if you multiply this top number by three secretly you're also multiplying the bottom number by three and we're going to do to get 15 and over here if you multiply the top number by five you're actually going to be multiplying secretly the bottom number by five and we're getting 15 and voila you have found a common denominator right so we can think of 2/3 we have made an equivalent fraction called 10 15s right we multiply the top and the Bottom by five and if we look at 4 fifths we' made an equivalent fraction called 12 15 and we have secretly figured out a common denominator or in a sneaky way right now so we know that since the denominator is the same we can just look at the numerator so which one of these is the smallest turns out that 10 15 I should change the color here 10 15 is less than 12 15 so 10 15 is in the middle and our Lar largest fraction is going to be 12 15 okay and that is the greatest okay 12 15 is the greatest so this cross multiplying trick finds a common denominator I I'm just going to show you that often we don't really bother figuring out the bottom part right because this trick shows us and I call it a trick because you're kind of cheating a little bit you're you're finding the answer without finding a common denominator um so this trick shows us the the the numerators right that one was 12 and that one was 10 so you look at the greatest numerator and that is the greater fraction okay let's try one more boom now uh first of all of course your first step always estimate um 6 out of 8 more than half 4 out of 12 less than half so let's write a L up here more 7 out of 9 definitely more than half okay so 4 12ths is the least and I'm left with uh 68 and 7 9ths 68 and 7 9ths so let's write these down I've got 68s and I've got seven nths and I'm going to use this cross multiplication trick right so here we go 7 8 is 56 and 9 6 is 54 so we can see if these are the numerators this is the bigger fraction 7 9ths is going to be the bigger fraction all right and so if that's the greatest one then obviously 68 is the one that's going to be in the middle 68 and there it is now I I want to say one other thing uh a lot of people are going to insist that they can just tell which fraction is the biggest say oh I knew 79ths was the biggest and that's a terrible mistake because let's check Che get this out um If I multiply this top number by eight and I multiply the bottom number by eight we get a denominator of 72 and If I multiply this top number by 9 to get 54 I multiply the bottom the denominator by 9 and we get 72 let's look at these two fractions 54 out of 72 versus 56 out of 72 70 seconds right if something is cut into 72 pieces those pieces are really small right really small pieces and this one only has two more of those tiny pieces right so if you think you can just look at 68 and 7 9ths and see a difference just mentally estimate a difference of two 70 seconds you're you're fooling yourself right so I don't want you to do that right being honest and owning what you know and what you don't know is going to help you be more successful right you don't want to guess and try and trick yourself into thinking that you really know what you're doing when fractions are this close you have to use math to check all right good luck |
189213 | http://www.hccs.edu/departments/division-of-instructional-services/hcc-texas-success-initiative-assessment/optional-math-review/section--7---polynomials/section-6-solving-polynomial-equations-by-factoring/ | About the College
Instructional Locations
Please select
Explore
Areas of Study
Extended Learning
Support Services
Apply & Enroll
Financial Aid & Scholarships
Paying for College
Academic Support
Career Planning
HCC's Career & Job Placement Services Center offers career resources, employment opportunities, and internship possibilities.
Additional Services
Health & Safety
About Our Campuses
With several locations throughout Houston, there's an HCC campus near you.
In Your Neighborhood
Meet HCC
Want to change your life? Hear from students, alumni, staff and faculty who've done just that at HCC, from culinary arts to engineering.
Student Life
Centers of Excellence
HCC's 14 Centers of Excellence focus on top-notch faculty and industry best practices to give students the skills they need for a successful career.
Learn more about our Centers -- from Energy and Consumer Arts & Sciences to Business and Manufacturing -- and partner with us today.
HCC Foundation
HCC Foundation empowers HCC student success through philanthropic support, aligned with key HCC institutional initiatives.
Business Community & Partnerships
Section 6: Solving Polynomial Equations by Factoring
It is important that you watch the video first.
Solve Quadratic Equations by the Zero-Factor Property
A quadratic equation is an equation that can be written in the form
ax2 + bx + c = 0
where a, b, and c are real numbers, with a ≠ 0.
ax2 + bx + c = 0 is called the standard form of a quadratic equation. Sometimes we may have to rearrange the terms of the equation to put it in standard form. The only requirement here is that we have an x2 in the equation. We guarantee that this term will be present in the equation by requiring a ≠ 0.
Note however, that it is okay if b and/or c is zero.
| The following equations are examples of quadratic equations. | | |
---
| 1. | 2k2 + 4k + 1 = 0 | a = 2, b = 4, c = 1 |
| 2. | x2 – 12x = 0 | a = 1, b = – 12, c = 0 |
| 3. | t2 = 81 | a = 1, b = 0, c = – 81 |
| 4. | (y + 1)(y – 7) = 1 | a = 1, b = – 6, c = – 8 (Hint: Expand and collect terms). |
The Zero-Factor Property
Solving by Factoring As the heading suggests, we will be solving quadratic equations here by factoring them. To do this we will need the following fact.
If ab = 0 then either a = 0 and/or b = 0.
This fact is called the zero factor property or zero factor principle. All that the property says is that if a product of two terms is zero then at least one of the terms had to be zero to start off with. Notice that this property will ONLY work if the product is equal to zero. Consider the following product.
ab = 6
In this case there is no reason to believe that either a or b will be 6.
We could have a = 2 and b = 3 for instance. So, do not misuse this fact!
To solve a quadratic equation by factoring we first must move all the terms over to one side of the equation.
Doing this serves two purposes. First, it puts the quadratics into a form that can be factored. Secondly, and probably more importantly, in order to use the zero factor property we MUST have a zero on one side of the equation.
If we don’t have a zero on one side of the equation we won’t be able to use the zero factor property.
Example 1: Solve the following equation by factoring. x2 – x = 12
Solution.
First, get everything on one side of the equation and then factor.
x2 – x – 12 = 0 (x – 4)(x + 3) = 0
Now at this point we’ve got a product of two terms that is equal to zero. This means that at least one of the following must
be true.
x – 4 = 0 OR x + 3 = 0
x = 4 OR x = – 3
Note that each if these is a linear equation that is easy enough to solve. What this tells us is that we have two solutions to the equation, x = 4 and x = – 3. As with linear equations we can always check our solutions by plugging the solution back into the equation. We will check x = – 3 and leave the other to you to check.
(– 3)2 – (– 3) = 12 ?
9 + 3 = 12 ?
12 = 12 OK
So, this was in fact a solution.
Example 2: Solve the following equation by factoring.
4m2 – 1 = 0
Solution.
Factor the equation. 4m2 – 1 = 0
(2m – 1)(2m + 1) = 0
Now apply the zero factor property. The zero factor property tells us that,
2m – 1 = 0 OR 2m + 1 = 0
2m = 1 OR 2m = – 1
Again, we will typically solve these in our head, but we needed to do at least one in complete detail. So we have two solutions to the equation.
Again, we will typically solve these in our head, but we needed to do at least one in complete detail. So we have two solutions to the equation.
Example 3: Solve the following equation by factoring.
5x³ – 5x2 – 10x = 0
Solution.
The first thing to do is factor this equation as much as possible. In this case that means factoring out the greatest common factor first. Here is the factored form of this equation.
5x(x2 – x – 2) = 0
5x(x – 2)(x + 1) = 0
Now, the zero factor property will still hold here. In this case we have a product of three terms that is zero. The only way this product can be zero is if one of the terms is zero. This means that,5x = 0 —› x = 0
x – 2 = 0 —› x = 2
x + 1 = 0 —› x = – 1
So, we have three solutions to this equation. So, provided we can factor a polynomial we can always use this as a solution technique. The problem is, of course, that it is sometimes not easy to do the factoring.
Test Your Knowledge by opening up the Test Yourself Activity.
Contact Us
For more information about the TSI Assessment and how to register, visit or email hcc.testing@hccs.edu
APPLY
VISIT
Events
Support
Strategic Planning
Institution
Work with HCC
Resources
College Links
Strategic Planning
Institution
Work with HCC
Resources
College Links
HCC is the official education partner of: |
189214 | https://space.stackexchange.com/questions/29891/how-would-the-rocket-differ-in-a-higher-atmospheric-or-ambient-pressure | engine design - How would the rocket differ in a higher atmospheric or ambient pressure? - Space Exploration Stack Exchange
Join Space Exploration
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Space Exploration helpchat
Space Exploration Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Space Exploration
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
How would the rocket differ in a higher atmospheric or ambient pressure?
Ask Question
Asked 7 years, 1 month ago
Modified3 years, 3 months ago
Viewed 1k times
This question shows research effort; it is useful and clear
1
Save this question.
Show activity on this post.
I imagine the both rocket nozzles in either environment underwater or Jupiter or underground would have to be different than that of standard rocket nozzles because of the back pressure.
In this answer a rocket engine is less efficient when launched out from under the ground without a exhaust vent because of the back pressure on the rocket engine. How would a rocket nozzle be changed to work efficiently at ambient pressures far higher than Earth's atmosphere? Realistic examples of such might include a launch from a high pressure gas environment such as an explorer of Jupiter's atmosphere or from below the surface of a liquid (.e.g under water)?
In other words: Similar to being fired from a gun but with a slow burn with a modified chamber, can back pressure be used with a different type of rocket nozzle?
Possibly related, Status-6 Oceanic Multipurpose System:
rockets
engine-design
high-pressure-environment
Share
Share a link to this question
Copy linkCC BY-SA 4.0
Improve this question
Follow
Follow this question to receive notifications
edited Aug 5, 2018 at 5:36
user12102
asked Aug 4, 2018 at 21:16
MuzeMuze
1
2
@Jack I imagine the both rocket nozzles in either environment underwater or Jupiter or underground would have to be different than that of standard rocket nozzles because of the back pressure.Muze –Muze 2018-08-04 22:23:36 +00:00 Commented Aug 4, 2018 at 22:23
This question seems clear-enough to me, and to at least two others who've been able to write answers to it. Nonetheless I've adjusted to wording to further clarify why Jupiter and under water are examples of higher pressure. @Jack does this look better?user12102 –user12102 2018-08-05 05:38:20 +00:00 Commented Aug 5, 2018 at 5:38
Add a comment|
2 Answers 2
Sorted by: Reset to default
This answer is useful
3
Save this answer.
Show activity on this post.
Assuming that the ambient pressure is constant, this is trivial - you simply design the nozzle so that the exit plane pressure matches ambient.
The problem in the other question arose because the engine was exhausting into a closed tube, where the ambient pressure would increase dramatically.
Changing ambient pressure is what causes trouble. Normally on a rocket launch the ambient pressure is dropping...if that really causes trouble you can use tricks like extendable nozzles.
If the ambient pressure is increasing rapidly...not sure. I suppose you could use a retractable nozzle! Although that makes my head hurt just thinking about it.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Improve this answer
Follow
Follow this answer to receive notifications
edited Aug 5, 2018 at 1:29
answered Aug 5, 2018 at 1:23
Organic MarbleOrganic Marble
198k 10 10 gold badges 694 694 silver badges 900 900 bronze badges
4
2 I don't think a retractable nozzle would be that hard--simply make a nozzle that burns off.Loren Pechtel –Loren Pechtel 2018-08-05 04:15:08 +00:00 Commented Aug 5, 2018 at 4:15
That's a clever idea! An aerospike might work too.Organic Marble –Organic Marble 2018-08-05 12:08:35 +00:00 Commented Aug 5, 2018 at 12:08
1 Is it so simple "make a nozzle that burns off"? For higher ambient pressure a shorter nozzle is needed, not a longer one. Extension is needed for the lower ambient pressure or vacuum.Uwe –Uwe 2018-08-05 13:46:45 +00:00 Commented Aug 5, 2018 at 13:46
2 I think the idea is that as the ambient pressure rises in the closed container the nozzle will burn off and get shorter. Getting the timing right might be tricky. Like I said, this makes my head hurt to think about. Definitely NOT simple.Organic Marble –Organic Marble 2018-08-05 13:49:04 +00:00 Commented Aug 5, 2018 at 13:49
Add a comment|
This answer is useful
5
Save this answer.
Show activity on this post.
Want to improve this answer? Add details and include citations to explain why this answer is correct. Answers without enough detail may be edited or deleted.
Taking a step back, the overall goal of the rocket engine is to convert very hot gas going in every direction (combustion products) into very hot gas going in primarily one direction.
The latter half of the process is a job for the rocket nozzle. The rocket's performance is dependent on the ambient air pressure.
An underexpanded exhaust is bad because the flow can become unstable and 'stick to' one side of the nozzle, kind of like when a shower is barely on and just dribbling water. An overexpanded flow is bad because the gases are spread out and not all going in the same direction, leading to wasted kinetic energy.
As the rocket goes higher and into lower air pressures, the optimal nozzle size changes. Most rockets get around this by having multiple stages, where the first stage is optimised for atmospheric flight, and has relatively small nozzles. The second stage has engines optimised for near vacuum conditions and has a much larger nozzle. The somewhat counterintuitive feature of this is that larger rocket nozzles don't necessarily indicate a more powerful rocket engine.
Below is the silhouette of the same engine, the Merlin, for vacuum, for 1 atm and a hypothetical one for 2××air pressure (that I just made by guessing). The higher the atmospheric pressure, the smaller the rocket nozzle needs to be.
Launching from under the sea, I suspect that using a rocket would be much less efficient than relying on, say, buoyancy or propeller engines. I do know that some nuclear subs can launch missiles, though I'd have to check if they can do that underwater. As to launching from the depths of Jupiter's atmosphere, well, that's just silly. The temperatures and pressures are just too high for anything to maintain a shape that humans would recognise. Besides, it takes about 33 000 m s 33 000 m s to get into low Jupiter orbit, which is more than what chemical rockets can generally provide.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Improve this answer
Follow
Follow this answer to receive notifications
edited Jun 4, 2022 at 14:51
No Nonsense
1
answered Aug 5, 2018 at 4:16
IngolifsIngolifs
6,579 30 30 silver badges 53 53 bronze badges
3
2 Boomers don't actually light their missiles underwater. There's a pre-stage that uses cold gas to lift it out of the water and throw it a bit into the sky. Before it falls back the main engine lights.Loren Pechtel –Loren Pechtel 2018-08-05 22:51:20 +00:00 Commented Aug 5, 2018 at 22:51
2 The first part of this answer, and the accompanying image, are just wrong. In an overexpanded nozzle (not pictured), the exhaust gases have expanded to such an extent that the pressure at the exit is lower than ambient. Beyond the exit, the exhaust pinches and forms pretty Mach diamonds. In a grossly overexpanded nozzle (labeled as underexpanded in the picture), the expansion is so great that the exhaust stream separates from the bell inside the rocket. (continued)David Hammen –David Hammen 2020-09-17 09:55:06 +00:00 Commented Sep 17, 2020 at 9:55
2 In an underexpanded nozzle (labeled as as overexpanded in the picture), the exhaust stream has not expanded enough to make the the pressure at the exit be equal to ambient pressure. The expansion continues beyond the exit, resulting in a blooming exhaust.David Hammen –David Hammen 2020-09-17 10:00:11 +00:00 Commented Sep 17, 2020 at 10:00
Add a comment|
Your Answer
Thanks for contributing an answer to Space Exploration Stack Exchange!
Please be sure to answer the question. Provide details and share your research!
But avoid …
Asking for help, clarification, or responding to other answers.
Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Draft saved
Draft discarded
Sign up or log in
Sign up using Google
Sign up using Email and Password
Submit
Post as a guest
Name
Email
Required, but never shown
Post Your Answer Discard
By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
rockets
engine-design
high-pressure-environment
See similar questions with these tags.
The Overflow Blog
The history and future of software development (part 1)
Getting Backstage in front of a shifting dev experience
Featured on Meta
Spevacus has joined us as a Community Manager
Introducing a new proactive anti-spam measure
Report this ad
Linked
11What benefits can be gained from launching below ground?
5Contracting rocket engine nozzles
Related
6How can phenolic (resin?) handle rocket engine nozzle temperatures?
5Static forces on a rocket engine
0Efficiency of rocket stages and converging/diverging nozzle
5Contracting rocket engine nozzles
3Do pulse detonation rocket engines require any type of nozzle?
2What expansion ratio is the best for mars surface level engines?
6How does a closed-cycle rocket engine keep the chamber pressure from stalling the pump turbine?
Hot Network Questions
Is there a way to defend from Spot kick?
Proof of every Highly Abundant Number greater than 3 is Even
Riffle a list of binary functions into list of arguments to produce a result
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
How different is Roman Latin?
Does a Linux console change color when it crashes?
How to home-make rubber feet stoppers for table legs?
How to start explorer with C: drive selected and shown in folder list?
Gluteus medius inactivity while riding
What can be said?
Lingering odor presumably from bad chicken
Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish?
Is it possible that heinous sins result in a hellish life as a person, NOT always animal birth?
Why do universities push for high impact journal publications?
Weird utility function
Why is a DC bias voltage (V_BB) needed in a BJT amplifier, and how does the coupling capacitor make this possible?
How do you emphasize the verb "to be" with do/does?
Fundamentally Speaking, is Western Mindfulness a Zazen or Insight Meditation Based Practice?
In the U.S., can patients receive treatment at a hospital without being logged?
Who is the target audience of Netanyahu's speech at the United Nations?
Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term?
On being a Maître de conférence (France): Importance of Postdoc
With line sustain pedal markings, do I release the pedal at the beginning or end of the last note?
How can the problem of a warlock with two spell slots be solved?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Space Exploration
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
189215 | http://cdn.kutasoftware.com/Worksheets/Grade6/15%20-%20Area%20Formula%20for%20Triangles.pdf | ©r [2K0W2f5r aKuuCtjaO AS\oTf]ttwRarrueo XLHLTCl.O U BAQlFlK mrJikgZhDtYsH vrtegs^ecrMvpepdq.Y G EMJapd[eG ^wAictkhS XIynbfjimnMigtdee cGmriakdFeI ^6fMOaWteh_.
Worksheet by Kuta Software LLC Grade 6 Math Area Formula for Triangles Name___________________________________ Date________________ Period____ Find the area of each triangle using a method you learned in class. Each square on the grid represents one square unit.
1) 2) 3) 4) ©y z2j0q2S5G jKSuttXaq DSOopfntwwLaWrxen RLMLJCW.e l SAHlJlO rrOiugshHtfsP [rGeLsIejrYv]e[dg.f K aM_azd[eZ \wXistChH eIanxf^iEnbiQtleZ JGNrBaEdgeV e6D qMcaNt^hg.
Worksheet by Kuta Software LLC 5) 6) 7) 8) ©W M2Z0m2A5U GKGuItnaM rSBoHfctuwDaCrdeYLILuCS.E F WAYlHljrZivgChntasp grEemsIeQrcvgendA.u s jMaacdBe AwBistOhS tIJn\fWiDniiutxeP QGlrHa]diep c6G KMnaytthL.
Worksheet by Kuta Software LLC Grade 6 Math Area Formula for Triangles Name_____ Date____ Period____ Find the area of each triangle using a method you learned in class. Each square on the grid represents one square unit.
1) rectangle area = 4 × 4 = 16 triangle area = 1 2 × 16 = 8 8 units² 2) parallelogram area = 7 × 6 = 42 triangle area = 1 2 × 42 = 21 21 units² 3) parallelogram area = 3 × 6 = 18 triangle area = 1 2 × 18 = 9 9 units² 4) parallelogram area = 2 × 8 = 16 triangle area = 1 2 × 16 = 8 8 units² ©P \2C0R2^5r OKRuYtBai gSLoafZtjwda[rweB cLSLmCq.Y H wAWl]lj grkiQgXhUtMsv irgeeseeOrsvJekdk.I A rMPaedeeU lwbiTtqhd sInn^fsiendiotUel QGKr`aadGe\ ^6N ^Mka[thhL.
Worksheet by Kuta Software LLC 5) rectangle area = 6 × 4 = 24 triangle area = 1 2 × 24 = 12 12 units² 6) rectangle area = 6 × 7 = 42 triangle area = 1 2 × 42 = 21 21 units² 7) left rectangle area = 2 × 4 = 8 left triangle area = 1 2 × 8 = 4 right rectangle area = 3 × 4 = 12 right triangle area = 1 2 × 12 = 6 triangle area = 4 + 6 = 10 10 units² 8) left rectangle area = 2 × 6 = 12 left triangle area = 1 2 × 12 = 6 right rectangle area = 3 × 6 = 18 right triangle area = 1 2 × 18 = 9 triangle area = 6 + 9 = 15 15 units² Create your own worksheets like this one with Infinite Grade 6 Math. Free trial available at KutaSoftware.com |
189216 | https://www.quora.com/How-do-you-show-that-the-triangle-with-vertices-A-2-2-B-2-2-and-C-2-2-is-an-isosceles-triangle | How to show that the triangle with vertices A (-2,-2), B(2,2) and C (2,- 2) is an isosceles triangle - Quora
Something went wrong. Wait a moment and try again.
Try again
Skip to content
Skip to search
Sign In
Geometry
Proofs (mathematics)
Distance Formula
Isosceles
Coordinate Systems
Triangles
Coordinate Plane
Geometric Mathematics
PLANE GEOMETRY
5
How do you show that the triangle with vertices A (-2,-2), B(2,2) and C (2,- 2) is an isosceles triangle?
All related (39)
Sort
Recommended
Bob Collier
Former EE Designed Specialized Computers for 33 Years. · Author has 3.1K answers and 1.6M answer views
·3y
As with any other such problem
ALWAYS DRAW A PICTUREE FIRST:
1 Plot those points.
2 Draw that triangle(s?)
w what you are supposed to know about triangles and isosceleses, you’ll need a mop to clean up the floor after you just look at this/your plot/picture.
3 Mop up and go on to the next one.
Continue Reading
As with any other such problem
ALWAYS DRAW A PICTUREE FIRST:
1 Plot those points.
2 Draw that triangle(s?)
w what you are supposed to know about triangles and isosceleses, you’ll need a mop to clean up the floor after you just look at this/your plot/picture.
3 Mop up and go on to the next one.
Upvote ·
Sponsored by Grammarly
Stuck on the blinking cursor?
Move your great ideas to polished drafts without the guesswork. Try Grammarly today!
Download
99 34
Related questions
More answers below
How do you show that the points (1,0,2), (4,3,2) and (0,7,6) are the vertices of an isosceles triangle?
How do you prove that the triangle with vertices (5, -2), (6, 5), (2, 2) is isosceles?
Are (5,-2), (6,4), and (7,-2) the vertices of an isosceles triangle?
How do you prove that (-1,4), (-3,-6), (3,-2) are vertices of an isosceles triangle?
How would you draw a triangle with vertices A (-1,-7), B(5,5), and C (-2,1), and show that the triangle is isosceles?
Mark Regan
Former Inmate Teacher's Aid and Tutor for G.E.D. at CDCR (1994–2014) · Author has 313 answers and 117.8K answer views
·3y
You only have to show that two of the sides are equal. There is a vertical side and a horizontal side. The vertical side is the one whose x-value does not change from point to point, BC. The length of BC is the distance between y-values of its end-points, or 4. The horizontal side is the side whose y-value does not change from point to point, AC. Its length is the distance between the x-values of its end points, also 4. Since two of its sides are equal, the triangle is an isosceles triangle.
Upvote ·
Dean Rubine
I am obsessed with triangles, especially their area. · Author has 10.6K answers and 23.7M answer views
·3y
This is pretty obvious from the symmetry, but the general answer is to show two of the three quadrances, squared distances between the points, are the same.
A(−2,−2),B(2,2),C(2,−2)A(−2,−2),B(2,2),C(2,−2)
A B 2=(−2−2)2+(−2−2)2=32 A B 2=(−2−2)2+(−2−2)2=32
B C 2=(2−2)2+(2−−2)2=16 B C 2=(2−2)2+(2−−2)2=16
A C 2=(−2−2)2+(−2−−2)2=16 A C 2=(−2−2)2+(−2−−2)2=16
We have A C 2=B C 2 A C 2=B C 2 so ABC is an isosceles triangle.
Upvote ·
Gary Ward
MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views
·3y
B and C have the same x-coordinate so the difference in their y-values is the length of side BC, 4.
A and C have the same y-coordinates so the difference in their x-values is the length of side AC, 4.
A triangle having two equal sides is isosceles by definition.
Upvote ·
9 8
Related questions
More answers below
How do you show that points A (3, 8), B (-11, 3) and C (-8, -2) are vertices of an isosceles triangle? How do you prove that triangle A (1, 4), B (10, 6) and B (2, 2) is a right triangle?
Is a triangle with vertices (2,2) (3,1) (7,6) an isosceles triangle?
The vertices of a triangle are (-2,0),(2,3) and (1,-3).Is the triangle equilateral, isosceles, or scalene?
How do you show that the points (0,2), (-3,-1) and (-4,3) are the vertices of an isosceles triangle in a graph form?
If A (-2,4), B(6,2), and C (1,-1), how do I prove that angle ABC is an isosceles right triangle?
Gary Ward
MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views
·1y
Related
A triangle is formed by A (2,-2), B(1,5) and C (-5,-1). How do you prove that the triangle is isosceles?
A triangle is formed by A (2,-2), B(1,5) and C (-5,-1). How do you prove that the triangle is isosceles?
Use the distance formula, d = √((x2-x1)²+(y2-y1)²) to find the length of AB, BC and AD.
AB = AD, so the triangle is isosceles.
Upvote ·
9 7
Sponsored by All Out
Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out.
Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association.
Shop Now
999 618
Ansh Keer
Former Owner · Author has 6K answers and 1.7M answer views
·May 19
Related
The vertices of an isosceles triangle are A (3, 6), B (7, 2), and C (4, 3). What is the equation of the triangle's line of symmetry?
First let's see which two sides are equal
AB = √32, BC = √10, CA = √10
So required line of symmetry is the median thro C
Midpoint of AB = (5,4)
Median on the unequal side is also the altitude in case of an isosceles triangle.
Slope of BC is -1
So reqd line thro A (5,4) is x-y = 1
.
Continue Reading
First let's see which two sides are equal
AB = √32, BC = √10, CA = √10
So required line of symmetry is the median thro C
Midpoint of AB = (5,4)
Median on the unequal side is also the altitude in case of an isosceles triangle.
Slope of BC is -1
So reqd line thro A (5,4) is x-y = 1
.
Upvote ·
9 7
Ajay Sreenivas
Former Aerospace Engineer/ Staff Consultant at Ball Aerospace (1980–2010) · Author has 5.3K answers and 1.7M answer views
·3y
AC = BC = 4 => ABC is an isosceles triangle with AB as the base.
Upvote ·
9 1
Sponsored by LPU Online
Career Ka Turning Point with LPU Online.
100% Online UGC-Entitled programs with LIVE classes, recorded content & placement support.
Apply Now
999 262
Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views
·May 18
Related
The vertices of an isosceles triangle are A (3, 6), B (7, 2), and C (4, 3). What is the equation of the triangle's line of symmetry?
Given 3 3 vertices of isosceles triangle A(3,6)A(3,6), B(7,2),B(7,2), and C(4,3)C(4,3)
Let us find side lengths
A B=√(7−3)2+(2−6)2=4√2 A B=(7−3)2+(2−6)2=4 2
B C=√(4−7)2+(3−2)2=√10 B C=(4−7)2+(3−2)2=10
A C=√(4−3)2+(3−6)2=√10 A C=(4−3)2+(3−6)2=10
We find that A C=B C A C=B C
For this isosceles triangle, the line of symmetry is the perpendicular bisector of side A B A B. It is also the median that passes though the the centroid and vertex C C.
Equation of A B A B
y−6 x−3=2−6 7−3⟹x+y=9 y−6 x−3=2−6 7−3⟹x+y=9
Equation of line of symmetry has the form,
x−y=k…(1)x−y=k…(1)
Since this line passes through C C, we write
4−3=k⟹k=1 4−3=k⟹k=1
From eqn. (1),
x−y=1 x−y=1
Continue Reading
Given 3 3 vertices of isosceles triangle A(3,6)A(3,6), B(7,2),B(7,2), and C(4,3)C(4,3)
Let us find side lengths
A B=√(7−3)2+(2−6)2=4√2 A B=(7−3)2+(2−6)2=4 2
B C=√(4−7)2+(3−2)2=√10 B C=(4−7)2+(3−2)2=10
A C=√(4−3)2+(3−6)2=√10 A C=(4−3)2+(3−6)2=10
We find that A C=B C A C=B C
For this isosceles triangle, the line of symmetry is the perpendicular bisector of side A B A B. It is also the median that passes though the the centroid and vertex C C.
Equation of A B A B
y−6 x−3=2−6 7−3⟹x+y=9 y−6 x−3=2−6 7−3⟹x+y=9
Equation of line of symmetry has the form,
x−y=k…(1)x−y=k…(1)
Since this line passes through C C, we write
4−3=k⟹k=1 4−3=k⟹k=1
From eqn. (1),
x−y=1 x−y=1
Upvote ·
99 10
Satyanarayana Mudunuri
Worked at ITC Limited (1979–2002) · Author has 3.4K answers and 1.8M answer views
·3y
AC= -2-(2)-4 (difference of x-coordinates and minus indicates direction) and BC = 2-(-2) =4 difference of y-coordinates) . Thus AC=BC=4 and the triangle is isosceles
Upvote ·
Promoted by JH Simon
JH Simon
Author of 'How To Kill A Narcissist'
·Updated Fri
Is there a way to prevent a narcissistic triangulation?
Stopping narcissistic triangulation requires first understanding how triangulation works, how it is maintained within us, and how we transcend it through a strong spiritual practice.
What Is Triangulation?
Triangulation is the conscious or unconscious introduction of a third person into the relationship dynamic between two people.
Examples of triangulation are:
Physical: Your significant other invites somone over or to mutual events, or spends increasingly more time with that person. In other cases, the other person is a core part of your significant other’s life.
Verbal: Your significant other tal
Continue Reading
Stopping narcissistic triangulation requires first understanding how triangulation works, how it is maintained within us, and how we transcend it through a strong spiritual practice.
What Is Triangulation?
Triangulation is the conscious or unconscious introduction of a third person into the relationship dynamic between two people.
Examples of triangulation are:
Physical: Your significant other invites somone over or to mutual events, or spends increasingly more time with that person. In other cases, the other person is a core part of your significant other’s life.
Verbal: Your significant other talks about the other person favorably or compares you to them unfavourably.
Triangulation can either be situational, or it can be manipulative. In the situational case, the presence of another person simply happens to be a part of relating with your significant other. For example, your parnter might still have a relationship with their ex-husband or ex-wife in order to keep things civilised for the children.
A third person can also be brought up in conversation. A partner can discuss or even compare you with an ex or a sexual competitor. A mother can compare you unfavourably with a sibling. Triangulation becomes manipulative when it is introduced as a way to impact you negatively or to pressure you into reacting from a place of threat or inferiority. The line between the two types of triangulation can quickly grow blurry in relationships.
In either case, whether it is situational or manipulative, triangulation can make you feel the following:
That you are less than.
Jealous.
Threatened.
That you are not as important to your significant other as you had hoped.
That your significant other is a valuable commodity who you must fight to keep or please.
Why Do People Triangulate?
As already stated, triangulation can be situational or manipulative. Therefore, there are many reasons people triangulate. Often the triangle simply exists without anyone doing anything. Other times your significant other actively feeds it. Then you have narcissistic triangulation, where a three-corner opposition is actively and manipulatively used against you to ensure your significant other’s sense of grandiosity, control and safety.
Some reasons people might triangulate are:
It makes them feel wanted and in demand: Having multiple people show interest makes a person feel like a valuable commodity. Telling the people about each other then creates competition and increases insecurity, making the ‘competitors’ feel inferior and unworthy. It communicates “You are not special, let alone the only one.”
They want to control you: Triangulation can trigger a person’s jealousy and sense of abandonment. This makes them feel needy and insecure. When someone is in that state, they become desperate to feel close to the person at the centre of the triangle. These feelings of insecurity make you reactive and panicky, and so easier to control. You are always on edge, doing everthing you can to ensure the person you need does not make the other person more important than you, or drop you altogether. A mother or father can also triangulate to get their way and direct their children how they want. In all such cases, it is about control.
It helps them win an argument: If so-and-so also thinks the same way as your significant other, then that’s two against one. Often it can be dozens against one, since “everyone” thinks the same way.
Fear of commitment and vulnerability: They keep multiple people around to avoid being ‘stuck’ with all their eggs in one basket.
Grandiosity: They can never make someone as special or more special than them. Triangulation keeps others feeling insignificant and tips the power distribution in your significant other’s favour.
They simply want that person in their life: People do not need to drop others out of their life when they meet you. They have a unique constellation of connections, some of which might not make sense to you, especially when exes become friends. That means you might have company in your intimate space.
Point 6 is the key to to this entire puzzle. You may well be in a relationship with a narcissist who simply wants to control you and make you feel small. It might be a protection or coping mechanism. Your significant other might diversify their emotional interests to ensure they don’t get too close to you, and therefore avoid being vulnerable. Or, they might just like having that peson in their life. It can be tempting to simply see all triangulation as manipulative, but that is not the case. In the last example, it has nothing to do with you. Other people simply exist in that person’s life. However, even if that person is purposely triangulating you, such as is the case with narcissism, you still have the choice to decide whether it affects you.
How To Stop Triangulation
Ultimately, triangulation exists only in your mind. It is created and reinforced by the thoughts, meanings and emotions you apply to the situation. Triangulation, above all, says less about the other person than it does about you. This is a difficult pill to swallow, but it is also a doorway to enlightenment. It is a way to break codependent habits, learn the art of detachment, and above all, to deepen your connection with yourself. Yes, a triangle has three points, but every single point leads directly back to you.
Triangulation stops with you. By shifting your paradigms, perspective, beliefs and focus, you can free yourself of the triangle for good.
While triangulation can be painful, it is also a chance to move closer toward spiritual growth, freedom and actualisation. Triangulation is also a natural psychological phenomenon which narcissists exploit constantly. This makes your spiritual journey a kind of holy war when the two forces clash.
Triangulation can be a great teacher. It can help you understand that:
We are born alone, and die alone: Accepting and leaning into this reality can help diminish the power of the triangle by bringing your focus deep into your greatest fear.
Nobody is in our keeping: You are your own Self, and others are their own Self. Your Self is sufficient onto itself, and answers only to God.
We always have the option of directing the focus back on ourselves: The power to shift your focus is always available to you. By focussing on your significant other and who they bring into your relationship dynamic, you are ceding control to them.
Enmeshment stunts growth: We must constantly let go of the other person and relate to them from a healthy distance if we are to grow with them. They will do or say what they must. What is it that you must attend to?
Control is an illusion: You cannot make the other person ‘yours’. You cannot make them give up anyone. You can only make sure you DO NOT GIVE UP YOURSELF in the panic, shame, jealousy and anger of triangulation.
The triangle is ultimately in your head: It only seeps into your heart and body if you let it. Reject it. Call that person out. Or simply see it for what it is, and divest your emotions away from it and into Self growth instead.
We should compare our current Self to our previous Self, and not to others: Every day is an opportunity to connect deeply with our True Self and let it guide us toward growth. Triangulation distracts us from this accessible reality by drawing us into the ego-based duality realm. Less than, more than. Special, not special. Mine, not mine.
You have an unhealthy desire to be the ‘special’ one: This comes from childhood, where the first triangle emerged between your mother, father and you. Understand that your emotions usually stem from this time. Yet you are no longer a child. Having the focus of your parent meant life and death when you were a baby. Now you are an adult with numerous resources. You can always develop more resources, both within and without. Your significant other is important, but they are not the difference between life and death. Those existential feelings are not from the present moment, but from the past. You needed to be special to survive. Now, you are inherently special, but you do not need another person’s complete focus and positive regard to access that. It is there either way. And there is a way to find it.
Inner Freedom From Triangulation
Overcoming triangulation begins within. A spiritual approach to breaking the triangle includes the following:
Self-Rememberance: Orient with your world. Notice the details around you. What objects are there in the room you are in? What can you hear, feel and smell? Orienting with the world around you allows you to shift your focus out of your mind, where duality rules. You stop comparing and analysing, and start being. From this place of neutral observing, you will notice your current emotional state. Allow those feelings to arise. Sadness, grief, shame, anger. Regardless of what you feel, allow it to come up. Now ask yourself: Who is noticing these details? Who is feeling these feelings? In this awareness you will remember your Self. There is no more triangle in your mind. There is simply you and your feeling Self, i.e. Your True Self. And by asking the question of who is feeling and noticing, you introduce your Higher Self into the fold. A new triangle emerges, one that serves you rather than crushing you.
Centering: Bring your focus inside your body. Allow your muscles to relax, breathe deeply into your belly and chest, and then allow the exhale to center you. Keep repeating this over and over until you get a sense of inner anchoring and calm.
Self-Comparison: Stop comparing yourself to the other person in the triangle. Start comparing your Self to your previous Self. What growth have you noticed in your life since a month ago? A year ago? Five years ago? Practice this often. Every time you feel less than or threatened by another person, bring your focus back within and celebrate where you are and where you have come from. Master this, and you learn the essence of spirituality.
Die before you die: Reflect on death. What might death be like? How do you feel about it? Does it terrify you? Intrigue you? Inspire you to live fully? Think about it. Read about it. Fall into it. All reaction to triangulation is a desperate clinging to escape death. Death of a relationship. Death of your self-worth. Death of your specialness. Transcend it all by meditating on and accepting death.
Outer freedom From Triangulation
Everyone has the freedom to do what they want. You cannot control them, and you cannot control every outcome. However, freedom is not free. It comes with consequences. A person can find someone ‘more special.’ A significant other can cheat and leave you behind. But if you drop the triangle regardless, at least you are living on your own terms. You are no longer being crippled and controlled by another person’s manipulation or circumstances.
With triangulation, you can only choose from two options:
Codependency
You control each other, and limit your relationships to each other. You use triangulation to control each other. You lash out whenever you feel insecure or threatened. You panic, feel anxious, and try to snoop around on the other person to feel secure again. You feel terrible about yourself and fight for the acceptance and approval of the other person.
OR
Freedom
There is another way: You let go and accept the consequences. Your parent might yap on about how great your sibling or cousin are. Your partner might have feelings for others, or have awkward close encounters before deciding to pull away. They might cheat. You might be left for someone else. Ultimatley, you need to accept that you have no control over that. We all have the freedom to live on our terms, and connect with whoever we want while agreeing to a set of boundaries and rules. Your significant other might slip. You might slip. You have to live with this possibility while trusting and hoping for the best. This is the nature of freedom.
Putting A Stop To Narcissistic Triangulation
Until you have done the inner work, you are always susceptible to narcissistic triangulation. When it is done with malignant intent, you will be bashed around by it. When it is situational, your sense of security, serenity and agency will be gradually worn down.
When a narcissist triangulates and you have done the work, it will cease to impact you. You can speak out firmly when they introduce another person into the relationship dynamic in a hurtful or manipualtive way. You can simply bask in your own serenity and peace while they focus on the other person. From this place of power, you can then look over what is left of the relationship with calmness, and simply act, rather than react.
Walking away is always an option. Either the person is being grossly manipulative, or the situation is simply untenable for you, especially when it comes to exes. But first ask yourself: What inner work can you do? Can you truly be fine with yourself alone? If not, then why? Can you tolerate the presence of ‘outsiders’ in your intimate space? If not, then why? Can you feel sufficient and worthy even when others enter your relationship dynamic? Why do you need to be ‘special’, i.e. the only one? Is it possible that such a rigid two-person world can grow stale, and become a breeding ground for resentment? Do you not think that some flexibility and trust in yourself and others can create space for wonder and growth to enter into your life?
Regardless of whether you are alone or in a relationship, in a narcissistic triangle or out of it, ask yourself: Can I direct my focus within every single day, and grow from that place? Do that, and you will discover the greatest triangle of all: You, your True Self, and your Higher Self. There is no more empowering dynamic that you can develop. And there is no other way to stop narcissistic triangulation.
If you have just started your narcissistic abuse recovery journey, check out How To Kill A Narcissist. Or if you wish to immunise yourself against narcissists and move on for good, take a look at How To Bury A Narcissist.
Upvote ·
999 302
99 33
99 13
George Ivey
Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views
·1y
Calculate the lengths of the sides. A B=√(−2−2)2+(−2−2)2=√4 2+4 2=4√2 A B=(−2−2)2+(−2−2)2=4 2+4 2=4 2.
A C=√(−2−2)2+(−2+2)2=√(4 2+0 2=4 A C=(−2−2)2+(−2+2)2=(4 2+0 2=4.
B C=√(2−2)2+(2+2)2=√0 2+4 2=4 B C=(2−2)2+(2+2)2=0 2+4 2=4.
Sides AC and BC are congruent but AC is different so ABC is an isosceles triangle but not an equilateral triangle.
Upvote ·
Reuven Harmelin
Lecturer at Technion - Israel Institute of Technology (1982–present) · Author has 2.3K answers and 1.9M answer views
·3y
Related
How do I prove that if A= (-n^2 , mn) B= (m^2 , mn) C= (n^2 , - mn) Three vertices of a triangle ABC, where n and m has parameters, it is an isosceles triangle?
On one hand, the point O(0,0) is CLEARLY the mid point of the side AC, so if we’ll show that the median OB of the side AC is perpendicular to AC, it would imply that the two right triangles AOB,COB are congruent and consequently AB=AC, that is, the triangle ABC is indeed an isosceles triangle.
Now, in order to show that OB and AB are perpendicular to each other, we just need to prove that the product of their slopes is -1.
Indeed, the slope of OB is
and the slope of AC, which is equal to the slope of OA, is
and now it is evident that the product of these two slopes is -1. Case closed.
Continue Reading
On one hand, the point O(0,0) is CLEARLY the mid point of the side AC, so if we’ll show that the median OB of the side AC is perpendicular to AC, it would imply that the two right triangles AOB,COB are congruent and consequently AB=AC, that is, the triangle ABC is indeed an isosceles triangle.
Now, in order to show that OB and AB are perpendicular to each other, we just need to prove that the product of their slopes is -1.
Indeed, the slope of OB is
and the slope of AC, which is equal to the slope of OA, is
and now it is evident that the product of these two slopes is -1. Case closed.
Upvote ·
9 1
Related questions
How do you show that the points (1,0,2), (4,3,2) and (0,7,6) are the vertices of an isosceles triangle?
How do you prove that the triangle with vertices (5, -2), (6, 5), (2, 2) is isosceles?
Are (5,-2), (6,4), and (7,-2) the vertices of an isosceles triangle?
How do you prove that (-1,4), (-3,-6), (3,-2) are vertices of an isosceles triangle?
How would you draw a triangle with vertices A (-1,-7), B(5,5), and C (-2,1), and show that the triangle is isosceles?
How do you show that points A (3, 8), B (-11, 3) and C (-8, -2) are vertices of an isosceles triangle? How do you prove that triangle A (1, 4), B (10, 6) and B (2, 2) is a right triangle?
Is a triangle with vertices (2,2) (3,1) (7,6) an isosceles triangle?
The vertices of a triangle are (-2,0),(2,3) and (1,-3).Is the triangle equilateral, isosceles, or scalene?
How do you show that the points (0,2), (-3,-1) and (-4,3) are the vertices of an isosceles triangle in a graph form?
If A (-2,4), B(6,2), and C (1,-1), how do I prove that angle ABC is an isosceles right triangle?
The vertices of an isosceles triangle are A (3, 6), B (7, 2), and C (4, 3). What is the equation of the triangle's line of symmetry?
How do you show that P (-1,3), Q(6,8) and R (11,1) are vertices of an isosceles triangle? How do you determine whether PQR is a right angle triangle?
How do we prove that the points A (-5,4), B (-1,-2) and C (5, 2) are the vertices of an isosceles right angle triangle (By section formulae)?
A triangle is formed by A (2,-2), B(1,5) and C (-5,-1). How do you prove that the triangle is isosceles?
If the vertices are (-2,-1), and (4,1), where might the third vertex of the isosceles triangle be?
Related questions
How do you show that the points (1,0,2), (4,3,2) and (0,7,6) are the vertices of an isosceles triangle?
How do you prove that the triangle with vertices (5, -2), (6, 5), (2, 2) is isosceles?
Are (5,-2), (6,4), and (7,-2) the vertices of an isosceles triangle?
How do you prove that (-1,4), (-3,-6), (3,-2) are vertices of an isosceles triangle?
How would you draw a triangle with vertices A (-1,-7), B(5,5), and C (-2,1), and show that the triangle is isosceles?
How do you show that points A (3, 8), B (-11, 3) and C (-8, -2) are vertices of an isosceles triangle? How do you prove that triangle A (1, 4), B (10, 6) and B (2, 2) is a right triangle?
Advertisement
About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press ·
© Quora, Inc. 2025
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Performance Cookies
Always Active
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Functional Cookies
Always Active
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Targeting Cookies
Always Active
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices |
189217 | https://www.thesaurus.com/browse/debilitate | Daily Crossword
Word Puzzle
Word Finder
All games
Word of the Day
Word of the Year
New words
Language stories
All featured
Slang
Emoji
Memes
Acronyms
Gender and sexuality
All culture
Writing tips
Writing hub
Grammar essentials
Commonly confused
All writing tips
Games
Featured
Culture
Writing tips
Advertisement
View definitions for debilitate
debilitate
verb as in incapacitate
Strongest matches
attenuate
cripple
disable
eviscerate
Strong matches
blunt
devitalize
enervate
enfeeble
exhaust
extenuate
harm
hurt
injure
mar
prostrate
relax
sap
spoil
unbrace
undermine
weaken
Weak matches
unstrengthen
wear out
Discover More
Example Sentences
Examples are provided to illustrate real-world usage of words in context. Any opinions expressed do not reflect the views of Dictionary.com.
Prior to her diagnosis, Ms Boocock - who also has endometriosis - had lived with debilitating symptoms that left her regularly needing to use a walking stick.
FromBBC
While schools were beginning to return to normal, Avalyn was struggling with a deep and debilitating fatigue, and eventually left school for home education.
FromBBC
It was reported that she suffered from a debilitating infection known as the Epstein Barr virus, and from the chronic fatigue syndrome, ME.
FromBBC
And he accomplished his most significant work without the use of his hands or legs, which became effectively paralyzed after he contracted a rare and debilitating autoimmune condition called Guillain-Barré syndrome in 2003.
FromLos Angeles Times
Inspired by her late father, motivated to overcome a debilitating shoulder injury, and fuelled by jelly beans, the 41-year-old ski instructor had made it to the top of K2.
FromBBC
Advertisement
Discover More
Related Words
Words related to debilitate are not direct synonyms, but are associated with the word debilitate. Browse related words to learn more about word associations.
blunt
verbas in make dull
attenuate
benumb
dampen
deaden
debilitate
desensitize
enfeeble
hebetate
numb
obtund
sap
soften
take the edge off
undermine
water down
weaken
cripple
verbas in physically disable
debilitate
disable
dislimb
dismember
enfeeble
hamstring
hurt
immobilize
impair
incapacitate
injure
lame
maim
mangle
mutilate
palsy
paralyze
prostrate
sap
sideline
unstrengthen
weaken
debase
verbas in degrade, shame
abase
bemean
cast down
cheapen
corrupt
debauch
debilitate
demean
demoralize
deprave
devaluate
devalue
disable
disgrace
dishonor
drag down
dump on
enfeeble
fluff off
humble
humiliate
lower
put away
put down
reduce
sap
shoot down
sink
take down
take down a peg
undermine
weaken
demoralize
verbas in depress, unnerve
abash
blow out
blow up
chill
damp
dampen
daunt
debilitate
deject
disarrange
disconcert
discountenance
discourage
dishearten
disorder
disorganize
disparage
dispirit
disturb
embarrass
enfeeble
get to
jumble
muddle
nonplus
psych out
rattle
sap
send up
shake
snarl
take apart
take steam out
undermine
unglue
unman
unsettle
unzip
upset
weaken
depress
verbas in deject, make despondent; exhaust
abase
afflict
ail
bear down
beat
beat down
bother
bug
bum out
cast down
chill
cow
damp
dampen
darken
daunt
debase
debilitate
degrade
desolate
devitalize
discourage
dishearten
dismay
dispirit
distress
disturb
drag
drain
dull
enervate
faze
keep under
lower
mock
mortify
oppress
perturb
press
put down
reduce
reduce to tears
run down
sadden
sap
scorn
slow
throw cold water on
torment
trouble
try
turn one off
upset
weaken
weary
weigh down
Viewing 5/35related words
From Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group.
Advertisement
Advertisement
Advertisement |
189218 | https://scholarlytraining.com/wp-content/uploads/2025/07/UK-11-Exam-Multi-Step-Word-Problem-Strategy-Handbook.pdf | UK 11+ Exam Multi-Step Word-Problem Strategy Handbook Master Complex Problem Solving for Grammar School and Independent School Success Table of Contents Chapter 1: Understanding 11+ Multi-Step Word Problems Chapter 2: The RUCSAC Problem-Solving Framework Chapter 3: Advanced Problem-Solving Strategies Chapter 4: Types of Multi-Step Problems Chapter 5: Worked Examples and Solutions Chapter 6: Time Management and Exam Technique Chapter 7: Common Mistakes and How to Avoid Them Chapter 8: Practice Problems and Self-Assessment Chapter 9: Preparation Strategies and Resources Chapter 1: Understanding 11+ Multi-Step Word Problems What Are Multi-Step Word Problems?
Multi-step word problems are mathematical questions that require students to perform multiple operations or reasoning steps to reach a solution. Unlike simple one-step problems, these questions test a student's ability to break down complex scenarios, identify the mathematical relationships involved, and apply various operations in the correct sequence.
In the context of 11+ examinations, multi-step word problems serve several important purposes: They assess logical reasoning and analytical thinking skills They test the ability to extract relevant information from complex scenarios They evaluate mathematical knowledge across multiple topics simultaneously They measure problem-solving persistence and systematic thinking Why Multi-Step Problems Are Crucial for 11+ Success Grammar schools and independent schools place significant emphasis on multi-step word problems because they reflect real-world mathematical thinking. These problems require students to demonstrate not just computational skills, but also: Key Skills Assessed: Critical Reading: Understanding complex problem statements Information Processing: Identifying relevant data and discarding irrelevant details Strategic Planning: Determining the sequence of operations needed Mathematical Fluency: Applying multiple mathematical concepts Verification Skills: Checking answers for reasonableness Common Characteristics of 11+ Multi-Step Problems Successful students learn to recognise the typical features of multi-step problems: Complex Scenarios: Problems often involve realistic situations such as shopping, travel, or sports, requiring students to navigate through multiple pieces of information.
Hidden Steps: Not all required operations are explicitly stated. Students must infer intermediate calculations needed to reach the final answer.
Multiple Mathematical Concepts: Problems frequently combine topics such as percentages, fractions, ratios, and basic algebra within a single question.
Distracting Information: Problems may include irrelevant data designed to test students' ability to focus on essential information.
The Challenge of Multi-Step Problems Many students find multi-step word problems challenging because they require simultaneous management of several cognitive processes. Common difficulties include: Feeling overwhelmed by the amount of text and information Struggling to identify where to begin the solution process Making arithmetic errors in intermediate steps Losing track of the original question during calculation Running out of time due to inefficient problem-solving approaches Success Tip: The key to mastering multi-step problems is developing a systematic approach that can be applied consistently to any problem type. This handbook will provide you with proven frameworks and strategies used by successful 11+ candidates.
Chapter 2: The RUCSAC Problem-Solving Framework Introduction to RUCSAC RUCSAC is a systematic problem-solving method that provides a clear structure for approaching multi-step word problems. This acronym stands for Read, Understand, Choose, Solve, Answer, and Check. This framework has been successfully used by thousands of students to improve their problem-solving accuracy and confidence.
The RUCSAC Framework R - Read the problem carefully U - Understand what is being asked C - Choose the appropriate method S - Solve the problem step by step A - Answer the question clearly C - Check your work and answer Step 1: Read the Problem Carefully The first step involves a thorough reading of the problem statement. This may seem obvious, but many students rush through this stage and miss crucial information. Effective reading strategies include: Multiple Read-Throughs: Read the problem at least twice. The first reading should focus on getting a general sense of the scenario, while the second reading should focus on identifying specific information and requirements.
Highlight Key Information: Identify and underline or circle important numbers, units, and key words that indicate mathematical operations (such as "total," "difference," "share," or "remaining").
Note Important Details: Pay attention to units of measurement, time periods, and any conditions or constraints mentioned in the problem.
Step 2: Understand What Is Being Asked Understanding the problem involves several components: Identify the Question: Clearly determine what the problem is asking you to find. Sometimes problems contain multiple pieces of information but only ask for one specific answer.
Recognise the Context: Understand the real-world scenario being described and how it relates to mathematical concepts.
Determine Required Information: Identify what information you have been given and what you need to find out.
Example: "Sarah has £45. She spends 2/5 of her money on books and then spends £8 on lunch. How much money does she have left?" Understanding Analysis: Starting amount: £45 First expense: 2/5 of £45 (fraction calculation needed) Second expense: £8 Question asks for: remaining money Step 3: Choose the Appropriate Method This step involves selecting the mathematical operations and approach needed to solve the problem. Consider: Required Operations: Determine whether you need to add, subtract, multiply, divide, or use more complex operations like percentages or ratios.
Order of Operations: Plan the sequence in which calculations should be performed, keeping in mind mathematical rules like BODMAS.
Estimation Strategy: Consider whether estimation might help verify your final answer.
Step 4: Solve the Problem Step by Step Execute your chosen method systematically: Work Step by Step: Break the problem into smaller, manageable parts and solve each part clearly.
Show Your Working: Write down all calculations, even simple ones. This helps prevent errors and makes checking easier.
Label Intermediate Results: Give meaningful labels to intermediate calculations to maintain clarity about what each number represents.
Continuing the previous example: Step 1: Calculate 2/5 of £45 2/5 × £45 = £18 (spent on books) Step 2: Calculate remaining after book purchase £45 - £18 = £27 Step 3: Subtract lunch cost £27 - £8 = £19 Step 5: Answer the Question Clearly Provide a complete answer that directly addresses the original question: Include Units: Always include appropriate units in your final answer (pounds, metres, minutes, etc.).
Answer in Context: Phrase your answer in terms of the original question.
Check Completeness: Ensure you have answered all parts of the question if it has multiple components.
Step 6: Check Your Work and Answer The final step involves verification: Reasonableness Check: Consider whether your answer makes sense in the context of the problem.
Calculation Check: Review your arithmetic to ensure accuracy.
Method Verification: Confirm that you used appropriate mathematical operations for the problem type.
Memory Aid: Create a mental checklist using RUCSAC. With practice, this systematic approach will become automatic, leading to fewer errors and greater confidence in problem-solving situations.
Chapter 3: Advanced Problem-Solving Strategies Beyond RUCSAC: Enhanced Techniques While RUCSAC provides an excellent foundation, advanced problem solvers employ additional strategies to tackle particularly complex multi-step problems. These techniques can significantly improve both accuracy and efficiency.
Visualisation Techniques Creating visual representations of problems can make complex relationships clearer and solution paths more obvious.
Diagram Drawing: For geometry problems, draw accurate diagrams with labeled measurements.
For word problems involving movement or arrangements, sketch the scenario.
Table Creation: Organise information in tables when dealing with multiple variables or comparing different scenarios.
Timeline Construction: For problems involving sequences of events or time-based calculations, create a timeline to track changes.
Visualisation Example: "A rectangular garden is 15m long and 8m wide. A path 2m wide runs around the inside perimeter. What is the area of the path?" Visual Approach: Draw the outer rectangle (15m × 8m) and inner rectangle (11m × 4m) to clearly see that the path area = total area - inner area.
Working Backwards Strategy Some problems are more easily solved by starting with the final result and working backwards to find unknown values.
When to Use: This strategy is particularly useful for problems that describe a series of operations leading to a final result, but ask you to find an initial value.
Implementation: Start with the given final result and reverse each operation mentioned in the problem.
Pattern Recognition Experienced problem solvers learn to recognise common problem patterns and apply proven solution methods.
Age Problems: Problems involving ages typically use algebraic thinking, setting up equations based on relationships between current and future ages.
Rate Problems: Problems involving speed, time, and distance follow predictable patterns using the formula: Distance = Speed × Time.
Percentage Problems: Learn to recognise whether a problem requires finding a percentage of a number, finding what percentage one number is of another, or finding the original number when given a percentage.
Estimation and Approximation Strategic use of estimation can both speed up problem-solving and provide a check for answer reasonableness.
Quick Calculation Checks: Round numbers to make mental calculation easier, providing a ballpark figure for comparison with your exact answer.
Elimination Strategy: In multiple-choice questions, use estimation to eliminate obviously incorrect answers before performing exact calculations.
The "Sense-Making" Test Before submitting any answer, ask yourself: Is this answer reasonable given the problem context?
Are the units correct?
Is the magnitude (size) of the answer logical?
Does it answer the actual question being asked?
Breaking Down Complex Problems Large, intimidating problems become manageable when broken into smaller components.
Identify Sub-Problems: Look for natural breaking points where one calculation must be completed before the next can begin.
Create Intermediate Goals: Set mini-objectives for each part of the problem, celebrating small successes along the way.
Use Logical Sequencing: Determine which calculations depend on others and create a logical order for your work.
The Strategic Pause Many errors occur when students rush into calculations without fully understanding the problem.
Implementing a "strategic pause" can prevent many mistakes.
Problem Analysis Time: Spend 20-30 seconds after reading a problem to think about the approach before beginning calculations.
Method Verification: Before executing your chosen method, quickly verify that it will lead to the information requested in the question.
Chapter 4: Types of Multi-Step Problems Classification of Problem Types Understanding different categories of multi-step problems helps students recognise patterns and apply appropriate strategies more quickly. The most common types appearing in 11+ examinations include: Money and Shopping Problems These problems involve calculations with currency, often including concepts like change, discounts, and multiple purchases.
Key Features: Multiple transactions or purchases Percentage discounts or mark-ups Calculating change or total costs Comparing prices or value for money Example: "Tom has £50. He buys 3 books costing £7.99 each and receives a 10% discount on his total purchase. He then buys lunch for £8.50. How much money does he have left?" Solution Approach: 1. Calculate cost of books: 3 × £7.99 = £23.97 2. Apply discount: 10% of £23.97 = £2.40 (approximately) 3. Discounted book cost: £23.97 - £2.40 = £21.57 4. Total spent: £21.57 + £8.50 = £30.07 5. Money remaining: £50.00 - £30.07 = £19.93 Time and Scheduling Problems These problems involve calculations with time intervals, schedules, and durations.
Key Features: Converting between different time units Calculating elapsed time Working with timetables or schedules Time zones or scheduling conflicts Measurement and Geometry Problems These problems combine geometric concepts with multi-step calculations.
Key Features: Area and perimeter calculations Volume and capacity problems Scale and proportion in measurements Combining different shapes or units Ratio and Proportion Problems These problems involve relationships between quantities and scaling.
Key Features: Sharing quantities in given ratios Scaling recipes or measurements Direct and inverse proportion Comparing rates or ratios Problem Type Recognition Tips Look for Keywords: Money problems: "costs," "change," "discount," "total price" Time problems: "hours," "minutes," "duration," "schedule" Geometry problems: "area," "perimeter," "volume," "dimensions" Ratio problems: "share," "proportion," "for every," "ratio" Percentage and Fraction Problems These problems require understanding of proportional relationships and conversions between different representations.
Common Scenarios: Percentage increases or decreases Finding fractions of quantities Converting between fractions, decimals, and percentages Compound percentage changes Data Analysis Problems These problems require extracting information from tables, graphs, or charts and performing calculations based on the data.
Key Skills Required: Reading and interpreting data from various formats Calculating averages, totals, or differences Making comparisons between data sets Drawing conclusions from numerical data Age and Relationship Problems These problems involve mathematical relationships between people's ages, often requiring algebraic thinking.
Typical Patterns: Current ages with future or past relationships Age differences that remain constant Proportional age relationships Multiple people with interconnected age relationships Chapter 5: Worked Examples and Solutions Comprehensive Problem-Solving Examples This chapter provides detailed solutions to representative multi-step problems, demonstrating the application of RUCSAC and advanced strategies.
Example 1: Complex Shopping Problem Problem: "Emma goes shopping with £75. She spends 1/3 of her money on clothes, then 40% of what remains on books. She then spends £12 on lunch. If she wants to save at least £15 from her original amount, does she have enough money left?" RUCSAC Solution: R - Read: Emma starts with £75, makes three purchases, and needs £15 left over.
U - Understand: We need to track her spending through three steps and determine if £15 remains.
C - Choose: Sequential calculation of each expense, then check final amount.
S - Solve: Step 1: Money spent on clothes = 1/3 × £75 = £25 Money remaining after clothes = £75 - £25 = £50 Step 2: Money spent on books = 40% × £50 = 0.4 × £50 = £20 Money remaining after books = £50 - £20 = £30 Step 3: Money remaining after lunch = £30 - £12 = £18 A - Answer: Emma has £18 left, which is more than the £15 she wanted to save, so yes, she has enough.
C - Check: Total spent = £25 + £20 + £12 = £57. Starting amount - spent = £75 - £57 = £18 ✓ Example 2: Time and Distance Problem Problem: "A train leaves London at 09:45 and travels at 80 km/h for 2 hours 30 minutes. It then travels at 60 km/h for another 1 hour 45 minutes. What time does it arrive at its destination and how far has it travelled?" RUCSAC Solution: R - Read: Train journey with two different speeds and times, asking for arrival time and total distance.
U - Understand: This requires both time calculations and distance calculations using Speed × Time = Distance.
C - Choose: Calculate each segment separately, then combine results.
S - Solve: First segment: Time = 2 hours 30 minutes = 2.5 hours Distance = 80 km/h × 2.5 h = 200 km Second segment: Time = 1 hour 45 minutes = 1.75 hours Distance = 60 km/h × 1.75 h = 105 km Total journey time = 2.5 + 1.75 = 4.25 hours = 4 hours 15 minutes Arrival time = 09:45 + 4:15 = 14:00 (2:00 PM) Total distance = 200 + 105 = 305 km A - Answer: The train arrives at 14:00 (2:00 PM) having travelled 305 km.
C - Check: Journey time seems reasonable for the distance, and arithmetic checks out.
Example 3: Geometry and Area Problem Problem: "A rectangular garden is 24m long and 15m wide. A square patio with sides of 6m is built in one corner, and a circular pond with radius 3m is built in the centre. What is the remaining grass area?" RUCSAC Solution: R - Read: Rectangle with a square and circle removed, asking for remaining area.
U - Understand: Need to calculate total area, then subtract patio and pond areas.
C - Choose: Area formulas: Rectangle = length × width, Square = side², Circle = πr² S - Solve: Garden area = 24 × 15 = 360 m² Patio area = 6² = 36 m² Pond area = π × 3² = 9π ≈ 28.3 m² Grass area = 360 - 36 - 28.3 = 295.7 m² A - Answer: The remaining grass area is approximately 295.7 m².
C - Check: Total removed area (64.3 m²) is much smaller than total area (360 m²), so answer is reasonable.
Example 4: Ratio and Proportion Problem Problem: "A recipe for 6 people requires 450g flour, 3 eggs, and 300ml milk. If I want to make this recipe for 10 people, but I only have 2 eggs, how much flour and milk should I use to maintain the correct proportions?" RUCSAC Solution: R - Read: Recipe scaling problem with a constraint (limited eggs).
U - Understand: Scale based on available eggs rather than desired serving size.
C - Choose: Find what portion of original recipe 2 eggs represents, then scale other ingredients accordingly.
S - Solve: Original recipe uses 3 eggs for 6 people With 2 eggs, I can make: (2/3) of the original recipe This serves: (2/3) × 6 = 4 people Flour needed: (2/3) × 450g = 300g Milk needed: (2/3) × 300ml = 200ml A - Answer: Use 300g flour and 200ml milk with the 2 eggs. This will serve 4 people, not 10.
C - Check: Ratios maintained: 300g flour, 2 eggs, 200ml milk for 4 people gives the same proportions as the original recipe.
Chapter 6: Time Management and Exam Technique Strategic Time Allocation Success in 11+ mathematics requires not only mathematical knowledge but also effective time management. Multi-step problems often carry higher marks but require more time, making strategic allocation crucial.
Understanding Question Values Different problems carry different point values, and students should allocate time proportionally: Question Type Typical Marks Suggested Time Strategy Simple calculations 1-2 marks 30-60 seconds Quick mental math Two-step problems 2-3 marks 1-2 minutes Clear working shown Multi-step problems 3-5 marks 3-5 minutes Systematic approach Complex word problems 4-6 marks 4-7 minutes Full RUCSAC method The Three-Pass Strategy Effective exam technique involves multiple passes through the paper: First Pass (Quick Win): Complete all questions you can solve immediately and confidently. This builds momentum and secures easy marks.
Second Pass (Moderate Challenge): Tackle problems requiring more thought but using familiar techniques. Apply RUCSAC systematically.
Third Pass (Difficult Problems): Use remaining time for the most challenging questions, employing advanced strategies and careful checking.
Speed Enhancement Techniques Several techniques can significantly improve solving speed without sacrificing accuracy: Mental Math Fluency: Practice common calculations (times tables, percentage calculations, fraction conversions) until they become automatic.
Estimation Skills: Use rounding and approximation to quickly check if answers are reasonable and to eliminate incorrect multiple-choice options.
Pattern Recognition: Learn to quickly identify problem types and apply appropriate standard methods.
Speed-Accuracy Balance Remember that accuracy is more important than speed. A few correct answers are better than many incorrect ones. Focus on: Getting questions right the first time Showing clear working that can earn partial marks Moving on from questions that are taking too long Returning to difficult questions only if time permits Managing Stress and Anxiety Multi-step problems can create anxiety, especially under exam conditions. Effective management techniques include: Breathing Techniques: If you feel overwhelmed, take three deep breaths before attempting a difficult problem.
Positive Self-Talk: Replace "I can't do this" with "I'll work through this step by step." Strategic Skipping: Don't get stuck on one problem. Move on and return if time allows.
Checking Strategies for Time-Pressured Situations When time is limited, prioritise these checking methods: Quick Reasonableness Check: Does the answer make sense in context?
Units Verification: Are the correct units included in the final answer?
Arithmetic Spot-Check: Quickly verify one or two calculations, especially the final step.
Chapter 7: Common Mistakes and How to Avoid Them Understanding Error Patterns Recognising common mistakes helps students avoid these pitfalls and develop more reliable problem-solving habits. Research into 11+ mathematics performance has identified several recurring error patterns.
Reading and Comprehension Errors Many errors occur before any mathematics is attempted, during the problem-reading phase.
Misreading Numbers: Confusing digits, decimal places, or units. Always re-read numerical information carefully.
Missing Information: Overlooking crucial details or constraints mentioned in the problem text.
Question Misinterpretation: Solving for the wrong quantity or misunderstanding what the question asks.
Prevention Strategy: After reading a problem, paraphrase the question in your own words to ensure understanding. Ask yourself: "What exactly am I trying to find?" Calculation Errors Arithmetic mistakes are among the most frustrating errors because they often involve correct methods applied incorrectly.
Order of Operations Mistakes: Forgetting BODMAS rules, especially when dealing with mixed operations.
Decimal Place Errors: Misplacing decimal points during multiplication or division.
Fraction Calculation Mistakes: Errors in adding, subtracting, or converting fractions.
Percentage Calculation Errors: Confusing percentage of a number with percentage increase/decrease calculations.
Common Error Example: Problem: "Find 15% of 240" Incorrect approach: 240 ÷ 15 = 16 Correct approach: 15% × 240 = 0.15 × 240 = 36 Error cause: Confusing percentage calculation with division Method Selection Errors Students sometimes choose inappropriate methods or fail to recognise the type of problem they're solving.
Wrong Operation Choice: Using addition when multiplication is needed, or vice versa.
Incomplete Method Application: Starting with a correct approach but failing to complete all necessary steps.
Formula Misapplication: Using incorrect formulas or applying correct formulas inappropriately.
Logic and Reasoning Errors These errors involve flawed thinking about problem structure or relationships.
Sequence Errors: Performing operations in the wrong order in multi-step problems.
Unit Confusion: Mixing different units without proper conversion.
Proportion Misunderstanding: Incorrect application of ratio and proportion concepts.
Presentation and Communication Errors Even correct mathematical thinking can lose marks due to poor presentation.
Missing Units: Forgetting to include units in final answers.
Unclear Working: Showing work in a way that's difficult to follow or verify.
Incomplete Answers: Providing intermediate results instead of final answers.
Error Prevention Strategies The Error Prevention Checklist Before Starting: Read the problem at least twice Identify what you're looking for Note all given information During Solving: Show all working clearly Label intermediate steps Use estimation to check reasonableness After Solving: Check units are included Verify the answer addresses the question Do a quick reasonableness check Learning from Mistakes Developing an effective error analysis process helps prevent repeated mistakes: Mistake Categorisation: When you make an error, identify whether it was a reading error, calculation error, method error, or presentation error.
Root Cause Analysis: Ask why the mistake happened. Was it due to rushing, misunderstanding, or lack of knowledge?
Prevention Strategy Development: For each type of mistake, develop a specific strategy to prevent its recurrence.
Chapter 8: Practice Problems and Self-Assessment Graduated Practice Problems This chapter provides practice problems of increasing difficulty, allowing students to build confidence and skills progressively.
Beginner Level Problems Problem 1: Sarah buys 4 pens costing £1.25 each and 3 notebooks costing £2.80 each. How much does she spend in total?
Key Skills: Multiplication, addition, money calculations Estimated Time: 2 minutes Problem 2: A recipe serves 6 people and requires 450g of flour. How much flour is needed to serve 9 people?
Key Skills: Ratio, proportion, scaling Estimated Time: 2 minutes Intermediate Level Problems Problem 3: A train travels 180 km in 2 hours 15 minutes. What is its average speed in km/h?
Key Skills: Time conversion, division, speed calculation Estimated Time: 3 minutes Problem 4: In a class of 28 students, 3/7 are boys. Of the boys, 2/3 play football. How many boys play football?
Key Skills: Fractions of quantities, multi-step calculation Estimated Time: 3 minutes Advanced Level Problems Problem 5: A rectangular swimming pool is 25m long, 12m wide, and 2m deep. If it is filled to 80% capacity, how many litres of water does it contain? (1 cubic metre = 1000 litres) Key Skills: Volume calculation, percentage, unit conversion Estimated Time: 4 minutes Problem 6: Emma saves money each week. In the first week she saves £5, in the second week £8, in the third week £11, and so on, increasing by £3 each week. How much will she have saved in total after 10 weeks?
Key Skills: Pattern recognition, arithmetic sequences, addition Estimated Time: 5 minutes Expert Level Problems Problem 7: A shopkeeper marks up his goods by 40% on the cost price. During a sale, he offers a discount of 25% on the marked price. If a customer pays £42 for an item during the sale, what was the original cost price to the shopkeeper?
Key Skills: Percentage increases and decreases, working backwards Estimated Time: 6 minutes Self-Assessment Framework Use this framework to evaluate your progress: Skill Level Time Target Accuracy Target Next Steps Beginner Within suggested time + 50% 80% correct Focus on method accuracy Intermediate Within suggested time + 25% 85% correct Improve speed and efficiency Advanced Within suggested time 90% correct Tackle expert problems Expert Within suggested time - 15% 95% correct Focus on exam technique Self-Assessment Questions After solving each problem, ask yourself: Did I understand the problem on first reading?
Did I choose an appropriate method quickly?
Were my calculations accurate?
Did I check my answer for reasonableness?
Would I be confident solving a similar problem in an exam?
Chapter 9: Preparation Strategies and Resources Creating an Effective Study Plan Success in multi-step problem solving requires consistent, structured practice. This chapter outlines evidence-based preparation strategies.
The 6-Week Preparation Schedule Week Focus Area Daily Practice Assessment 1 RUCSAC Method 3-4 basic problems Method checklist 2 Problem Types Mixed problem types Type recognition quiz 3 Speed Building Timed practice Speed accuracy test 4 Advanced Strategies Complex problems Strategy application 5 Exam Technique Mock exam sections Full practice paper 6 Review & Refinement Error correction Final assessment Daily Practice Routine Establish a consistent daily routine that builds skills systematically: Warm-up (5 minutes): Practice mental arithmetic and basic calculations to build fluency.
Focused Practice (15-20 minutes): Work on 3-5 problems using the current week's focus area.
Review and Reflection (5 minutes): Analyse any errors and note areas for improvement.
Resource Recommendations Practice Papers: Use official past papers from your target schools when available. If not available, use papers from schools with similar standards.
Online Resources: Utilise reputable websites that offer 11+ practice materials with detailed solutions.
Textbooks: Choose books specifically designed for 11+ preparation that include multi-step problem-solving strategies.
Quality over Quantity: It's better to solve fewer problems thoroughly, understanding each step completely, than to rush through many problems superficially.
Building Mathematical Confidence Confidence plays a crucial role in problem-solving success. Build confidence through: Progressive Difficulty: Start with problems you can solve comfortably, then gradually increase difficulty.
Success Tracking: Keep a record of problems solved correctly to visualise progress.
Strategy Mastery: Focus on mastering one strategy at a time rather than trying to learn everything simultaneously.
Exam Day Preparation Final Week: Focus on light practice and review of strategies rather than learning new concepts.
Night Before: Review your strategy summaries but avoid intensive practice that might increase anxiety.
Morning of Exam: Do light warm-up problems to activate mathematical thinking, but don't attempt anything challenging.
Post-Exam Analysis Regardless of exam outcomes, conduct a thoughtful analysis to inform future mathematical learning: Strategy Effectiveness: Which problem-solving approaches worked best under exam conditions?
Time Management: How well did time allocation strategies work in practice?
Confidence Factors: What factors contributed to feeling confident or anxious during problem solving?
Final Success Reminders Multi-step problems become manageable with systematic approaches RUCSAC provides a reliable framework for any problem type Regular practice builds both skill and confidence Error analysis accelerates learning and prevents repeated mistakes Time management skills are as important as mathematical knowledge Success comes from consistent effort rather than last-minute cramming Conclusion Multi-step word problems represent one of the most challenging aspects of 11+ mathematics, but they also offer the greatest opportunity for students to demonstrate sophisticated mathematical thinking. By mastering the RUCSAC framework, developing advanced problem-solving strategies, and maintaining consistent practice, students can transform these challenging problems from obstacles into opportunities for success.
Remember that becoming proficient at multi-step problem solving is a journey that requires patience, persistence, and systematic effort. The strategies outlined in this handbook have been successfully used by thousands of students to achieve their grammar school and independent school goals. With dedication and proper preparation, you too can master the art of multi-step problem solving and approach your 11+ examinations with confidence.
The key to success lies not just in mathematical knowledge, but in developing the thinking skills, organisational abilities, and confidence needed to tackle complex problems systematically. Use this handbook as your guide, practice regularly, learn from your mistakes, and trust in your ability to improve. Your 11+ success is within reach.
Best wishes for your 11+ examination success! |
189219 | https://math.stackexchange.com/questions/3172918/why-can-we-modular-reduce-arguments-of-sums-and-products-in-modular-arithmetic | Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Why can we modular reduce arguments of sums and products in modular arithmetic?
Ask Question
Asked
Modified 2 years, 1 month ago
Viewed 376 times
0
$\begingroup$
Suppose,I have a number 1256. I want (1256 % 11)! That is 2 . But, here, if i try in this way, ( (125 % 11 ) 10 + 6 ) % 11 = 2. That is exactly the same answer 2 as above.
I am confused , how This two process gives the similar answer ??? Even if , I try this method for any number n,m to find (n mod m) . How this works ?? Can anybody explain ?
number-theory
modular-arithmetic
Share
edited Dec 1, 2019 at 15:13
Bill Dubuque
284k4242 gold badges339339 silver badges1k1k bronze badges
asked Apr 3, 2019 at 7:47
Ovi PoddarOvi Poddar
322 bronze badges
$\endgroup$
8
$\begingroup$ What is your exact process? In your example, you took something mod $2$, and that thing was clearly an even number (since it's multiple of $10$, plus $6$), so of course the final answer is congruent to $2$ mod $2$. $\endgroup$
Minus One-Twelfth
– Minus One-Twelfth
2019-04-03 07:53:16 +00:00
Commented Apr 3, 2019 at 7:53
1
$\begingroup$ Welcome to Maths SX! What does 1256%11 denote? $\endgroup$
Bernard
– Bernard
2019-04-03 07:58:53 +00:00
Commented Apr 3, 2019 at 7:58
$\begingroup$ % @Bernard is often used in place of Mod(n,y) in Pari $\endgroup$
user645636
– user645636
2019-04-03 09:41:20 +00:00
Commented Apr 3, 2019 at 9:41
$\begingroup$ if i want a number mod by another number, say, n mod m , and suppose n consists of 4 digits ,,,, if a take mod of the number consisting left 3 digits and then multiply by 10 , then add to remaining last digit of right side , and finally take mod of the number then i actually get exactly the same (n mod m) ! I want (1256 % 11)! if i try in this way, ( (125 % 11 ) 10 + 6 ) % 11 = 2. That is exactly the same answer 2 ! how this works ? $\endgroup$
Ovi Poddar
– Ovi Poddar
2019-04-03 09:54:23 +00:00
Commented Apr 3, 2019 at 9:54
$\begingroup$ See en.wikipedia.org/wiki/Modular_arithmetic#Properties. From these rules you can see that $$ ax+b \equiv c\mod{m} \qquad \Leftrightarrow (a\mod{m}) x + b \equiv c \mod{m}$$ $\endgroup$
Matti P.
– Matti P.
2019-04-03 10:05:17 +00:00
Commented Apr 3, 2019 at 10:05
| Show 3 more comments
2 Answers 2
Reset to default
1
$\begingroup$
The Congruence Sum & Product Rules imply that replacing $\rm\color{#c00}{arguments}$ of sums and products by $\rm\color{#0a0}{congruent}$ arguments yields a congruent sum or product. Applying this inductively (cf. Note below) shows the same holds true for arbitrary "polynomial" expressions composed of sums and products, yielding a multivariate form of the linked (univariate) Polynomial Congruence Rule. In particular this implies that for any such arithmetical expression we obtain a congruent expression if we replace (some or all) arguments of its sums and products by their $\rm\color{#0a0}{(congruent)}$ remainders.
Yours is the special case below for the polynomial $\,10a+b,\,$ for modulus $\, n = 11.\,$ For completeness, below we give a direct proof of this special case of the above sketched proof.
$\left.\begin{align}{\bf Theorem}\ \ \bmod n!:\,\ \color{#c00}{a'}\equiv \color{#0a0}a\ b'\equiv b\end{align}\right}\, $ $\Rightarrow$ $\,\ \begin{align} &10\,\color{#c00}{a'}+b'\ \ \ \ [\:!x' = x\bmod 11 = x\% 11\,\ \rm in\ OP]\ \equiv\ &10\,\color{#0a0}a\,+\,b\end{align}$
$\begin{align}{\bf Proof}\qquad a'&\equiv a\qquad\quad\ \, \text{by hypothesis}\ 10a'&\equiv 10a\qquad\ \ \text{by the Congruence Product Rule}\ b'&\equiv b\qquad\quad\ \ \text{by hypothesis}\ \Rightarrow\ 10a'+b'&\equiv 10a+b\ \ \ \text{by the Congruence Sum Rule} \end{align}$
Remark $ $ To get the exact form of your result apply a final $\bmod 11\,$ to the above to convert it from a congruence relation to a mod operation (remainder), using the following $$ a\equiv b!!!\pmod{n}\iff (a\bmod n) = (b\bmod n) $$
Generally this is the easiest way to prove identities about mod operations, i.e. use more flexible congruences to first prove the analogous congruence relation, then apply a final mod operation to get (canonical / normal) remainders (or residues).
See this answer for another worked example: $\,(g^b \bmod m)^a \bmod m = (g^a \bmod m)^b \bmod m$
Note we induct on the "height" of the expression = the total number of sum & product operations. The base case of height $= 0\,$ has no operations so the expression is an integer, so any replacement by a congruent integer yields a congruent result. Else the expression is a sum or product. Its arguments have smaller height (since they exclude this operation), so by induction any replacements yield a congruent argument, so this sum or product remains congruent by the congruence sum or product rule.
Share
edited Aug 20, 2023 at 19:17
answered Apr 3, 2019 at 15:23
Bill DubuqueBill Dubuque
284k4242 gold badges339339 silver badges1k1k bronze badges
$\endgroup$
Add a comment |
-1
$\begingroup$
Two modular arithmetic rules you are missing:
The product of remainders, is congruent to the remainder of the unmodded product.
The sum of remainders, is congruent the remainder of their unmodded sum.
that means you can turn both into:$$1000\bmod 11+200\bmod 11+50\bmod 11+6\bmod 11 = (10+2+6+6)\bmod 11=10+2+12 \bmod 11 =10+2+1\bmod 11 =11+2\bmod 11= 2\bmod 11$$
Share
edited Apr 3, 2019 at 14:54
answered Apr 3, 2019 at 10:14
user645636user645636
$\endgroup$
2
$\begingroup$ Both 1 and 2 are incorrect - you need to apply a final mod to the product & sum of the remainders. Generally it is better to express these laws in congruence form, then take the remainder only at the final step, using $\,a\equiv b\pmod{n}\iff (a\bmod n) = (b\bmod n)\ \ $ $\endgroup$
Bill Dubuque
– Bill Dubuque
2019-04-03 14:43:08 +00:00
Commented Apr 3, 2019 at 14:43
$\begingroup$ fine corrected the statements. $\endgroup$
user645636
– user645636
2019-04-03 14:54:35 +00:00
Commented Apr 3, 2019 at 14:54
Add a comment |
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
number-theory
modular-arithmetic
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Linked
1 How come modulo gives same result while applying on each step or applying once in end?
0 Suppose that a and b are integers, a $\equiv$ 11 (mod 19) and b $\equiv$ 3 (mod 19). Find the integer c such that c $\equiv$ a - b (mod 19).
0 Finding all solutions for congurence when m equals $2$
-1 Split two integers of module operation $\pmod{m}$
0 Does congruence mod k work for exponents?
0 Binomial expansion and congruence?
0 Mod Arithmetic, involving multiplication.
107 Congruence Arithmetic Laws, e.g. in divisibility by $7$ test
mod [= remainder] operation (and relation), name and meaning
solve for $20x+15 \equiv 47 \pmod{4},$ if there are no solution why?
See more linked questions
Related
3 Why modular arithmetic in secret sharing?
0 Find $x \in{Z_{250}}$ so that $ x\equiv{248^{156,454,638}} \pmod{250}$?
0 Modular Arithmetic Related Question
0 Modular arithmetic finding the lowest number which solve equation
Help me understand division in modular arithmetic
0 Modular Arithmetic.......
0 Modular-arithmetic proofs
Hot Network Questions
What do christians believe about morality and where it came from
Singular support in Bezrukavnikov's equivalence
Is this commentary on the Greek of Mark 1:19-20 accurate?
The altitudes of the Regular Pentagon
Why weren’t Prince Philip’s sisters invited to his wedding to Princess Elizabeth?
Converting hypergeometric function to Struve form
An odd question
What is the name of the 1950’s film about the new Scots lord whose relative is a frog like creature living in the ancestral home?
How to use cursed items without upsetting the player?
Find non-trivial improvement after submitting
Are there any world leaders who are/were good at chess?
Mishearing Monica’s line in Friends: “beacon that only dogs…” — is there a “then”?
What "real mistakes" exist in the Messier catalog?
Do sum of natural numbers and sum of their squares represent uniquely the summands?
Calculating the node voltage
Can a state ever, under any circumstance, execute an ICC arrest warrant in international waters?
Riffle a list of binary functions into list of arguments to produce a result
What can be said?
Fix integral lower bound kerning in textstyle or smaller with unicode-math
Bypassing C64's PETSCII to screen code mapping
Can a cleric gain the intended benefit from the Extra Spell feat?
How can I show that this sequence is aperiodic and is not even eventually-periodic.
Can a GeoTIFF have 2 separate NoData values?
How to home-make rubber feet stoppers for table legs?
more hot questions
Question feed |
189220 | https://www.hamilton.edu/academics/centers/writing/style/essentials/tense | Utility
You are here:
Tense
Shorthand: “t”
Navigation
To tell the story of a work of literature or a historical sequence, most writers use the “historical present”; they discuss the events as if they were happening in the present.
At the end of the movie Love Story, Jennifer dies a tragic death.
Historical present correctly used. It is permissible to use either the historical present or the simple past tense in such cases, but whichever one you use, use it consistently.
Frequently, however, errors of tense are due to standards that vary across disciplines. Check with your professors to find out what tenses their departments prefer.
For more information see Verb Tense.
Find Your Voice
What good is having a great idea if you can’t communicate it effectively? We’ll teach you how to express yourself through writing and speaking, of course, but also through digital communications and artistic expression — all of which will help you stand out no matter what path you choose after graduation.
What good is having a great idea if you can’t communicate it effectively? We’ll teach you how to express yourself through writing and speaking, of course, but also through digital communications and artistic expression — all of which will help you stand out no matter what path you choose after graduation.
Contact
Office / Department Name
Nesbitt-Johnston Writing Center
Contact Name
Jennifer Ambrose
Writing Center Director
Help us provide an accessible education, offer innovative resources and programs, and foster intellectual exploration.
Social
Social
Footer
©
2025
Hamilton College.
All Rights Reserved.
Site Menu
Site
Secondary
Actions
Support
Audience
Site Search |
189221 | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/45f39f52d5aeef8a25b805b1726cfa60_MIT6_01SCS11_swLab08.pdf | Software Lab 8 6.01 – Fall 2011 Describing Circuits Goals: This software lab seeks to develop a method for describing circuits at a high level of abstraction, and to convert that description into linear equations which can then be solved. This is the method that CMax, and many other systems, use when simulating electrical circuits. You will: • Understand methods for representing and solving sets of linear equations • Understand the node-voltages-with-component-currents (NVCC) analy sis method for circuits • Expore a high-level Python representation of circuits, the Circuit class • Implement representations of resistors and opamps • Implement the NodeToCurrents class, to solve for circuit voltages and currents 1 Setup Using your own laptop Be sure you have the 6.01 software libraries ins talled.
• Download and unzip swLab08.zip into a convenient folder (e.g., ~/Desktop/6.01/swLab08).
• This week’s work files are circSkeleton.py and swLab08Work.py. 2 Specifying and solving linear equations Consider the problem of finding values for x and y that satisfy the two equations: 5x − 2y = 3, and 3x + 4y = 33. You would probably approach this with the substitution method, in which you solve the first equa tion for x, getting x = 2 5y + 3 5, and then substituting that into the second equation, getting Some of the software and design labs contain the command athrun 6.01 getFiles. Please disregard this instruction; the same files are available on the 6.01 OCW Scholar site as a .zip file, labeled Code for [Design or Software Lab number]. 1 Software Lab 8 6.01 Fall 2011 6 9 y + + 4y = 33. 5 5 Then, solving for y yields y = 6, and substituting y = 6 into the x expression yields x = 3. Solving two equations in two unknowns is easy, but humans are not generally good at solving larger systems of the type that we encounter in circuits. Therefore, we will use a computer. We could try to write a computer program to perform the substitution method, but this method is complicated and computationally inefficient. By contrast, Gaussian elimination is efficient and relatively easy to implement on a computer. We’ll use a standard implementation of it from the Python numpy library. We will use the 6.01 software module le to represent sets of equations.1 An equation is repre sented with an instance of class le.Equation, which takes, at initialization time, three arguments: coeffs: a list of numerical coefficients for the variables mentioned in the linear equation • variableNames: a list of strings, naming the variables in the equation; the variable names must • be listed in the same order as the coefficients constant: the numerical constant in the equation, with the sign chosen so that the constant is • the only term on one side of the equality. For example, we could represent the equation −3x + 9.2z = 4 as le.Equation([-3, 9.2], [’x’, ’z’], 4) or by any of several alternative (and equally valid) expressions, including these two: le.Equation([9.2, -3], [’z’, ’x’], 4) le.Equation([3, -9.2], [’x’, ’z’], -4) We can represent a set of equations using an instance of the class le.EquationSet. This class takes no parameters at initialization time, and returns an object that represents an empty set of equations. Equations can then be added using the addEquation method, whose input is an instance of le.Equation, described above. You can find the solution to a set of equations by using the solve method of an le.EquationSet. The solve method returns an instance of le.Solution. You can then look up the value of any of the variables by using the translate method, which takes, as input, a string that represents the variable of interest, and returns the value of that variable in the solution. Using these classes, you can describe, then solve, our simple example like this: >>> small = le.EquationSet() >>> small.addEquation(le.Equation([5, -2], [’x’, ’y’], 3)) >>> small.addEquation(le.Equation([3, 4], [’x’, ’y’], 33)) >>> sol = small.solve() >>> sol.translate(’x’) 3.0 >>> sol.translate(’y’) 6.0 1 see the 6.01 Software Documentation under the Reference Material tab of the 6.01 web page. 2 Software Lab 8 6.01 Fall 2011 3 NVCC circuit equations We now customize our equation solving software to solve circuit equations. Our method is based on the node-voltages-with-component-currents (NVCC) method of analysis (see Section 6.4.1 in the Course Notes), which we describe by way of the following example, where ’n1’, ’n2’, and ’n3’ represent node voltages and ’i1’, ’i2, ’i3’, and ’i4’ represent component currents. +10 A +15 V 3 Ω 2 Ω n1 n2 n3 i1 i2 i3 i4 We start by creating an empty equation set: ckt = le.EquationSet() Now, we add one equation for each of the four components: n1 − n3 = 15 n1 − n2 = 3i1 n2 − n3 = 2i2 i3 = 10 as follows: ckt.addEquation(le.Equation([1.0, -1.0], [’n1’, ’n3’], 15.0)) ckt.addEquation(le.Equation([1.0, -1.0, -3], [’n1’, ’n2’, ’i1’], 0.0)) ckt.addEquation(le.Equation([1.0, -1.0, -2], [’n2’, ’n3’, ’i2’], 0.0)) ckt.addEquation(le.Equation([1.0], [’i3’], 10.0)) Next, we need to specify an equation that sets the voltage of the ground node to be zero. We have chosen n3 as ground, n3 = 0 but any other choice would also lead to a valid (but different) solution. We add the ground equa tion to the equation set with ckt.addEquation(le.Equation([1.0],[’n3’], 0.0)) Finally, we specify KCL equations for all of the nodes except the ground node: − i4 − i1 = 0 i1 − i2 + i3 = 0 which are added to the equation set with ckt.addEquation(le.Equation([-1.0, -1.0], [’i4’, ’i1’], 0.0)) ckt.addEquation(le.Equation([1.0, -1.0, 1.0], [’i1’, ’i2’, ’i3’], 0.0)) 3 Software Lab 8 6.01 Fall 2011 Now, we can solve the circuit, with the following result: >>> ckt.solve() i1 = -1.0 i2 = 9.0 i3 = 10.0 i4 = 1.0 n1 = 15.0 n2 = 18.0 n3 = 0.0 This is convenient, because it saves us from our own algebra errors. Unfortunately, it can be hard to remember to construct exactly the right set of equations (do I have one for each constituent component? do I have all of the current equations for each node? did I use the right names and coefficients for the currents?). 4 Describing circuits: the Circuit class A higher-level way to specify a circuit is illustrated as follows: c = circ.Circuit([circ.VSrc(15, ’n1’, ’n3’), circ.Resistor(3, ’n1’, ’n2’), circ.Resistor(2, ’n2’, ’n3’), circ.ISrc(10, ’n3’, ’n2’) ]) Here we specify each component (e.g., a 15V voltage source, a 3Ω resistor, a 2Ω resistor, and and 10A current source) along with strings that represent the nodes to which that component is connected. The solve method then takes the symbolic name for the ground node as its single input and solves the resulting equations using the le module discussed previously: >>> c.solve(’n3’) Solving equations +n1-n3 = 15 +n1-n2-3i_n1->n2_14 = 0 +n2-n3-2i_n2->n3_15 = 0 +i_n3->n2_16 = 10 +i_n1->n3_13+i_n1->n2_14 = 0.0 -i_n1->n2_14+i_n2->n3_15-i_n3->n2_16 = 0.0 +n3 = 0 i_n1->n2_14 = -1.0 i_n1->n3_13 = 1.0 i_n2->n3_15 = 9.0 i_n3->n2_16 = 10.0 n1 = 15.0 n2 = 18.0 n3 = 0.0 The first four equations (listed above between the borders) describe the components, the next two are KCL equations, and the last specifies the ground. The solution (listed after the borders) tells us the currents through the components and the voltages at the nodes. 4 Software Lab 8 6.01 Fall 2011 Notice that the Circuit class provides two major simplifications. First, the equations for each component were automatically generated from the specified element type (e.g., resistor, voltage source, etc.). Second, we don’t have to specify KCL equations at all. Also notice that names are automatically assigned to the component currents. For example, i_n1->n2_14 is a current that flows between nodes n1 and n2. We append an additional unique number (in this case 14) to the name, because there could be multiple components connected in parallel between n1 and n2. Wk.8.1.1 Write the EquationSet for the circuit shown in the tutor problem. Also, write the more abstract representation (described above) for that circuit. You can debug this in idle using the file swLab08Work.py. 5 Implementing components: resistors and opamps The Circuit class keeps track of all of the components of a circuit as instances of classes that rep resent each type of component (e.g., the Resistor class represents resistors, the VSrc represents voltage sources, etc.). Each component class is a subclass of the Component class, and must supply two methods: getEquation, which returns an instance of le.Equation that constrains the voltage across the • terminals of the component, and getCurrents, which returns the list of currents that this component introduces to the nodes to • which it is connected. Each current is represented as a list [i, node, sign], where i is the name of a current variable, node is the name of a node, and sign is the sign of that current at that node, either +1 or -1 (indicating whether the current going in or out of the node). All two-input components have the same pattern of currents: they make a new current variable when created, and then assert that it flows into their node n1 and out of their node n2. So, we have implemented this pattern as the default getCurrents method in the Component class. class Component: def getCurrents(self): return Here is how the Resistor component is implemented. class Resistor(Component): def init(self, r, n1, n2): self.current = util.gensym(’i_’+n1+’->’+n2) self.n1 = n1 self.n2 = n2 self.r = r def getEquation(self): # your code here The util.gensym procedure takes a string as an argument and returns a string which is the ar gument with a unique integer appended to it. This procedure is useful for implementing names for currents. 5 Software Wk.8.1.2 This problem guides you through implementing the getEquation method for the Resistor class. Wk.8.1.3 This problem guides you through implementing the OpAmp class as a voltage-controlled voltage source; see Section 6.6.1 of the Course Notes. Lab 8 6.01 Fall 2011 6 Implementing the NodeToCurrents class The heart of the method of NVCC analysis method for solving circuit equations is assembling the correct set of equations, and using a linear equation solver to obtain the node voltages and circuit currents. In this part of the lab, you build the NodeToCurrents class which performes these steps to solve for circuit parameters. The Circuit class has two methods; class Circuit: def init(self, components): self.components = components def solve(self, gnd): es = le.EquationSet() n2c = NodeToCurrents() for c in self.components: es.addEquation(c.getEquation()) n2c.addCurrents(c.getCurrents()) es.addEquations(n2c.getKCLEquations(gnd)) return es.solve() A circuit is just a list of instances of the Component class. When we ask the circuit to solve itself, we provide the name of a node, passed in as parameter gnd, which will be the ground node and have voltage 0. Calling the solve method then does the following: 1. Makes a new empty equation set es. 2. Makes a new instance, n2c, of the NodeToCurrents class. This class keeps track of which cur rents are flowing into and out of each node. 3. For each component, adds the equation that describes the relationship between voltage and current that the component induces, and it adds the currents to the appropriate nodes in NodeToCurrents. 4. Adds the KCL equations that result from the node-current relationships stored in n2c, and one that sets the node named by the gnd variable to have voltage 0. 5. Solves the equations. You can read about the NodeToCurrents class and its methods in the software documentation. 6 Software Lab 8 6.01 Fall 2011 Wk.8.1.4 Implement the NodeToCurrents class. Note that the getKCLEquations method should return a list of equations (not an EquationSet). There should be one KCL equation in the list for each unique node other than ground. There should also be one equation in the list to specify that the voltage associated with the ground node is zero. You should debug your code in the circSkeleton.py file and then paste it into the Tutor. 7 MIT OpenCourseWare 6.01SC Introduction to Electrical Engineering and Computer Science Spring 2011 For information about citing these materials or our Terms of Use, visit: |
189222 | https://cdn.intratec.us/docs/reports/previews/cyclohexane-e11a-b.pdf | Cyclohexane from Benzene and Hydrogen (Liquid-Phase Process) Report Cyclohexane E11A Cost Analysis United States CYCLOHEXANE FROM BENZENE AND HYDROGEN (LIQUID-PHASE PROCESS) REPORT CYCLOHEXANE E11A Published by www.intratec.us Intratec All rights reserved. Reproduction of any part of this work by any process whatsoever without written permission of Intratec is expressly forbidden. See 'Appendix B. Terms & Conditions' and 'Appendix C. Report Sharing Terms'.
The information in this report is given in good faith and belief in its accuracy, but does not imply the acceptance of any legal liability or responsibility whatsoever, by Intratec for the consequences of its use or misuse in any particular circumstances.
Analysis developed by Intratec Copyright 2021 Intratec Solutions LLC i ABSTRACT This report presents a cost analysis of Cyclohexane from benzene and hydrogen. The process examined is a liquid phase hydrogenation process. In this process, benzene is hydrogenated in a liquid-phase reactor in the presence of a nickel-based catalyst. The reactor's overhead is sent to a second reactor to convert the non-reacted benzene. Cyclohexane product is obtained.
The report examines one-time costs associated with the construction of a plant and the continuing costs associated with the daily operation of such a plant.
The analysis assumes a United States-based plant capable of producing 200 kt of Cyclohexane per year and includes: Capital Investment, broken down by: -Total fixed capital required, divided in process unit (ISBL); infrastructure (OSBL), contingency and owner's cost - Working capital and costs incurred during industrial plant commissioning and start-up Operating cost, broken down by: - Variable operating costs (raw materials, utilities) - Fixed operating costs (maintenance, operating charges, plant overhead, local taxes and insurance) - Depreciation This report was developed based essentially on the following reference(s): US Patent 5,668,293, issued to Institut Francais du Petrole in 1997 Keywords: Benzene Hexahydride, Cycloalkane, Liquid-Phase Reaction, Hydrogenation Reaction, IFP Raw materials consumption, products generation and labor requirements Process block flow diagram and description of industrial site installations (process unit and infrastructure) ii BASIC ANALYSIS About This Report............................................................................................................................................................ 1 Study Objective..................................................................................................................................................... 1 Report Overview....................................................................................................................................................1 How to Use this Report?.................................................................................................................................... 2 Executive Summary.........................................................................................................................................................3 About Cyclohexane..............................................................................................................................................3 Cyclohexane Production Process.................................................................................................................... 4 Economic Analysis...............................................................................................................................................5 About Cyclohexane......................................................................................................................................................... 6 Introduction............................................................................................................................................................6 Commercial Forms & Applications...................................................................................................................7 Cyclohexane Production Pathways................................................................................................................. 8 Process Overview............................................................................................................................................................ 9 Product(s) Generated..........................................................................................................................................9 Process Inputs.......................................................................................................................................................9 Technology Maturity Assessment...................................................................................................................11 Highlights & Remarks........................................................................................................................................13 Industrial Site...................................................................................................................................................................14 Introduction.......................................................................................................................................................... 14 Process Unit Description...................................................................................................................................16 Site Infrastructure Description........................................................................................................................20 Key Process Input and Output Figures........................................................................................................ 24 Labor Requirements..........................................................................................................................................24 Capital Investment.........................................................................................................................................................25 TABLE OF CONTENTS Available in 'Advanced' version only Available in 'Extended' and 'Advanced' versions only iii Fixed Capital....................................................................................................................................................... 26 Working Capital..................................................................................................................................................28 Additional Capital Requirements....................................................................................................................29 Total Capital Investment.................................................................................................................................. 30 Operating Costs............................................................................................................................................................. 31 Operating Variable Costs.................................................................................................................................33 Operating Fixed Costs......................................................................................................................................34 Depreciation........................................................................................................................................................34 Total Operating Cost.........................................................................................................................................34 Product Value Analysis................................................................................................................................................ 36 Corporate Overhead......................................................................................................................................... 37 Return on Capital Employed (ROCE)............................................................................................................37 Process Economics Summary....................................................................................................................................38 Product Value Composition.............................................................................................................................38 Economic Datasheet.........................................................................................................................................38 Economic Remarks........................................................................................................................................... 40 References.......................................................................................................................................................................41 Analysis Methodology.................................................................................................................................................. 42 Introduction......................................................................................................................................................... 42 Bibliographical Research................................................................................................................................. 42 Process Overview..............................................................................................................................................42 Technology Maturity Assessment................................................................................................................. 44 Examining an Industrial Site............................................................................................................................44 Capital Investment Estimating........................................................................................................................ 46 Operating Cost Estimating.............................................................................................................................. 53 Product Value Estimating.................................................................................................................................56 Estimates Limitation..........................................................................................................................................58 Available in 'Advanced' version only Available in 'Extended' and 'Advanced' versions only iv APPENDIX A - ABOUT INTRATEC................................................................................................................................A-1 APPENDIX B - TERMS & CONDITIONS.......................................................................................................................B-1 APPENDIX C - REPORT SHARING TERMS................................................................................................................C-1 APPENDIX D - EXTENDED ANALYSIS...................................................................................................................... D-1 Utilities Consumption Breakdown........................................................................................................................... D-2 Utilities Consumption Figures....................................................................................................................... D-2 Impact of Utilities on Operating Cost.......................................................................................................... D-3 Economic Analysis for Different Capacities..........................................................................................................D-4 Introduction........................................................................................................................................................D-4 Capital Investment Comparison................................................................................................................... D-4 Operating Cost & Product Value Comparison..........................................................................................D-5 Project Implementation & Construction Schedule...............................................................................................D-7 Materials & Utilities Pricing Data..............................................................................................................................D-9 Analysis Pricing Basis.....................................................................................................................................D-9 Historical Prices................................................................................................................................................D-9 APPENDIX E - ADVANCED ANALYSIS....................................................................................................................E-1 Process Unit (ISBL) Construction Cost by Functional Unit............................................................................... E-2 Infrastructure Construction Cost by Piece of Equipment.................................................................................. E-4 Introduction........................................................................................................................................................E-4 Construction Cost: Area 90 - Storage Installations.................................................................................E-5 Construction Cost: Area 91 - Utilities Facilities.........................................................................................E-7 Construction Cost: Area 92 - Support & Auxiliary Buildings................................................................ E-9 Site Infrastructure Cost Summary...............................................................................................................E-11 Plant Cost Breakdown per Discipline....................................................................................................................E-12 Introduction......................................................................................................................................................E-12 Direct Costs Breakdown...............................................................................................................................E-13 Indirect Costs Breakdown............................................................................................................................E-14 Available in 'Advanced' version only Available in 'Extended' and 'Advanced' versions only v Plant Cost Breakdown Summary............................................................................................................... E-15 Process Flow Diagrams & Equipment List...........................................................................................................E-16 Available in 'Advanced' version only Available in 'Extended' and 'Advanced' versions only vi Economic Analysis Summary............................................................................................................................................. 5 Process Technology Maturity Scale................................................................................................................................12 Raw Materials Consumption.............................................................................................................................................24 Labor Requirements...........................................................................................................................................................24 Plant Cost Estimate Accuracy Range (USD Million)...................................................................................................27 Owner's Cost Summary.....................................................................................................................................................27 Fixed Capital Summary......................................................................................................................................................28 Working Capital Breakdown.............................................................................................................................................28 Additional Capital Requirements Breakdown...............................................................................................................29 Operating Variable Costs Breakdown............................................................................................................................33 Operating Fixed Costs Breakdown.................................................................................................................................34 Operating Cost Summary..................................................................................................................................................34 Corporate Overhead Costs Breakdown........................................................................................................................37 Cyclohexane from Benzene and Hydrogen (Liquid-Phase Process) - Economic Datasheet...........................39 Process Contingency Factor Estimation Methodology..............................................................................................47 Project Contingency Factor Estimation Methodology................................................................................................48 Fixed Capital Estimate Accuracy Range.......................................................................................................................52 Maintenance Cost Estimation Methodology.................................................................................................................55 Marketing & Advertising Cost Estimation Methodology............................................................................................57 Research & Development Cost Estimation Methodology......................................................................................... 57 Expected ROCE Factor Estimation Methodology.......................................................................................................58 Net Utility Consumption Rates (per mt of Cyclohexane).........................................................................................D-2 Utilities Share in Total Operating Cost.........................................................................................................................D-3 Capital Investment Analysis for Different Capacities (MM USD)...........................................................................D-5 Operating Cost & Product Value Analysis for Different Capacities (USD/mt)................................................... D-6 LIST OF TABLES Available in 'Advanced' version only Available in 'Extended' and 'Advanced' versions only vii Project Phases Schedule.................................................................................................................................................D-7 Materials & Utilities Prices (United States, Q1'21)..................................................................................................... D-9 Process Unit (ISBL) Construction Cost by Functional Unit..................................................................................... E-2 Area 90 - Storage Installations: Scope Description.................................................................................................. E-5 Area 91 - Utilities Facilities: Scope Description.......................................................................................................... E-7 Area 92 - Support & Auxiliary Buildings: Scope Description..................................................................................E-9 Site Infrastructure (OSBL) Construction Cost by Piece of Equipment................................................................E-11 Plant Construction Cost by Discipline.........................................................................................................................E-15 Available in 'Advanced' version only Available in 'Extended' and 'Advanced' versions only viii Process Schematic Diagram...............................................................................................................................................4 Cyclohexane Production Pathways Diagram..................................................................................................................8 Industrial Site Configuration.............................................................................................................................................. 15 Block Flow Diagram - Manufacturing Process............................................................................................................. 17 Plant Cost Summary (USD Million)..................................................................................................................................26 Capital Investment Summary (USD Million).................................................................................................................. 30 Operating Cost Breakdown (USD/mt)........................................................................................................................... 35 Product Value Composition (USD/mt)............................................................................................................................38 Production Cost Report Development Methodology..................................................................................................43 Cost Contribution of Each Utility................................................................................................................................... D-3 Capital Investment (USD Million) Versus Plant Capacity.........................................................................................D-4 Operating Cost (USD/mt) Versus Plant Capacity......................................................................................................D-5 Implementation & Construction Schedule...................................................................................................................D-8 Process Unit (ISBL) Construction Cost by Functional Unit..................................................................................... E-3 Site Infrastructure (OSBL) Construction Cost by Area............................................................................................ E-4 Storage Installations Construction Cost per Piece of Equipment..........................................................................E-5 Utilities Facilities Construction Cost per Piece of Equipment.................................................................................E-7 Support & Auxiliary Buildings Construction Cost per Piece of Equipment......................................................... E-9 Plant Construction Cost Summary.............................................................................................................................. E-12 Direct Construction Costs by Discipline.....................................................................................................................E-13 Indirect Costs Summary.................................................................................................................................................E-14 Process Flow Diagram (1 of 1).......................................................................................................................................E-19 LIST OF FIGURES Available in 'Advanced' version only Available in 'Extended' and 'Advanced' versions only ix ABOUT THIS REPORT This report presents the economics of Cyclohexane from benzene and hydrogen. The process examined is a liquid phase hydrogenation process.
The primary objective of this study is to explain the cost structure of the aforementioned process, encompassing capital investment and operating cost figures.
The process design and economics in this report are based on an industrial facility with a capacity of 200,000 metric ton of Cyclohexane per year, a nominal capacity that is globally competitive.
Study Objective This report is structured into eight main parts which follow a logical sequence.
Each of these parts is described below.
By way of introduction, the first part – the current chapter – briefly explains the report itself, its structure and objective. Readers are encouraged to spend a few minutes reading this chapter, so as to make the most of the study.
In the second part, About Cyclohexane, the reader will learn the basics of Cyclohexane itself. This chapter also covers its applications and major production pathways.
The third part, Process Overview, presents basic aspects of the process studied: products generated, process inputs, and physico-chemistry highlights.
The fourth part, Industrial Site, describes an industrial plant based on the process under analysis, in terms of the process unit and infrastructure required. This technical analysis underlies the entire study.
The fifth part, Capital Investment, presents all capital costs associated with the process examined, from design and erection of an industrial site to plant startup.
Operating Costs of the process are examined in the sixth part. Ongoing costs related to the operation of a unit based on the process are studied, including operating fixed costs, operating variable costs and depreciation.
The seventh part, Product Value, targets to estimate the gate cost of the plant final product, by adding corporate overhead costs and a parcel that will guarantee an expected Return On Capital Employed (ROCE). It provides an idea of the minimum price at which the product may be sold, and how competitive it is.
The eighth part, Process Economics Summary, summarizes all economic figures presented throughout the Report Overview In addition, the economic assessment, developed for the period Q2 2017, assumes the construction of a United States-based industrial facility that includes the infrastructure typically required for such a project.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
1 report.
Finally, to address any questions or concerns about the methodologies and procedures adopted in the development of this report, the reader is referred to the eighth part, Analysis Methodology.
How to Use this Report?
The main purpose of this Report is to assist readers in a preliminary economic evaluation of the production process approached. It is a valuable support tool for a myriad of activities and studies, such as screening and assessment of investment options, preliminary evaluation of the economic potential of emerging production processes, rough assessment of the economic feasibility of industrial ventures, cost estimates double-checking, preliminary budget approval, research planning, and so on.
Readers must always bear in mind the nature of this report and the resulting limitations on how to properly use it. Limitations that apply to both technical data and economic assessment presented in this study are explained below.
Technical Data The preliminary design of the process, presented in the part Industrial Site, is based on fast techniques that rely on reduced design efforts. The goal of such preliminary design is exclusively to represent the process in sufficient detail for supporting capital and production costs estimation within the accuracy expected: class 4 budgetary estimates.
Therefore the technical data presented must not be confused with an actual conceptual process design, and must not be used as such.
Economic Assessment The economic assessment presented in this report (parts Capital Investment, Operating Cost, Product Value Analysis and Process Economic Summary), developed for the period 2017 Q2, assumes the construction of a United States-based industrial facility.
This means that capital and production costs estimates presented are based on several general assumptions (e.g. average market figures for raw materials, chemicals and utilities prices, labor costs, taxes and duties), believed to suitably portray local conditions for the period of analysis informed, on a country-level basis.
Accordingly, the economic assessment provided in this report is not meant to fit any specific industrial venture, which would involve a wealth of specific data and assumptions not herein considered.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
2 EXECUTIVE SUMMARY Cyclohexane is a relatively stable cycloalkane, present in crude oils in concentrations of 0.1–1.0%. This cycloparaffin is a colourless, flammable liquid, comprised of a ring of six carbon atoms and has a gasoline-like odor. Cyclohexane is soluble in ether, alcohol, chlorinated/aromatic solvents and insoluble in water.
Cyclohexane must be stored in cool, dry, well-ventilated area out of direct sunlight, away from open flames, hot surfaces and sources of ignition. Explosion-proof ventilating equipment is recommended. Cyclohexane should not be stored near oxidizing agents or acidic material. Carbon steel, stainless steel, glass, Teflon and Viton can be used as packaging materials. Polypropylene tanks or containers should be avoided. It is transported by barge, pipeline and rail for long distances, and by truck for short distances.
Almost all cyclohexane produced commercially is employed as intermediate in the production of precursors to nylon. A very small fraction of the cyclohexane produced is consumed in minor miscellaneous uses, such as solvents, polymer reaction diluents and fungicidal formulations.
About Cyclohexane THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
3 The present analysis approaches Cyclohexane from benzene and hydrogen.
The process under analysis comprises three major sections: (1) Reaction; (2) Hydrogen Recovery; and (3) Purification.
Reaction.
Initially, benzene is fed to the primary reactor alongside fresh and recycled hydrogen. The hydrogenation reaction is carried out in a bubble column reactor, in the presence of a nickel based catalyst.
The catalyst is maintained in a suspension with the aid of an external circulation loop. Most of the heat of reaction is removed by the vaporization of the product stream; the remaining reaction heat is removed in the external loop, by passing the reactor reflux stream through a heat exchanger against boiler feedwater, producing low pressure steam. Most of the benzene feed is converted in this step.
The top gaseous product stream is directed to a fixed-bed reactor, where the remaining benzene content is converted to cyclohexane.
The finishing hydrogenation reaction is conducted in the presence of a solid nickel based catalyst supported on alumina.
Hydrogen Recovery. The product stream is fed to a knock-out drum operated at high pressure. Most of the cyclohexane in the feed condenses producing two streams, one gaseous hydrogen-rich stream and a liquid cyclohexane-rich stream. The hydrogen stream is routed to the Recycle Compressor, where it is compressed to the Primary Reactor reaction pressure and recycled.
Purification. In a distillation column, the liquid cyclohexane-rich stream is stripped of lighter contaminants, such as methane, ethane and soluble hydrogen. Light-ends recovered from column’s top are used for fuel, while a cyclohexane stream with a residual benzene content that is lower than 100 ppm is obtained from column’s bottom.
Cyclohexane Production Process Process Schematic Diagram THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
4 Economic Analysis The next table provides a summary of all capital, operating and non-operating costs related to the process described in the report, based on a mt/y plant.
Also, it presents some remarks about the key aspects surrounding the economic analysis performed.
% QUANTITY PER MT PRICE MM USD/YR DESCRIPTION Economic Analysis Summary CAPITAL INVESTMENT SUMMARY BASIS: UNITED STATES, Q1 2021 Fixed capital Working capital Additional capital MM USD TOTAL CAPITAL INVESTMENT PLANT CAPACITY & OPERATION (IC INDEX: 161.8) Net raw materials cost Net utilities cost OPERATING VARIABLE COSTS OPERATING FIXED COSTS OPERATING CASH COST Depreciation TOTAL OPERATING COST PRODUCT VALUE Corporate Overhead USD/MT % for ISBL & Owner's cost / % for OSBL The above table shows the impact of variable costs in the cost of production - it represents approximately % of the total operating cost.
Regarding the capital investment it is worth mentioning that, in order to fulfill the infrastructure requirements assumed in the present analysis, OSBL investment represents about % of the plant cost.
Nominal capacity (mt/y) Operating rate (h/y) Actual production (mt/y) 5 ABOUT CYCLOHEXANE Introduction Cyclohexane is a relatively stable cycloalkane, present in crude oils in concentrations of 0.1–1.0%. This cycloparaffin is a colourless, flammable liquid, comprised of a ring of six carbon atoms and has a gasoline-like odor. Cyclohexane is soluble in ether, alcohol, chlorinated/aromatic solvents and insoluble in water.
The structure of Cyclohexane is presented below: On a commercial scale, Cyclohexane production is almost entirely based on the catalytic hydrogenation of benzene, which can be conducted in the liquid phase or in the vapor phase in the presence of hydrogen.
Nickel, platinum, or palladium have been used as catalyst in such Cyclohexane processes.
To a lesser extent, Cyclohexane is produced from the naphtha fraction from crude oil, by superfractionation.
Cyclohexane must be stored in cool, dry, well-ventilated area out of direct sunlight, away from open flames, hot surfaces and sources of ignition. Explosion-proof ventilating equipment is recommended. Cyclohexane should not be stored near oxidizing agents or acidic material. Carbon steel, stainless steel, glass, Teflon and Viton can be used as packaging materials. Polypropylene tanks or containers should be avoided. It is transported by barge, pipeline and rail for long distances, and by truck for short distances.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
6 Commercial Forms & Applications Almost all cyclohexane produced commercially is employed as intermediate in the production of precursors to nylon. A very small fraction of the cyclohexane produced is consumed in minor miscellaneous uses, such as solvents, polymer reaction diluents and fungicidal formulations.
Cyclohexane can be used in the manufacture of other products, including: adipic acid; caprolactam; cyclohexanol; and cyclohexanone.
The uses and applications of Cyclohexane may vary according to the product grade. The main forms of Cyclohexane are: THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
7 Cyclohexane has been primarily manufactured from hydrogenation of benzene, which can be carried out in both liquid and gaseous phases, in the presence of different metal-based catalysts. This chemical may also be obtained from naphtha fractionation, or by isomerization of methylcyclopentane.
The diagram below presents different production pathways of Cyclohexane, starting from different raw materials.
Cyclohexane Production Pathways Cyclohexane Production Pathways Diagram THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
8 The main product obtained in the process under analysis is high purity Cyclohexane, with less than 500 ppm of total impurities.
Product(s) Generated PROCESS OVERVIEW Process Inputs Benzene Benzene (C6H6) is a flammable and volatile liquid. As the most simple aromatic hydrocarbon, it is used as an intermediate to produce many important chemicals, such as styrene (raw material for polystyrene and synthetic rubber), cyclohexane (used in nylon production), alkylbenzenes (used in detergent industry), aniline (used to produce dyes and polyurethanes), and chlorobenzenes. Also, benzene is used to produce pharmaceuticals, specialty chemicals, plastics, and pesticides. It occurs naturally in crude oil and coal tar.
In the present analysis, it is assumed the use of a high purity grade of benzene, with a minimum purity of 99.9 wt% and a maximun sulfur content of 1 ppm.
Hydrogen Hydrogen (H2) is an important chemical feedstock, mainly applied in the manufacture of ammonia and methanol, and for hydroprocessing operations in petroleum refineries.
H2 is produced on a commercial scale primarily from steam reforming and partial oxidation of natural gas, coal, coke, or residue, and electrolysis of water.
In the present analysis, it is assumed that hydrogen is supplied at 30 bar.
Catalysts and Chemicals In the process under analysis, benzene hydrogenation is conducted in two steps, and in both steps the hydrogenation reaction is catalyzed by a nickel-based catalyst.
In the first step, conducted in the liquid phase, the catalyst used is a suspension of nickel 2-ethylhexanoate and sodium 2-ethylhexanoate in cyclohexane, reduced by triethylaluminium.
In the finishing step, conducted in the gas phase in a fixed bed reactor, a solid catalyst is used to promote the hydrogenation reaction. The catalyst contains nickel, both in the oxidized and metal forms, supported on alumina. Almost 95 wt% of the total nickel present in the catalyst is available on its oxidized form.
Raw Material(s) This chapter presents technical aspects of the process under analysis.
More specifically, the current chapter describes the outputs generated, the process inputs, and highlights about the physico-chemistry related to this process.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
9 Utilities The main utilities consumed in the process are cooling water, electricity, steam, and boiler feed water (BFW). The process also generates steam, which is exported to nearby facilities.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
10 The process technology under study was categorized according to its maturity.
The technical maturity, while a measure of performance, reliability, and operating experience associated with the technology being assessed, serves as an important input in the definition of assumptions that have a relevant impact on process economics (e.g. process contingency, project contingency, costs related to start-up inefficiencies and R&D, etc).
The process technology maturity is defined by Intratec team through a method adapted from the so-called Technology Readiness Level (TRL) method, developed by NASA and nowadays used in a broad range of sectors/industries.
There are nine TRLs, which describe the maturity of a technology, from basic technology reasearch to system test, launch and operations.
Originally intended to supporting decision-making over research and development activity, the nine technology readiness levels were divided into five major classes to portray the maturity level of chemical process technologies, from ‘concept’ to ‘established technology’. The table in the next page describes such five classes according to which Intratec team classifies technologies being studied, as well as the TRLs included within each class.
Technology Maturity Assessment THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
11 Process Technology Maturity Scale DESCRIPTION Established (Outdated) TECHNOLOGY STATUS Existing plants being shut down No longer adopted in new plants Obsolete technology Commercial (at least 1 plant) -SCALE TRL Established (In Use) 2+ commercial plants Proven technology (successful operations) Commercial (2+ plants) 9 Emerging 1 commercial plant Basic data for commercial plant Performance validation Demonstration plant Prototype near or at planned operational system Commercial (1 plant) 7-8 Embryonic Pilot-scale demonstration Engineering-scale models / prototypes Basic data for scale-up "Proof-of-Concept" validation Bench-scale demonstration Lab-scale technology definition Process modeling Analytical studies Active R&D Pilot 4-6 Conceptual Unproven idea/proposal No analysis or testing Paper concept/studies Concept Idea 1 Bench Lab 2-3 Demonstration THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
12 Highlights & Remarks THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
13 INDUSTRIAL SITE The information presented in this chapter is based on commonly utilized concepts related to the type of installations found within a typical industrial site. These concepts include: Process unit. Also known as inside battery units, these installations comprise all main units of the site required to modify the input stream and obtain the target output. These units are located Inside the Battery Limits (ISBL).
Infrastructure. Also known as outside battery units or offsite facilities, these installations do not directly enter into the modification of the process input stream. They are support buildings, auxiliary units used for providing and distributing utilities and storage facilities. These units are located Outside the Battery Limits (OSBL).
In order to make a better distinction between these types of installation, a diagram is presented in the next page. It provides an insightful overview of the industrial site as whole and helps to clarify how raw materials and utilities are supplied to the process unit. In addition, it shows how any products or utilities generated in the process are discharged from the process unit.
Introduction This chapter presents all installations that comprise an industrial site for Cyclohexane from benzene and hydrogen. The process examined is a liquid phase hydrogenation process.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
14 Industrial Site Configuration THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
15 The process unit is the core of an industrial site.
Comprising the site’s battery limits (ISBL), it may be complex and involve several pieces of equipment. In this context, the most didactic approach to present a process unit is through the use of a block flow diagram. Visual information is, in fact, the clearest way to present a chemical process and is least likely to be misinterpreted.
In general, these diagrams consist of a series of blocks, representing unit operations or groups of equipment, connected by input and output streams.
In fact, there are no strict standards according to which such diagrams are made.
To facilitate the presentation of the process unit under analysis, Intratec developed a block flow diagram according to some standards.
The process areas represented (in light blue) correspond to a “functional unit”. Basically, a “functional unit” is a significant step in the process in which a particular physico-chemical operation (i.e., distillation, reaction, evaporation) occurs. According to this definition, a given functional unit is not associated with a single piece of equipment, but rather with a group of equipment and ancillaries required to perform a particular operation.
Blocks representing process areas also show key technical parameters related to these areas, including: highest operating temperature and pressure, representative material of construction, and other parameters.
As to the process streams represented, there is an indication of their phase when leaving/entering a block.
Also, such streams provide a global material balance of the process, normalized by the mass flow rate of the product considered in the analysis. In other words, the number near each stream represents the ratio between its mass flow rate and the output flow rate of the product under analysis.
It is worth noting that areas having no significant impact on the economics of the process may not be included in the diagram. Similarly, some streams may also not be represented. Nevertheless, the diagram presented is still extremely useful in providing readers with an overall understanding of the process studied.
For more information on how the process examined was divided into functional units, the reader is referred to the section on Process Unit in the “Analysis Methodology” chapter.
Process Unit Infrastructure requirements comprise the offsite facilities, or the units located Outside the Battery Limits (OSBL). The OSBL usually have a significant impact on the capital cost estimates associated with any new industry venture.
This impact is largely dictated by, among other things: specific conditions where the industrial site will be erected; the level of integration the new site will have with nearby facilities or industrial complexes; and assurance and promptness in the supply of chemicals.
Site Infrastructure The functional units related to the process under analysis are described based on the above explanation.
On the next page, a block flow diagram illustrates the functional units examined.
Process Unit Description THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
16 Block Flow Diagram - Manufacturing Process THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
17 The process under analysis is briefly described below. For clarity, the description was divided according to the process areas indicated in the diagram.
For a more detailed diagram presenting pieces of equipment and more process streams, reader is referred to the Process Flow Diagram & Equipment List, available in the 'Advanced Analysis' (Appendix E).
It is important to mention that some aspects of the process examined are either industrial secrets, not published in patents, or have changed and were not reported in the literature at the time this report was developed.
That being the case, the design herein presented is partially based on Intratec process synthesis knowledge such that there may be some differences between the industrial process actually employed and the process described in this study. Nevertheless, the design presented suitably represents the technology examined in sufficient detail to estimate the economics of the technology within the degree of accuracy expected from conceptual evaluations.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
18 THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
19 Site Infrastructure Description The infrastructure requirements of the industrial site examined are defined based on the following assumptions: Finally, offsite facilities were divided into areas according to their type/function. These areas are listed in the following pages, as well as the major equipment, systems and facilities included in each of them.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
20 Area 90 - Storage Installations THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
21 Area 91 - Utilities Facilities THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
22 Area 92 - Support & Auxiliary Buildings THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
23 Key Process Input and Output Figures In accordance with the block flow diagram and the global material balance previously presented, the following tables show key process indicators of the technology examined in this report. These indicators reflect the raw material consumption and the products generation rates per metric ton of Cyclohexane produced.
Raw Materials Consumption It should be noted that estimation of raw material requirements in the conceptual design phase is usually reasonably accurate but tends to be somewhat understated compared to real operations.
Losses from vessel vents, unscheduled equipment, inerting systems, physical property inaccuracies, startup, shutdown and other process operations not typically addressed in this phase may increase raw materials consumption.
Labor Requirements The following table presents the number of operators per shift required to run the equipment of the process examined, as well as the personnel per shift required to directly supervise the operating labor.
Labor Requirements QUANTITY PER MT OF PRODUCT RAW MATERIAL UNIT For detailed figures regarding utilities consumption, reader is referred to the the 'Extended Analysis' found in Appendix D.
PERSONNEL REQUIRED WORKERS PER SHIFT Operators Supervisors THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
24 CAPITAL INVESTMENT This chapter details all capital costs associated with Cyclohexane from benzene and hydrogen, from design to industrial plant startup.
The costs that comprise the total capital investment are grouped under three main headings: Fixed capital. Depreciable capital invested in erecting the industrial plant and making it operational Working capital. Funds for getting the plant into operation, and meeting subsequent obligations Additional capital requirements. Costs incurred during industrial plant start-up The graph below illustrates the composition of total capital investment.
IC Index-United States at the period of analysis: 161.8 The IC Index stands for Intratec Chemical Plant Construction Index, an indicator published monthly by Intratec to scale capital costs from one time period to another. It reconciles price trends of key components The estimates included in this chapter are based on the following assumptions: Plant nominal capacity: 200,000 metric ton of Cyclohexane per year Industrial plant location: United States Construction on a cleared, level site Period of analysis: 2021 Q1 THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
25 Fixed Capital Breakdowns for both ISBL and OSBL investments are provided in the 'Advanced Analysis' (Appendix E).
of chemical plant construction (e.g. labor, material, energy), providing historical and forecast data for readers.
The plant cost is related to the erection of the industrial site itself. It includes Total Process Capital (TPC) and Project Contingency, as described below.
Total Process Capital encompasses the investment required for the construction of: (1) process areas previously presented in the "Process Block Flow Diagram" (ISBL investment); (2) a process contingency reflecting technical uncertainties associated with limited design data, which may incur capital cost increases (e.g., additional equipment not included in the preliminary design); and (3) the site infrastructure (OSBL Investment), also previously discussed.
Project Contingency, in turn, is included to cover the costs that may arise as the project evolves.
Such costs include: project errors or incomplete specifications, labor costs changes, strikes, problems caused by weather; inflation, etc.
The chart below summarizes all items that make up the plant cost.
Plant Cost Also referred as “capital expenditures” (CAPEX), fixed capital constitutes the fraction of the capital investment which is depreciable. It includes Plant Cost and the Owner's Cost, further detailed below.
Plant Cost Summary (USD Million) THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
26 Plant Cost Estimate Accuracy Range (USD Million) COMPONENT % Outside battery limits (OSBL) Project contingency ( % of TPC) Inside battery limits (ISBL) PLANT COST Total process capital (TPC) The above table presents the lower and upper limits for the plant cost figures, according to the accuracy range expected from conceptual evaluations presented in this report. The presented range is associated with a confidence level of 90%. In other words, a 90% confidence level means that, for every 100 times the project is actually implemented, the plant cost required will fall into the range predicted with our estimates 90 times.
Process contingency ( % of ISBL) UPPER LIMIT Prepaid royalties % of plant cost Besides the plant cost, there are other costs that the owner must account for, such as: Initial catalyst load in reactors (if relevant) Purchase of technology through paid-up royalties or licenses Miscellaneous costs (pre-feasibility/environmental studies, regulatory and permits, long distance pipelines, etc.) The aforementioned costs are included as owner's cost.
The table below presents a breakdown of the owner's cost.
Unless otherwise indicated, all figures presented are in US million dollars (MM USD).
OWNER'S COST MM USD COMPONENT Owner's Cost Summary % ASSUMPTION Miscellaneous costs % of plant cost Owner's Cost For more information about the components of owner's cost, the reader is referred to section Capital Investment Estimating in the “Analysis Methodology” chapter.
ESTIMATE LOWER LIMIT THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
27 Working Capital Working Capital Breakdown For the purposes of this study, working capital is defined as the funds, in addition to the fixed capital, that a company must contribute to a project. Those funds must be adequate to bringing the plant into operation and meeting subsequent obligations.
Working capital includes: raw materials inventory, products inventory, in-process inventory, supplies and stores, accounts receivable and accounts payable.
The table below presents a breakdown of working capital. Unless otherwise indicated, all figures presented are in US million dollars (MM USD).
Raw materials inventory Products inventory In-process inventory Cash on hand Accounts receivable Accounts payable Net accounts receivable TOTAL WORKING CAPITAL Supplies and stores day(s) of total operating cost + corporate overhead day(s) of operating cash cost + corporate overhead day(s) of raw materials costs day(s) of total operating cost + corporate overhead day(s) of operating cash cost + corporate overhead % of annual operating labor and maintenance costs day(s) of operating cash cost + corporate overhead MM USD ASSUMPTION COMPONENT % Fixed Capital Summary The table below summarizes the fixed capital components discussed thus far.
Owner's cost TOTAL FIXED CAPITAL MM USD COMPONENT Fixed Capital Summary Plant cost % THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
28 Additional Capital Requirements MM USD ASSUMPTION COMPONENT Additional Capital Requirements Breakdown % Unscheduled plant modifications Start-up costs Land & site development Operator training Commercialization costs Start-up inefficiencies TOTAL ADDITIONAL CAPITAL day(s) of operating + supervision labor costs % of annual operating cash cost + corporate overhead % of plant cost % of annual operating cash cost + corporate overhead % of plant cost Several expenses are incurred during commissioning and start-up of an industrial site.
These expenses may be related to: Employee training Initial commercialization costs Operating inefficiencies and unscheduled plant modifications (equipment, piping, instruments, etc.) In addition, expenses with land acquisition and site development must also be accounted for.
Such additional costs are not addressed in most studies, but can become a significant expenditure.
In the current analysis, these costs are represented by additional capital requirements.
The table below presents a breakdown of additional capital investment.
Unless otherwise indicated, all figures presented are in US million dollars (MM USD).
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
29 Total Capital Investment The chart below summarizes all major capital costs discussed thus far, from the design and erection of an industrial site to plant startup.
Capital Investment Summary (USD Million) For more information about how the capital costs were estimated, the reader is referred to section on Capital Investment Estimating in the “Analysis Methodology” chapter.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
30 OPERATING COSTS This chapter details all ongoing costs required for Cyclohexane from benzene and hydrogen. Also referred as operational expenditures (OPEX), these encompass costs associated with the plant operation and depreciation. In the current analysis, the operating cost was grouped under three main headings: Operating variable costs. Costs directly proportional to the actual operating rate of the industrial site (i.e.
raw materials and utilities consumption) Operating fixed costs. Operating costs directly tied to the plant capacity, but which do not change with the operating level (i.e., operating labor, supervision labor, maintenance costs, plant overhead) Depreciation. Refers to the decrease in value of industrial assets with passage of time It should be kept in mind that the sum of operating fixed costs and operating variable costs is referred as “cash cost”. The sum of cash cost with depreciation, in turn, is referred to as “total operating cost”.
The graph below illustrates the composition of total operating cost.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
31 The estimates included in this chapter are based on the following assumptions: Industrial plant location: United States Period of analysis: 2021 Q1 Plant operating rate (hours per year): Plant nominal capacity: 200,000 metric ton of Cyclohexane per year THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
32 All costs presented in this table are derived from unit consumptions, detailed in the previous chapter, and pricing information.
Operating Variable Costs Variable costs change in direct proportion to changes in the operating level.
Such costs include raw materials and utilities (i.e., steam, electricity, fuel, and refrigeration).
QUANTITY PER MT PRICE MM USD/YR COMPONENT Operating Variable Costs Breakdown Net raw materials cost Net utilities cost OPERATING VARIABLE COSTS % In this study, the operating variable costs are considered to be negligible in comparison to the magnitude of the fixed costs.
Those costs correspond to the cost of purchasing the minor chemicals and generating minor utilities required.
USD/MT THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
33 Operating fixed costs Operating cash cost Depreciation TOTAL OPERATING COST COMPONENT Operating Cost Summary Operating variable costs MM USD/YR Operating Fixed Costs Total Operating Cost Supervision Maintenance cost Operating charges OPERATING FIXED COSTS COMPONENT Operating Fixed Costs Breakdown Operating labor MM USD/YR % Operating fixed costs are the costs primarily related to the capacity of an industrial site, but which do not change with operating rate.
Such costs include maintenance costs, operating charges, plant overhead, local taxes and insurance.
The table below presents a breakdown of operating fixed costs.
Plant overhead Property taxes and insurance Depreciation refers to the decrease in value of industrial assets with passage of time, primarily because of wear and tear. While not a true operating cost, depreciation is considered to be a operating expense for accounting purposes – it allows the recovery of the cost of an asset over a time period.
In this study, the depreciation unit cost corresponds to USD/metric ton of Cyclohexane produced. This calculation was based on the straight-line method and a project economic life of 10 years for both the core production unit (ISBL assets) and owner's assets, and 20 years for the site infrastructure (OSBL assets).
Depreciation The table below summarizes all operating cost components discussed thus far.
% of operating labor costs % of operating labor and maintenance costs % of fixed capital per year % of plant cost per year ASSUMPTION % USD/MT USD/MT operators/shift ; supervisors/shift ; THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
34 For more information about how the operating cost components were estimated, the reader is referred to the section on Operating Cost Estimating in the “Analysis Methodology” chapter.
Operating Cost Breakdown (USD/mt) The chart below presents a graphical representation of the operating cost breakdown.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
35 PRODUCT VALUE ANALYSIS Heretofore, the capital cost and operating cost related to the process examined were described. In order to provide a more consistent economic analysis of the process examined, all such costs are combined in a single item: “Full Cost”.
“Full Cost” is a term commonly used wherein all costs associated with a process are combined.
More specifically, it includes the operating cost (operating variable costs, operating fixed costs, and depreciation), as well as corporate overhead costs.
The graph below illustrates the composition of the "Total Cost".
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
36 Corporate Overhead Administration costs CORPORATE OVERHEAD COMPONENT Corporate Overhead Costs Breakdown MM USD/YR % Corporate overhead is associated with costs incurred by a company’s head office such as general administrative costs, information technology, research & development activities and marketing.
The table below presents a breakdown of corporate overhead costs.
Marketing & advertising ASSUMPTION Return on Capital Employed (ROCE) The ROCE is included to pay the investment made to manufacture target product.
This component is based on the expected return on capital employed typically aimed by chemical companies.
This “Expected ROCE Amount” component is, in fact, a measure of the cost of investment required to construct the plant, in terms of US dollars per amount of product.
Most chemical companies aim to achieve a ROCE percentage ranging from 5% to 25% for the construction of a new plant.
In this context, the Intratec team usually assumes an expected ROCE percentage in the range of 7% to 25%, depending on the type of product manufactured and the readiness of the technology employed (early-stage industrial processes inherently involve a larger amount of risk and cost uncertainty). For this specific process a ROCE percentage of 0% was assumed.
This results in an increment of 0 USD/mt in the product value.
USD/MT Research & development Information & technology of operating labor and maintenance costs of operating cash cost at full capacity of operating cash cost at full capacity of fixed capital THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
37 The chart below shows the impact of each cost component on the product value.
Product Value Composition PROCESS ECONOMICS SUMMARY This chapter provides a summary of all capital, operating and non-operating costs related to the process described so far. Also, it presents some remarks about the key aspects surrounding the economic analysis The table on the next page condenses the analysis developed in this report.
Economic Datasheet Product Value Composition (USD/mt) It is important to emphasize that product value should not be confused with product price.
The product value should be seen as a minimum price for which the product could be sold, so as the plant owner can get the expected ROCE according to the assumptions on which the economic analysis was based.
In other words, the product value is a variable calculated based on the costs associated with the manufacture of a product, which is something different from the actual product price seen in the market.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
38 % QUANTITY PER MT PRICE MM USD/YR DESCRIPTION Cyclohexane from Benzene and Hydrogen (Liquid-Phase Process) - Economic Datasheet CAPITAL INVESTMENT SUMMARY BASIS: UNITED STATES, 2021 Q1 Fixed capital Working capital Additional capital MM USD TOTAL CAPITAL INVESTMENT PLANT CAPACITY & OPERATION Net raw materials cost Net utilities cost Supervision Maintenance cost Operating charges OPERATING FIXED COSTS Operating labor Plant overhead Property taxes and insurance OPERATING CASH COST Depreciation TOTAL OPERATING COST % of operating labor costs % of operating labor and maintenance costs % of fixed capital per year PRODUCT VALUE Corporate Overhead % of plant cost per year oper./shift sup./shift USD/MT Nominal capacity (mt/y) Operating rate (h/y) Actual production (mt/y) THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
39 The "Return On Capital Employed" (ROCE) percentage usually ranges from 5% to 25% for the construction of a new plant.
A ROCE of 7% is more commonly expected for established industrial processes in the basic chemicals sector, while higher percentages are expected for less mature industrial processes or in more risky sectors, such pharmaceuticals or consumer products.
It should be noted that the risk taken into account in this analysis is limited to the technical risks associated with the process uncertainties or inherent risks associated with the venture's industry sector.
Other venture risks, such as business environment, raw materials and product prices variations, change in government policy, shall be evaluated case by case, and are not taken into account here, since this is a general analysis.
It is also important to mention that product value must not be confused with product price.
While the product value is calculated based on operating cost and expected ROCE, the product price is the actual value practiced in market transactions.
For more information about ROCE calculation, the reader is referred to the Product Value Estimating section of the chapter "Analysis Methodology".
For further clarification about the pricing assumptions used in this analysis, the reader is referred to the Operating Cost Estimating section, also in the chapter "Analysis Methodology".
Economic Remarks THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
40 REFERENCES THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
41 ANALYSIS METHODOLOGY Introduction Intratec distilled its expertise, gained from more than a decade of supporting companies worldwide in the analysis of chemical markets and process economics, and developed a consistent report development methodology.
The methodology ensures a holistic, coherent and consistent techno-economic evaluation, guiding the development of a report that allows readers to fully understand a specific chemical process technology.
In addition to being based on a common methodology, all Intratec reports that approach industrial processes have a common structure, i.e., indexes, tables and charts share similar standards. This ensures that Intratec’s readers know upfront what they will get and, more than that, will be able to compare technologies addressed in different reports.
Our methodology is continuously tested and proven by the many chemical and oil corporations, R&D centers, EPC companies, financial institutions and government agencies that rely on our reports.
The methodology used in the development of this report is illustrated in the diagram presented on the next page.
The report is based on a comprehensive bibliographical research, entirely focused on the industrial process to be examined. Our research encompasses patents, encyclopedias, text books, technical papers and non-confidential information disclosed by licensors, duly reviewed by the Intratec team.
The main goal of this research is to provide a solid understanding of the process examined, which in fact underlies the entire study. During this research, Intratec team identifies the maturity of the process under analysis.
Basically, established processes are mature industrial processes, i.e., several plants employing these processes have been constructed worldwide, while new industrial processes are those that have only been employed in a few plants constructed around the world. Finally, early-stage industrial processes are the processes still under development; currently, either no plants have employed such technologies or the designs of the processes themselves have yet to be completed.
Process Overview Bibliographical Research The Intratec team’s first goal is to understand the chemical, biological and/or physical transformations occurring in the target process, as well as reactants required and products formation.
Thus, initially, bibliographical research focuses on stoichiometry, conversions, yields and/or selectivity of processes’ main reactions or biological processes, while also addressing the occurrence of side reactions and relevant information about catalyst employed.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
42 Production Cost Report Development Methodology THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
43 Process Unit The Intratec team compiles all knowledge acquired around the process into a process block flow diagram, showing major process areas and main process streams, accompanied by a process description.
The process areas correspond to what Intratec defines as “functional units”. Basically, a “functional unit” is a significant step in the process in which a particular physico-chemical operation (i.e., distillation, reaction, evaporation, etc.) occurs. According to this definition, a given functional unit is not associated with a single piece of equipment, but rather with a group of equipment and ancillaries required to perform a particular operation.
Examining an Industrial Site At this point, the Intratec team examines how an industrial site based on the process under analysis would be, in terms of process units and infrastructure required.
In this step, Intratec team defines a preliminary design of the process under analysis, based on fast techniques for process and capital cost estimation, which rely on reduced design efforts. The main goal is to represent the technology examined in sufficient detail to estimate the economics of the technology within the degree of accuracy expected from conceptual evaluations.
It is important to highlight that some specific technical data are not taken into account neither in the preliminary design defined nor in the economic estimates further calculated. In fact, Intratec Reports are meant to be tools to assist the preliminary economic evaluation of emerging or consolidated industrial processes for producing chemicals, and must NOT be viewed as process design packages, design basis or front end engineering design (FEED) packages.
Technology Maturity Assessment As part of Intratec’s report development methodology, the process technology under study is categorized according to its maturity at the moment the report is being developed. The technical maturity serves as a measure of performance, reliability, durability, and operating experience associated with the technology being assessed.
Such assessment is crucial in the development of each Intratec report, since important parameters explained later on, which actually impact on process economics (e.g. process contingency, project contingency, costs related to start-up inefficiencies and R&D, etc), are defined based on it.
The process technology maturity is defined by Intratec team through a method adapted from the so-called Technology Readiness Level (TRL) method, developed by NASA and nowadays used in a broad range of sectors/industries.
Originally intended to supporting decision-making over research and development activity, technology readiness levels were modified by Intratec team to portray, on a scale with five divisions, the maturity level of chemical process technologies, from ‘concept’ to ‘established technology’.
Regarding raw materials, the Intratec team identifies minimum quality requirements (e.g. minimum purity, maximum presence of specific contaminants), as well as typical industrial sources.
For products, the Intratec team gathers information regarding possible uses and applications, as well as the usual specifications necessary to ensure their suitability for those applications.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
44 Site Infrastructure The Intratec team also examines the industrial site in terms of the infrastructure (OSBL facilities) required.
More specifically, this analysis identifies installations that are required but do not directly enter into the manufacture of a product (e.g., storage, utilities supply, auxiliary and administrative buildings).
The first step in identifying the required infrastructure is to define the level of integration the industrial site under analysis will have with nearby facilities or industrial complexes. Integration levels primarily impact storage requirements – e.g., a plant that is not integrated needs storage for all raw materials and products, while a plant that is fully integrated with nearby complexes does not need such installations.
The Intratec team assumes a level of integration based on what is most typical for the type of industrial plant examined.
So, based on the process analysis previously developed and on how integrated the industrial site will be, the Intratec team defines the OSBL facilities requirements.
Defining Site Requirements Key Process Inputs & Outputs At this point, the main processing steps have been identified and global material balance calculations are performed. This preliminary global material balance leads to the identification of key process indicators, which reflect raw material consumption, utilities consumption rates and products generation rates per amount of the main product manufactured.
It is worth mentioning that estimation of raw materials and utilities requirements in the conceptual design phase is generally reasonably accurate but tends to be somewhat understated compared to real operations.
Losses from vessel vents, unscheduled equipment, inerting systems, physical property inaccuracies, startup, shutdown and other process operations not typically addressed in conceptual design may increase raw materials consumption.
Labor Operating labor is associated with the number of operators per shift actually required to run the equipment, while supervision labor is the personnel per shift required to directly supervise the operating labor.
The number of operators and supervisors estimated is based on the type and number of functional units included in the process examined.
Also, it is important to mention that in addition to the operating and supervision labor considered, chemical plants also require other types of labor, not included as an operating cost item. Examples of such labor Such division in process areas not only facilitates process understanding, but also serves as the basis for further economic analysis development.
While outlining process block flow diagram, the Intratec team also maps key technical parameters related to each process area portrayed, including: highest operating temperature and pressure, representative material of construction of equipment, and other parameters.
These parameters serve as inputs for the cost estimating methods used by Intratec, further described in this methodology.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
45 Plant Cost The Plant Cost, together with the Owner's Cost (described further), composes the fixed capital, which is related to the erection of the industrial site itself. It constitutes the fraction of the capital investment which is depreciable.
The Plant Cost comprises the costs directly, or indirectly, associated with the construction of the plant itself. It can be broken down in many ways according to specific goals. In the present report, two different breakdowns are available. They are described below.
Plant Cost Summarized Breakdown The summarized plant cost breakdown presented includes (1) Inside Battery Limits (ISBL) Investment, (2) Process Contingency, (3) Outside Battery Limits (OSBL) Investment and (4) Project Contingency, estimated as follows.
(1) Inside Battery Limits (ISBL) Investment The ISBL investment is the fraction of the fixed capital associated with the construction of all process areas (functional units) portrayed in the process block flow diagram.
Initially, to calculate ISBL investment, the Intratec team approaches each process area individually.
The construction cost of a given area is estimated based on the respective process parameters detailed in the block flow diagram (flow rates, pressure and temperature conditions, materials of construction, complexity), through the use of specific preliminary cost estimation models.
It is worth noting that the Intratec cost models were founded on a number of established cost estimating methods, based on mathematical and statistical processing of an extensive volume of actual cost data of well-known industrial processes and functional units.
In fact, such a massive refining of established methods has led to robust cost models, continuously tested and proven for more than a decade by major global companies that rely on Intratec’s cost estimates of industrial processes.
So, from the process parameters identified, the output of Intratec cost models is the construction cost for each functional unit, including all costs associated with the erection of those units: direct material and labor Capital Investment Estimating The costs that comprise the capital investment are grouped under three main headings: fixed capital; working capital; and additional capital requirements.
Before estimating such capital investment figures, the Intratec team defines plant nominal capacity according to the process under analysis, considering that the plant should be competitive on a global scale.
Once this assumption has been made, the Intratec team begins the actual estimation of the capital investment figures as follows.
are: maintenance labor, outsourced labor, technical assistance to operation, plant engineers, restaurant, purchasing, employee relations department, etc.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
46 Process Contingency Factor Estimation Methodology INFORMATION AVAILABILITY Established (Outdated) Established (In Use) Emerging Embryonic Conceptual High Average Low TECHNOLOGY READINESS 15% 15% 18% 22% 26% 10% 10% 15% 5% 5% costs, as well as indirect costs, such as construction overheads, including: payroll burdens, field supervision, equipment rentals, tools, field office expenses, temporary facilities, etc.
In the case of nonstandard functional unit, additional research is conducted and the construction cost is estimated from the use of specialized engineering design software or through quotations provided by equipment suppliers.
Finally, the sum of all construction cost figures, associated with the functional units examined, leads to the total ISBL investment figure.
NOTE: a detailed assessment of the ISBL investment, showing the share of each functional unit inside battery limits in the total ISBL investment, is presented in Appendix E, available exclusively in the "Advanced" version of the report.
(2) Process Contingency Process contingency is utilized in an effort to lessen the impact of absent technical information or the uncertainty of that which is obtained.
That being the case, the reliability of the information gathered, its amount and the inherent complexity of the process are significant to its evaluation. Errors that occur may be related to: a. Addition and integration of new process steps b. Uncertainty in process parameters, such as severity of operating conditions and quantity of recycles c. Estimation of cost through scaling factors d. Off-the-shelf equipment Hence, process contingency is a function of the readiness of the technology and the availability of information about this technology. This value typically falls between 5% and 30% of ISBL investment and is estimated according to the table below.
It is important to highlight that different assumptions may be adopted in particular analyses due to specific conditions of the process or the context approached in the economic analysis.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
47 Project Contingency Factor Estimation Methodology DEGREE OF COMPLEXITY Established (Outdated) Established (In Use) Emerging Embryonic Conceptual High Average Low TECHNOLOGY READINESS 15% 15% 20% 25% 30% 20% 20% 25% 30% 35% 25% 25% 30% 35% 40% It is worth noting that the amount of information about less mature processes is small in comparison to established processes, mainly because of the inherent uncertainties surrounding its development.
Therefore, it is not coherent to define a process contingency value for technologies in the conceptual or embryonic phases when information availability is different from low, because this situation does not occur.
(3) Outside Battery Limits (OSBL) Investment The OSBL investment is the fraction of the plant cost associated with the construction of all infrastructure (storage, utilities, auxiliary units and buildings) required.
The Intratec team employs cost estimation models similar to those previously described for estimating ISBL investment, i.e., by approaching the components of each process area individually.
The cost of a given functional unit or building associated with plant infrastructure is estimated based on a preliminary design of OSBL equipment, facility or building, according to the process requirements. As with ISBL functional units, this preliminary design information serves as an input to Intratec's cost estimation models, with which Intratec team calculates the fixed capital for each OSBL functional unit. The fixed costs include all costs associated with the erection of those units.
The sum of all construction cost figures, associated with the functional units examined, leads to the total area investment figure. Finally, the sum of the investment figures for all areas associated with plant infrastructure give the final OSBL investment.
NOTE: a detailed assessment of the OSBL investment, showing the share of each functional unit outside battery limits in the total OSBL investment, is presented in Appendix E, available exclusively in the 'Advanced' version of the report.
(4) Project Contingency Project Contingency is included to cover the costs which may arise as the project evolves, related to: project errors or incomplete specifications, labor costs changes, strikes, problems caused by weather; inflation, etc.
Project contingency is largely dependent on the plant complexity and technology maturity, identified during initial research. The following table shows how project contingency varies according to such aspects.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
48 Plant Cost Breakdown per Discipline For a better understanding of the total plant cost previously calculated, the construction costs for all functional units (process areas, storage, utilities, auxiliary units and buildings) are rearranged into a different cost breakdown: direct process costs, indirect process costs and project contingency. This alternative breakdown is commonly adopted for the assessment of construction costs, in a range of industries.
Fundamentally, the direct process costs are the total installed equipment cost (from purchase to installation, including the required installation bulks). They include bare equipment, equipment setting, piping civil, steel, instrumentation & control, electrical, insulation, painting, as described below: Bare Equipment. This is the cost associated with the purchase of process equipment Equipment Setting. Those are costs related to the labor cost for setting the purchased equipment in place.
Piping. The costs related to piping include materials, such as valves, fittings, pipe and supports used in the erection of the piping used directly in the process (for raw materials, intermediate-products, finished-products, steam, water, air, as well as any other process piping). The labor for piping erection and installation is also covered in this topic.
Civil. This topic covers costs associated with material and labor required for equipment foundations and civil work related to any building required in the industrial site.
Steel. Costs associated with material and labor required for equipment platforms erection, as well as any supports needed during equipment installation.
Instrumentation & Control. Those costs relate to instruments, controllers and industrial networks material, and labor required to install it.
Electrical. The costs related to electrical system cover power wiring, instrument wiring, lighting, as well as transformation and service.
Insulation. Costs related to any labor or material required to insulate process equipment, either for process needs or for operators safety.
Painting. Those costs are related to labor and material required to paint and/or coat equipment according to process requirements.
The indirect costs account for field indirects, engineering costs, overhead, and contract fees, as described below: Engineering & Procurement. Engineering expenses include process and project engineers involved in process and construction design, as well as associated overhead. Development of computer-based drawings and cost engineering are also costs included in this topic. Procurement costs are those related to the purchase team, associated home office and overhead, and accounting professionals.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
49 The Owner's Cost is defined as those expenses that, despite not being associated with the construction of the plant itself, are required to make the plant operational.
The Owner's Cost comprises the (1) initial charge of chemicals & catalysts (if required), (2) Prepaid Royalties and (3) Miscellaneous Costs, estimated as follows.
(1) Initial Charge of Chemicals & Catalyst This cost may occur if the process requires an inventory of a specific chemical and/or catalyst that will last more than a year and represents a significant expense.
In this case, it should not be included in the working capital (described further), which, in turn, corresponds to the funds used in its day-to-day operation.
(2) Prepaid Royalties Royalty charges on portions of the plant are usually levied for proprietary processes. A value ranging from 0.5 to 1% of the plant cost is generally used.
(3) Miscellaneous Costs A value ranging from 5% to 10% of the plant cost is generally used to account for: a. Preliminary planning studies, HAZOP studies and environmental reviews b. Legal costs, rights of way, permits and fees c. Long distance pipelines, transport equipment and plant vehicles d. Initial stock of maintenance Owner's Cost Construction Material & Indirects. Those costs relate to field temporary buildings and their operation, construction tools, rentals, home office personnel located at the construction site, construction payroll, burdens and benefits.
General & Administrative Overheads. General and administrative costs are associated with construction management and general costs incurred during construction, such as construction supervision, taxes and insurance, internal and licensed software, communications and travel & living.
Contract Fee. Expenses related to contract fees for engineering, equipment purchase and construction work.
NOTE: The Plant Construction Cost Breakdown per Discipline as described above, including direct costs, indirect costs and project contingency, is presented in Appendix E, available exclusively in the 'Advanced' version of the report. This analysis includes a direct costs breakdown (bare equipment, equipment setting, piping civil, steel, instrumentation & control, electrical, insulation, and painting) and an indirect costs breakdown (engineering & procurement, construction material & indirects, general & administrative overheads and contract fee).
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
50 Fixed Capital Estimate Accuracy The accuracy range for a fixed capital cost estimate is mainly influenced by: Reliability and amount of the information available Examined technology readiness Degree of extension of the study As previously explained, the estimate within this analysis is based on the preliminary design of functional units which, in turn, relies on a process scheme.
The greatest essential uncertainty lies in the basic conception of this process scheme. The level of uncertainty varies broadly among published information and from steps of a process in a given research.
In some instances, sufficient information may not be available to support rigorous estimation, thus, only basic design methods are warranted.
The maturity of the examined technology, in turn, also plays an important role in the fixed capital estimates.
Processes that are still on a conceptual stage require an extra level of caution.
In addition, the extension of the analysis helps enormously to reduce uncertainties and improve the accuracy of the cost estimation. Detailed studies are crucial to achieving more precise estimates.
Finally, the accuracy range for the fixed capital estimate obtained according to the methods hereby presented is -15% to -40% on the low side and +25% to +70% on the high side, depending on the readiness of the technology under analysis and the amount of information available, in accordance with the table on next page.
The absence of factors for emerging, embryonic and conceptual technologies when there is high availability of information is explained by the inherent nature of such processes, which, while in the development / scale up phases, present a lot of uncertainties.
Therefore, the amount and reliability of the information about such processes is not comparable to established technologies in operation for several years.
e. Owner's engineering (staff paid by owner to evaluate the work of the company in charge of plant construction) f. Owner's contingency Fixed Capital Estimates Validation Depending on the availability of information about the process examined, the Intratec team utilizes three different methods to double-check fixed capital estimates: (1) Published investment data, related to the construction of industrial plants of that process worldwide (adjusted in time, location and capacity); and/or (2) Fixed capital of similar plants (adjusted in time, location and capacity); and/or (3) Reverse engineering methods, i.e., the fixed capital is calculated based on the known profitability of the process examined.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
51 Working Capital For the purposes of this report, working capital is defined as the funds, in addition to the fixed capital, that a company must contribute to a project. Those funds must be adequate to getting the plant into operation and meeting subsequent obligations.
The initial amount of working capital is regarded as an investment item.
The Intratec team uses the following items/assumptions for working capital estimation: Accounts receivable.
Products shipped to but not paid for by the customer; represents the extended credit given to customers. It is estimated as a certain period – in days – of total operating cost (including depreciation and excluding by-product credits, if any) plus corporate overhead.
Accounts payable.
A credit for accounts payable such as feedstock, chemicals, and packaging materials received but not paid to suppliers. It is estimated as a certain period – in days – of operating cash cost (excluding by-product credits, if any) plus corporate overhead.
Product inventory.
Products in storage tanks. The total amount depends on sales flow for each plant, which is directly related to plant conditions of integration to the manufacturing of the product‘s derivatives.
It is estimated as a certain period – in days – of total operating cost (including depreciation and excluding by-product credits, if any) plus corporate overhead.
Raw material inventory.
Raw materials in storage tanks.
The total amount depends on raw material availability, which is directly related to plant conditions of integration to raw material manufacturing (estimated as a certain period – in days – of raw material delivered costs).
In-process inventory.
Material contained in pipelines and vessels, except for the material inside the storage tanks, assumed to be 1 day of cash cost (excluding by-product credits, if any) plus corporate overhead.
Established (Outdated) Established (In Use) Emerging Embryonic Conceptual Average Low -25% / 40% -25% / 40% -30% / 50% -35% / 60% -40% / 70% -20% / 30% -20% / 30% -25% / 40% Fixed Capital Estimate Accuracy Range High -15% / 25% -15% / 25% INFORMATION AVAILABILITY TECHNOLOGY READINESS The non-uniform spread of accuracy ranges (+50 to – 30 %, rather than ±40%, e.g.) is justified by the fact that a lack of available information usually results in underestimating rather than overestimating project costs.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
52 Additional Capital Requirements There are certain one-time expenses related to bringing a process on stream. From a time standpoint, a variable undefined period exists between the nominal end of construction and the correct operation of the plant (e.g.production of quality product in the quantity required).
This period is commonly referred to as start-up.
During the start-up period, expenses are incurred for operator and maintenance employee training, temporary construction, auxiliary services, testing and adjustment of equipment, piping, and instruments, etc. Intratec’s method of estimating start-up expenses may consist of the following components: Labor training. Represents costs of plant crew training for plant start-up, estimated as a certain number of days of total plant labor costs (operators, supervisors, maintenance personnel and laboratory labor).
Commercialization costs. Commercialization costs are those associated with marketing the product and include developing a market plan, establishing a distribution network and devising a customer support strategy.
Those costs are dependent on how integrated the plant is with consumer facilities and on the maturity of the product – how established and well-known it is. These costs range from 0.5% to 5% of annual cash cost (excluding by-product credits, if any).
Start-up inefficiency. Takes into account those operating runs when operation cannot be maintained or there are false starts.
Start-up inefficiency varies according to the process maturity: 1% for established processes and up to 5% for less mature technologies, based on annual cash cost (excluding by-product credits, if any).
Unscheduled plant modifications. A key fault that can occur during the start-up of the plant is the risk that the product(s) may not meet market specifications. Then, equipment modifications or additions may be required.
Land & Site Development.
Site preparation, including roads and walkways, parking, railroad sidings, lighting, fencing, sanitary and storm sewers, and communications.
Pricing & Wage Rates Definition In order to calculate fixed and variable operating costs, the Intratec team collects average transaction prices of raw materials and average operators’ wage rates in the region examined in the study.
Operating Cost Estimating Supplies and stores.
Parts inventory and minor spare equipment (estimated as a percentage of operating labor and supervision and maintenance cost).
Cash on hand.
An adequate amount of cash on hand to give plant management the necessary flexibility to cover unexpected expenses. It is estimated as a certain period – in days – of cash cost (excluding by-product credits, if any) plus corporate overhead.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
53 Operating Variable Cost Variable costs change in direct proportion to changes in the operating rate. Examples of common variable costs include raw materials and utilities.
The Intratec team calculates the operating variable costs of the plant under analysis from previously identified process input and output figures and historical pricing data, as follows: Operating Variable Costs = Net Raw Material Costs + Net Utilities Costs Net Raw Materials Costs “Net raw material costs” are the difference between raw materials costs and credits from by-products generation, as expressed in the formula below.
Net Raw Material Costs = Raw Material Costs – By-product Credits The raw materials costs, in turn, are estimated by multiplying process’ consumption figures by the respective raw material prices in the region considered.
The formula below illustrates the raw materials costs calculation: Raw Material Costs = Sum ( Raw Material Price Raw Material Consumption ) By-products credits were estimated in a similar way, based on process’ input and output figures and pricing data.
Net Utilities Cost In this report, the utilities cost component encompasses costs related to a plant’s consumption of steam, electricity, fuel, and refrigeration. These utilities requirements, in turn, are estimated through correlations internally developed by the Intratec team that were refined from a well-established method reported in technical literature by Mardsen et al. related to chemical process industries. (See “References” chapter) Through the use of these correlations, utilities consumption figures can be quickly estimated with basic information, related to chemical properties of components involved in the process and parameters presented in the block flow diagram.
Such parameters include: number of functional units; type of each functional unit according to its energy consumption (i.e., if it involves phase changes, endothermic or exothermic reactions, negligible use of energy, if it is a nonstandard functional unit, etc.); flow rates; heats The prices are based on trade statistics issued by official government agencies, over the time period considered.
Pricing information is checked to verify consistency, but issues like differences in product qualities, discounts related to volumes, or contractual negotiations are not considered.
However, for some chemicals, there are no trade statistics (e.g., intermediate chemicals that are not traded because of transportation issues, but are usually generated and consumed onsite).
In those cases, the Intratec team assumes a transfer price that considers all the costs related to the manufacturing of that product plus an amount to pay the investment made to manufacture it.
The operators’ wage rates are based on data published by official government agencies.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
54 Operating Fixed Cost Operating fixed costs are all the costs related to the plant operation that are not proportional to the plant operating rate. They are estimated as the sum of the following items: Operating labor.
This item accounts for the total costs of plant operators actually required to run the equipment.
This cost includes wages, burdens and benefits.
The annual operator cost is obtained according to the formula: number of operators per shift x number of shifts per day x operator hourly wage rate x hours worked per week x weeks per year.
Supervision.
Accounts for the costs of field supervision labor, including wages, burdens and benefits.
The annual supervision cost is obtained according to the formula: number of supervisors per shift x number of shifts per day x supervisor hourly wage rate x hours worked per week x weeks per year.
NOTE: a detailed assessment of utilities consumption, presented per utility (e.g., steam, cooling water, electricity.) is presented in Appendix D, available exclusively in versions 'Extended' and 'Advanced' of the report.
Maintenance cost. This item accounts for the costs related both to the labor and material costs related to the maintenance of the plant. It is calculated as a percentage of the fixed capital, ranging between from 1 to 4% of TFC per year.
This figure is primarily based on the type of equipment employed (intimately associated with the kind of fluid handled in the plant) and the industry sector. The percentages assumed are based on average industry values and are defined according to the following table.
Operating charges.
This category includes operating supplies (i.e., consumable items such as charts, lubricants, test chemicals, etc.); packaging; laboratory supplies and laboratory labor. It is calculated as a percentage of the total labor cost (item operating labor + item supervision).
Plant overhead. This item comprises all other non-maintenance (labor and materials) and non-operating site labor costs for services associated with the manufacture of the product, including: outsourced labor; technical assistance to operation; plant engineers; restaurant; recreation; purchasing; employee relations department; and janitorial. It is calculated as a percentage of the sum of total labor and maintenance costs.
Property taxes and insurance.
This cost is associated with the local property taxes charged by governments on commercial land or buildings as well as the cost of insurance to cover third party liabilities and potential plant damages. It is calculated as a percentage of the fixed capital per year.
Maintenance Cost Estimation Methodology Solids Gas-Liquid-Solids Gas-Liquid 3% 2.5% 2% 4% 3.5% 3% 2% 1.5% 1% INDUSTRY SECTOR Basic Chemicals Consumer Product Pharmaceutical s FLUIDS HANDLED Specialty 2% 1.5% 1% of reactions involved in the process; molecular weight and approximate boiling points of the components.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
55 Depreciation Depreciation refers to the decrease in value of industrial assets with the passage of time, primarily due to wear and tear. While not a true operating cost, depreciation is considered to be a operating expense for accounting purposes – it allows the recovery of the cost of an asset over a time period.
In this report, depreciation is calculated based on the straight-line method, according to which the cost of an asset is uniformly distributed over its lifetime. A 10-year lifetime is assumed for the main production unit (ISBL units) and assets derived from owner's costs, while the site infrastructure (OSBL facilities) is assumed to have a total life-time of 20 years.
Therefore, depreciation adopted for ISBL facilties and owner's costs is 10% of respective capital investment per year, and, for OSBL assets, 5% of respective capital investment per year.
Corporate Overhead Corporate overhead represents costs incurred by a company’s head office not directly related to the process operation and is estimated as the sum of the following items: Administration costs. This item comprises the executive and administrative activities. It includes salaries and wages for administrators, accountants, secretaries, legal costs, communications, office maintenance and other costs associated with the company’s head office. It is calculated as a percentage of the sum of total labor and maintenance costs.
Information technology.
Information technology (IT) expenses refers to the total cost related to information processing (e.g.
computer software, hardware, personnel, data communications, miscellaneous). The total IT expense is estimated as 1.4% of the fixed capital per year.
Product Value Estimating Heretofore, capital investment and operating cost of the process examined were estimated.
If the examined process targets to produce a chemical, the next step in the methodology is the development of a more consistent analysis, encompassing all costs estimated so far, and aiming to estimate the value of this target product generated.
In this context, all costs estimated are combined in a single item: the “Product Value”. More specifically, the product value results from the sum of operating costs (i.e., operating variable costs, operating fixed costs, and depreciation) with corporate overhead, and a return on capital employed (ROCE), a parcel which reflects the capital investment. The formula below expresses the product value calculation.
Product Value = Operating Variable Costs + Operating Fixed Costs + Depreciation + Corporate Overhead + Expected ROCE Amount where all components are expressed in US dollars per amount of product.
The corporate overhead and the ROCE are estimated as follows.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
56 Return on Capital Employed (ROCE) Research & development.
This is associated with the research activities related to the process and products. It includes salaries and wages for personnel and funds for machinery, equipment, materials and supplies related to the research and development activities. This cost is calculated as a percentage of the operating cash cost (excluding by-product credits, if any), considering the plant operating at full capacity and will vary according to the process maturity and the industry sector.
Research & Development Cost Estimation Methodology INDUSTRY SECTOR Established Under Development Basic Chemicals Consumer Product Pharmaceutical s TECHNOLOGY READINESS 12% 17% 2% 2.5% 2% 3% Specialty 3% 5% The above factor values are based on industry average values according to the plant industry segment and employed technology readiness.
Different assumptions may be adopted in particular analyses due to specific conditions of the process or the context approached in the economic analysis.
Marketing & Advertising Cost Estimation Methodology INDUSTRY SECTOR Assumption Basic Chemicals Consumer Product Pharmaceutical s 5% 6% .8% Specialty 1.6% The expected ROCE amount is a component which reflects the capital costs of a given process into its product value.
This component is based on the expected return on capital employed typically aimed by chemical companies.
It is calculated by multiplying capital costs by the expected ROCE percentage, divided by the total amount of product manufactured: Expected ROCE Amount = Capital Costs Expected ROCE Percentage / Product Annual Production This “Expected ROCE Amount” component is, in fact, a measure of the cost of investment required to construct the plant, in terms of US dollars per amount of product.
Most chemical companies aim to achieve a ROCE percentage ranging from 5% to 25% for the construction of a new plant. In this context, the Intratec team assumes an expected ROCE percentage of 7% for established industrial processes in the basic chemicals sector.
Marketing & advertising. This is related to the costs associated with the sales (sales personnel, advertising, technical sales service) of the products manufactured in the plant. This cost is calculated as a percentage of the operating cash cost (excluding by-product credits, if any), considering the plant operating at full capacity. The costs associated with marketing and advertising is intimately related to the industry sector (basic chemicals, specialties, pharmaceuticals or consumer products).
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
57 Estimates Limitation The cost estimates presented refer to a process technology based on a standardized design practice, typical of major chemical companies.
The specific design standards employed can have a significant impact on capital and operating costs. In this context, cost estimates calculated by Intratec team naturally have limitations.
In fact, the accuracy range for operating cost estimated in the present study is -10% to -20% on the low side and +10% to +20% on the high side, depending on the maturity level of the process examined. The presented accuracy considers a confidence level of 90%, which is consistent with the type of conceptual evaluation that this study aims to provide.
Also, it is to be noted that the basis for capital and operating costs estimation is that the plant is considered to be built in a clear field with a typical large single-line capacity, unless explicitly stated otherwise.
In comparing the cost estimates presented with actual plant costs and/or contractor's estimate, the following must be considered: Technologies under development are those that are not yet established on commercial scale, i.e., a technology that is either on a conceptual, embryonic or emerging phase.
The above percentages are based on industry average values according to the plant industry segment and employed technology readiness.
Different assumptions may be adopted in particular analyses due to specific conditions of the process or the context approached in the economic analysis.
Finally, it is also important to mention that product value must not be confused with product price. While the product value is calculated based on operating cost, corporate overhead and expected ROCE, the product price is the actual value practiced in market transactions.
Expected ROCE Factor Estimation Methodology INDUSTRY SECTOR Established Under Development Basic Chemicals Consumer Product Pharmaceutical s TECHNOLOGY READINESS 20% 25% 15% 18% 7% 10% Specialty 12% 15% In contrast, a 25% expected ROCE is assumed for early-stage industrial processes in the pharmaceuticals business, as such processes inherently involve a larger amount of risk and cost uncertainty. It should be noted that the risk taken into account here is limited to the technical risk associated with the process uncertainties.
Other venture risks were not considered, such as business environment, product market changes, increased competition, raw materials and product prices variations, change in government policy, etc.
The ROCE assumptions, according to the industry sector and technology readiness, are presented in the following table.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
58 Minor differences or details (many times, unnoticed) between similar processes can noticeably affect cost.
The omission of process areas in the design considered may invalidate comparisons with the estimated cost presented.
Industrial plants may be overdesigned for particular objectives and situations.
Rapid fluctuation of equipment or construction costs may invalidate cost estimate.
Market price fluctuations may invalidate operating cost estimate.
Equipment vendors or engineering companies may provide goods or services below profit margins during economic downturns.
Specific locations may impose higher taxes and fees, which can impact costs considerably.
Furthermore, no matter how much time and effort are devoted to accurately estimating costs, errors may occur due to the aforementioned factors, as well as cost and labor changes, construction problems, weather-related issues, strikes, or other unforeseen situations.
This is partially considered in the project contingency. Finally, it must be said that an estimated project cost is not an exact number, but is rather a projection of the probable cost.
THIS PAGE'S CONTENT IS INCLUDED IN ALL VERSIONS.
59 A-1 APPENDIX A – ABOUT INTRATEC OUR BUSINESS .................................................................................................................................................................. A-2 PRODUCTION COST REPORTS .................................................................................................................................... A-3 A-2 Our Business In operation since 2002, Intratec is a leading provider of data & analysis focused on commodities, energy and production processes. In a nutshell, we produce independent, readily accessible data and analyses, so that our customers can focus on what they do best and take better decisions. We serve our valued customers from all over the world through our website, an e-commerce where they easily learn about our offerings – reports and online databases – and promptly order what they need. Our portfolio of reports and databases provide key information surrounding commodities and energy, such as prices, trends and forecasts, as well as cost structures of production processes. Our expertise spans the fields of petrochemicals; plastics & polymers; inorganic chemicals; fertilizers & food; renewables & recyclables; oil, gas & derivatives; metals & mining; specialty chemicals and energy & utilities. With well-designed offerings, we serve a diverse group of customers, from Fortune 500 companies and local manufacturers to biotech startups and consultants. Our studies and data have been used in multiple ways, such as: To evaluate transaction prices in international markets To identify monthly/yearly price trends To assess procurement planning performance To obtain estimates of capital and operating costs of industrial plants To learn about the economic potential of R&D breakthroughs To screen and assess industrial investment options A-3 Production Cost Reports Intratec offers more than 900 up-to-date reports examining production costs of chemicals and utilities. Our portfolio covers +300 chemicals and utilities, including basic chemicals and inorganics, plastics, fibers and rubbers, green chemicals and biofuels, fertilizers, specialties and more. In short, each report examines the economics of one specific production process, presenting key information such as raw materials consumption, capital investment and operating costs. Intratec Reports may be acquired individually or through one of our subscription plans. A-4 Find below the chemicals covered in Intratec reports. For a more complete and updated list, reader is encouraged to visit our online store at 3-Hydroxypropionic Acid Acetone Acetylene Acetyls Acrylic Acid and Derivatives Acrylic/Maleic Copolymer Acrylonitrile Adipic Acid Aldehydes Alkylbenzenes Amino Acids Ammonia Aniline Biodiesel Bisphenol A BTX Butadiene and C4's C6's Caprolactam Carbon Monoxide Chlorine and Derivatives Chloroprene Citric Acid Cosmetics Cumene Detergents Dicyclopentadiene Diesel Dimethyl Carbonate (DMC) Dimethyl Terephthalate Diols Diphenyl Carbonate Dyes & Pigments Electricity Epichlorohydrin Ethanol Ethylene Ethylene Oxide Fertilizers Fibers Fire Retardants Food Additives Furans and Derivatives Glycerol Glycols Hydrogen Hydrogen Cyanide Hydrogen Peroxide Industrial Gases Insecticides Isocyanates Isophthalic Acid Isoprene Lactic Acid Linear Alpha Olefins Methacrylic Acid & Derivatives Methanol MTBE Nitric Acid Nitro Aromatics Nylon Oxalic Acid Oxo Alcohols Pentaerythritol PET Pharmaceuticals Phenol Phosgene Phthalic Anhydride Polyacrylate Polyacrylonitrile Polyalphaolefins Polycarbonates Polyesters Polyethers Polyethylenes Polylactic Acid (PLA) Polypropylene Polyurethanes Propanol and Isopropanol Propylene Propylene Oxide PVC Reformate Resins Silanes Silicones Siloxanes Sodium Hydroxide Speciality Polymers Styrenics Succinic Acid Sulfuric Acid Synhetic Rubbers Synthesis Gas Vitamins Terephthalic Acid Trimethylolpropane Urea Vinyls B-1 APPENDIX B – TERMS & CONDITIONS B-2 Terms & Conditions Data, information, reports, tools, analyses and/or models (“products”) herein presented are prepared on the basis of publicly available information and non-confidential information disclosed by third parties. Third parties, including, but not limited to technology licensors, trade associations or marketplace participants, may have provided some of the information on which the analyses or data are based. The products herein presented are developed independently by Intratec and, as such, are the opinion of Intratec and do not represent the point of view of any third parties nor imply in any way that they have been approved or otherwise authorized by third parties that are mentioned in this report. Intratec conducts analyses and prepares products for readers in conformance with generally accepted professional standards. Although the statements in this product are derived from or based on several sources that Intratec believe to be reliable, Intratec does not guarantee their accuracy, reliability, or quality; any such information, or resulting analyses, may be incomplete, inaccurate or condensed. All estimates included in this product are subject to change without notice. This product is for informational purposes only and is not intended as any recommendation of investment. Reader agrees it will not, without prior written consent of Intratec, represent, directly or indirectly, that its products have been approved or endorsed by the other parties. In no event shall Intratec, its employees, representatives, resellers or distributors be liable to readers or any other person or entity for any direct, indirect, special, exemplary, punitive, or consequential damages, including lost profits, based on breach of warranty, contract, negligence, strict liability or otherwise, arising from the use of this product, whether or not they or it had any knowledge, actual or constructive, that such damages might be incurred. Reader agrees that Intratec retains all rights, title and interest, including copyright and other proprietary rights, in this product and all material, including but not limited to text, images, and other digital files, provided or made available as part of this product. The Reader is responsible for ensuring that copyright laws are adhered to at all times. Digital products may be supplied watermarked with the Purchaser’s name, email, and/or company. Reader may, in the course of its business, create its own analyses incorporating information and/or data presented in Intratec products, provided that: (i) Reader acknowledges Intratec as a data source in relation to such analyses created; and (ii) Reader does not use or authorize the use of such analyses created in products that are competitive with Intratec products. The Reader further agrees to refrain from any general release of the information presented in this product, so as to constitute passage of title into the public domain or otherwise jeopardize common law or statutory copyright. Full Reports Terms and Conditions are available at www.intratec.us/legal#general-terms . C-1 APPENDIX C – REPORT SHARING TERMS C-2 Report Sharing Terms Reader may, in the course of its business, create studies, analyses and/or assessments incorporating pricing data, price forecasts, market data, industry data, technology data and any other data, information and editorial content contained in Intratec Production Cost Reports for internal use only, provided that; (i) the Reader acknowledges Intratec as a data source in relation to all studies, analyses and/or assessments developed; and (ii) the Reader does not use or authorize the use of studies, analyses and/or assessments in products that are competitive with Intratec Reports. The Reader shall have a limited right in the ordinary course of its business to share a Report purchased and/or its content, according to the version of the Report, on a confidential basis. The Reader who ordered Basic Version(s) of Report(s) is not allowed to share data from Report(s.) The Reader who ordered Extended or Advanced Version(s) of Report(s) shall have a limited right in the ordinary course of its business to share data from Report(s), on a confidential basis, according to the report version, as follows: Extended Version. The Reader who ordered an Extended Version of a Report is allowed to share it and/or its content exclusively with CO-WORKERS FROM THE SAME INSTITUTION OR CORPORATION AS THE READER, who also work at the SAME COUNTRY of the Reader. Advanced Version. The Reader who ordered an Advanced Version of a Report is allowed to share it and/or its content exclusively with CO-WORKERS FROM THE SAME INSTITUTION OR CORPORATION AS THE READER. The use of Intratec Reports data for any purposes not expressly permitted by these terms would be subject to a separate additional license or agreement and additional or different fees or payment arrangements. The Reader further agrees to refrain from any general release of the information presented in Intratec Reports and/or Extra Analysis, so as to constitute passage of title into the public domain or otherwise jeopardize common law or statutory copyright. Full Report Sharing Terms are available at www.intratec.us/legal. APPENDIX D - EXTENDED ANALYSIS CONTENTS Utilities Consumption Breakdown........................................................................................................................... D-2 Utilities Consumption Figures....................................................................................................................... D-2 Impact of Utilities on Total Operating Cost................................................................................................D-3 Economic Analysis for Different Capacities..........................................................................................................D-4 Introduction........................................................................................................................................................D-4 Capital Investment Comparison................................................................................................................... D-4 Operating Cost & Product Value Comparison..........................................................................................D-5 Project Implementation & Construction Schedule...............................................................................................D-7 Materials & Utilities Pricing Data..............................................................................................................................D-9 Analysis Pricing Basis.....................................................................................................................................D-9 Historical Prices................................................................................................................................................D-9 THIS PAGE'S CONTENT IS INCLUDED IN 'EXTENDED' AND 'ADVANCED' VERSIONS ONLY.
D-1 UTILITIES CONSUMPTION BREAKDOWN Find below the key utilities consumption indicators of the technology examined in the report.
These indicators reflect the net utilities consumption rates per unit of product manufactured.
It should be noted that estimation of utility requirements in the conceptual design phase is usually fairly accurate but tends to be somewhat low compared to real operations.
Losses from vessel vents, unscheduled equipment, inerting systems, physical property inaccuracies, startup, shutdown and other process operations not typically addressed in this phase may increase utilities consumption.
Net Utility Consumption Rates (per mt of Cyclohexane) NET CONSUMPTION RATES UTILITY Utilities Consumption Figures THIS PAGE'S CONTENT IS INCLUDED IN 'EXTENDED' AND 'ADVANCED' VERSIONS ONLY.
D-2 Cost Contribution of Each Utility Impact of Utilities on Operating Cost Operating fixed costs Operating cash costs Depreciation TOTAL OPERATING COST COMPONENT Utilities Share in Total Operating Cost Net raw material costs % The table below summarizes utility costs share in total operating cost.
Net utilities The following chart indicates the utilities with the greatest impact on the economics of the process.
THIS PAGE'S CONTENT IS INCLUDED IN 'EXTENDED' AND 'ADVANCED' VERSIONS ONLY.
D-3 ECONOMIC ANALYSIS FOR DIFFERENT CAPACITIES Capital Investment Comparison Introduction This analysis presents the impact of a plant capacity change on the economic analysis presented in this report. Additional capacity scenarios were analyzed using the same methodology and compared with the base case presented in the report.
The analysis is divided into two parts: (1) a capital investment comparison, examining fixed investment, working capital and additional capital requirements; and (2) an operating costs & product value Comparison.
From the range depicted above, the minimum, the mid-range and the maximum capacities are compared in details on the next page. The following table presents detailed capital cost figures to better portray how the economy of scale impacts the process under analysis.
The economic analysis presented in this report was reproduced for a range of plant capacities, in such a way as to estimate a curve representing the ways in which capital investment varies with the plant nominal output. This curve is presented in the chart below.
Capital Investment (USD Million) Versus Plant Capacity THIS PAGE'S CONTENT IS INCLUDED IN 'EXTENDED' AND 'ADVANCED' VERSIONS ONLY.
D-4 Capital Investment Analysis for Different Capacities (MM USD) BASIS: UNITED STATES, Q1 2021 FIXED CAPITAL Working capital Additional capital TOTAL CAPITAL INVESTMENT BASE (IC INDEX:161.8) SMALLER PLANT LARGER PLANT Inside battery limits (ISBL) Process contingency ( % of ISBL) Outside battery limits (OSBL) Total process capital (TPC) Project contingency ( % of TPC) PLANT COST Owner's cost Operating Cost & Product Value Comparison The operating costs and the product value were also estimated for a range of plant capacities, resulting in the chart below. A summary of the datasheet presented in the section "Process Economics Summary" is reproduced on the next page, and also includes the two additional scenarios evaluated in this analysis.
Operating Cost (USD/mt) Versus Plant Capacity Capacity (mt/y) THIS PAGE'S CONTENT IS INCLUDED IN 'EXTENDED' AND 'ADVANCED' VERSIONS ONLY.
D-5 Net raw materials cost Net utilities cost OPERATING VARIABLE COSTS Supervision Maintenance cost Operating charges OPERATING FIXED COSTS Operating labor Plant overhead Property taxes and insurance OPERATING CASH COST Depreciation TOTAL OPERATING COST ROCE PRODUCT VALUE Operating Cost & Product Value Analysis for Different Capacities (USD/mt) BASIS: UNITED STATES, Q1 2021 BASE (IC INDEX:161.8) Annual production (mt/y) SMALLER PLANT LARGER PLANT Operating rate (h/y) Corporate Overhead Capacity (mt/y) OPERATING COSTS THIS PAGE'S CONTENT IS INCLUDED IN 'EXTENDED' AND 'ADVANCED' VERSIONS ONLY.
D-6 PROJECT IMPLEMENTATION & CONSTRUCTION SCHEDULE The primary objective of this analysis is to present a preliminary project implementation schedule, encompassing the period from the decision to invest to the start of commercial production.
This is divided in five major steps: (1) Basic Engineering; (2) Detailed Engineering; (3) Procurement; (4) Construction; and (5) Start-up.
The duration of each project phase is detailed in the table below: Project Phases Schedule Basic engineering DURATION PHASE START Since the project phases overlap, the total project duration is not equal to the sum of each phase duration.
The Engineering, Procurement & Construction (EPC) period - from the basic engineering start until the end of construction - is about 0 months. The total project duration, also including commissioning and start-up, is approximately 0 months.
The bar chart below illustrates the project implementation and construction schedule and clarifies the overlaps among the distinct project phases.
Months Months After Project Start Detailed engineering Procurement Construction Commissioning & start-up THIS PAGE'S CONTENT IS INCLUDED IN 'EXTENDED' AND 'ADVANCED' VERSIONS ONLY.
D-7 Implementation & Construction Schedule THIS PAGE'S CONTENT IS INCLUDED IN 'EXTENDED' AND 'ADVANCED' VERSIONS ONLY.
D-8 MATERIALS & UTILITIES PRICING DATA The economic analysis presented within this report is based on the prices seen in the table below.
Materials & Utilities Prices (United States, Q1'21) DESCRIPTION Historical Prices The charts on the following pages depict how the costs of some materials have evolved during the last three years.
Analysis Pricing Basis Raw Materials Utilities UNIT REMARK PRICE THIS PAGE'S CONTENT IS INCLUDED IN 'EXTENDED' AND 'ADVANCED' VERSIONS ONLY.
D-9 THIS PAGE'S CONTENT IS INCLUDED IN 'EXTENDED' AND 'ADVANCED' VERSIONS ONLY.
D-10 APPENDIX E - ADVANCED ANALYSIS CONTENTS Process Unit (ISBL) Construction Cost by Functional Unit............................................................................... E-2 Infrastructure Construction Cost by Piece of Equipment.................................................................................. E-4 Introduction........................................................................................................................................................E-4 Construction Cost: Area 90 - Storage Installations.................................................................................E-5 Construction Cost: Area 91 - Utilities Facilities.........................................................................................E-7 Construction Cost: Area 92 - Support & Auxiliary Buildings................................................................ E-9 Site Infrastructure Cost Summary...............................................................................................................E-11 Plant Cost Breakdown per Discipline....................................................................................................................E-12 Introduction......................................................................................................................................................E-12 Direct Costs Breakdown...............................................................................................................................E-13 Indirect Costs Breakdown............................................................................................................................E-14 Plant Cost Breakdown Summary............................................................................................................... E-15 Process Flow Diagrams & Equipment List...........................................................................................................E-16 THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-1 In accordance with all the assumptions previously presented, a cost estimate was developed for each functional unit inside battery limits (ISBL). The following table shows the share of each functional unit.
Process Unit (ISBL) Construction Cost by Functional Unit DESCRIPTION % PROCESS UNIT (ISBL) CONSTRUCTION COST 100.0 PROCESS UNIT (ISBL) CONSTRUCTION COST BY FUNCTIONAL UNIT On the next page, the pie charts present an illustration of the construction cost breakdown.
THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-2 Process Unit (ISBL) Construction Cost by Functional Unit THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-3 Site Infrastructure (OSBL) Construction Cost by Area The investment estimated for each area is further detailed in the next chapters This analysis provides a more detailed explanation of the fixed capital associated with the process described in the report.
More specifically, it is focused on the investment required for erecting the site surrounding infrastructure, also referred to as Outside Battery Limits (OSBL), comprising support buildings, auxiliary units used for providing and distributing utilities and storage facilities.
In accordance with the configuration previously presented, a cost estimate was developed for each facility outside battery limits. The following pie chart shows OSBL investment broken down into each area.
INFRASTRUCTURE CONSTRUCTION COST BY PIECE OF EQUIPMENT Introduction THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-4 Storage Installations Construction Cost per Piece of Equipment Construction Cost: Area 90 - Storage Installations COMPONENT This chapter details the cost estimate associated with Area 90 - Storage Installations.
The components included in the estimate are detailed in the table below.
DESCRIPTION Area 90 - Storage Installations: Scope Description The pie chart below illustrates how each component impacts the construction cost estimate for this area.
THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-5 The cost of each component was based on the following assumptions: THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-6 Utilities Facilities Construction Cost per Piece of Equipment Construction Cost: Area 91 - Utilities Facilities COMPONENT This chapter details the cost estimate associated with Area 91 - Utilities Facilities.
The components included in the estimate are detailed in the table below.
DESCRIPTION Area 91 - Utilities Facilities: Scope Description The pie chart below illustrates how each component impacts the construction cost estimate for this area.
THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-7 The percentage associated with the remaining facilities is divided as follows: The cost of each component was based on the following assumptions: % % % % % THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-8 Support & Auxiliary Buildings Construction Cost per Piece of Equipment Construction Cost: Area 92 - Support & Auxiliary Buildings COMPONENT This chapter details the cost estimate associated with Area 92 - Support & Auxiliary Buildings.
The components included in the estimate are detailed in the table below.
DESCRIPTION Area 92 - Support & Auxiliary Buildings: Scope Description The pie chart below illustrates how each component impacts the construction cost estimate for this area.
THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-9 The percentage associated with the remaining facilities is divided as follows: The cost of each component was based on the following assumptions: % % THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-10 Site Infrastructure Cost Summary Site Infrastructure (OSBL) Construction Cost by Piece of Equipment COMPONENT SITE INFRASTRUCTURE CONSTRUCTION COST % OF TOTAL Area 90 - Storage Installations Area 91 - Utilities Facilities Area 92 - Support & Auxiliary Buildings THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-11 Plant Construction Cost Summary The two charts in the next page detail the composition of direct field costs and indirect costs, respectively.
The primary objective of this analysis is to provide an alternative perspective on the Plant Construction Cost.
This analysis presents the Plant Construction Cost divided in three categories: (1) direct costs (all material and labor costs associated with the process equipment); (2) indirect costs (defined by the American Association of Cost Engineers (AACE) Standard Terminology as those "costs which do not become a final part of the installation but which are required for the orderly completion of the installation"); and (3) contingency.
It is important to highlight that the breakdown presented within this analysis refers exclusively to the Plant Cost figure included in the report. Other fixed capital components, such as Owner's Cost, are not included in this breakdown.
The chart below presents the plant cost divided in each category described above.
PLANT COST BREAKDOWN PER DISCIPLINE Introduction THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-12 Direct Construction Costs by Discipline Direct Costs Breakdown Fundamentally, the direct process costs are the total installed equipment cost (from purchase to installation, including the required installation bulks). They include the following disciplines: bare equipment, equipment setting, piping civil, steel, instrumentation & control, electrical, insulation, painting.
Accordingly, the chart below presents the direct costs broken down by aforementioned discipline.
THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-13 Indirect Costs Summary Indirect Costs Breakdown The indirect costs account for field indirects, engineering costs, overhead, and contract fees.
Accordingly, the chart below presents the plant cost broken down by direct process costs, indirect costs and project contingency.
THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-14 The absolute cost of the plant is presented in the 'Plant Cost Summary' chart (chapter Capital Investment).
It is worth noting that the process contingency presented in the aforementioned table is included within each component listed in the table above.
For further information about the components included in the plant cost breakdown, reader is referred to the chapter “Methodology”.
Plant Construction Cost by Discipline COMPONENT Piping Project contingency ( % of TPC) Bare equipment (BEQ) TOTAL PLANT COST % OF BEQ % OF TOTAL Direct costs Equipment setting Civil Steel Instrumentation & control Electrical Insulation Painting Engineering & procurement Construction material & indirects General & administrative overheads Contract fee Indirect costs Total process capital (TPC) The next table presents the detailed plant cost breakdown, based on the direct and indirect costs approach. Two alternative views are presented in the table: (1) % of BEQ. Each component is presented as a percentage of the bare equipment (BEQ) cost; (2) % of Total. Each component is presented as a percentage of total plant cost.
Plant Cost Breakdown Summary THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-15 This chapter comprises a schematic representation of relevant operations of the process examined in the report.
It indicates process operations, main process streams, main pieces of equipment and utilities consumed.
Please find below the standards adopted in the development of the Process Flow Diagrams.
PROCESS FLOW DIAGRAMS & EQUIPMENT LIST Equipment Tags Streams Equipment Symbols Reactors & Vessels THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-16 Heat Exchangers Columns THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-17 Compressors & Pumps THIS PAGE'S CONTENT IS INCLUDED IN 'ADVANCED' VERSION ONLY.
E-18 PROVIDER PROCESS FLOW DIAGRAM REPORT ID PAGE IMPORTANT NOTICE THIS PFD IS BASED ON NON-CONFIDENTIAL INFORMATION. DETAILS (INDUSTRIAL SECRETS AND/OR CHANGES NOT REPORTED IN THE LITERATURE) ARE BASED ON INTRATEC PROCESS SYNTHESIS KNOWLEDGE, IN SUCH A WAY THAT DIFFERENCES MAY EXIST. NEVERTHELESS, THIS PFD RELIABLY PORTRAYS THE PROCESS, CONCEPTUALLY.
FOR A BETTER UNDERSTANDING, SEE THE RELATED REPORT.
1/1 www.intratec.us Process Flow Diagram (2 of 1) |
189223 | https://www.geeksforgeeks.org/maths/linear-equations-in-one-variable/ | Linear Equations in One Variable
Linear equation in one variable is the equation that is used for representing the conditions that are dependent on one variable. It is a linear equation i.e. the equation in which the degree of the equation is one, and it only has one variable.
A linear equation in one variable is a mathematical statement that involves a first-degree polynomial, and it can be expressed in the form: ax + b = c
Example: We can take any variable such as x, y, a, b, etc. Some examples of linear equations in one variable are,
The above equations are linear in one variable as they only have one variable and the highest degree of the variable is 1.
Graph of Linear Equation in One Variable
These linear equation can be easily represented on the graphs and they represent the straight line which can be horizontal to the coordinate axis or vertical to the coordinate axis.
We represent the mathematical condition using these equations as the condition "a number which is 5 less than twice of itself" is represented by the linear equation,
x = 2x - 5 , Where x is the unknown number.
In the above equation there is only one variable (x) and the degree of the variable is one thus, it is a linear equation in one variable.
Standard Form of Linear Equation in One Variable
A linear equation in one variable can be expressed in the standard form
ax+b = 0
where x is the variable and a and b are the constants involved. These constants (a and b) are non-zero real numbers. These equations have only one possible solution for the value of the variable.
Solving Linear Equations in One Variable
Linear Equations in One Variable can easily be solved by following the steps discussed below,
Let's understand the above steps with the help of the following example.
Examples on Linear Equation in One Variable
Solve the following linear equation in one variable.
Example: Solve 3x + 1/2 = (1/2)x - 13/2
Solution:
Step 1: Arranging in standard form and taking LCM
3x - (1/2)x + 1/2 + 13/2 = 0
⇒ (6x - x + 1 + 13)/2 = 0
⇒ 6x - x + 1 + 13 = 0
Step 2: Transposing the constant to the right-hand side
6x - x = -1 - 13
Step 3: Simplification
5x = -14
Step 4: Make coefficient of x to 1
Dividing both sides by 5 we get,
5x/5 = -14/5
x = -14/5
This is the required solution.
Linear Equations and Non-Linear Equations
As we have already learned that linear equations are the equations with the highest degree of "one" and they represent a straight line on the coordinate plane. Examples of linear equations are,
All these linear equations represent the straight line in coordinate planes.
All the equations which have a degree greater than one are called non-linear equations they represent curves in the coordinate plane. Various types of non-linear equations represent various types of curves in the 2-D plane. Some of the common curves which we study are,
Circle: x2 + y2 = 49, this equation represents a circle in an x-y coordinate plane with a center at (0, 0) and a radius of 7 units.
Similarly,
All these equations discussed above are non-linear equations.
Difference between Linear Equations and Non-Linear Equations
The key differences between Linear Equations and Non-Linear Equations are:
| Feature | Linear Equations | Non-Linear Equations |
---
| Definition | Equations of the form ax + by = c, where a, b, and c are constants, and x and y are variables raised to the power of 1. | Equations where the highest power of the variables is greater than 1, or variables are multiplied together. |
| Graph | Represented by straight lines. | Represented by curves or irregular shapes. |
| Solution(s) | Always forms a straight line when graphed. | Can form curves, loops, or irregular shapes when graphed. |
| Number of Solutions | Always one solution (if the lines intersect). | Can have multiple solutions or none. |
| Solution Method | Solvable using algebraic methods like substitution, elimination, or graphing. | Often require numerical or iterative methods for solutions. |
Must Read
Also Check
Linear Equation in One Variable Examples
Example 1: Solve for y, 8y - 4 = 0
Solution:
Solving for value of y,
Adding 4 to both sides of the equation ,
8y -4 + 4 = 4
8y = 4
Dividing both sides of equation by 8
y = 4/8
Simplifying the equation ,
y = 1/2
Example 2: Solve the equation in x, 3x +10 = 55
Solution:
Taking constants to RHS,
3x = 45
x = 15
Example 3: Solve the equation in x, 4/x5 -5 = 15
Solution:
4/x5 -5 = 15
⇒ 4x/5 = 15 + 5
⇒ 4x/5 = 20
⇒ x = 20×5/4
⇒ x = 25
Example 4: The age of Ravi is twice the age of his sister Kiran if the sum of their age is 24 find their individual age.
Solution:
Let the age of Kiran is x, then the age of Ravi is 2x
Given, the sum of their ages is 24
x + 2x = 24
⇒ 3x = 24
⇒ x = 8
Thus, the age of Kiran = x = 8 years
Age of Ravi = 2x = 2×8 = 16 years
Example 5: Akshay earns three times more than Abhay and if the difference in their salaries is Rs 5000 find their individual salaries.
Solution:
Let the salary of Abhay be x, then the salary of Akshay is 3x
Given, the difference in their salaries is Rs 5000
3x - x = 5000
⇒ 2x = 5000
⇒ x = 2500
Thus, the salary of Abhay = x = Rs 2500
The salary of Akshay = 3x = 3×2500 = Rs 7500
Y
Explore
Maths
Basic Arithmetic
What are Numbers?
Arithmetic Operations
Fractions - Definition, Types and Examples
What are Decimals?
Exponents
Percentage
Algebra
Variable in Maths
Polynomials| Degree | Types | Properties and Examples
Coefficient
Algebraic Identities
Properties of Algebraic Operations
Geometry
Lines and Angles
Geometric Shapes in Maths
Area and Perimeter of Shapes | Formula and Examples
Surface Areas and Volumes
Points, Lines and Planes
Coordinate Axes and Coordinate Planes in 3D space
Trigonometry & Vector Algebra
Trigonometric Ratios
Trigonometric Equations | Definition, Examples & How to Solve
Trigonometric Identities
Trigonometric Functions
Inverse Trigonometric Functions | Definition, Formula, Types and Examples
Inverse Trigonometric Identities
Calculus
Introduction to Differential Calculus
Limits in Calculus
Continuity of Functions
Differentiation
Differentiability of Functions
Integration
Probability and Statistics
Basic Concepts of Probability
Bayes' Theorem
Probability Distribution - Function, Formula, Table
Descriptive Statistic
What is Inferential Statistics?
Measures of Central Tendency in Statistics
Set Theory
Practice
NCERT Solutions for Class 8 to 12
RD Sharma Class 8 Solutions for Maths: Chapter Wise PDF
RD Sharma Class 9 Solutions
RD Sharma Class 10 Solutions
RD Sharma Class 11 Solutions for Maths
RD Sharma Class 12 Solutions for Maths
Thank You!
What kind of Experience do you want to share? |
189224 | https://www.math.utah.edu/~anna/Fall12/LessonPlans/Section106.pdf | Section 10.6, Graphs of Polar Equations Homework: 10.6 #1–37 odds 1 Symmetry in Graphing with Polar Coordinates Symmetry is a helpful tool when graphing in Polar Coordinates.
• If replacing (r, θ) by (r, −θ) gives an equivalent equation, the graph is symmetric with respect to the polar axis (the horizontal axis). For example, if r = cos θ and we replace θ by −θ, we get r = cos(−θ) = cos θ since cosine is an even function. Since this is what we started with, we know that the graph is symmetric with respect to the polar axis.
• If replacing (r, θ) by (r, π −θ) or (−r, −θ) gives an equivalent equation, the graph is symmetric with respect to the line θ = π/2 (the vertical axis). For example, if r = sin θ, replacing r by −r and θ by −θ gives −r = sin(−θ) = −sin θ. After we cancel out the negative signs, this is exactly what we started with, so we know that the graph of r = sin θ is symmetric with respect to the line θ = π/2.
• If replacing (r, θ) by (r, π + θ) or (−r, θ) gives an equivalent equation, the graph is symmetric with respect to the pole (origin). For example, r = 5 and θ = π/4 satisfy this criterion.
2 Types of Polar Graphs • The equation for the graph of a lima¸ con has the form r = a ± b cos θ or r = a ± b sin θ. If a = b, the graph is called a cardioid. The graphs of some lima¸ cons look like: a > b a = b a < b Note: These graphs may be symmetric with respect to the x-axis instead of the y-axis. See Figure 4 on page 543 of the book for examples.
• Figure-eight-shaped curves are called lemniscates. The equation has the form r2 = ±a cos 2θ or r2 = ±a sin 2θ.
• The equation for the graph of a rose has the form r = a cos nθ or r = a sin nθ. If n is odd, the graph has n leaves. If n is even, the graph has 2n leaves. In general, the graph looks like • Spirals have equations of the form r = aθ.
Examples Sketch the graph of each of the following functions.
Identify what kind of graph the equation represents, as well as what kind of symmetry exists.
1. r = 4 sin θ This is a lima¸ con that is symmetric about the vertical axis. Specifically, it is a circle of radius 2 centered at the point (0, 2). The graph looks like: -6 &% '$ 2. r2 = −16 cos 2θ This is a lemniscate that is symmetric with respect to the horizontal axis. The graph looks like: -6 3. r = 4 −3 sin θ This is the graph of a lima¸ con that is symmetric with respect to the vertical axis. The graph looks like: 4. r = 2θ This is the graph of a spiral. There is no symmetry. The graph was sketched in class.
5. r = √ 2 − √ 2 sin θ This is the graph of a cardioid that is symmetric with respect to the vertical axis.
6. r = 4 cos 2θ This is the graph of a rose with 4 petals. It is symmetric with respect to the horizontal axis.
The graph looks like: |
189225 | https://www.geeksforgeeks.org/maths/dependent-and-independent-events-probability/ | Dependent and Independent Events
Last Updated :
23 Jul, 2025
Suggest changes
4 Likes
Dependent and Independent Events are the types of events that occur in probability. Suppose we have two events say Event A and Event B then if Event A and Event B are dependent events then the occurrence of one event is dependent on the occurrence of other events if they are independent events then the occurrence of one event does not affect the probability of other events.
We can learn about dependent and independent events with the help of examples such as the event of tossing two coins simultaneously the outcome of one coin does not affect the outcome of another coin then they are independent events. Suppose we take other experiments where we toss a coin only when we get a six in the throw of dice, where the outcome of one event is affected by other events then they are dependent events.
In this article, we will learn about dependent events, independent events, their formulas, examples, and others in detail.
Table of Content
Dependent and Independent Events in Probability
Dependent Events
Examples of Dependent Events
Independent Events
Examples of Independent Events
Difference Between Independent Events and Dependent Events
Mutually Exclusive Events
Conditional Probability Formula
Dependent and Independent Events in Probability
An event in probability falls under two categories,
Dependent Events
Independent Events
Dependent Events
Dependent events are those events that are affected by the outcomes of events that had already occurred previously. i.e. Two or more events that depend on one another are known as dependent events. If one event is by chance changed, then another is likely to differ. Thus, If whether one event occurs does affect the probability that the other event will occur, then the two events are said to be dependent.
When the occurrence of one event affects the occurrence of another subsequent event, the two events are dependent events. The concept of dependent events gives rise to the concept of conditional probability which will be discussed in the article further.
Examples of Dependent Events
For Example, let's say three cards are to be drawn from a pack of cards. Then the probability of getting a king is highest when the first card is drawn, while the probability of getting a king would be less when the second card is drawn.
In the draw of the third card, this probability would be dependent upon the outcomes of the previous two cards. We can say that after drawing one card, there will be fewer cards available in the deck, therefore the probabilities after each drawn card changes.
Independent Events
Independent events are those events whose occurrence is not dependent on any other event. If the probability of occurrence of an event A is not affected by the occurrence of another event B, then A and B are said to be independent events.
Examples of Independent Events
Various examples of Independent events are:
Tossing a Coin
Sample Space(S) in a Coin Toss = {H, T}
Both getting H and T are Independent Events
Rolling a Die
Sample Space(S) in Rooling a Die = {1, 2, 3, 4, 5, 6}, all of these events are independent too.
Both of the above examples are simple events. Even compound events can be independent events. For example:
Tossing a Coin and Rolling a Die
If we simultaneously toss a coin and roll a die then the probability of all the events is the same and all of the events are independent events,
Sample Space(S) of such experiment = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T) (5, T) (6, T)}.
These events are independent because only one can occur at a time and occurring of one event does not affect other events.
Note:
A and B are two events associated with the same random experiment, then A and B are known as independent events if
P(A ∩ B) = P(B).P(A)
Difference Between Independent Events and Dependent Events
The difference between independent events and dependent events is discussed in the table below,
| Independent Events | Dependent Events |
--- |
| Independent events are events that are not affected by the occurrence of other events. | Dependent events are events that are affected by the occurrence of other events. |
| The formula for the Independent Events is, P(A∩B)=P(A)⋅P(B) | The formula for the Dependent Events is, P(A∩B)=P(A)⋅P(B∣A) |
| Examples of Independent Events are, Tossing one coin was not affected by the tossing of other coins Raining for a day and getting six in dice are independent events. | Examples of Dependent Events are, The probability of finding a red ball from a box of 4 red balls and 3 green balls changes if we take out two balls from the box. |
Mutually Exclusive Events
Two events A and B are said to be mutually exclusive events if they cannot occur at the same time. Mutually exclusive events never have an outcome in common. If we take two events A and B as mutually exclusive events where the probability of event A is P(A) and the probability of event B is P(B) then the probability of happening both events together is,
P(A∩B) = 0
Then the probability of occurring any one event is,
P(AUB) = P(A) or P(B) = P(A) + P(B)
Conditional Probability Formula
Conditional probability formula tells the formula for the probability of the event when an event has already occurred. If the probability of events A and B are P(A) and P(B) respectively then the conditional probability of A such that B has already occurred is denoted as P(A/B).
If P(A) > 0, then the P(A/B) is calculated by using the formula,
P(A/B) = P(A ∩ B)/P(B)
In the case of P(A) = 0 means A is an impossible event, in this case, P(A/B) does not exist.
Article Related Dependent and Independent Events:
Bayes Theorem
Permutations and Combinations
Binomial Theorem
Solved Examples on Dependent and Independent Events
Example 1: An instructor has a question bank with 300 Easy T/F, 200 Difficult T/F, 500 Easy MCQ, and 400 Difficult MCQ. If a question is selected randomly from the question bank, What is the probability that it is an easy question given that it is an MCQ?
Solution:
Total question in the question bank = 300 + 200 + 500 + 400
P(Easy) = (300+500)/1400 = 800/1400 = 4/7
P(MCQ) = (400+500)/1400 = 900/1400 = 9/14
P(Easy ∩ MCQ) = (500)/1400 =5/14
P(Easy|MCQ) = P(Easy ∩ MCQ)/P(MCQ)
P(Easy|MCQ) = (5/14)/(9/14) = 5/9
Thus, probability of an easy question given it is an MCQ is 5/9.
Example 2: In a shipment of 20 apples, 3 are rotten. 3 apples are randomly selected. What is the probability that all three are rotten if the first and second are not replaced?
Solution:
Total Apple = 20
Rotten Apple = 3
Possibility of the first apple being rotten = 3/20
Possibility of the second apple being rotten = 2/19
Possibility of the third apple being rotten = 1/18
Probability of all three apples being rotten = P(3 Rotten) = (3/20 × 2/19 × 1/18) = 6/6840 = 1/1140
Thus, probability that all three apples are rotten is, 1/1140
Example 3: John has to select two students from a class of 10 girls and 15 boys. What is the probability that both students chosen are boys?
Solution:
Total number of students = 10 + 15 =25
Probability of choosing the first boy
P (Boy 1) = 15/25
P (Boy 2) = 14/24
P (Boy 1 and Boy 2) = P (Boy 1) and P (Boy 2)
P (Boy 1 and Boy 2) = (15/25) × (14/24) = 7/20
Thus, probability of choosing both boys is 7/20
Example 4: A multiple-choice test consists of two problems. Problem 1 has 5 options and Problem 2 has 4 options. Each problem has only one correct answer. What is the probability of randomly guessing the correct answer to both problems?
Solution:
Here, the probability of the correct answer to Problem 1 = P(A) and the probability of the correct answer to Problem 2 = P(B) are independent events.
Thus the probability of a correct answer to Problem 1 and Problem 2 both = P(A ∩ B) = P(A).P(B)
P(A) = 1/5
P(B) = 1/4
P(A ∩ B) = (1/5) × (1/4) = 1/20
Thus, probability of getting both answers correct is 1/20.
Also Check:
Coin Toss Probability
Events in Probability
Types of Events
S
saadshashi25
Improve
Article Tags :
Mathematics
School Learning
Class 10
Probability
Maths-Class-10
Explore
Maths
4 min read
Basic Arithmetic
What are Numbers?
15+ min readArithmetic Operations
9 min readFractions - Definition, Types and Examples
7 min readWhat are Decimals?
10 min readExponents
9 min readPercentage
4 min read
Algebra
Variable in Maths
5 min readPolynomials| Degree | Types | Properties and Examples
9 min readCoefficient
8 min readAlgebraic Identities
14 min readProperties of Algebraic Operations
3 min read
Geometry
Lines and Angles
9 min readGeometric Shapes in Maths
2 min readArea and Perimeter of Shapes | Formula and Examples
10 min readSurface Areas and Volumes
10 min readPoints, Lines and Planes
14 min readCoordinate Axes and Coordinate Planes in 3D space
6 min read
Trigonometry & Vector Algebra
Trigonometric Ratios
4 min readTrigonometric Equations | Definition, Examples & How to Solve
9 min readTrigonometric Identities
7 min readTrigonometric Functions
6 min readInverse Trigonometric Functions | Definition, Formula, Types and Examples
11 min readInverse Trigonometric Identities
9 min read
Calculus
Introduction to Differential Calculus
6 min readLimits in Calculus
12 min readContinuity of Functions
10 min readDifferentiation
2 min readDifferentiability of Functions
9 min readIntegration
3 min read
Probability and Statistics
Basic Concepts of Probability
7 min readBayes' Theorem
13 min readProbability Distribution - Function, Formula, Table
13 min readDescriptive Statistic
5 min readWhat is Inferential Statistics?
7 min readMeasures of Central Tendency in Statistics
11 min readSet Theory
3 min read
Practice
NCERT Solutions for Class 8 to 12
7 min readRD Sharma Class 8 Solutions for Maths: Chapter Wise PDF
5 min readRD Sharma Class 9 Solutions
10 min readRD Sharma Class 10 Solutions
9 min readRD Sharma Class 11 Solutions for Maths
13 min readRD Sharma Class 12 Solutions for Maths
13 min read
Improvement
Suggest Changes
Help us improve. Share your suggestions to enhance the article. Contribute your expertise and make a difference in the GeeksforGeeks portal.
Create Improvement
Enhance the article with your expertise. Contribute to the GeeksforGeeks community and help create better learning resources for all. |
189226 | https://physics.stackexchange.com/questions/360894/work-done-by-a-variable-force-in-two-dimensions | homework and exercises - Work done by a variable force in two dimensions - Physics Stack Exchange
Join Physics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Physics helpchat
Physics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Work done by a variable force in two dimensions
Ask Question
Asked 7 years, 11 months ago
Modified7 years, 11 months ago
Viewed 5k times
This question shows research effort; it is useful and clear
0
Save this question.
Show activity on this post.
In physics class today, our teacher ended class with an example problem. It essentially asked (perhaps with different numbers):
Given a variable force F⃗=3 x i^+4 y j^F→=3 x i^+4 y j^ acting on an object moving from r⃗=0 r→=0 to r⃗=5 i^r→=5 i^, calculate the work done on the object by F⃗F→.
Our teacher wrote "∫5 0 3 x i^+4 y j^d x i^∫0 5 3 x i^+4 y j^d x i^" on the board, which looked wrong. While the solution to this problem is fairly intuitive, what is the correct notation for the integral and solution in general?
homework-and-exercises
work
notation
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Improve this question
Follow
Follow this question to receive notifications
asked Oct 4, 2017 at 4:03
devin7devin7
103 1 1 silver badge 4 4 bronze badges
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
3
Save this answer.
Show activity on this post.
The total work done is
W=∫γ F⃗⋅d r⃗W=∫γ F→⋅d r→
where d r⃗d r→ is the infinitesimal, tangential line element along some path γ γ.
We need to figure out the path we'd like to take, and the associated parameterization we would like to use. One possible choice (which mirrors what your instructor used) is the following:
r⃗(t)=⟨x(t),y(t)⟩=⟨t,0⟩t∈[0,5]r→(t)=⟨x(t),y(t)⟩=⟨t,0⟩t∈[0,5]
so
F⃗=⟨3 x(t),4 y(t)⟩=⟨3 t,0⟩F→=⟨3 x(t),4 y(t)⟩=⟨3 t,0⟩
d r⃗=⟨1,0⟩d t d r→=⟨1,0⟩d t
and the integral becomes
∫5 0 3 t d t=3 2 t 2∣∣∣5 0=75 2∫0 5 3 t d t=3 2 t 2|0 5=75 2
Your instructor chose to parameterize the path by one of its coordinates. That's a perfectly good choice for that particular path, but it isn't always possible to do this - in particular, if the path has a squiggle or a loop such that it no longer passes the so-called "vertical line test", you can't use the x x coordinate as a valid parameter. Similarly, if the path doesn't pass the "horizontal line test", then you can't use the y y coordinate as a valid parameter. I like to use a totally separate parameter t t which circumvents these issues and makes the parameterization clearer.
For a bit of enrichment, notice that we could have chosen a totally different path. What if we picked a crazy, loopy path like this?
r⃗(t)=⟨t+cos(2 π t)−1,sin(2 π t)⟩t∈[0,5]r→(t)=⟨t+cos(2 π t)−1,sin(2 π t)⟩t∈[0,5]
Notice that this beast does not pass the horizontal or vertical line tests, so I certainly cannot parameterize this path by x x or y y. However, I can use my parameter t t and everything works out just as before. It's a bit messy, but if you work it out you'll find that you get the same answer as before, and for good reason.
An arbitrary line integral needs to be equipped with a path along which the integration is taking place. However, for a special class of integrands, the value of the integral is independent of the path chosen. This is true whenever F⃗F→ can be written as the gradient of a scalar function,
F⃗=−∇U F→=−∇U
where the minus sign is just a convention commonly used in physics. In such cases, F⃗F→ is called a conservative vector field, and
∫B A F⃗⋅d r⃗=U(A)−U(B)∫A B F→⋅d r→=U(A)−U(B)
regardless of the path chosen. For your F⃗F→, we could pick
U=−3 2 x 2−2 y 2 U=−3 2 x 2−2 y 2
You can check to see that this yields the correct answer. When F⃗F→ is interpreted as a force, we call an associated function U U a potential energy function corresponding to that force.
But once again, vector fields are not generally conservative. When they are not, a line integral is not even defined until you've chosen a path (though you can parameterize that path in essentially whatever way you choose).
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Improve this answer
Follow
Follow this answer to receive notifications
answered Oct 4, 2017 at 5:54
AlbatrossAlbatross
74.2k 5 5 gold badges 105 105 silver badges 210 210 bronze badges
1
I up voted your answer since I believe it'll be very useful to the OP. The important here are the gradient, the potential and that the integral is independent of the path. But I think that you might omit the full solutions because this is a homework like question, for example: the value of the integral and the potential U U leaving the OP to find out them by him(/her)self.VoulKons –VoulKons 2017-10-04 12:27:08 +00:00 Commented Oct 4, 2017 at 12:27
Add a comment|
Your Answer
Thanks for contributing an answer to Physics Stack Exchange!
Please be sure to answer the question. Provide details and share your research!
But avoid …
Asking for help, clarification, or responding to other answers.
Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Draft saved
Draft discarded
Sign up or log in
Sign up using Google
Sign up using Email and Password
Submit
Post as a guest
Name
Email
Required, but never shown
Post Your Answer Discard
By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
homework-and-exercises
work
notation
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Related
1Why is the formula written this way?
0Solution to diffusion equation of a random walk
4Work Done by a time-variable Force
0How to rigorously treat integrals where both the integrand and the variable of integration are vectors and there's a dot product?
Hot Network Questions
What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left?
How can the problem of a warlock with two spell slots be solved?
On Chern roots and the Chern polynomial
What's the expectation around asking to be invited to invitation-only workshops?
Direct train from Rotterdam to Lille Europe
Xubuntu 24.04 - Libreoffice
For every second-order formula, is there a first-order formula equivalent to it by reification?
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish?
Interpret G-code
A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man
Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations?
How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done?
Implications of using a stream cipher as KDF
Analog story - nuclear bombs used to neutralize global warming
Can a cleric gain the intended benefit from the Extra Spell feat?
Lingering odor presumably from bad chicken
What is a "non-reversible filter"?
Do sum of natural numbers and sum of their squares represent uniquely the summands?
Matthew 24:5 Many will come in my name!
в ответе meaning in context
Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done?
Making sense of perturbation theory in many-body physics
How different is Roman Latin?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Physics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
189227 | https://www.jmap.org/htmlstandard/A.APR.C.5.htm | | | | | |
--- --- |
| | | | | --- | | | | | 20 Years of JMAP www.jmap.org went online on March 1, 2005. Please consider a $20 donation to acknowledge this website's impact on high school mathematics education for the last 20 years! | | |
| | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- ---
| | | | RESOURCES BY STANDARD AI GEO AII PLUS or www.commoncorestatestandards.org RESOURCES BY TOPIC 6-8 AI GEO AII PRECALCULUS CALCULUS QUICK TOPICS REGENTS EXAMS WORKSHEETS JMAP ON JUMBLED An online platform for the above Algebra I resources REGENTS BOOKS WORKSHEET GENERATORS EXTRAS REGENTS EXAM ARCHIVES 1866-now JMAP RESOURCE ARCHIVES AI/GEO/AII (2015-now) IA/GE/A2 (2007-17) Math A/B (1998-2010) REGENTS RESOURCES INTERDISCIPLINARY EXAMS NYC TEACHER RESOURCES City Tech Math Education Program For students considering a career in teaching math | | | | | STANDARD A.APR.C.5 Precalculus Use the Binomial Theorem for the expansion of (x+y)n for a positive integer n. | | REGENTS WORKSHEETS | Regents-Binomial Expansions 1 A2/B | 10/8 | TST PDF DOC | | Regents-Binomial Expansions 2 SIII (1980-1991) | 25 | TST PDF DOC | | Regents-Binomial Expansions 3 SIII (1992-2004) | 19 | TST PDF DOC | | PRACTICE WORKSHEETS & JOURNALS | Practice-Binomial Expansions expand | | WS PDF | | Journal-Binomial Expansions | | WS PDF | |
| | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | | | --- --- --- | HOME | REVIEW | REGENTS EXAM ARCHIVES | JMAP ON JUMBLED An online platform for JMAP's Algebra I Resources below | EXAMVIEW | JMAP ARCHIVES A/B 2005 CCSS | REGENTS RESOURCES | | ABOUT | ROSTER | REGENTS BOOKS | EXAMS | TOPIC AI GEO AII PRE-C CALC | STANDARD AI GEO AII PLUS | REGENTS BANKS | NYC | | DONATE | DONORS | WORKSHEETS | EXTRAS | IJMAP | | Questions should be directed to JMAP's Editor, Steve Sibol or Cofounder, Steve Watson Copyright © 2004-now JMAP, Inc. - All rights reserved JMAP, Inc. is a 501(c)(3) New York Not-for-Profit Corporation | | |
189228 | https://www.osmosis.org/learn/Haemophilus_influenzae | Fall in love with Osmosis at 20% off! Save now until September 30 at 11:59 PM PT.
Learn more
Skip to the video
Haemophilus influenzae
Foundational SciencesMicrobiologyBacteriologyGram negative bacteriaCoccobacilli
8,896views
00:00 / 00:00
Haemophilus influenzae
Introduction to bacteria
Bacterial structure and functions
Gram positive bacteria
Bacillus anthracis (Anthrax)
Bacillus cereus (Food poisoning)
Corynebacterium diphtheriae (Diphtheria)
Listeria monocytogenes
Clostridium botulinum (Botulism)
Clostridium difficile (Pseudomembranous colitis)
Clostridium perfringens
Clostridium tetani (Tetanus)
Actinomyces israelii
Nocardia
Staphylococcus aureus
Staphylococcus epidermidis
Staphylococcus saprophyticus
Streptococcus agalactiae (Group B Strep)
Streptococcus pneumoniae
Streptococcus pyogenes (Group A Strep)
Streptococcus viridans
Enterococcus
Gram negative bacteria
Bacteroides fragilis
Bartonella henselae (Cat-scratch disease and Bacillary angiomatosis)
Enterobacter
Escherichia coli
Klebsiella pneumoniae
Legionella pneumophila (Legionnaires disease and Pontiac fever)
Proteus mirabilis
Pseudomonas aeruginosa
Salmonella (non-typhoidal)
Salmonella typhi (typhoid fever)
Serratia marcescens
Shigella
Yersinia enterocolitica
Yersinia pestis (Plague)
Campylobacter jejuni
Helicobacter pylori
Vibrio cholerae (Cholera)
Moraxella catarrhalis
Neisseria gonorrhoeae
Neisseria meningitidis
Bordetella pertussis (Whooping cough)
Brucella
Francisella tularensis (Tularemia)
Haemophilus ducreyi (Chancroid)
Haemophilus influenzae
Pasteurella multocida
Mycobacteria
Mycobacterium tuberculosis (Tuberculosis)
Mycobacterium avium complex (NORD)
Mycobacterium leprae
Other bacteria
Chlamydia pneumoniae
Chlamydia trachomatis
Gardnerella vaginalis (Bacterial vaginosis)
Mycoplasma pneumoniae
Coxiella burnetii (Q fever)
Ehrlichia and Anaplasma
Rickettsia rickettsii (Rocky Mountain spotted fever) and other Rickettsia species
Borrelia burgdorferi (Lyme disease)
Borrelia species (Relapsing fever)
Leptospira
Treponema pallidum (Syphilis)
Assessments
Flashcards
0 / 26 complete
USMLE® Step 1 questions
0 / 2 complete
PANCE® questions
0 / 1 complete
High Yield Notes
9 pages
Flashcards
Haemophilus influenzae
0 of 26 complete
Questions
USMLE® Step 1 style questions USMLE
0 of 2 complete
A 28-year-old man with polysubstance use disorder presents to the emergency department with fever, dyspnea, and stridor. The patient is diagnosed with epiglottitis. He is started on intravenous antibiotics and later undergoes endotracheal intubation for airway protection. He did not receive most of the childhood immunizations. Blood and sputum cultures are collected. Which of the following agars is most likely to grow the organism responsible for this patient’s infection?
Transcript
Watch video only
Content Reviewers
Viviana Popa, MD
Contributors
Alexandru Duhaniuc, MD,
Evan Debevec-McKenney,
Elizabeth Nixon-Shapiro, MSMI, CMI
Haemophilus influenzae is a small Gram-negative coccobacillus which can normally colonize the human respiratory tract. There are two major categories of H. influenzae - encapsulated strains and unencapsulated strains.
Encapsulated strains are classified into six serotypes based on their capsular antigens - a, b, c, d, e and f, and unencapsulated strains are called nontypable, because they lack the polysaccharide capsule, and, consequently, capsular antigens.
The strains that cause disease in humans are most often Haemophilus influenzae type b, or Hib for short and Haemophilus influenzae nontypable.
Now, Haemophilus influenzae has a thin peptidoglycan layer, so it doesn’t retain the crystal violet dye during Gram staining. Instead, like any other Gram-negative bacteria, it stains pink with safranin dye.
And since it’s a coccobacillus, it’s shaped somewhere between round, like a coccus, and linear, like a bacillus. Haemophilus influenzae is non-motile, so it doesn’t move, and facultative anaerobic which means it can survive both in aerobic and anaerobic environments. It’s also catalase and oxidase positive which means it produces both these enzymes.
Finally, Haemophilus influenzae can be cultivated on chocolate agar, because this medium contains essential nutrients that Haemophilus influenzae needs to grow, like factor X, also called hemin, and factor V, also called nicotinamide adenine nucleotide.
Another way to grow it is to grow it with Staphylococcus aureus colonies, on blood agar, which provides factor V via red blood cells hemolysis.
On both blood agar and chocolate agar, Haemophilus influenzae grows into convex, smooth, gray or transparent colonies. Now, Haemophilus influenzae has a number of virulence factors, that are like assault weaponry that help it attack and destroy the host cells, and evade the immune system.
So first, encapsulated strains of Haemophilus influenzae are covered by a polysaccharide layer called a capsule. Now, this capsule is a major virulence factor for Haemophilus influenzae because of its antiphagocytic ability, meaning that it protects the bacteria against phagocytosis by macrophages or neutrophils.
This allows Haemophilus influenzae to escape destruction and attach to epithelial cells in the airways. Also, on the capsule, there are pili, which are hair-like extensions, and adhesion proteins such as HMW1 and HMW2 that help the bacteria attach to host cells.
Both encapsulated and unencapsulated strains have an outer membrane - which consist of lipooligosaccharides, or LOS.
LOS inhibits mucociliary clearance, or the self-clearing mechanism of the bronchi, that would normally sweep the bacteria out of the respiratory tract. With the help of LOS, Haemophilus influenzae can colonize the respiratory tract.
What is more, both encapsulated and unencapsulated strains make IgA protease, a toxic protein that destroys Immunoglobulin A, or IgA.
IgA is an immune system protein found in the nasopharyngeal mucosa secretions that normally opsonizes invading bacteria, meaning it tags them so neutrophils can recognize and destroy them. So IgA protease neutralizes the first line of mucosal defense!
Interestingly, unencapsulated strains have two other abilities that allow it to evade the immune system. First, these strains can change the oligosaccharides they express on their outer membrane each time they infect a cell, and this process is called phase variation.
Because of phase variation, the immune system can’t remember the infecting strain, so it can’t mount a quick specific immune response against it when it encounters the same strain in the future.
Finally, unencapsulated strains have the ability to produce biofilms. A biofilm is basically a layer of goop-like material made of exopolysaccharides or EPS, within which Haemophilus influenzae bacteria live and reproduce.
Comparing a biofilm to strawberry jam, the seeds would be the bacteria and the rest of the jam would be the EPS. The bacteria in the biofilm hide from the host's immune system and antibiotics.
Ok, now, encapsulated strains tend to cause more invasive disease on account of their capsule. One way is to spread directly from the nasopharynx to the epiglottis, causing epiglottitis, or to the soft skin tissues in the face, causing cellulitis.
Encapsulated strains can also breach through the epithelium of the nasopharynx and invade the blood capillaries directly, causing bacteremia.
From the bloodstream, it can spread to distant sites, like the meninges, causing meningitis, the bones, causing osteomyelitis or the joints, causing septic arthritis.
Nonencapsulated strains of Haemophilus influenzae, on the other hand, are less invasive, and they mostly cause mucosal infections by direct extension.
For example, they may go up the eustachian tubes and reach the middle ear, causing otitis media. If they spread to the paranasal sinuses, they cause sinusitis.
Finally, if they go down the respiratory tract, they may cause bronchitis, which is the inflammation of the bronchi, or pneumonia, which is the inflammation of the lungs.
Now, nonencapsulated strains of Haemophilus influenzae colonize the nasopharynx of 40 to 80% of children and adults, whereas encapsulated strains, such as Hib, colonizes 3 to 5% of children aged 2 to 5 years.
Either way, Haemophilus influenzae is mostly transmitted via respiratory droplets and respiratory secretions. Risk factors vary depending on the infecting strain.
Summary
Haemophilus influenzae, or just H. influenzae, is a gram-negative, facultatively anaerobic, non-motile coccobacillus. Haemophilus influenzae bacteria are classified into encapsulated strains and unencapsulated strains. Based on their capsular antigen type, encapsulated strains are further classified into six serotypes (serotypes a, b, c, d, e, and f). Unencapsulated strains are also referred to as nontypable because they lack the polysaccharide capsule, and capsular antigens.
Nontypeable strains of Haemophilus influenzae are known to cause relatively simple mucosal infections, like otitis media, sinusitis, and pneumonia. On the other hand, encapsulated strains such as Haemophilus influenza type b can cause more severe infections such as meningitis and epiglottitis, but such severe infections are not so common nowadays due to vaccination.
We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. For more information, see ourCookie Policy
Cookie Preference Center
We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors.
You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings.
You may also be able to exercise your privacy choices as described in our Privacy Policy
Manage Consent Preferences
Strictly Necessary Cookies
Always active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site.
Targeting Cookies
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. If you do not allow these cookies, you will experience less targeted advertising. |
189229 | https://quotation.io/quote/drunkard-gutter-just-according-fitness-the | A drunkard in the gutter is just where he ought to be, according - William Graham Sumner | Quotation.io
Home
Quote of the Day
Random Quotes
Topics
Authors
Quote by William Graham Sumner
A drunkard in the gutter is just where he ought to be, according to the fitness and tendency of things. Nature has set upon him the process of decline and dissolution by which she removes things which have survived their usefulness.
Summary
This quote suggests that a person who is drunk and in the gutter is in their rightful place in accordance with the natural cycle of decline. The quote implies that nature has designed a process to eliminate things that have outlived their purpose, and it is implied that the drunkard falls under this category. It highlights a fatalistic viewpoint that sees the downward spiral of the drunkard as an inevitable consequence of nature's way of removing the no longer functional elements in society.
Topics
Fitness
By William Graham Sumner
Liked the quote? Share it with your friends.
Random Quotations
Truth is a risky proposition. It's the nature of mediocre human beings to believe that lies are necessary, that they serve a purpose, that truth is subversive, that candor is dangerous, that the very scaffold of communal life is supported by lies.
Anne Rice
TruthLies
i don't know. i don't care. all i know is when you pay for something that's supposed to give you a cup of coffee, you deserve to get your fucking cup of coffee.
Jodi Picoult
Coffee
He who can have patience can have what he will.
Benjamin Franklin
A half-truth is usually less than half of that.
Bern Williams
Truth
Great things are not done by impulse, but by a series of small things brought together.
Vincent Van Gogh
Success & Failure
Bankers know that history is inflationary and that money is the last thing a wise man will hoard.
Will Durant
Inflation
We no longer dare to believe in beauty and we make of it a mere appearance in order the more easily to dispose of it.
Hans Urs von Balthasar
Beauty
It's really necessary for the United States to continue to give strong leadership to the Middle East peace process, supported by European countries at the same time.
William Hague
Leadership
Everything you need is already inside.
Bill Bowerman
SportsInspirationalRunningBowermanPre
I'm pretty quiet. But I love to play sports. I like playing all sports. I'll act goofy at times around my wife and my son, around my own family. I like to have fun in general.
Steve Blake
TimeLoveFamilySportsThetimesFun
Unlike my grandfather or my brother, I've actually been able to make some money at a racetrack.
John Malkovich
Money
Make hunger thy sauce, as a medicine for health.
Thomas Tusser
Health
© 2025 Quotation.io. All rights reserved. |
189230 | https://www.youtube.com/watch?v=zf8vnCcETqM | Transformers - working & applications (step up and step down) | A.C. | Physics | Khan Academy
Khan Academy India - English
548000 subscribers
9602 likes
Description
397294 views
Posted: 21 Oct 2020
Transformers step up (increase) or step down (decrease) AC voltage using the principle of electromagnetic induction - mutual induction. A changing current in the primary coil induces an e.m.f in the secondary. Since the e.m.f generated depends on the number of turns, the voltage induced in the secondary can be changed - stepped up or down - by altering the turn's ratio.
Khan Academy is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We offer quizzes, questions, instructional videos, and articles on a range of academic subjects, including math, biology, chemistry, physics, history, economics, finance, grammar, preschool learning, and more. We provide teachers with tools and data so they can help their students develop the skills, habits, and mindsets for success in school and beyond. Khan Academy has been translated into dozens of languages, and 15 million people around the globe learn on Khan Academy every month. As a 501(c)(3) nonprofit organization, we would love your help!
Donate here:
Created by Mahesh Shenoy
277 comments
Transcript:
how do you run a microwave oven that requires thousands of volts and a phone which requires very tiny amount of voltage by plugging them both into the same supply 230 volts main supply i mean the voltage is too low to run a microwave so how does it work and the voltage is too high for our cell phones so why don't they blow up the secret is a transformer so what's a transformer a transformer is a device that can either step up which means increase or step down which means decrease step down and ac voltage okay what does that mean how does it work well at the core the transformer is basically just two coils kept close to each other one which is connected to the supply and ac supply is called the primary coil and the other which is connected to some device which we want to run we'll call that the secondary coil now the basic principle is the supply voltage is going to generate an alternating current so the current will keep fluctuating back and forth and current passing through a coil will generate a magnetic field and that magnetic field will also fluctuate because magnetic field depends solely on the current and because of that mr faraday comes and says ah changing magnetic flux through a coil there will be an emf induced so an emf gets induced in the primary what is also important to understand is that that same magnetic field will also get linked because the second coil is kept very close to the first one it will also pass through the secondary and as a result as the flux changes over here as the magnetic field changes over there the flux also changes in the secondary and again an induction takes place in the secondary and because of that there will be an emf generator in the secondary and as a result there will be a current generated in the secondary and that's how the bulb starts glowing now at this point you might say hold on a second what a waste of time why can't i just directly connect the supply to the bulb wouldn't that be just easier well if you do that then the voltage at the bulb will be exactly the same as the voltage that you are providing over here and that's not what we want we want to be able to either increase this voltage or decrease the voltage and that's the idea behind a transformer either step it up or step it down so how do these coils help do that well let's analyze the voltages at the primary and the secondary to do that let's start by looking at a single coil over here we know from faraday's law that the emf induced in any coil due to the changes in the flux is given by e if emf e is the negative rate of change of the magnetic flux it basically means if the flux changes very fast then there'll be a higher emf induced that's the whole idea behind this so this is the emf generated in one coil now if there are total i don't know let's say np number of coils where p stands for primary then what is the total emf generated well the total emf generated that will be the voltage in the primary that will be just np times e np times e what about the secondary coil well we can do similar calculation we can say that through each coil the emf generated must be the same we can because the flux here and the flux here must be the same now you might say at this point hold on wouldn't the flux decrease because we're going farther away wouldn't the magnetic field lines go farther away and become weaker you're right but there is a way in which we can make sure that the magnetic field lines over here and the magnetic field lines over here have the same strength and we'll talk a little bit about how that happens a little bit later how we can make sure of that but if we assume that the flux here is exactly the same as the flux here at any moment then the flux through each coil the emf through each coil sorry is going to be e and from that we can now figure out what the secondary voltage is going to be i want you to pause the video at this point and see if you can using that figure out a relationship between the primary voltage and secondary voltage go ahead give it a try all right if the number of turns in the secondary is let's call that as n s where s stands for secondary then the voltage in the secondary is going to be well one coil has emf e ns number of coils will have ns times e notice the voltages in the secondary and primary are not the same if i divide them we'll get the relationship between them we get vs divided by vp equals ns divided by np this means that if the number of turns in the secondary is more than the number of turns in the primary like shown over here then notice the voltage in the secondary would be higher than the voltage of the primary or the voltage of the supply and we call this the step up transformer increasing the voltage that's what happens in your microwaves your microwave oven requires thousands of volts to run but you might know that our ac mains supplies only to about 230 volts so roughly around 200 volts let's say so if you want to increase the voltage say by 10 times as an example then all you have to do is make sure that the number of turns in the secondary is 10 times more than the number of turns in the primary step up transformer on the other hand if the number of turns in the secondary is smaller than the number of turns in the primary notice the voltage in the secondary would be smaller than the supply voltage or the voltage in the primary we get step down a transformer and that's what you would use if you wanted to charge your mobile phone because it requires a very tiny voltage the ac supply gives you a lot so you step it down appropriately by reducing the number of turns and as a bonus notice we're able to charge your electric phones without having a direct connection between these two circuits wireless charging that's right that's how wireless charging works the secondary coil would be inside the phone the primary will be connected to your it would be inside the charging pad you keep the phone on the charging pad wireless charging beautiful right before we wrap up we still have to answer how do i make sure that the flux here and here remains exactly the same right now before i do that i have one question for you do you think this transformer would work on a dc supply what if i used a battery here instead of ac what do you think would the transformer work pause and think about this all right hopefully you've tried so does it work on dc well let's see the main principle is electromagnetic induction and for induction to happen the flux needs to keep changing and that can only happen if the current keeps changing and that does not happen in dc and that's why you cannot use transformers for dc you can only use it for ac okay lastly how do we make sure the flux here and here remains exactly the same otherwise the equation won't work well a way to do that is by introducing a ferromagnetic core a ferro magnet has the ability to sort of suck in magnetic field lines and as a result almost all the field lines from the primary passes through the secondary making sure the flux through each coil is exactly the same of course in a real transformer there will be some flux leakage and the equation will not be valid but for our purposes we can assume ideal transformers and work with it and we have just touched the basics we still have to dig deeper and think about what happens to the currents or the energy when voltage gets stepped up or stepped down and explore how electric power transmission would be impossible today without transformers hopefully you're getting a sense that you can't live without a transformer they are more than meets the eye |
189231 | https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Mannich_Reaction | Skip to main content
Mannich Reaction
Last updated
: Jan 23, 2023
Save as PDF
Jacobsen Rearrangement
Periodic Acid Oxidation
Page ID
: 66861
Organic Reactions Wiki
( \newcommand{\kernel}{\mathrm{null}\,})
The Mannich reaction is an organic reaction which consists of an amino alkylation of an acidic proton placed next to a carbonyl functional group by formaldehyde and a primary or secondary amine or ammonia. The final product is a β-amino-carbonyl compound also known as a Mannich base. Reactions between aldimines and α-methylene carbonyls are also considered Mannich reactions because these imines form between amines and aldehydes. The reaction is named after chemist Carl Mannich.
The Mannich reaction is an example of nucleophilic addition of an amine to a carbonyl group followed by dehydration to the Schiff base. The Schiff base is an electrophile which reacts in the second step in an electrophilic addition with a compound containing an acidic proton (which is, or had become an enol). The Mannich reaction is also considered a condensation reaction.
In the Mannich reaction, primary or secondary amines or ammonia, are employed for the activation of formaldehyde. Tertiary amines lack an N–H proton to form the intermediate enamine. α-CH-acidic compounds (nucleophiles) include carbonyl compounds, nitriles, acetylenes, aliphatic nitro compounds, α-alkyl-pyridines or imines. It is also possible to use activated phenyl groups and electron-rich heterocycles such as furan, pyrrole, and thiophene. Indole is a particularly active substrate; the reaction provides gramine derivatives.
Reaction Mechanism
The mechanism of the Mannich reaction starts with the formation of an iminium ion from the amine and the formaldehyde.
The compound with the carbonyl functional group (in this case a ketone) can tautomerize to the enol form, after which it can attack the iminium ion.
Asymmetric Mannich reactions
Progress has been made towards asymmetric Mannich reactions. When properly functionalized the newly formed ethylene bridge in the Mannich adduct has two prochiral centers giving rise to two diastereomeric pairs of enantiomers. The first asymmetric Mannich reaction with an unmodified aldehyde was carried with (S)-proline as a naturally occurring chiral catalyst.
The reaction taking place is between a simple aldehyde, such as propionaldehyde, and an imine derived from ethyl glyoxylate and p-methoxyaniline (PMP = paramethoxphenyl) catalyzed by (S)-proline in dioxane at room temperature. The reaction product is diastereoselective with a preference for the syn-Mannich reaction 3:1 when the alkyl substituent on the aldehyde is a methyl group or 19:1 when the alkyl group the much larger pentyl group. Of the two possible syn adducts (S,S) or (R,R) the reaction is also enantioselective with a preference for the (S,S) adduct with enantiomeric excess larger than 99%. This stereoselectivity is explained in the scheme below.
Proline enters a catalytic cycle by reacting with the aldehyde to form an enamine. The two reactants (imine and enamine) line up for the Mannich reaction with Si facial attack of the imine by the Si-face of the enamine-aldehyde. Relief of steric strain dictates that the alkyl residue R of the enamine and the imine group are antiperiplanar on approach which locks in the syn mode of addition. The enantioselectivity is further controlled by hydrogen bonding between the proline carboxylic acid group and the imine. The transition state for the addition is a nine-membered ring with chair conformation with partial single bonds and double bonds. The proline group is converted back to the aldehyde and a single (S,S) isomer is formed.
By modification of the proline catalyst to it is also possible to obtain anti-Mannich adducts.
An additional methyl group attached to proline forces a specific enamine approach and the transition state now is a 10-membered ring with addition in anti-mode. The diastereoselectivity is at least anti:syn 95:5 regardless of alkyl group size and the (S,R) enantiomer is preferred with at least 97% ee.
Applications
The Mannich-Reaction is employed in the organic synthesis of natural compounds such as peptides, nucleotides, antibiotics, and alkaloids (e.g. tropinone). Other applications are in agro chemicals such as plant growth regulators, paint- and polymer chemistry, catalysts and main mechanism of formalin tissue crosslinking. The Mannich reaction is also used in the synthesis of medicinal compounds e.g. rolitetracycline (Mannich base of tetracycline), fluoxetine (antidepressant), tramadol, and tolmetin (anti-inflammatory drug)and azacyclophanes. The Mannich reaction is employed to synthesize alkyl amines, converting non-polar hydrocarbons into soap or detergents. This is used in a variety of cleaning applications, automotive fuel treatments, and epoxy coatings. Similar methods of substituted branched chain alkyl ethers into polyetheramines are achieved via a number of reactions.
Jacobsen Rearrangement
Periodic Acid Oxidation |
189232 | https://www.khanacademy.org/math/cc-third-grade-math/represent-and-interpret-data/imp-bar-graphs/e/reading_bar_charts_2 | Use of cookies
Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account.
For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session.
Functional Cookies
These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences.
Targeting Cookies
These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission.
We do not use cookies to serve third party ads on our Service.
Performance Cookies
These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates.
For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users.
We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
Cookie List
Consent Leg.Interest
label
label
label |
189233 | https://www.youtube.com/watch?v=PdhmIgH6teM | The end correction of a resonance column is 1.0 cm. If the shortest length resonating with the t...
PW Solutions
508000 subscribers
8 likes
Description
811 views
Posted: 31 Jul 2022
The end correction of a resonance column is 1.0 cm. If the shortest length resonating with the tuning fork is 15.0 cm, the next resonating length will be
(a) 31 cm
(b) 45 cm
(c) 46 cm
(d) 47 cm
📲PW App Link -
🌐PW Website -
📌 PHYSICS WALLAH OTHER CHANNELS :
🌐 PhysicsWallah -Alakh Pandey:
🌐 Alakh Pandey:
🌐 JEE Wallah:
🌐 Competition Wallah:
🌐 PW Foundation:
🌐 NCERT Wallah :
🌐 Defence Wallah-NDA:
🌐 PhysicsWallah English:
🌐 PW Vidyapeeth:
🌐 Commerce Wallah by PW:
🌐 CA Wallah bt PW:
🌐 PW Sarkari Exam:
🌐 PW - Hindi Medium:
🌐 PW Bangla:
🌐 PW Maharashtra:
🌐 PW Telugu:
🌐 PW Kannada:
🌐 PW Gujarati:
🌐 PW Facts:
🌐 PW Insiders:
🌐 PW Little Champs:
🌐 PW Pathshala:
🌐 Banking Wallah :
🌐 SSC Wallah :
🌐 JEE Challengers by PW :
🌐 UPSC Wallah :
🌐 UPSC Wallah हिन्दी :
🌐 GATE Wallah :
🌐 GATE Wallah - EC,EE & CS :
🌐 GATE Wallah - ME, CE & XE :
🌐 GATE English :
🌐 College Wallah :
🌐 PW IIT JAM & CSIR NET :
🌐 MBA Wallah :
🌐 Teaching Wallah :
📌 PHYSICS WALLAH SOCIAL MEDIA PROFILES :
🌐 Telegram :
🌐 Facebook :
🌐 Instagram :
🌐 Twitter :
🌐 Linkedin :
📌 For any queries or complaints Visit :
Transcript:
हेलो बच्चों यहां लिखा है एंड करेक्शन ऑफ रेजोनेंस कॉलम इ 1 सेंटीमीटर इफ द शॉर्ट वेवलेंथ फॉर द रेजोनेंस शॉर्ट लेंथ रेसोनेट विद द ट्यूनिंग फर्क इ 15 सेंटीमीटर द नेक्स्ट रेजोनेंस लेंथ विल बी क्लियर तो सबसे पहले तो एंड करेक्शन समझ लीजिए क्या होता है तो जो रेजोनेंस ट्यूब होती है बटा ये वन एंड क्लोज ट्यूब होती है ऐसी इसमें एक तरफ में पानी भरा होता है जिसके लेवल को हम लोग चेंज कर सकते हैं और जब हम लोग ट्यूनिंग फर्क को रखते हैं ना तो ट्यूनिंग फॉक जो होता है ये इससे थोड़ा सा ऊपर होता है तो ट्यूनिंग फॉर को हम लोग जितना ऊपर रखते हैं तो एयर के वाइब्रेशन यहीं से स्टार्ट हो जाते हैं तो यह जो एडिशनल वेवलेंथ एडिशनल लेंथ आती है इसको हम लोग बोलते हैं एंड करेक्शन तो फंडामेंटल मोड में यहां पर केवल और केवल यह पैटर्न बनता है मतलब आपके पास केवल लडा बा फो बनेगा तो जो लडा बा फो होता है ये लेंथ ऑफ द पाइप के बराबर होता है तो इस यहां से हमने लिखा लड बा 4 = l प् e बराबर है एंड करेक्शन e होता है क्लोज ऑर्गन पाइप के लिए एक e होगा अगर ओपन होता तो वो प् 2e हो जाता बिकॉज वो दोनों तरफ में हम लोगों को मिलता यहां से l इटू दिया था 15 प्सव तो ये आपके पास हो गया 16 तो लडा बा 4 इ अपने को मिल गया 16 सो लडा इक्वल टू हो जाएगा 64 और लडा को हम लोग लिख सकते थे v बा f तो ये बा ए इ अपने को मिल गया 64 य पहला रिलेशन मिल गया अच्छा अगला जो रेसोनेट लेंथ होगा वो हमको समझ में आ रहा है अगले रेजोनेंस के केस में जैसे यहां पर एक नोड बना था यहां पर अब दो नोड बन जाएंगे मतलब यह वाला पैटर्न आपको दिखाई देगा अच्छा इस पैटर्न में अगर मैं टोटल नंबर ऑफ लडा काउंट करूंगा तो ये हम लोगों को यहां से मिलेगा 3 लडा बा फ तो 3 ल बा फ इ वापस से हम लिखेंगे एल प् e यहां से मैंने लिख दिया इसको बा फ एक काम करता हूं लडा की वैल्यू में यही से पुट कर देता हूं तो लडा की वैल्यू हो गई 64 इ एल प् एंड करेक्शन दिया था वन यहां से अगर मैं इसको कैंसिल करू तो कितने टाइम जा रहा है 16 टाइम्स देखिए हा 16 टाइम्स चला जाएगा तो 16 इन 3 मिलेगा 48 इक्वल ए प्व तो जो नया वाला लेंथ आपको मिलेगा ट इ 47 सेंटीमीटर लेथ कैसे जैसे मैंने आपसे बोला था कि जो रेसोनेंस ट्यूब होती है उसमें नीचे पानी जो भरा होता उसके लेवल को हम कम कर लेते हैं तो जो एयर कॉलम की लेंथ होती व बढ़ जाती है तो नेक्स्ट लेंथ मिलेगी 47 सेंटीमीटर प जो की डी ऑप्शन में दी गई है सो फाइनली इ डी ऑप्शन इज करेक्ट क्लियर बेटा ओके ऑल द बेस्ट |
189234 | https://www.youtube.com/watch?v=j1FPga9gd9s | AP Physics 1 - Unit 1 - Lesson 14 - Projectile Motion
Allen Tsao The STEM Coach
11600 subscribers
5 likes
Description
289 views
Posted: 8 Aug 2025
Master projectile motion in AP Physics 1! This video simplifies 2D kinematics, helping high school physics students tackle complex problems with confidence.
Learn how to break down projectile motion into manageable horizontal and vertical components. We'll cover essential kinematic variables, problem-solving strategies, and work through example problems including horizontal and angled launches. Understanding these concepts is crucial for success in AP Physics 1.
Chapter Titles
Understanding 2D Motion (0:00)
Key Kinematic Variables (0:14)
Problem-Solving Steps (0:55)
Example 1: Horizontal Launch (2:09)
Calculating Time in Projectile Motion (4:46)
Example 2: Angled Launch (6:12)
Decomposing Initial Velocity (7:28)
Using Desmos for Calculations (9:09)
Key Takeaways
Projectile motion deals with objects moving in two dimensions.
Split motion into horizontal (x) and vertical (y) parts.
Each part is treated as a separate 1D motion problem.
Time is the only variable shared between horizontal and vertical motions.
Gravity only affects vertical motion (acceleration is 9.8 m/s² downwards); horizontal acceleration is zero (ignoring air resistance).
Use kinematic equations for both directions after decomposing initial velocity and displacement.
What you'll practice:
Sketching motion and listing kinematic variables for horizontal and vertical directions.
Assigning positive and negative directions.
Decomposing displacement and initial velocity vectors into x and y components.
Applying kinematic equations to solve for unknowns in projectile motion problems.
Using Desmos for quadratic equation solving in physics contexts.
AP Physics 1 Projectile Motion, 2D Kinematics Explained, How to Solve Projectile Motion, Physics Projectile Problems, Horizontal and Vertical Motion Physics, Gravity and Projectile Motion, AP Physics 1 Exam Help, Projectile Motion Examples, Physics Cliff Problem, Angle Launch Projectile, Physics 2D Motion Tutorial, Kinematic Equations Projectile, Physics kinematic variables, Velocity decomposition, Acceleration due to gravity
APPhysics1 #ProjectileMotion #physicstutorial
If you want to learn more resources we offer, you can check it out at:
1 comments
Transcript:
Understanding 2D Motion All right, so let's take a look at the culmination of this is projectile motion, which is our 2D motion. So now that we have the vector information out of the way, we can talk about what are how do we solve projectile motion problems. So our vector quantities, Key Kinematic Variables displacement, initial velocity, final velocity, accelerating, our normal acceleration, our normal vector quantities and kinematics variables um same kinematic variables. However, because now we have 2D vectors, they may be pointing up and to the right, down to the right, straight up, straight down. They may have x and y components. Now, in general, the magnitude of the vectors, the length of the vector. Direction is which way it's pointing. And one thing to note is speed. In one dimension of motion, speed was the absolute value of the velocity because there was no direction. But now speed is the magnitude or length of the velocity vector. So, we keep that in mind. That's a slight twist when we go and move into two-dimensional vectors. Now, our Problem-Solving Steps problem solving steps are very very very very similar to what we did where the whole idea is we have these vectors and we think of them as just split into x and y components and then treat them as two separate one-dimensional vectors. So, our problem solving process steps are going to be we're going to sketch out the motion. We're going to list the five kinematic variables, but you're going to do it for both the horizontal and vertical directions separately. We're going to assign positive and negative directions for the horizontal and vertical directions. And then for each kinematic vector v variable vector, we're going to draw the arrow and decompose them into the x and y component. So that means we're going to find the displacement, decompose into the x and y, find the initial velocity, decompose it into the x and y. So, we're splitting up all of the kinematic variables, the vectors into x and y components. Now, couple of tips and we'll get to those tips when we go through some problems. Now, time is the only variable that's shared between the x and y components. We want to and then the rest is the same. Identify the unknowns where look and what we're trying to solve for. We apply the kinematic equations for the vertical and horizontal based on the three known quantities and the unknown we're looking for. Um, so let's just take a look at an Example 1: Horizontal Launch example problem. So we have a ball that's thrown horizontally from a cliff. So it's thrown horizontally here at 10 meters/s from a cliff that's a height 100 m. Okay, so this is 100 mters here. How far does it ball how far does the ball land from the base of the cliff? So it's going to go down like this and it's going to land right around here. And I want to know what this distance here is. I would like to know what that is. This is what I would like to know. Right? So, let's list out. First step is to list out in the x and the y all of the kinematic variables that like we did in one dimensional motion. I know it's been a while ago. We did grass motion. We did all kinds of stuff, but just remember we're going to list out those kinematic variables. And now we're going to look at each one of them. We want to assign positive directions. I'm going to make right positive. You can make up positive or down positive. It doesn't really matter. We'll make up positive in this case. And then now we're going to look at the displacement vector first. So remember the displacement vector is an arrow from where it starts to where it ends or where we're interested in it. So that's our displacement vector and we want to decompose it into components, right? So we'll make a component goes down and then a component that goes to the right. Okay? So in the horizontal displacement, notice the horizontal displacement, that length, that's what I want to know. So this is what I'm looking for right here. The vertical displacement is down 100 meters. It's pointing down and that length is 100 meters. So that is negative 100. And that displacement that's like distance, right? So it's this vertical distance. That's going to be negative 100 because down is the negative direction we set up was positive. Okay. So now we look at okay. So that handles the displacement vector. Now let's handle the initial velocity vector. The velocity at the start of the motion. It's 10 meters/s to the right. Okay, so that means there's no vertical component. The vertical component is zero because the vector is only pointing to the right. And then um it's pointing to the right at 10. So we put 10 here. Final velocity is the velocity down here. Don't know what that is. So not given in the problem. And the acceleration, the rule is still the same. When we are in the air, the acceleration is downward at 9.8 m/s squared or 10. Not you can use round to 10. That's fine. It's downward at 10. Okay. Or downward at 9.8. So that's zero in the horizontal direction. It's not pointing to the right or left at all. It's only pointing straight down. And that's going to be negative 9.8 here. Okay. And then we don't know the time. Calculating Time in Projectile Motion Now the thing about the time vector, okay, this is the one thing I was going to that's what this part is here. Time is the same between the x and y components because time is not a vector. So the time variable here and the time variable here are the same value. Now on this side I want to know this but I need three known quantities and I don't have it. However on this side I have three known quantities and if I solve for the time here I can then plug it into here and then I'll have three known quantities and I can solve for this guy. So what kinematic equation is going to relate that right? That's going to be -100 this is going to be 0 is 129.8 8 t ^2 and then you can solve for t. That's going to be um 100 4.9 and then square root of that. I get 4.52 seconds. And then you can plug that in over here. 4.52 because it's this it's the same time for the x and the y. That's the only variable that's shared. Everything else is completely separate. That's shared. And then we use the same kinematic equation but now on the x side because now we have enough information. We're going to say the displacement here is going to be 10 4.52 and then the a is zero. So we don't need to include that term there. So that's going to be 45.2 m. Okay. So that's it. That's the process. That's what we're Example 2: Angled Launch going to do. Let's look at another one. So ball is thrown at an angle now 30° above the horizontal. So this is 30° with a velocity of 10 meters/s from a cliff. That's a height of 100 meters. So, same cliff. Uh 100 meters. Okay. How far does it land from the base of the cliff? So, step one, draw the motion. So, we we're drawing it out and it's going to land here. We would like to know this horizontal distance from the base. That's what we're looking for. And so, now we're going to list out all our kinematic variables. delta x v 0 v a t and then we're going to go we're going to assign our directions. So we'll make right positive and up positive again. And now let's talk about the displacement vector first. Displacement vector is a vector from there to there. Okay? And we decompose it into because you remember displacement is a vector from where it starts to where it ends. It's a straight arrow and then you decompose it to going down and then to the right. Okay? So it goes down 100 meters. So that's -100 m on the vertical displacement. Horizontal displacement again is what I'm looking for. But now what's different in this Decomposing Initial Velocity problem is the initial velocity. We got to decompose the initial velocity vector now. Okay. So that's got to decompose. And here we're going to make it to the right and up. That's our vector decomposition. This is going to be 10 cosine 30° and then 10 sin 30° because the opposite side because it make that right triangle, right? So it's pointing to the right. Right is positive in the x direction. So that's 10 cosine 30°. Vertical direction it's pointing upwards 10 sin 30°. And we said up was the positive direction. Final velocity we don't know. The acceleration is pointing downward at 9.8 m/s squared. So the acceleration is zero. There's no horizontal component, but there's a vertical component of 9.8. And that's why I put that in the notes in the problem solving steps. Note horizontal direction if there's no air resistance the acceleration zero in the x direction and in the vertical direction it's downward at 9.8 m/s squared. Okay. That's what that part is telling you there. Okay. So again, similar thing. We have three known quantities um on this side. So we could solve for the final velocity. We can solve for the time. We're going to use the time. Plug that into here. That will then give us three pieces of information which we can then use to solve for the horizontal displacement. So we're going to use the same kinematic equation, right? Which one relates B it's based on the known quantities and the one unknown we're looking for right. So this is -100 is 10 sin 30° t + 12 9.8 t ^2. Now this is not a Using Desmos for Calculations math class. That is to say that your goal is to just be able to solve it. And so we're going to use Desmos given that that's the common tool on your AP exam for Blue Book. So it's very easy to calculate. You could just say you could just make it move the 100 over to the other side. Set it equal to zero. So we're just going to say y is equal to 100 + 10 sine of 30. Now you got to switch it to degree mode when you're doing this. So if you guys are that's how we switch to degree mode times then 30 um + 12 oops times9.8 t^2 and so we have time zero look we have two numbers the negative number doesn't really make sense okay because we want to know when the y is zero so that's going to occur at 4.63 seconds so that's going to be t is equal to 4.63 63 seconds. And so then we're going to look at the horizontal side. Use the same equation. It's going to be 10 cosine 30°. That's the v 0 times the time. We're just using the x column variables now. And this part is zero in the x column because the a is zero here. And so then we're just going to say, oh, we're going to take that number. So, we're going to do 10 cosine oops cosine of 30 4.63 and I get about 40. We'll just say that 40.1 m like that. Thanks for checking out this video. I hope you found it really helpful. If you'd like more support, maybe you need more multiple choice practice, maybe you just need more guidance and things like that, I have plenty of information on my website. If you look in the description below and go to www.blstemcoach.com, bottlstemcoach.com. |
189235 | https://turkmenportal.com/en/compositions/1796 | A mark on history: the great four of the Renaissance
Famous People of Turkmenistan
Events
Special publications
Education
Sport
Hi-tech
Tourism
Auto
The property
Job
Economy
Culture and Leisure
Services
The shops
health and beauty
Latest News
In Trend
Timeline
20 September 2025
123883
A member of the Council of Elders of the Halk Maslahaty proposed stopping the increase...
20 September 2025
102076
Large-scale construction of educational, medical, and infrastructure facilities continues in Turkmenistan
05 September 2025
96305
Students from Turkmenistan sent to the Republic of Korea for a year-long internship
05 September 2025
94486
Turkish Airlines published Ashgabat-Istanbul-Ashgabat flight schedule for September 2025
20 September 2025
80885
New etraps have been formed in four velayats of Turkmenistan
08 September 2025
74550
Astronomers report about three upcoming supermoons in 2025
09 September 2025
70049
New settlements formed in two regions of Turkmenistan
94686 94685 94282 94288 94682 94360 94371
29 September 2025
2762
In Turkmenistan, winners of horse races were awarded in honor of Independence Day
28 September 2025
2222
The President of Turkmenistan and the head of a Japanese company discussed new areas of...
28 September 2025
2236
The head of Turkmenistan attended the horse races held in honor of Independence Day
28 September 2025
4528
UNFPA held a training session for demographic research specialists in Ashgabat
28 September 2025
6251
Two of Ankara's tallest buildings have been painted in the colors of the Turkmen flag
28 September 2025
11117
The Chairman of the Mejlis of Turkmenistan will visit Russia in October
28 September 2025
22539
A military parade was held in Ashgabat to commemorate the 34th anniversary of Turkmenistan's independence
94896 94894 94893 94885 94883 94880 94879
News Archive
| | September, 2025 | |
---
| Su | Mo | Tu | We | Th | Fr | Sa |
| | 1 | 2 | 3 | 4 | 5 | 6 |
| 7 | 8 | 9 | 10 | 11 | 12 | 13 |
| 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| 21 | 22 | 23 | 24 | 25 | 26 | 27 |
| 28 | 29 | 30 |
A mark on history: the great four of the Renaissance
16:57 02.02.2024
0
69903
Leonardo, Michelangelo, Raphael and Donatello are the creators of one of the greatest cultural and historical eras, the Renaissance. This period gave the world many brilliant inventors, artists, architects and sculptors. However, the names of this famous four are known, perhaps, to everyone primarily thanks to the characters of the same name in the popular animated series. If for you Leonardo, Michelangelo, Raphael and Donatello are primarily associated with a team of mutant turtles, then you will be interested to know about the real people after whom they were named.
Photo:ru.wikipedia.org
Leonardo da Vinci- Italian painter, sculptor, architect, scientist and engineer was born on April 15, 1452 in the town of Anchiano near the city of Vinci, not far from Florence. His father was the notary Pierrot, and his mother was the peasant woman Katerina. Leonardo spent the first years of his life with his mother. Soon his father married a rich and noble girl, but this marriage turned out to be childless. Therefore, Piero took his three-year-old son to be raised. Leonardo yearned for his mother all his life and tried to recreate her image in his works of art.
Online art courses
At the age of 15, Leonardo da Vinci was apprenticed to the artist Andrea del Verrocchio. In Verrocchio's workshop he received a multifaceted education, including painting, sculpture, and the technical and mechanical arts. Leonardo da Vinci owes his outstanding achievements in various fields to his teacher, who helped him develop analytical skills and instilled in him a love of universalism.
In 1472, Leonardo da Vinci became a member of the guild of painters of Florence.
One day Verrocchio received an order for the painting “The Baptism of Christ”. He commissioned Leonardo to paint one of the two angels. This was a common practice of the time: the teacher created the picture together with the students. Two Angels, painted by Leonardo and Verrocchio, clearly demonstrated the superiority of the student over the teacher. As the Italian painter Vasari writes, the amazed Verrocchio abandoned his brush and never returned to painting.
Online art courses
It is believed that about 15 paintings by Leonardo da Vinci (in addition to frescoes and drawings) have survived. Of these, five are kept in the Louvre, one each in the Uffizi, Alte Pinakothek, Czartoryski Museum, London and Washington National Galleries, as well as in other museums.
Paintings by Leonardo da Vinci: “Mona Lisa”, “The Last Supper”, “Portrait of a Musician”, “Savior of the World”, “Madonna and Child”, “Madonna of the Rocks”, “Portrait of Ginevra de Benci”, “Benois Madonna”, “The Beautiful Ferroniere”, “John the Baptist”, “Annunciation”.
As a scientist and inventor, Leonardo da Vinci considered experience to be the only criterion of truth. He studied mathematics, mechanics, hydraulics, anatomy and physiology of humans and animals, botany, geology, etc.
Leonardo da Vinci was the first to describe the equilibrium of liquids in communicating vessels, approaching Pascal's law, which was discovered only in 1653. He also studied binocular vision and human and animal anatomy. He created detailed anatomical drawings that formed the basis of modern scientific illustration, was the first to describe a number of bones and nerves, and also studied questions of embryology and comparative anatomy. A brilliant scientist proposed a way to determine the age of trees by tree rings.
Leonardo da Vinci's inventions were ahead of their time. The only invention that gained recognition during his lifetime was the wheel lock for a pistol. Leonardo da Vinci was involved in the design of aircraft, creating a model of an ornithopter; his notes also contain diagrams of a glider and a parachute. He also developed a design for a diving suit, a lifebuoy, webbed gloves (prototype flippers), a bicycle, a metal cart for transporting soldiers (prototype tank), a catapult, a searchlight, a rotating crane, various machines, etc.
In 1502, Leonardo da Vinci, commissioned by the Ottoman Sultan Bayezid II, created a design for a stone bridge across the Golden Horn in Istanbul. The bridge was supposed to have a span of 240 meters and consist of three arches resembling a bird's tail.
The Sultan considered the project too complex and risky, so he abandoned it. In 2001, a pedestrian bridge with a span of 100 meters was built in Norway based on da Vinci's sketches.
Leonardo da Vinci died on May 2, 1519 in Amboise, France. He was buried in the Church of Saint-Florentin, but his tomb was destroyed during the French Revolution. In 1874, da Vinci's supposed remains were reburied in the Saint-Hubert chapel.
Photo:ru.wikipedia.org
Michelangelo Buonarroti, one of the most famous sculptors, artists, poets, painters and architects of all times, was born in 1475 in the city of Caprese in the family of an impoverished Florentine nobleman, Lodovico Buonarroti. Michelangelo's mother died from exhaustion due to frequent pregnancy in the year of his sixth birthday.
In elementary school, Michelangelo did not show much interest in his studies. He preferred communicating with artists and copying icons and frescoes to his studies.
In 1488, the boy’s father apprenticed in the studio of the artist Domenico Ghirlandaio, and a year later he moved to the school of the sculptor Bertoldo di Giovanni, which existed under the patronage of Lorenzo de’ Medici. The talent of young Michelangelo attracted the attention of Lorenzo de' Medici. He accepted him into his home and financially helped Michelangelo develop.
Michelangelo is considered not only one of the greatest masters of the High Renaissance, but also the true “father of Baroque” in architecture. His works were recognized as masterpieces of the Renaissance during his lifetime. The Italian sculptor was the first representative of Western European art whose biography was published during his lifetime.
Online art courses
Michelangelo lived almost 89 years. During this period, thirteen popes were replaced - he carried out orders for nine of them.
The ceiling frescoes of the Sistine Chapel are one of the artist’s most monumental works. Goethe wrote: “Without seeing the Sistine Chapel, it is difficult to form a clear idea of what one person can do”. Among his architectural achievements are the design of the dome of St. Peter's Basilica, the stairs of the Laurentian Library, Campidoglio Square and others.
The most famous sculptural works include “David”, “Bacchus”, “Pieta”, statues of Moses, Leah and Rachel for the tomb of Pope Julius II. Giorgio Vasari, Michelangelo's first official biographer, wrote that “David” “robbed the glory of all statues, modern and ancient, Greek and Roman”.
Michelangelo was not only a sculptor, but also a poet. About 300 poems of the genius have survived to this day. The main themes are the glorification of man, the bitterness of disappointment and the loneliness of the artist.
Michelangelo died in February 1564 in Rome, just short of his 89th birthday. Before his death, he dictated his will in his characteristic laconic manner: “I give my soul to God, my body to the earth, my property to my relatives”. At the beginning of March, the sculptor's body was secretly transported to Florence and solemnly buried on July 14, 1564 in the Franciscan church of Santa Croce.
Photo:ru.wikipedia.org
Rafael Santiwas born in 1483, in the city of Urbino. The Italian painter, draftsman and architect of the Umbrian, Florentine, and then Roman schools lived only 37 years and during his short life became one of the most famous and wealthy artists in Italy.
Rafael lost his parents at an early age. His mother, Margie Charla, died when he was 8 years old, and his father, Giovanni Santi, died three years later.
Raphael Santi's earliest work, the fresco of the Madonna and Child, is still in the Raphael House Museum in Urbino. Among the artist’s first works are “The Banner with the Image of the Holy Trinity” and the altar image “The Coronation of St. Nicholas of Tolentino” for the Church of Sant'Agostino in Città di Castello.
Raphael's works are admired for their clarity of form, simplicity of composition, and embodiment of the Neoplatonic ideal of human greatness and beauty. Together with Michelangelo and Leonardo da Vinci, Raphael forms the traditional triad of great masters of the High Renaissance. Raphael was thirty-one years younger than da Vinci, eight years younger than Michelangelo, he became a representative of the new generation of the era, and with his premature death in 1520, the decline of the High Renaissance in Italy began.
Raphael was an incredibly productive artist. He led a large workshop and left behind a rich creative heritage.
One of Raphael's most recognizable paintings is the Sistine Madonna. The artist’s most famous works include: the fresco “The School of Athens”, “The Transfiguration”, the frescoes “Stanzas”, “Donna Velata”, “The Three Graces”, “The Triumph of Galatea”, etc.
Like many other Renaissance artists, Raphael was not only an excellent painter, but also a talented poet. His drawings, accompanied by sonnets, have survived.
Raphael died in Rome on April 6, 1520 at the age of 37. Modern researchers are inclined to believe that the cause of death was a respiratory disease and the associated high temperature. A fatal role was played by a medical error - the doctors sent by the Pope, having made an incorrect diagnosis, tried to reduce the temperature by bloodletting, which is dangerous for severe colds (this was already known in the XVI century).
Photo:ru.wikipedia.org
Donatello, full name Donato di Niccolo di Betto Bardi, was an Italian Renaissance sculptor of the Florentine school, considered one of the founders of Renaissance monumental sculpture and relief, as well as the genre of sculptural portraiture.
Donatello was born in Florence in 1386 into a simple family. His father was a wool carder. Since childhood, he showed talent for sculpture and painting. Donatello studied in one of the many Florentine sculpture workshops, where he met the architect Brunelleschi. Together they went to Rome to study with the famous master Lorenzo Ghiberti.
After returning from Rome, Donatello immediately began working on numerous orders. He created sculptures for aristocrats, bankers and even Casimo de' Medici himself. During this time, Donatello experimented with two styles: classical and realistic. His realistic sculptures were so expressive that they shocked the imagination of his contemporaries, accustomed to the idealized images of Greek and Roman sculpture. Donatello's works still adorn many architectural landmarks in Florence, including Giotto's Tower, the Bargello Museum and the Old Palace of Casimo de' Medici.
In 1444 Donatello moved to Padua, where he was commissioned to decorate the Church of St. Anthony. He managed to revive the technique that ancient Roman masters once owned. Thanks to this, his works in Padua are considered one of the most outstanding in the master’s work.
They did not want to let Donatello leave Padua. The master was inundated with orders just so that he would not go anywhere, but he was too attached to Florence.
After 13 years of work in Padua, Donatello returned to Florence, where in 1457 he began work on the sculpture of John the Baptist. it is now in the Bargello Museum in Florence. After its completion, he created bas-reliefs to decorate the Church of St. Lawrence and tombstones for the church aristocracy.
Donatello was known for his hard work and unpretentiousness. He took on any orders, even the most insignificant ones. He was not interested in money, he created for the sake of art. An interesting fact is that he kept all the proceeds in a basket in his workshop. Each student could take from there as much as he needed.
Online art courses
The statue of “David”, created for the garden of Casimo de' Medici, was revolutionary for its time. Before Donatello, no one could dare to sculpt a completely nude sculpture.
Very little is known about Donatello's personal life. It is unknown whether he had a family. However, he was very respected and loved in Florence. After his death, the townspeople decided to preserve his sculptures for posterity and prohibited their removal from the city.
The great sculptor worked until he was very old. The master died in 1466 and was buried in the Church of San Lorenzo, which he once decorated with bas-reliefs.
Resources: ru.wikipedia.org; obrazovaka.ru; ria.ru; dzen.ru; artifex.ru
Also read:
8 modern neoclassical composers for fans of quality instrumental music
Share on social network:
VKontakte
Twitter
Odnoklassniki
Moi Mir
Comments
To leave a comment, log in or register
Related Publications
What would happen if the Earth lost oxygen for 5 seconds?
Speech development and gadgets: how technology affects a child’s vocabulary
Seven reasons why your child has become worse in school
6 large-scale and tragic man-made disasters in the XX century
Four mistakes in managing money
1998 1951 1946 1945 1889
Poster
Афиша кинотеатра «Ашхабад»
Articles
OGT 2025: A Gateway for Turkmen Entrepreneurs to Global Partnerships
From Natural Gas to Hydrogen: Key Trends of Gastech 2025
Concluding remarks from the XXIV International exhibition and conference “White City Ashgabat”
How to protect your skin from the summer sun
x
Application Turkmenportal
Reading the news is much more easier.
DownloadNo, Thanks
Advertise
Cooperation
Jobs
Contacts
Information support
Placement of PR articles
Download our app
All rights reserved . When using materials from the site link to turkmenportal.com is required.
Запись о регистрации средства массовой информации ЭЛ № ФС 77 - 89056 от 07.02.2025 г. Роскомнадзор | 18+
September 29, 2025 06:26:23
0° Ashgabat
| Central Bank Currency Rate | ATM list | 18+
0°
Select language:
All categoriesnews Articles Media Business Information programm "Watan" UIET - members Tourism Business Directory Culture and Leisure Establishments The shops health and beauty Education Partnership Turkmenistan tenders Sport Jobs Poster The property Auto Job Directory Services Adverts Discounts and promotions
news
Articles
Media
Business
Information programm "Watan"
UIET - members
Add advert
Search
Login |
189236 | https://www.nationaljewish.org/for-professionals/diagnostic-testing/advanced-diagnostic-laboratories/cd20-assay | Advanced Diagnostic Laboratories | National Jewish Health | Clinical Laboratory | ISO 15189 | Central Laboratory Services
Skip to content
For Professionals
Skip Navigation
Advanced Diagnostic Laboratories
Overview
Requisition
Diagnostic Testing
Test Catalog
Featured Tests
Alpha-1 Antitrypsin
C3a
C4a Level
Donor Bone Tissue TB Testing
Myositis Panel
TRIKAFTA Drug Monitoring
Requisitions & Consent Forms
Specimen Submission
Shipping Guidelines
Order Supplies
Test Updates
Holiday Schedule
Biotech / Pharmaceutical
Academic Research
Our Laboratories
Laboratory Directors
Certifications & Licenses
Clinical Chemistry & Hematology
Complement
Dr. Giclas Founder of the National Jewish Health Complement Laboratory
Immunology
Anti-GMCSF Autoantibodies
Metal Hypersensitivity
Beryllium LPT
Nickel LPT
Microbiology
Strategic Innovations
Mycobacteriology
AST for NTM & Aerobic Actinomycetes
AST for TB
Isolate Identification
Molecular & Growth Detection
Therapeutic Drug Monitoring
Contact Us
Billing
Careers
Hours of Operation
Shipping Guidelines
Client Bill Pay
Home
For Professionals
Diagnostic Testing
Advanced Diagnostic Laboratories
Advanced Diagnostic Laboratories
Overview
Diagnostic Testing
Biotech / Pharmaceutical
Academic Research
Our Laboratories
Contact Us
Skip Navigation
Advanced Diagnostic Laboratories
Overview
Requisition
Diagnostic Testing
Test Catalog
Featured Tests
Alpha-1 Antitrypsin
C3a
C4a Level
Donor Bone Tissue TB Testing
Myositis Panel
TRIKAFTA Drug Monitoring
Requisitions & Consent Forms
Specimen Submission
Shipping Guidelines
Order Supplies
Test Updates
Holiday Schedule
Biotech / Pharmaceutical
Academic Research
Our Laboratories
Laboratory Directors
Certifications & Licenses
Clinical Chemistry & Hematology
Complement
Dr. Giclas Founder of the National Jewish Health Complement Laboratory
Immunology
Anti-GMCSF Autoantibodies
Metal Hypersensitivity
Beryllium LPT
Nickel LPT
Microbiology
Strategic Innovations
Mycobacteriology
AST for NTM & Aerobic Actinomycetes
AST for TB
Isolate Identification
Molecular & Growth Detection
Therapeutic Drug Monitoring
Contact Us
Billing
Careers
Hours of Operation
Shipping Guidelines
Client Bill Pay
CD20 + B Cell Enumeration Assay
Download Requisition
Test Code
CD20
Description
CD19 and CD20 are both B cell receptors that play a central role in cell maturation and function. CD19 is expressed at all stages of B cell maturation and thus is the preferred marker for lymphocyte phenotyping assays. The exact biological function of CD20 is not known. It may be involved in the regulation of activation, proliferation and differentiation of B cells. Its expression is more variable than CD19 and is increased following activation.
Patients with autoimmune diseases or cancer who are being treated with drugs like rituximab, which destroy CD20+ B cells, may benefit from flow cytometric detection of residual CD20+ B cells as an indication of treatment efficacy. This assay may be insufficiently sensitive to detect minimal residual disease (MRD) following chemotherapy for malignancy.
Collection/Transport
Collect
Draw one green top (sodium or lithium heparin), lavender top (EDTA) or ACD (acid citrate dextrose) vacutainer tube.
Please Note: For absolute counts, a complete blood count (CBC) is also required, so an additional EDTA lavender tube should also be submitted. Alternatively, CBC results from a blood collection within 3 days of the phenotyping blood collection can be submitted.
Patient Prep
Handling: Do not centrifuge! Keep at room temp (18-22°C).
Preferred volume: 4 mL
Pediatric Collection
Minimum volume: 1mL blood
Unacceptable Conditions
Frozen, clotted, hemolyzed, spun down, citrate anticoagulant (blue top), past stability
Storage Transport Temp
Maintain and transport specimens at room temperature (18-22°C).
Ship specimens Priority Overnight via Fed Ex in container sufficiently insulated to avoid temperature extremes (<10°C and >37°C).
Stability
48 hours at room temperature (18-22°C).
Notes
CBC with Differential results from within 3 days of draw must also be provided or a lavender top included with required specimen to run CBC at ADx.
Samples are accepted Monday through noon on Fridays.
Test Detail
Performed
Tuesday - Friday
Samples accepted Monday through noon on Friday
Methodology
Flow Cytometry
Turnaround Time
2-4 business days
Department
Immunology Lab - Phenotyping Assays
Synonyms
B Cell Enumeration, CD20 B Cells, Flow Cytometry (CD20), B Cell Marker;CD20M
Study Offerings
CAP/CLIA
Related Tests
TBSBS, CD3SB, CD4M, CD48, NKMK, MEMB, XBCP
Result Interpretation
Reference Interval
Lymphocyte Subset Normal Range Chart
Interpretation
This test is useful for assessing B Cell percentages and numbers and assists in the evaluation of humoral immunodeficiences. It is also useful for monitoring depletion of CD20 B cells following anti-CD20 monoclonal (e.g. Rituximab) therapy.Sensitivity threshold is 1% of lymphocytes, which may be insufficient for the detetion of minimal residual disease (MRD) following chemotherapy for malignancy.
Fees/Coding
CPT Code
86355
Licensing & Certification
New York Approved
No: Run with Waiver
FDA
LDT
Search for Another Test:
Search Text is Required.
Search
Advanced Diagnostic Laboratories
1400 Jackson St
Denver, CO 80206
800.550.6227 Toll-free
1.800.652.9556 Fax
+1.303.398.1339 International
© 2025 National Jewish HealthAbout Us
Careers
Certifications & Licenses
Laboratory Directors
Laboratory Expertise
Contact UsOrdering Tests
Test Catalog
Test Updates
Specimen Guidelines
Specimen Shipping
Services
Complement Analysis
Immune System
Molecular Genomics
Mycobacteriology
HomeBack To Top
×
Find a Test |
189237 | https://www.math-salamanders.com/converting-decimals-to-fractions-worksheet.html | Home |
About |
Contact Us |
Privacy |
Newsletter |
Shop |
Donate
Home
🔍 Search Site
Preschool
Kindergarten
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th Grade
All Generated Sheets
Place Value Generated Sheets
Addition Generated Sheets
Subtraction Generated Sheets
Multiplication Generated Sheets
Division Generated Sheets
Money Generated Sheets
Negative Numbers Generated Sheets
Fraction Generated Sheets
All Calculators
Fraction Calculators
Percent calculators
Area & Volume Calculators
Age Calculator
Height Calculator
Roman Numeral Calculator
All Zones
Place Value Zones
Number Bonds
Addition & Subtraction
Times Tables
Fraction & Percent Zones
All Math Quizzes
Kindergarten Math Quizzes
1st Grade Quizzes
2nd Grade Quizzes
3rd Grade Quizzes
4th Grade Quizzes
5th Grade Quizzes
6th Grade Math Quizzes
Coloring Pages
Fun Math Sheets
Math Puzzles
Math Riddles
Math Games
Dot to Dot
Mental Math Sheets
Times Tables
Math Facts
Online Times Tables
Online Addition & Subtraction
Math Grab Packs
Place Value
Counting
Rounding Numbers
Comparing Numbers
Number Lines
Prime Numbers
Negative Numbers
Roman Numerals
Addition
Subtraction
Add & Subtract
Multiplication
Division
Times Tables
Fraction Worksheets
Learning Fractions
Fraction Printables
Fraction Calculators
Percent Worksheets & Help
Percent calculators
Decimals
All Geometry
2d Shapes Worksheets
3d Shapes Worksheets
Shape Properties
Geometry Cheat Sheets
Printable Shapes
Nets
Symmetry
Coordinates
Triangles
Measurement
Money
Math Conversion
Area
Perimeter
Time
Statistics Worksheets
Bar Graph Worksheets
Venn Diagrams
All Word Problems
Finding all possibilities
Logic Problems
Ratio Word Problems
All UK Maths Sheets
Year 1 Maths Worksheets
Year 2 Maths Worksheets
Year 3 Maths Worksheets
Year 4 Maths Worksheets
Year 5 Maths Worksheets
Year 6 Maths Worksheets
All AU Maths Sheets
Kindergarten Maths Australia
Year 1 Maths Australia
Year 2 Maths Australia
Year 3 Maths Australia
Year 4 Maths Australia
Year 5 Maths Australia
Christmas
Easter
Halloween
Thanksgiving
Math Help
What's New
Meet the Sallies
Certificates
About us
Shop
Sitemap
Converting Decimals to Fractions Worksheet
Welcome to our Converting Decimals to Fractions Worksheets page. Here you will find our selection of worksheets to help you to practice converting decimals into fractions.
Looking for help? We also have some worked examples and links to our support page which will help you to nail this concept.
Try our NEW quick quiz at the bottom of this page to test your skills online.
Converting Decimals to Fractions Worksheets
How to convert decimals into fractions?
Converting a decimal into a fraction is a fairly straightforward process.
It involves looking at the number of decimal places the decimal has and putting the decimal part as the numerator and the denominator as a '1' followed the the same number of zeros as there are decimal places.
This converts the decimal into a decimal fraction (a fraction where the denominator is a power of 10.
After that, you will probably need to simplify the fraction.
Example 1) Convert 0.37 into a fraction.
This decimal has two decimal places.
So the numerator is the decimal number, which is 37.
The denominator is a '1' followed by 2 zeros, which is 100.
This give us [ 0.37 = {37 \over 100} ]
Example 2) Convert 0.416 into a fraction.
This decimal has 3 decimal places.
So the numerator is the decimal number, which is 416.
The denominator is a '1' followed by 3 zeros, which is 1000.
This give us [ 0.416 = {416 \over 1000} ]
Example 3) Convert 0.9 into a fraction.
This decimal has 1 decimal places.
So the numerator is the decimal number, which is 9.
The denominator is a '1' followed by 1 zero, which is 10.
This give us [ 0.9 = {9 \over 10} ]
Looking for more help converting decimals to fractions?
Take a look at our dedicated help page on how to convert a decimal into a fraction.
This page has lots of worked examples and a video to watch!
Convert Decimal to Fraction Support
Converting Decimals to Fractions Worksheets
Here are our worksheets to help you practice converting decimals to fractions.
The sheets within each section are graded with the easiest ones first.
The harder sheets involve both converting the decimals and then simplifying the fractions.
We also have separate sheets involving converting mixed decimals (with a value greater than 1) into fractions.
These sheets are aimed at students in 5th and 6th grade.
Converting Decimals to Fractions Sheet 1
Answers
PDF version
Converting Decimals to Fractions Sheet 2
Answers
PDF version
Converting Decimals to Fractions Sheet 3
Answers
PDF version
Converting Mixed Decimals to Fractions Worksheets
These sheets are similar to those in the section above, but involve mainly mixed decimals greater than one to convert.
Converting Mixed Decimals to Fractions Sheet 1
Answers
PDF version
Converting Mixed Decimals to Fractions Sheet 2
Answers
PDF version
Converting Mixed Decimals to Fractions Sheet 3
Answers
PDF version
Converting Decimals to Fractions Walkthrough Video
This short video walkthrough shows several problems from our Convering Mixed Decimals to Fractions Sheet 1 being solved and has been produced by the West Explains Best math channel.
If you would like some support in solving the problems on these sheets, check out the video!
Fraction to Decimal Riddles
We have some fun fraction - decimal worksheets involving working your way through clues to solve a riddle.
The riddles all involve converting between fractions and decimals.
Fraction - Decimal Riddles Sheet 1
Answers
PDF version
Fraction - Decimal Riddles Sheet 2
Answers
PDF version
More Recommended Math Worksheets
Take a look at some more of our worksheets similar to these.
Simplifying Fractions
Take a look at our Simplifying Fractions Practice Zone or try our worksheets for finding the simplest form for a range of fractions.
You can choose from proper fractions, improper fractions or both.
You can print out your results or benchmark your scores against future achievements.
Good for practising equivalent fractions as well as converting to simplest form.
Great for using with a group of children as well as individually.
Simplify Fractions Practice Zone
Simplifying Fractions Worksheet page
Learning Fractions
Are you looking for free fraction help or fraction support?
Here you will find a range of fraction help on a variety of fraction topics, from simplest form to converting fractions.
There are fraction videos, worked examples and practice fraction worksheets.
Learning Fractions
Converting between Fractions, Decimals & Percentages
The weblinks below contains more support, examples and practice converting between fractions, decimals and percents.
Convert Fractions to Decimal
Converting Fractions to Percentages
Convert Percent to Fraction
Back to Top
Convert Improper Fractions Online Quiz
This quick quiz tests your knowledge and skill at converting a range of decimals into fractions.
The link below will open the quiz in a dedicated quiz page or just simply start using the quiz underneath.
Convert Decimals to Fractions Quiz
Convert Fractions to Decimals Quiz
Our quizzes have been created using Google Forms.
At the end of the quiz, you will get the chance to see your results by clicking 'See Score'.
This will take you to a new webpage where your results will be shown. You can print a copy of your results from this page, either as a pdf or as a paper copy.
For incorrect responses, we have added some helpful learning points to explain which answer was correct and why.
We do not collect any personal data from our quizzes, except in the 'First Name' and 'Group/Class' fields which are both optional and only used for teachers to identify students within their educational setting.
We also collect the results from the quizzes which we use to help us to develop our resources and give us insight into future resources to create.
For more information on the information we collect, please take a look at our Privacy Policy
We would be grateful for any feedback on our quizzes, please let us know using our Contact Us link, or use the Facebook Comments form at the bottom of the page.
Convert Decimals to Fractions Quiz
This quick quiz tests your knowledge and skill at converting a range of decimals into fractions.
How to Print or Save these sheets 🖶
Need help with printing or saving?Follow these 3 steps to get your worksheets printed perfectly!
How to Print support
How to Print or Save these sheets 🖶
Need help with printing or saving?Follow these 3 steps to get your worksheets printed perfectly!
How to Print support
Subscribe to Math Salamanders News
Sign up for our newsletter to get free math support delivered to your inbox each month. Plus, get a seasonal math grab pack included for free!
Newsletter Signup
Return to 4th Grade Math Worksheets
Return to Fraction Worksheets
Return from Converting Decimals to Fractions Worksheet to Math Salamanders Homepage
Math-Salamanders.com
The Math Salamanders hope you enjoy using these free printable Math worksheets and all our other Math games and resources.
If you have any questions or need any information about our site, please get in touch with us using the 'Contact Us' tab at the top and bottom of every page.
Share this page
FacebookXPinterestWhatsApp
TOP OF PAGE
Home
🔍 Search Site
Preschool
Kindergarten
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th Grade
All Generated Sheets
Place Value Generated Sheets
Addition Generated Sheets
Subtraction Generated Sheets
Multiplication Generated Sheets
Division Generated Sheets
Money Generated Sheets
Negative Numbers Generated Sheets
Fraction Generated Sheets
All Calculators
Fraction Calculators
Percent calculators
Area & Volume Calculators
Age Calculator
Height Calculator
Roman Numeral Calculator
All Zones
Place Value Zones
Number Bonds
Addition & Subtraction
Times Tables
Fraction & Percent Zones
All Math Quizzes
Kindergarten Math Quizzes
1st Grade Quizzes
2nd Grade Quizzes
3rd Grade Quizzes
4th Grade Quizzes
5th Grade Quizzes
6th Grade Math Quizzes
Coloring Pages
Fun Math Sheets
Math Puzzles
Math Riddles
Math Games
Dot to Dot
Mental Math Sheets
Times Tables
Math Facts
Online Times Tables
Online Addition & Subtraction
Math Grab Packs
Place Value
Counting
Rounding Numbers
Comparing Numbers
Number Lines
Prime Numbers
Negative Numbers
Roman Numerals
Addition
Subtraction
Add & Subtract
Multiplication
Division
Times Tables
Fraction Worksheets
Learning Fractions
Fraction Printables
Fraction Calculators
Percent Worksheets & Help
Percent calculators
Decimals
All Geometry
2d Shapes Worksheets
3d Shapes Worksheets
Shape Properties
Geometry Cheat Sheets
Printable Shapes
Nets
Symmetry
Coordinates
Triangles
Measurement
Money
Math Conversion
Area
Perimeter
Time
Statistics Worksheets
Bar Graph Worksheets
Venn Diagrams
All Word Problems
Finding all possibilities
Logic Problems
Ratio Word Problems
All UK Maths Sheets
Year 1 Maths Worksheets
Year 2 Maths Worksheets
Year 3 Maths Worksheets
Year 4 Maths Worksheets
Year 5 Maths Worksheets
Year 6 Maths Worksheets
All AU Maths Sheets
Kindergarten Maths Australia
Year 1 Maths Australia
Year 2 Maths Australia
Year 3 Maths Australia
Year 4 Maths Australia
Year 5 Maths Australia
Christmas
Easter
Halloween
Thanksgiving
Math Help
What's New
Meet the Sallies
Certificates
About us
Shop
Sitemap
© 2010-2025 Math Salamanders Limited. All Rights Reserved.
Home
About
Privacy Policy
Copyright Policy
Sitemap
Contact Us
Home
About
Privacy Policy
Copyright Policy
Sitemap
Contact Us |
189238 | https://www.codewithc.com/closed-under-multiplication-exploring-polynomial-sets-in-programming/?amp=1 | Blog
Closed Under Multiplication: Exploring Polynomial Sets in Programming
CodeLikeAGirl
Share
10 Min Read
Understanding Polynomial Sets
Contents
Definition of Polynomial SetsProperties of Polynomial SetsClosed Under MultiplicationExamples of Multiplication in Polynomial SetsImplementing Polynomial Sets in ProgrammingApplications of Polynomial Sets in ProgrammingStability of Polynomial SetsEfficiency of Multiplication in Polynomial SetsComplexities in Polynomial Set OperationsOvercoming Challenges in Polynomial Set ProgrammingProgram Code – Closed Under Multiplication: Exploring Polynomial Sets in ProgrammingCode Output:Code Explanation:
Alright, folks, hold on to your hats, because we’re about to unravel the mystery behind polynomial sets! 🎩✨ So, what in the world are polynomial sets? Well, let’s break it down. 🤔
Definition of Polynomial Sets
In the world of mathematics, a polynomial set is a collection of polynomials. Yeah, those funky math expressions with a bunch of terms involving variables raised to different powers. But don’t fret, I’ve got your back! 🤓 These polynomials belong to the same set if they share similar properties, and we can perform operations like addition, subtraction, and multiplication on them. It’s like having a crew of polynomials that hang out together in an exclusive math club! 😎
Properties of Polynomial Sets
Now, let’s take a gander at the properties that make polynomial sets so special. These sets exhibit closure under addition, subtraction, and multiplication. In simple terms, when you add, subtract, or multiply two polynomials from the same set, the result will still be within that same set. It’s like a mathematical version of “what happens in polynomial set stays in polynomial set.” 💫
Multiplication in Polynomial Sets
Alright, time to talk turkey about multiplication in polynomial sets! 🦃
Closed Under Multiplication
One of the most fascinating aspects of polynomial sets is that they are closed under multiplication. What does that mean exactly? It means that when you multiply any two polynomials within a set, the result will always be another polynomial within that same set. Talk about keeping it all in the family, right? 🤯
Examples of Multiplication in Polynomial Sets
Let’s spice things up with a couple of examples to drive the point home. Consider two polynomials f(x) and g(x) belonging to a polynomial set. When we multiply f(x) by g(x), the resulting polynomial will also belong to the same set. It’s like a never-ending cycle of polynomial goodness! 🌀
Exploring Polynomial Sets in Programming
Alright, here’s where things get even more exciting—diving into the world of programming! 🖥️
Implementing Polynomial Sets in Programming
You’d be surprised, my fellow tech enthusiasts, at how we can implement polynomial sets in our code. Using data structures and algorithms, we can create classes or structures to represent polynomial sets and define operations like addition and multiplication. It’s like bringing the magic of math into the digital realm! ✨
Applications of Polynomial Sets in Programming
Now, let’s talk real-world applications. Polynomial sets find their way into various domains, such as cryptography, signal processing, and error-correcting codes. Pretty nifty, huh? It’s like unleashing the power of polynomial sets to tackle real-world challenges head-on! 💥
Advantages of Closed Under Multiplication
Why does “closed under multiplication” matter, you ask? Well, buckle up, because we’re about to uncover its hidden gems! 💎
Stability of Polynomial Sets
The closed property under multiplication ensures that the integrity of polynomial sets is maintained. No matter how many multiplications we throw at them, they stay true to their mathematical roots. It’s like having a steadfast companion in the ever-changing world of computation! 🌱
Efficiency of Multiplication in Polynomial Sets
Another perk of the closed property is the efficiency it brings to the table. With the guarantee that multiplication will always result in a polynomial within the same set, we can streamline our calculations and trust that the outcomes will be consistent. Efficiency for the win! 🏆
Challenges and Limitations of Polynomial Sets
Of course, nothing in this world is without its quirks and challenges. Let’s shine a light on the underbelly of polynomial sets.
Complexities in Polynomial Set Operations
One of the upsides—also a downside—is that polynomial sets can get pretty complex when it comes to performing operations. The sheer volume of terms and varying degrees among polynomials can make our lives a tad more interesting—sometimes too interesting! 🙃
Overcoming Challenges in Polynomial Set Programming
But hey, we didn’t come this far to back down from a challenge, did we? With the right algorithms, optimizations, and a sprinkle of perseverance, we can tame the complexities and make polynomial set programming a piece of cake! 🍰
In Closing…
And there you have it, folks! From unraveling the mystique of polynomial sets to diving headfirst into their programming prowess, we’ve relished every bit of this rollercoaster ride. Polynomial sets might throw some curveballs, but they certainly add that extra zing to our mathematical and computational escapades. So, embrace the closed property under multiplication, conquer the complexities, and let polynomial sets light up your coding world! 💻🚀
Program Code – Closed Under Multiplication: Exploring Polynomial Sets in Programming
Copy Code
Copied
Use a different Browser
```
```
def is_closed_under_multiplication(polynomials):
'''
Determines if a set of polynomials is closed under multiplication.
This function checks every pair of polynomials to see if their product
is still in the set.
Parameters:
polynomials (list of lists): A list of polynomials, where each polynomial
is represented as a list of coefficients. E.g., [1, 2, 3] represents
the polynomial 1x^2 + 2x + 3.
Returns:
bool: True if the set is closed under multiplication, False otherwise.
'''
# Function to multiply two polynomials
def multiply_poly(poly1, poly2):
# Initialize the result list with zeros, size = (n-1)+(m-1)+1
result = (len(poly1)+len(poly2)-1)
# Multiply each term in the first polynomial with each term in the second
for i in range(len(poly1)):
for j in range(len(poly2)):
result[i+j] += poly1[i] poly2[j]
return result
# Generate all possible products between the polynomials
for i in range(len(polynomials)):
for j in range(i, len(polynomials)):
product = multiply_poly(polynomials[i], polynomials[j])
# Check if the resulting product is within the set
if product not in polynomials:
return False
return True
Example: Checking closure for a set of polynomials
polynomials_set = [
[1, 0, -1], # x^2 - 1
[0, 1], # x
, # 1
[1, 0, 0, -1] # x^3 - 1
]
print(is_closed_under_multiplication(polynomials_set))
```
```
Code Output:
False
Code Explanation:
The program contains a function named is_closed_under_multiplication that aims to verify if a given set of polynomials is closed under multiplication. The closure property in the context of a set of polynomials means that multiplying any two polynomials in the set yields a result that is also within the set. Here’s the walkthrough:
The function accepts one parameter, which is a list of lists. Each inner list represents a polynomial. For example, [1, 2, 3] would correspond to the polynomial (1x^2 + 2x + 3).
Inside the function, there’s a nested function, multiply_poly, which multiplies two polynomials together. It follows standard polynomial multiplication, where each term of one polynomial is multiplied by every term of the second polynomial.
The main part of the program is a double loop that iterates through each possible pair of polynomials in the given set. It does not repeat pairs; it calculates the product only once for each unique pair.
For each pair, it calls the multiply_poly function to get the product, which is itself another polynomial represented as a list of coefficients.
It then checks whether this resulting product is present in the original set of polynomials. If the product is not found, the function returns False, indicating that the set is not closed under multiplication.
If the function completes the loops without finding any products outside the set, it returns True, but this is not the case in the provided example.
The provided set contains polynomials representing (x^2 – 1), (x), (1), and (x^3 – 1). The function returns False because not all possible products of these polynomials are present in the set. For example, if we multiply (x) by itself, we get (x^2), which is not in the set.
Share This Article
Facebook
Twitter
Previous Article
Multiclassing in Coding: Utilizing BG3 for Pragmatic Solutions
Next Article
Dynamic Memory Allocation: An Essential Skill for Coders
Leave a comment
Leave a Reply Cancel reply
We use cookies on our website to give you the most relevant experience by remembering your preferences and repeat visits. By clicking “Accept”, you consent to the use of ALL the cookies.
Do not sell my personal information.
Cookie settingsACCEPT
Privacy & Cookies Policy
Privacy Overview
This website uses cookies to improve your experience while you navigate through the website. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. We also use third-party cookies that help us analyze and understand how you use this website. These cookies will be stored in your browser only with your consent. You also have the option to opt-out of these cookies. But opting out of some of these cookies may have an effect on your browsing experience.
Necessary cookies are absolutely essential for the website to function properly. This category only includes cookies that ensures basic functionalities and security features of the website. These cookies do not store any personal information.
Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. It is mandatory to procure user consent prior to running these cookies on your website.
Exit mobile version |
189239 | https://fiveable.me/analytic-number-theory/unit-4/partial-summation-techniques/study-guide/p1HGUqoaez5upkAc | Partial summation techniques | Analytic Number Theory Class Notes | Fiveable | Fiveable
new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom
ap study content toolsprintablespricing
my subjectsupgrade
🔢Analytic Number Theory Unit 4 Review
4.2 Partial summation techniques
All Study Guides
Analytic Number Theory
Unit 4 – Arithmetic Function Averages & Summation
Topic: 4.2
🔢Analytic Number Theory Unit 4 Review
4.2 Partial summation techniques
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
print study guide
copy citation
APA
🔢Analytic Number Theory
Unit & Topic Study Guides
Intro to Analytic Number Theory
Arithmetic Functions & Dirichlet Convolution
Asymptotic Notation and Summations
Arithmetic Function Averages & Summation
4.1 Average order of arithmetic functions
4.2 Partial summation techniques
4.3 Dirichlet's divisor problem and related estimates
Prime Numbers and Sieve of Eratosthenes
Dirichlet Series & Euler Products
The Riemann Zeta Function
Zeta Function & Prime Number Theorem
Dirichlet Characters and L-functions
Dirichlet's Theorem on Prime Progressions
Riemann Zeta Function's Functional Equation
Zeta Function Zeroes & Riemann Hypothesis
Multiplicative Theory: Analytic Proofs
Advanced Topics in Analytic Number Theory
print guide report error
Partial summation techniques are powerful tools for transforming complex sums into more manageable forms. These methods, including Abel's summation and Euler-Maclaurin formula, help analyze sums where one sequence varies slowly while another oscillates rapidly.
These techniques connect to the broader theme of averages of arithmetic functions by providing ways to approximate and analyze sums. They bridge the gap between discrete sums and continuous integrals, offering insights into the behavior of arithmetic functions over large ranges.
Partial Summation Formulas
Fundamental Partial Summation Techniques
Partial summation transforms sums into more manageable forms by rewriting them in terms of partial sums
Abel's summation formula expresses a sum as a product of two sequences minus an integral
Formula for Abel's summation: ∑n=1 N a n b n=A N b N−∫1 N A(t)d b(t)\sum_{n=1}^{N} a_n b_n = A_N b_N - \int_{1}^{N} A(t) db(t)∑n=1 Na nb n=A Nb N−∫1 NA(t)d b(t)
$A_n$ represents the partial sum of $a_n$, and $b(t)$ is a continuous extension of $b_n$
Useful for analyzing sums where one sequence varies slowly while the other oscillates rapidly
Euler-Maclaurin summation formula connects sums to integrals using Bernoulli numbers
General form: ∑k=m n f(k)=∫m n f(x)d x+f(m)+f(n)2+∑k=1 p B 2 k(2 k)!(f(2 k−1)(n)−f(2 k−1)(m))+R p\sum_{k=m}^n f(k) = \int_m^n f(x) dx + \frac{f(m)+f(n)}{2} + \sum_{k=1}^p \frac{B_{2k}}{(2k)!} (f^{(2k-1)}(n) - f^{(2k-1)}(m)) + R_p∑k=m nf(k)=∫m nf(x)d x+2 f(m)+f(n)+∑k=1 p(2 k)!B 2 k(f(2 k−1)(n)−f(2 k−1)(m))+R p
Bernoulli numbers ($B_{2k}$) play a crucial role in the formula's accuracy
Provides a powerful tool for approximating sums and deriving asymptotic expansions
Advanced Summation Techniques
Summation by parts generalizes integration by parts to discrete sums
Formula: ∑k=m n a k b k=A n b n+1−A m−1 b m−∑k=m n A k(b k+1−b k)\sum_{k=m}^n a_k b_k = A_n b_{n+1} - A_{m-1} b_m - \sum_{k=m}^n A_k (b_{k+1} - b_k)∑k=m na kb k=A nb n+1−A m−1b m−∑k=m nA k(b k+1−b k)
$A_k$ represents the partial sum of $a_k$ up to index $k$
Particularly useful when dealing with products of sequences
Can simplify complex sums by relating them to simpler, more manageable sums
Often applied in conjunction with other summation techniques for enhanced results
Provides a discrete analogue to integration by parts, bridging continuous and discrete analysis
Integral Representations
Stieltjes Integral and Its Applications
Stieltjes integral generalizes the Riemann integral, allowing integration with respect to functions
Defined as ∫a b f(x)d g(x)=limn→∞∑i=1 n f(x i)(g(x i)−g(x i−1))\int_a^b f(x) dg(x) = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^) (g(x_i) - g(x_{i-1}))∫a bf(x)d g(x)=lim n→∞∑i=1 nf(x i)(g(x i)−g(x i−1))
Partition points $a = x_0 < x_1 < ... < x_n = b$, with $x_i^$ in $[x_{i-1}, x_i]$
Provides a powerful tool for analyzing sums and integrals simultaneously
Applications include probability theory (expectation calculations) and functional analysis
Enables the study of discontinuous integrators, expanding the scope of integration theory
Serves as a bridge between discrete sums and continuous integrals in analytic number theory
Asymptotic Series and Their Properties
Asymptotic series approximate functions for large values of a variable
General form: f(x)∼∑n=0∞a n x−n as x→∞f(x) \sim \sum_{n=0}^{\infty} a_n x^{-n} \text{ as } x \to \infty f(x)∼∑n=0∞a nx−n as x→∞
Does not necessarily converge, but provides increasingly accurate approximations as $x$ grows
Truncating the series at a finite number of terms often yields highly accurate estimates
Poincaré's definition: $f(x) = \sum_{n=0}^{N-1} a_n x^{-n} + O(x^{-N})$ for any fixed $N$
Useful in analyzing the behavior of complex functions and sequences for large arguments
Frequently employed in conjunction with summation formulas to study arithmetic functions
Error Analysis
Truncation Error Estimation
Truncation error arises from approximating infinite series or integrals with finite sums
For Euler-Maclaurin formula, error term $R_p$ bounds the truncation error
Error typically decreases as more terms are included in the approximation
Magnitude of error often expressed using big O notation
Taylor's theorem with remainder provides a framework for estimating truncation errors
Error bounds crucial for determining the accuracy and reliability of numerical approximations
Techniques include Lagrange remainder, integral remainder, and series remainder forms
Advanced Smoothing Techniques
Smoothing techniques reduce oscillations and noise in data or functions
Moving averages smooth data by replacing each point with the average of nearby points
Kernel smoothing uses weighted averages with a kernel function determining the weights
Savitzky-Golay filters preserve higher moments while smoothing data
Wavelets decompose signals into different frequency components for multi-scale analysis
Regularization methods (Tikhonov regularization) balance smoothness and fidelity to data
Spline smoothing fits piecewise polynomial functions to data, ensuring smoothness at joints
Applications include signal processing, statistical analysis, and numerical approximations in number theory
4.1 BackNext 4.3
Study Content & Tools
Study GuidesPractice QuestionsGlossaryScore Calculators
Company
Get $$ for referralsPricingTestimonialsFAQsEmail us
Resources
AP ClassesAP Classroom
every AP exam is fiveable
history
🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history
social science
✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾⚖️ ap us government
english & capstone
✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar
arts
🎨 ap art & design🖼️ ap art history🎵 ap music theory
science
🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics
math & computer science
🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p
world languages
🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature
go beyond AP
high school exams
✏️ PSAT🎓 Digital SAT🎒 ACT
honors classes
🍬 honors algebra II🐇 honors biology👩🏽🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history
college classes
👩🏽🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽🔬 science💶 social science
RefundsTermsPrivacyCCPA
© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
every AP exam is fiveable
Study Content & Tools
Study GuidesPractice QuestionsGlossaryScore Calculators
Company
Get $$ for referralsPricingTestimonialsFAQsEmail us
Resources
AP ClassesAP Classroom
history
🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history
social science
✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾⚖️ ap us government
english & capstone
✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar
arts
🎨 ap art & design🖼️ ap art history🎵 ap music theory
science
🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics
math & computer science
🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p
world languages
🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature
go beyond AP
high school exams
✏️ PSAT🎓 Digital SAT🎒 ACT
honors classes
🍬 honors algebra II🐇 honors biology👩🏽🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history
college classes
👩🏽🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽🔬 science💶 social science
RefundsTermsPrivacyCCPA
© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
0 |
189240 | https://web.pdx.edu/~veerman/0_mainfile.pdf | April 30, 2023 An Introduction to Number Theory J. J. P. Veerman Preface This work contains a one year (three terms or two semesters) first course in number theory at the advanced undergraduate to intermediate graduate level. No prior knowledge of number theory is assumed. We do require knowledge and familiarity with the notions of basic linear algebra (matrices, determinants), calculus (integrals), and proof-writing. But beyond that, this text aims to be completely self-contained. Part I covers what students might learn in an advanced undergraduate first course on number theory. Part II discusses the basics of the different branches of number theory (algebraic, analytic, and ergodic or probabilistic) and is pitched at a slightly higher level. Part III discusses some more advanced topics.
Too often, in our opinion, it seems that introductory texts skimp on ei-ther the algebraic aspects or the geometric aspects of number theory (some-times both). We have tried to cover all areas of introductory number the-ory. In contrast, algebraic texts are often very hard to read for the non-algebraically inclined students due to the barrage of definitions one needs to digest to get to the “meat”. Our approach is pragmatic: we want to un-derstand a broad outline of number theory, so how do we do that most effi-ciently? For example, when dealing with continued fractions, a geometrical approach greatly clarifies the subject by making it easy to visualize results, and so we adopt that approach. In the study of algebraic integers, we skip or summarize many of the commonly used definitions. For instance, we do not dwell very long on the fact that the integers form a ring. We merely note 1 2 Preface that the crucial part is the distinction between ring (where multiplication does not have an inverse) and field (where it does). This — together with some other observations — is sufficient to arrive in very few pages at accu-rate definitions of, say, Euclidean domains. Our approach, then, is to try to uncover number theory by any means, regardless of whether the subject at hand is traditionally seen as part of algebra, analysis, or geometry.
Each chapter consists of some basic material followed by exercises.
The exercises are of two types. Some are simple computational routines, to get the students used to the notions described in the text. Others develop the theory a little further. All exercises, including the ones of the latter type, should be easily doable for graduate students. Nonetheless, my recommen-dation is to have the students work out all the exercises in groups to lessen even more the chance of getting stuck. Nothing discourages self-study more than being stuck in some exercise. If students tend to get stuck in some par-ticular exercise, I hope I will be notified, so I can add more hints in the next edition.
You can’t really learn mathematics without doing it. But on the other hand, you can’t possible invent everything from scratch either. So a good way to design a course, is to give the lay of the land in each chapter, and let students figure out some of the details and corollaries through exercises at the end of the chapter. I wrote these notes aiming for a division of roughly equal parts between blackboard type lecturing and interactive practice for the student of about half-half. This has the additional advantage that the basic and most important material is concise and contained in few pages, and therefore can be consulted with great ease to find important results.
A word on alternate uses of this work. Because of the crossdisciplinary (within mathematics) approach taken, parts of the book can easily be used in other courses. For example, a graduate course on dynamical systems or ergodic theory could contain an 8 week segment consisting of Chapters 6, 9, 10, and 14. Chapters 4, 2, 11, 12, and 13 could easily constitute a short analytic number theory course. Chapters 5, 7, and 8 could start off a course in algebraic number theory.
This work can be used at the intermediate undergraduate level, where students still need to train how to write proofs. In that case, I recommend a modest program consisting roughly of the following content. Start with Chapters 1 and 2, followed by Sections 3.1 through 3.4 and 3.6, 4.1 and Preface 3 4.2, and finally Chapter 5. If this is the course taken, then the instructor would probably want to gather some extra material on how to write proofs, starting with mathematical induction. Care should be taken that some of the exercises may be a little too advanced for the students at this level.
We advise the reader that underlined words and clauses are indexed.
J. J. P. Veerman Portland, Oregon September, 2022.
List of Figures 1 Eratosthenes’ sieve up to n = 30. All multiples of a less than √ 31 are cancelled. The remainder are the primes less than n = 31.
4 2 f is a minimal polynomial for the irrational number r. By minimality f ′(p/q) is not zero. On the interval (r −1,r), the absolute value of the derivative of f attains its maximum at t.
10 3 A directed path γ passing through all points of Z2.
13 4 Lowest member of nth twin prime pair less than 1000 (left) and less than 10000 (right), horizontal axis is n.
15 5 The orbits of n under f of exercise 1.11 for n ∈{1,··· ,10}.
17 6 A rectangle of 30 by 12 squares can be subdivided into squares non larger than 6 by 6.
20 7 Two meshing gear wheels have 30, resp. 12 teeth. Each tiny square represents the turning of one tooth in each wheel. After precisely 5 turns of the first wheel and 2 of the second, both are back in the exact same position.
21 8 The division algorithm: for any two integers r1 and r2, we can find an integer q and a real e ∈[0,1) so that r1/r2 = q2 +e.
24 9 On the left, the function R x 2 lnt dt in blue, π(x) in red, and x/lnx in green. On the right, we have R x 2 lnt dt −x/lnx in blue, π(x)−x/lnx in red. Note the different scales.
34 5 6 List of Figures 10 Proof that ∑∞ n=1 f(n) is greater than R ∞ 1 f(x)dx if f is positive and (strictly) decreasing.
41 11 Proof that ∑∞ n=1 f(n) (shaded in blue and green) minus f(1) (shaded in blue) is less than R ∞ 1 f(x)dx if f is positive and (strictly) decreasing to 0.
42 12 The origin is marked by “×”. The red dots are visible from ×; between any blue dot and × there is a red dot. The picture shows exactly one quarter of {−4,··· ,4}2(0,0) ⊂Z2.
42 13 The general solution of the inhomogeneous equation ( ⃗ r,⃗ x) = c in R2.
49 14 A ‘greedy’ (or locally best) algorithm to tile the the 188 × 158 rectangle by squares. The 3 smallest — and barely visible — squares are 2×2. Note how the squares spiral inward as they get smaller. See exercise 3.13.
54 15 The greedy algorithm of exercise 3.2 (a) applied to the golden mean rectangle. The spiral connecting the corners of the square is known as the golden spiral. (In actual fact we used a 55 by 34 rectangle as an approximation. An approximation to a true spiral was created by fitting circular segments to the corners.) 57 16 The totient function ϕ(n) versus n. Its subtle structure is clearly visible, see also exercise 4.7.
67 17 A one parameter family ft of maps from the circle to itself. For every t ∈[0,1] the map ft is constructed by truncating the map x →2x mod 1 as indicated in this figure.
71 18 Two ways of computing the volume of a big box: add the volumes of the small boxes, or compute the dimensions of the big box.
73 19 The fraction of numbers in {1,··· ,n} relatively prime to n, or ϕ(n)/n, versus n.
74 20 Left, the orbits in Z7 under addition of 3; middle, the orbits under multiplication by 3; and right, the orbit under multiplication by 2.
Observe that the multiplicative graphs are not connected and have less symmetries.
82 List of Figures 7 21 Left: the relation a is an additive inverse mod 6 of b. Right: the relation a is an multiplicative inverse mod 6 of b.
90 22 Left: the relation a is an additive inverse mod 7 of b. Right: the relation a is an multiplicative inverse mod 7 of b.
91 23 The group of symmetries of an equilateral triangle, S3, is not commutative. The group operation is composition. In the top row, we perform first a rotation given by 1 →2, 2 →3, and 3 →1 and then a refection in the dotted line through 1. In the bottom row, the reverse.
92 24 The part to the right of the intersection with ℓ: y = x+1 (dashed) of a bad path (in red) is reflected. The reflected part in indicated in green. The path becomes a monotone path from (0,0) to (n−1,n+1).
102 25 Four branches of the Gauss map.
106 26 This figure illustrates how the convergents pn/qn approach their limit x.
111 27 Successive approximations to the line y = wx. In green, the vectors ei. In blue, aiei−1 for i = 1 and 2. Note that a1 = a2 = 2 in this figure. Note that a1 = ⌊1/w⌋.
114 28 The line y = ωx and (in red) successive iterates of the rotation Rω.
Closest returns in this figure are q in {2,3,5,8}.
116 29 The geometry of successive closest returns. In this figure, an+1 = 3. Note that |dn−1| > 4dn+1.
117 30 Drawing of the shaded parallelogram P, the line y = ωx, and a few successive approximants. The green arrows correspond to en−1, en, and en+1. Here, an+1 is taken to be 3.
118 31 A few branches of the twice iterated Gauss map T 2. The points of T −2({0}) (the complete inverse image of 0 under T 2) are marked in red. The reader should compare this plot to Figure 25.
119 32 Black: thread from origin with golden mean slope; red: pulling the thread down from the origin; green: pulling the thread up from the origin.
122 33 The placement of x between its convergents pn/qn and pn+1/qn+1.122 8 List of Figures 34 Plots of the points (n,n) in polar coordinates, for n ranging from 1 to 50, 180, 330, and 3000, respectively.
125 35 Plots of the prime points (p, p) (p prime) in polar coordinates with p ranging between 2 and 3000, and between 2 and 30000, respectively.
126 36 Left, the functions f(x) = √2+3x and x. The fixed point x∗is the solution of f(x) = x is the value sought in exercise 6.23. Show that for positive x, x∗−f(x) x∗−x , and that that implies that xn →x∗.
127 37 The functions f(x) = 1 + 3/x (red), f( f(x) (green), and x (black). Since f(f(x)) is increasing slowly, the sequence {0, f(0), f(f(0)),···} must converge to the fixed point.
128 38 ABCar+s is the sum of the BbiCcj along the green line in the i−j diagram. The red lines indicate where p ∤Bbi and p ∤Ccj. So all contributions exceptBbrCcs are divisible by p. Thus p ∤ABCar+s. 133 39 Intuitively we wrap R around a circle of length 1, so that points that differ by an integer land on the same point.
138 40 The Gaussian integers are the lattice points in the complex plane; both real and imaginary parts are integers. For an arbitrary point z ∈C — marked by x in the figure, a nearby integer is k1 + ik2 where k1 is the closest integer to Re(z) and k2 the closest integer to Im(z). In this case that is 2+3i.
167 41 A depiction of Z[ √ −6] in the complex plane; real parts are integers and imaginary parts are multiples of √ 6.
169 42 Left, the elements of the ring Z[√−3]. Right, the ring Z[ 1 2(1+√−3)]. The units of each ring are indicated in green and the ideals ⟨2,1+√−3⟩on the left and ⟨2⟩on the left are indicated in red. Fundamental domains (Definition 8.17) are shaded in blue. 171 43 Left, the fundamental domain of Z[√−3]. Here, h = i √ 3. Right, one of the 2 isosceles triangles that constitute the fundamental domain of Z[ 1 2(1 + √−3)]. Its height d equals 1 2 √ 3. The point that maximizes the distance to the closest of the 3 corner points lies on the bisector of the top angle at height y.
172 List of Figures 9 44 Points in the area red shaded are a distance less than from an integer in Z. The blue area maps into the red under x →2x − √ 19/4 indicated by the arrow. We note that √ 19/4 ≈1.09 and √ 3/2 ≈0.87.
175 45 Left: the relation a is a multiplicative inverse mod 8 of b. Right: the multiplication table of Z8.
178 46 Possible values of ργ−1 in the proof of Proposition 8.16.
179 47 The Gaussian primes described in Proposition 8.30. There are approximately 950 within a radius 40 of the origin (left figure) and about 3300 within a radius 80 (right figure).
181 48 A comparison between approximating the Lebesgue integral (left) and the Riemann integral (right).
193 49 The pushforward of a measure µ.
194 50 This map has many ergodic measures 197 51 The first two stages of the construction of the singular measure νp.200 52 The left figure illustrates that (∩Ai)c = ∪i Ac i . The right figure illustrates that for an open set O containing Sc, S\Oc = O\Sc (shaded in red).
202 53 The first two stages of the construction of the middle third Cantor set. The shaded parts are taken out.
203 54 An impression of the Cantor function of exercises 9.10 and 9.11.
The first four stages are drawn in black, the red segments are affine interpolations.
204 55 The inverse image of a small interval dy is T −1(dy) 210 56 The interval [0,x) (shaded red) and its pre-image under the Gauss map (shaded blue).
212 57 r is irrational and p q is a convergent of r. Then x+qr modulo 1 is close to x. Thus adding qr modulo 1 amounts to a translation by a small distance.
214 58 ℓ(I) is between 1 3 and 1 2 of ℓ(J). So there are two disjoint images of I under R−1 ω that fall in J.
215 10 List of Figures 59 An example of the transformation T : [0,1) →[0,1) described in Theorem 10.9.
216 60 The pre-image in the nth level interval of A (in red) is the blue interval. This branch of T n is affine and so there is no distortion.
As a consequence, the proportions that A and its pre-image occupy are the same.
217 61 The pre-image in the nth level interval of A (in red) is the blue interval. This branch of T n is far from affine and so the distortion is large. As a consequence, the proportions that A and its pre-image occupy are very different.
218 62 Plot of the function ln(x)ln(1+x) 228 63 If f is analytic on the closed set S, then f must be analytic on some open set containing S.
236 64 Left, a curve. Then two simple, closed curves with opposite orientation. The curve on the right is a union of two simple, closed curves.
238 65 In the interior of the curve obtained by concatenating γ, p, c, and −p, f is analytic. Therefore H γ f dz − H c f dz = 0. If f is also bounded inside c, we also have H c f dz = 0.
239 66 The curve γ goes around z exactly once in counter-clockwise direction. If d is small enough, z+d also lies inside γ.
242 67 F(z) does not depend on the path. So F(z + d) −F(z) = R c f ≈ f(z)d 243 68 The curve w goes around z0 exactly once in counter-clockwise direction. .
245 69 g is analytic in DR := {Rez ≥−dR}∩{|z| ≤R} (shaded). The red curve is given by C+(s) = Reis with s ∈(−π 2 , π 2 ). The green curve is given by C+(s) = Reis with s ∈( π 2 , 3π 2 ). The blue L−consists of 2 small circular segments plus the segment connecting their left endpoints at a distance 0 < d < dR to the left of the the imaginary axis.
246 70 In the proof of Proposition 11.20, |p(z)| must have a minimum z0 in the interior of the disk |z| < 2R and it cannot have a minimum unless at z0 unless it is zero.
249 List of Figures 11 71 The complex plane with eit, −e−it and e−it on the unit circle. cost is the average of eit and e−it and isint as the average of eit and −e−it.
252 72 Moving around the origin once in the positive direction increases ϕ, and thus lnz, by 2π. Discontinuities can be avoided if we agree never to cross the half line or branch cut L.
252 73 The functions gk and hk of exercise 11.19 for k ∈{2,8,15,30}.
256 74 The contour C is the concatenation of c (celeste), b1 (blue), r1 (red), g (green), p (purple), −g, r2, and b2. The path r is a semi-circle of radius R. The path p is a small circle of radius r.
See exercise 11.20.
257 75 The Riemann-Stieltjes integral (12.1) near x = 5 picks up the value f(5)(θ(xi+1)−θ(xi)).
260 76 Integration over the shaded triangle of area 1/2 in equation (12.11).
266 77 The prime gaps pn+1−pn divided by ln pn+1 for n in {1,··· ,1000}.275 78 The functions θ(x)/x (green), ψ(x)/x (red), and π(x)lnx/x (blue) for x ∈[1,1000]. All converge to 1 as x tends to infinity. The x-axis is horizontal.
277 79 Plot of the function f(n) := (lcm(1,2,··· ,n)) 1 n for n in {1,··· ,100} (left) and in {104,··· ,105} (right). The function converges to e indicated in the plots by a line.
279 80 The two characters modulo 3 illustrate the orthogonality of the Dirichlet characters.
286 81 The set S consists of the natural numbers contained in intervals shaded in the top figure of the form [22n−1,22n). The bottom picture is the same but with a logarithmic horizontal scale.
301 82 Proof that R 1 0 f(x)dx is between ∑k j=1 f( jdx)dx and ∑k−1 j=0 f(jdx)dx if f is strictly decreasing.
302 83 The function ln(ln(x)) for x ∈[1,1040].
305 12 List of Figures 84 Illustration of the fact that if h+(x) = sup{ f(x),g(x)}, then h−1 + ((y,∞)) = f −1((y,∞)) ∪g−1((y,∞)). Similarly, if h−(x) = inf{f(x),g(x)}, then h−1 −((z,∞)) = f −1((z,∞))∩g−1((z,∞)).
309 85 A sequence of functions fn(x) = x1/n that converge almost everywhere pointwise to f(x) = 1 on [0,1]. The convergence is uniform on U = [ε,1] for any ε ∈(0,1).
311 86 The definition of the Lebesgue integral. Let {yi} be a countable partition of the range of g. We approximate R gdµ from below by ∑i µ g−1 ({y : y ≥yi+1}) (yi+1 −yi). Then g is integrable if the limit converges as the mesh of the partition goes to zero. The function y in the proof of Lemma 14.9 (ii) is indicated in red.
(Here µ is the Lebesgue measure.) 312 87 The non-zero values of the function f4 in red.
314 88 The functions 1[a,b] in black and gε in red. ε is the width (indicated by two-sided arrows) of the regions where gε and the step function do not agree.
317 89 A plot of Sn f (x0) for some fixed x0 for n ∈{0,··· ,N}.
319 90 The functions µ(X− c ) and µ(X+ c ).
322 91 Left: an impression of the function g(x) = c(x)+x where c(x) is taken from figure 54. Left: its inverse h := g−1 is well-defined but not Lebesgue measurable. Linear interpolated segments are colored red.
324 92 The function fn (in red) in exercise 14.5 is a sum of very thin triangles with height 1. Each triangle is given by hn( j,k,x) (in black).
326 93 The functions fn in exercise 14.6 and the function g (in red) that dominates them.
326 94 In this figure r = 2. We show the function gk(x) (red) on [0,1] and its intersections. The sum of the rectangles like the one shaded in red give a lower bound for R Gk dx while the sum of the red and green rectangles give an upper bound.
327 95 A schematic illustration of the quantities defined in exercises 14.10 and 14.11.
330 List of Figures 13 96 A few branches of the L¨ uroth map T of exercise 14.14. The names of the branches are as indicated in the figure. The L¨ uroth expansion of x ∈)0,1] is [a1,a2,···], where the ith digit ai equals k if T i−1(x) falls in the domain of the kth branch.
331 97 A few branches of the k+1st iterate of the L¨ uroth map T restricted to the interval Ik. In red a branch of T k and in black a few branches of T k+1.
333 98 Left: the symmetric difference of two sets A1 and A2 is A1△A2 := (A1 ∪A2)(A1 ∩A2). On the right side, we illustrate that A3△A1 is contained in (A3△A2)∪(A2△A1) (green).
334 99 The permutation σ = r◦g consist of first applying the green cycles g and then the red cycle r.
340 100 Denote the red 3-cycle by r and the green by r. The commutator r−1g−1rg equals the 3-cycle (123).
341 101 The effect of conjugating a permutation p by the reflection s in the dotted line.
342 102 Two loops in C{0} based at 1 that wind around 0 exactly once.
The red arrows indicate how the outer curve can be continuously deformed in C{0} to the inner curve.
347 103 The curious behavior of lifts loops based at 1 through a 4th root.
The fiber of 1 is {1,i,−1,−i}. The green loop on the left has winding number 1 and lifts to curves whose endpoints are not the same. We exhibit one lift based at 1 and one based at -1. The red loop on the left has winding number 0 and lifts to loops: we exhibit the ones based at 1 and -1. The blue neighborhood on the left lifts to disjoint blue neighborhoods on the right. Restricted to these local neighborhoods, p is invertible.
348 104 Left, the coefficients c1(t) = 0 and c0(t) = e2πit of a quadratic polynomial. Right, its solutions s1(t) and s2(t). Between t = 0 and t = 1, the solutions are permuted.
350 105 At some point in the evaluation of q, we must take a multi-valued nth radical n √.. : C →C.
351 106 Right, paths s1(t) and s2(t) that swap the first two roots in green.
This results in a green loop g(t) on the left, upon taking the 3rd 14 List of Figures power; Similarly, in red, paths that swap the second and third loop and their 3rd power.
352 107 Four loops numbered from 1 to 4 make a commutator of commutators, namely .
353 108 A permutation on n symbols can be decomposed into transpositions graphically, and thus its parity can be determined.
356 109 The graph of p3(z) of exercise 15.10 showing its three roots at approximately {1.52,16,16.48}.
358 110 The graph of p4(z) of exercise 15.11 showing its roots are approximately {−3.41,−0.59,−0.65,4.65}.
359 111 The graph of the Bring radical as function of (real) a.
360 Contents Preface 1 List of Figures 5 Part 1. Introduction to Number Theory Chapter 1.
A Quick Tour of Number Theory 3 §1.1.
Divisors and Congruences 4 §1.2.
Rational and Irrational Numbers 6 §1.3.
Algebraic and Transcendental Numbers 8 §1.4.
Countable and Uncountable Sets 10 §1.5.
Exercises 14 Chapter 2.
The Fundamental Theorem of Arithmetic 23 §2.1.
B´ ezout’s Lemma 24 §2.2.
Corollaries of B´ ezout’s Lemma 26 §2.3.
The Fundamental Theorem of Arithmetic 27 §2.4.
Corollaries of the Fundamental Theorem of Arithmetic 29 §2.5.
The Riemann Hypothesis 31 §2.6.
Exercises 35 Chapter 3.
Linear Diophantine Equations 43 15 16 Contents §3.1.
The Euclidean Algorithm 43 §3.2.
A Particular Solution of ax+by = c 45 §3.3.
Solution of the Homogeneous equation ax+by = 0 47 §3.4.
The General Solution of ax+by = c 48 §3.5.
Recursive Solution of x and y in the Diophantine Equation 50 §3.6.
The Chinese Remainder Theorem 51 §3.7.
Polynomials 52 §3.8.
Exercises 53 Chapter 4.
Number Theoretic Functions 61 §4.1.
Multiplicative Functions 61 §4.2.
Additive Functions 64 §4.3.
M¨ obius inversion 64 §4.4.
Euler’s Phi or Totient Function 66 §4.5.
Dirichlet and Lambert Series 68 §4.6.
Exercises 72 Chapter 5.
Modular Arithmetic and Primes 81 §5.1.
Euler’s Theorem and Primitive Roots 81 §5.2.
Fermat’s Little Theorem and Primality Testing 85 §5.3.
Fermat and Mersenne Primes 88 §5.4.
A Divisive Issue: Rings and Fields 90 §5.5.
Wilson’s Theorem 95 §5.6.
Exercises 96 Part 2. Currents in Number Theory: Algebraic, Probabilistic, and Analytic Chapter 6.
Continued Fractions 105 §6.1.
The Gauss Map 105 §6.2.
Continued Fractions 106 §6.3.
Computing with Continued Fractions 112 §6.4.
The Geometric Theory of Continued Fractions 113 Contents 17 §6.5.
Closest Returns 116 §6.6.
Another Interpretation of the Convergents 119 §6.7.
Exercises 120 Chapter 7.
Fields, Rings, and Ideals 129 §7.1.
Rings of Polynomials 129 §7.2.
Ideals 135 §7.3.
Fields and Extensions 138 §7.4.
The Algebraic Integers 142 §7.5.
Rings of Quadratic Numbers and Modules 145 §7.6.
Exercises 148 Chapter 8.
Factorization in Rings 159 §8.1.
So, How Bad Does It Get?
159 §8.2.
Integral Domains 161 §8.3.
Euclidean Domains 164 §8.4.
Example and Counter-Example 166 §8.5.
Ideal Numbers 169 §8.6.
Principal Ideal Domains 173 §8.7.
ED, PID, and UFD are Different 174 §8.8.
Exercises 177 Chapter 9.
Ergodic Theory 187 §9.1.
The Trouble with Measure Theory 188 §9.2.
Measure and Integration 190 §9.3.
The Birkhoff Ergodic Theorem 194 §9.4.
Examples of Ergodic Measures 196 §9.5.
The Lebesgue Decomposition 200 §9.6.
Exercises 202 Chapter 10.
Three Maps and the Real Numbers 209 §10.1.
Invariant Measures 209 §10.2.
The Lebesgue Density Theorem 213 18 Contents §10.3.
Rotations and Multiplications on R/Z 214 §10.4.
The Return of the Gauss Map 218 §10.5.
Number Theoretic Implications 220 §10.6.
Exercises 224 Part 3. Topics in Number Theory Chapter 11.
The Cauchy Integral Formula 235 §11.1.
Analyticity versus Isolated Singularities 235 §11.2.
The Cauchy Integral Formula 238 §11.3.
Corollaries of the Cauchy Integral Formula 241 §11.4.
A Tauberian Theorem 245 §11.5.
A Polynomial Must Have a Root 248 §11.6.
Exercises 249 Chapter 12.
The Prime Number Theorem 259 §12.1.
Preliminaries 259 §12.2.
Chebyshev’s Theorem 262 §12.3.
Zeroes and Poles of the Zeta Function 265 §12.4.
The Function Φ(z) 268 §12.5.
The Prime Number Theorem 269 §12.6.
Exercises 270 Chapter 13.
Primes in Arithmetic Progressions 281 §13.1.
Finite Abelian Groups 281 §13.2.
The Hermitian Inner Product 283 §13.3.
Characters of Finite Abelian Groups 284 §13.4.
Dirichlet Characters and L-functions 288 §13.5.
Preliminary Steps 290 §13.6.
Primes in Arithmetic Progressions 292 §13.7.
Exercises 295 Chapter 14.
The Birkhoff Ergodic Theorem 307 §14.1.
Measurable Sets 307 Contents 19 §14.2.
Measurable Functions 309 §14.3.
Dominated Convergence 310 §14.4.
Littlewood’s Three Principles 314 §14.5.
Weyl’s Criterion 316 §14.6.
Proof of Birkhoff’s Ergodic Theorem 318 §14.7.
Exercises 323 Chapter 15.
The Unsolvability of the Quintic 337 §15.1.
Solvable Groups 337 §15.2.
The Derived Series of Sn 340 §15.3.
Solving Cubic and Quartic Equations 343 §15.4.
Monodromy 346 §15.5.
The General Quintic is Not Soluble via Radicals 349 §15.6.
Exercises 354 Bibliography 363 Index 367 Part 1 Introduction to Number Theory Chapter 1 A Quick Tour of Number Theory Overview. We give definitions of the following concepts of congruence and divisor in the integers, of rational and irrational number, and of countable versus uncountable sets. We also discuss some of the elementary properties of these notions.
Before we start, a general comment about the structure of this book may be helpful. Each chapter consists of a “bare bones” outline of a piece of the theory followed by a number of exercises. These exercises are meant to achieve two goals. The first is to get the student used to the mechanical or computational aspects of the theory. For example, the division algorithm in Chapter 2 comes back numerous times in slightly different guises. In Chap-ter 3, we use solve equations of the type ax+by = c for given a, b, and c, and in Chapter 6, we take that even further to study continued fractions. To rec-ognize and understand the use of the algorithm in these different contexts, it is therefore crucial that the student sufficient practice with elementary ex-amples. Thus, even if the algorithm is “more or less” clear or familiar, a wise student will carefully do all the computational problems in order for it to become “thoroughly” familiar. The second goal of the exercises is to extend the bare bones theory, and fill in some details covered in most text-books. For instance, in this Chapter we explain what rational and irrational numbers are. However, the proof that the number e is irrational is left to the 3 4 1. A Quick Tour of Number Theory exercises. In summary, as a rule the student should spend at least as much time on the exercises as on the theory.
The natural numbers starting with 1 are denoted by N, and the collec-tion of all integers (positive, negative, and 0) by Z. Elements of Z are also called integers .
1.1. Divisors and Congruences Definition 1.1. Given two numbers a and b. A multiple b of a is a number that satisfies b = ac. A divisor a of b is an integer that satisfies ac = b where c is an integer. We write a | b. This reads as a divides b or a is a divisor of b.
Definition 1.2. Let a and b non-zero. The greatest common divisor of two integers a and b is the maximum of the numbers that are divisors of both a and b. It is denoted by gcd(a,b). The least common multiple of a and b is the least of the positive numbers that are multiples of both a and b. It is denoted by lcm(a,b).
Note that for any a and b in Z, gcd(a,b) ≥1, as 1 is a divisor of every integer. Similarly lcm(a,b) ≤|ab|.
Definition 1.3. A number p > 1 is prime1 in N if its only divisors in N are a and 1 (the so-called trivial divisors). A number a > 1 is composite or reducible if it has more than 2 divisors in N. (The number 1 is neither.) 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Figure 1. Eratosthenes’ sieve up to n = 30. All multiples of a less than √ 31 are cancelled. The remainder are the primes less than n = 31.
1In a more general context — see Chapter 8 — these are called irreducible numbers, while the term prime is reserved for numbers satisfying Corollary 2.9.
1.1. Divisors and Congruences 5 An equivalent definition of prime is a natural number with precisely two (distinct) divisors. Eratosthenes’ sieve is a simple and ancient method to generate a list of primes for all numbers less than, say, 225. First, list all integers from 2 to 225. Start by circling the number 2 and crossing out all its remaining multiples: 4, 6, 8, etcetera. At each step, circle the smallest unmarked number and cross out all its remaining multiples in the list. It turns out that we need to sieve out only multiples of √ 225 = 15 and less (see exercise 2.5). This method is illustrated if Figure 1. When done, the primes are those numbers that are circled or unmarked in the list.
It will turn out that it is more natural to work in Z where all elements have an additive inverse. We therefore introduce extend the definition of primes to Z and introduce units.
Definition 1.4. A (multiplicative) unit in Z is a number a such that there is b ∈Z with the property that ab = 1. The only units in Z are 1 and −1. All other numbers are non-units. A number n ̸= 0 in Z is called composite or reducible if it can be written as a product of two non-units. If n is not 0, not a unit, and not composite, it is a prime or irreducible .
Remark 1.5. A concise way to characterize a unit is saying that it is an invertible element.
Definition 1.6. Let a and b in Z. Then a and b are relatively prime if gcd(a,b) = 1.
Definition 1.7. Let a and b in Z and m ∈N. Then a is congruent to b modulo m if a+my = b for some y ∈Z or m | (b−a). We write a =m b or a = b mod m or a ∈b+mZ .
Definition 1.8. The residue of a modulo m is the (unique) integer r in {0,···m−1} such that a =m r. It is denoted by Resm (a).
These notions are cornerstones of much of number theory as we will see. But they are also very common in all kinds of applications. For in-stance, our expressions for the time on the clock are nothing but counting modulo 12 or 24. To figure out how many hours elapse between 4pm and 3am next morning is a simple exercise in working with modular arithmetic, that is: computations involving congruences.
6 1. A Quick Tour of Number Theory 1.2. Rational and Irrational Numbers We start with a few results we need in the remainder of this subsection.
Theorem 1.9 (well-ordering principle). Any non-empty set S in N ∪{0} or in N has a smallest element.
Proof. Suppose this is false. Pick s1 ∈S. Then there is another natural number s2 in S such that s2 ≤s1 −1. After a finite number of steps, we pass zero, implying that S has elements less than 0 in it. This is a contradiction and so the supposition is false.
■ Note that any non-empty set S of integers with a lower bound can be transformed by addition of a integer b ∈N0 into a non-empty S + b in N0.
Then S+b has a smallest element, and therefore so does S. Furthermore, a non-empty set S of integers with a upper bound can also be transformed into a non-empty −S + b in N0. Here, −S stands for the collection of elements of S multiplied by −1. Thus we have the following corollary of the well-ordering principle.
Corollary 1.10. Let be a non-empty set S in Z with a lower (upper) bound.
Then S has a smallest (largest) element.
Definition 1.11. i) An element x ∈R is called rational if it a root of a degree 1 polynomial with integer coefficients, that is: qx−p = 0 where p and q ̸= 0 are integers.
ii) Otherwise it is called an irrational number.
Remark 1.12. Note that the integers themselves are also considered to be rational numbers: they satisfy part (i) above with the leading coefficient, q, equal to 1.
The set of integers is denoted by Z, and the rational numbers are de-noted by Q. The usual way of expressing a rational number is that it can be written as p q. The advantage of expressing a rational number as the solu-tion of a degree 1 polynomial, however, is that it naturally paves the way to Definitions 1.16 and 1.17.
Theorem 1.13. Any interval in R contains an element of Q. We say that Q is dense in R.
1.2. Rational and Irrational Numbers 7 The crux of the following proof is that we take an interval and scale it up until we know there is an integer in it, and then scale it back down.
Proof. Let I = (a,b) with b > a any interval in R. From Corollary 1.10 we see that there is an n such that n > 1 b−a. Indeed, if that weren’t the case, then N would be bounded from above, and thus it would have a largest element n0. But if n0 ∈N, then so is n0 + 1. This gives a contradiction and so the above inequality must hold.
It follows that nb −na > 1. Thus the interval (na,nb) contains an in-teger, say, p. So we have that na < p < nb. The theorem follows upon dividing by n.
■ Theorem 1.14.
√ 2 is irrational.
Proof. Suppose √ 2 can be expressed as the quotient of integers r s. We may assume that gcd(r,s) = 1 (otherwise just divide out the common factor).
After squaring, we get 2s2 = r2 .
The right-hand side is even, therefore the left-hand side is even. But the square of an odd number is odd, so r is even. But then r2 is a multiple of 4.
Thus s must be even. This contradicts the assumption that gcd(r,s) = 1.
■ It is pretty clear who the rational numbers are. But who or where are the others? We just saw that √ 2 is irrational. It is not hard to see that the sum of any rational number plus √ 2 is also irrational. Or that any rational non-zero multiple of √ 2 is irrational. The same holds for √ 2, √ 3, √ 5, etcetera. We look at this in exercise 1.7. From there, is it not hard to see that the irrational numbers are also dense (exercise 1.7). In exercise 1.13, we prove that the number e is irrational. The proof that π is irrational is a little harder and can be found in [section 11.17]. In Chapter 2, we will use the fundamental theorem of arithmetic, Theorem 2.11, to construct other irrational numbers. In conclusion, whereas rationality is seen at face value, irrationality of a number may take some effort to prove, even though they are much more numerous as we will see in Section 1.4.
If you think about it, we cannot express the exact numerical value of an irrational number! The only way to do that would be in a decimal (or any other base) expansion. But if such an expansion were finite, of course, the 8 1. A Quick Tour of Number Theory number would be rational! Thus the question of how well we can approxi-mate irrational numbers by rational ones arises (see exercise 1.16). Here is an important general result which we will have occasion to prove in Chapter 6.
Theorem 1.15. Let ρ ∈R be irrational. Then there are infinitely many p q ∈Q such that ρ −p q < 1 q2 .
1.3. Algebraic and Transcendental Numbers The set of polynomials with coefficients in Z, Q, R, or C is denoted by Z[x], Q[x], R[x], and C[x], respectively.
Definition 1.16. An element x ∈C is called an algebraic integer if it satis-fies p(x) = 0, where p is a non-zero polynomial in Z[x] with leading coeffi-cient 1.
Definition 1.17. An element x ∈C is called an algebraic number if it sat-isfies p(x) = 0, where p is a non-zero polynomial in Z[x]. Otherwise it is called a transcendental number.
The transcendental numbers are even harder to pin down than the gen-eral irrational numbers. We do know that e and π are transcendental, but the proofs are considerably more difficult (see ). We’ll see below that the transcendental numbers are far more abundant than the rationals or the alge-braic numbers. In spite of this, they are harder to analyze and, in fact, even hard to find. This paradoxical situation where the most prevalent numbers are hardest to find, is actually pretty common in number theory.
The most accessible tool to construct transcendental numbers is Liou-ville’s Theorem. The setting is the following. Given an algebraic number y, it is the root of a polynomial with integer coefficients f(x) = ∑d i=0 aixi, where we always assume that the coefficient ad of the highest power is non-zero. That highest power is called the degree of the polynomial and is denoted by deg( f) . Note that we can always find a polynomial of higher degree that has y as a root. Namely, multiply f by any other polynomial g.
Definition 1.18. We say that f(x) = ∑d i=0 aixi in Z[x] is a minimal polynomial in Z[x] for ρ if f is a non-zero polynomial in Z[x] of minimal degree, say d, such that f(ρ) = 0. We say that the degree of ρ is d.
1.3. Algebraic and Transcendental Numbers 9 Theorem 1.19 (Liouville’s Theorem). Let f be a minimal polynomial of degree d ≥2 for r ∈R. Then ∃c(r) > 0 such that ∀p q ∈Q : r −p q > c(r) qd .
Proof. Clearly, if r −p q ≥1, the inequality is satisfied. So assume that r −p q < 1. Now let f be a minimal polynomial for r (see Figure 2), and set K = max t∈[r−1,r+1] f ′(t) .
We know that f (p/q) is not zero, because otherwise f would have a factor (x−p/q). In that case, the quotient g of f and (x−p/q) would not neces-sarily have integer coefficients, but some integral multiple mg of g would.
However, mg would be of lower degree, thus contradicting the minimality of f. This gives us that qd f(p/q) is an integer, because qd f p q = d ∑ i=0 aipiqd−i ≥1, because it is a non-zero integer. Thus |f(p/q)| ≥q−d. Finally, we use the mean value theorem which tells us that for K as above, there is a t between r and p q such that K ≥ f ′(t) = f p q −f(r) p q −r ≥ q−d p q −r , since f(r) = 0. This gives us the desired inequality.
■ Definition 1.20. A real number ρ is called a Liouville number if for all n ∈N, there is a rational number p q such that ρ −p q < 1 qn .
10 1. A Quick Tour of Number Theory ( ) r−1 r r+1 p/q t f Figure 2. f is a minimal polynomial for the irrational number r. By minimality f ′(p/q) is not zero. On the interval (r −1,r), the absolute value of the derivative of f attains its maximum at t.
It follows directly from Liouville’s theorem that such numbers must be transcendental. Liouville numbers can be constructed fairly easily. The number ρ = ∞ ∑ k=1 10−k!
is an example. If we set p q equal to ∑n k=1 10−k!, then q = 10n!. Then ρ −p q = ∞ ∑ k=n+1 10−k! .
(1.1) It is easy to show that this is less than q−n (exercise 1.15).
It is worth noting that there is an optimal version of Liouville’s Theo-rem. We record it here without proof.
Theorem 1.21 (Roth’s Theorem). Let ρ ∈R be algebraic. Then for all ε > 0 ∃c(ρ,ε) > 0 such that ∀p q ∈Q : ρ −p q > c(ρ,ε) q2+ε , where c(ρ,ε) depends only on ρ and ε.
This result is all the more remarkable if we consider it in the context of Theorem 1.15.
1.4. Countable and Uncountable Sets Definition 1.22. i) A set S is finite if there is a bijection f : {1,··· ,n} →S for some n > 0.
ii) A set S is countably infinite if there is a bijection f : N →S.
1.4. Countable and Uncountable Sets 11 iii) A set S is countable if it is finite or if it is countably infinite. iv) An infinite set for which there is no bijection as in (ii) is called uncountable.
Proposition 1.23. Every infinite set S contains a countable subset.
Proof. Choose an element s1 from S. Now S−{s1} is not empty because S is not finite. So, choose s2 from S−{s1}. Then S−{s1,s2} is not empty be-cause S is not finite. In this way, we can remove sn+1 from S−{s1,s2,···sn} for all n. The set {s1,s2,···} is countable and is contained in S.
■ So countable sets are the smallest infinite sets in the sense that there are no infinite sets that contain no countable set. But there certainly are larger sets, as we will see next.
Theorem 1.24. The set R is uncountable.
Proof. The proof is one of mathematics’ most famous arguments: Cantor’s diagonal argument . The argument is developed in two steps .
Let T be the set of semi-infinite sequences formed by the digits 0 and 2. An element t ∈T has the form t = t1t2t3 ··· where ti ∈{0,2}. The first step of the proof is to prove that T is uncountable. So suppose it is countable. Then a bijection t between N and T allows us to uniquely define the sequence t(n), the unique sequence associated to n. Furthermore, they form an exhaustive list of the elements of T. For example, t(1) = 0,0,0,0,0,0,0,0,0,0,0··· t(2) = 2,0,2,0,2,0,2,0,2,2,2··· t(3) = 0,0,0,2,2,2,2,2,2,2,2··· t(4) = 2,2,2,2,2,2,0,0,0,0,0··· t(5) = 0,0,0,2,0,0,2,0,0,2,0··· t(6) = 2,0,0,0,0,2,0,0,0,2,2··· .
.
.
.
.
.
.
.
.
Construct t∗as follows: for every n, its nth digit differs from the nth digit of t(n). In the above example, t∗= 2,2,2,0,2,0,···. But now we have a contradiction, because the element t∗cannot occur in the list. In other words, there is no surjection from N to T. Hence there is no bijection between N and T.
12 1. A Quick Tour of Number Theory The second step is to show that there is a subset K of R such that there is no surjection (and thus no bijection) from N to K. Let t be a sequence with digits ti. Define f : T →R as follows f(t) = ∞ ∑ i=1 ti3−i .
If s and t are two distinct sequences in T, then for some k they share the first k −1 digits but tk = 2 and sk = 0. So f(t)−f(s) = 2·3−k + ∞ ∑ i=k+1 (ti −si)3−i ≥2·3−k −2 ∞ ∑ i=k+1 3−i = 3−k .
Thus f is injective. Therefore f is a bijection between T and the subset K = f(T) of R. If there is a surjection g from N to K = f(T), then, N g − →K f ← −T .
And so f −1g is a surjection from N to T. By the first step, this is impossible.
Therefore, there is no surjection g from N to K, much less from N to R.
■ The crucial part here is the diagonal step, where an element is con-structed that cannot be in the list. This really means the set T is strictly larger than N. The rest of the proof seems an afterthought, and perhaps needlessly complicated. You might think that it is much more straightfor-ward to just use the digits 0 and 1 and the representation of the real numbers on the base 2, as opposed to the digits 0 and 2 and the base 3. But if you do that, you run into a problem that has to be dealt with. The sequence t∗ might end with an infinite all-ones subsequence such as t∗= 1,1,1,1,···.
This corresponds to the real number x = 1.0... which might be in the list.
To circumvent that problem leads to slightly more complicated proofs (see exercise 1.9).
Meanwhile, this gives us a very nice corollary which we will have occasion to use in later chapters. For b an integer greater than 1, denote by {0,1,2,···b −1}N the set of sequences a1a2a3 ··· where each ai is in {0,1,2,···b−1}. Such sequences are often called words.
Corollary 1.25. (i) The set of infinite sequences in {0,1,2,···b −1}N is uncountable. (ii) The set of finite sequences (but without bound) in {0,1,2,···b− 1}N is countable.
1.4. Countable and Uncountable Sets 13 Proof. The proof of (i) is the same as the proof that T is uncountable in the proof of Theorem 1.24. The proof of (ii) consists of writing first all b words of length 1, then all b2 words of length 2, and so forth. Every finite string will occur in the list.
■ Theorem 1.26. (i) The set Z2 is countable. (ii) Q is countable.
Proof. (i) The proof relies on Figure 3. In it, a directed path γ is traced out that passes through all points of Z2. Imagine that you start at (0,0) and travel along γ with unit speed. Keep a counter c ∈N that marks the point (0,0) with a “1”. Up the value of the counter by 1 whenever you hit a point of Z2. This establishes a bijection between N and Z2.
Figure 3. A directed path γ passing through all points of Z2.
(ii) Again travel along γ with unit speed. Keep a counter c ∈N that marks the point (0,1) with a “1”. Up the value of the counter by 1. Con-tinue to travel along the path until you hit the next point (p,q) that is not a multiple of any previous and such q is not zero. Mark that point with the value of the counter. Q contains N and so is infinite. Identifying each marked point (p,q) with the rational number p q establishes the countability of Q.
■ Notice that this argument really tells us that the product (Z × Z)of a countable set (Z) and another countable set is still countable. The same 14 1. A Quick Tour of Number Theory holds for any finite product of countable set. Since an uncountable set is strictly larger than a countable, intuitively this means that an uncountable set must be a lot larger than a countable set. In fact, an extension of the above argument shows that the set of algebraic numbers numbers is count-able (see exercises 1.8 and 1.24). And thus, in a sense, it forms small subset of all reals. All the more remarkable, that almost all reals that we know anything about are algebraic numbers, a situation we referred to at the end of Section 1.4.
It is useful and important to have a more general definition of when two sets “have the same number of elements”.
Definition 1.27. Two sets A and B are said to have the same cardinality if there is a bijection f : A →B. It is written as |A| = |B|. If there is an injection f : A →B, then |A| ≤|B|.
Definition 1.28. An equivalence relation on a set A is a (sub)set R of or-dered pairs in A×A that satisfy three requirements.
- (a,a) ∈R (reflexivity).
- If (a,b) ∈R, then (b,a) ∈R (symmetry).
- If (a,b) ∈R and (b,c) ∈R, then If (a,c) ∈R (transitivity).
Usually (a,b) ∈R is abbreviated to a ∼b. The mathematical symbol “=” is an equivalence.
It is easy to show that having the same cardinality is an equivalence relation on sets (exercise 1.22). Note that the cardinality of a finite set is just the number of elements it contains. An excellent introduction to the cardinality of infinite sets in the context of naive set theory can be found in .
1.5. Exercises Exercise 1.1. Apply Eratosthenes’ Sieve to get all prime numbers between 1 and 200. (Hint: you should get 25 primes less than 100, and 21 between 100 and 200.) Exercise 1.2. Factor the following into prime numbers (write as a product of primes): 393, 16000, 5041, 1111, 1763, 720.
Exercise 1.3. Find pairs of primes that differ by 2. These are called twin primes. Do you think there there infinitely many such pairs? See Figure 4.
1.5. Exercises 15 Figure 4. Lowest member of nth twin prime pair less than 1000 (left) and less than 10000 (right), horizontal axis is n.
Conjecture 1.29 (Twin Prime Conjecture). There are infinitely many twin prime pairs2.
Exercise 1.4. Show that all even integers greater than 3 but smaller than 21 can be written as the sum of two primes. Is this always true?
Conjecture 1.30 (Goldbach Conjecture). Every even number greater than two is a sum of two (positive) primes3.
Exercise 1.5. Comment on the types of numbers (rational, irrational, tran-scendental) we use in daily life.
a) What numbers do we use to pay our bills?
b) What numbers do we use in computer simulations of complex pro-cesses?
c) What numbers do we use to measure physical things?
d) Give examples of the usage of the “other” numbers.
Exercise 1.6. Let a and b be rationals and x and y irrationals.
a) Show that ax is irrational if and only if a ̸= 0.
b) Show that b+x is irrational.
c) Show that ax+b is irrational if and only if a ̸= 0.
d) Conclude that a √ 2+b is irrational if and only if a ̸= 0.
Exercise 1.7. a) Show that √ 3, √ 5, et cetera (square roots of primes) are irrational. (Hint: use Corollary 2.9.) b) Show that for p prime, the numbers {a√p + b : a,b ∈Z} are dense in the reals.
2Still unsolved in 2022.
3Still unsolved in 2022.
16 1. A Quick Tour of Number Theory Lemma 1.31. The countable union of countable sets is countable.
Exercise 1.8. a) Use an pictorial argument similar to that of Figure 3 to show that N × N (the set of lattice points (n,m) with n and m in N) is countable.
b) Suppose Ai are countable sets where i ∈I and I countable. Show that there is a bijection {1,··· ,n} →I or N →I.
c) Define A′ 1 = A1, A′ 2 = A2\A′ 1, A′ 3 = A3{A′ 1 ∪A′ 2}, et cetera. Show that there is a bijection fi : {1,··· ,ni} →Ai or fi : N →Ai for each i.
c) Show there is a bijection F : N×N →S i∈I Ai given by F(n,m) = fn(m).
(Hint: place the elements of A′ 1 on (1,1), (1,2), (1,3), ...; the elements of A′ 2 on (2,1), (2,2), (2,3), ... and so on. Now use the argument in item (a).) d) Conclude that Lemma 1.31 holds.
Exercise 1.9. What is wrong in the following attempt to prove that [0,1] is uncountable?
Assume that [0,1] is countable, that is: there is a bijection f between [0,1] and N. Let r(n) be the unique number in [0,1] assigned to n. Thus the infinite array (r(1),r(2),···) forms an exhaustive list of the numbers in [0,1], as follows: r(1) = 0.00000000000··· r(2) = 0.10101010111··· r(3) = 0.00011111111··· r(4) = 0.11111100000··· r(5) = 0.00010010010··· r(6) = 0.10000100011··· .
.
.
.
.
.
.
.
.
(Written as number on the base 2.) Construct r∗as the string whose nth digit differs from that of r(n). Thus in this example: r∗= 0.111010··· , which is different from all the other listed binary numbers in [0,1].
(Hint: what if r∗ends with an infinite all ones subsequence?) Exercise 1.10. The set f(T) in the proof of Theorem 1.24 is called the middle third Cantor set. Find its construction. What does it look like?
(Hint: locate the set of numbers whose first digit (base 3) is a 1; then the set of numbers whose second digit is a 1.) 1.5. Exercises 17 Exercise 1.11. The integers exhibit many, many other intriguing patterns.
Given the following function: n even: f(n) = n 2 n odd: f(n) = 3n+1 2 .
a) (Periodic orbit) Show that f sends 1 to 2 and 2 to 1.
b) (Periodic orbit attracts) Show that if you start with any positive integer less than 18 and apply f repeatedly, eventually you fall on the orbit in (a).
See Figure 5.
c) Replace “+1” by “−1” and show that now 1 is a fixed point.
d) Show that the system is (c) has another periodic orbit.
Conjecture 1.32 (Collatz conjecture or 3n + 1 problem). The orbit of every positive integer under the map f defined in exercise 1.11 ends in 2 ↔ 14.
1 2 3 4 5 6 7 8 9 10 11 13 17 14 20 26 Figure 5. The orbits of n under f of exercise 1.11 for n ∈{1,··· ,10}.
Exercise 1.12. This exercise prepares for Mersenne and Fermat primes, see Definition 5.13.
a) Use ∑a−1 i=0 2ib = 2ab−1 2a−1 to show that if 2p −1 is prime, then p must be prime.
b) Use ∑a−1 i=0 (−2b)i = (−2b)a−1 (−2)a−1 to show that if 2p +1 is prime, then p has no odd factor. (Hint: assume a is odd.) 4Still unsolved in 2022. For an interesting survey, see M. Chamberland, A 3x+ 1 survey: Number theory and dynamical systems in 18 1. A Quick Tour of Number Theory Exercise 1.13. In what follows, we assume that e −1 = ∑∞ i=1 1 i! = p q is rational and show that this leads to a contradiction.
a) Show that the above assumption implies that q ∑ i=1 q!
i! + ∞ ∑ i=1 q!
(q+i)! = p(q−1)! .
(Hint: multiply both sides of by q! .) b) Show that ∑∞ i=1 q!
(q+i)! < ∑∞ i=1 1 (q+1)i . (Hint: write out a few terms of the sum on the left.) c) Show that the sum on the left hand side in (b) cannot have an integer value.
d) Show that the other two terms in (a) have an integer value.
e) Conclude there is a contradiction unless the assumption that e is rational is false.
Exercise 1.14. Show that Liouville’s theorem (Theorem 1.19) also holds for rational for rational numbers ρ = r s as long as p q ̸= r s.
Exercise 1.15. a) Show that for all positive integers p and n, we have p(n+1)n! ≤(n+ p)! .
b) Use (a) to show that ∞ ∑ k=n+1 10−k! ≤ ∞ ∑ p=1 10−p(n+1)n! = 10−(n+1)n! 1−10−(n+1)n!−1 .
c) Show that b) implies the affirmation after equation (1.1).
Exercise 1.16. a) Use a calculator to write down the decimal expansion of √ 2 in 10 decimal places.
b) How close to √ 2 is the decimal approximation 1414/1000?
c) Compute 1393/985 is 10 decimal places. How close is it to √ 2? (Hint: compare with Theorem 1.15.) Exercise 1.17. Show that the inequality of Roth’s theorem does not hold for all numbers. (Hint: Let ρ be a Liouville number.) Definition 1.33. Let A be a set. Its power set P(A) is the set whose elements are the subsets of A. This always includes the empty set denoted by / 0.
In the next two exercises, the aim is to show something that is obvious for finite sets, namely: Theorem 1.34. The cardinality of a power set is always (strictly) greater than that of the set itself.
1.5. Exercises 19 Exercise 1.18. a) Given a set A, show that there is an injection f : A → P(A). (Hint: for every element a ∈A there is a set {a}.) b) Conclude that |A| ≤|P(A)|. (Hint: see Definition 1.27.) Exercise 1.19. Let A be an arbitrary set. Assume that that there is a surjec-tion S : A →P(A) and define R = {a ∈A|a ̸∈S(a)} .
(1.2) a) Show that there is a q ∈A such that S(q) = R.
b) Show that if q ∈R, then q ̸∈R. (Hint: equation (1.2).) c) Show that if q ̸∈R, then q ∈R. (Hint: equation (1.2).) d) Use (b) and (c) and exercise 1.18, to establish that |A| < |P(A)|. (Hint: see Definition 1.27.) In the next two exercises we show that the cardinality of R equals that of P(N). This implies that that |R| > |N|, which also follows from Theorem 1.24.
Exercise 1.20. Let T be the set of sequences defined in the proof of Theo-rem 1.24. To a sequence t ∈T, associate a set S(t) in P(N) as follows: i ∈S if t(i) = 2 and i ̸∈S if t(i) = 0 .
a) Show that there is a bijection S : T →P(N).
b) Use the bijection f in the proof of Theorem 1.24 to show there is a bi-jection K →P(N).
c) Show that (a) and (b) imply that |P(N)| = |K| = |T|. (Hint: see Defini-tion 1.27.) d) Find an injection K →R and conclude that |P(N)| ≤|R|.
Exercise 1.21. a) Show that there is a bijection R →(0,1).
b) Show that there is an injection (0,1) →T. (Hint: use usual binary (base 2) expansion of reals.) c) Use (a), (b), and exercise 1.20 (a), to show that |R| ≤|P(N)|.
d) Use (c) and exercise 1.20 (d) to show that |R| = |P(N)|.
Exercise 1.22. Show that having the same cardinality (see Definition 1.27) is an equivalence relation on sets.
Exercise 1.23. a) Fix some n > 0. Show that having the same remainder modulo n is an equivalence relation on Z. (Hint: for example, -8, 4, and 16 have remainder 4 modulo 12.) b) Show that addition respects this equivalence relation. (Hint: If a+b = c, a ∼a′, and b ∼b′, then a′ +b′ = c′ with c ∼c′.) c) The same question for multiplication.
20 1. A Quick Tour of Number Theory Exercise 1.24. a) Show that the set of algebraic numbers is countable.
(Hint: use Lemma 1.31.) b) Conclude that the transcendental numbers form an uncountable set.
Exercise 1.25. Base 60 number systems have a long history and are still used (think of the number of minute in an hour). Suppose you do not have a good theory of fractions. Why is base 60 convenient? (Hint: what is the least common multiple of the numbers 1 through 6?) Exercise 1.26. a) Show that rectangular grid of n by m squares can be divided into squares of size d by d where d is a common divisor of n and m.
b) Show that in (a) the largest d equals gcd(n,m), see Figure 6.
30 12 Figure 6. A rectangle of 30 by 12 squares can be subdivided into squares non larger than 6 by 6.
Exercise 1.27. Suppose two meshing gear wheels have n and m teeth, re-spectively. Each wheel has one marked tooth.
a) Show that the positions of the wheels after ℓteeth are traversed is in-dicated by the projection of the point (ℓ,ℓ) on both in a rectangular coor-dinate system with n by m units. See Figure 7. (Hint: each small square corresponds to the turn through one tooth on both wheels. Show that the first time the marked teeth return exactly to their original position occurs when the first wheel has made lcm(n,m)/n = m/gcd(n,m) complete turns and the second lcm(n,m)/n = n/gcd(n,m).
1.5. Exercises 21 12 12 30 12 30 12 12 Figure 7. Two meshing gear wheels have 30, resp. 12 teeth. Each tiny square represents the turning of one tooth in each wheel. After precisely 5 turns of the first wheel and 2 of the second, both are back in the exact same position.
Chapter 2 The Fundamental Theorem of Arithmetic Overview. We derive the Fundamental Theorem of Arithmetic. The most important part of that theorem says every integer can be uniquely written as a product of primes up to re-ordering of the factors, and up to factors -1.
We discuss two of its most important consequences, namely the fact that the number of primes is infinite and the fact that non-integer roots are irrational.
On the way to proving the Fundamental Theorem of Arithmetic, we need B´ ezout’s Lemma and Euclid’s Lemma. The proofs of these well-known lemma’s may appear abstract and devoid of intuition. To have some intuition, the student may assume the Fundamental Theorem of Arithmetic and derive from it each of these lemma’s (see Exercise 2.9) and things will seem much more intuitive. The reason we do not do it that way in this book is of course that indirectly we use both results to establish the Fundamental Theorem of Arithmetic.
The principal difference between Z and N is that in Z addition has an inverse (subtraction). This makes Z a into a ring, a type of object we will encounter in Chapter 5. It will thus save us a lot of work and is not much more difficult to work in Z instead of in N.
23 24 2. The Fundamental Theorem of Arithmetic 2.1. B´ ezout’s Lemma Definition 2.1. The floor of a real number θ is defined as follows: ⌊θ⌋is the greatest integer less than than or equal to θ. The fractional part {θ} of the number θ is defined as θ −⌊θ⌋. Similarly, the ceiling of θ, ⌈θ⌉, gives the smallest integer greater than or equal to θ.
By the well-ordering principle, Corollary 1.10, the number ⌊θ⌋and ⌈θ⌉ exist for any θ ∈R. Given a number ξ ∈R, we denote its absolute value by |ξ|.
Lemma 2.2. Given r1 and r2 with r2 > 0, then there are q2 and r3 with |r3| < |r2| such that r1 = r2q2 +r3.
Proof. Noting that r1 r2 is a rational number, we can choose the integer q2 = ⌊r1 r2 ⌋so that r1 r2 = q2 +e , where e ∈[0,1) (see Figure 8). The integer q2 is called the quotient. Multi-plying by r2 gives the result.
■ 0 q e r /r 1 2 Figure 8. The division algorithm: for any two integers r1 and r2, we can find an integer q and a real e ∈[0,1) so that r1/r2 = q2 +e.
Note that in this proof, in fact, r3 ∈{0,···r2 −1} and is unique. Thus among other things, this lemma implies that every integer has a unique residue (see Definition 1.8). More generally, we just require |r3| < |r2|, and there is more than one choice for q2. This is typical in the more general context of rings (Chapter 8).
If |r1| < |r2|, then we can choose q2 = 0. In this case, ε = r1 r2 and we learn nothing new. But if |r1| > |r2|, then q2 ̸= 0 and we have written r1 as a multiple of r2 plus a remainder r3.
2.1. B´ ezout’s Lemma 25 Definition 2.3. Given r1 and r2 with r2 > 0, the computation of q2 and r3 in Lemma 2.2 is called the division algorithm. Note that r3 = Resr2 (r1) (see Definition 1.8).
Remark 2.4. Lemma 2.2 is also called Euclid’s division lemma. This is not to be confused with the Euclidean algorithm of Definition 3.3.
Lemma 2.5. (B´ ezout’s Lemma) Let a and b be such that gcd(a,b) = d.
Then ax + by = c has integer solutions for x and y if and only if c is a multiple of d.
Proof. Let S and ν(S) be the sets: S = {ax+by : x,y ∈Z, ax+by ̸= 0} ν(S) = {|s| : s ∈S} ⊆N∪{0} .
Then ν(S) ̸= / 0 (it contains |a| and |b|) and is bounded from below. Thus by the well-ordering principle of N, it has a smallest element n. Then there is an element d ∈S that has that norm: |d| = n.
For that d, we use the division algorithm to establish that there are q and r ≥0 such that a = dq+r and |r| < |d| .
(2.1) Now substitute d = ax+by. A short computation shows that r can be rewrit-ten as: r = a(1−qx)+b(−qy) .
Suppose r ̸= 0. Then this shows that r ∈S. But we also know from (2.1) that |r| is smaller than |d|. This is a contradiction because of the way d is defined. But r = 0 implies that d is a divisor of a. The same argument shows that d is also a divisor of b. Thus d is a common divisor of both a and b.
Now let e be any divisor of both a and b. Then e | (ax + by), and so e | d. But if e | d, then |e| must be smaller than or equal to |d|. Therefore, d is the greatest common divisor of both a and b.
By multiplying x and y by f, we achieve that for any multiple fd of d that afx+bfy = fd .
26 2. The Fundamental Theorem of Arithmetic On the other hand, let d be as defined above and suppose that x, y, and c are such that ax+by = c.
Since d divides a and b, we must have that d | c, and thus c must be a multiple of d.
■ 2.2. Corollaries of B´ ezout’s Lemma Lemma 2.6. (Euclid’s Lemma) Let a and b be such that gcd(a,b) = 1 and a | bc. Then a | c.
Proof. By B´ ezout, there are x and y such that ax+by = 1. Multiply by c to get: acx+bcy = c .
Since a | bc, the left-hand side is divisible by a, and so is the right-hand side.
■ Euclid’s lemma is so often used, that it will pay off to have a few of the standard consequences for future reference.
Theorem 2.7 (Cancellation Theorem). Let gcd(a,b) = 1 and b positive.
Then ax =b ay if and only if x =b y.
Proof. The statement is trivially true if b = 1, because all integers are equal modulo 1.
If ax =b ay, then a(x−y) =b 0. The latter is equivalent to b | a(x−y).
The conclusion follows from Euclid’s Lemma. Vice versa, if x =b y, then (x−y) is a multiple of b and so a(x−y) is a multiple of b.
■ Used as we are to cancellations in calculations in R, it is easy to un-derestimate the importance of this result. As an example, consider solving 21x =35 21y. It is tempting to say that this implies that x =35 y. But in fact, gcd(21,35) = 7 and the solution set is x =5 y, as is easily checked. This example is in fact a special case of the following corollary.
Corollary 2.8. Let gcd(a,b) = d and b positive. Then ax =b ay if and only if x =b/d y.
2.3. The Fundamental Theorem of Arithmetic 27 Proof. Divide by d to get a d x = b d a d y and apply the cancellation theorem.
■ For the following results, recall the definition of primes in Z (Definition 1.4).
Corollary 2.9. For any n ≥1, p is prime and p | ∏n i=1 ai, if and only if there is j ≤n such that p | aj.
Proof. If p | aj, then p | ∏n i=1 ai. We prove the other direction by induction on n, the number of terms in the product. Let S(n) be the statement of the corollary. S(1) says: If p is prime and p | a1, then p | a1, which is trivially true.
For the induction step, suppose that for any k > 1, S(k) is valid and let p | ∏k+1 i=1 ai. Then p | k ∏ i=1 ai !
ak+1 !
.
Applying Euclid’s Lemma, it follows that p | k ∏ i=1 ai or, if not, then p | ak+1 .
In the former case S(k + 1) holds because S(k) does. In the latter, we see that S(k +1) also holds.
■ Corollary 2.10. If p and qi are prime and p | ∏n i=1 qi, then there is j ≤n such that p = q j.
Proof. Corollary 2.9 says that if p and all qi are primes, then there is j ≤n such that p | qj. Since qj is prime, its only divisors are 1 and itself. Since p ̸= 1 (by the definition of prime), p = qj.
■ 2.3. The Fundamental Theorem of Arithmetic The last corollary of the previous section enables us to prove the most im-portant result of this chapter.
Theorem 2.11 (The Fundamental Theorem of Arithmetic). Every non-zero integer n ∈Z i) is a product of powers of primes (up to multiplication by units) and 28 2. The Fundamental Theorem of Arithmetic ii) that product is unique (up to the order of multiplication and up to multi-plication by the units).
Remark 2.12. The theorem is also called the unique factorization theorem.
Its statement means that up to re-ordering of the pi and factors ±1, every integer n can be uniquely expressed as n = ±1· r ∏ i=1 pℓi i , where the pi are distinct primes.
Proof. First we prove (i). Define S to be the set of integers n that are not products of primes times a unit, and the set ν(S) their absolute values. If the set S is non-empty, then by the well-ordering principle (Theorem 1.9), ν(S) has a smallest element. Let a be one of the elements in S that minimize ν(S).
If a is prime, then it can be factored into primes, namely a = a, which contradicts the assumption. Thus a is a composite number, a = bc and both b and c are non-units. Thus |b| and |c| are strictly smaller than |a|. By assumption, both b and c are products of primes. Then, of course, so is a = bc. But this contradicts the assumptions on a.
Next, we prove (ii). Let S be the set of integers that have more than one factorization and ν(S) the set of their absolute values. If the set S is non-empty, then, again by the well-ordering principle, ν(S) has a smallest element. Let a be one of the elements in S that minimize ν(S).
Thus we have a = u r ∏ i=1 pi = u′ s ∏ i=1 p′ i , where at least some of the pi and p′ i do not match up. Here, u and u′ are units. Clearly, p1 divides a. By Corollary 2.10, p1 equals one of the p′ i, say, p′ 1. Since primes are not units, a p1 is strictly less than |a|. Therefore, by hypothesis, a p1 is uniquely factorizable. But then the primes in a p1 = u r ∏ i=2 pi = u′ s ∏ i=2 p′ i , all match up (up to units).
■ 2.4. Corollaries of the Fundamental Theorem of Arithmetic 29 Remark 2.13. It is interesting to note that the proof of this theorem depends on two distinct characterizations of primes. In part (i), we use Definition 1.4, which essentially says that primes are numbers that cannot be factored into smaller numbers (the literal meaning of “irreducible”). But for part (ii), we essentially use the fact that if a prime p divides ab, then it divides a or b (or both). Now (through Corollary 2.10) we know both characterizations hold in Z, but it will turn out that they are not equivalent in general (see Proposition 8.3).
If the reader investigates the arguments carefully, it will become clear that underneath it all lurks the division algorithm in Z. To wit, we use Corollary 2.10 which Corollary 2.9 which uses Euclid’s lemma which uses B´ ezout which finally uses the division algorithm. It is precisely this division algorithm that is not available in all rings, and which plays an important role in algebraic number theory, see Chapter 8).
Remark 2.14. The student might reflect on this and conclude that one can-not write 1 as a product of primes. So how come that in Theorem 2.11 we do not make an exception for the number 1 (or -1 for that matter). The answer is this: 1 is a unit times “the empty product” of primes, and this is unique. This piece of apparent sophistry actually turns out to be useful as we will see in Chapter 8 (corollary 8.14).
2.4. Corollaries of the Fundamental Theorem of Arithmetic The unique factorization theorem is intuitive and easy to use. It is very effective in proving a great number of results. Some of these results can be proved with a little more effort without using the theorem (see exercise 2.6 for an example). We start with two somewhat technical results that we need for later reference.
Lemma 2.15. We have ∀i ∈{1,···n} : gcd(ai,b) = 1 ⇐ ⇒ gcd( n ∏ i=1 ai,b) = 1.
Proof. The easiest way to see this uses prime power factorization. If gcd(∏n i=1 ai,b) = d > 1, then d contains a factor p > 1 that is a prime.
Since p divides ∏n i=1 ai, at least one of the ai must contain (by Corollary 30 2. The Fundamental Theorem of Arithmetic 2.9) a factor p. Since p also divides b, this contradicts the assumption that gcd(ai,b) = 1.
Vice versa, if gcd(ai,b) = d > 1 for some i, then also ∏n i=1 ai is divisible by d.
■ Corollary 2.16. For all a and b in Z not both equal to 0, we have that gcd(a,b)· lcm(a,b) = ab up to units.
Proof. Given two numbers a and b, let P = {pi}k i=1 be the list of all prime numbers occurring in the unique factorization of a or b. We then have: a = u s ∏ i=1 pki i and b = u′ s ∏ i=1 pℓi i , where u and u′ are units and ki and ℓi in N∪{0}. Now define: mi = min(ki,ℓi) and Mi = max(ki,ℓi) , and let the numbers m and M be given by m = s ∏ i=1 pmi i and M = s ∏ i=1 pMi i .
Since mi +Mi = ki +ℓi, it is clear that the multiplication m·M yields ab.
Now all we need to do, is showing that m equals gcd(a,b) and that M equals lcm(a,b). Clearly m divides both a and b. On the other hand, any integer greater than m has a unique factorization that either contains a prime not in the list P and therefore divides neither a nor b, or, if not, at least one of the primes in P in its factorization has a power greater than mi. In the last case m is not a divisor of at least one of a and b. The proof that M equals lcm(a,b) is similar.
■ A question one might ask is: how many primes are there? In other words, how long can the list of primes in a factorization be? Euclid provided the answer around 300BC.
Theorem 2.17 (Infinitude of Primes). There are infinitely many primes.
Proof. Suppose the list P of all primes is finite, so that P = {pi}n i=1. Define the integer d as the product of all primes (to the power 1): d = n ∏ i=1 pi .
2.5. The Riemann Hypothesis 31 If d +1 is a prime, we have a contradiction. So d +1 must be divisible by a prime pi in P. But then we have pi | d and pi | d +1 .
(2.2) But since (d +1)(1)+d(−1) = 1, B´ ezout’s lemma implies that gcd(d,d + 1) = 1, which contradicts equation (2.2).
■ One the best known consequences of the fundamental theorem of arith-metic is probably the theorem that follows below. A special case, namely √ 2 is irrational (see Theorem 1.14), was known to Pythagoras in the 6th century BC.
Theorem 2.18. Let n > 0 and k > 1 be integers. Then n 1 k is either an integer or irrational.
Proof. Assume n 1 k is rational. That is: suppose that there are integers a and b such that n 1 k = a b = ⇒ n·bk = ak .
Divide out any common divisors of a and b, so that gcd(a,b) = 1. Then by the fundamental theorem of arithmetic, b = ∏s i=1 pmi i and a = ∏r i=s+1 pℓi i (a and b share no prime factors) and so n s ∏ i=1 pkmi i = r ∏ i=s+1 pkℓi i .
The primes pi on the left and right side are distinct. This is only possible if ∏s i=1 pkmi i equals 1. But then n is the k-th power of an integer.
■ 2.5. The Riemann Hypothesis Analytic continuation will be discussed in more detail in Chapter 11. For now, we note that it is akin to replacing ex where x is real by ez where z is complex. A better example is the series ∑∞ j=0 zj. This series diverges for |z| > 1. But as an analytic function, it can be replaced by (1−z)−1 on all of C except at the pole z = 1 where it diverges.
Analytic continuations are meaningful because they are unique. The reason this is true is roughly as follows (for details, see Theorem 11.22).
32 2. The Fundamental Theorem of Arithmetic Analytic functions are functions that are differentiable, that is to say, wher-ever the derivative is non-zero, the derivative equals a scaling times a rota-tion. Equivalently, they are locally given by a convergent power series. If f and g are two analytic continuations to a region U of a function h given on a region V ⊂U, then the difference f −g is zero on V. One can then show that the power series of f −g must be zero on the entire region U. Hence, analytic continuations f and g are unique.
Definition 2.19. The Riemann zeta function ζ(z) is a complex function de-fined on {z ∈C|Rez > 1} by ζ(z) = ∞ ∑ n=1 n−z .
On other values of z ∈C it is defined by the analytic continuation of this function (except at z = 1 where it has a simple pole).
In analytic number theory, it is common to denote the argument of the zeta function by s, while in other branches of complex analysis z is the go-to complex variable. We will stick to the latter. Note that n−z = e−lnn Rez−ilnn Imz , and so |n−z| = n−Rez. Therefore for Rez > 1 the series is absolutely con-vergent. More about this in Chapter 11. At this point, the student should remember – or look up in – the fact that absolutely convergent series can be re-arranged arbitrarily without changing the sum. This leads to the following proposition.
Proposition 2.20 (Euler’s Product Formula). For Rez > 1 we have ζ(z) := ∞ ∑ n=1 n−z = ∏ p prime (1−p−z)−1 .
There are two common proofs of this formula. It is worth presenting both.
proof 1. The first proof uses the Fundamental Theorem of Arithmetic. First, we use the geometric series (1−p−z)−1 = ∞ ∑ k=0 p−kz 2.5. The Riemann Hypothesis 33 to rewrite the right-hand side of the Euler product. This gives ∏ p prime (1−p−z)−1 = ∞ ∑ k1=0 p−k1z 1 ! ∞ ∑ k2=0 p−k2z 2 ! ∞ ∑ k3=0 p−k3z 3 !
··· Re-arranging terms yields ··· = ∑ k1,k2,k3,···≥0 pk1 1 pk2 2 pk3 3 ··· −z .
By the Fundamental Theorem of Arithmetic, the expression pk1 1 pk2 2 pk3 3 ··· runs through all positive integers exactly once. Thus upon re-arranging again we obtain ∑∞ n=1 n−z.
■ proof 2. The second proof, the one that Euler used, employs a sieve method.
This time, we start with the left-hand side of the Euler product. If we mul-tiply ζ by 2−z, we get back precisely the terms with n even. So 1−2−z ζ(z) = 1+3−z +5−z +··· = ∑ 2∤n n−z .
Subsequently we multiply this expression by (1−3−z). This has the effect of removing the terms that remain where n is a multiple of 3. It follows that eventually 1−p−z ℓ ··· 1−p−z 1 ζ(z) = ∑ p1∤n,···pℓ∤n n−z .
The argument used in Eratosthenes’ sieve (Section 1.1) now serves to show that in the right-hand side of the last equation all terms other than 1 disap-pear as ℓtends to infinity. Therefore, the left-hand side tends to 1, which implies the proposition.
■ The most important theorem concerning primes is probably the follow-ing. We will give a proof in Chapter 12.
Theorem 2.21 (Prime Number Theorem). Let π(x) denote the prime counting function, that is: the number of primes less than or equal to x with x ≥2. Then 1) lim x→∞ π(x) (x/lnx) = 1 and 2) lim x→∞ π(x) R x 2 lnt dt = 1, where ln is the natural logarithm.
34 2. The Fundamental Theorem of Arithmetic Figure 9. On the left, the function R x 2 lnt dt in blue, π(x) in red, and x/lnx in green. On the right, we have R x 2 lnt dt −x/lnx in blue, π(x)− x/lnx in red. Note the different scales.
The first estimate is the one we will prove directly in Chapter 12. It turns out the second is equivalent to it (exercise 12.10). However, it is this one that gives the better estimate of π(x). In Figure 9 on the left, we plotted, for x ∈[2,1000], from top to bottom the functions R x 2 lnt dt in blue, π(x) in red, and x/lnx. In the right-hand figure, we augment the domain to x ∈[2,105]. and plot the difference of these functions with x/lnx. It now becomes clear that R x 2 lnt dt is indeed a much better approximation of π(x).
From this figure one may be tempted to conclude that R x 2 lnt dt −π(x) is always greater than or equal to zero. This, however, is false. It is known that there are infinitely many n for which π(n) > R n 2 lnt dt. The first such n is called the Skewes number. Not much is known about this number1, except that it is less than 10317.
Perhaps the most important open problem in all of mathematics is the following. It concerns the analytic continuation of ζ(z) given above.
Conjecture 2.22 (Riemann Hypothesis). All non-real zeros of ζ(z) lie on the line Rez = 1 2.
In his only paper on number theory , Riemann realized that the hypothesis enabled him to describe detailed properties of the distribution 1In 2020.
2.6. Exercises 35 of primes in terms of of the location of the non-real zero of ζ(z). This completely unexpected connection between so disparate fields — analytic functions and primes in N — spoke to the imagination and led to an enor-mous interest in the subject2 In further research, it has been shown that the hypothesis is also related to other areas of mathematics, such as, for exam-ple, the spacings between eigenvalues of random Hermitian matrices , and even physics [12,15].
2.6. Exercises Exercise 2.1. Apply the division algorithm to the following number pairs.
(Hint: replace negative numbers by positive ones.) a) 110 , 7.
b) 51 , −30.
c) −138 , 24.
d) 272 , 119.
e) 2378 , 1769.
f) 270 , 175560.
Exercise 2.2. In this exercise we will exhibit the division algorithm applied to polynomials x+1 and 3x3 +2x+1 with coefficients in Q, R, or C.
a) Apply long division to divide 3021 by 11. (Hint: 3021 = 11·275−4.) b) Apply the exact same algorithm to divide 3x3 +2x+1 by x+1. In this algorithm, xk behaves as 10k in (a). (Hint: at every step, cancel the highest power of x.) c) Verify that you obtain 3x3 +2x+1 = (x+1)(3x2 −3x+5)−4.
d) Show that in general, if p1 and p2 are polynomials such that the degree of p1 is greater or equal to the degree of p2, then p1 = q2p2 + p3 , where the degree of p3 is less than the degree of p2. (Hint: perform long division as in (b). Stop when the degree of the remainder is less than that of p2.) e) Why does this division not work for polynomials with coefficients in Z?
(Hint: replace x+1 by 2x+1.) Exercise 2.3. a) For a, b in Z, let gcd(a,b) = 1. Show that if a | c and b | c, then ab | c. (Hint: observe that a | by and use Euclid’s lemma.) b) Show that ax =m c has a solution if and only gcd(a,m) | c. (Hint: note that ax+my = c for some y and use B´ ezout.) 2This area of research, complex analysis methods to investigate properties of primes, is now called analytic number theory. We take this up in Chapters 11 and 12.
36 2. The Fundamental Theorem of Arithmetic Exercise 2.4. a) Compute by long division that 3021 = 11·274+7.
b) Conclude from exercise 2.2 that 3021 = 11(300−30+5)−4. (Hint: let x = 10.) c) Conclude from exercise 2.2 (b) that 3·163 +2·16+1 = 17(3·162 −3·16+5)−4 .
(Hint: let x = 16.) Exercise 2.5. a) Use unique factorization to show that any composite num-ber n must have a prime factor less than or equal to √n.
b) Use that fact to prove: If we apply Eratosthenes’ sieve to {2,3,···n}, it is sufficient to sieve out numbers less than or equal to √n.
Exercise 2.6. We give an elementarya proof of Corollary 2.16.
a) Show that a· b gcd(a,b) is a multiple of a.
b) Show that a gcd(a,b) ·b is a multiple of b.
c) Conclude that ab gcd(a,b) is a multiple of both a and b and thus greater than or equal to lcm(a,b).
d) Show that a/ ab lcm(a,b) = lcm(a,b) b is an integer. Thus ab lcm(a,b) is a divisor of a.
e) Similarly, show that ab lcm(a,b) is a divisor of b.
f) Conclude that ab lcm(a,b) ≤gcd(a,b).
h) Finish the proof.
aThe word elementary has a complicated meaning, namely a proof that does not use some at first glance unrelated results. In this case, we mean a proof that does not use unique factor-ization. It does not imply that the proof is easier. Indeed, the proof in the main text seems much easier once unique factorization is understood.
Exercise 2.7. It is possible to extend the definition of gcd and lcm to more than two integers (not all of which are zero). For example gcd(24,27,54) = 3.
a) Compute gcd(6,10,15) and lcm(6,10,15).
b) Give an example of a triple whose gcd is one, but every pair of which has a gcd greater than one.
c) Show that there is no triple {a,b,c} whose lcm equals abc, but every pair of which has lcm less than the product of that pair. (Hint: consider lcm(a,b)·c.) Exercise 2.8. a) Give the prime factorization of the following numbers: 12, 392, 1043, 31, 128, 2160, 487.
b) Give the prime factorization of the following numbers: 12 · 392, 1043 · 31, 128·2160.
c) Give the prime factorization of: 1,250000, 633, 720, and the product of the last three numbers.
2.6. Exercises 37 Exercise 2.9. Use the Fundamental Theorem of Arithmetic to prove: a) B´ ezout’s Lemma.
b) Euclid’s Lemma.
Exercise 2.10. For positive integers m and n, suppose that mα = n. Show that α = a b with gcd(a,b) = 1 if and only if m = s ∏ i=1 pki i and n = s ∏ i=1 pℓi i with ∀i : aki = bℓi .
Exercise 2.11. Let E be the set of even numbers. Let a, c in E, then c is divisible by a if there is a b ∈E so that ab = c. Define a prime p in E as a number in E such that there are no a and b in E with ab = p.
a) List the first 30 primes in E.
b) Does Euclid’s lemma hold in E? Explain.
c) Factor 60 into primes (in E) in two different ways.
Exercise 2.12. See exercise 2.11. Show that any number in E is a product of primes in E. (Hint: follow the proof of Theorem 2.11, part (i).) Exercise 2.13. See exercise 2.11 which shows that unique factorization does not hold in E = {2,4,6,···}. The proof of unique factorization uses Euclid’s lemma.
In turn, Euclid’s lemma was a corollary of B´ ezout’s lemma, which depends on the division algorithm. Where exactly does the chain break down in this case?
Exercise 2.14. Let L = {p1, p2,···} be the list of all (infinitely many) primes, ordered according ascending magnitude.
Show that pn+1 ≤ ∏n i=1 pi. (Hint: consider d = ∏n i=1 pi and let pn+1 be the smallest prime divisor of d −1. See the proof of Theorem 2.17.) A much stronger version of exercise 2.14 is the so-called Bertrand’s Pos-tulate. That theorem says that for every n ≥1, there is a prime in {n + 1,··· ,2n}. It was proved by Chebyshev. Subsequently the proof was sim-plified by Ramanujan and Erd¨ os .
Exercise 2.15. Let p and q be primes greater than 3.
a) Show that Res12 (p) = r with r ∈{1,5,7,11}. (The same holds for q.) b) Show that 24 | p2 −q2. (Hint: use (a) to show that p2 = 24x + r2 and check all cases.) 38 2. The Fundamental Theorem of Arithmetic Exercise 2.16. A square full integer is an integer n that has a prime factor and each prime factor occurs with a power at least 2. A square free integer is an integer n such that each prime factor occurs with a power at most 1.
a) If n is square full, show that there are positive integers a and b such that n = a2b3.
b) Show that every integer greater than one is the product of a square free number and a square full number.
Exercise 2.17. Let L = {p1, p2,···} be the list of all primes, ordered ac-cording ascending magnitude. The numbers En = 1 + ∏n i=1 pi are called Euclid numbers.
a) Check the primality of E1 through E6.
b) Show that En =4 3. (Hint: En −1 is twice an odd number.) c) Show that for n ≥3 the decimal representation of En ends in a 1. (Hint: look at the factors of En.) Exercise 2.18. Twin primes are a pair of primes of the form p and p+2.
a) Show that the product of two twin primes plus one is a square.
b) Show that p > 3, the sum of twin primes is divisible by 12. (Hint: see exercise 2.15) Exercise 2.19. Show that there arbitrarily large gaps between successive primes. More precisely, show that every integer in {n!+2,n!+3,···n!+n} is composite for any n ≥2.
The usual statement for the fundamental theorem of arithmetic includes only natural numbers n ∈N (i.e. not Z) and the common proof uses in-duction on n. We review that proof in the next two problems.
Exercise 2.20. a) Prove that 2 can be written as a product of primes.
b) Let k > 2. Suppose all numbers in {1,2,···k} can be written as a product of primes (or 1). Show that k +1 is either prime or composite.
c) If in (b), k+1 is prime, then all numbers in {1,2,···k+1} can be written as a product of primes (or 1).
d) If in (b), k + 1 is composite, then there is a divisor d ∈{2,···k} such that k +1 = dd′.
e) Show that the hypothesis in (b) implies also in this case, all numbers in {1,2,···k +1} can be written as a product of primes (or 1).
f) Use the above to formulate the inductive proof that all elements of N can be written as a product of primes.
2.6. Exercises 39 Exercise 2.21. The set-up of the proof is the same as in exercise 2.20. Use induction on n. We assume the result of that exercise.
a) Show that n = 2 has a unique factorization.
b) Suppose that if for k > 2, {2,···k} can be uniquely factored. Then there are primes pi and qi, not necessarily distinct, such that k +1 = s ∏ i=1 pi = r ∏ i=1 qi .
c) Show that then p1 divides ∏r i=1 qi and so, Corollary 2.10 implies that there is a j ≤r such that p1 = qj.
d) Relabel the qi’s, so that p1 = q1 and divide n by p1 = q1. Show that k +1 q1 = s ∏ i=2 pi = r ∏ i=2 qi .
e) Show that the hypothesis in (b) implies that the remaining pi equal the remaining qi. (Hint: k q1 ≤k.) f) Use the above to formulate the inductive proof that all elements of N can be uniquely factored as a product of primes.
Here is a different characterization of gcd and lcm. We prove it as a corol-lary of the prime factorization theorem.
Corollary 2.23. (1) A common divisor d > 0 of a and b equals gcd(a,b) if and only if every common divisor of a and b is a divisor of d.
(2) Also, a common multiple d > 0 of a and b equals lcm(a,b) if and only if every common multiple of a and b is a multiple of d.
Exercise 2.22. Use the characterization of gcd(a,b) and lcm(a,b) given in the proof of Corollary 2.16 to prove Corollary 2.23.
40 2. The Fundamental Theorem of Arithmetic Exercise 2.23. We develop the proof of Theorem 2.17 as it was given by Euler. We start by assuming that there is a finite list L of k primes. We will show in the following steps how that assumption leads to a contradiction.
We order the list according to ascending order of magnitude of the primes.
So L = {p1, p2,··· , pk} where p1 = 2, p2 = 3, p3 = 5, and so forth, up to the last prime pk.
a) Show that ∏k i=1 pi pi−1 is finite, say M.
b) Show that for r > 0, k ∏ i=1 pi pi −1 = k ∏ i=1 1 1−p−1 i > k ∏ i=1 1−p−r−1 i 1−p−1 i = k ∏ i=1 r ∑ j=0 p−j i !
.
c) Use the fundamental theorem of arithmetic to show that there is an α(r) > 0 such that k ∏ i=1 r ∑ j=0 1 pj i !
= α(r) ∑ ℓ=1 1 ℓ+R, where R is a non-negative remainder.
d) Show that for all K there is an r such that α(r) > K.
e) Thus for any K, there is an r such that k ∏ i=1 r ∑ j=0 1 pj i !
≥ K ∑ ℓ=1 1 ℓ.
f) Conclude with a contradiction between a) and e). (Hint: the harmonic series ∑1 ℓdiverges or see exercise 2.24 c).) Exercise 2.24. In this exercise we consider the Riemann zeta function for real values of z greater than 1.
a) Show that for all x > −1, we have ln(1+x) ≤x.
b) Use Proposition 2.20 and a) to show that lnζ(z) = ∑ p prime ln 1+ p−z 1−p−z ≤∑ p prime p−z 1−p−z ≤∑ p prime p−z 1−2−z .
c) Use the following argument to show that limz↘1 ζ(z) = ∞.
∞ ∑ n=1 n−1 > ∞ ∑ n=1 n−z > Z ∞ 1 x−z dx.
(Hint: for the last inequality, see Figure 10.) d) Show that b) and c) imply that ∑p prime p−z diverges as z ↘1.
e) Use (d) to show that — in some sense — primes are more frequent than squares in the natural numbers. (Hint: ∑∞ n=1 n−2 converges.) 2.6. Exercises 41 Figure 10. Proof that ∑∞ n=1 f(n) is greater than R ∞ 1 f(x)dx if f is posi-tive and (strictly) decreasing.
Exercise 2.25. a) Let p be a fixed prime. Show that the probability that two independently chosen integers in {1,··· ,n} are divisible by p tends to 1/p2 as n →∞. Equivalently, the probability that they are not divisible by p tends to 1−1/p2.
b) Make the necessary assumptions, and show that the probability that two two independently chosen integers in {1,··· ,n} are not divisible by any prime tends to ∏p prime 1−p−2 . (Hint: you need to assume that the probabilities in (a) are independent and so they can be multiplied.) c) Show that from (b) and Euler’s product formula, it follows that for 2 random (positive) integers a and b to have gcd(a,b) = 1 has probability 1/ζ(2) ≈0.61.
d) Show that for d > 1 and integers {a1,a2,···ad} that probability equals 1/ζ(d). (Hint: the reasoning is the same as in (a), (b), and (c).) e) Show that for real d > 1: 1 < ζ(d) < 1+ Z ∞ 1 x−d dx = 1+ 1 d For the middle inequality, see Figure 11.
f) Show that for large d, the probability that gcd(a1,a2,···ad) = 1 tends to 1.
Exercise 2.26. This exercise in based on exercise 2.25.
a) In the {−4,··· ,4}2(0,0) grid in Z2, find out which proportion of the lattice points is visible from the origin, see Figure 12.
b) Use exercise 2.25 (c) to show that in a large grid, this proportion tends to 1/ζ(2).
c) Use exercise 2.25 (d) to show that as the dimension increases to infin-ity, the proportion of the lattice points Zd that are visible from the origin, increases to 1.
42 2. The Fundamental Theorem of Arithmetic Figure 11. Proof that ∑∞ n=1 f(n) (shaded in blue and green) minus f(1) (shaded in blue) is less than R ∞ 1 f(x)dx if f is positive and (strictly) decreasing to 0.
X Figure 12. The origin is marked by “×”. The red dots are visible from ×; between any blue dot and × there is a red dot. The picture shows exactly one quarter of {−4,··· ,4}2(0,0) ⊂Z2.
Exercise 2.27. We note here that ζ(2) = π2 6 .
a) Show that the irrationality of π implies that ζ(2) is irrational.
b) Show that (a) and Proposition 2.20 yield another proof of the infinity of primes.
Chapter 3 Linear Diophantine Equations Overview. A Diophantine equation is a polynomial equation in two or more unknowns and for which we seek to know what integer solutions it has. We determine the integer solutions of the simplest linear Diophantine equation ax + by = c. The central element this reasoning is the Euclidean algorithm. That algorithm has much wider applications. We discuss a few of those.
3.1. The Euclidean Algorithm Lemma 3.1. In the division algorithm of Lemma 2.2, we have gcd(r1,r2) = gcd(r2,r3).
Proof. On the one hand, we have r1 = r2q2+r3, and so any common divisor of r2 and r3 must also be a divisor of r1 (and of r2). Vice versa, since r1 −r2q2 = r3, we have that any common divisor of r1 and r2 must also be a divisor of r3 (and of r2).
■ Thus by calculating r3, the residue of r1 modulo r2, we have simplified the computation of gcd(r1,r2). This is because r3 is strictly smaller (in ab-solute value) than both r1 and r2. In turn, the computation of gcd(r2,r3) can be simplified similarly, and so the process can be repeated. Since the ri form 43 44 3. Linear Diophantine Equations a monotone decreasing sequence in N, this process must end when rn+1 = 0 after a finite number of steps. We then have gcd(r1,r2) = gcd(rn,0) = rn.
Corollary 3.2. Given r1 > r2 > 0, apply the division algorithm until rn > rn+1 = 0. Then gcd(r1,r2) = gcd(rn,0) = rn. Since ri is decreasing, the algorithm always ends.
Definition 3.3. The repeated application of the division algorithm to com-pute gcd(r1,r2) is called the Euclidean algorithm.
We now give a framework to reduce the messiness of these repeated computations. Suppose we want to compute gcd(188,158). We do the following computations: 188 = 158·1+30 158 = 30·5+8 30 = 8·3+6 8 = 6·1+2 6 = 2·3+0 , We see that gcd(188,158) = 2. The numbers that multiply the ri are the quotients of the division algorithm (see the proof of Lemma 2.2). If we call them qi, the computation looks as follows: r1 = r2 q2 +r3 r2 = r3 q3 +r4 .
.
.
.
.
.
.
.
.
rn−3 = rn−2 qn−2 +rn−1 rn−2 = rn−1 qn−1 +rn rn−1 = rn qn +0 , (3.1) where we use the convention that rn+1 = 0 while rn ̸= 0. Observe that with that convention, (3.1) consists of n −1 steps. A much more concise form (in part based on a suggestion of Katahdin ) to render this computation is as follows.
| qn | qn−1 | ··· | q3 | q2 | 0 | rn | rn−1 | ··· | r3 | r2 | r1 | (3.2) Thus, each step ri+1 | ri | is similar to the usual long division, except that its quotient qi+1 is placed above ri+1 (and not above ri), while its remainder 3.2. A Particular Solution of ax+by = c 45 ri+2 is placed all the way to the left of of ri+1. The example we worked out before, now looks like this: | 3 | 1 | 3 | 5 | 1 | 0 | 2 | 6 | 8 | 30 | 158 | 188 | (3.3) There is a beautiful visualization of this process outlined in exercise 3.2.
3.2. A Particular Solution of ax+by = c Another interesting way to encode the computations done in equations (3.1) and (3.2), is via matrices.
ri−1 ri = qi 1 1 0 ri ri+1 .
(3.4) Denote the matrix in this equation by Qi. Its determinant equals −1, and so it is invertible. In fact, Qi = qi 1 1 0 and Q−1 i = 0 1 1 −qi .
These matrices Qi are very interesting. We will use them again to study the theory of continued fractions in Chapter 6. For now, as we will see in Theorem 3.4, they give us an explicit algorithm to find a solution to the equation r1x+r2y = rgcd(r1,r2). Note that from B´ ezout’s lemma (Lemma 2.5), we already know this has a solution. But the next result gives us a simple way to actually calculate a solution. In what follows Xij means the (i, j) entry of the matrix X.
Theorem 3.4. Give r1 and r2, a solution for x and y of r1x+r2y = rgcd(r1,r2) is given by x = r Q−1 n−1 ···Q−1 2 2,1 and y = r Q−1 n−1 ···Q−1 2 2,2 .
46 3. Linear Diophantine Equations Proof. Let ri, qi, and Qi be defined as above, and set rn+1 = 0. From equa-tion (3.4), we have ri ri+1 = Q−1 i ri−1 ri = ⇒ r rn−1 rn = rQ−1 n−1 ···Q−1 2 r1 r2 .
Observe that rn+1 = 0 and so gcd(r1,r2) = rn and rn−1 rn = xn−1 yn−1 xn yn r1 r2 .
The theorem follows immediately by setting x = xn and y = yn.
■ In practice, rather than multiplying all these matrices, it may be more convenient to solve equation (3.1) or (3.2) “backward”, as the expression goes. This can be done as follows. Start with gcd(r1,r2) = rn = rn−2 −rn−1 qn−1 , which follows from equation (3.1). The line above it in that same equation gives rn−1 = rn−3 −rn−2 qn−2. Use this to eliminate rn−1 in favor of rn−2 and rn−3. So, gcd(r1,r2) = rn = rn−2 −(rn−3 −rn−2 qn−2) qn−1 = rn−2 (1+qn−1qn−2)+rn−3(−qn−1) .
This computation can be done still more efficiently by employing the notation of equation (3.2) again.
| + | − | + | − | + | ··· | qn | qn−1 | qn−2 | qn−3 | qn−4 | ··· 0 | rn | rn−1 | rn−2 | rn−3 | rn−4 | ··· | 1 | | | | | | 0 | −qn−1 | 1 | | | | | | qn−1qn−2 | −qn−1 | | | | | | −qn−3(1+qn−1qn−2) | 1+qn−1qn−2 | ··· The algorithm proceeds as follows. Number the columns from right to left, so that ri (in row 1) and qi (in row 2) are in the ith column. (The signs in row “0” serve only to keep track of the signs of the coefficients in row 3 and below.) In the first two rows, the algorithm proceeds from right to left.
3.3. Solution of the Homogeneous equation ax+by = 0 47 From ri−1 and ri determine qi and ri+1 by ri−1 = riqi + ri+1. The division guarantees that these exist, but they may not be unique (see exercise 7.17).
In rows 3 and below, the algorithm proceeds from left to right. Each column has at most two non-zero entries. Start with column n + 1 which has only zeroes and column n which has one 1. The bottom non-zero entry of column i equals the sum of column i+1 times qi times (-1). The top non-zero entry of column i equals the sum of the entries in column i+2. Finally, we obtain that rn = r2x + r1y, where x is the sum of the entries in the 2nd column (rows 3 and below) and y, the sum of the entries (row 3 and below) of the 1st column.
Applying this to the example gives | + | − | + | − | + | − | 3 | 1 | 3 | 5 | 1 | 0 0 | 2 | 6 | 8 | 30 | 158 | 188 | 1 | | | | | | | −1 | 1 | | | | | | 3 | −1 | | | | | | −20 | 4 | | | | | | 21 | −21 (3.5) Adding the last two lines gives that 2 = 158(25)+188(−21).
3.3. Solution of the Homogeneous equation ax+by = 0 Proposition 3.5. The general solution of the homogeneous equation r1x+ r2y = 0 is given by x = k r2 gcd(r1,r2) and y = −k r1 gcd(r1,r2) , where k ∈Z.
Proof. On the one hand, by substitution the expressions for x and y into the homogeneous equation, one checks they are indeed solutions. On the other hand, x and y must satisfy r1 gcd(r1,r2) x = − r2 gcd(r1,r2) y .
48 3. Linear Diophantine Equations The integers ri gcd(r1,r2) (for i in {1,2}) have greatest common divisor equal to 1. Thus Euclid’s lemma applies and therefore r1 gcd(r1,r2) is a divisor of y while r2 gcd(r1,r2) is a divisor of x.
■ A different proof of this lemma goes as follows. The set of all solu-tion in R2 of r1x + r2y = 0 is given by the line ℓ:= t r2 −r1 : t ∈R orthogonal to r1 r2 . To obtain all its lattice points (i.e., points that are also in Z2), both tr2 and −tr1 must be integers. The smallest positive number t for which this is possible, is t = 1 gcd(r1,r2).
3.4. The General Solution of ax+by = c Definition 3.6. Let r1 and r2 be given.
The equation r1x + r2y = 0 is called homogeneous1. The equation r1x + r2y = c when c ̸= 0 is called inhomogeneous. An arbitrary solution of the inhomogeneous equation is called a particular solution. By general solution, we mean the set of all possible solutions of the full (homogeneous or inhomogeneous) equation.
It is useful to have some geometric intuition relevant to the equation r1x + r2y = c. In R2, we set⃗ r = r1 r2 , ⃗ x = x y , etcetera. The standard inner product is written as (·,·). The set of points in R2 satisfying the above inhomogeneous equation thus lie on the line m ⊂R2 given by ( ⃗ r,⃗ x) = c.
This line is orthogonal to the vector⃗ r and its distance to the origin (mea-sured along the vector⃗ r) equals |c| √ ( ⃗ r, ⃗ r). The situation is illustrated in Figure 13.
It is a standard result from linear algebra that the problem of finding all solutions of a inhomogeneous equation comes down to to finding one 1The word “homogeneous” in daily usage receives the emphasis often on its second syllable (“ho-MODGE-uhnus”). However, in mathematics, its emphasis is always on the third syllable (“ho-mo-GEE-nee-us”). A probable reason for the daily variation of the pronunciation appears to be conflation with the word “homogenous” (having the same genetic structure). For details, see wiktionary.
3.4. The General Solution of ax+by = c 49 solution of the inhomogeneous equation, and finding the general solution of the homogeneous equation.
Lemma 3.7. Let (x(0),y(0)) be a particular solution of r1x + r2y = c. The general solution of the inhomogeneous equation is given by (x(0) +z1,y(0) + z2) where (z1,z2) is the general solution of the homogeneous equation r1x+ r2y = 0.
(r ,r ) (x ,x ) x x 1 1 2 2 2 1 m (0) (0) (z ,z ) 1 1 (z +x ,z +x ) 2 2 1 (0) (0) 2 m’ Figure 13. The general solution of the inhomogeneous equation ( ⃗ r,⃗ x) = c in R2.
Proof. Let x(0) y(0) be that particular solution. Let m be the line given by ( ⃗ r,⃗ x) = c. Translate m over the vector −x(0) −y(0) to get the line m′. Then an integer point on the line m′ is a solution z1 z2 of the homogeneous equation if and only if x(0) +z1 y(0) +z2 on m is also an integer point (see Figure 13).
■ 50 3. Linear Diophantine Equations B´ ezout’s Lemma says that r1x + r2y = c has a solution if and only if gcd(r1,r2) | c. Theorem 3.4 gives a particular solution of that equation (via the Euclidean algorithm). Putting those results and Proposition 3.5 together, gives our final result.
Corollary 3.8. Given r1, r2, and c, the general solution of the equation r1x + r2y = c, where gcd(r1,r2) | c, is the sum of a particular solution of Theorem 3.4 and the general homogeneous solution of r1x + r2y = 0 of Proposition 3.5.
3.5. Recursive Solution of x and y in the Diophantine Equation Theorem 3.4 has two interesting corollaries. The first is in fact stated in the proof of that theorem, and the second requires a very short proof. We will make extensive use of these two results in Chapter 6 when we discuss continued fractions.
Corollary 3.9. Given r1, r2, and the successive quotients q2 through qn as in equation (3.1). Then for i ∈{3,··· ,n}, the solution for (xi,yi) in ri = r1xi +r2yi is given by: ri ri+1 = Q−1 i ···Q−1 2 r1 r2 .
Corollary 3.10. Given r1, r2, and the successive quotients q2 through qn as in equation (3.1). Then xi and yi of Corollary 3.9 can be solved as follows: xi yi xi+1 yi+1 = 0 1 1 −qi xi−1 yi−1 xi yi with x1 y1 x2 y2 = 1 0 0 1 .
Proof. The initial condition follows, because r1 = r1 ·1+r2 ·0 r2 = r1 ·0+r2 ·1 .
3.6. The Chinese Remainder Theorem 51 Notice that, by definition, ri ri+1 = r1xi +r2yi r1xi+1 +r2yi+1 = xi yi xi+1 yi+1 r1 r2 .
From Corollary 3.9, we now have that ri ri+1 = Q−1 i ri−1 ri = ⇒ xi yi xi+1 yi+1 = Q−1 i xi−1 yi−1 xi yi .
From this, one deduces the equations for xi+1 and yi+1.
■ We remark that the recursion in Corollary 3.10 can also be expressed as xi+1 = −qixi +xi−1 yi+1 = −qiyi +yi−1 .
3.6. The Chinese Remainder Theorem We now present an important generalization of these ideas. First we need a small update of Definition 1.2.
Definition 3.11. Let {bi}k i=1 be non-zero integers. Their greatest common divisor, gcd(b1,··· ,bk), is the maximum of the numbers that are divisors of every bi; their least common multiple, lcm(b1,··· ,bk), is the least of the positive numbers that are multiples of of every bi.
Surprisingly, for this more general definition, the generalization of Corol-lary 2.16 is false. For an example, see exercise 2.7. However, other impor-tant properties do generalize.
Lemma 3.12. Let {bi}k i=1 be non-zero integers.
(i) If m is a common divisor of the bi, then m | gcd(b1,···bk).
(ii) If M is a common multiple of the bi, then lcm(b1,···bk) | M.
Proof. The proof follows from unique factorization and is similar to that of Corollary 2.16. Suppose bj = ∏s i=1 p kij i , where kij ≥0. Set mi = min j kij and Mi = max j kij , 52 3. Linear Diophantine Equations Then gcd(b1,···bk) = s ∏ i=1 pmi i and lcm(b1,···bk) = s ∏ i=1 pMi i .
Any common divisor of the bi must be equal to ∏s i=1 pℓi i with ℓi ≤mi and similar for common multiples.
■ Theorem 3.13 (Chinese Remainder Theorem). Let n = ∏k i=1 bi, where bi are positive integers such that gcd(b j,bi) = 1 for i ̸= j. The set of solutions of ∀i ∈{1,··· ,k} : z =bi ci is given by z =n k ∑ j=1 n bj xjcj where xi satisfies n bi xi =bi 1.
Proof. Note that gcd(n/bi,bi) = 1. So by B´ ezout, there are xi and yi (for i ∈{1,··· ,k}) so that n bi xi +biyi = 1 ⇐ ⇒ n bi xi =bi 1.
For these xi, we have k ∑ j=1 n bj xj =bi 1.
Thus z = ∑k j=1 n b j xjcj is a particular solution. By Lemma 3.12, the homo-geneous equation has solution z =n 0. The proof is completed by observing that the general solution is the sum of a particular solution plus the solutions to the homogeneous equation.
■ 3.7. Polynomials In this section, we illustrate that the division and Euclidean algorithms have much wider applications than just the integers, see also exercises 2.2 and 2.4.
Definition 3.14. A polynomial f in Q[x] of positive degree is irreducible over Q if it cannot be written as a product of two polynomials in Q[x] with positive degree. Recall (Definition 1.18) that f is minimal polynomial in 3.8. Exercises 53 Q[x] for ρ if f is a non-zero polynomial in Q[x] of minimal degree such that f(ρ) = 0.
Definition 3.15. Let f and g in Q[x]. The greatest common divisor of f and g, or gcd(f,g), is a polynomial in R[x] with maximal degree that is a factor of both f and g. The least common multiple of f and g, or lcm( f,g), is a polynomial in Q[x] with minimal degree that has both f and g as factors.
Remark 3.16. If p is minimal for ρ, it must be irreducible, because if not, one of its factors with smaller degree would also have ρ as a root.
We mention without proof (but see exercise 2.2) that in Q[x] the divi-sion algorithm holds: given r1 and r2, then there are q2 and r3 such that r1 = r2q2 +r3 such that degree(r3) < degree(r1).
Remark 3.17. To make this valid without exceptions, we adopt the con-vention that the degree of a non-zero constant equals 0, while the degree of 0 equals −∞. For example, if r1 = r2 = 1, the inequality for r3 still holds.
The student is likely already familiar with these facts.
It is important to understand that for this to work, division of coeffi-cients is essential. For example, with coefficients in Z, we cannot express 2x2 +1 as a multiple of 3x+1 plus a remainder of smaller degree. However, in Q[x] we can divide coefficients and thus follow the reasoning of Section 2.1 and show the following. See also exercise 3.22).
The gcd of two polynomials can be computed in the same two ways we have seen before, and the proofs are the same. One is done by factor-ing both polynomials and multiplying together the common factors to the lowest power as in the proof of Corollary 2.23. Note though that factoring polynomials is hard. The other is applying the Euclidean Algorithm as in equation (3.1). An example is given in exercise 3.22. The relation between lcm and gcd of two polynomials is the same as in the proof of Corollary 2.23.
3.8. Exercises Exercise 3.1. Let ℓbe the line in R2 given by y = ρx, where ρ ∈R.
a) Show that ℓintersects Z2 if and only if ρ is rational.
b) Given a rational ρ > 0, find the intersection of ℓwith Z2. (Hint: set ρ = r1 r2 and use Proposition 3.5.) 54 3. Linear Diophantine Equations Exercise 3.2. This problem was taken (and reformulated) from .
a) Tile a 188 by 158 rectangle by squares using what is called a greedy algorithm a. The first square is 158 by 158. The remaining rec-tangle is 158 by 30. Now the optimal choice is five 30 by 30 squares.
What remains is an 30 by 8 rectangle, and so on. Explain how this is a visualization of equation (3.3). See Figure 14.
b) Consider equation (3.1) or (3.2) and use a) to show that r1r2 = n ∑ i=2 qir2 i .
(Hint: assume that r1 > r2 > 0, rn ̸= 0, and rn+1 = 0.) aBy “greedy” we mean that at every step, you choose the biggest square possible and as many of them as possible. In general a greedy algorithm always makes a locally optimal choice.
158 158 30 30 30 30 30 30 8 8 8 8 6 6 Figure 14. A ‘greedy’ (or locally best) algorithm to tile the the 188 × 158 rectangle by squares.
The 3 smallest — and barely visible — squares are 2 × 2.
Note how the squares spiral inward as they get smaller. See exercise 3.13.
Exercise 3.3. In (3.1), assume that r1 > r2 > 0. What happens if you start the Euclidean algorithm with r2 = r1 ·0+r3 instead of r1 = r2 ·q2 +r3?
3.8. Exercises 55 Exercise 3.4. Apply the Euclidean algorithm to find the greatest common divisor of the following number pairs. (Hint: replace negative numbers by positive ones. For the division algorithm applied to these pairs (r1,r2), see exercise 2.1) a) 110 , 7.
b) 51 , −30.
c) −138 , 24.
d) 272 , 119.
e) 2378 , 1769.
f) 270 , 175560.
Exercise 3.5. Determine if the following Diophantine equations admit a solution for x and y. If yes, find a (particular) solution. (Hint: Use one of the algorithms in Section 3.2.) a1) 110x+7y = 13.
a2) 110x+7y = 5.
b1) 51x−30y = 6.
b2) 51x−30y = 7.
c1) −138x+24y = 7.
c2) −138x+24y = 6.
d1) 272x+119y = 54.
d2) 272x+119y = 17.
e1) 2378x+1769y = 300.
e2) 2378x+1769y = 57.
f1) 270x+175560y = 170.
f2) 270x+175560y = 150.
Exercise 3.6. Find all solutions for x and y of the following (homogeneous) Diophantine equations. (Hint: Use one of the algorithms in Section 3.2.) a) 110x+7y = 0.
b) 51x−30y = 0.
c) −138x+24y = 0.
d) 272x+119y = 0.
e) 2378x+1769y = 0.
f) 270x+175560y = 0.
Exercise 3.7. Find the general solution for x and y in all problems of exer-cise 3.5 that admit a solution. (Hint: use Corollary 3.8.) Exercise 3.8. Use Corollary 3.10 to express xi and yi in the successive remainders ri in each of the items in exercise 3.4. (Hint: you need to know the qi for each item in exercise 3.4.) 56 3. Linear Diophantine Equations Exercise 3.9. Consider the line ℓin R3 defined by ℓ(ξ) = r1 r2 r3 ξ, where ξ ∈R and the ri are integers.
a) Show that ℓ(ξ) ∈Z3{ ⃗ 0} if and only if ξ = t gcd(r1,r2,r3) and t ∈Z.
b) Show that this implies that if any of the ri is irrational, then ℓhas no non-zero points in common with Z3.
Definition 3.18. The sequence {Fi}∞ i=0 of Fibonacci numbers Fi is defined as follows F0 = 0 , F1 = 1 , ∀i > 1 : Fi+1 = Fi +Fi−1 .
Exercise 3.10. Denote the golden mean , or 1+ √ 5 2 ≈1.618, by g.
a) Show that g2 = g+1 and thus for n ∈Z: gn+1 = gn +gn−1.
b) Show that F3 ≥g1 and F2 ≥g0.
c) Use induction to show that Fn+2 ≥gn for n > 0.
d) Use the fact that 5 log10 1+ √ 5 2 ≈1.045, to show that F5k+2 > 10k for k ≥0.
Exercise 3.11. Consider the equations in (3.1) and assume that rn+2 = 0 and rn+1 > 0.
a) Show that rn+1 ≥F2 = 1 and rn ≥F3 = 2. (Hint: r(i) is strictly increas-ing.) b) Show that r1 ≥Fn+2.
c) Suppose r1 and r2 in N and max{r1,r2} < Fn+2. Show that the Eu-clidean Algorithm to calculate gcd(r1,r2) takes at most n −1 iterates of the division algorithm.
Exercise 3.12. Use exercises 3.10 and 3.11 to show that the Euclidean Algorithm to calculate gcd(r1,r2) takes at most 5k −1 iterates where k is the number of decimal places of max{r1,r2}.
(This is known as Lam´ e’s theorem.) 3.8. Exercises 57 Exercise 3.13. Apply the greedy algorithm of exercise 3.2 (a) to the rec-tangle whose sides have length 1 and g (see exercise 3.10 (a)). At step 0, we start with the 1×1 square.
a) Use exercise 3.10 (a) to show at that step i, you get one g−i ×g−i square (see Figure 15).
b) Use exercise 3.2 (b) to show that g = ∑∞ i=0 g−2i.
c) Use this construction, but now with a Fn+1 ×Fn Fibonacci rectangle, to show that Fn+1Fn = ∑n i=1 F2 i . For Fi, see Definition 3.18.
d) Show that in polar coordinates (r,θ) the red spiral connecting the cor-ners of the squares in Figure 15 is given by r = Cg2θ/π for some C.(Note: this is called the golden spiral.) 1 1 1/g 1/g 1/g 1/g 1/g 1/g 2 2 3 3 Figure 15. The greedy algorithm of exercise 3.2 (a) applied to the golden mean rectangle. The spiral connecting the corners of the square is known as the golden spiral. (In actual fact we used a 55 by 34 rectan-gle as an approximation. An approximation to a true spiral was created by fitting circular segments to the corners.) Exercise 3.14. a) Write the numbers 287, 513, and 999 in base 2, 3, and 7, using the division algorithm. Do not use a calculating device. (Hint: start with base 10. For example: 287 = 28·10+7 28 = 2·10+8 2 = 0·10+2 Hence the number in base 10 is 2·102 +8·101 +7·100.) b) Show that to write n in base b takes about logb n divisions.
58 3. Linear Diophantine Equations Exercise 3.15. Use Theorem 3.13 to solve: z =2 1 z =3 2 z =5 3 z =7 5.
Exercise 3.16. The Fibonacci numbers Fn are defined in Definition 3.18.
a) Use the method of equation (3.1) to show that gcd(Fn,Fn+1) = 1.
b) Determine the qi in (a).
c) Use recursion to show that 0 1 1 1 n = Fn−1 Fn Fn Fn+1 .
d) Show that (c) implies that Fn+1Fn−1 −F2 n = (−1)n. (Hint: in (c) take the determinant.) Exercise 3.17. Use Theorem 3.13 to solve: z =Fn Fn−1 z =Fn+1 Fn .
where Fn are the Fibonacci numbers of Definition 3.18. (Hint: you need to use exercise 3.16 (a) and (d).) Exercise 3.18. (The Chinese remainder theorem generalized.) Suppose {bi}n i=1 are positive integers. We want to know all z that satisfy z =bi ci for i ∈{1,···n} .
a) Set B = lcm(b1,b2,···bn) and show that the homogeneous problem is solved by z =B 0 .
b) Show that if there is a particular solution then ∀i ̸= j : ci =gcd(bi,bj) cj .
c) Formulate the general solution when the condition in (b) holds.
Exercise 3.19. Use exercise 3.18 to solve: z =6 15 z =10 6 z =15 10.
See also exercise 2.7.
3.8. Exercises 59 Exercise 3.20. There is a reformulation of the Euclidean algorithm that will be very useful in Chapter 6.
a) Rewrite the example in Section 3.1 as follows.
30 158 = 188 158 −1 8 30 = 158 30 −5 6 8 = 30 8 −3 2 6 = 8 6 −1 .
Note that the right hand side is a fraction minus its integer part.
b) Now rewrite this again as 30 158 = 1 158/188 −1 8 30 = 1 30/158 −5 6 8 = 1 8/30 −3 2 6 = 1 6/8 −1 .
Exercise 3.21.
a) Apply the Euclidean algorithm to (r1,r2) = (14142,10000).
(Hint: you should get (q2,··· ,q10) = (1,2,2,2,2,2,1,1,29).) b) Show that for i ∈{2,··· ,8}: ri+1 ri = 1 ri−1/ri − 1 ri−1/ri , where ⌊x⌋is the greatest integer less than or equal to x. (Hint: see also exercise 3.20.) 60 3. Linear Diophantine Equations Exercise 3.22. For this exercise, read Section 3.7 carefully. All polynomi-als are in Q[x] (that is: with coefficients in Q). Let p1(x) = x7 −x2 + 1, p2(x) = x3 +x2, and e(x) = 2−x.
a) Use the Euclidean Algorithm to determine gcd(p1, p2). Hint: We list the steps of the Euclidean algorithm: (x7 −x2 +1) = (x3 +x2) (x4 −x3 +x2 −x+1) + (−2x2 +1) (x3 +x2) = (−2x2 +1) (−1 2x−1 2) + ( 1 2x+ 1 2) (−2x2 +1) = ( 1 2x+ 1 2) (−4x+4) + (−1) ( 1 2x+ 1 2) = (−1) (−1 2x−1 2) + (0) , b) Explain why there are polynomials gp and hp such that p1(x)gp(x)+ p2(x)hp(x) = e(x).
c) Use “backward solving” to find a particular solution of the equation in (b).
d) Find the general (homogeneous) solution of p1(x)g0(x)+ p2(x)h0(x) = 0.
e) Use (c) and (d) to give the general solution of the inhomogeneous equa-tion (the one in (b)).
Exercise 3.23. All polynomials are in Q[x]. Let p(x) be a polynomial and p′(x) its derivative.
a) Show that if p(x) has a multiple root λ of order k > 1, then p′(x) has that same root of order k −1. (Hint: Differentiate p(x) = h(x)(x−λ)k.) b) Use exercise 3.22, to give an algorithm to find a polynomial q(x) that has the same roots as p(x), but all roots are simple (i.e. no multiple roots).
(Hint: you need to divide p by gcd(p, p′).) Exercise 3.24. Assume that every polynomial f of degree d ≥1 has at least 1 root, prove the fundamental theorem of algebra. (Hint: let ρ be a root and use the division algorithm to write f(x) = (x −ρ)q(x)+r where r has degree 0.) In Proposition 11.20, we will prove that every polynomial with complex coefficients has at least one zero in C. Together with the result of exercise 3.24, this establishes the following important theorem.
Theorem 3.19 (Fundamental Theorem of Algebra). A polynomial in C[x] (the set of polynomials with complex coefficients) of degree d ≥1 has ex-actly d roots, counting multiplicity.
Exercise 3.25. Let f and p be polynomials in Q[x] with root ρ and suppose that p is minimal (Definition 1.18). Show that p | f. (Hint: use the division algorithm and 2.4 to write f(x) = p(x)q(x)+r(x) where r has degree less than g.) Chapter 4 Number Theoretic Functions Overview. We study number theoretic functions. These are functions de-fined on the positive integers with values in C. In the context of number theory, the value typically depends on the arithmetic nature of its argument (i.e. whether it is a prime, and so forth), rather than just on the size of its argument. An example is τ(n) which equals the number of positive divisors of n.
4.1. Multiplicative Functions Definition 4.1. Number theoretic functions, arithmetic functions, or sequences are functions defined on the positive integers (i.e. N) with values in C.
Note that outside number theory, the term sequence is the one that is most commonly used. We will use these terms interchangeably.
Definition 4.2. A multiplicative function is a sequence such that gcd(a,b) = 1 implies f(ab) = f(a) f(b). A completely multiplicative function is one where the condition that gcd(a,b) = 1 is not needed.
Note that completely multiplicative implies multiplicative (but not vice versa).
The reason this definition is interesting, is that it allows us to evaluate the 61 62 4. Number Theoretic Functions value of a multiplicative function f on any integer as long as we can com-pute f(pk) for any prime p. Indeed, using the fundamental theorem of arithmetic, if n = r ∏ i=1 pℓi i then f(n) = r ∏ i=1 f(pℓi i ), as follows immediately from Definition 4.2.
Proposition 4.3. Let f be a multiplicative function on the integers. Then F(n) = ∑ d|n f(d) is also multiplicative.
Proof. Let n = ∏s i=1 pℓi i . The summation ∑d|n f(d) can be written out using the previous lemma and the fact that f is multiplicative: F(n) = ∑ℓ1 a1=0 ··· ∑ℓs as=0 f(pa1 1 )··· f(par r ) = ∏s i=1 ∑ℓi ai=0 f(pai i ) .
Exercise 4.3 provides a visual explanation for the second equality.
Now let a and b two integers greater than 1 and such that gcd(a,b) = 1 and ab = n. Then by the unique factorization theorem a and b can be written as: a = r ∏ i=1 pℓi i and b = s ∏ i=r+1 pℓi i Applying the previous computation to a and b yields that f(a) f(b) = f(n).
■ Perhaps the simplest multiplicative functions are the ones where f(n) = nk for some fixed k. Indeed, f(n) f(m) = nkmk = f(nm). In fact, this is a completely multiplicative function. Thus Proposition 4.3 implies that the functions σk defined below are multiplicative.
Definition 4.4. Let k ∈R. The multiplicative function σk : N →R gives the sum of the k-th power of the positive divisors of n. Equivalently: σk(n) = ∑ d|n dk .
4.1. Multiplicative Functions 63 Note that the multiplicativity of σk follows directly from Proposition 4.3.
Special cases are when k = 1 and k = 0. In the first case, the function is simply the sum of the positive divisors and the subscript ‘1’ is usually dropped. When k = 0, the function is usually called τ, and the function’s value is the number of positive divisors of its argument.
Theorem 4.5. Let n = ∏r i=1 pℓi i where the pi are primes. Then for k ̸= 0 σk(n) = r ∏ i=1 pk(ℓi+1) i −1 pk i −1 !
, while for k = 0 σ0(n) = τ(n) = r ∏ i=1 (ℓi +1) .
Proof. By Proposition 4.3, σk(n) is multiplicative, so it is sufficient to com-pute for some prime p: σk(pℓ) = ℓ ∑ i=0 pik = pk(ℓ+1) −1 pk −1 .
Thus σk(n) is indeed a product of these terms.
■ However, there are other interesting multiplicative functions beside the powers of the divisors. The M¨ obius function defined below is one of these, as we will see.
Definition 4.6. The M¨ obius function µ : N →Z is given by: µ(n) = 1 if n = 1 0 if ∃p > 1 prime with p2 | n (−1)r if n = p1 ··· pr and pi are distinct primes .
Definition 4.7. We say that n is square free if there is no prime p such that p2 | n.
Lemma 4.8. The M¨ obius function µ is multiplicative.
Proof. By unique factorization, we are allowed to assume that n = ab where a = r ∏ i=1 pℓi i and b = s ∏ i=r+1 pℓi i .
64 4. Number Theoretic Functions If a equals 1, then µ(ab) = µ(a)µ(b) = 1µ(b), and similar if b = 1. If either a or b is not square free, then neither is n = ab, and so in that case, we again have µ(ab) = µ(a)µ(b) = 0. If both a and b are square free, then r (in the definition of µ) is strictly additive and so (−1)r is strictly multiplicative, hence multiplicative.
■ 4.2. Additive Functions Also important are the additive functions to which we will return in Chapter 12.
Definition 4.9. An additive function is a sequence such that gcd(a,b) = 1 implies f(ab) = f(a)+ f(b). A completely addititive function is one where the condition that gcd(a,b) = 1 is not needed.
Here are some examples.
Definition 4.10. Let ω(n) denote the number of distinct prime divisors of n and let Ω(n) denote the total number of prime divisors of n. These functions are called the prime omega functions.
So if n = ∏s i=1 pℓi i , then ω(n) = s and Ω(n) = s ∑ i=1 ℓi .
The additivity of ω and the complete additivity of Ωshould be clear. By way of example, since 72 = 23 ·32, ω(72) = 2 while Ω(72) = 5.
4.3. M¨ obius inversion Lemma 4.11. Define ε(n) ≡∑d|n µ(d). Then ε(1) = 1 and for all n > 1, ε(n) = 0.
Proof. Lemma 4.8 says that µ is multiplicative. Therefore, by Proposition 4.3, ε is also multiplicative. It follows that ε(∏r i=1 pℓi i ) can be calculated by evaluating a product of terms like ε(pℓ) where p is prime. For example, when p is prime, we have ε(p) = µ(1)+ µ(p) = 1+(−1) = 0 and ε(p2) = µ(1)+ µ(p)+ µ(p2) = 1−1+0 = 0.
4.3. M¨ obius inversion 65 Thus one sees that ε(pℓ) is zero unless ℓ= 0.
■ Lemma 4.12. For n ∈N, define Sn ≡ (a,b) ∈N2 : ∃d > 0 such that d | n and ab = d and Tn ≡ n (a,b) ∈N2 : b | n and a | n b o .
Then Sn = Tn.
Proof. Suppose (a,b) is in Sn. Then ab | n and so ab = d d | n o = ⇒b | n and a | n b .
And so (a,b) is in Tn. Vice versa, if (a,b) is in Tn, then by setting d ≡ab, we get b | n a | n b = ⇒d | n and ab = d .
And so (a,b) is in Sn.
■ Theorem 4.13. (M¨ obius inversion) Let F : N →C be any number theoretic function and µ the M¨ obius function. Then the following equation holds F(n) = ∑ d|n f(d) if and only if f : N →C satisfies f(d) = ∑ a|d µ(a)F d a = ∑ {(a,b):ab=d} µ(a)F (b) .
Proof. ⇐ =: We show that substituting f gives F. Define H as H(n) ≡∑ d|n f(d) = ∑ d|n ∑ a|d µ(a)F d a .
Then we need to prove that H(n) = F(n). This proceeds in three steps. For the first step we write ab = d, so that now H(n) ≡∑ d|n f(d) = ∑ d|n ∑ ab=d µ(a)F (b) .
(4.1) 66 4. Number Theoretic Functions For the second step we apply Lemma 4.12 to the set over which the sum-mation takes place. This gives: H(n) = ∑ b|n ∑ a| n b µ(a)F(b) = ∑ b|n ∑ a| n b µ(a) F(b) .
(4.2) Finally, Lemma 4.11 implies that the term in parentheses equals ε n b . This equals 0, except when b = n when it equals 1. The result follows.
= ⇒: By the previous part, we already know one solution for f if we are given that F(n) = ∑d|n f(d). So suppose there are two solutions f and g. We have: F(n) = ∑ d|n f(d) = ∑ d|n g(d) .
We show by induction on n that f(n) = g(n).
Clearly F(1) = f(1) = g(1). Now suppose that for i ∈{1,···k}, we have f(i) = g(i). Then F(k+1) = ∑ d|(k+1), d≤k f(d) !
+ f(k+1) = ∑ d|(k+1), d≤k g(d) !
+g(k+1) .
The desired equality for k +1 follows from the induction hypothesis.
■ Remark 4.14. It is important that multiplicativity plays no role in this ar-gument.
4.4. Euler’s Phi or Totient Function Definition 4.15. Euler’s phi function, also called the totient function is de-fined as follows: ϕ(n) equals the number of integers in {1,···n} that are relative prime to n (see Figure 16).
Lemma 4.16 (Gauss’ Theorem). For n ∈N: n = ∑d|n ϕ(d).
Proof. Define S(d,n) as the set of integers m between 1 and n such that gcd(m,n) = d: S(d,n) = {m ∈N : m ≤n and gcd(m,n) = d} .
4.4. Euler’s Phi or Totient Function 67 Figure 16. The totient function ϕ(n) versus n. Its subtle structure is clearly visible, see also exercise 4.7.
Since every for natural number m ≤n has a unique gcd(m,n) which is a divisor of n, we get n = ∑ d|n |S(d,n)| .
Because the definition of Sn can be rewritten as S(d,n) = n m ∈N : m ≤n and gcd m d , n d = 1 o , the cardinality |S(d,n)| of S(d,n) is given by ϕ n d . Thus we obtain: n = ∑ d|n |S(d,n)| = ∑ d|n ϕ n d .
As d runs through all divisors of n in the last sum, so does n d . Therefore the last sum is equal to ∑d|n ϕ(d), which proves the lemma.
■ Theorem 4.17. Let ∏r i=1 pℓi i be the prime power factorization of n. Then ϕ(n) = n∏r i=1 1−1 pi .
Proof. 1 Apply M¨ obius inversion to Lemma 4.16: ϕ(d) = ∑ a|d µ(a)d a = d∑ a|d µ(a) a .
(4.3) 1There is a conceptually simpler — but in its details much more challenging — proof if you are familiar with the inclusion-exclusion principle. We review that proof in exercise 4.13.
68 4. Number Theoretic Functions The functions µ and a →1 a are multiplicative. It is easy to see that the product of two multiplicative functions is also multiplicative. Therefore ϕ is also multiplicative (Proposition 4.3). Thus for n as given, ϕ(n) = ϕ r ∏ i=1 pℓi i !
= r ∏ i=1 ϕ pℓi i .
(4.4) So it is sufficient to evaluate the function ϕ on prime powers. Noting that the divisors of the prime power pℓare {1, p,··· pℓ}, we get from equation (4.3) ϕ(pℓ) = pℓ ℓ ∑ j=0 µ(pj) p j = pℓ 1−1 p .
Substituting this into equation (4.4) completes the proof.
■ From this proof we obtain the following corollary.
Corollary 4.18. Euler’s phi function is multiplicative.
4.5. Dirichlet and Lambert Series We will take a quick look at some interesting series without worrying too much about their convergence, because we are ultimately interested in the analytic continuations that underlie these series. For that, it is sufficient that there is convergence in any open non-empty region of the complex plane.
Definition 4.19. Let f, g, and F be arithmetic functions (see Definition 4.1). Define the Dirichlet convolution of f and g, denoted by f ∗g, as ( f ∗g)(n) ≡∑ ab=n f(a)g(b) .
This convolution is a very handy tool. Similar to the usual convolution of sequences, one can think of it as a sort of multiplication. It pays off to first define a few standard number theoretic functions.
Definition 4.20. We use the following notation for certain standard se-quences. The sequence ε(n) is 1 if n = 1 and otherwise returns 0, 1(n) always returns 1, and I(n) returns n (so I(n) = n).
4.5. Dirichlet and Lambert Series 69 The function ε acts as the identity of the convolution. Indeed, (ε ∗g)(n) = ∑ ab=n ε(a)g(b) = g(n).
Note that I(n) is the identity as a function, but should not be confused with the identity of the convolution (ε). In other words, I(n) = n but I ∗f ̸= f.
We can now do some very cool2 things of which we can unfortunately give but a few examples. As a first example, the M¨ obius inversion of Theo-rem 4.13 F(n) = ∑ d|n f(d) ⇐ ⇒ f(d) = ∑ {(a,b):ab=d} µ(a)F (b) , can be more succinctly translated as follows: F = 1∗f ⇐ ⇒ f = µ ∗F .
(4.5) This leads to the next example. The first of the following equalities holds by Lemma 4.16, the second follows from M¨ obius inversion (4.5).
I = 1∗ϕ ⇐ ⇒ ϕ = µ ∗I .
(4.6) And the best of these examples is gotten by substituting the identity ε for F in equation (4.5): ε = 1∗f ⇐ ⇒ f = µ ∗ε = µ .
(4.7) Thus µ is the convolution inverse of the sequence (1,1,1···). This imme-diately leads to an unexpected3 expression for 1/ζ(z) of equation (4.8).
Definition 4.21. Let f(n) is an arithmetic function (or sequence). A Dirichlet series is a series of the form F(z) = ∑∞ n=1 f(n)n−z. Similarly, a Lambert series is a series of the form F(x) = ∑∞ n=1 f(n) xn 1−xn .
The prime example of a Dirichlet series is – of course – the Riemann zeta function of Definition 2.19, ζ(z) = ∑1(n)n−z.
Lemma 4.22. For the product of two Dirichlet series we have ∑ n=1 f(n)n−z ! ∑ n=1 g(n)n−z !
= ∞ ∑ n=1 ( f ∗g)(n)n−z .
2A very unusual word in mathematics textbooks.
3The fact that this follows so easily, justifies the use of the word referred to in the previous footnote 70 4. Number Theoretic Functions Proof. This follows easily from re-arranging the terms in the product: ∞ ∑ a,b≥1 f(a)g(b) (ab)z = ∞ ∑ n=1 ∑ ab=n f(a)g(b) !
n−z .
We collected the terms with ab = n.
■ Can we find f(n) such that 1 ζ(z) = ∑f(n)(n)n−z? Yes! Because Lemma 4.22 translates 1 = ζ(z)· 1 ζ(z) as ε = 1∗f .
And equation (4.7) gives that f = µ, or 1 ζ(z) = ∑ n≥1 µ(n) nz .
(4.8) Recall from Chapter 2 that one of the chief concerns of number theory is the location of the non-real zeros of ζ. At stake is Conjecture 2.22 which states that all its non-real zeros are on the line Rez = 1/2. The original definition of the zeta function is as a series that is absolutely convergent for Rez > 1 only. Equation (4.8) converges in that same region, and so establishes that at least in Rez > 1 there are no zeroes. A (weak) partial result in the direction of the Riemann Hypothesis!
It is also important to establish that the analytic continuation of ζ is valid for all z ̸= 1. The next result serves as a first indication that ζ(z) can indeed be continued for values Rez ≤1.
Corollary 4.23. Let ζ be the Riemann zeta function and σk as in Definition 4.4, then ζ(z−k)ζ(z) = ∞ ∑ n=1 σk(n) nz .
Proof.
ζ(z−k)ζ(z) = ∑ a≥1 a−z ∑ b≥1 bk b−z = ∑ n≥1 n−z ∑ b|n bk .
■ Lemma 4.24. A Lambert series can re-summed as follows: ∞ ∑ n=1 f(n) xn 1−xn = ∞ ∑ n=1 (1∗f)(n)xn .
4.5. Dirichlet and Lambert Series 71 Proof. First use that xb 1−xb = ∞ ∑ a=1 xab .
This gives that ∑ b=1 f(b) xb 1−xb = ∞ ∑ b=1 ∞ ∑ a=1 f(b)xab .
Now set n = ab and collect terms. Noting that (1∗f)(b) = ∑b∤n f(b) yields the result.
■ Corollary 4.25. The following equality holds ∑ n≥1 ϕ(n) xn 1−xn = x (1−x)2 .
Proof. We have ∑ n≥1 ϕ(n) xn 1−xn = ∑ n≥1 (1∗ϕ)(n)xn = ∑ n≥1 I(n)xn .
The first equality follows from Lemma 4.24 and the second from Lemma 4.16. The last sum can be computed as x d dx(1−x)−1 which gives the desired expression.
■ t t/2 Figure 17. A one parameter family ft of maps from the circle to itself.
For every t ∈[0,1] the map ft is constructed by truncating the map x → 2x mod 1 as indicated in this figure.
The last result is of importance in the study of dynamical systems. In figure 17, the map ft is constructed by truncating the map x →2x mod 1 72 4. Number Theoretic Functions for t ∈[0,1]. Corollary 4.25 can be used to show that the set of t for which ft does not have a periodic orbit has measure (“length”) zero [71,72], even though that set is uncountable.
4.6. Exercises Exercise 4.1. Decide which functions are not multiplicative, multiplica-tive, or completely multiplicative (see Definition 4.2).
a) f(n) = 1.
b) f(n) = 2.
c) f(n) = ∑n i=1 i.
d) f(n) = ∏n i=1 i.
e) f(n) = n.
f) f(n) = nk.
g) f(n) = ∑d|n d.
h) f(n) = ∏d|n d.
Exercise 4.2. a) Let h(n) = 0 when n is even, and 1 when n is odd. Show that h is multiplicative.
b) Now let H(n) = ∑d|n h(d). Show without using Proposition 4.3 that H is multiplicative. (Hint: write a = 2k ∏r i=1 pℓi i by unique factorization, where the pi are odd primes. Compute the number of odd divisors. Similarly for b.) c) What does Proposition 4.3 say?
Exercise 4.3. In Figure 18 a large volume in R3 with coordinates x, y, and z is chopped up into smaller rectangular boxes of dimensions xi by yj by zk as indicated. See the proof of Proposition 4.3.
a) Show that the volume of the big box equals ∑n1 i=1 ∑n2 j=1 ∑n3 k=1 xiyjzk.
(Hint: add the volumes of the small boxes.) b) Show that the volume of the big box equals ∑n1 i=1 xi ∑n2 j=1 yj ∑n3 k=1 zk .
(Hint: compute the dimensions of the big box.) Exercise 4.4. a) Compute the numbers σ1(n) = σ(n) of Definition 4.4 for n ∈{1,··· ,30} without using Theorem 4.5.
b) What is the only value n for which σ(n) = n?
c) Show that σ(p) = p+1 whenever p is prime.
d) Use (c) and multiplicativity of σ to check the list obtained in (a).
e) For what values of n in the list of (a) is n | σ(n)? (Hint: 6 and 28.) 4.6. Exercises 73 x y z y x 1 1 1 x2 z z2 y x z 3 2 3 4 3 4 Figure 18. Two ways of computing the volume of a big box: add the volumes of the small boxes, or compute the dimensions of the big box.
Exercise 4.5. a) Compute the numbers σ0(n) = τ(n) of Definition 4.4 for n ∈{1,··· ,30} without using Theorem 4.5.
b) What is the only value n for which τ(n) = 1?
c) Show that τ(p) = 2 whenever p is prime.
d) Use (c) and multiplicativity of τ to check the list obtained in (a).
Exercise 4.6. a) Compute the numbers ϕ(n) of Definition 4.15 for n ∈ {1,··· ,30} without using Theorem 4.17.
b) What is ϕ(p) when p is a prime?
c) How many positive numbers less than pn are not divisible by p?
d) Use (c) and multiplicativity of ϕ to check the list obtained in (a).
Exercise 4.7. Consider Figure 19 and prove the following statements.
a) limi→∞ϕ(pi)/pi = 1. (Hint: use Theorem 4.17.) b) limi→∞ϕ(2pi)/2pi = 1/2, where pi are the odd primes.
c) If ni = ∏k 1 pi, then limi→∞ϕ(ni)/ni = 0. (Hint: use Theorem 4.17 and Proposition 2.20 plus the fact that ∑n−1 diverges. Note that the ni form a useful basis for number systems if you want to minimize division, see exercise 1.25) Exercise 4.8. a) Compute the numbers µ(n) of Definition 4.6 for n ∈ {1,··· ,30}.
b) What is µ(p) when p is a prime?
c) Use (c) and multiplicativity of µ to check the list obtained in (a).
Exercise 4.9. Let τ(n) be the number of distinct positive divisors of n.
Answer the following question without using Theorem 4.5.
a) Show that τ is multiplicative.
b) If p is prime, show that τ(pk) = k +1.
c) Use the unique factorization theorem, to find an expression for τ(n) for n ∈N.
74 4. Number Theoretic Functions Figure 19. The fraction of numbers in {1,··· ,n} relatively prime to n, or ϕ(n)/n, versus n.
Exercise 4.10. Two positive integers a and b are called amicable if σ(a) = σ(b) = a+b. The smallest pair of amicable numbers is is formed by 220 and 284.
a) Use Theorem 4.5 to show that 220 and 284 are amicable.
b) The same for 1184 and 1210.
Exercise 4.11. A positive integer n is called perfect if σ(n) = 2n.
a) Show that n is perfect if and only if the sum of its positive divisors less than n equals n.
b) Show that if p and 2p −1 are primes, then n = 2p−1(2p −1) is perfect.
(Hint: use Theorem 4.5 and exercise 4.4(c).) c) Use exercise 1.12 to show that if 2p −1 is prime, then p is prime, and thus n = 2p−1(2p −1) is perfect.
d) Check that this is consistent with the list in exercise 4.4.
4.6. Exercises 75 Exercise 4.12. Draw the following directed graph G: the set of vertices V represent 0 and the natural numbers between 1 and 50. For a,b ∈V, a directed edge ab exists if σ(a) −a = b. Finally, add a loop at the vertex representing 0. Notice that every vertex has 1 outgoing edge, but may have more than 1 incoming edge.
a) Find the cycles of length 1 (loops). The non-zero of these represent per-fect numbers.
b) Find the cycles of length 2 (if any). A pair of numbers a and b that form a cycle of length 2 are called amicable numbers. Thus for such a paira , σ(b)−b = a and σ(a)−a = b.
c) Find any longer cycles. Numbers represented by vertices in longer cy-cles are called sociable numbers.
d) Find numbers whose path ends in a cycle of length 1. These are called aspiring numbers.
e) Find numbers (if any) that have no incoming edge. These are called un-touchable numbers.
f) Determine the paths starting at 2193 and at 562. (Hint: both end in a cycle (or loop).) aAs of 2017, about 109 amicable number pairs have been discovered.
A path through this graph is called an aliquot sequence. The so-called Catalan-Dickson conjecture says that every aliquot sequence ends in some finite cycle (or loop). However, even for a relatively small number such as 276, it is unknown (in 2017) whether its aliquot sequence ends in a cycle.
76 4. Number Theoretic Functions Exercise 4.13. In this exercise, we give a different proof of Theorem 4.17.
It uses the principle of inclusion-exclusion . We state it here for com-pleteness. Let S be a finite set with subsets A1, A2, and so on through Ar.
Then, if we denote the cardinality of a set A by |A|, S− r [ i=1 Ai = |S|−|S1|+|S2|−···+(−1)r|Sr| , (4.9) where |Sℓ| is the sum of the sizes of all intersections of ℓmembers of {A1,···Ar}.
Now, in the following we keep to these conventions. Using prime factor-ization, write n = ∏r i=1 pki i , Ai = {z ∈S| pi divides z} .
S = {1,2···n} and R = {1,2···r} , Iℓ⊆R such that |Iℓ| = ℓ.
a) Show that ϕ(n) = S−Sr i=1 Ai . (Hint: any number that is not co-prime with n is a multiple of at least one of the pi.) b) Show that Ai = n pi .
c) Show that T i∈IℓAi = n ∏i∈Iℓ 1 pi . (Hint: use Lemma 3.12.) d) Show that Sℓ = n ∑Iℓ⊆R ∏i∈Iℓ 1 pi .
e) Show that the principle of inclusion-exclusion implies that S − Sr i=1 Ai = n+n ∑r ℓ=1(−1)ℓ∑Iℓ⊆R ∏i∈Iℓ 1 pi .
f) Show that n + n ∑r ℓ=1(−1)ℓ∑Iℓ⊆R ∏i∈Iℓ 1 pi = n ∏r i=1(1 −1 pi ) . Notice that this implies Theorem 4.17. (Hint: write out the product ∏r i=1(1−1 pi ).) Exercise 4.14. Let F(n) = n = ∑d|n f(n). Use the M¨ obius inversion for-mula (or f(n) = ∑d|n µ(d)F( n d )) to find f(n). (Hint: substitute the M¨ obius function of Definition 4.6 and use multiplicativity where needed.) Exercise 4.15. a) Compute the sets Sn and Tn of Lemma 4.12 explicitly for n = 4 and n = 12.
b) Perform the resummation done in equations 4.1 and 4.2 explicitly for n = 4 and n = 12.
4.6. Exercises 77 Exercise 4.16. Recall the definition of Dirichlet convolution f ∗g of the arithmetic functions f and g (Definition 4.19).
a) Show that the set A of arithmetic functions with addition forms an Abelian group (see Definition 5.19).
b) Show that Dirichlet convolution is associativea, that is: (f ∗g)∗h = f ∗(g∗h) .
c) Show that Dirichlet convolution is distributive over addition, that is: f ∗(g+h) = f ∗g+ f ∗h .
d) The binary operation Dirichlet convolution has an identity ε (Definition 4.20), defined by f ∗ε = ε ∗f = f .
Show that the function ε of Lemma 4.11 is the identity of the convolution.
e) Show that Dirichlet convolution is commutative, that is: f ∗g = g∗f .
(Note: In this exercise we proved that the set of arithmetic functions with addition and convolution is a commutative ring, see Definitions 5.20 and 5.26. This ring is sometimes called the Dirichlet ring .) aAssociativity is a property whose importance is sometimes hush-hushed a bit. We chose to elaborate it, see exercise 5.23 Exercise 4.17. Use exercise 4.16 to prove the following: a) Show that the Dirichlet convolution of two multiplicative functions is multiplicative.
b) Show that the sum of two multiplicative functions is not necessarily multiplicative. (Hint: ε +ε.) Exercise 4.18. See Definition 4.10. Define f(n) ≡τ(n2) and g(n) ≡2ω(n).
a) Compute ω(n), f(n), and g(n) for n equals 10n and 6!.
b) For p prime, show that τ(p2k) = ∑d|pk 2ω(d) = 2k +1. (Hint: use The-orem 4.5.) c) Show that f is multiplicative. (Hint: use that τ is multiplicative.) d) Use (d) to show that g is multiplicative.
e) Show that τ(n2) = ∑ d|n 2ω(d) .
78 4. Number Theoretic Functions Exercise 4.19. Let S(n) denote the number of square free divisors of n with S(1) = 1 and ω(n) the number of distinct prime divisors of n. See also Definition 4.10.
a) Show that S(n) = ∑d|n |µ(d)|. (Hint: use Definition 4.6.) b) Show that S(n) = 2ω(n). (Hint: let W be the set of prime divisors of n. Then every square free divisor corresponds to a subset — product — of those primes. How many subsets of primes are there in W?) c) Conclude that ∑ d|n |µ(d)| = 2ω(n) .
Exercise 4.20. Define the Liouville λ-function by λ(1) = 1 and λ(n) = (−1)Ω(n).
a) Compute λ(10n) and λ(6!).
b) Show that λ is multiplicative. (Hint: Ω(n) is completely additive.) c) Use Proposition 4.3 to show that F(n) = ∑d|n λ(d) is multiplicative.
d) For p prime, show that ∑ d|pk λ(d) = k ∑ i=0 (−1)i which equals 1 if k is even and 0 if k is odd.
e) Use (c) and (d) to conclude that F(n) = ∑ d|n λ(d) = n1 if n = m2 0 else .
Exercise 4.21. Let f be a multiplicative function.
Define q(n) ≡∑d|n µ(d) f(d), where µ is the M¨ obius function.
a) Show that f(1) = 1.
b) Show that f µ (their product) is multiplicative.
c) Use Proposition 4.3 to show that q(n) is multiplicative.
d) Show that if p is prime, then q(pk) = f(1)−f(p) = 1−f(p).
e) Use (c) and (d) to show that q(n) = ∑ d|n µ(d) f(d) = ∏ p prime, p|n (1−f(p)) .
Exercise 4.22. Use exercise 4.21 (e) and the definition of ω in exercise 4.18 and λ in exercise 4.20 to show that ∑ d|n µ(d)λ(d) = 2ω(n) .
4.6. Exercises 79 Exercise 4.23. a) Show that for all n ∈N, µ(n)µ(n+1)µ(n+2)µ(n+3) = 0. (Hint: divisibility by 4.) b) Show that for any integer n ≥3, ∑n k=1 µ(k!) = 1. (Hint: use (a).) Exercise 4.24. a) Use Euler’s product formula and the sequence µ of Def-inition 4.6 to show that 1 ζ(z) = ∏ p prime 1−p−z = ∏ p prime ∑ i≥0 µ(pi)p−iz !
.
b) Without using equation (4.7), prove that the expression in (a) equals ∑n≥1 µ(n)n−z. (Hint: since µ is multiplicative, you can write a proof re-arranging terms as in the first proof of Euler’s product formula.) Exercise 4.25. a) Use equation (4.8) to show that ζ(z−1) ζ(z) = ∑ a≥1 a az ∑ b≥1 µ(b) bz .
b) Show that I ∗µ = ϕ.
c) Use Lemma 4.22, (a), and (b) to show that ζ(z−1) ζ(z) = ∑ n≥1 ϕ(n) nz .
Exercise 4.26. a) Use Corollary 4.23 to show that ζ(z−k) = ∑ a≥1 σk(a) az ∑ b≥1 µ(b) bz .
b) Show that ζ(z−k) = ∑ n≥1 (σk ∗µ)(n)n−z , where ∗means the Dirichlet convolution (Definition 4.19).
Exercise 4.27. Show that ζ(z) has no zeroes and no poles in the region ℜ(z) > 1. (Hint: use that ζ(z) converges for ℜ(z) > 1 and (4.8).) Chapter 5 Modular Arithmetic and Primes Overview. We return to the study of primes in N. This is related to the study of modular arithmetic (the properties of addition and multiplication in Zb), because a ∈N is a prime if and only if there are no non-trivial divi-sors or, expressed differently, there is no 0 < b < a so that a =b 0. Modular arithmetic concerns itself with computations involving addition and mul-tiplication in Z modulo b, denoted by Zb, i.e. calculations with residues modulo b (see Definition 1.8). One common way of looking at this is to consider integers x and y that differ by a multiple of b as equivalent (see exercise 5.1). We write x ∼y. One then proves that the usual addition and multiplication is well-defined for these equivalence classes. This is done in exercise 5.2.
5.1. Euler’s Theorem and Primitive Roots The order of an element g is the smallest positive integer k such that g∗g∗ ··· ∗g, repeated k times and usually written as gk, equals e. One can show that the elements {e,g,g2,··· ,gk−1} also form a group (Definition 5.19).
More details can be found in , , or . In the case at hand, Zb, we have a structure with two operations, namely addition with identity element 0 and multiplication with identity element 1. We could therefore define the order of an element in Zb with respect to addition and with respect to 81 82 5. Modular Arithmetic and Primes multiplication. As an example, we consider the element 3 in Z7 (see Figure 20): 3+3+3+3+3+3+3 =7 0 and 3·3·3·3·3·3 =7 1 .
The first gives 7 as the additive order of 3, and the second gives 6 for the multiplicative order. For our current purposes, however, it is sufficient to work only with the multiplicative version.
1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 Figure 20. Left, the orbits in Z7 under addition of 3; middle, the orbits under multiplication by 3; and right, the orbit under multiplication by 2.
Observe that the multiplicative graphs are not connected and have less symmetries.
Definition 5.1. The (multiplicative) order of a modulo b, written as Ord× b (a), is the smallest positive number k such that ak =b 1. (If there is no such k, the order is ∞.) Recall that ϕ denotes Euler’s phi or totient function (Definition 4.15).
Definition 5.2. i) A complete set of residues modulo b is a set of b integers in Z that has exactly one integer in each congruence class (modulo b).
ii) A reduced set of residues modulo b is the subset of (i) of elements that are relatively prime to b (gcd(a,b) = 1).
As an example, the set {0,1,2,··· ,11} is a complete set of residues modulo 12, while {1,5,7,11} is a reduced set of residues modulo 12.
Lemma 5.3. Suppose gcd(a,b) = 1. If the numbers {xi} form a complete set of residues modulo b (reduced set of residues modulo b), then {axi} is a complete set of residues modulo b (reduced set of residues modulo b).
5.1. Euler’s Theorem and Primitive Roots 83 Proof. Let {xi} be a complete set of residues modulo b. Then the b numbers {axi} form complete set of residues unless two of them are congruent. But that is impossible by Theorem 2.7.
Let {xi} be a reduced set of residues modulo b. Then, as above, no two of the ϕ(b) numbers {axi} are congruent modulo b. Furthermore, Lemma 2.15 implies that if gcd(a,b) = 1 and gcd(xi,b) = 1, then gcd(axi,b) = 1.
Thus the set {axi} is a reduced set of residues modulo b.
■ Theorem 5.4 (Euler). Let a,b > 1 and gcd(a,b) = 1. Then aϕ(b) =b 1.
Proof. Let {xi}ϕ(b) i=1 be a reduced set of residues modulo b. Then by Lemma 5.3, {axi}ϕ(b) i=1 is a reduced set of residues modulo b. Because multiplication is commutative, we get ϕ(b) ∏ i=1 xi =b ϕ(b) ∏ i=1 axi =b aϕ(b) ϕ(b) ∏ i=1 xi Since gcd(xi,a) = 1, Lemma 2.15 implies that gcd ∏ ϕ(b) i=1 xi,a = 1. The cancelation theorem applied to the equality between the first and third terms proves the result.
■ Euler’s theorem says that ϕ(b) is a multiple of Ord× b (a). But it does not say what multiple. In fact, in practice, that question is difficult to decide.
It is of theoretical importance to decide when the two are equal.
Definition 5.5. Let a and b positive integers with gcd(a,b) = 1. If Ord× b (a) = ϕ(b), then a is called a primitive root modulo b.
For example, the smallest integer k for which 3k =7 1 is 6 (see Figure 20). Since ϕ(7) = 6, we see that 3 is a primitive root of 7. Since multi-plication is well-defined in Z7, it follows that (3 + 7k)6 =7 36 =7 1. Thus {···−4,3,10,···} are all primitive roots of 7. The only other non-congruent primitive root of 7 is 5. Not all numbers have primitive roots. For instance, 8 has none.
The importance of the notion of primitive root is perhaps more easily remembered via the next lemma.
Lemma 5.6. a is a primitive root modulo b if and only if the orbit {ai mod b}ϕ(b) i=1 contains all reduced residues modulo b.
84 5. Modular Arithmetic and Primes Proof. If a is a primitive root, then all values of {ai mod b}ϕ(b) i=1 must be distinct, because if ai = a j for some i > j in {1,··· ,ϕ(b)}, then ai−j =b 1, contradicting that a is a primitive root.
We prove the contrapositive1 of the other direction. If ai =b 1 for some positive i less than ϕ(b), then ai+1 =b a and the numbers start repeating so that {ai mod b}ϕ(b) i=1 cannot contain all reduced residues modulo b.
■ The salient fact about prime roots is that we know exactly when they occur. An accessible proof of Theorem 5.7 (i) can be found in chapter 8 and part (ii) in chapter 10.
Theorem 5.7. i) An integer n has a primitive root if and only if n equals 1, 2, 4, pk, or 2pk, where p is an odd prime and k ≥1.
ii) If n has a primitive root g, then it has ϕ(ϕ(n)) primitive roots given by gi for every i such that gcd(i,ϕ(n)) = 1.
The primitive root also has interesting connections with day-to-day arithmetic, namely the expression of rational numbers in any base. We use base 10 as an example.
Proposition 5.8. Let a and n greater than 0 and gcd(a,n) = gcd(10,n) = 1.
The expansion of a/n in base 10 is non-terminating and eventually periodic with period p, where (i) p = Ord× n (10) and (ii) p | ϕ(n).
Proof. The proof proceeds in steps, each of which uses the division algo-rithm. Start by reducing a modulo n and call the result r0.
a = nq0 +r0 , where r0 ∈{0,···n−1}. Lemma 3.1 implies that gcd(a,n) = gcd(r0,n) = 1.
So in particular, r0 ̸= 0. The integer part of a/n is q0. The next step is: r1 n := 10r0 n or 10r0 = nq1 +r1 , where again r1 ∈{0,···n−1}.
Note that 0 ≤10r0 < 10n and so q1 ∈{0,···9}. We now record the first digit “after the decimal point” of the decimal expansion: q1. By Lemma 3.1, 1The contrapositive of (P ⇒Q) is (⌝Q ⇒⌝P) (or: not Q implies not P) and holds if and only if the former holds.
5.2. Fermat’s Little Theorem and Primality Testing 85 we have gcd(10r0,n) = gcd(r1,n). In turn, this implies via Lemma 2.15 that gcd(r0,n) = gcd(r1,n). And again, we see that r1 ̸= 0.
The process now repeats itself.
r2 n := 10r1 n or 10(10r0 −nq1) | {z } r1 = nq2 +r2 , and we record the second digit after the decimal dot, q2 ∈{0,···9}. By the same reasoning, gcd(r2,n) = 1 and so r2 ̸= 0. One continues and proves by induction that gcd(ri,n) = 1. In particular, ri ̸= 0, so the expansion does not terminate.
Since the remainders ri are in {1,···n−1}, the sequence must be even-tually periodic with (least positive) period p. At that point, we have 10k+pr0 =n 10kr0 .
By Theorem 2.7, we can cancel the common factors 10k and r0, and we obtain that 10p =n 1. Since p is the least such (positive) number, we have proved (i). Item (ii) follows directly from Euler’s Theorem.
■ Of course, this proposition easily generalizes to computations in any other base b. As an en example, we mention that if gcd(a,n) = 1 and b is a primitive root of n, then the expansion of a/b has period ϕ(n).
The next result follows by setting y = x + kϕ(b) in ay and applying Euler’s theorem. It has important applications in cryptography.
Corollary 5.9. Let a and b be coprime with b > 1.
x =ϕ(b) y = ⇒ ax =b ay .
5.2. Fermat’s Little Theorem and Primality Testing Euler’s theorem has many other important consequences. It implies what is known as Fermat’s little theorem, although it was not proved by Fermat himself, since, as he writes in the letter in which he stated the result, he feared “its being too long” [Section 5.2]. Not an isolated case, it would appear!
Corollary 5.10 (Fermat’s little theorem). If p is prime and gcd(a, p) = 1, then ap−1 =p 1.
86 5. Modular Arithmetic and Primes This follows from Euler’s Theorem by noticing that for a prime p, ϕ(p) = p −1. There is an equivalent formulation which allows p to be a divisor of a. Namely, if p is prime, then ap =p a. Notice that if p | a, then both sides are congruent to 0.
Primes are of great theoretical and practical value (think of encryption, for example). Algorithms for primality testing are therefore very useful.
The simplest test to find out if some large number n is prime, consists of course of applying some version of Eratosthenes’ sieve to the positive inte-gers less than or equal to √n. To carry this out, we will have to perform on the order of √n divisions.
Another possibility is to use the converse of Fermat’s little theorem (Corollary 5.10). If n and p are distinct primes, we know that pn−1 =n 1.
The Fermat primality test for n consists of testing, for example, whether 2n−1 =n 1. If that fails, we know that n is not prime. However, the converse of Fermat’s little theorem is not true! So even if 2n−1 =n 1, it could be that n is not prime; we will discuss this possibility at the end of this section. As it turns out, primality testing via Fermat’s little theorem can be done much faster than the naive method, provided one uses fast modular exponentiation algorithms. We briefly illustrate this technique by computing 11340 modulo 341.
Start by expanding 340 in base 2 as done in exercise 3.14, where it was shown that this takes on the order of log2 340 (long) divisions.
340 = 170·2+0 170 = 85·2+0 85 = 42·2+1 42 = 21·2+0 21 = 10·2+1 10 = 5·2+0 5 = 2·2+1 2 = 1·2+0 1 = 0·2+1 And so 340 = 101010100 in base 2 .
Next, compute a table of powers 112i modulo 341, as done below. This can be done using very few computations. For instance, once 118 =341 143 5.2. Fermat’s Little Theorem and Primality Testing 87 has been established, the next up is found by computing 1432 modulo 341, which gives 330, and so on. So this takes about log2 340 multiplications.
0 111 =341 11 0 112 =341 121 1 114 =341 319 0 118 =341 143 1 1116 =341 330 0 1132 =341 121 1 1164 =341 319 0 11128 =341 143 1 11256 =341 330 The first column in the table thus obtained now tells us which coefficients in the second we need to compute the result.
11340 =341 330·319·330·319 =341 132.
Again, this takes no more than log2 340 multiplications. Thus altogether, for a number n and a computation in base b, this takes on the order of 2logb n multiplications plus logb n divisions2. For large numbers, this is much more efficient than the √n of the naive method.
As mentioned, the drawback is that we can get false positives. While there are partial converses to Fermat’s little theorem, they do not yield com-putationally efficient improvements (see exercise 5.20).
Definition 5.11. The number n ∈N is called a pseudoprime to the base b if gcd(b,n) = 1 and bn−1 =n 1 but nonetheless n is composite. (When the base is 2, the clause to the base 2 is often dropped.) Some numbers pass all tests to every base and are still composite.
These are called Carmichael numbers. The smallest Carmichael number is 561. It has been proved that there are infinitely many of them.
Definition 5.12. The number n ∈N is called a Carmichael number if it is composite and it is a pseudoprime to every base.
2Divisions take more computations than multiplications. We do not pursue this here.
88 5. Modular Arithmetic and Primes The smallest pseudoprime is 341, because 2340 =341 1 while 341 = 11 · 31. In this case, one can still show that 341 is not a prime by using a different base: 3340 =341 56. Thus by Fermat’s little theorem, 341 cannot be prime.
The reason that the method sketched here is still useful is that pseu-doprimes are very much rarer than primes. The numbers below 2.5 · 1010 contain on the order of 109 primes. At the same time, this set contains only 21853 pseudoprimes to the base 2. There are only 1770 integers below 2.5·1010 that are pseudoprime to the bases 2, 3, 5, and 7. Thus if a number passes these four tests, it is overwhelmingly likely that it is a prime.
5.3. Fermat and Mersenne Primes Through the ages, back to early antiquity, people have been fascinated by numbers, such as 6, that are the sum of their positive divisors other than itself, to wit: 6=1+2+3. Mersenne and Fermat primes, primes of the form 2k ± 1, have also attracted centuries of attention. Note that if p is a prime other than 2, then pk ±1 is divisible by 2 and therefore not a prime.
Definition 5.13. (i) The Mersenne numbers are Mk = 2k −1. Mersenne prime is a Mersenne number that is also prime.
(ii) The Fermat numbers are Fk = 22k + 1. A Fermat prime is a Fermat number that is also prime.
(iii) The number n ∈N is called a perfect, if σ(n) = 2n.
Lemma 5.14. (i) If ab = k, then (2b −1) | (2k −1).
(ii) If ab = k and a is odd, then (2b +1) | (2k +1).
Proof. We only prove (ii); (i) can be proved similarly. So suppose that a is odd, then 2b =2b+1 −1 = ⇒2ab =2b+1 (−1)a =2b+1 −1 = ⇒2ab +1 =2b+1 0 which proves the statement. Notice that this includes the case where b = 1.
In that case, we have 3 | (2a +1) (whenever a odd).
■ A proof using geometric series can be found in exercise 1.12. This lemma immediately implies the following.
Corollary 5.15. i) If 2k −1 is prime, then k is prime.
ii) If 2k +1 is prime, then k = 2r.
5.3. Fermat and Mersenne Primes 89 So candidates for Mersenne primes are the numbers 2p −1 where p is prime. This works for p ∈{2,3,5,7}, but 211 −1 = 2047 is the monkey-wrench. It is equal to 23·89 and thus is composite. After that, the Mersenne primes become increasingly sparse. For example, 8 of the first 11 Mersenne numbers are prime (M11, M23, M29 are not prime). However, among the first approximately 2.3 million Mersenne numbers, only 45 give Mersenne primes. As of this writing (2021), it is not known whether there are infin-itely many Mersenne primes. In 2020, a very large Mersenne prime was discovered: 282,589,933 −1. Mersenne primes are used in pseudo-random number generators.
Turning to primes of the form 2k +1, the only candidates are Fr = 22r + 1. Fermat himself noted that Fr is prime for 0 ≤r ≤4}, and he conjectured that all these numbers were primes. Again, Fermat did not quite get it right!
It turns out that the 5-th Fermat number, 232 + 1, is divisible by 641 (see exercise 5.11). In fact, as of this writing in 2017, there are no other known Fermat primes among the first 297 Fermat numbers! Fermat primes are also used in pseudorandom number generators.
Lemma 5.16. If 2k −1 is prime, then k > 1 and 2k−1(2k −1) is perfect.
Proof. If 2k −1 is prime, then it must be at least 2, and so k > 1. Let n = 2k−1(2k −1). Since σ is multiplicative and 2k −1 is prime, we can compute (using Theorem 4.5): σ(n) = σ(2k−1)σ(2k −1) = k−1 ∑ i=0 2i !
2k = (2k −1)2k = 2n which proves the lemma.
■ Theorem 5.17 (Euler’s Theorem). Suppose n > 0 is even. Then n is of the form 2k−1(2k −1) where 2k −1 is prime if and only if n is perfect.
Proof. One direction follows from the previous lemma. Thus we only need to prove that if an even number n is perfect, then it is of the form stipulated.
Since n is even, we may assume n = q2k−1 where k ≥2 and q is odd.
Using multiplicativity of σ and the fact that σ(n) = 2n: σ(n) = σ(q)(2k −1) = 2n = q2k .
90 5. Modular Arithmetic and Primes Thus (2k −1)σ(q)−2k q = 0 .
(5.1) Since 2k −(2k −1) = 1, we know by B´ ezout that gcd((2k −1),2k) = 1. Thus Proposition 3.5 implies that the general solution of the above equation is: q = (2k −1)t and σ(q) = 2kt , (5.2) where t > 0, because we know that q > 0.
Assume first that t > 1. The form of q, namely q = (2k −1)t, allows us to identify at least four distinct divisors of q. This gives that σ(q) ≥1+t +(2k −1)+(2k −1)t = 2k (t +1) .
This contradicts equation (5.2), and so t = 1.
Now use equation (5.2) again (with t = 1) to get that n = q2k−1 = (2k −1)2k−1 has the required form. Furthermore, the same equation says that σ(q) = σ(2k −1) = 2k which proves that 2k −1 is prime.
■ It is unknown at the date of this writing (2021) whether any odd perfect numbers exist.
5.4. A Divisive Issue: Rings and Fields 1 2 3 4 5 0 1 2 3 4 5 0 Figure 21. Left: the relation a is an additive inverse mod 6 of b. Right: the relation a is an multiplicative inverse mod 6 of b.
The next result is a game changer! It tells us that there is a unique element a−1 such that aa−1 =b 1 if and only if a is in the reduced set of residues (modulo b). Thus division is only well-defined in the reduced set of residues modulo b. So, for example, the reduced set of residues modulo 15 equals {1,2,4,7,8,11,13,14}. In this group, we can multiply and divide all we want. For example, the inverse of 8 in Z15 is 2 because 8 · 2 =15 1.
5.4. A Divisive Issue: Rings and Fields 91 In fact, this set forms a nice Abelian group (defined below) under multipli-cation. Another illustration is given in Figure 21 for Z6. On the left, we see that every element has an additive inverse. So, for example 4+2 =6 0, and so 4 and 2 are additive inverses. Notice that the relation is symmetric, so the edges in the graph have no arrows on them. However, on the right, we see that only 5 and 1 have a multiplicative inverse. So 52 =6 1 and similar for 1. Since 7 is a prime, all elements of Z7 except 0 have multiplicative inverses modulo 7 (see Figure 22).
1 2 3 4 5 6 0 1 2 3 4 5 6 0 Figure 22. Left: the relation a is an additive inverse mod 7 of b. Right: the relation a is an multiplicative inverse mod 7 of b.
Proposition 5.18. Let R be a reduced set of residues modulo b. Then i) for every a ∈R, there is a unique a′ in R such that a′a =b aa′ =b 1, ii) for every a ̸∈R, there exists no x ∈Zb such that ax =b 1, iii) let R = {xi}ϕ(b) i=1 , then also R = {x−1 i }ϕ(b) i=1 .
Proof. Statement (i): Since gcd(a,b) = 1, the existence of a solution fol-lows immediately from B´ ezout’s Lemma. Namely a′ solves for x in ax + by = 1. This solution must be in R, because a, in turn, is the solution of a′x+by = 1 and thus B´ ezout’s Lemma implies that gcd(a′,b) = 1. Suppose we have two solutions ax =b 1 and ay =b 1, then uniqueness follows from applying the cancelation Theorem 2.7 to the difference of these equations.
Statement (ii): By hypothesis, gcd(a,b) > 1. We have that ax =b 1 is equivalent to ax+by = 1, which contradicts B´ ezout’s lemma.
Statement (iii): This is similar to Lemma 5.3. By (1), we know that all inverses are in R. So if the statement is false, there must be two elements of R with the same inverse: ax =b cx. This is impossible by cancellation (Theorem 2.7).
■ 92 5. Modular Arithmetic and Primes What this means is that in structures like Zb addition and multiplication have a complicated relationship. Under addition, they form a group.
Definition 5.19. A group is defined as a set G with an operation satisfy-ing: i) G is closed under the operation, or all a, b in G, a∗b ∈G.
ii) The operation is associative or (a∗b)∗c = a∗(b∗c).
iii) R has an identity element e and for all a in G, a∗e = e∗a = a.
iv) Each a ∈G has an inverse a−1 such that a∗a−1 = a−1 ∗a = e.
The group is called Abelian group if the operation is commutative or a∗b = b∗a).
It is important to realize that not all groups are commutative. The small-est non-commutative group is the group of symmetries of an equilateral tri-angle, S3, with the composition as the group operation (see Figure 23). This group is isomorphic to the group of permutations of the symbols {1,2,3}.
Thus the refection in the line through 1 in the triangle corresponds to swap-ping 2 and 3, while one possible rotation corresponds to 1 →2, 2 →3, and 3 →1. The figure that the order in which we carry out the operations affects the outcome.
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 Figure 23. The group of symmetries of an equilateral triangle, S3, is not commutative. The group operation is composition. In the top row, we perform first a rotation given by 1 →2, 2 →3, and 3 →1 and then a refection in the dotted line through 1. In the bottom row, the reverse.
The additive group Zb is generated by the element 1, because repeated addition of 1 gives the entire group. This also makes it clear that we can not leave any elements out and still obtain an additive group. But under multiplication, the story is more complicated. There is no multiplicative inverse of 0. But even if we exclude 0, then according to Proposition 5.18, 5.4. A Divisive Issue: Rings and Fields 93 we only get a multiplicative group if b is prime. Indeed, in general we only get a multiplicative group if we further restrict to the reduced set of residues modulo b. Let us illustrate the point by showing the tables for multiplication in Z5 and Z6. In the latter case, the only multiplicative group consists of the elements 1 and 5.
Z5(×) 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1 Z6(×) 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 1 2 3 4 5 2 0 2 4 0 2 4 3 0 3 0 3 0 3 4 0 4 2 0 4 2 5 0 5 4 3 2 1 The optimistic reader might be inclined to think that maybe not all is lost, as long as things work for the most important number system, Z itself.
Alas, a moment’s thought reveals that multiplication in Z, like multiplica-tion in Zb for b non-prime, does not have an inverse. Thus our hand is forced, and we define a structure where addition has all the nice properties — in particular, it has an inverse — and where we are a bit more prudent in assigning the characteristics of multiplication.
Definition 5.20. A ring is defined as a set R which is closed under two op-erations, usually called addition and multiplication, and has the following properties: i) R with addition is an Abelian group (with additive identity 0).
ii) Multiplication in R is associative (see exercise 5.23).
iii) Multiplication is distributive over addition (that is: a(b + c) = ab + bc and (b+c)a = ba+ca).
iv) R has a (multiplicative) identity denoted by 1 and 0 ̸= 1.
A commutative ring is a ring in which multiplication is commutative.
Remark 5.21. Note that N is not a ring, because addition is not invertible.
We will from here on out consider the primes as a subset of Z.
Remark 5.22. We will assume rings to be commutative and drop the ad-jective “commutative” for brevity, unless needed for clarity.
Remark 5.23. The requirement that 0 ̸= 1 only excludes the 0 ring (R = {0}).
94 5. Modular Arithmetic and Primes Remark 5.24. An important example of an “almost ring” are the multiples nZ in Z for n > 1. Indeed, that set satisfies all the requirements of a ring except that it does not have a multiplicative identity. This is sometimes called a rng.
Definition 5.25. A unit in a ring is an element that has a multiplicative inverse in the ring. This is also called an invertible element.
On the other hand, other important sets, such as Q, R, or C, do have a well-defined multiplicative inverse (again excepting 0) much like Zp for p prime. Thus we also need to define a structure where multiplication is treated on more equal footing with addition — it has an inverse.
Definition 5.26. A field is a commutative ring for which multiplication by a non-zero number has an inverse. Equivalently, considered as a ring, all non-zero elements are units.
But in generally, the words division and multiplicative inverse have to be used carefully in a ring.
Definition 5.27. Let a, b, and x in a ring. We say that b is a divisor of a and write b | a if there is a solution x of bx = a.
The sets Z, Q, and Zb are all examples of rings, but of these only Q and Zp with p prime are fields, because all elements are invertible as we saw in Proposition 5.18. The field of the integers modulo a prime p will be from now be denoted by Fp, where p is understood to be a prime.
Rings and fields occur in all kinds of other situations and applications.
We already looked at one interesting example of a ring, namely the arith-metic functions with addition and convolution as operations (exercise 4.16).
Here are some other examples of rings that are not fields. Real numbers of the form a+b √ 3 where a and b in Z, complex numbers of the form a+ib or those of the form a+ib √ 6 where a and b in Z. Other examples are the n by n matrices (n ≥2). We have already seen the polynomials with ra-tional coefficients exercise 3.22. They also form a ring. All of these rings have different properties. For instance, the ring of n by n matrices is not commutative. We will see later that not all rings (that are not fields) have primes.
It is useful to reflect a moment on how the absence of division influ-ences how we think about such sets. It is precisely that curious absence that 5.5. Wilson’s Theorem 95 brings us to the study of primes, integers that have no non-trivial divisors at all. The situation in fields like Zp (for prime p) or R is very different! Here multiplication does have an inverse, and thus given a and b not equal to 0, we can always write a as a non-trivial product as follows: a =p (ab)b−1 .
Here is another interesting observation. If we extend the integers to the rationals Q, we obtain a field. Thus the problem of where the primes are goes away: in Q (or R) we can always divide (except by 0), and there are no primes. Of course, since, even in mathematics, nothing is perfect, in the rationals we have other problems. If we allow the integers to be arbitrarily divided by other integers, we obtain the field of the rational numbers. It was a source of surprise and mystery to the ancients, that within the rational numbers we still cannot solve for x in x2 = 2, although we can get arbi-trarily good approximations. Those ‘gaps’ in the rational numbers, are the irrational numbers. We are then left with the thorny question of whether the reals containing both the rational and the irrational numbers still have gaps. How can we approximate irrational numbers using rational numbers?
How can we calculate with the reals? Well, among other things you have to learn how to take limits, which is a whole other can of worms.
5.5. Wilson’s Theorem We end this chapter with one important application of division in Zp.
Lemma 5.28. Let p be prime. Then a2 =p 1 if and only if a =p ±1. Equiv-alently, a ∈Zp is its own multiplicative inverse if and only if a =p ±1.
Proof. We have a2 =p 1 ⇐ ⇒a2 −1 =p (a+1)(a−1) =p 0 ⇐ ⇒p | (a+1)(a−1).
Because p is prime, Corollary 2.9 says that the last statement holds if and only if either p | a+1 (and so a =p −1) or p | a−1 (and so a =p +1).
■ Perhaps surprisingly, this last lemma is false if p is not prime. For example, 42 =15 1, but 4 ̸=15 ±1.
Theorem 5.29 (Wilson’s theorem). If p prime in Z, then (p−1)! =p −1.
If b is composite, then (b−1)! ̸=b ±1.
96 5. Modular Arithmetic and Primes Proof. This is true for p is 2 and 3. If p > 3, then Proposition 5.18 (3) and Lemma 5.28 imply that every factor ai in the product (p−1)! other than -1 or 1 has a unique inverse a′ i different from itself. The factors a′ i run through all factors 2 through p −2 exactly once. Thus in the product, we can pair each ai different from ±1 with an inverse a′ i distinct from itself. This gives (p−1)! =p (+1)(−1)∏aia′ i =p −1 .
The second part is easier. If b is composite, there are least residues a and d greater than 1 so that ad =b 0. Now either we can choose a and d distinct and then (b −1)! contains the product ad, and thus it equals zero mod b. Or else this is impossible and there exists a such that a2 =b 0. But then still gcd((b−1)!,b) is a multiple of a. Then, by B´ ezout, (b−1)! mod b cannot be equal to ±1.
■ Wilson’s theorem could be used to test primality of a number n. How-ever, this takes n multiplications, which in practice is more expensive than trying to divide n by all numbers less than √n. Note, however, that if you want to compute a list of all primes between 1 and N, Wilson’s theorem can be used much more efficiently. After computing (k −1)! mod k to de-termine whether k is prime, it takes only 1 multiplication and 1 division to determine whether k +1 is prime.
5.6. Exercises Exercise 5.1. a) Let m > 0. Show that a =m b is an equivalence relation on Z. (Use Definitions 1.7 and 1.28.) b) Describe the equivalence classes of Z modulo 6. (Which numbers in Z are equivalent to 0? Which are equivalent to 1? Et cetera.) c) Show that the equivalence classes are identified by their residue, that is: a ∼b if and only if Resm (a) = Resm (b).
Note: If we pick one element of each equivalence class, such an element is called a representative of that class. The smallest non-negative represen-tative of a residue class in Zm, is called the least residue (see Definition 1.8). The collection consisting of the smallest non-negative representative of each residue class is called a complete set of least residues.
5.6. Exercises 97 Exercise 5.2. This exercise relies on exercise 5.1. Denote the set of equiva-lence classes of Z modulo m by Zm (see Definition 1.7). Prove that addition and multiplication are well-defined in Zm, using the following steps.
a) If a =m a′ and b =m b′, then Resm (a) + Resm (b) =m Resm (a′) + Resm (b′). (Hint: show that a + b = c if and only if a + b =m c. In other words: the sum modulo m only depend on Resm (a) and Resm (b) and not on which representative in the class (see exercise 5.1) you started with.) b) Do the same for multiplication.
Exercise 5.3. Let n = ∑k i=1 ai10i where ai ∈{0,1,2,··· ,9}.
a) Show that 10k =3 1 for all k ≥0. (Hint: use exercise 5.2.) b) Show that n =3 ∑k i=1 ai.
c) Show that this implies that n is divisible by 3 if and only the sum of its digits is divisible by 3.
Exercise 5.4. Let n = ∑k i=1 ai10i where ai ∈{0,1,2,··· ,9}. Follow the strategy in exercise 5.3 to prove the following facts.
a) Show that n is divisible by 5 if and only if a0 is. (Hint: Show that n =5 a0.) b) Show that n is divisible by 2 if and only if a0 is.
c) Show that n is divisible by 9 if and only if ∑k i=1 ai is.
d) Show that n is divisible by 11 if and only if ∑k i=1 (−1)iai is.
e) Find the criterion for divisibility by 4.
f) Find the criterion for divisibility by 7. (Hint: this is a more complicated criterion!) Exercise 5.5. a) Determine the period of the decimal expansion of the following numbers: 100/13, 13/77, and 1/17 through long division.
b) Use Proposition 5.8 to determine the period.
c) Check that this period equals a divisor of ϕ(n).
d) The same questions for expansions in base 2 instead of base 10.
Exercise 5.6. a) Compute 2n−1 mod n for n odd in {3···40}.
b) Are there any pseudo-primes in the list?
Exercise 5.7. Assume that n is a pseudoprime to the base 2.
a) Show that 2n −2 =n 0.
b) Show from (a) that n | Mn −1. (See Definition 5.13.) c) Use Lemma 5.14 to show that (b) implies that Mn | 2Mn−1 −1.
d) Conclude from (c) that if n is a pseudoprime in base 2, so is Mn.
98 5. Modular Arithmetic and Primes Exercise 5.8. a) List (n−1)! mod n for n ∈{2,··· ,16}.
b) Where does the proof of the first part of Wilson’s theorem fail in the case of n = 16?
c) Does Wilson’s theorem hold for p = 2? Explain!
d) Characterize the set of n ≥2 for which (n−1)! mod n is not in {0,−1}.
Exercise 5.9. a) Compute 772 mod 13, using modular exponentiation.
b) Similarly for 484187 mod 1189.
c) Find 100!+102! mod 101. (Hint: Wilson.) d) Show that 1381! =1382 0. (Hint: Wilson.) Exercise 5.10. a) For i in {1,2,···11} and j in {2,3,···11}, make a table of Ord× j (i), i varying horizontally. After the jth column, write ϕ( j).
b) List the primitive roots i modulo j for i and j as in (a). (Hint: the smallest primitive roots modulo j are: {1,2,3,2,5,3, / 0,2,3,2}.) Exercise 5.11. We show that the 5-th Fermat number, 232 + 1, is a com-posite number.
a) Show that 24 =641 −54.(Hint: add 24 and 54.) b) Show that 275 =641 −1.
c) Show that 232 +1 = (27)424 +1 =641 0.
d) Conclude that F5 is divisible by 641.
Exercise 5.12. a) Compute ϕ(100). (Hint: use Theorem 4.17.) b) Show that 179121 =100 79121.
c) Show that 79121 =100 791. (Hint: use Theorem 5.4) d) What are the last 2 digits of 179121?
The following 5 exercises on basic cryptography are based on .
First some language. The original readable message is called the plain text.
Encoding the message is called encryption. And the encoded message is often called the encrypted message or code. To revert the process, that is: to turn the encrypted message back into plain text, you often need a key. Below we will encode the letters by 0 through 25 (in alphabetical order). We encrypt by using a multiplicative cipher. This means that we will encrypt our text by multiplying each number by the cipher modulo 26, and then return the corresponding letter. For example, if we use the cipher 3 to encrypt the plain text bob, we obtain the encrypted text as follows 1.14.1 →3.42.3 →3.16.3.
5.6. Exercises 99 Exercise 5.13. a) Use the multiplicative cipher 3 to decode DHIM.
b) Show that an easy way to decode is multiplying by 9 (modulo 26). The corresponding algorithm at the number level is called division by 3 modulo 26.
c) Suppose instead that our multiplicative cipher was 4. Encode bob again.
d) Can we invert this encryption by using multiplication modulo 26? Ex-plain why.
Exercise 5.14. Suppose we have an alphabet of q letters and we encrypt using the multiplicative cipher p ∈{0,···q−1}. Use modular arithmetic to show that the encryption can be inverted if and only gcd(p,q) = 1. (Hint: Assume the encryption of j1 and j2 are equal. Then look up and use the Unique Factorization theorem in Chapter 2.) Exercise 5.15. Assume the setting of exercise 5.14. Assume p and q are such that the encryption is invertible. What is the decryption algorithm?
Prove it. (Hint Find r ∈{0,···q−1} such that rp =q 1. Then multiply the encryption by r.) Exercise 5.16. Work out the last two problems if we encrypt using an affine cipher (a, p) . That is, the encryption on the alphabet {0,···q−1} is done as follows: i →a+ pi mod q Work out when this can be inverted, and what the algorithm for the inverse is.
Exercise 5.17. Decrypt the code V’ir Tbg n Frperg.
Theorem 5.30 (Binomial Theorem). If n is a positive integer, then (a+b)n = n ∑ i=0 n i aibn−i where n i = n!
i!(n−i)! .
Exercise 5.18. a) If p is prime, show that p i mod p equals 0 if 1 ≤i ≤ p−1 and equals 1 if i = 0 or i = p.
b) Evaluate 4 i mod 4 and 6 i mod 6. So where in (a) did you use the fact that p is prime?
c) Use (a) and the binomial theorem to show that if p is prime, then we have (a+b)p =p ap +bp.
100 5. Modular Arithmetic and Primes Exercise 5.19. Let p be prime.
a) Show that 1p =p 1.
b) Use exercise 5.18 (c) to show that for k > 0, if kp =p k, then (k+1)p =p k +1.
c) Conclude from (b) that for for all n ∈N, np =p n. (Hint: use induction.) d) Prove that for for all n ∈Z, np =p n. (Hint: (−n)p =p (−1)pnp and assume p odd. Prove separately for p = 2.) e) Use (d) to prove Fermat’s little theorem. (Hint: use cancellation.) There are partial converses to Fermat’s little theorem. But if our aim is testing for primality, these do not yield computationally efficient improve-ments. We give the simplest of these results here.
Lemma 5.31. Suppose a and n in N such that an−1 =n 1 and that for all primes that divide n−1 we have a(n−1)/p ̸=n 1. Then n is a prime.
Exercise 5.20. In this exercise, we prove Lemma 5.31. For this purpose, abbreviate Ord× n (a) by o and assume the condition of the lemma.
a) Show that n−1 = oj for some j ∈N.
b) Show that if j > 1 in (a), there is a prime p dividing j such that a(n−1)/p =n ao( j/p) =n 1.
c) Show that j = 1 and so o = Ord× n (a) = n−1.
d) Show that (c) implies the lemma. (Hint: use Euler.) e) Use the lemma to show that 997 is prime. (Hint: 996 has prime divisors 2, 3, and 83.) Theorem 3.13 and exercise 3.18 show how to solve linear congruences gen-erally. Quadratic congruences are much more complicated. As an example, we look at the equation x2 =p ±1 in the following exercise.
Exercise 5.21. a) Show that Fermat’s little theorem gives a solution of x2 −1 =p 0 whenever p is an odd prime. (Hint: consider x p−1 2 .) b) Use Lemma 5.28 to show that x p−1 2 =p ±1.
c) Show that Wilson’s theorem implies that for odd primes p (−1) p−1 2 p−1 2 !
2 =p −1.
(Hint: the left-hand side gives all reduced residues modulo p.) d) Use (c) to show that if p =4 1 (examples are 13, 17, 29, etc), then h p−1 2 !
i satisfies the quadratic congruence x2 +1 =p 0.
e) Show that if p =4 3 (examples are 7, 11, 19, etc), then the quadratic congruence x2 + 1 =p 0 has no solutions. (Hint: we have x4 =p 1 and by Euler xϕ(p) =p 1; derive a contradiction if p =4 3.) 5.6. Exercises 101 Exercise 5.22. Given b > 2, let R ⊆Zb be the reduced set of residues and let S ⊆Zb be the set of solutions in Zb of x2 =b 1 (or self inverses).
a) Show that S ⊆R. (Hint:B´ ezout.) b) Show that ∏ x∈R x =b ∏ x∈S x (=b 1 if S is empty).
c) Show that if S contains a, then it contains −a.
d) Show that if a =b −a, then a and −a are not in S.
e) Show that ∏ x∈R x =b (−1)m some m.
f) Show that ∏ x∈R x =b (−1)|S|/2 .
g) Compute ∏x∈R x in a few cases (b = 6, 8), and verify that (f) holds.
Definition 5.32. The nth Catalan numberCn equals 1 n+1 2n n = 2n n − 2n n+1 .
Exercise 5.23. Many common operations in R are not associative.
a) Compute 234, 4−3−2, 4/3/2. (Hint: depending on how you place the parentheses, you get different answers.) In the last two cases, the problem disappears if we recast the computation in terms of the (associative) oper-ators + and ×: compute 4+(−3)+(−2) and 4× 1 3 × 1 2.
b) Show that the number of monotone lattice paths from (0,0) to (a,b) where a,b > 0 equals a+b a . (Hint: place a+b edges of which a are hori-zontal and b are vertical in any order.) c) For notational ease, indicate the non-associative operation by ∗. Show that the number of ways ∗n+1 i=1 ai can be interpreted equals the number of “good paths”, that is: monotone lattice paths in R2 from (0,0) to (n,n) that do not go above the diagonal. (Hint: write the expression so that it has n opening parentheses “(” in it; there are n operations to be performed; reading from left to right, each ( corresponds to a “right” move, each to an “up” move.) d) Show that there is a bijection from the set of “bad paths”, that is: mono-tone lattice paths in R2 from (0,0) to (n,n) that touch the line ℓ: y = x+1, to the set of monotone paths in R2 from (0,0) to (n−1,n+1). (Hint: re-flect the bad path in ℓas indicated in Figure 24 and show this is invertible.) e) Use (c) and (d) to show that the number of good paths equals the num-ber of monotone paths from (0,0) to (n,n) minus the number of monotone paths from (0,0) to (n−1,n+1).
f) Use (e) to show that the number of interpretations in (c) equals Cn of Definition 5.32.
102 5. Modular Arithmetic and Primes (0,0) (n,n) Figure 24. The part to the right of the intersection with ℓ: y = x + 1 (dashed) of a bad path (in red) is reflected. The reflected part in in-dicated in green. The path becomes a monotone path from (0,0) to (n−1,n+1).
Exercise 5.24. Show that the following sets with the usual additive and multiplicative operations are not fields: a) The numbers a+b √ 3 where a and b in Z.
b) The numbers of the form a+ib √ 6 where a and b in Z.
c) Z6.
d) The 2 by 2 real matrices.
e) The polynomials with rational coefficients.
f) The Gaussian integers, i.e. the numbers a+bi where a and b in Z.
(Hint: in each case, exhibit at least one element that does not have a mul-tiplicative inverse.) Exercise 5.25. We revisit the Dirichlet ring of exercise 4.16.
a) Show that given an arithmetic function f, we have that if f(1) ̸= 0 g∗f = ε ⇐ ⇒ g(1) = 1 f(1) if n = 1 g(n) = −1 f(1) ∑d|n,d 1 (Note: g is called the Dirichlet inverse of f.) b) Show that f is a unit if and only if f(1) ̸= 0.
c) Compute the first 12 terms of the Dirichlet inverse of the Fibonacci sequence (Definition 3.18).
(Hint: (1,−1,−2,−2,−5,−4,−13,−16,−30,−45,−89,−122).) d) Show g(n) = −f(n) if n is prime.
e) What is the Dirichlet inverse of the (non-zero) constant function? (Hint: Equation (4.7).) Part 2 Currents in Number Theory: Algebraic, Probabilistic, and Analytic Chapter 6 Continued Fractions Overview. The algorithm for continued fractions is really a reformulation of the Euclidean algorithm. However, the reformulated algorithm has had such a spectacular impact on mathematics that it deserves its own name and a separate treatment. One of the best introductions to this subject is the classic . A generalization of these ideas can be found in .
6.1. The Gauss Map Definition 6.1. The Gauss map (see Figure 25) is the transformation T : [0,1] →[0,1) defined by T(ξ) = 1 ξ − 1 ξ = 1 ξ and T(0) = 0 , where we have used the notation of Definition 2.1.
Lemma 6.2. Set qi = j ri−i ri k as in equation (3.1). Then the sequence {ri} defined by the Euclidean algorithm of Definition 3.3 satisfies: ri+1 ri = 1 ri/ri−1 −qi = T ri ri−1 and ri ri−1 = 1 qi + ri+1 ri .
105 106 6. Continued Fractions 0 0 1 1 1/2 1/3 Figure 25. Four branches of the Gauss map.
Proof. From equation (3.1) or (3.4), we recall that that {ri} is a decreasing sequence and that ri−1 = riqi +ri+1 Upon dividing by ri, we get ri−1 ri = qi + ri+1 ri where qi = ri−1 ri , The first equation of the lemma is obtained by subtracting qi from both sides, and the second equation is obtained by taking the reciprocal of both sides of the above equation.
■ In the exercises 3.20 and 3.21, we indicated by example how the Gauss map is related to the Euclidean algorithm. In particular, the lemma implies that ri+1 ri = 1 ri/ri−1 − 1 ri/ri−1 .
(6.1) 6.2. Continued Fractions The beauty of the relation in Lemma 6.2 is that, having sacrificed the value of gcd(r1,r2) — whose value we therefore may as well set at 1, we have a procedure that applies to rational numbers! There is no reason why this recursive procedure should be restricted to rational numbers. Indeed, very interesting things happen when we extend the procedure to also allow irra-tional starting values.
Definition 6.3. Here is a modified version of the Euclidean algorthm. In the second equation of Lemma 6.2, write ωi = ri+1 ri and ai = 1 ωi (or,equivalently, ai = ℓif ωi ∈ 1 ℓ+1, 1 ℓ ). Extend ω to allow for all values in [0,1).
6.2. Continued Fractions 107 It is important to note that, in effect, we have set ai equal to qi+1. This very unfortunate bit of redefining is done so that the qi mesh well with the Euclidean algorithm (see equation (3.2)) while making sure that the sequence of the ai in Definition 6.4 below starts with a1.
At any rate, with these conventions, the equations of Lemma 6.2 be-come (see also (6.1)): ωi+1 = 1 ωi −ai = T (ωi) , ai = ⌊1/ωi⌋and ωi = 1 ai +ωi+1 .
(6.2) The way one thinks of this is as follows. The first equation defines a dynamical system1. Namely, given an initial value ω1 ∈[0,1), the repeated application of T gives a string of positive integers {a1,a2,···}. The string ends if and only if after n steps ωn = 1 ℓ, and so ωn+1 = 0. We show in Theorem 6.5 that this happens if and only if ω1 is rational. The ℓth branch of T, depicted in Figure 25, has Iℓ= ( 1 ℓ+1, 1 ℓ] as its domain. It is easy to see that ai = ℓprecisely if ωi ∈Iℓ.
If, on the other hand, the {ai} are given, then we can use the second equation to formally2 derive a possibly infinite quotient that characterizes ω1. For, in that case, we have ω1 = 1 a1 + 1 a2 + 1 a3 +··· .
(6.3) The expression stops after n steps, if ωn+1 = 0. Else the expression contin-ues forever, and we can only hope that converges to a limit. We now give some definitions.
Definition 6.4. Let ω1 ∈[0,1]. The expression ω1 = 1 a1 + 1 a2 + 1 a3 +··· def ≡[a1,a2,a3,···] .
1A dynamical system is basically a rule that describes short term changes. Usually the purpose of studying such a system is to derive long term behavior, such as, in this case, deciding whether the sequence {ai} is finite, periodic, or neither.
2Here, “formally” means that we have an expression for ω1, but (1) we don’t yet know if the actual computation of that expression converges to that number, and on the other hand (2) we “secretly” do know that it converges, or we would not bother with it.
108 6. Continued Fractions is called the continued fraction expansion of ω1. The finite truncations pn qn = 1 a1 + 1 a2 +··· 1 an def ≡[a1,a2,··· ,an] .
are called the continued fraction convergents (or continued fraction approximants) of ω1. The coefficients ai are called the continued fraction coefficients .
Let us illustrate this definition with a few examples of continued frac-tion expansions: π −3 = [7,15,1,292,1,1,1,2,···] , e−2 = [1,2,1,1,4,1,1,6,1,1,8,···] , θ ≡ √ 2−1 = [2,2,2,2,2,···] , g ≡ √ 5−1 2 = [1,1,1,1,1,···] .
For example, π −3 the sequence of continued fraction convergents starts out as: 1 7, 15 106, 16 113, 4687 33102, 4703 33215, ···. The number g is also well-known. It is usually called the golden mean. Its continued fraction convergents are formed by the Fibonacci numbers defined in Definition 3.18 and given by {1,1,2,3,5,8,13,21,···}. Namely, the convergents are 1 1, 1 2, 2 3, 3 5, 5 8, and so forth.
We have defined continued fraction expansion only for numbers in ω in [0,1). This can be easily be remedied by adding a “zeroth” digit a0 — signifying the floor of ω — to it. Thus the expansion of π would then become [3;7,15,1,292,1,···]. We do not pursue this further.
Theorem 6.5. The continued fraction expansion of ω ∈[0,1) is finite if and only if ω is rational.
Proof. If ω is rational, then by Lemma 6.2 and Corollary 3.2, the algorithm ends. On the other hand, if the expansion is finite, namely [a1,a2,··· ,an], then, from equation (6.3), we see that ω is rational.
■ Theorem 6.6. For the continued fraction convergents, we have pn = anpn−1 + pn−2 qn = anqn−1 +qn−2 with q0 = 1 , p0 = 0 q−1 = 0 , p−1 = 1 , 6.2. Continued Fractions 109 or, in matrix notation, qn pn qn−1 pn−1 = An qn−1 pn−1 qn−2 pn−2 = An ···A2A1 , where Ai = ai 1 1 0 and A−1 i = 0 1 1 −ai .
Remark. We encountered Ai in Chapter 3 where it was called Qi+1. We changed the name so we have convenient subscript that agrees with the standard notation. Note that the variables qi are not the same as the qi of Chapter 3.
Proof. From Definition 6.4, we have that q1 = a1 and p1 = 1 and thus q1 p1 q0 p0 = a1 1 1 0 = A1 .
We proceed by induction. Suppose that the recursion holds for all n ≤k, then pk = akpk−1 + pk−2 qk = akqk−1 +qk−2 .
(6.4) The definition of the convergents gives: pk qk = 1 a1 + ··· 1 ak and pk+1 qk+1 = 1 a1 +··· 1 ak+ 1 ak+1 .
Thus pk+1 qk+1 is obtained from pk qk by replacing ak by ak + 1 ak+1 or pk+1 = ak + 1 ak+1 pk−1 + pk−2 qk+1 = ak + 1 ak+1 qk−1 +qk−2 .
Using equation (6.4) gives pk+1 = pk + 1 ak+1 pk−1 qk+1 = qk + 1 ak+1 qk−1 .
110 6. Continued Fractions The quotient pk+1 qk+1 does not change if if we multiply only the right-hand side of these equations by ak+1 to insure that both pk+1 and qk+1 are integers.
This gives the result.
■ Corollary 6.7. We have (i) qn+1pn −qnpn+1 = (−1)n+1 (ii) pn qn −pn+1 qn+1 = (−1)n+1 qnqn+1 .
Proof. The left-hand side of the expression in (i) equals the determinant of qn+1 pn+1 qn pn , which, by Theorem 6.6, must equal the determinant of An+1 ···A2A1. Finally, each Ai has determinant -1. To get the second equa-tion, divide the first by qn+1qn.
■ Corollary 6.8. We have (i) pn ≥2 n−1 2 and qn ≥2 n−1 2 (ii) gcd(pn,qn) = 1 .
Proof. i) Iterating the recursion in Theorem 6.6 twice, we conclude that pn+1 = (anan−1 +1)pn−1 +anpn−2 ≥2pn−1 + pn−2 , while p1 = 1 and p2 ≥2. The same holds for qn.
ii) By Corollary 6.7 (i) and B´ ezout.
■ Theorem 6.9. For irrational ω, the limit limn→∞ pn qn exists and equals ω.
In fact, |ω −pn/qn| < 1/(qnqn+1).
Remark 6.10. The way the convergents approxmate ω is illustrated in Fig-ure 26.
6.2. Continued Fractions 111 Proof. If we replace n by n−1 in the equality of Corollary 6.7(ii), we get another equality. Adding it to the original equality gives: pn−1 qn−1 −pn+1 qn+1 = (−1)n qn−1qn + (−1)n+1 qnqn+1 or pn+1 qn+1 −pn−1 qn−1 = (−1)n+1 qn 1 qn−1 − 1 qn+1 .
By Theorem 6.6, the qi are positive and strictly increasing, and so the right-hand side of the last equality is positive if n is even, and negative if n is odd.
Thus the sequence { pn qn }n even is increasing while the sequence { pn qn }n odd is decreasing.
In addition, by substituting 2k for n in Corollary 6.7(ii), we see that the decreasing sequence (n odd) is bounded from below by the increas-ing sequence, and vice versa. Since a bounded monotone sequence of real numbers has a limit3, the decreasing sequence has a limit ω−. Similarly, the increasing sequence must have a limit ω+. Now we use Corollary 6.7(ii) again to see that for all n, the difference between the two cannot exceed 1 qn−1qn . So ω+ = ω−= ω.
Clearly, ω lies between pn qn and pn+1 qn+1 , and so the final estimate of the theorem follows from Corollary 6.7(ii).
■ x p /q p /q 2n 2n 2n−2 2n−2 2n−1 2n−1 2n+1 2n+1 p /q p /q Figure 26. This figure illustrates how the convergents pn/qn approach their limit x.
Corollary 6.11. Suppose ω is irrational. For every n > 0, we have p2n q2n < ω < p2n+1 q2n+1 . If ω is rational, the same happens, until we obtain equality of ω and the last convergent.
3This is the monotone convergence theorem, see for example 112 6. Continued Fractions 6.3. Computing with Continued Fractions Suppose we have a positive real ω0 and want to know its continued fraction coefficients ai. By the remark just before Theorem 6.5, we start by setting a0 = ⌊ω0⌋ and ω1 = ω0 −a0 .
After that, we use Lemma 6.2, and get ai = 1 ωi and ωi+1 = 1 ωi −ai .
For example, we want to compute the ai for ω1 = 1+ √ 6 5 ≈0.6898979··· .
(6.5) If you do this numerically, bear in mind that to compute all the ai you need to know the number with infinite precision. This is akin to computing, say, the binary representation of ω1: if we want infinitely many binary digits, we need to know all its decimal digits. To circumvent this issue, we keep the exact form of ω1. This involves some careful manipulations with the square root. Here are the details. Since ω1 ∈(1/2,1), we have a1 = 1. Thus ω2 = 5 1+ √ 6 −1 = 4− √ 6 1+ √ 6 .
To get rid of the square root in the denominator, we multiply both sides by the “conjugate” 1− √ 6 of the denominator. Note that (1+ √ 6)(−1+ √ 6) gives −1+6 = 5. So we obtain ω2 = 4− √ 6 1+ √ 6 · −1+ √ 6 −1+ √ 6 = −2+ √ 6 ≈0.45 ∈ 1 3, 1 2 = ⇒ a2 = 2.
Subsequently, we repeat the same steps to get ω3 = 1 −2+ √ 6 −2 = ··· = −2+ √ 6 2 ≈0.225 ∈ 1 5, 1 4 = ⇒ a3 = 4.
This is beginning to look desperate, but rescue is on the way: ω4 = 2 −2+ √ 6 −4 = −2+ √ 6 = ω2 .
Now everything repeats, and thus we know the complete representation of ω1 in terms of its continued fraction coefficients: ω1 = 1+ √ 6 5 = [1,2,4,2,4,2,4···] = [1,2,4].
6.4. The Geometric Theory of Continued Fractions 113 The reverse problem is also interesting. Suppose we just know the con-tinued fraction coefficients {ai}∞ i=1 of ω1. We can compute the continued fraction convergents by using Theorem 6.6 qn pn qn−1 pn−1 = An ···A2A1 where Ai = ai 1 1 0 .
Theorem 6.9 assures us that the limit of the convergents { pn qn }∞ i=1 indeed equals ω1 = [a1,a2,···]. If also a0 > 0, add a0 to ω1 in order to obtain ω0.
So in our example ω1 = [1,2,4], this is easy enough to do: i : 0 1 2 3 4 5 ··· ai : -1 2 4 2 4 ··· pi : 0 1 2 9 20 89 ··· qi : 1 1 3 13 29 129 ··· But, because the ai are eventually periodic, we can also opt for a more explicit representation of ω1. The periodic tail can be easily analyzed. In-deed, let x = 1 2 + 1 4+ 1 2+··· = ⇒ x = 1 2 + 1 4+x After some manipulation, this simplifies to a quadratic equation for x with one root in [0,1).
x2 +4x−2 = 0 = ⇒ x = −2± √ 6.
Select the root in [0,1) as answer. Now we compute ω1 as follows.
ω1 = 1 1 + 1 2+ 1 4+··· = 1 1 + x = 1 −1+ √ 6 = 1+ √ 6 5 , which agrees with our earlier choice of ω1 in equation (6.5).
6.4. The Geometric Theory of Continued Fractions We now give a brief description of the geometric theory of continued frac-tions. This description allows us to prove one of the most remarkable char-acteristics of the continued fraction convergents (Theorem 6.14). Another geometric description can be found in exercise 6.11.
114 6. Continued Fractions y = w x e e e e −1 0 1 2 a e 1 0 a e 2 1 1/w Figure 27. Successive approximations to the line y = wx. In green, the vectors ei. In blue, aiei−1 for i = 1 and 2. Note that a1 = a2 = 2 in this figure. Note that a1 = ⌊1/w⌋.
The theory consists of constructing successive line segments that ap-proximate the line y = ω1x in the Cartesian plane. The construction is in-ductive. Here is the first step.
Start with e−1 = 0 1 and e0 = 1 0 .
(6.6) Although at first sight a little odd, it is the convention that e−1 is the basis vector along the y-axis and e0 the one along the x-axis. To get the first new approximation, define e1 = a1e0 + e−1 = a1 1 , (6.7) where we choose a1 to be the largest integer so that e1 and e−1 lie on the same side of y = ω1x (see Figure 27). From the figure, it is easy to see that in particular e1 = a1 1 and a1 = ⌊1/ω1⌋, the same as in Definition 6.3.
This way, we can construct en as segments whose slopes are pn/qn. In that sense, en cam be considered the approximants of ω1.
6.4. The Geometric Theory of Continued Fractions 115 There is another way of seeing this. Note that ω1 lies between the slopes of e0 and e1. Now define the two by two matrix A1 as the ma-trix corresponding to the coordinate change T1 such that T1(e−1) = e0 and T1(e0) = e1. Thus from equations (6.6) and (6.7), one concludes that the matrix A1 satisfies A1 x2 x1 = x1e0 +x2e1 and x2 x1 = A−1 1 (x1e0 +x2e1) .
The first equation implies that, indeed, A1 is the matrix we defined earlier (in Theorem 6.6). The second equation says that A−1 1 is the coordinate transform that gives the coordinates of a point in terms of the new basis e0 and e1. The new coordinates of the line x ω1x become A−1 1 x ω1x = ω1x x−a1ω1x = t 1 ω−1 1 −a1 , upon reparametrizing t = ω1x. Thus the slope of that line in the new co-ordinates, ω2, is the one given by equation (6.2). Since a1 was chosen the greatest integer so that the new slope is non-negative, we obtain that ω1 is contained in [0,1).
Since ω2 > 0, the construction now repeats itself, so that we get en+1 = an+1en + en−1 , as long as ωn > 0. By construction, ω1 always lies between pn qn and pn+1 qn+1 .
We obtain the same relations for the components of en as those in Theorem 6.6, and so en = qn pn .
116 6. Continued Fractions Consider the parallelogram p(en,en−1) spanned by en and en−1. The oriented area of p(en,en−1) is exactly the determinant of the matrix qn pn qn−1 pn−1 .
One now obtains Corollary 6.7 again4.
6.5. Closest Returns Consideration of the line ωx in the plane gives us another insight, see Figure 28. The successive intersections with the vertical unit edges are in fact the iterates of the rotation Rω : x →x + ω mod 1 on the circle starting with initial condition 0. A natural question that arises is: when do these iterates return close to their starting point?
Considering Theorem 6.9, we see that Rqn ω (0) = ωqn −pn is very small.
In this section, we prove something much better (Theorem 6.14), namely that these iterates are the closest returns.
Definition 6.12 (Closest Returns). Rq ω is a closest return if Rq ω(0) is closer to 0 (on the circle) than Rn ω(0) for any 0 < n < q.
p q Figure 28. The line y = ωx and (in red) successive iterates of the rota-tion Rω. Closest returns in this figure are q in {2,3,5,8}.
4Geometrically, the proof of that corollary is most easily expressed in the language of exterior or wedge products. These follow the rules of determinant computations. Thus the relevant induction step is the following computation.
en ∧en−1 = (anen−1 + en−2) ∧en−1 = −en−1 ∧en−2 .
6.5. Closest Returns 117 Lemma 6.13. Define dn ≡ωqn −pn. Then the sequence {dn} is alter-nating and its absolute value decreases monotonically. In fact, |dn+1| < 1 1+an+1 |dn−1|.
0 d d dn n−1 n+1 dn−1 dn−1 dn dn + +2 Figure 29. The geometry of successive closest returns. In this figure, an+1 = 3. Note that |dn−1| > 4dn+1.
Proof. The sequence {ω −pn qn } alternates in sign by construction (see Fig-ure 26). Therefore, so does {dn}. Recall that an+1 is the largest integer such that ωqn+1 −pn+1 = ω(an+1qn +qn−1)−(an+1pn + pn−1) = (ωqn−1 −pn−1)+an+1(ωqn −pn) , has the same sign as ωqn−1 −pn−1. This says that dn+1 = dn−1 +an+1dn .
Together with the fact that the dn alternate, this implies that dn is decreasing.
So from Figure 29, one concludes that (1+an+1)|dn+1| < |dn−1|.
■ Theorem 6.14 (The closest return property).
p q is a continued fraction convergent of ω if and only if q = 1 or |ωq −p| < |ωq′ −p′| for all q′ such that 0 < q′ < q and for all p′.
Proof. Note that if a1 > 1, then q = 1 is not a continued fraction denomi-nator, but it still (trivially) satisfies the above inequality.
We will first show by induction that the parallelogram p(en+1,en) spanned by en+1 and en contains no integer lattice points except on its four vertices.
Clearly, this is the case for p(e−1,e0). Suppose p(en,en−1) has the same property. The next parallelogram p(en+1,en) is contained in a union of an+1 +1 integer translates of the previous and careful inspection of Figure 30 shows that it inherits this property.
118 6. Continued Fractions n+1 q en−1 a e = a e + e n+1 n n+1 n−1 y = w x b c en Figure 30. Drawing of the shaded parallelogram P, the line y = ωx, and a few successive approximants. The green arrows correspond to en−1, en, and en+1. Here, an+1 is taken to be 3.
Next we show, again by induction, that the Rqn ω are closest returns, and that there are no others. It is trivial that Ra1 ω is a closest return, because by definition, a1ω < 1 and (a1 +1)ω1 > 1 (see (6.2)). Now suppose that up to q = qn the only closest returns are ei, i ≤n. We have to prove that the next closest return is en+1. By Lemma 6.13, dn+1 < dn. Now we only need to prove that there are no closest returns for q in {qn +1,qn +2,··· ,qn+1 −1}.
To that purpose we consider Figure 30. Consider the shaded parallel-ogram P bounded by the vertical lines x = 0, x = qn+1, and two lines with slope ω, one through en and one through en+1 −en. From the first paragraph it follows that the only lattice points in P are the origin and the endpoints of en, and en+1 −en, en+1. Since the segment c is parallel to and larger than en, we also have that b > a. Thus the distance of en+1 −en to ωx is greater than the distance of en and ωx. And so there is a band of width dn around y = ωx in P that contain no points in Z2 except the origin, en, and en+1.
■ Corollary 6.15. If pn qn is a continued fraction convergent of ω, then |ωqn −pn| ≤|ωq′ −p′| 6.6. Another Interpretation of the Convergents 119 for all q′ such that 0 < q′ < qn+1 and all p′, with equality only if q′ = qn and p′ = pn.
Proof. This follows from the statement of the previous theorem and the fact that there is no closest return for q′ (strictly) between qn and qn+1.
■ 6.6. Another Interpretation of the Convergents Given a number x1 ∈[0,1), we easily see that the first convergent 1/a1 maps to zero under the Gauss map T, that is: T(p1/q1) = 0. Furthermore, since x1 = 1 a1 + 1 a2 + 1 a3 +··· = 1 a1 +x2 , and x2 ∈[0,1), we can conclude that x1 lies in the domain ( 1 a1+1, 1 a1 ] of the a1-branch of T, see Figure 25. More precisely, if b1 : I1 →[0,1) is the branch of T such that x ∈I1, then the end point of I1 that maps to zero under T is the first convergent. It is this statement we wish to generalize.
0 0 1 1 1/2 1/3 ...
...
2/3 ...
...
Figure 31. A few branches of the twice iterated Gauss map T 2. The points of T −2({0}) (the complete inverse image of 0 under T 2) are marked in red. The reader should compare this plot to Figure 25.
To get an idea what iterates of T look like, let’s have a look at T 2 in Figure 31. T has a countable collection of branches with negative slope, each one mapping onto [0,1]. Thus T 2 has countably many branches (with positive slope) for every single branch b : I →[0,1] of T. In turn, each of the branches of T 2 also maps onto [0,1]. And so forth.
120 6. Continued Fractions Proposition 6.16. Given x ∈(0,1). Let bk : Ik →[0,1] be the branch of T k such that x ∈Ik, then the kth convergent pk/qk of x is the (unique) end point of Ik that maps to zero under T k.
Proof. From the expression given in Definition 6.4 for pn qn = [a1,a2,··· ,an], we see that T([a1,a2,··· ,an]) = [a2,··· ,an]. Continuing by induction, we find T n([a1,a2,··· ,an]) = T n−1([a2,··· ,an]) = ··· = T([an]) = 0.
So the nth convergent is indeed an nth pre-image of 0 under T.
Similarly, (6.2) implies that x = [a1,a2,···] = [a1,a2,··· ,an,an+1,···] = [a1,a2,··· ,(an +xn+1)].
Since xn+1 ∈[0,1), this is a single branch whose domain contains x.
■ By way of example, we look briefly at the golden mean g = [1,1,···] ≈ 0.61803··· in this context. The first convergent is 1/1 = 1. We immediately remark something perhaps a little unexpected: while this convergent is pre-image of 0 that belongs to the same branch as g, it is not that element of T −1(0) that is closest to 0 under T, The next convergent of g is 1/2. The same thing happens: again, the element of T −2(0) closest to g is in fact 2/3.
This characterization of convergents is in fact very familiar. Indeed in the usual decimal expansion, based on the map T : [0,1) →[0,1) given by T(x) = {10x}, the third convergent of the golden mean mentioned above is p3/q3 = [6,1,8], more commonly written as 0.618. Note that T 3(p3/q3) = 0 and that g lies in the domain of the 618 branch of T 3.
Another interesting observation is that the fact that all slopes of T are negative, means that the signs of the slopes of T k equal (−1)k. So, for odd k the convergents (the zeroes of the branches) are on the right of the interval of definition of the branch they belong to, and for the k even they are on the left side. This is convenient, because it implies that x is always ‘sandwiched’ between two successive convergents.
6.7. Exercises Exercise 6.1. Give the continued fraction expansion of 13 31, 21 34, 34 21, n−1 n for n > 1, n−1 n2 for n > 1 by following the steps in Section 6.3.
6.7. Exercises 121 Exercise 6.2. a) Find the continued fraction expansion of the fixed points (i.e. solutions of T(x) = x for T in Definition 6.1) of the Gauss map.
b) Use the continued fractions in (a) to find quadratic equations for the fixed points in (a).
c) Derive the same equations from T(x) = x.
d) Give the positive solutions of the quadratic equations in (b) and (c).
Exercise 6.3. Compute the continued fraction expansion for √n for n be-tween 1 and 15.
Exercise 6.4. Given the following continued fraction expansions, deduce a quadratic equation for x. (Hint: see Section 6.3.) a) x = = [8,8,8,8,8,···].
b) x = [3,6] = [3,6,6,6,6,···].
c) x = [1,2,3] = [1,2,3,1,2,3,···].
d) x = [4,5,1,2,3] = [4,5,1,2,3,1,2,3,···].
Exercise 6.5. In exercise 6.4: a) solve the quadratic equations (leaving roots intact).
b) give approximate decimal expressions for x.
c) give the first 4 continued fraction convergents.
Exercise 6.6. Derive a quadratic equation for the number with continued fraction expansion: [n], [m,n], [n,m], [a,b,n,m].
Exercise 6.7. From the expressions given in Section 6.2, use Theorem 6.6 to compute the first 6 convergents of π −3, e−2, θ, and g.
Exercise 6.8. a) In exercise 6.7, numerically check how close the nth con-vergent of ω is to the actual value of ω.
b) Compare your answer to (a) with the decimal expansion approximation using i digits.
c) In exercise 6.7, check that the increasing/decreasing patterns of the ap-proximants satisfies the one described in the proof of Theorem 6.9.
Exercise 6.9. What does the matrix in Theorem 6.6 correspond to in terms of the Euclidean algorithm of Chapter 3?
Exercise 6.10. Use Lemma 6.13 to show that ω −p2n+1 q2n+1 < 1 q2n+1 n ∏ i=1 1 1+a2i+1 .
122 6. Continued Fractions Figure 32. Black: thread from origin with golden mean slope; red: pulling the thread down from the origin; green: pulling the thread up from the origin.
Exercise 6.11. (Adapted from ) Consider the line ℓgiven by y = ωx with ω ∈(0,1) an irrational number. Visualize a thread lying on the line ℓ fastened at infinity on one end and at the origin at the other. An infinitely thin pin is placed at every lattice point in the positive quadrant. Since the slope of the thread is irrational, the thread touches none of the pins (except the one at the origin). Now remove the pin at the origin and pull the free end of the thread downward towards e0 (as defined in the text).
The thread will touch the pin at e0 and certain other pins with slopes less than ω. Mark the nth of those pins as f2n for n ∈N. Denote the points of the positive quadrant now lying on or below the thread by A. Next, pull the thread up towards e−1. Mark the pins the thread touches, starting with e−1 as f2n−1 for n ∈N. Denote the points of the positive quadrant now lying or or above the thread by B. See Figure 32.
a) Show that A∪B contain all the lattice points of the positive quadrant.
b) Show that for all n ∈N, fn = (qn, pn) where (qn, pn) are as defined in the text.
c) Compute the slopes of the upper boundary of the region A. The same for the lower boundary of the region B.
d) Show that A and B are convex sets.
p /q p /q n n n+1 n+1 x Figure 33. The placement of x between its convergents pn/qn and pn+1/qn+1.
6.7. Exercises 123 Exercise 6.12. Assume x is irrational.
a) Use Corollary 6.7(ii) and Corollary 6.11 to show that x−pn qn < 1 qnqn+1 .
b) Use Lemma 6.13 to show that x−pn+1 qn+1 < x−pn qn .
c) Use (a), (b), and Figure 33 to show that 1 2qnqn+1 < x−pn qn < 1 qnqn+1 .
(Hint: note that x is closer to pn+1 qn+1 than to pn qn .) Exercise 6.13. Use exercise 6.12 to generate bounds for the errors com-puted in exercise 6.8. Compare your answers.
Exercise 6.14. Use exercise 6.12 (a) to prove Theorem 1.15.
Exercise 6.15. a) Let x ∈[0,1) have periodic coefficients ai. Show that x satisfies x = ax+b cx+d where a, b, c, and d are integers. (Hint: see Section 6.3.) b) Show that x in (a) is an algebraic number of degree 2 (See Definition 1.17).
c) Show that if x ∈[0,1) has eventually periodic coefficients ai, then x is an algebraic number of degree 2.
This is one direction of the following Theorem.
Theorem 6.17. The continued fraction coefficients {ai} of a number x are eventually periodic if and only if x is an algebraic number of degree 2.
It is not known (in 2022) whether the continued fraction coefficients of algebraic numbers of degree 3 exhibit a recognizable pattern.
124 6. Continued Fractions Exercise 6.16. A natural question that arises is whether you can formulate continued fractions for polynomials in Q[x] or R[x] (suggested to us by ). We try this for the rational function f(x) = x3+x2 x7−x2+1. Referring to exercise 3.22 and the definition of ai in the remark after Definition 6.3, we see that a1 = (x4 −x3 +x2 −x+1) a2 = (−1 2x−1 2) a3 = (−4x+4) and a4 = (−1 2x−1 2) a) Compute the continued fraction convergents pn qn for n ∈{1,···4} of f(x).
(Hint: perform the computations as given in Theorem 6.6.) b) In (a), you obtained the polynomials of exercise 3.22 up to a factor -1. Why? (Hint: The gcd we computed in that exercise is actually -1. As stated in that exercise, we neglect constants when using the algorithm for polynomials. At any rate, in the quotient, the constant cancels.) c) Is there a theorem like the one in exercise 6.14?
d) Solve for y: y = [x]. (Hint: check exercise 6.6) e) Any ideas for other non-rational functions? (Hint: check the web for Pad´ e approximants.) Exercise 6.17. What is the mistake in the following reasoning?
We “prove” that countable = uncountable. First we show that a countably infinite product of countably infinite sets is countable.
n = ∏r i=1 pℓi i and there are infinitely many primes. Thus we can encode the natural numbers as infinite sequences (ℓ1,ℓ2,ℓ3,···) of natural numbers.
That gives a bijection of infinite product of N’s to N. Therefore an infinite product of N is countable.
On the other hand, an infinite number of natural numbers [q1,q2,···] can be used to give the real numbers in (0,1) in terms of their continued fraction expansion. This gives of bijection on to the interval. Therefore the infinite product of N is uncountable.
Exercise 6.18. Consider Figure 34.
The first plot contains the points {(n,n)}50 n=1 in standard polar coordinates, the first coordinate denoting the radius and the second, the angle with the positive x-axis in radians. The next plots are the same, but now for n ranging from 1 to 180, 330, and 2000, respectively.
a) Determine the first 4 continued fraction convergents of 2π.
b) Use (a) to explain why we appear to see 6, 19, 25, and 44 spiral arms.
c) Why does the curvature of the individual spiral arms appear to (a) alter-nate in sign and (b) decrease?
6.7. Exercises 125 Figure 34. Plots of the points (n,n) in polar coordinates, for n ranging from 1 to 50, 180, 330, and 3000, respectively.
Exercise 6.19. The exercise depends on exercise 6.18. Suppose we restrict the points plotted in that exercise to primes (in N) only. Consider the last plot (with 44 spiral arms) of Figure 34.
a) Show that each spiral arm corresponds to a residue class i modulo 44.
b) Show that if gcd(i,44) > 1, that arm contain no primes (except possibly i itself), see the left plot of Figure 35.
c) Use Theorem 6.18 below to show that the primes tend (as max p →∞) to be equally distributed over the co-prime arms.
d) Use Theorem 4.17 to determine the number of co-prime arms. Confirm this in the left plot of Figure 35.
e) Explain the new phenomenon occurring in the right plot of Figure 35.
126 6. Continued Fractions Figure 35. Plots of the prime points (p, p) (p prime) in polar coordi-nates with p ranging between 2 and 3000, and between 2 and 30000, respectively.
The following result will be proved in Chapter 13.
Theorem 6.18 (Prime Number Theorem for Arithmetic Progressions).
For given n, denote by r any of its reduced residues. Let π(x;n,r) stand for the number of primes p less than or equal to x such that Resn(p) = r. Then lim x→∞ π(x;n,r) π(x) = 1 ϕ(n) .
Exercise 6.20. a) Visualize the continued fraction expansion of another irrational number ρ ∈(0,1) by plotting a polar plot of the numbers (n, 2π ρ n) for various ranges of n as in exercise 6.18.
b) Visualize Theorem 6.18 as in exercise 6.19 (e).
Call p q is a closest approximation of ω if |ω −p q | < |ω −p′ q′ | for all q′ such that 0 < q′ < q and for all p′. In the following two exercises, we show that it is not true that p/q is a closest approximation if and only if it is a continued fraction. See Theorem 6.14.
6.7. Exercises 127 Exercise 6.21. Set ω = e−2 ≈0.71828.
a) Compute a1 through a4 numerically.
b) From (a), compute the convergents pi/qi for i ∈{1,2,3}.
c) Show that 1/2 (which is not a convergent) is a closest approximant. d) Show that 1/2 is not a closest return in the sense of Theorem 6.14.
Exercise 6.22. Set ω = = −1 2(−3+ √ 13) ≈0.302775638.
a) Check that ω = and that p0 q0 = 0 1 , p1 q1 = 1 3 , p2 q2 = 3 10 , p3 q3 = 10 33 , p4 q4 = 33 109 .
b) For a ∈{1,2} and i ∈{1,2,3}, (numerically) check which ones of the numbers api+1+pi aqi+1+qi approximate ω better than pi qi .
c) Use (b) to show that the continued fraction convergents are not the only closest approximants.
Exercise 6.23. Let x2 −bx−c = 0, where b and c are positive integers.
a) Show that x = √ c+bx gives the unique positive solution.
b) Use (a) to show that formally x = r c+b q c+b √ c+···.
c) Set f(x) := √ c+bx. Show that the sequence x1 = √c, x2 = p c+b√c, et cetera, corresponds to x1 = f(0), x2 = f(f(0)), and so on.
d) Show that these repeated images of 0 under f converge to the positive solution of (a) by studying figure 36.
Figure 36. Left, the functions f(x) = √2+3x and x. The fixed point x∗is the solution of f(x) = x is the value sought in exercise 6.23. Show that for positive x, x∗−f(x) x∗−x , and that that implies that xn →x∗.
128 6. Continued Fractions Exercise 6.24. Let x2 −bx−c = 0, where b and c are positive integers. See exercise 6.23.
a) Show that x = b+c/x for x > 0, gives the unique positive solution.
b) Use (a) to show that formally x = b+ c b+ c b+··· .
(Note: this is an example of what is known as a generalized continued fraction . In turn, these can be generalized to arbitrary (finite) dimension .) c) Set f(x) := b + c/x. Show that the sequence x1 = b, x2 = b + c/b, x3 = b+ c b+ c b et cetera, corresponds to x1 = b, x2 = f(b), x3 = f( f(b)), and so on.
d) Show that these repeated images of 0 under f converge to the positive solution of (a) by studying Figure 37. The functions f(x) = 1 + 3/x (red), f( f(x) (green), and x (black).
Since f( f(x)) is increasing slowly, the sequence {0, f(0), f(f(0)),···} must converge to the fixed point.
Chapter 7 Fields, Rings, and Ideals Overview. The characteristics of Z are so familiar to us, that it is hard to break through that familiarity to understand what makes things like unique factorization tick. Algebraic number theory and with it large swaths of algebra were developed to deal with more general number systems in or-der to overcome this problem. So in this chapter, we initially move away from numbers a little to study concepts of abstract algebra. This disci-pline of mathematics seems to start with a daunting barrage of definitions or nomenclature1. Here, we look at some of these and relate them as much as possible to their origins in number theory. An excellent introduction to abstract algebra is , while is a standard among the more advanced texts.
7.1. Rings of Polynomials Since one of our aims is to study factorization properties in certain sets of algebraic integers — which are defined through polynomials — we need to start by studying sets of polynomials. Broadly speaking, there are two important cases. The coefficients of the polynomials belong either to a ring such as Z or — an important special case — they belong to a field such as Q. In what follows we denote a ring by R and a field by F.
1From Latin nomen or ‘name’ and calare or ‘to call’. So — taken quite literally — name-calling.
129 130 7. Fields, Rings, and Ideals Definition 7.1. A ring R[x] of polynomials is the set of polynomials with co-efficients in a (commutative) ring R without zero divisors2 (unless otherwise mentioned).
Without the extra requirements, the resulting ring would have very strange properties indeed. For example, if R consists of the integers modulo 6, then, indeed, very strange factorizations can happen: (2x−3)(3x+2) =6 6x2 −5x−6 =6 x.
So, in particular, the degree of the product is not equal to the sum of the de-grees of the factors. Dropping commutativity would lead to another strange problem. Given f ∈R[x], we may want to evaluate f at c ∈R by substituting the value c for x. Suppose for example that R is the non-commutative ring of 2 by 2 matrices. Set for some a ∈R, f(x) = (x−a)(x+a) = x2 −a2 .
But if we substitute another 2 by 2 matrix c for x such that the matrices a and c do not commute, then the above equality does not hold anymore.
However, if R satisfies the two requirements, one can prove that the result-ing polynomial ring has no zero divisors, evaluations are safe, and that the degree of a product is additive (see [sections 8.5 and 8.6] for details).
Definition 7.2. Recall (Definition 1.18) that f is minimal polynomial in R[x] for ρ if f is a non-zero polynomial in R[x] of minimal degree such that f(ρ) = 0. A polynomial f in R[x] of positive degree is irreducible over R if it cannot be written as a product of two polynomials in R[x] with positive degree. A polynomial f in R[x] is prime over R if if whenever f divides gh (g and h in R[x]), it must divide g or h.
Definition 7.3. Let f and g in R[x]. The greatest common divisor of f and g, or gcd(f,g), is a polynomial in R[x] with maximal degree that is a factor of both f and g. The least common multiple of f and g, or lcm( f,g), is a polynomial in R[x] with minimal degree that has both f and g as factors.
Remark 7.4. If p is minimal for ρ, it must be irreducible, because if not, one of its factors with smaller degree would also have ρ as a root.
2This means that if for a, b in R, we have that ab = 0, then a = 0 or b = 0, see Definition 8.4.
7.1. Rings of Polynomials 131 It turns out that in the special case where the coefficients of the poly-nomials are taken from a field F, the result is a ring F[x] that is very rem-iniscent of the trusty old ring Z. The underlying reason for this similarity is that in F[x], the division algorithm works (see exercise 7.1): given r1 and r2, then there are q2 and r3 such that3 r1 = r2q2 +r3 such that deg(r3) < deg(r1).
Recall that the gcd of two polynomials in F[x] can be computed by factor-ing both polynomials and multiplying together the common factors to the lowest power as in the proof of Corollary 2.23. Since factoring polynomials is hard, it is often easier to just use the Euclidean algorithm. An example is given in exercise 3.22. The relation between lcm and gcd of two polynomi-als is the same as in the proof of Corollary 2.23. The minimal polynomials of F[x] are “like” the primes in Z. We will see later that this implies unique factorization, and that primes and irreducibles are the same4. We give a few properties that will be immediately useful5 Proposition 7.5. Given ρ ∈C and p ∈F[x] so that p(ρ) = 0 and p′ ̸= 0.
i) p is minimal for ρ if and only if p is irreducible.
ii) If p is minimal, it has no repeated roots.
Proof. If p is minimal, see Remark 7.4. On the other hand, if f is irre-ducible and p is minimal for ρ, then the division algorithm tells us that there are polynomials q and r such that f = pq+r, where r has degree strictly less than p. Since p(ρ) = f(ρ) = 0, we have r(ρ) = 0. But since p is minimal, we must have r(x) = 0. Thus p | f. But f is irreducible, so q must be a constant and f is also minimal. This proves (i).
To prove (ii), suppose that p has a repeated root α. Since p ∈F[x], we have that also p′ (its derivative) in F[x]. But if p(x) = (x−α)2r(x) then p′(x) = 2(x−α)r(x)+(x−α)2r′(x).
3Since remark 3.17, we adopt the convention that the degree of a non-zero constant equals 0, while the degree of 0 equals −∞.
4In fact, the fact that the division algorithm works, makes this ring a Euclidean domain (Definition 8.11).
5But in Corollary 8.13 we will get much more: irreducibles equal primes and unique factorization.
132 7. Fields, Rings, and Ideals The latter is non-zero and of lower degree and still has a root α. This contradicts the minimality of p.
■ An even simpler argument gives the following result.
Lemma 7.6. Given ρ ∈C and p minimal for ρ in F[x]. If f ∈F[x] has a root ρ, then p | f.
Proof. We use again the division algorithm to establish that f = pq+r, where r has degree less than p. Since f(ρ) = p(ρ) = 0, also r(ρ) must be zero, contradicting the minimality of p unless r(x) = 0. The lemma follows.
■ Theorem 7.7. Given a(x) and b(x) in F[x], there are g and h in F[x] satis-fying a(x)g(x)+b(x)h(x) = c(x) if and only if c is a multiple of gcd(a,b).
Proof. We paraphrase the proof of Lemma 2.5 with “degree” replacing “ab-solute value”. Let S and ν(S) be the sets: S = {a(x)g(x)+b(x)h(x) : a(x)g(x)+b(x)h(x) ̸= 0} ν(S) = {deg(s) : s ∈S} ⊆N∪{−∞,0} .
Again ν(S) is non-empty, and so by well-ordering, it must have a smallest element, say δ, the degree of a polynomial d(x). If δ = 0, then d(x) is a constant γ ∈F. After dividing by γ, we see that gcd(a,b) = 1 since no common factor can have degree less than 0.
If δ > 0, we use the division algorithm exactly as in the proof of Lemma 2.5 and conclude that d(x) is a divisor (or factor) of both a(x) and b(x).
Suppose e is a factor of both a and b. Since d(x) = a(x)g(x)+b(x)h(x), we see that e must also be a factor of d. And thus d is the greatest common divisor.
The proof is finished by repeating the last paragraph of the proof of Lemma 2.5 to show that a(x)g(x) + b(x)h(x) = c(x) has a solution if and only if c is a multiple of d.
■ 7.1. Rings of Polynomials 133 Next, we present a result that holds for more general rings of the form R[x] (though not for all). For simplicity, however, we give the result for Z[x].
It says that if we can factor a polynomial in Z[x] as a product of polynomials with rational coefficients, then, in fact, those coefficients are integers.
Lemma 7.8 (Gauss’ Lemma). Let Aℓ∈Z, and bi,cj ∈Q. If m+n ∑ ℓ=0 Aℓxℓ= m ∑ i=0 bixi ! n ∑ j=0 cjx j !
, then bi,cj ∈Z.
Proof. Let A := gcd({Aℓ}) and set aℓ= Aℓ/A. In addition, we fix integers B and C such that Bbi and Cc j are integers and gcd({Bbi}) = gcd({Cc j}) = 1.
We then get m+n ∑ ℓ=0 ABCaℓxℓ= m ∑ i=0 Bbixi ! n ∑ j=0 Ccjx j !
.
We now show that ABC = ±1 and so all three are ±1. Given any prime p in Z, let r be the minimum of the index i such that p ∤Bbi, and s the minimum of the index j such that p ∤Ccj. From the way the coefficient ABCar+s is computed, see Figure 38, it immediately follows that p ∤ABCar+s. Since we can find such r and s for every prime p, the result follows.
■ 0 1 2 r−1 r 0 1 s−1 s m n r+s i j Figure 38. ABCar+s is the sum of the BbiCcj along the green line in the i −j diagram. The red lines indicate where p ∤Bbi and p ∤Cc j. So all contributions except BbrCcs are divisible by p. Thus p ∤ABCar+s.
A similar argument gives another useful result.
134 7. Fields, Rings, and Ideals Lemma 7.9 (Eisenstein’s Criterion). Let a(x) = ∑d i=0 aixi be a polynomial in Z[x] of degree d ≥2. If there is a prime p such that p | ai if i < d and p ∤ad and p2 ∤a0 , then a is irreducible over Q.
Proof. Suppose a is reducible over Q, then by Lemma 7.8, it is reducible over Z. So let a(x) = b(x)c(x) where b has degree m > 0 and c has degree n > 0 and d = m+n. Since ad = bmcn and a0 = b0c0, the requirements on the ai give, without loss of generality, p ∤bm and p ∤cn and p ∤b0 .
There is a smallest integer s ≤n < d such that p ∤cs. We compute as: as = min{s,m} ∑ i=0 bics−i .
All terms on both sides are divisible by p except b0cs, which is impossible.
■ We end this section with a note on some notation that can be confusing.
We can “adjoin” x to a ring R in two ways. If we use square brackets [·], we take R[x] to be the minimal (smallest) ring that contains both R and x. On the other hand, parentheses (·) are used to indicate the minimal (smallest) field that contains both R and x. On the other hand, A little reflection leads to the following definition.
R[x] := {f(x) : f is a polynomial over R} , R(x) := f(x) g(x) : f,g are polynomials over R .
(7.1) Here, x can be a place holder or an actual number. In the former case, R(x) denotes the rational functions in x, and R[x] are the polynomials.
The ring of power series (not just polynomials of finite degree) is in-dicated by R . For a field F, the field of quotients or fractions F is written as F((x)). This field consists of the quotients of power series. Con-sider f(x) = ∑∞ i=0 aixi and g(x) = ∑∞ i=0 bixi. Then if b0 ̸= 0, the quotient f/g can be formally reduced to a power series: f(x) g(x) := c0 +c1x+c2x2 +··· = a0 b0 + a1 b0 −a0b1 b2 0 x+··· .
(7.2) 7.2. Ideals 135 If b0 = 0 and b1 ̸= 0, then employing the same method (exercise 7.2) we get f(x) g(x) = 1 x a0 b1 + a1 b1 −a0b2 b2 1 x+··· .
(7.3) Continuing this way, we see that the F((x)) is the set of formal Laurent series (which is how it is usually defined): F((x)) = ( ∞ ∑ i=n cixi : n ∈Z and ci ∈F ) .
For a ring R, the notation R((x)) is best avoided because it is ambiguous: in this case the field of quotients is not the same as the set of Laurent series over R.
7.2. Ideals Definition 7.10. A non-empty subset I of a ring R is called an ideal6 if i) For all i and j in I, i± j is in I (closed under addition and negatives).
ii) For all x in R and i in I, xi and7 ix are in I (it “absorbs” products).
The smallest ideal containing the elements i and j will be indicated8 by ⟨i, j⟩.
A principal ideal is an ideal that is generated by a single element, that is: it is of the form Ri. An ideal I is a maximal ideal if there is no other ideal L so that I ⊊L ⊊R.
To guide our considerations, we look at Z first. In Z it is clear that for any j ∈Z, the corresponding ideal ⟨j⟩is given by the set jZ of integer multiples of j. The relation 3 | 15 can now be replaced by ⟨3⟩⊇⟨15⟩.
Addition of ideals is defined as in the following example ⟨6⟩+⟨15⟩:= {n+m : n ∈⟨6⟩, m ∈⟨15⟩} = ⟨gcd(6,15)⟩.
Notice that the last equality is not trivial. It in fact encodes B´ ezout’s lemma (Lemma 2.5). In turn, this says that ⟨6⟩+⟨15⟩is the smallest ideal contain-ing both ⟨6⟩and ⟨15⟩. We also say that ⟨6⟩+⟨15⟩is the ideal generated by 6Usually “fraktur” letters (a, b, c ...) are used for ideals. On a blackboard or whiteboard, these are hard to distinguish from normal letters. So instead we will use capital letters to indicate ideals.
7One of the two is sufficient if R is commutative.
8In most texts parentheses (·) are used. We want to avoid ambiguity with the notation for an n tuple (i, j,···).
136 7. Fields, Rings, and Ideals 6 and 15. This is more conveniently written as ⟨6,15⟩. More generally, for ideals A and B, we have that A+B := {a+b : a ∈A, b ∈B} .
(7.4) This example also illustrates the fact that ⟨6⟩+⟨15⟩is a principal ideal.
In fact, in Z, it is easy to see that every ideal I is principal. One can use B´ zout to show that I is generated by its least positive element. Another non-trivial example of a principal ideal is the set of polynomials q satisfying q(ρ) = 0 in the ring of polynomials over a field F. Indeed, we need to refer to Lemma 7.6 to establish that this is the case (work out the details in exercise 7.8).
Next, we look at multiplication of ideals . If ideals are to behave like numbers, then the product of two ideals should also be an ideal. At first glance, one would think the collection of products of one element in ⟨6⟩and one in ⟨15⟩would do the trick. This is indeed the case in Z (exercise 7.4).
However, in general this construct is not closed under addition (exercise 7.5). Thus we define AB as the smallest ideal containing the products of one element in A and one in B, or AB := ( k ∑ i=1 aibi : ai ∈A, bi ∈B, k ∈N ) .
(7.5) The relation between ring and ideal is very similar to the one between group and normal subgroup (Definition 7.30). In fact, since a ring R is an Abelian group with respect to addition, any ideal in R is a normal subgroup.
There is one interesting difference: a normal subgroup is also a group. In contrast an ideal (like the even numbers) may not have a multiplicative iden-tity and so it is not a ring (see Remark 5.24). The remainder of this section spells out the relation between rings and their ideals.
Definition 7.11. Given two rings I and J, a ring homomorphism is a map f : I →J that preserves addition and multiplication and their respective identities 0 and 1. The kernel of a ring homomorphism is the pre-image of the additive identity 0. A ring isomorphism is a ring homomorphism that is a bijection. The word “‘ring” is often omitted.
Proposition 7.12. i) The quotient R/K of a ring R by an ideal K is a ring.
ii) The kernel K of a ring homomorphism f : R →H is an ideal.
7.2. Ideals 137 Proof. K is an ideal and thus a normal subgroup of the Abelian additive group R. Thus R/K is a group under addition (exercise 7.6). We have to show that multiplication is well-defined and is associative, distributive, and has an identity (Definition 5.20).
Multiplication in R/K is well-defined if for all a, a′, b, and b′ in R such that a−a′ and b−b′ are in K, we have (a+K)(b+K)−(a′ +K)(b′ +K) ⊆K .
The left hand side can be expanded as ab−a′b′ +(a−a′)K +K(b−b′)+K ·K = (a−a′)b+a′(b−b′)+(a−a′)K +K(b−b′)+K ·K .
The absorption property of the product does the rest.
Associativity and distributivity now follow easily. For example, since [ab]c = a[bc] in R and multiplication is well-defined, we must have (a+K)(b+K) = (a+K)[(b+K)(c+K)] .
Similarly for distributivity. Again, by absorption, (a+K)(1+K) ⊆(a+K) and so 1+K is the multiplicative identity. This proves (i).
The proof of (ii) is rather trivial. Just use Definitions 7.10 and 7.11.
Choose x and y in the kernel of f and conclude that f(x ± y) = 0 and that for any r ∈G, f(rx) also equals 0.
■ Theorem 7.13 (Fundamental Homomorphism Theorem). If f : R →H is a surjective ring homomorphism with kernel K, then H is (ring) isomorphic to R/K.
Proof. Define the map ϕ : R/K →H by ϕ(K +x) := f(x).
We need to prove that (a) ϕ is a bijection, that (b) it preserves addition and that (c) it preserves multiplication.
To prove (a), note that ϕ is onto because f is. So next suppose that ϕ(K+x) = ϕ(K+y). Because f preserves addition, we get f(x−y) = 0 and therefore x−y ∈K. Injectivity follows: because K +x = (K +(x−y))+y and K +K = K, we get K +x = K +y.
138 7. Fields, Rings, and Ideals The proofs of (b) and (c) are almost identical. We prove only (c).
ϕ(K +x)ϕ(K +y) = f(x) f(y) = f(xy) = ϕ(K +xy).
But by the absorbing property of ideals, (K +x)(K +y) = K +xy.
■ The idea that quotients of certain structures are isomorphic to structures they map onto, is important not only in algebra (groups, modules) but also in topology and analysis (quotient spaces). For instance, R/Z with the right topology is homeomorphic to the standard (unit) circle. See Figure 39 Figure 39. Intuitively we wrap R around a circle of length 1, so that points that differ by an integer land on the same point.
Theorem 7.13 has the surprising consequence, for example, that there are no non-trivial (ring) homomorphisms C →R (see exercise 7.7).
Corollary 7.14. A ring homomorphism f : F →R where F is a field is either trivial (zero) or injective.
Proof. If f is not injective, it has a non-trivial kernel, which by Theorem 7.13, is an ideal I in the field F. So I contains a a non-zero element i. Now pick any x ∈F. Then by Definition 7.10 (ii), xi−1 ·i = x is in I. Thus I = F, and hence f(F) = 0.
■ In many common cases, the conclusion if the fundamental homomor-phism theorem is intuitively obvious. For example, we did not need it to prove that Z/5Z is isomorphic to Z5. However, in Theorem 7.16 below the conclusion is not self-evident and we make essential use of it.
7.3. Fields and Extensions As a first example, let us consider the field Q and adjoin the number π (or any other transcendental number). We denote the smallest field containing both by Q(π). The pair of fields (Q(π),Q) in this example is called a field 7.3. Fields and Extensions 139 extension. Q(π) is the extension field of Q. By equation (7.1), it consists of all quotients of polynomials. Since π is transcendental, there exists no polynomial p with rational coefficients so that p(π) = 0. Thus none of these expressions simplify. Therefore this set is isomorphic to Q(x). An extension of this nature is also called a transcendental extension .
In order to get something both new and manageable, we should adjoin a number α to the field Q that requires us to take only finitely many powers of α into account. This is done by taking α to be an algebraic number. Such as extension is called finite or algebraic .
A simple example tells the whole story. Let us take α = √ 2 and study Q( √ 2). Clearly, α2+i = 2αi, so any polynomial over Q in α can be rewrit-ten as a + b √ 2 with a and b in Q. Any quotient of polynomials in α can therefore be written as a+b √ 2 c+d √ 2 = (a+b √ 2)(c−d √ 2) (c−d √ 2)(c−d √ 2) = ac−2bd c2 −2d2 + (bc−ad) √ 2 c2 −2d2 .
Since 2 is square free, the denominator is not zero and hence every element in the field Q( √ 2) can be written as e+ f √ 2 with e and f in Q. This holds is general as the next result shows.
Proposition 7.15. Let F(ρ) a finite extension of a field F. Suppose p is a minimal polynomial for ρ and has degree d. Then, as sets, F(ρ) = ( d−1 ∑ i=0 aiρi : ai ∈F ) .
Proof. Clearly, {1,ρ,··· ,ρd−1} are independent over F (otherwise the min-imal polynomial would have degree less than d) and since a field is closed under addition, subtraction, and multiplication, and so F(ρ) must contain all expressions ∑d−1 i=0 aiρi.
We only need to check that F(ρ) is closed under (multiplicative) inver-sion. So choose bi ∈F such that f(x) := ∑d−1 i=0 bixi is not 0. The minimal polynomial p for ρ is irreducible (Proposition 7.5); it can have only trivial factors in common with f. Thus by Theorem 7.7, there are polynomials s and t such that f(x)s(x)+ p(x)t(x) = 1.
140 7. Fields, Rings, and Ideals Using the minimal polynomial, s can be reduced to have degree less than d.
Substitute ρ for x in this equation to obtain (since p(ρ) = 0 and f(ρ) ̸= 0) s(ρ) = 1/f(ρ).
Thus F(ρ) is indeed closed under (multiplicative) inversion.
■ All we are doing in this last proof, really, is taking an arbitrary quo-tient f/g of polynomials f and g in ρ and reducing it using the minimal polynomial. That insight leads to a sharper result.
Theorem 7.16. Let F(ρ) a finite extension of a field F. Suppose p is a minimal polynomial for ρ. Then F(ρ) is ring isomorphic to F[x]/⟨p(x)⟩.
Proof. Define a map σρ : F[x] →F(ρ) as follows. Given a polynomial f, σρ(f) = f(ρ).
Clearly, σρ is a ring homomorphism, because σρ(f ·g) = σρ( f)σρ(g) and σρ( f +g) = σρ( f)+σρ(g).
Since σρ( d−1 ∑ i=0 aixi) = d−1 ∑ i=0 aiρi , Proposition 7.15 shows that σρ is onto. By Proposition 7.12, the kernel of σρ is an ideal, and by Lemma 7.6 it is the ideal ⟨p(x)⟩generated by p(x).
Thus by the fundamental homomorphism theorem, F(ρ) is isomorphic to F[x]/⟨p(x)⟩.
■ Remark 7.17. The map σρ is called an evaluation map .
This is all very well, but what if we adjoin another algebraic element, β, to Q(α)? What does Q(α,β) look like? Are the results we just proved still useful? The answer, miraculously, is yes. And the reason is the primitive element theorem below (Theorem 7.19).
Let us look at an example again. Adjoin β = √ 3 to the previous exam-ple Q(α) = Q( √ 2), and consider Q(α,β). Since the squares of α and β are integers, it is clear that every element of Q( √ 2, √ 3) can be written as a+b √ 2+c √ 3+d √ 6 where a,b,c,d ∈Q.
7.3. Fields and Extensions 141 What is not immediately obvious is that 1, √ 2, √ 3, and √ 6 are linearly independent over the rationals, but let us assume that for now (see Lemma 7.33 in the exercises).
Remark 7.18. We obtain a 4 dimensional vector space over Q with a basis formed by the vectors {1, √ 2, √ 3, √ 6}.
Now we make the “inspired guess9” that in this example Q(α +β) is iden-tical to Q(α,β)! To verify that, set γ = √ 2+ √ 3. Clearly, γ ∈Q(α,β) and so Q(γ) ⊆Q(α,β).
A simple computation indeed yields γ2 = 5+2 √ 6 , γ3 = 11 √ 2+9 √ 3 , γ4 = 49+20 √ 6.
(7.6) And so γ3−9γ generates √ 2, γ3−11γ generates √ 3, while γ2−5 generates √ 6. Thus Q(α,β) ⊆Q(γ).
We have established that Q(γ) = Q(α,β). That we can do this in general, is the content of the primitive element theorem.
Theorem 7.19 (Primitive Element Theorem). Every finite extension of an infinite field is a simple extension.
Remark 7.20. Any single element that generates the finite extension is called a primitive element.
Proof. Let F be an infinite field and K := F(α,β,γ,··· ,δ) a finite (alge-braic) extension. Suppose we can find a single generator ϕ for α and β. We can then repeat the argument to find a generator θ for ϕ and γ, and so forth.
Thus it is sufficient to prove this result for F(α,β).
Let p and q be minimal polynomials in F[x] for α and β, respectively.
The roots of p are {αi}m i=1 with α1 ≡α and those of q are {βi}n i=1 with β1 ≡β. Now define for c ̸= 0 in F r(x) := p(α +cβ −cx).
9“Inspired guess” is pretentious way of saying that I do not want to say where I got this (but see the proof of Theorem 7.19).
142 7. Fields, Rings, and Ideals This polynomial has several intriguing properties. First, it is a member of the field F(α + cβ)[x], for it has coefficients in F(α + cβ). Furthermore, its roots are given by α1 +cβ −cx = αi ⇐ ⇒ xi = α1 −αi c +β1 .
For i = 1, we of course get β = β1 as a root. But now, since F is infinite, we fix a value of c∗of c such that none of the other roots equals βi for i > 1.
Since both q ∈F[x] ⊆F(α + c∗β)[x] and r ∈F(α + c∗β)[x] and both have β as a root, Lemma 7.6 implies that the minimal polynomial d for β in F(α +c∗β)[x] must be a divisor of both q and r. But these two share only one root, and therefore d ∈F(α +c∗β)[x] has degree one: s(x) = a1x+a0 = a1(x−β).
Clearly, the ai are in F(α +c∗β), but then so does β = a0/a1, and the same holds for α = (α +c∗β)−c∗β. Thus α +c∗β generates F(α,β).
■ Thus a primitive element generates the whole field extension through ad-dition and multiplication (and their inverses). In contrast, a primitive root (Definition 5.5) is an element of Fp (the elements of Zp with addition and multiplication as operations) whose powers generate Fp.
As mentioned in our last example, Q(γ) is in fact a vector space over Q. From (7.6), it is clear that γ4 −10γ2 + 1 = 0. Therefore Q(γ) has four basis vectors, like Q(α,β), namely {1,γ,γ2,γ3} span the space Q(γ). The scalars are elements of Q. As such, it is somewhat confusingly denoted by Q(γ)/Q in the literature, though this is not to be interpreted as a quotient.
The dimension of the vector space is denoted by [Q(γ) : Q] and is also commonly called the degree of the extension. Notice that [Q(α,β) : Q] = [Q(α,β) : Q(α)]·[Q(α) : Q].
This holds much more generally (see or ).
7.4. The Algebraic Integers We look at the ring of all algebraic integers and show that it unsuitable for the study factorization into primes or irreducibles for it has neither primes nor irreducibles.
7.4. The Algebraic Integers 143 Theorem 7.21. The set A of algebraic integers forms a ring with no zero divisors.
We can take advantage of the fact that algebraic integers are complex numbers, which in turn form a commutative field (and thus a ring) without zero divisors. Many of the properties mentioned in Definition 5.20 as well as the absence of zero divisors are thus automatically satisfied. To make a long story short, we only need to prove that A is closed under additive in-version, under addition, and under multiplication. The first is easy. Suppose that θ ∈A is a root of xd +ad−1xd−1+···+a0, where the ai are in Z. Then, of course, −θ is a root of the same polynomial with the odd ai replaced by −ai. The remaining two criteria have a very interesting constructive proof.
To understand it, we need to define the Kronecker product.
Definition 7.22. Given two matrices A and B, their Kronecker product A⊗ B is given by A⊗B := A11B A12B A13B ··· A21B A22B ··· ··· .
.
.
.
.
.
.
Lemma 7.23. Suppose that A and B be square matrices of dimension a and b, respectively and denote by Ia and Ib the identity matrices of the appropri-ate dimension. If A has eigenpair10 (α,x) and B, (β,y). Then i) A⊗B has eigenpair (αβ,x⊗y), and ii) A⊗Ib +Ia ⊗B has eigenpair (α +β,x⊗y) .
Proof. We have that A11B A12B A13B ··· A21b A22B ··· ··· .
.
.
.
.
.
x1y x2y .
.
.
= A11x1By+A12x2By+··· A21x1By+A22x2By+··· .
.
.
, which equals Ax⊗By or αx⊗βy. Using Definition 7.22 again, it is easy to check that this in turn equals αβx⊗y. This proves item (i).
10This means that Ax = αx.
144 7. Fields, Rings, and Ideals By (i), A⊗Ib has eigenpair (α,x⊗y), and Ia ⊗B has eigenpair (β,x⊗ y). Adding the two gives item (ii).
■ As an example, consider the matrices A = 0 2 1 0 and B = 0 3 1 0 , with eigenvalues ± √ 3 and ± √ 2. We obtain: A⊗B = 0 0 0 6 0 0 2 0 0 3 0 0 1 0 0 0 and A⊗Ib +Ia ⊗B = 0 3 2 0 1 0 0 2 1 0 0 3 0 1 1 0 , with eigenvalues ± √ 6 (of multiplicity 2) and ± √ 3± √ 2, respectively. The characteristic polynomials are (x2−6)2 and x4−10x2+1, respectively. The polynomial was obtained earlier from equation (7.6).
Proof of Theorem 7.21. We only need to prove that A is closed under ad-dition and under multiplication. So let α and β be in A . Then α is a root of a monic11 polynomial pA(x) of degree a and the same for β and pB(x) of degree b. Suppose pA(x) = xa +∑a−1 i=0 aixi. The so-called companion matrix , that is: the a×a matrix whose characteristic polynomial equals pA, is A = 0 0 0 ··· −a0 1 0 0 ··· −a1 0 1 0 ··· −a2 .
.
.
.
.
.
0 ··· 0 1 −aa−1 , and similarly a matrix B can be defined for pB. Now form the matrices mentioned in Lemma 7.23 (i) and (ii). Then αβ and α + β are roots of these polynomials or of factors of these polynomials. Since a factor (over 11A monic polynomial is a polynomial whose leading term has coefficient 1.
7.5. Rings of Quadratic Numbers and Modules 145 Z) of a monic polynomial is monic, we see that both αβ and α + β are algebraic integers.
■ While, like Z, the algebraic integers A form a ring, that ring does not “look” like Z at all! We will take this up later when we prove that A is dense in the complex numbers and has no irreducibles and no primes (Theorem 8.6). So to study factorization, we must look at more restricted collections of algebraic integers.
Examples of more restricted rings of integers are Z(γ), the ring con-sisting of numbers of the form ∑d−1 i=0 ciγi with ci ∈Z, where γ is algebraic of degree d. To see that Z(γ) is a ring is trivial, since we do not have to worry about multiplicative inverses, which was the only complication in Proposition 7.15.
We end this section with a slightly confusing definition and a warning in the form of a Lemma.
Definition 7.24. Consider the field Q(γ). The integers of Q(γ) are those elements in Q(γ) that are algebraic integers.
This is not necessarily the same as the set Z(γ)! As an example we will prove the lemma below in exercise 7.20.
Lemma 7.25. Let j be square free. The integers of Q(√j) are precisely the elements of the ring Z( 1 2 (1+√j)) if j =4 1, and Z(√j) else.
7.5. Rings of Quadratic Numbers and Modules Let j be a non-zero square free integer j ∈Z (see exercise 2.16) not equal to 0 or 1. Then √j is a algebraic integer of degree 2. If j is negative, we can think of Z[√j] and Q[√j] as subsets of the complex plane. If j is positive, then they are subset of the real line. In both cases Z[√j] and Q[√j] are countable (see Theorem 1.26). All elements of Z[√j] are algebraic integers of degree 2, because they are roots of (x−a−b p j)(x−a+b p j) = x2 −2ax+a2 −b2 j = 0, (7.7) and that degree 2 polynomial cannot be factored over the integers.
146 7. Fields, Rings, and Ideals We can look at Z[√j] as having two basis vectors 1 0 = 1 and 0 1 = p j .
The elements of Z[√j] are precisely the linear combinations a · 1 + b · √j.
Just like a the vector space of remark 7.18! The only difference is that the “scalars” now belong to a ring and not a field. The resulting construction is called a module.
Definition 7.26. A module M (or left module ) is a set with the same struc-ture as a finite-dimensional vector space, except that its scalars form a com-mutative ring R (and not a field as in a vector space). Scalars multiply the elements of M from the left. (If in a non-Abelian ring, scalars multiply from the right, the result is called a right module.) Next, we interpret multiplication by α = a + b√j in Z[√j] when √j is an algebraic integer of degree 2. Clearly, it is linear, because α(c+d p j) = cα1+dα p j .
Therefore, α can be seen as a matrix. Identify 1 with 1 0 and √j with 0 1 . Then the equations (a+b√j)1 = a+b√j and (a+b√j)√j = bj+ a√j can be rewritten as A 1 0 = a b and A 0 1 = bj a .
Thus we can use elementary linear algebra to see that A = a bj b a .
(7.8) What is interesting here, is that the determinant of A detA = a2 −b2 j (7.9) 7.5. Rings of Quadratic Numbers and Modules 147 is clearly an integer and cannot be zero, because if jb2 −a2 = 0, because j is square free. The beauty of this is that this allows us to study factorization in complicated rings like Z[√j] using the tools of a simpler ring, namely Z. All we have to do is to phrase factorization in Z[√j] in terms of the determinant of A. In number theory, this is known as the norm of α.
Definition 7.27. The field norm, or simply norm12, of an element α of Z[√j] or Q[√j] is the determinant of the matrix that represents multiplica-tion by α. It will be denoted by N(α). The trace of that matrix will be call the trace of α and is denoted by T(α).
A fundamental result about determinants from linear algebra (detAB = detAdetB) gives a handy rule.
Corollary 7.28. The norm of a ring of quadratic integers is a completely multiplicative function: N(αβ) = N(α)N(β). (See Definition 4.2.) Remark 7.29. Suppose α = a + b√j in Z[√j]. From equation (7.9), we also get N(α) = αα where α = a −b√j. α is called the conjugate of α.
Note that if j is negative, the conjugate α corresponds to the usual complex conjugate of α and so the norm N(α) corresponds to the usual absolute value squared |α|2.
All this can be seamlessly generalized to Z[β] where β is some alge-braic number of degree d > 2. We then get a d-dimensional module.
12This is another case of assigning a name that gives rise to confusion: the “norm” as defined here can be negative! Nonetheless, this seems to be the most common name for this notion, and so we’ll adhere to it.
148 7. Fields, Rings, and Ideals 7.6. Exercises Exercise 7.1. The reader might want to review exercises 3.22 to 3.25 first.
Let f and g in F[x]. We will show that there are polynomials q and r in F[x] such that f = gq+r and deg(r) < deg(g).
(7.10) a) Show that this holds if deg(g) > deg( f).
b) Now let n = deg( f) ≥deg(g) = m and f(x) = ∑n i=0 aixi and g(x) = ∑m i=0 bixi. Define f j(x) = f(x)−an bm xn−mg(x), where f j has degree j. Show that j ≤n−1. (Hint: by assumption, an and bm are not zero.) c) Show that the computation in (b) can be repeated with f replaced by f j as long as j ≥m. (Hint: we are just formalizing long division here.) d) Show that r(x) = fi(x), where fi is the first of the f j to have degree less than m.
e) Show that the leading term of q(x) in (7.10) is an bm xn−m.
Exercise 7.2. We perform long division to divide f(x) = a0 +a1x+a2x2 + ··· by g(x) = b0 + b1x + b2x2 + ···. In contrast to exercise 7.1, now con-sider the constant term as the leading term, the next leading term is the one linear in x, and so on.
a) Assume b0 ̸= 0, then f −a0 b0 g cancels the constant term. So the first term of the quotient equals a0 b0 . Find the next two terms. (Hint: see equation 7.2.) b) Assume b0 = 0 and b1 ̸= 0. Divide f by xg(x) = b1 + b2x + ··· using the method in (a). Find the first three terms of the quotient. (Hint: see equation 7.3.) Exercise 7.3. We prove that Z(i) = Q[i].
a) Show that Z[i] is the set {a+bi : a,b ∈Z}. (Hint: i2 ∈Z.) b) Show that Z(i) equals {(a+bi)/(c+di) : a,b,c,d ∈Z}.
c) From (b), rewrite Z(i) as {r +si : r,s ∈Q}.
d) Show that R[i] = C.
Exercise 7.4. a) Given two ideals ⟨a⟩and ⟨b⟩in Z. Show that ⟨a⟩·⟨b⟩= ( k ∑ i=1 nimiab : ni, mi ∈Z, k ∈N ) .
b) Use (a) to prove that in Z ⟨a⟩·⟨b⟩= ⟨ab⟩.
7.6. Exercises 149 Exercise 7.5. Consider the ideals I = ⟨2,x⟩and J = ⟨3,x⟩in Z[x].
a) Show that I = g1(x)2+g2(x)x and J = h1(x)3+h2(x)x, where gi and hi are arbitrary elements of Z[x].
b) Show that 3x and −2x can be written as (g1(x)2 + g2(x)x)(h1(x)3 + h2(x)x).
c) Use (b) to show that x must be in the ideal IJ.
d) Show that x cannot be written as (g1(x)2+g2(x)x)(h1(x)3+h2(x)x).
e) Use (d) to show that IJ is not equal to the “naive” definition of the product of ideals, {ab : a ∈I,b ∈J}.
Definition 7.30. Let G be a group and N a group contained in G. Then N is a normal subgroup of G if for every n ∈N and every x ∈G, also x−1nx ∈N.
In other words, if n ∈N, then every conjugate x−1nx of n is also in N.
Exercise 7.6. a) Show that for a not necessarily Abelian additive group G with a subgroup I, we have (a+I)+(b+I) = (a+b−b+I)+(b+I) = (a+b)+(−b+I +b)+I .
b) Show that (a) implies that addition of cosets is well-defined if I is nor-mal.
c) Let I be a normal subgroup of a group R, then R/I is a group. Where in the proof do you need normality. (Hint: check the items in Definition 5.20 (1).) d) Let h : R →H be a homomorphism of groups. Show that kerh is a nor-mal subgroup. (Hint: write h(x−1nx) = h(x−1)h(n)h(x). What is h(n)?) We have proved the fundamental homomorphism for groups. This is really a slightly weaker version of Theorem 7.13. So we will record it here as a corollary.
Corollary 7.31 (Fundamental Homomorphism Theorem for Groups).
If f : R →H is a surjective group homomorphism with kernel K, then H is group isomorphic to R/K and K is a normal subgroup.
Exercise 7.7. Show that there is no non-trivial (ring) homomorphism C → R. (Hint: use Corollary 7.14 to show that the kernel of f is {0}. Use i2 = −1 to see f(i) is undefined.) Exercise 7.8. Let ρ be an algebraic number with minimal polynomial p.
a) Show that the set of polynomials q in Q[x] such that q(ρ) = 0 form an ideal. (Hint: use only Definition 7.10.) b) Show that this is a principal ideal. (Hint: Lemma 7.6.) 150 7. Fields, Rings, and Ideals Exercise 7.9. a) Solve the polynomial γ4 −10γ2 +1 = 0 using the standard quadratic formula and then taking a square root again. Show that γ = ± q 5±2 √ 6.
b) Show that the root with the two ‘+’ signs equals √ 2+ √ 3 .
Exercise 7.10. a) Show that −1 2 + i 2 √ 3 is an algebraic integer. (Hint: compute (x+ 1 2 −i 2 √ 3)(x+ 1 2 + i 2 √ 3).) b) Use a computation similar to (a) to show that −1 2 + 1 2 √ 3 satisfies x2 + x−1 2 = 0.
c) Show that (b) implies that −1 2 + 1 2 √ 3 is not a algebraic integer. (Hint: what if that number also satisfied x2 +bx+c = 0 with b and c in Z?) Exercise 7.11. Consider primes p and q (in Z). Use Lemma 7.23 to find minimal polynomials for √p√q and √p+√q.
Exercise 7.12. Let ρ be algebraic integer with minimal polynomial p(x) = xd +∑d−1 i=0 cixi (ci ∈Z).
a) Use Lemma 7.23 to show that for all a and b in Z, a + bρ is also an algebraic integer of degree at most d. (Hint: Let C be the companion matrix for the minimal polynomial for ρ; the lemma leads to considering the characteristic polynomial of aI +bC.) b) Show that q(a+bρ) = 0 if q is the polynomial given by q(x) = (x−a)d + d−1 ∑ i=0 cibd−i(x−a)i .
c) Show that if b ̸= 0, then if q(x) can be factored over the integers by f(x)g(x), then p(x) can be factored by b−d f(bx+a)g(bx+a).
d) Conclude that q is the minimal polynomial for a+bρ (b ̸= 0).
Theorem 7.21 and the next two exercises imply the following. Theorem 8.6 provides more information.
Proposition 7.32. The set A forms a integral domain but not a field and A is dense in C.
7.6. Exercises 151 Exercise 7.13. a) Show that the algebraic numbers are closed under mul-tiplicative inversion. (Hint: let d be the degree of the polynomial p and consider the polynomial q(x) := xd p(x−1).) b) Show that if the degree d polynomial p ∈Z[x] is irreducible, then so is q(x) := xd p(x−1).
(Hint: q(x) = f(x)g(x) implies p(x−1) = f(x−1)g(x−1).) c) Use (b) to prove the following. An algebraic integer α is a unit (is in-vertible) if and only if α has minimal polynomial p(x) = xd + ∑d−1 i=0 aixi with a0 = ±1. (Hint: in a minimal polynomial, a0 cannot be zero.) d) Conclude that the algebraic integers do not form a field.
Exercise 7.14. a) For any real α > 1, and any n ∈N, we can choose k = ⌊αn⌋. Show that k 1 n ≤α < (k +1) 1 n .
b) Use (a) to show that (k +1) 1 n −k 1 n < α 2 1 n −1 .
c) Show that the algebraic integers are dense in {x ∈R : x ≥1}. (Hint: k1/n is an algebraic integer.) d) Extend the conclusion in (c) to all of R by using exercise 7.12 (a).
e) Use (d) and Lemma 7.23 to prove that A is dense in C.
Exercise 7.15. a) Use the method of Section 7.3 to find the minimal poly-nomial in Z[x] for √ 2 + √ 3 + √ 5. (Hint: x8 −40x6 + 352x4 −960x2 + 576.) Exercise 7.16. Suppose ρ ∈A is not a unit and has minimal (monic) poly-nomial p in Z[x].
a) Show that q(x) = p(x2) has root √ρ.
b) Show that any factor in Z[x] of q is monic.
c) Show that √ρ is not a unit. (Hint: if it is, then its square must be too.) d) Conclude that ρ is not irreducible.
152 7. Fields, Rings, and Ideals Exercise 7.17. We apply the Euclidean algorithm in Z[√−1] to 17 + 15i and 7 + 5i. Compare with the computations in Section 3.2 and exercise 3.22.
a) Check all computations in the following diagram.
| + | − | + | − | | 2+2i | −2−i | 3 | 0 | 0 | −1+i | −4 | 7+5i | 17+15i | | 1 | | | | | | 2+i | 1 | | | | | −6−3i | −2−i | b) Check all computations in the following diagram.
| + | − | + | − | + | | 1+2i | 1−i | 1−i | 2 | 0 | 0 | 1+i | −1+3i | 3+5i | 7+5i | 17+15i | | 1 | | | | | | | −1+i | 1 | | | | | | −2i | −1+i | | | | | | −2+4i | 1−2i | c) From the diagram in (a), compute values for x and y in Z[√−1] such that −1+i = (7+5i)x+(17+15i)y .
(Hint: follow instructions in Section 3.2.) d) From the diagram in (b), compute values for x and y in Z[√−1] such that 1+i = (7+5i)x+(17+15i)y .
e) Compute gcd(17+15i,7+5i) (up to invertible elements).
f) Compute lcm(17 + 15i,7 + 5i) (up to invertible elements). (Hint: see Corollary 2.16.) Exercise 7.18. Find a greatest common divisor and a least common multi-ple for each of the following pairs of Gaussian integers. (Hint: see exercise 7.17.) a) 7+5i and 3−5i.
b) 8+38i and 9+59i.
c) −9+19i and 52+68i.
7.6. Exercises 153 Exercise 7.19. a) Show that the arithmetic functions (Definition 4.1) with the operations addition and Dirichlet convolution (Definition 4.19 form a commutative ring. (Hint: see exercise 4.16).
b) Show that the same does not hold for the multiplicative (Definition 4.1) arithmetic functions. (Hint: see exercise 4.17).
c) Show that the functions f : R →R together with the operations addition and multiplication form a commutative ring.
d) Is the ring in (c) a domain?
e) Show that the square integrable functions f : [0,∞) →[0,∞) together with the operations addition and convolution form almost a commutative ring. (Hint: only the multiplicative identity is missing.) f) Look up Titchmarsh’s convolution theorem and show that it implies that the ring in (e) (with the “Dirac delta function” added) is a domain.
Exercise 7.20. a) Show that all elements of Q[√j], j ∈Z, are algebraic numbers. (Hint: see equation (7.7).) b) Now let j be square free and show that if a+b√j is an integer of Q[√j], then 2a ∈Z and a2 −b2 j ∈Z.
c) Show (b) implies that if a ∈Z, then b ∈Z. (Hint: set b = p q where gcd(p,q) = 1.) d) Show that (b) implies that if a ∈Z+ 1 2, then 4b2 j ∈4Z+1.
e) Show that in (d) we obtain that b ∈Z+ 1 2 and j =4 1. (Hint: set b = p q where gcd(p,q) = 1 and conclude that q = 2. Then show that p2 j = 2n+1 implies that j =4 1.) f) Use (c) and (e) to show that if j =4 1, the integers of Q[√j] are given by I = n a+b p j : a,b ∈Z o ∪ a+ 1 2 + b+ 1 2 p j : a,b ∈Z .
g) Use (f) to prove Lemma 7.25.
In the remainder of these exercises, we use the following notation. We let {m1,··· ,mn} denote a collection of distinct, square free (non-zero) integers in Z. In the interest of brevity of notation, we write ε1q1 + ··· + εnqn as (ε ·q)(n) in exercise 7.21 and ε1a1√m1 +···+εnan√mn as (ε ·a√m)(n) in exercise 7.22. (If m < 0, take the root in the upper half plane of C.) We also set Sn = {−1,+1}n, n > 0. Our ultimate aim is proving Proposition 7.34.
Along the way, though, we will find some interesting gems. Our approach is inspired by ; more general results can be found in .
154 7. Fields, Rings, and Ideals Exercise 7.21. Define the following polynomial of degree 2n in x with parameters q1 through qn : P n(x) := ∏ ε∈Sn x+(ε ·q)(n) , a) Show that for n > 1, P n(x) is equal to P n(x) = ∑ ℓ, j1,···, jn κℓ, j1,···,jn xℓqj1 1 ···qjn n where the κ’s are integers.
b) Show from the definition of P n that replacing qi by −qi leaves P n invari-ant.
c) Show that (b) implies that the powers ji are even.
Exercise 7.22. a) Define the polynomial of degree 2n P n(x) := ∏ ε∈Sn x+ ε ·a√m (n) , and show that for n ≥1, P n(x) satisfies P n(x) = P n−1(x+an √mn)·P n−1(x−an √mn).
b) Use exercise 7.21 (c) to show that P n ∈Z[x].
c) Use that P n−1 ∈Z[x] to show that P n−1 (x±an √mn) = n−1 ∑ i=0 bi(x±an √mn)i = ±√mn On−1(x)+En−1(x), where On−1 and En−1 are in Z[x].
d) Show that En−1(x) is not the zero polynomial. (Hint: x2n.). Then show that if On−1(x) is zero, we obtain a contradiction as follows.
P n−1 (x+an √mn) = P n−1 (x−an √mn) implies that the roots of those two polynomials are equal. But [ ε∈Sn−1 n ε ·a√m (n−1) +an √mn o = [ ε∈Sn−1 n ε ·a√m (n−1) −an √mn o .
The following lemma (proved in Exercise 7.23) is interesting in its own right, but, as we will see, it also has important applications.
Lemma 7.33. Let {m1,··· ,mℓ} be a collection of distinct, square free in-tegers in Z. Then for ai ∈Z, ∑ℓ i=1 ai√mi = 0 if and only if ai = 0 for all i.
7.6. Exercises 155 Exercise 7.23. This exercise relies heavily on exercise 7.22. Given ℓdis-tinct square free integers in Z, {m1,m2,··· ,mℓ}. Assume the lemma is false. Thus let r > 1 be the smallest integer such that upon re-ordering the mi, the first r terms are linearly dependent over Z: r ∑ i=1 ai √mi = 0 and ∀i ai ̸= 0.
(7.11) Define the polynomials Pj, O j, and Ej as in exercise 7.22.
a) Show that r > 2. (Hint: this is trivial if one of the mi is negative and one is positive; if they have the same sign, square the relation a1 √m1 = −a2 √m2.) b) Show that (7.11) implies that Or−1(0) = Er−1(0) = 0. (Hint: show that 0 is a root of P r, so one of its factors P r−1(x±a√mr) has a root 0.) c) Show that (b) implies that there is an ε ∈Sr−1 so that ε ·a√m (r−1) +ar √mr = 0 and ε ·a√m (r−1) −ar √mr = 0.
d) Show that this proves Lemma 7.33. (Hint: adding the equalities in (c) contradicts minimality of r.) The next exercise is another one of the promised ‘gems’. We find the mini-mal polynomial for ∑n i=1 √pi where the pi are distinct primes in N.
156 7. Fields, Rings, and Ideals Exercise 7.24. Let {p1,··· , pn} be a collection of positive primes in Z, and for ε ∈Sn, set γε := (ε ·√p)(n) Furthermore, denote by {mi}2n i=1 the 2n distincta values of ∏n i=1 pδi i for δi ∈{0,1} including 1 = ∏p0.
a) Show that γℓ ε = 2n ∑ i=1 bℓ,iηi √mi .
where bℓ,i ∈Z and ηi = ±1 is determined exclusively by ε and mi. (Hint:if mi = p2p5p9, say, then ηi = ε2ε5ε9.) b) Let P = ∑k i=1 aixi in Z[x]. Show that P(γε) = 2n ∑ i=1 cℓ,iηi √mi .
where bℓ,i ∈Z and ηi as in (a).
c) Use Lemma 7.33 to show that (b) implies: ∃ε ∈Sn such that P(γε) = 0 = ⇒∀ε ∈Sn such that P(γε) = 0.
d) Define the polynomial of degree 2n P n(x) := ∏ ε∈Sn x+(ε ·√p)(n) .
Show that P n defined in (b) is the minimal polynomial for any of the γε.
(Hint: by (c) the minimal polynomial must have degree at least 2n, and by exercise 7.22 (c), P n ∈Z[x].) aHere we use positivity of the primes, because, for instance, (−2)·(−3) = 2·3. However, if the collection of primes contains no pair {−p, p}, this difficulty goes away.
Now we come to the main result of these exercises.
Proposition 7.34. Let {p1,··· , pn} denote any collection of distinct primes in N.
i) Let γn := γε=1 = ∑n i=1 √pi. Then Q(√p1,√p2,··· ,√pn) = Q(γn) ii) Denote Fn = Q(√p1,√p2,··· ,√pn). The degree [Fn : Fn−1] equals 2 and so [Fn : Q] = 2n.
iii) Items (i) and (ii) also hold for {ε1p1,··· ,εnpn} where ε ∈Sn.
The first part of this proposition actually says that γn is a primitive element (see Theorem 7.19) for Fn. Part (ii) says that the fields Fn form an infinite tower of fields (see definition 7.35), each having degree 2 with respect to the previous field.
7.6. Exercises 157 Definition 7.35. A tower of fields is a (finite or infinite) sequence of succes-sive field extensions F1 ⊆F2 ⊆··· ⊆Fn ⊆···.
Exercise 7.25. a) Show that [Fn : Q] ≤2n. (Hint: see preamble of exercise 7.24.) b) Show that [Fn : Q] = 2n. (Hint: use Lemma 7.33.) The next exercise proves part (i) of Proposition 7.34. Note that by definition of γn, Q(γn) ⊆Fn. So all we need to show is the reverse inclusion.
Exercise 7.26. Define P n as in exercise 7.24 (d) and γn as in item (i) of the proposition.
a) Proceed as in exercise 7.22, but with ai√mi replaced by √pi, to show that P n−1 (x±√pn) = En−1(x)±√pn On−1(x), where On−1 and En−1 are non-zero polynomials in Z[x].
b) Use (a) and exercise 7.24 (d) to show that On−1(γn) ̸= 0.
c) Show that P n−1(γn −√pn) = 0. (Hint: from the definition of P n.) d) Show that En−1(γn)±√pn On−1(γn) = 0, and thus Q(γn) contains √pn.
f) Use that fact that the order of the primes is arbitrary to show that Q(γn) contains √pi for any i ∈{1,··· ,n}.
Finally, we carefully checking the proofs and making sure that adjoining negative primes does not cause any problems. See for example the footnote to exercise 7.24. Note that in this case, we generally get a tower in C, not in R.
Exercise 7.27. Check the proofs in exercises 7.22 through 7.26 to make sure that adjoining negative pi does not cause problems.
Chapter 8 Factorization in Rings Overview. We now get back to factorization. It is instructive to go back to the discussion of the proof of unique factorization in Z (Section 2.3) at this point. Our familiarity with Z may hide underlying structures from us. To circumvent this familiarity, we study factorization in rings. Perhaps unex-pectedly, at this level of generality, pretty much anything can happen, as we show in the first section below. We then add various ingredients to rings in an effort to end up with an abstract structure that guarantees unique factor-ization. Unless mentioned otherwise, we restrict to commutative rings.
8.1. So, How Bad Does It Get?
Recall that even in Z, we have unique factorization up to factor -1 (see remark 2.12). So the best we can reasonably hope for in a general ring is to have unique factorization up to multiplication by units (Definition 5.25) and up to re-ordering. In this section, we dash those hopes. Let us start by revising our basic notions to this more general context.
Definition 8.1. Given a ring R and an element r that is not zero or a unit.
Then r is reducible if it is a product of two non-units (or non-invertible elements). If r is not equal to a product of two non-invertible elements it is called irreducible (or not reducible). If whenever r | ab, then r | a or r | b (or both), then it is called a prime.
159 160 8. Factorization in Rings The important observation here is that the two characteristics of primes in Z that mentioned in remark 2.13 have been separated, because they do not coincide in general rings: irreducibles and primes become two different things1.
Next we must realize that in general rings, we cannot necessarily order divisors according to their absolute value as we do in Z (see Definition 1.2).
Instead, in the new definition we order divisors according to the partial order given by the division relation.
Definition 8.2. Let R be a integral domain and α and β non-zero el-ements. A greatest common divisor g = gcd(α,β) is a common divisor of both α and β such that for any common divisor γ we have γ | g. A least common multiple ℓ= lcm(α,β) is a common multiple of both α and β such that for any common multiple γ we have ℓ| γ.
So given a general ring, pick an arbitrary element, what different iden-tities can it have? Well, it can be irreducible, reducible, a unit, or 0. These categories are mutually exclusive. In addition, every non-zero, non-unit el-ement can also be prime or non-prime. But the primes and irreducibles are not necessarily the same. The next result gives a sample of the truly bizarre behaviors of factorizations in general (commutative) rings.
Proposition 8.3. i) In a ring that is also a field, there are no primes or irreducibles.
ii) The set of algebraic integers A form a proper ring (i.e. not a field) that has no irreducibles and no primes.
iii) In the ring Z6, the element 2 is prime, but not irreducible.
iv) In the ring Z[ √ −5], the element 3 is irreducible, but not prime.
v) In Z[√−3], the gcd of 4 and 2+2√−3 does not exist.
vi) In Z6[x], 2x(1+3h(x))n divides 4x2 for any polynomial h and any n ≥0.
Proof. (i) Recall that in a field, every non-zero element is a unit, and so there are no primes or irreducibles.
(ii) Pick any non-zero, non-unit ρ ∈A . According to exercise 7.13 (c), ρ has minimal polynomial p(x) = d ∑ i=0 aixi with ad = 1 and a0 ̸= −1,0,+1.
1And the meaning of “prime” has changed to confuse non-algebraists. But we’re not falling for it!
8.2. Integral Domains 161 Set a = b = √ρ. Then a is a root of q(x) = p(x2) by exercise 7.13 (b).
Now q ∈Z[x] is monic and any polynomial factor of q must also be monic.
Therefore is in A . Since ρ = ab, ρ is reducible. Clearly, we also have ρ | ab. But if ρ divides a (or b), then a/ρ is in A . Since A is closed under multiplication, its square, which equals ρ−1 would then also be in A . This contradicts our initial choice of ρ. Hence ρ cannot divide a or b, and so ρ is not prime.
(iii) Suppose 2 | ab in Z6. Then in Z, 2 divides ab+6m. But that means that ab is even and thus a (or b) has a factor 2. But then in Z6, 2 divides a (or b). Therefore 2 is prime in Z6. On the other hand, 2·4 =6 2. Since both 2 and 4 are non-invertible, 2 is reducible.
(iv) Suppose the number 3 equals the product xy, where x and y in Z[ √ −5]. Clearly, x and y cannot both be real, because 3 is prime and irre-ducible in Z. If both are non-real, then b ̸= 0 and each has absolute value at least √ 5, and |xy| ≥5, a contradiction. If one of them is non-real, then so is their product, another contradiction. Therefore, one of x or y must be a unit. This proves that 3 is irreducible in Z[ √ −5]. But on the other hand, (2+i √ 5)(2−i √ 5) = 9 = ⇒ 3 | (2+i √ 5)(2−i √ 5) .
But since (2±i √ 5) 3 ̸∈Z[ √ −5], 3 does not divide either of these factors.
(v) Since 4 = 2·2 = (1+ √ −3)(1− √ −3), both 2 and (1 + √−3) are divisors of 4. They are also divisors of (2 + 2√−3). however, it is a simple check to see that 2 and (1 + √−3) do not divide each other. In other words, there is no mximal common divisor in this case.
(vi) Using the binomial theorem, we see that modulo 6 2x·2x·(1+3h(x))n =6 4x2 n ∑ i=0 n i 3ih(x)i =6 4x2 , because 4·3i =6 0 for i > 0.
■ 8.2. Integral Domains In order to “tame” factorizations, the first thing to do is to require the ab-sence of zero divisors.
162 8. Factorization in Rings Definition 8.4. An integral domain or domain is a commutative ring R with no zero divisors (i.e. if a ̸= 0 and b ̸= 0, then ab ̸= 0).
Thus, in an integral domain, if we have ab = 0, then we can conclude that either a = 0 or b = 0 or both. This applies to the situation where we have a(x−y) = 0. If a ̸= 0, we must have x = y. This immediately implies (see Theorem 2.7) the following.
Theorem 8.5 (Cancellation Theorem). In an integral domain, if a ̸= 0, then ax = ay if and only if x = y. (See also Theorem 2.7.) Polynomials whose coefficients form an integral domain are themselves an integral domain (see Section 3.7 and Definition 7.1). Other examples are the fields Fp of the integers modulo a prime p. In this context, Lagrange’s theorem (Theorem8.32) is interesting: it says that an degree n polynomial over a field has at most n roots. So, x2 +5x+6 =11 0 = ⇒ (x+2)(x+3) =11 0 .
And this implies that x =11 −2 =11 9 or x =11 −3 =11 8. If we work modulo 12, factoring does not solve the problem. For example, x2 +5x+6 modulo 12 has roots {1,6,9,10}.
A (non-Abelian) ring that does have zero divisors are the 2 by 2 matri-ces with coefficients in Z. In fact, if N = 0 1 0 0 , then N2 = 0.
Note that C does not have zero divisors. Therefore, the same holds for any subset of C, such as the set A of all algebraic integers. Propositions 7.32 and 8.3 (ii) imply the following remarkable facts.
Theorem 8.6. The set A forms a integral domain but not a field and i) A is dense in C, ii) A has no irreducibles, iii) A has no primes.
It might seem that we have not done much to tame the factorization process. However, the following result indicates that we on the right track.
Theorem 8.7. Any prime p in an integral domain R is irreducible.
Proof. Suppose that the prime p satisfies p = ab. We need to show that a or b is a unit. Certainly p ̸= 0 divides ab, and so, from Definition 8.1, p | a or p | b. Assume the former. So there is a c such that pc = abc = a. We then 8.2. Integral Domains 163 get The cancellation theorem gives, of course, that bc = 1, and so b has an inverse and therefore is a unit. Similar if we assume p | b.
■ Like any ring in C, Z[ √ −5] is an integral domain. So Proposition 8.3 (iv) shows that the converse is false. However, here is an interesting lemma that implies (once again) that Fp is a field (see Proposition 5.18). The proof is essentially the same as that of Lemma 5.3.
Lemma 8.8. A finite integral domain is a field.
Proof. Fix some a ̸= 0 in the integral domain R. Consider the (finitely many) elements {ax}x∈R. Either all these elements are all distinct, or two are the same. But if ax = ay, the cancellation theorem gives a contradiction.
If they are all distinct, then there is an x such that ax = 1 ∈R. Thus a has a multiplicative inverse.
■ Theorem 8.9. Let R be an integral domain in which every element has a factorization into irreducibles. Every irreducible is a prime if and only if factorization into irreducibles is always unique.
Proof. First, suppose that every irreducible is a prime and assume that the following are two factorizations of x ∈R into irreducibles.
x = up1 ··· pk = u′q1 ···qℓ.
Now if p1 is a prime, upon relabeling the qi, it must divide q1. Since q1 is irreducible, we must have p1 = q1 up to units. Doing finitely many steps, one proves that the factorization is unique.
Next, suppose that q is irreducible and that there are non-zero a and b such that q | ab. This implies qc = ab. We factor both sides of this last equation into irreducibles.
uq(p1 ··· pk) = u′(q1 ···qℓ)(qℓ+1 ···qm).
By unique factorization, q must equal to q1 (upon relabeling and up to units) and thus it divides a or b.
■ Definition 8.10. An integral domain R is a unique factorization domain2 if every element admits a unique factorization into irreducibles. This is often abbreviated to UFD.
2The word “domain” serves as a reminder that R must be an integral domain.
164 8. Factorization in Rings By Theorems 8.7 and 8.9, in a UFD, “prime” and “irreducible” are syn-onymous. In a UFD, the notions of greatest common divisor and least com-mon multiple are well-defined. The reason these notions are well-defined can be found in the proof of Corollary 2.16. To repeat that argument, sup-pose that α = u s ∏ i=1 pki i and β = u′ s ∏ i=1 pℓi i , where u and u′ are units and ki and ℓi in N∪{0}. Now define: mi = min(ki,ℓi) and Mi = max(ki,ℓi) .
Then, of course, we have gcd(α,β) = s ∏ i=1 pmi i and lcm(α,β) = s ∏ i=1 pMi i .
The pi are unique up to a unit. And so are the gcd and lcm, since the product of units is a unit.
We still need to be slightly cautious. For instance, in Z[i], which is a UFD, the units are ±1 and ±i. The gcd of 2i and −4 is 2 up to units, that is: ±2 or ±2i.
8.3. Euclidean Domains The next step in the taming process, is to make sure there is a division algorithm.
Definition 8.11. A Euclidean function on a ring R is a function E : R{0} → N∪{0} that satisfies: i) For all ρ1 and ρ2 in R, there are κ and ρ3 in R such that ρ1 = κρ2 + ρ3 and E(ρ3) < E(ρ2) and ii) For all α and γ in R{0}, we have E(αγ) ≥E(α) .
A Euclidean ring or Euclidean domain is an integral domain R for which there is a Euclidean function.
In a Euclidean domain, we can perform the division algorithm of Lemma 2.23. All statements and proofs in Chapter 2 from B´ ezout’s Lemma (Lemma 2.5) on, up to and including Corollary 2.16, hold with minor modifications.
For example, we need to use E(n) instead of the norm of n. Theorem 2.7 3The name “Euclidean domain” derives from the alternative name of that algorithm, see remark 2.4.
8.3. Euclidean Domains 165 needs the reformulation given by Theorem 8.5. Corollary 2.8 would need to be reformulated (which we omit). Among other things, the unique fac-torization, and the Euclidean algorithm of Chapter 3, which in turn led us to continued fractions, follow from these. So the consequences of having a Euclidean function are indeed staggering! Exercise 8.12 investigates the relation between the two chapters.
In Euclidean domains the notions of prime and irreducible are again happily reunited.
Proposition 8.12. Let R be a Euclidean domain. If p ∈R is irreducible, then p is prime.
Proof. Suppose p is irreducible and p | ab and let g be a gcd(a, p). Then there are h and k such that p = gh and a = gk. Since p is irreducible, either g or h is a unit. Suppose first that h is a unit. Then a = ghh−1k = ph−1k and so p | a. If, on the other hand, g is a unit, then g divides 1 (the multiplicative identity). Of course, 1 is a common divisor of a and p, and thus we also have gcd(a, p) = 1. Euclid’s lemma (Lemma 2.6) gives that p | b.
■ Corollary 8.13. Let R be a Euclidean domain. Then i) p ∈R is prime if and only if p is irreducible.
ii) Every element admits a unique factorization into powers of primes up to re-ordering and products of units.
Proof. Item (i) follows from the previous proposition together with Theo-rem 8.7. Theorem 8.9 implies item(ii).
■ Polynomial rings over a field, such as Q[x] or R[x], are a great examples of Euclidean domains. We already saw in Section 7.1, that the degree is a Euclidean function in these rings.
We finally come to the reason to introduce empty products in Remark 2.14.
Corollary 8.14. A field F is a Euclidean domain and therefore has unique factorization.
Namely, every non-zero x ∈F is a unit times the empty product of primes. In particular, there are no primes and no irreducible numbers in F.
166 8. Factorization in Rings Proof. We take x and y in F and write x = yq+0, where q = y−1x. So every remainder maps to zero4.
■ Thus the results in Chapter 2 starting with Theorem 2.17 (the infinitude of primes) do not generalize to all Euclidean domains. The problem in the proof of Theorem 2.17 is that it crucially depends on adding “1” to some number in order to get a “bigger” number. The rest of that Chapter depends on the embedding of the integers in the real numbers (or even C).
The last result, together with Definitions 8.11, 8.4, and 5.20, immedi-ately implies the following.
Corollary 8.15. We have the following inclusions: fields ⊊Eucl. domains ⊊UFDs ⊊domains ⊊comm. rings ⊊rings .
8.4. Example and Counter-Example As an example we consider the elements of the set Z[√−1]. These are usually called the Gaussian integers (see Figure 40). From equations 7.8 and 7.9, we can infer that α = a+bi can be represented in matrix form as: α = a −b b a with N(α) = a2 +b2 .
It is easy to check that multiplication of these matrices is commutative — after all, multiplication of the underlying complex numbers is commutative.
Proposition 8.16. The Gaussian integers form a Euclidean domain with the norm as Euclidean function.
Proof. For j a square free integer, N(α) is the square of the absolute value of α, and so it is a positive integer. So the second requirement of Definition 8.11 follows immediately from Corollary 7.28. It remains to prove that the first requirement is satisfied.
Given any ρ1 and ρ2 in Z[i], we can certainly choose κ and ρ3 so that ρ1 = κρ2 +ρ3 .
4This is one of reasons we added 0 to the image of E in Definition 8.11 8.4. Example and Counter-Example 167 x Im Re Figure 40. The Gaussian integers are the lattice points in the complex plane; both real and imaginary parts are integers. For an arbitrary point z ∈C — marked by x in the figure, a nearby integer is k1 + ik2 where k1 is the closest integer to Re(z) and k2 the closest integer to Im(z). In this case that is 2+3i.
(For example, κ = 0 and ρ3 = ρ1.) Dividing by ρ2 gives ρ1ρ−1 2 = κ +ρ3ρ−1 2 .
(8.1) We choose κ to be the closest5 Gaussian integer to ρ1ρ−1 2 (indicated by “x” in Figure 40). Recalling that in this case, the norm corresponds to the usual absolute value squared, we immediately see from the figure that we can choose κ so that N(ρ3ρ−1 2 ) ≤1/2. And thus with that choice, using Corollary 7.28, N(ρ3) = N(ρ3ρ−1 2 )N(ρ2) ≤1 2N(ρ2) (8.2) which proves the first requirement.
■ The computation that leads from equation (8.1) to equation (8.2) can also be done explicitly. Let ρ1 = a + bi and ρ2 = c + di. It is an easy computation to see that ρ1ρ−1 2 = ac+bd c2 +d2 +i −ad +bc c2 +d2 .
We want to express this as a Gaussian integer κ = k1 +ik2 plus a remainder ρ3ρ−1 2 = ε1 +iε2 whose norm is less than 1. We choose k1 to be the integer closest (or one of the integers closest) to ac+bd c2+d2 , and k2, the integer closest to −ad+bc c2+d2 . With those choices, the remainders ε1 = ac+bd c2 +d2 −k1 and ε2 = −ad +bc c2 +d2 −k2 5If there is more than one closest Gaussian integer, pick any one of them.
168 8. Factorization in Rings are each not greater than 1 2 in absolute value. Thus ρ3 = (ε1 +iε2)(c+id) , with norm (ε2 1 +ε2 2)(c2 +d2) by Corollary 7.28. Since the εi are no greater than 1 2, (8.2) follows.
The computation in the foregoing proof will be important, and so it is useful to summarize it even more succinctly.
Definition 8.17. A fundamental domain of Z[i] is a simply connected region in C such that it contains exactly one representative of every set z + Z[i].
Usually one takes the unit square as a fundamental domain for Z[i].
Remark 8.18. For j negative and square free, N is a Euclidean function on Z[√j] if and only if in a fundamental domain, the distance to the nearest algebraic integer is strictly less than 1.
Note that in Z, to get a small remainder we simply choose the floor of ρ1ρ−1 2 for the equivalent of κ (see the proof of Lemma 2.2). But in the above proof — working the Gaussian integers — it is clear that in general there is no obvious natural choice for κ = k1 +ik2 that makes N(ε) less than 1. In exercise 8.3, we look in some more detail at the possible choices for k1 and k2. So the Euclidean algorithm applied to, say, 17 + 15i and 7 + 5i may lead to different computations. We gave an example of this in exercise 7.17.
Proposition 8.19. The ring Z[ √ −6] does not have the unique factorization (into irreducibles) property. Therefore this ring is not a Euclidean domain.
Proof. Z[ √ −6] (see Figure 41) is an integral domain, because it is a sub-ring of C. We show that Z[ √ −6] does not have unique factorization in two steps. The first step is to observe that 10 = 2·5 = (2+i √ 6)(2−i √ 6) .
We are done if we show that 2, 5, and 2 ± i √ 6 are irreducible. Assume 2 = αγ, both non-units. Taking the norm6 (always using Corollary 7.28), we get 4 = N(α)N(γ) .
6This part of this proof illustrates how to use the norm to reduce the question whether a number in a Euclidean domain R is irreducible to the same question in Z.
8.5. Ideal Numbers 169 Im Re Figure 41. A depiction of Z[ √ −6] in the complex plane; real parts are integers and imaginary parts are multiples of √ 6.
Thus each of the norms equals 2. But 2 = a2 +6b2 has no integer solutions, hence 2 is irreducible. The exact same argument applied to 5 gives that 25 = N(α)N(γ) .
Each of the norms now must equal 5. But again 5 = a2 +6b2 has no integer solutions. If we apply the argument to 2±i √ 6, we obtain 10 = N(α)N(γ) .
Thus either α must have norm 2 and β must have norm 5, or vice versa.
But the previous arguments show that both are impossible.
■ 8.5. Ideal Numbers In this section, we explain how ideals arise from the study of factorization into primes in rings of algebraic integers. We base this description loosely on the historical record as described in chapter 21 of the excellent book . For the definition and basic properties of ideals, we refer to Section 7.2. We start by reformulating gcd and lcm in the language of ideals.
Definition 8.20. Let A and B ideals. The greatest common divisor of A and B is the smallest ideal that contains both of these. It is denoted by gcd(A,B).
The least common multiple of A and B is the largest ideal that is contained in both A and B. It is denoted by lcm(A,B).
Recall that an ideal ⟨j⟩in Z is maximal if and only if j is prime. For if j is not prime, the ideal generated by a divisor of j contains ⟨j⟩. On the other 170 8. Factorization in Rings hand, consider the ideal ⟨p, j⟩. The fact that it is generated by gcd(p, j) is non-trivial: it follows from B´ ezout (Lemma 2.5).
Now let us see how this pans out in some examples of ideals in rings of algebraic integers. Start by considering the ring Z[√−3] of algebraic integers (see equation (7.7)) displayed in the left of Figure 42. We start by showing that this ring does not have the unique factorization property.
Knowing that 4 = 2·2 = (1+i √ 3)(1−i √ 3), (8.3) the proof of that statement is almost verbatim that of Proposition 8.19 (see exercise 8.21. This exercise goes on to show that 4 admits no factorization at all into primes!).
What is interesting here is that the numbers 2 and (1±i √ 3) belong to the same maximal ideal.
Lemma 8.21. I = ⟨2,1+i √ 3⟩is a maximal ideal in R = Z[√−3].
Proof. I is depicted in red in the left of Figure 42. It clearly contains both 2 and 1+i √ 3. It clearly forms a lattice and so is closed under addition. Next we check the absorption property of the ideal. Denote the two generators by x and y for brevity. For any elements α, β, and γ of R, we must have α(βx+γy) = δx+εy ∈I .
It is an easy but tedious exercise to check that for any integers a, b, c, and d (a+ib √ 3)·2+(c+id √ 3)·(1+i √ 3) = (a−b−2d)·2+(c+d+2b)·(1+i √ 3).
And so all these elements lie in the lattice I.
If we add I any element not in I, then the resulting set contains the differences 1 and i √ 3 (see Figure 42). Taking the closure under addition, it immediately follows that we obtain all of Z[√−3]. Thus I is maximal.
■ The upshot is that we are tempted (or, rather, Kummer was ) to think of the set I as the set of multiples of some hidden or “ideal”7 prime Q. Then both 2 and (1 ± i √ 3) are multiples of this “ideal” number Q (up to units at least). This way, lo and behold, unique factorization into irre-ducibles or primes is restored!
7Hence the name “ideal”.
8.5. Ideal Numbers 171 Im Re Im Re Figure 42. Left, the elements of the ring Z[√−3].
Right, the ring Z[ 1 2 (1 + √−3)]. The units of each ring are indicated in green and the ideals ⟨2,1 + √−3⟩on the left and ⟨2⟩on the left are indicated in red.
Fundamental domains (Definition 8.17) are shaded in blue.
There is more than a grain of truth in this. Recall that the ring R′ = Z[ 1 2(1+i √ 3)] is the ring of integers in Q(√−3) (Lemma 7.25). This ring, depicted on the right of Figure 42, contains the units (drawn in green) 1±i √ 3 2 .
Clearly, 2 and 1 + i √ 3 are now the same up to a unit. Therefore, this time around 2 generates I. In other words, R′ contains R, and has the same set I as an ideal, only now it is a principal ideal. Indeed, in R′, equation (8.3) does not represent distinct factorizations of 4, precisely because in this ring, 2 and 1+i √ 3 differ by a unit.
Finally, we finish this section by checking that indeed the norm is not a Euclidean function for Z[√−3], while it is for Z[ 1 2(1+i √ 3)]. Thus this ring is a Euclidean domain and so, by Corollary 8.13, primes and irreducibles are the same, and factorization is unique. This ring is an important example and has its own name; its elements are called the Eisenstein integers .
Proposition 8.22. i) The norm in Z[√−3] is not a Euclidean function.
ii) The norm in Z[ 1 2(1+√−3)] is a Euclidean function.
Proof. According to Remark 8.18, the norm — which in these two cases is positive — is a Euclidean function if and only if it is less than 1 in a fundamental domain. In both cases, the norm of a number is simply the square of the usual absolute value of that number. The fundamental domains are shaded in Figure 42.
Proof of (i). The fundamental domain D is given by a rectangle of height |h| = √ 3 and width 1 (see Figure 43). The diagonals in D have 172 8. Factorization in Rings length √1+3 = 2 and so we have that the distance to the nearest algebraic integer is between 0 and 1. It equals 1 at the intersection of the diagonals.
Thus N fails to be a Euclidean function.
Proof of (ii). The fundamental domain consists of two isosceles trian-gles, one of which is depicted on the right of Figure 43. Its height d is 1 2 √ 3 and its base has length 1. We are looking for the point that maximizes the distance to the nearest point of the triangle. That point lies at height y on the bisector of the top-angle and its its distance d −y to the three points of the triangle is the same. Thus we compute 1 22 +y2 = (d −y)2 = ⇒ y = 4d2 −1 8d = ⇒ d −y = 4d2 +1 8d .
This evaluates to d −y = √ 3 3 which is less than 1.
■ 1/2 1/2 y 0 1 h 1+h d Figure 43. Left, the fundamental domain of Z[√−3]. Here, h = i √ 3.
Right, one of the 2 isosceles triangles that constitute the fundamental domain of Z[ 1 2 (1 + √−3)]. Its height d equals 1 2 √ 3. The point that maximizes the distance to the closest of the 3 corner points lies on the bisector of the top angle at height y.
It is surprising that in the first part of the proof, the criterion of Eu-clidean fails at only 1 point in the fundamental domain. An analyst might suspect that somehow we can get around the exception because it has mea-sure zero. Note, however, that (8.3) shows that Z[√−3] does not have have unique factorization and thus there is no Euclidean function (Proposition 8.13).
8.6. Principal Ideal Domains 173 8.6. Principal Ideal Domains Definition 8.23. A principal ideal domain is an integral domain in which every ideal is a principal ideal. This is usually abbreviated to PID.
We now complete the containments given in Corollary 8.15.
Theorem 8.24. We have the following inclusions: fields ⊊ED’s ⊊PID’s ⊊UFD’s ⊊domains ⊊comm. rings ⊊rings .
Proof. In view of Corollary 8.15, we only need to prove (i) that a Euclidean domain is a PID, (ii) that a PID is a UFD, and (iii) that the three categories are not equal. We leave (iii) for the next section.
i) In a Euclidean domain, the trivial ideal {0} is of course a principal ideal (as it has only one element). Let E be the Euclidean function in D.
Fix a non-trivial ideal I and pick x ∈I that minimizes E on I{0}. Pick any other y ∈I. Then by the division algorithm y = xq+r and E(r) < E(x).
But since y −xq ∈I, r is in I, and so E(r) must be zero by the minimality of x. Hence x generates y.
ii) Suppose x0 is an element of a principal ideal domain D that cannot be written as a a product of irreducibles. Then, clearly, there are non-zero non-units x1 and y1 so that x0 = x1y1. But by definition of x0, at least one of x1 and y1 cannot be written as a product of irreducibles. Suppose that is x1. Now x1 divides x0, and we get ⟨x0⟩⊊⟨x1⟩. We can apply the same ar-guments to x1, and so on. Thus we get what is called an (infinite) ascending chain of ideals: ⟨x0⟩⊊⟨x1⟩··· ⊊⟨xn⟩··· .
We define I = ∪∞ i=0 ⟨xi⟩. It is easy to see that I is an ideal (Definition 7.10).
But because D is a PID, I must have a single generator p. The element p must reside in ⟨xn⟩for some n. Since p generates ⟨xn⟩it must in fact be equal to xn. Thus the ascending chain must end, contradicting the hypothe-sis on x0, which implies that every element in D can be written as a product of irreducibles.
It is then sufficient by Theorem 8.9 to show that every irreducible p is also prime. Let element a not in ⟨p⟩and consider the ideal ⟨p,a⟩. Because 174 8. Factorization in Rings D is a PID, there is a q that generates this ideal: ⟨q⟩= ⟨p,a⟩. But then we must have ⟨q⟩= D, because if not, p has a non-trivial divisor q. In particular, we get that there are x and y so that 1 = px+ay = ⇒ ∀b ∈D : b = pxb+ayb But this implies that if p | ab and p ∤a, then we must have p | b. Thus p is prime.
■ Common PID’s are Z and F[x], but these are also Euclidean domains.
8.7. ED, PID, and UFD are Different PID’s that are not Euclidean domains are a not so easy to come by. Here we show, following , that Z[ 1+√−19 2 ] is an example of this. Recall that by Lemma 7.25, Z[ 1+√−19 2 ] is the set of integers of Q[√−19].
Lemma 8.25. In Z[ 1+√−19 2 ], the units are ±1, while 2 and 3 are irreducible.
Proof. For brevity, we set θ = 1+√−19 2 and denote R = Z[θ]. The norm of a+bθ satisfies (see, for example, remark 7.29) N(a+bθ) = a+ b 2 2 + 19b2 4 = a2 +ab+5b2 ∈N∪{0}.
We have that the norm of units must be ±1, so a+ b 2 2 + 19b2 4 = 1.
Clearly, the only solutions are a = ±1 and b = 0.
By the multiplicative property of the norm, if 2 is reducible we have 2 = xy = ⇒ N(2) = 4 = N(x)N(y).
N(x) and N(y) are natural numbers and not equal to 0 or ±1. The only solution is N(x) = N(y) = 2 which is easily seen to be impossible. Hence 2 is irreducible. The same reasoning works for 3.
■ Proposition 8.26. Z[ 1+√−19 2 ] is not a Euclidean domain.
8.7. ED, PID, and UFD are Different 175 Proof. We start by assuming that R is a Euclidean domain (ED) and derive a contradiction. So let E denote the Euclidean function8 of Definition 8.11 and let m be an element of R that minimizes E over the set of non-zero, non-unit elements. In a ED, we are allowed to use the division algorithm, so 2 = mq+r with E(r) < E(m).
From the inequality and the assumption on m, we see that r must be zero or a unit. So by Lemma 8.25, r ∈{0,±1}. Now if r = 1, then mq = 2−r = 1 and so m is invertible, contradicting the assumption on m.
If r equals 0 or −1, we need to do one more step. In this case, mq equals 2 or 3. By Lemma 8.25 these numbers are irreducible, and thus q must be a unit (since m is not), whence m ∈{±2,±3}. We apply the division algorithm to θ: θ = mq′ +r′ with E(r′) < E(m).
So θ −r′ is divisible by m where r′ ∈{0,±1}, that is to say: by 2 or 3. But it is easy to see that any of these numbers divided by 2 or 3 are not in R.
■ Theorem 8.27. The set R of the integers of Q[√−19] is a PID but not a Euclidean domain.
3 /2 − 3 /2 1 1 i i+1 i+1/2 19 /4 − 19 /4 Figure 44. Points in the area red shaded are a distance less than from an integer in Z. The blue area maps into the red under x →2x− √ 19/4 indicated by the arrow. We note that √ 19/4 ≈1.09 and √ 3/2 ≈0.87.
8Some unknown function, not necessarily the norm.
176 8. Factorization in Rings Proof. Of course, the second part is settled by the previous result. So here we just prove that R is a PID. So consider any non-zero ideal I in R and pick a b in I which minimizes the norm N(b) on the non-zero elements of I. Now assume I is not principal, then certainly bR will not equal I. In this case, we choose any a ∈I\bR and investigate what happens. By the absorption property, we have that ∀p,q ∈R : ap−bq ∈I .
We will show, however, that ∃p,q ∈R : ap−bq ̸= 0 and N(ap−bq) < N(b), (8.4) which contradicts our choice of b, and therefore disproves the assumption that I is not principal. By the multiplicativity of norms, (8.4) will be proved if N (ap/b−q) < 1. By remark 7.29 then, (8.4) is equivalent to ∃p,q ∈R : ap−bq ̸= 0 and a b p−q < 1.
(8.5) Clearly, we can choose q so that the real part of ap/b−q is not zero. Then add a multiple of i √ 19/2 to q so that the imaginary part of ap/b −q is in (− √ 19/4, √ 19/4]. Note that ap/b−q ̸= 0. If in fact the imaginary part is in (− √ 3/2, √ 3/2) (shaded red in Figure 44), then by subtracting an integer (in Z) from q we are done. If, however, the imaginary part of ap/b−q lands in [ √ 3/2, √ 19/4], then we multiply both p and q by 2 and subtract i √ 19/2.
One can check (see exercise 8.24) that the complex map g : z →2z−i √ 19/4 maps the top blue shaded area in Figure 44 into the area shaded in red. The argument for the lower blue area is identical.
■ Theorem 8.28. The set Z[x] is a UFD but not a PID.
Proof. We start by showing that I = ⟨2,x⟩is not a principal ideal in Z[x].
Let p ∈I. Then p(x) = 2 f(x)+xg(x) and so p(0) = 2n for some n ∈Z. If p generates the ideal I, then we must also have 2 = p(x)a(x) and x = p(x)b(x).
The first equality implies that p has degree 0 and p(x) = 2n, while the second then yields that x = 2nb(x) which is impossible. So I is not principal.
Given f ∈Z[x]. It is not surprising that any factorization of f in Z[x] is also a factorization in Q[x]. However, the reverse is also true by Gauss’ 8.8. Exercises 177 lemma (Lemma 7.8). Now f as an element of Q[x] has a unique factoriza-tion by Corollary 8.13 and the fact that the degree is a Euclidean function in Q[x]. Thus the same holds in Z[x].
■ Many results about factorization of rings of quadratic integers are known.
We mention a few without proof.
Proposition 8.29. For d square free, the norm in Q( √ d) is a Euclidean functions if and only if d is an element of {−11,−7,−3,−2,−1,2,3,5,6,7,11,13,17,19,21,29,33,37,41,57,73}.
There are exist quadratic fields, such as Q[ √ 69], that are Euclidean but whose norm is not a Euclidean function .
Furthermore (Baker-Heegner-Stark Theorem , see ), for negative square free d, the integers of Q( √ d) form a PID and not a Euclidean domain if and only if d ∈{−163,−67,−43,−19}.
Is has been conjectured that for positive (square free) d, there are infinitely many values for which the integers of Q( √ d) have unique factorization.
For d square free, if the integers of Q( √ d) are a UFD, then they are also a PID.
8.8. Exercises Exercise 8.1. Use Definition 8.1 (see also Proposition 5.18) to find the units, irreducibles (see Proposition 8.3 (iii), and primes in: a) Z6, b) Z7, and c) Z8.
(Hint: the multiplicative inverses and multiplication tables of Z6 and Z7 are given in Figures 21 and 22 and the tables after Definition 5.19. For Z8, the information is in Figure 45.) 178 8. Factorization in Rings 6 1 2 3 4 5 0 Z mod 8: 7 1 2 3 4 5 6 7 0 1 2 4 5 6 7 0 2 3 4 5 6 0 1 2 3 4 5 6 7 0 1 2 3 4 6 7 0 1 2 3 4 5 6 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 4 4 4 4 4 4 2 6 2 6 4 2 6 Figure 45. Left: the relation a is a multiplicative inverse mod 8 of b.
Right: the multiplication table of Z8.
Exercise 8.2. Let R be an integral domain. Consider the set R×{R{0}} = {(a,b) : a,b ∈R,b ̸= 0}.
Define an equivalence relation ∼as follows.
(a,b) ∼(c,d) if ad = bc.
Frac(R) is the collection of equivalence classes with addition and multipli-cation: (a,b)+(c,d) = (ad +bc,bd) and (a,b)·(c,d) = (ac,bd).
It is not hard (but tedious) to show [Chapter 8] that ∼is indeed an equivalence and that Frac(R) is the minimal field containing R. Frac(R) is called the field of fractions or field of quotients of R.
a) Show that addition and multiplication are well-defined in Frac(R).
b) What is the field of fractions of Z?
c) The identity is not used in the definition of Frac(R). What is the “field of fractions” of the “rng” (see remark 5.24) mZ where m > 1 in N?
d) Why is it necessary to require that R has no zero divisors?
8.8. Exercises 179 Exercise 8.3. We apply the Euclidean algorithm in Z[√−1] as in Section 8.4. For the notation, see the proof of Proposition 8.16. Suppose ργ−1 falls in the unit square depicted in Figure 46. We have drawn four quarter circles of radius 1 in the unit square, denoted by a, b, c, and d.
a) Show that we cannot always choose κ = κ1 +iκ2 where κ1 is the floor of the real part of κ +ργ−1 and κ2 the floor of the imaginary part. (Hint: Consider the region “northeast” of the quarter circle a.) b) Compute the coordinates of the points A, B, C, and D indicated in the figure. (Hint: Because of the symmetries of the figure, the x coordinate of A equals 1/2. et cetera.) c) Show that if ργ−1 falls in the interior of the convex shape FACE, then there are four possible choices for κ so that N(ρ) < N(γ).
d) Estimate the area of the convex shape FACE. (Hint: It is contained in a square with sides of length BD and it contains a square with sides of length AC.) e) Is it possible that there is only one value for κ so that N(ρ) < N(γ)?
a b c d A B C D E F 0 1 i 1+i Figure 46. Possible values of ργ−1 in the proof of Proposition 8.16.
In the following proposition and in exercises 8.4, 8.5, and 8.6, we study the primes in Z[√−1] — called Gaussian primes. Recall that the Gaussian integers from a Euclidean domain (Proposition 8.16), and so we have unique factorization and primes and irreducibles are the same (Corollary 8.13). We use the following notation. C (for “cross”) denotes the set Z ∪iZ minus the origin. Recall that the units in Z[√−1] are {±1,±i} and those in Z are {±1}. The notation π means a prime in Z[√−1], whereas p means a positive prime in Z.
180 8. Factorization in Rings Proposition 8.30 (Gaussian Primes). A number π ∈Z[√−1] is prime if: i) π ∈C and |π| equals a prime p in Z with p =4 3, ii) π ̸∈C and |π|2 equals a prime p in Z with p = 2 or p =4 1.
iii) Furthermore, if π is reducible then (i) and (ii) cannot hold. (So “if” can be replaced by “if and only if”.) For an illustration of the Gaussian primes, see Figure 47.
Exercise 8.4. a) Show that N(π) = ππ = ∏ i pki i .
b) Show that N(π) equals p or p2 (up to units). (Hint: π must divide one of the primes, say p, in (a).) c) Use (b) to show that if π ∈C, then N(π) = p2 and so |π| = p.
d) Use unique factorization in Z[√−1] to show that if π ̸∈C, then N(π) = p. (Hint: can p· p = π ·π?) Exercise 8.5. a) Use exercise 5.21 (c) to show that if p =4 1 and p prime in Z, then there is m such that p | m2 +1.
b) Show that if p =4 1, then p is not a prime in Z[√−1]. (Hint: use that p | (m+i)(m−i).) Also show that 2 is not a prime in Z[√−1].
c) Show that a2 +b2 ̸=4 3. (Hint: compute modulo 4.) d) Show that if a prime p in Z does not have residue 1 or 3 modulo 4, then p = 2.
e) Use exercise 8.4 (c) and (b) of this exercise to prove Proposition 8.30 (i).
f) Then use exercise 8.4 (d) and (c) and (d) of this exercise to prove part (ii).
Exercise 8.6. a) Show that for a reducible γ in Z[√−1], N(γ) is not prime in Z. (Hint: use Corollary 7.28.) b) Use (a) to show that a reducible γ cannot satisfy Proposition 8.30 (ii).
c) Assume γ in C and γ = αβ up to units. Show that if α and β are in C, then |γ| is not prime in Z.
d) Assume γ in N and γ = αβ up to units and that α and β are not in C.
Show that if γ = p, then |α| = |β|, and therefore are conjugates (Hint: use Corollary 7.28.). Show that this implies that N(γ) has the form a2 +b2.
e) Show that (c) and (d) and exercise 8.5 (c) imply that γ cannot satisfy Proposition 8.30 (i).
f) Extend the reasoning in (d) and (e) to all of C.
8.8. Exercises 181 Figure 47. The Gaussian primes described in Proposition 8.30. There are approximately 950 within a radius 40 of the origin (left figure) and about 3300 within a radius 80 (right figure).
Exercise 8.7. Again, we consider numbers in the ring R = Z[√−1].
a) Show that if bn −1 is prime in Z[√−1], then b−1 is a unit.
b) Use (a) to show that b must be 2 or 1±i.
c) Use Proposition 8.30 (i) to show that if b = 2, we obtain the usual Mersenne primes (Definition 5.13) as primes in Z[√−1].
d) Show that if n is not prime, then bn −1 is not prime. (Hint: as in exer-cise 1.12 (i).) e) Show that N((1±i)n −1) = 2n −21+ n 2 cos nπ 4 +1.
(Hint: (1±i) = 21/2e±iπ/4 and eiϕ +e−iϕ = 2cosϕ.) f) Show that (1 ± i)n −1 is prime if and only if its norm is prime and n is odd. (Hint: use (d) to show that n must be odd, and then Proposition 8.30.) The primes in exercise 8.7 are a generalization of the Mersenne primes of Definition 5.13. These primes in Z[√−1] of the form (1±i)n −1 are called Gaussian Mersenne primes . A similar construction works also in the Eisen-stein integers Z[ 1 2(1 + √−3)]; the resulting primes are called Eisenstein Mersenne primes . For more details, see .
182 8. Factorization in Rings Exercise 8.8. Given the ring R = Z[ √ −5].
a) Show that 2 is irreducible. (Hint: suppose 2 = βγ, where β and γ are non-units. Use Corollary 7.28 to see that N(β) = N(γ) = 2. Solve for the coefficients of β and γ.) b) Show that 3 is irreducible. (Hint: as (a).) c) Use (a) and (b) to show that 1±i √ 5 are irreducible.
d) Show that Z[ √ −5] is a not Euclidean domain. (Hint: Show it does not have unique factorization.) Exercise 8.9. Given the ring R = Z[ √ 2].
a) Show that R has no zero divisors. (Hint: If αβ = 0, then one of the norms must be zero by Corollary 7.28. Solve for the coefficients.) b) Suppose ρ1 = κρ2 +ρ3 , where ρ1 = a+b √ 2, ρ2 = c+d √ 2, κ = k1 +k2 √ 2, and ρ3 = ε1 +ε2 √ 2.
Show that ρ1ρ−1 2 = ac−2bd c2 −2d2 + −ad +bc c2 −2d2 √ 2 .
c) Choose k1 to be the integer closest to ac−2bd c2−2d2 and k2 the one closest to −ad+bc c2−2d2 . Show that the remainder has norm with absolute value less than 1. (Hint: recall that the norm is a2 −2b2!) d) Show that the ring Z[ √ 2] is a Euclidean domain (Hint: use Corollary 7.28.) Exercise 8.10. a) Show that in Z[ √ −5], (2 + i √ 5) is irreducible but not prime. (Hint: follow the proof of Proposition 8.3 (iv), except now start with 3·3 = 9 to prove non-primality.) b) Show that in Z6, 3 is prime but not irreducible. (Hint: follow the proof of Proposition 8.3 (iii).) c) Find other counterexamples.
Exercise 8.11. a) Solve 3x =b 6x where b is 11, 12, 13, 14, 15.
b) If b is such that Zb is an integral domain, solve by factoring. c) Use a result in Chapter 5 to show that Zb is an integral domain and hence a field if and only if p is prime.
d) Give a direct proof that a field is an integral domain. (Hint: if a and b are non-zero elements of F, then abb−1a−1 = 1.) 8.8. Exercises 183 Exercise 8.12. a) Prove Lemmas 2.5 and 2.6 for a Euclidean domain.
b) Theorem 8.5 follows immediately from the absence of zero divisors (Definition 8.4). In Chapter 2, we take the absence of zero divisors in Z for granted. Why do we need Euclid’s Lemma (Lemma 2.6) — whose proof uses that division algorithm — to prove Theorem 2.7? (Hint: does the cancellation take place in Z?) In the next exercise, we prove: Lemma 8.31. Let d ∈Z be square free. α ∈Z[ √ d] is a unit if and only if N(α) = ±1.
Exercise 8.13. a) Show that if α is a unit, N(α−1) = 1 N(α).
b) Use (a) to show that the norm of a unit must be ±1.
c) Vice versa, show that if N(α) = ±1, then α is invertible. (Hint: a matrix with determinant ±1 is invertible. Show that the inverse matrix corresponds to an element of Z[ √ d].) Exercise 8.14. Consider Z[ √ −6] and define a± = 2± √ −6.
a) Show that a−a+ = 10 = 2·5.
b) Show that a±, 2, and 5 are irreducible in Z[ √ −6]. (Hint: if a+ = αβ is reducible, then N(a+) = N(α)N(β). By Lemma 8.31, we may assume N(α) = 2. Solve that equation. And so forth.) c) Show that a±, 2, and 5 are not primes. (Hint: for a±, use (a)).
d) Show that unique factorization does not hold. (Hint: see (a)).
e) Show that Euclid’s lemma 2.6 does not hold here. (Hint: use Definition 8.2.) Exercise 8.15. a) Which ones of the sets in exercise 5.24 are integral do-mains?
b) Euclidean domains?
Exercise 8.16. a) Show that ±1 and ±1± √ 2) are units of Z[ √ 2]. (Hint: see Lemma 8.31.) b) Show if α is a unit, then for all n ∈Z, αn is a unit.
c) Show that Z[ √ 2] has infinitely many units.
d) Find solutions of the quadratic equation a2 −2b2 = ±1. (Note: an equation of the form a2 −db2 = 0 where d is square free, is called Pell’s equation .) One can show that the set of units of Z[ √ 2] is {±(1+ √ 2)n : n ∈Z}.
184 8. Factorization in Rings Exercise 8.17. Given the ring R = Z[ √ 10].
a) Show that there is no α ∈R with N(α) = ±2. (Hint: write α = a+b √ 10 and try to solve for the coefficients of α in Z10.) b) Show that there is no α ∈R with N(α) = ±5. (Hint: write α = a + b √ 10. Then in Z5, show that a =5 0. It follows that 25k2 −10b2 = ±5.
Divide by 5 and solve in Z5.) c) Use (a) and (b) to show that 2 and 5 are irreducible. (Hint: assume that 2 = αβ, show that then N(α) = ±2, et cetera.) d) Use (a) and (b) to show that √ 10 is irreducible.
e) Show that Z[ √ 10] is a not Euclidean domain. (Hint: Show that 10 does not have unique factorization.) Exercise 8.18. Given a field F, we form the ring F[x] of polynomials. For this exercise, read Section 3.7 again.
a) Use exercise 7.1 to show that the ring F[x] is a Euclidean domain with the degree d (of the polynomial) as a Euclidean function.
b) What goes wrong in (a) if F = Z? (Hint: give a counter-example.) c) What are the “primes” in F[x]. (Hint: see Proposition 7.5 and Corollary 8.13.) d) p1(x) = x2 +1 is reducible over C, R, or Q? What about p2(x) = x2 −2?
e) Show that the degree in R[x] is an additive function if R is a domain.
Exercise 8.19. Given a field F.
a) Show that for any α ∈F and p in F[x], there are q and r in F[x] such that p(x) = (x−α)q(x)+r(x), where r(x) is a constant. (Hint: the degree is a Euclidean function.) b) Show that in (a), p(α) = 0 if and only if r = 0. (Hint: Substitute x = α.) c) Use (b) to show that if pn ∈F[x] of degree n has a root, then pn(x) = pn−1(x)(x−α) where pn−1 has degree n−1.
d) Use (c) to show that a degree n polynomial in F[x] has at most n roots.
(Compare with exercises 3.22 and 7.17) We state the last result of exercise 8.19. It has important applications.
Theorem 8.32 (Lagrange’s Theorem). If f is a degree n polynomial with coefficients in a field F, then f(x) = 0 has at most n solutions.
Exercise 8.20. Define the product of ideals A and B as the smallest ideal containing {aibi : ai ∈A, bi ∈B}.
a) Show that AB must contain ∑k i=1 aibi : ai ∈A, bi ∈B k ∈N .
b) Show that the set in (a) is an ideal.
c) Suppose A is generated by {xi} and B by {yj}. Show that AB is the ideal generated by {aiyj}.
d) Use (c) to show that for I and J as in exercise 7.5, IJ = ⟨6,x⟩. (Hint: x2 is in ⟨x⟩, and so forth.) 8.8. Exercises 185 Exercise 8.21. a) Show that 2 and 1 ± i √ 3 are irreducible in Z[√−3].
(Hint: follow the proof of Proposition 8.19.) b) Use (a) to show that up to units, there are two factorizations in Z[√−3] of 4 (see equation (8.3)).
c) Use equation (8.3) to show that 4 is not prime.
d) Show that 2 and (1±i√−3) are not prime. (Hint: see Proposition 8.3.) e) Conclude that 4 does not admit any factorization into primes in Z[√−3].
f) Show that 2 and (1±i√−3) are prime in Z[ 1 2(1+√−3)].
Exercise 8.22. a) Modify the first part of the proof of Proposition 8.22 to show that the norm is a Euclidean function for Z[√−1] and Z[√−2] but not for Z[√−n] for n ≥3.
b) Modify the second part of the proof of Proposition 8.22 to show that the distance to the nearest lattice point of Z[ 1 2(1 + √j)] is less than 1 if j ∈{−11,··· ,−1}. (Hint: the height y of the equidistant point in triangle on the left of Figure 43 must be such that d −y < 1 where d = 1 2 p |j|.) c) Show that with Lemma 7.25, this implies that the norm is a Euclidean function for the integers of Q[√j] where j ∈{−11,−7,−3,−2,−1}.
Exercise 8.23. Use Definition 7.10 to show that I in part (ii) of the proof of Theorem 8.24 is an ideal.
Exercise 8.24. Consider the map g : C →C, defined in the proof of part (ii) of Theorem 8.27. a) Show that g √ 19 4 = 0.
b) Show that − √ 3 2 < g √ 3 2 < 0.
c) Show that (a) and (b) imply that g maps the blue region in Figure 44 into the red region.
Exercise 8.25. Consider the ring Z[x].
a) Show that the ideal I := ⟨3,5x⟩is not principal. (Hint: see proof of Theorem 8.28.) b) Show by direct computation that I does not generate Z[x]. (Hint: solve 1 = 3f(x)+5xg(x).) c) Show that (b) also follows directly from (a). (Hint: 1 generates all of Z[x] ⊇I.) d) Find gcd(3,5x) and lcm(3,5x).
e) Show that B´ ezout does not hold in this ring.
Chapter 9 Ergodic Theory Overview. This time we venture seemingly very distant from number the-ory. The reason is that we wish to investigate what properties “typical” real numbers have. By “typical” we mean “almost all”; and to define “almost all”, we have to delve fairly deeply into measure theory, one of the back-bones of abstract analysis. In this chapter, we will point to the technical problems that need to be addressed, and then quickly state the most impor-tant result (the Birkhoff ergodic theorem). In Chapter 10 we will then move to the implications for number theory. The proof of the Birkhoff ergodic theorem will be postponed to Chapter 14.
The ergodic theorem (there are various versions) is arguably one of the most important theorems in mathematics. In essence, it offers a means to re-place the study of long-term behavior of complex systems by much simpler statistical reasoning, allowing quantitative predictions for the long term be-havior of such a system. Considerations of this nature gave arose out of, and contributed to, an important branch of theoretical physics, namely statistical physics [31,49]. It is also widely applied in number theory, probability the-ory, functional analysis, and other fields of study. The discussion whether or not ‘physical’ systems tend to be ergodic has had a profound impact on science, in particular physics [31, 49]. The use of probabilistic methods to study number theory is often referred to as probabilistic number theory.
187 188 9. Ergodic Theory 9.1. The Trouble with Measure Theory In analysis we can distinguish short intervals from long ones by looking at their “length” even though both have the same cardinality (see Definition 1.27). The notion of length works perfectly well for simple sets such as intervals. But if we want to consider more general sets – such as Cantor sets — it is definitely very useful to have a more general notion of length, which we denote by measure. However, there is a difficulty in formulating a rigorous mathematical theory of measure for arbitrary sets. The source of the difficulty is that there are, in a sense, too many sets. Recall that the real line is uncountable (see Theorem 1.24). The collection of subsets of the line is in fact the same as the power set (Definition 1.33) P(R) of the the real line. And thus the cardinality of the collection of subsets is strictly larger than that of the real numbers (Theorem 1.34), making it a truly very big set.
A reasonable theory of measure for arbitrary subsets of R should have some basic properties that are consistent with with intuitive notions of “length”.
If we denote the measure of a set A by µ(A), then we would like µ to have the following properties.
1) µ : P(R) →[0,∞].
2) For any interval I: µ(I) equals the length of I.
3) µ is translation invariant.
4) For a countable collection of disjoint sets Ai: µ(∪∞ i=1Ai) = ∑∞ i=1 µ(Ai).
The problem is that no such function exists. Among all the possible sets, we can construct an — admittedly pretty weird — set for which the last three properties cannot simultaneously hold.
To explain this more easily, let us replace R by the circle S = R/Z.
Now define an equivalence relation (Definition 1.28) in S as follows: a ∼b if a−b is rational. Each element of S clearly belongs to some equivalence class (it is equivalent to itself), and cannot belong to two distinct equiva-lence classes, because if a ∼b and a ∼c, then also the difference between b and c is rational, and hence they belong to the same class. Note that each equivalence class is countable, and so (see exercise 1.8) there are uncount-ably many equivalence classes.
9.1. The Trouble with Measure Theory 189 For every one of these equivalence classes, we pick exactly one rep-resentative. The union of these representatives forms a set V. A set con-structed this way is known as a Vitali set . Now by requirement (1), any set, no matter how exotic its construction, should have a measure that is a real number. We choose V as our set and let its “measure” be equal to some ε ∈[0,∞]. Let r : N →Q be a bijection between N and the rationals in S.
Consider the union of the translates ∪∞ i=1 (V +ri) .
By definition of V, this union covers the entire circle without any overlaps.
So by requirement (2) above, the measure of the union is 1. By requirement (3), each of the translates of V must have the same measure, ε. Since the translates of V are disjoint, requirement (4) implies that 1 = ∞ ∑ i=1 ε , which is clearly impossible!
The construction of the setV just outlined is admittedly a little vague. It is not clear at all how exactly we could choose an individual representative, much less how we could achieve that feat for each of the uncountably many equivalence classes. If we wanted to draw a picture of the set V, we’d get nowhere1. Does this construction V really exist as an honest set? It turns out that one needs to invoke the axiom of choice2 to make sure that V exists.
The consensus in current mathematics (2020) is to accept the axiom of choice (though the consensus is not unanimous ). One consequence of that is that if we want to define a measure, then at least one of those four requirements above needs to be dropped or weakened. The measure theoretic answer to this quandary is to restrict the collection sets for which we can determine a measure. This means, that of the properties (1) through (4), we restrict property (1) to hold only for certain sets. These are called the Lebesgue measurable sets . More generally, not all measures are “length-like”, and so we may drop the second requirement. In that case, we speak of measurable sets .
1I tried.
2The axiom of choice states that for any set A, there exists a function f : P(A) →A that assigns to each non-empty subset of A assigns an element of that subset. For more details, see .
190 9. Ergodic Theory 9.2. Measure and Integration To surmount the difficulty sketched in the previous section so that we can define measure and integration unambiguously turns to be technically very involved. This section serves just to give an idea of that complication and its resolution. The interested student should consult the literature, such as the excellent introduction . In chapter 14, we provide some more details.
Recall that a set O ⊆R is an open set usually3 means that for all x ∈O there is a positive ε so that (x −ε,x + ε) is also contained in O. Closed sets are defined as sets whose complement is an open set. Vice versa, the complement of a closed set is open.
Definition 9.1. Consider the smallest collection of sets closed under com-plementation, countable intersection, and countable union that contains the open sets. These are called the Borel sets.
This is simply a way of saying that the Borel sets are the open sets plus all sets that can be obtained from these by complementations, countable intersections, and countable unions.
Definition 9.2. The outer measure of a set S is µout(S) = inf∑ k ℓ(Ik).
where the infimum is over the countable covers of S by disjoint open inter-vals Ik.
It turns out that the outer measure is a measure on the Borel sets. This takes some effort to prove and we refer to the literature ( and for a slightly different formulation of essentially those same ideas). To give an idea, one ingredient is that every open sets in R is a countable union of disjoint open intervals4 (see exercise 9.4), so the outer measure of an open sets can be calculated easily. After establishing that the outer measure is a measure on the Borel sets, the theory then proceeds by augmenting the Borel sets by arbitrary sets of outer measure zero.
3This is called the standard topology on R. It is possible to have different conventions for what the open sets in R are.
4Open sets in R2 are countable disjoint unions of open rectangles, and so forth in Rn, n > 2.
9.2. Measure and Integration 191 Definition 9.3. A set S is called Lebesgue measurable if it contains a Borel set B whose outer measure equals µout(S). That outer measure is the Lebesgue measure of S. (Thus the measure of an interval equals its length.) More informally, Lebesgue measurable sets are Borel sets modulo sets of outer measure zero. One can work out that the collection of Lebesgue measurable sets is also closed under complementation, countable intersec-tion, and countable union. As a consequence of these facts, we have the following result.
Proposition 9.4. i) A set S ⊂R is Lebesgue measurable if and only if there exist closed sets Ci ⊆S such that µout(S\∪∞ i=1 Ci) = 0.
ii) A set S ⊂R is Lebesgue measurable if and only if there exist open sets Oi ⊇S such that µout(∩∞ i=1 Oi\S) = 0.
Proof. For the first part of this proof, see also exercise 9.1. Observe that every closed set Ci ⊆S is the complement of an open set Oi ⊇S and vice versa. So if ∪n i=1Ci contains nearly all of S, then its complement ∩∞ i=1 Oi very little more than the complement Sc of S and vice versa. Since comple-mentation preserves the Lebesgue measurable sets (by definition 9.3), (i) and (ii) are equivalent.
A countable union of closed sets is Borel. Since the outer measure is a measure on Borel sets, (i) says that S is a Borel set plus something of outer measure zero. This implies Definition 9.3.
Vice versa, Definition 9.3 says that a Lebesgue measurable set consists of a Borel set contained in a countable collection of disjoint open intervals Ii (by Definition 9.2) plus possibly a set Z of outer measure zero. The latter set can be covered by a collection of intervals of arbitrarily small outer measure.
■ Finally, we can define a measure more generally — i.e. not “length-like” or Lebesgue — as follows and show that it satisfies the above charac-teristics, if one limits the definition to measurable sets.
192 9. Ergodic Theory Definition 9.5. A measure µ is a non-negative function from a collection5 of µ-measurable sets to [0,∞] such that µ(/ 0) = 0 and for every countable sequence of disjoint (measurable) sets Si: µ(∪∞ i=1 Si) = ∞ ∑ i=1 µ(Si).
A couple of remarks are in order. The first is the observation that the measurable sets for some arbitrary measure are not constructed in this the-ory (except for the Lebesgue measure). Rather, they are part of the def-inition of measure. Which sets are measurable? The sets on which µ is defined.
We remark further that this definition implies that in general sub-additivity holds: µ(∪∞ i=1Ai) ≤ ∞ ∑ i=1 µ(Ai).
(9.1) The reason is that the measure of the union equals the sum of the measures of the disjoint “new” parts A′ i of Ai, i.e. Ai minus the intersection of Ai with the A j where j < i. Since A′ i ∪(Ai\A′ i) = Ai and this is a disjoint union, we have µ(A′ i) ≤µ(Ai). Hence the sub-additivity.
Thus a Lebesgue measure µ is a sub-additive function from the (Lebesgue) measurable sets to the positive reals (including infinity) and the measurable sets are constructed so that properties (2), (3), and (4) in Section 9.1 hold, while a more general measure does not have to satisfy (2) and (3). We summarize this as follows.
Corollary 9.6. The Lebesgue measure µ on R or R/Z satisfies the follow-ing properties 1) µ : measurable sets →[0,∞] and µ(/ 0) = 0.
2) For any interval I: µ(I) equals the length of I.
3) µ is translation invariant.
4) For a countable collection of disjoint sets Ai: µ(∪∞ i=1Ai) = ∑∞ i=1 µ(Ai).
More generally, a (non Lebesgue) measure satisfies (1) and (3).
We need some more technical terms.
5The collection of measurable sets must be closed under complementation and countable unions and intersections 9.2. Measure and Integration 193 Definition 9.7. If we have a space X and a collection Σ of measurable sets, then the pair (X,Σ) is called a measurable space. A function T : X →X is called measurable if the inverse image under T of any measurable set is measurable. A triple (X,Σ,µ) is called a measure space. A probability measure is a measure that assigns a measure 1 to the entire space.
The Lebesgue integral of a measurable function f : X →R with respect to the measure µ is written as I = Z f dµ .
Assume f(x) is non-negative. To approximate the Lebesgue integral I, one Figure 48. A comparison between approximating the Lebesgue inte-gral (left) and the Riemann integral (right).
partitions the range of f into small pieces [yi,yi+1]. For each such layer, the contribution is the measure of the inverse image f −1 ({y : y ≥yi+1}) times yi+1 −yi. Sets of measure zero are neglected. Summing all contributions, one obtains an approximation of the Lebesgue integral (see Figures 48 and 86). The Lebesgue integral itself is defined as the limit (if it exists) of these. The Lebesgue integral of a not necessarily non-negative function f is computed by splitting up f into its non-negative part f + and its negative part f −, so that f = f + + f −. The integral of f is then defined as I = Z f + dµ − Z (−f −)dµ .
We’ll see in Section 14.2 that the domains of f + and f −are measurable so that this operation is well-defined. A function f is called integrable , or µ-integrable for clarity, if R |f|dµ exists and is finite. It turns out that the Lebesgue integral generalizes the Riemann integral6 we know from calculus (see exercise 9.6).
6Recall that the Riemann integral is approximated by partitioning the domain of f, see Figure 48.
194 9. Ergodic Theory This level of technical sophistication means that the fundamental the-orems in measure theory require a substantial mastery of the formalism.
Since pursuing all the technicalities would take a considerable effort and would lead us well and far away from number theory, we have suppressed some details in this chapter.
9.3. The Birkhoff Ergodic Theorem The context here is that we have a measurable transformation T from a mea-sure space (X,Σ,µ) to itself. The situation is quite general. The measure µ is not necessarily the Lebesgue measure, but we will assume that it is a probability measure, that is: R X dµ = µ(X) = 1.
F X Y −1 mu(F (B)) (F mu)(B) Figure 49. The pushforward of a measure µ.
Definition 9.8. Let F : X →Y be a measurable transformation and µ a measure on X. The pushforward F∗µ of the measure µ is a measure on Y defined as (F∗µ)(B) := µ F−1(B) , for every measurable set B in Y (see Figure 49).
Definition 9.9. Let T : X →X be measurable. We say that T preserves the (probability) measure µ, or, equivalently, that µ is an invariant measure , if T∗µ = µ. That is to say, if for every measurable set B, µ T −1(B) = µ(B).
Theorem 9.10 (Birkhoff or Pointwise Ergodic Theorem). Let T : X →X be a transformation that preserves the probability measure µ. If f : X →R is an integrable function, the limit of the time average ⟨f⟩(x) := lim n→∞ 1 n n−1 ∑ i=0 f(T i(x)) 9.3. The Birkhoff Ergodic Theorem 195 is defined on a set of full measure. It is an integrable function and satisfies (wherever defined) Z X ⟨f⟩(x)dµ = Z X f(x)dµ .
The proof of this theorem requires a more substantial technical mastery of analysis and we will postpone it to Chapter 14.
Definition 9.11. Let T be transformation T of a measure space (X,Σ,µ) to itself. A set S ∈Σ is called (weakly) invariant if T −1(S) = S except possibly for a set of measure zero. We will use the term strictly invariant if T −1(S) = S.
If no misunderstanding is likely, we may drop the word “weakly”.
Definition 9.12. A transformation T of a measure space X to itself is called ergodic (with respect to µ) if it preserves the measure µ and if every (weakly) T-invariant set has measure 0 or 1.
This is a slight departure from most texts. Usually, ergodicity means that only strictly invariant sets have measure 0 or 1. It turns out that these notions are equivalent (see exercise 14.18). This slight change allows us to give some interesting examples of ergodicity in Section 9.4.
Corollary 9.13. A measure preserving transformation T : X →X is ergodic with respect to a probability measure µ if and only if for every integrable function f lim n→∞ 1 n n−1 ∑ i=0 f(T i(x)) = Z X f(x)dµ for all x except possibly on a set of measure 0.
The proof of this corollary will also be given in Chapter 14. Somewhat confusingly, this last result is often also called the Birkhoff ergodic theo-rem. We will also adhere to that usage, just so that we can avoid saying “the corollary to the Birkhoff ergodic theorem” on many occasions. This corollary really says that a transformation is ergodic if and only if time av-erages equal spatial averages. This is a very important result because, as we will see, spatial averages are often much easier to compute. This has major implications in physics.
196 9. Ergodic Theory One needs to be careful, because it can happen that a transformation is ergodic with respect to two (or more) different measures.
Definition 9.14. Two probability measures µ and ν are mutually singular if there is a measurable set S with µ(S) = 1 and ν(S) = 0, and vice versa.
Corollary 9.15. If T is ergodic with respect to two distinct probability mea-sures µ and ν, then those measures are mutually singular.
Proof. If µ and ν are distinct measures, we can choose f such that c1 = Z X f dµ ̸= Z X f dν = c2 .
By Corollary 9.13, the time average ⟨f⟩(x) must be c1 for µ almost every x and so the x for which the average is c2 has µ measure 0. The reverse also holds.
■ One can furthermore prove that the set of invariant probability mea-sures is non empty and every invariant measure is a convex combination of ergodic measures [chapter 8]. This says that, in a sense, ergodic measures are the building blocks of chaotic dynamics. If we find ergodic behavior with respect to some measure µ, then we understand the statisti-cal behavior for almost all points with respect to µ. There may be other complicated behavior but this is “negligible” if you measure it with µ.
9.4. Examples of Ergodic Measures In this section, we consider the piecewise linear map T with derivative equal to 2, depicted in Figure 50. To fix our thoughts, we set A = [0,1] and B = [1,2]. In this section, we will exhibit uncountably many invari-ant probability measures µ with respect to which T is ergodic. Note that any two such measures must be mutually singular (Definition 9.14). This situation is by no means exceptional.
We start with the measure δ0 that assigns (full) measure 1 to the point 0 and measure 0 to any (measurable) set not containing 0. As we can see in Figure 50, for any set S 0 ∈S ⇐ ⇒ 0 ∈T −1(S).
9.4. Examples of Ergodic Measures 197 A B A B J Figure 50. This map has many ergodic measures Thus δ0(S) = δ0 T −1(S) , that is: δ0 is (weakly7) T-invariant. Since any T-invariant set either contains the point 0 or not, such a set trivially has measure either zero or one. By Definition 9.12, T is ergodic with respect to δ0. Let us check the conclusion of Corollary 9.13. For some very small ε > 0, set f(x) = 0 if x ∈[0,ε) α if x ∈[ε,2] Take some arbitrary x. Under iteration by T, it will most likely randomly bounce around either in the interval [0,1] or in the other interval, [1,2]. If x0 ∈(0,1), then f(T i(x0)) will nearly always be α, and if x0 ∈[1,2], it will always be α. Either way, the time-average is close to α. However, the integral R X f(x)dδ0) gives f(0) = 0! What is going on? See this footnote8.
Note that, in this example, the set {0} has pre-image {0}∪{1/2}. The second point (1/2) carries no measure. Thus {0} is invariant but not strictly invariant! Similarly, we can put a discrete measure on any q-periodic orbit by giving each point of the orbit a measure 1/q. With respect to that mea-sure, the transformation will be ergodic, because the only invariant with positive measure is the entire orbit. Again, the invariant set is not strictly invariant (see exercise 14.19).
7But not strictly!
8The set (0,2] has measure 0 with respect to δ0. Corollary 9.13 tells us to neglect such sets. Thus we must take x = 0, and then the summation also gives 0.
198 9. Ergodic Theory The next example is the uniform measure µA in A = [0,1]. Each mea-surable subset of A has a measure equal to its Lebesgue measure. It is easy to see that this is a probability measure (one that integrates to 1). From Figure 50, we see that the inverse image of an interval J ⊆A equals two disjoint intervals of half its length. This shows that µA is invariant under T.
We will show in Chapter 10 that each T-invariant set has µA measure either 0 or 1 (that is: µA is ergodic), but here is a partial result.
Proposition 9.16. Let T be the map x →2x mod 1 on A = [0,1] and suppose µ is the Lebesgue measure. If S ⊆A is a T-invariant set with µ(S) > 0, then S must be dense in A.
Proof. Note that T restricted to the interval A = [0,1] is just the doubling map. By the discussion above, the Lebesgue measure µ is T-invariant.
The inverse image T −1(S) is: S0 ∪S1 := S+0 2 ∪S+1 2 , According to Proposition 9.4 (ii), we can cover S with an open set O of measure less than µ(S) + ε for any ε > 0. Using exercise 9.4, we see that we can cover S with disjoint open intervals of measure less than µ(S) + ε.
Thus we can cover each of S0 and S1 with open intervals of length no more than half that. By Definition 9.3, µ(S0) and µ(S1) are at most 1 2µ(S). Since, however, S is invariant and so µ(S0) + µ(S1) = µ(S), we conclude that µ(S0) = µ(S1) = 1 2µ(S).
Iterating this procedure, we get T −2(S) = S00 ∪S01 ∪S10 ∪S11 := S+0 4 ∪S+2 4 ∪S+1 4 ∪S+3 4 .
Each of these contains 2−2 of the measure of S. Similarly, the nth iterate gives a collection of 2n regularly spaced copies of 2−nS. Clearly, the union of these over n is dense and each little copy must contain a set of positive measure belonging to S.
■ Note that the complement of an invariant set is also invariant. Thus result implies that if S ⊂[0,1] is an invariant set whose complement Sc has positive measure, then both are dense. This is equivalent to the following.
Corollary 9.17. Let T be the map x →2x mod 1 on A = [0,1] and suppose µ is the invariant Lebesgue measure. If T is not ergodic, then there must be 9.4. Examples of Ergodic Measures 199 an invariant S such that both it and its complement have positive measure and are dense in A.
We will see in Chapter 10 that in fact this is not possible, and so T is ergodic with respect to µA.
For now note that both A and B are T-invariant sets and µA(A) = 1 while µA(B) = 0. We check Corollary 9.13 again. Let f be f(x) = α if x ∈[ 1 2,1) 0 else For arbitrary x in [0,1], we expect T i(x) to hit the interval [0, 1 2] half the time on average. So the sum should give α 2 . Indeed, if we compute the integral R f dµA, that is what we obtain.
Now we turn to an at first sight very strange and counter-intuitive ex-ample. In the unit interval, we consider the set of x with all possible bi-nary expansions, but now we construct a measure νp that assigns a measure p ∈(0,1) to “0”, and 1 −p to “1”. In effect this amounts to assigning a measure p to the interval [0, 1 2] and 1−p to [ 1 2,1]. The interesting case is of course when p ̸= 1 2. So that is what we will assume.
Continuing the construction of the measure νp, the set of sequences starting with 00 get assigned a measure p2; the ones starting with 01, a measure p(1−p); 10, a measure (1−p)p; and 11, a measure (1−p)2. The sum of these is 1. We now keep going ad infinitum, always keeping the sum of the measures equal to 1, see Figure 51. So νp is a probability measure.
The same reasoning as in Proposition 9.16 shows that an interval I consisting of points whose binary expansion starts with a = a1a2 ···an has pre-image I0 ∪I1, where I0 consists of the points whose expansion starts with 0a and I1, those starting with 1a.
νp(I0)+νp(I1) = pνp(I)+(1−p)νp(I) = νp(I), and so the measure νp is T-invariant.
This gives us an uncountable set of T-invariant measures νp (one for each p ∈(0,1)). For each of these measures, we are in the same situation as Corollary 9.17: if νp is not ergodic, there must be very strange invariant sets. (And in fact, those measures are ergodic.
200 9. Ergodic Theory 0 1/2 1 p.p p(1−p) (1−p)p (1−p)(1−p) p 1−p Figure 51. The first two stages of the construction of the singular mea-sure νp.
9.5. The Lebesgue Decomposition The examples of invariant measures of Section 9.4 also help to illustrate the following fact which we mention without proof (but see ).
Theorem 9.18 (Lebesgue Decomposition). Let µ be a given measure. An arbitrary measure ν has a unique representation as the sum ν = νac +νd +νsc .
where νac absolutely continuous with respect to the Lebesgue measure µ, νd is a discrete measure, and νsc is singular continuous.
We now define these notions somewhat informally. A measure νac is absolutely continuous with respect to µ if for all measurable sets A, µ(A) = 0 implies that νac(A) = 0. It is usually written as νac ≪µ. The Radon-Nikodym theorem theorem then implies that νac has a non-negative, integrable density with respect to µ. This means that if νac ≪µ, we can write dνac = ρ(x)dµ (see ). The density ρ is also called the Radon-Nikodym derivative of νac (relative to µ) and it is often written as dνac dµ = ρ .
We can use the density to change variables under the integral. For any integrable f Z f(x)dνac(x) = Z f(x)ρ(x)dµ(x).
Thus ρ is the density of νac (with respect to µ). Often, µ is the Lebesgue measure so that dµ(x) = dx. This is usually the case when we think of common probability measures in statistics, such as the Beta distribution on [0,1], dν(x) = Cxa−1(1−x)b−1 dx.
9.5. The Lebesgue Decomposition 201 This is an example of a measure that is absolutely continuous with respect to the Lebesgue measure. In this case, ρ is called the probability density, and its integral is ν(x)−ν(0), the cumulative probability distribution. The constant C is needed to normalize the integral R dν = 1.
The discrete measure νd is concentrated on a finite or countable set of µ-measure zero. The measure δ0 is an example of this.
Finally, the measure νp for p ̸= 1 2 is an example of a singular continu-ous measure with respect to the Lebesgue measure µ. This is a measure that is singular with respect to µ, but, still, single element sets {x} that satisfy µ({x}) = 0 also have νp-measure zero.
Recall that Corollary 9.15 says that if p ̸= q are two numbers in [0,1], then the measures νp and νq are mutually singular, even though they are clearly continuous with respect to one another by the above informal defi-nition. Since this is maybe more than a little counter-intuitive, let us verify that again.
Lemma 9.19. Let p, q distinct numbers in [0,1]. The measures νp and νq are mutually singular.
Proof. As we saw in Section 9.4, the angle doubling transformation given by T restricted to the interval [0,1] is ergodic with respect to each of the two measures. So let f(x) = 1 on [0, 1 2] and 0 elsewhere. Birkhoff’s theorem implies that for x in a set of full νp-measure, we have lim n→∞ 1 n n−1 ∑ i=0 f(T i(x)) = Z X f(x)dνp = p.
This means that νp-almost all x land in [0, 1 2] a fraction p of the time on average. Thus the set of points that land in [0, 1 2] on average a fraction q of the time has νp measure zero. But those have full νq measure.
■ Note that the binary expansion of the νp typical (that is: in a subset having full measure) x has on average a fraction of exactly p zeroes.
202 9. Ergodic Theory 9.6. Exercises Exercise 9.1. Let S, An be sets in a space X, and I any (possibly uncount-able) index set. See Figure 52.
a) Show that (T n∈I An)c = S n∈I Ac n.
b) Show that (S n∈I An)c = T n∈I Ac n.
c) Show that every closed set C ⊆S is the complement of an open set O ⊇Sc.
d) Show that for O and S as in (c): S\Oc = O\Sc.
(Note: the first two statements are known as the De Morgan laws .) X A1 A2 S O Sc Figure 52. The left figure illustrates that (∩Ai)c = ∪i Ac i . The right figure illustrates that for an open set O containing Sc, S\Oc = O\Sc (shaded in red).
Exercise 9.2. Reformulate the counter example in Section 9.1 as a counter example in R. (Hint: two numbers in [0,1] are equivalent if their difference is rational. Let V ⊆[0,1] be a set that contains exactly one representative of each class. Let R be the set of rationals in [−1,1]. Then consider the union U := ∪r∈RV +r. Show that [0,1] ⊆U ⊆[−1,2].) Exercise 9.3. a) Show that the rational numbers in [0,1] can be contained in an open set of arbitrarily small measure. (Hint: for some λ > 1, put the number p/q in an open interval of length Cϕ(q)−1λ −q, where ϕ is the totient function.) b) Use (a) to show that the rational numbers in R an be contained in an open set of arbitrarily small measure. (Hint: in each unit interval, choose an appropriate C as defined in (a).) c) Show there is a closed set in [0,1] of measure greater than 1−ε that contains only irrational numbers.
Exercise 9.4. Show that any open set O in R is a finite or countable union of disjoint open intervals. (Hint: for every x ∈O there is an open interval (a,b) ⊆O that contains x. Now let α = inf{a : (a,b) ⊆O , x ∈(a,b)} and similar for β. This way we obtain a partitioning of O into open inter-vals. Each such interval must contain a rational number of which there are countably many.) 9.6. Exercises 203 In the next exercise, we prove the following Lemma.
Lemma 9.20. i) Any set in a probability space X with outer measure zero is Lebesgue measurable with Lebesgue measure zero.
ii) A countable union of measure 0 sets has measure 0.
Exercise 9.5. a) Show that the empty set has measure zero. (Hint: see Definition 9.5.) b) Prove part (i) of the lemma for a non empty set. (Hint: a non empty set contains a point which is a Borel set; now apply Definition 9.3.) c) Prove part (ii) of the lemma. (Hint: use equation (9.1).) Exercise 9.6. Let X = [0,1], E the set of irrational numbers in X, and µ the Lebesgue measure.
a) Show that R E dµ = 1. (Hint: approximate the Lebesgue integral as in Section 9.1.) b) Show that the Riemann integral R E dx is undefined. (Hint: look up the exact definition of Riemann integral) Exercise 9.7. Construct the middle third Cantor set C ⊆[0,1] in the follow-ing way (Figure 53). At stage 0, take out the open middle third interval of the unit interval. At stage 1, take out the open middle third interval of the two remaining intervals. At stage n, take out the open middle third interval of each of the 2n remaining intervals. The set C consists of the points that are not removed. See also exercise 1.10.
a) Show that C consists of all points x = ∑∞ i=1 ai3−i where {ai}∞ i=1 are ar-bitrary sequences in {0,2}N.
b) Show that the Lebesgue measure of C is zero.
c) Show that C is uncountable. (Hint: look at the proof of Theorem 1.24.) 0 1 1/3 2/3 1/9 2/9 7/9 8/9 Figure 53. The first two stages of the construction of the middle third Cantor set. The shaded parts are taken out.
Exercise 9.8. Construct the set C ⊆[0,1] in the same way as in exercise 9.7, but now at stage n, take out (open intervals of) an arbitrary fraction mn ∈(0,1) from the middle of each of the remaining intervals.
a) Show that C is non-empty. (Hint: find a point that is never taken out.) b) Let mi = 1−e−αi for some α ∈(0,1). Compute the Lebesgue measure of C and its complement. (Hint: at every stage, consider the length of the set that is left over. You should get that µ(C) = e−1/(1−α). ) 204 9. Ergodic Theory We remark that Cantor sets with positive measure such as those in exercise 9.8 are sometimes called fat Cantor sets.
Exercise 9.9. a) Show that the Borel sets contain the closed sets. (Hint: a closed set is the complement of an open set.) b) Show that the middle third Cantor set (see exercise 9.7) is a Borel set.
c) Show that the Cantor sets of exercise 9.8 are Borel sets.
d) Show the sets in (c) are Lebesgue measurable.
Exercise 9.10. Construct the Cantor function c : [0,1] →[0,1], also called Devil’s staircase as follows. See also exercise 9.7. See Figure 54.
Start with stage 0: c(0) = 0 and c(1) = 1. At stage 1, set c(x) = 1 2 if x ∈[ 1 3, 2 3]. At stage 2, set c(x) = 1 4 if x ∈[ 1 9, 2 9] and c(x) = 3 4 if x ∈[ 7 9, 8 9].
Use a computer program to draw 5 or more stages. c(x) is the continuous function that is the limit of this process.
1/3 2/3 1/9 2/9 7/9 8/9 1/2 1/4 3/4 0 0 1 1 Figure 54. An impression of the Cantor function of exercises 9.10 and 9.11. The first four stages are drawn in black, the red segments are affine interpolations.
9.6. Exercises 205 Exercise 9.11. Recall the definition of the Cantor function, c : [0,1] →[0,1] (exercise 9.10).
a) Use exercise 9.7 (a) to show that for x in the Cantor set x = ∞ ∑ i=1 ai3−i = ⇒ c(x) = ∞ ∑ i=1 ai 2 2−i .
b) Show that on any interval not intersecting the Cantor set c is constant.
c) Show that c : [0,1] →[0,1] is onto.
d) Show that c is non-decreasing.
e) Show that c(x), is continuous. (Hint: find a proof that a non-decreasing function from an interval onto itself is continuous.) Since c is increasing from 0 to 1, we can interpret it as a cumulative dis-tribution function. The measure µ of [a,b] ⊆[0,1] equals c(b) −c(a). If [a,b] is inside any of the flat parts, then its measure equals zero. Thus the measure of the complement of the Cantor set is zero, and all measure is concentrated on the Cantor set.
Exercise 9.12. Find the Lebesgue decomposition (Theorem 9.18) of c in exercise 9.11 interpreted as a measure. Explain!
Exercise 9.13. a) Show that the derivative c′ of the Cantor function c of exercise 9.10 equals 0 almost everywhere.
b) Show that Lebesgue integration gives R 1 0 c′(t)dt = 0. (Hint: c′(t) = 0 on a set of full measure. Then use the informal definition of Lebesgue integration in Section 9.2.) c) Conclude that in this case c(1)−c(0) = R 1 0 c′(t)dt is false.
The equation in item (c) of exercise 9.13 holds in the case where the func-tion c admits a derivative everywhere. The interested student should recall the second fundamental theorem of calculus (e.g. [Section 7.1]).
Exercise 9.14. Consider the map t : [0,1] →[0,1] given by T(x) = {10x}, the fractional part of 10x.
a) Show that the Lebesgue measure dx is invariant under T.
b) Prove Corollary 9.17 for this map.
c) Show that the frequency with T n(x) visits the interval I = [0.358,0.359) equals the frequency with which 358 occurs (if that average exists).
d) Assuming ergodicity of T, show that for Lebesgue almost every x, that average equals 10−3. (Hint: use the corollary to Birkhoff’s theorem with f(x) = 1 on I and 0 elsewhere.) 206 9. Ergodic Theory Exercise 9.15. a) Show that there exist x in whose decimal expansion the word “358” occurs more often than in almost all other numbers (see exer-cise 9.14 (d)).
b) Show that the frequency of occurrences of “358” in the decimal expan-sion of a number x does not necessarily exist.
c) What is the Lebesgue measure of of set of numbers referred to in (a) and (b). (Hint: see exercise 9.14 (d).) Exercise 9.16. In an interview, Yakov Sinai explained ergodicity as fol-lows. Suppose you live in a city above a shoe store. One day you decide you want to buy a perfect pair of shoes. Two strategies occur to you. You visit the shoe store downstairs every day until you find the perfect pair. Or you can rent a car to visit every shoe store in the city and find the best pair that way. The system is ergodic if both strategies give the same result.
Explain Sinai’s reasoning.
Exercise 9.17. a) Fix an integer b > 1 and let w be any finite word in {0,1,···b −1}N of length n. Show that for almost all x, the frequency with which that word occurs in the expansion in base b equals b−n. (Hint: assume ergodicity of x →{bx} and follow the reasoning in exercise 9.14.) b) The measure of the set of x for which that frequency is not b−n is zero.
Definition 9.21. Let b ≥2 an integer. A real number in [0,1] is called normal in base b if its infinite expansion in the base b has the property that all words of length n occur with frequency b−n. A number is called absolutely normal if the property holds for every integer b > 2.
Exercise 9.18. a) Use exercise 9.17, Corollary 1.25, and Lemma 9.20 to show that the set of words not normal in base b has measure 0.
b) Show that the set of absolutely normal numbers has full measure.
Exercise 9.19. a) Show that the set of numbers that are not normal in base b > 2 is uncountable. (Hint: words with a missing digit are a subset of these; see exercise 9.7.) b) Repeat (a), but now for base 2. (Hint: rewrite in base 4 with digits 00, 01, 10, and 11; follow (a).) Exercise 9.20. a) Show that the set of absolutely normal numbers is dense.
(Hint: follows from exercise 9.18.) b) Show that numbers with finite expansion in base b are non-normal in base b.
c) For any b > 1, show that the set of non-normal numbers in base b is also dense. (Hint: rational numbers.) 9.6. Exercises 207 Exercise 9.21. Show that a rational number is non-normal in any base.
(Hint: generalize proposition 5.8 to show that the expansion of a rational number in base b is eventually periodic.) Exercise 9.22. a) In base 2, construct a number C2 whose expansion is the list of all finite words. Start with the string s1 consisting of all length 1 words in ascending order, “0” and “1”. So s1 = 01. Then obtain s2 consist-ing of all length 2 words in ascending order. So s2 = 00011011. And so forth. The binary expansion of C2 is: 0.s1s2s2s3 ···. (Note: this number in base 2 and its generalizations to base b are usually called Champernowne numbers [Chapter 4].) b) Challengea: Show that the number Cb whose expansion in base b is the list of all finite words constructed following the method in (a) is normal in base b. (Hint: pick a word w of length n. Show that w occurs in 1 out of bn times in every “level” k ≥n.) c) Numerically compute the first 6 continued fraction convergents for C10 = 0.1234567890···. (Hint: the fourth continued fraction coefficient equals 135678.) aThough intuitively “obvious”, the details of this proof are very tricky! If you know a simple proof, let me know.
Definition 9.22. A real sequence {xi}∞ i=1 is equidistributed modulo 1 (with respect to Lebesgue measure) if its fractional values {ai}∞ i=1 are such that for each subinterval [a,b] of R/Z lim n→∞ |{a1,a2,···an}∩[a,b]| n = b−a.
In other words: the frequency of hitting a set is proportional to the Lebesgue measure of that set.
Exercise 9.23. Show that x is normal in base b > 2 (b ∈N) if and only if the sequence an = {xbn} is equidistributed modulo 1, where {·} means fractional part. (Hint: one direction is obvious; for the other direction, note that for all epsilon > 0 and any interval [c,d], there are b-adic intervals I and J such that I ⊆[c,d] ⊆J and |j\I| < ε.) As with so many issues in number theory, for any of the numbers we care about — such as e, π, √ 2, et cetera — it is not known (in 2021) whether they are normal in any base.
Exercise 9.24. a) Show that a rotation on R/Z preserves the Lebesgue measure. (Hint: Corollary 9.6 (iii).) b) Show that a rotation on R/Z by a rational number is not ergodic. (Hint: start with the identity which is a rotation by 0.) Chapter 10 Three Maps and the Real Numbers Overview. In this chapter, we consider the three maps from [0,1) to itself that are most important for our understanding of the statistical properties of real numbers. They are: multiplication by an integer n modulo 1, rotation by an irrational number, and the Gauss map that we discussed in Chapter 6.
In doing this, we review three standard techniques to establish ergodicity.
In this chapter we restrict all measures, transformations, and so on to live in one dimension ([0,1) or R/Z).
10.1. Invariant Measures If we wish to prove that a measurable transformation T : X →X is ergodic, we first need to find an invariant measure. Recall the notions of pushforward of a measure (Definition 9.8) and invariant measure (Definition 9.9).
Lemma 10.1. Let T : X →X a measurable transformation and µ a T-invariant measure on X. Then for every µ-integrable function f, we have Z f(x)dµ(x) = Z f(T(x))dµ(x).
209 210 10. Three Maps and the Real Numbers Proof. Setting y = Tx, the definition of the pushforward gives Z X f(y)dT∗µ(y) = Z X f(T(x))dµ(x).
On the other hand, since µ is invariant, we also have Z X f(y)dT∗µ(y) = Z X f(y)dµ(y).
Putting the two together gives the lemma.
■ In most cases, and certainly in this text, we are interested in invari-ant measures ν that are absolutely continuous with respect to the Lebesgue measure (see Section 9.3). Thus dν = ρ(x)dx. It is often easier to compute with densities than it is with measures. We formulate the pushforward for densities.
Lemma 10.2. The pushforward T∗ρ by T : [0,1) →[0,1) of a density ρ is given by T∗ρ(y) = ∑ Tx=y ρ(x) |T ′(x)| .
This is called the Perron-Frobenius operator.
Proof. The measure of the pushforward T∗ρ contained in the small interval dy is ˜ ρ(y)dy. By Definition 9.8, it is equal to ∑Tx=y ρ(x)dx where dx is the length of the interval T −1(dy) (see Figure 55). Now the length of T −1(dy) dy −1 −1 T (dy) T (dy) Figure 55. The inverse image of a small interval dy is T −1(dy) is of course equal to the length of d dyT −1(y) dy. Since d dyT −1(y) dy = dy |T ′(x)| , 10.1. Invariant Measures 211 the result follows.
■ Thus T preserves an absolute continuous (with respect to the Lebesgue measure) measure with density ρ if and only if ρ(y) = ∑ Tx=y ρ(x) |T ′(x)| .
(10.1) The first, and simplest, of the three transformations are the rotations. A rotation T is invertible and T ′(x) = 1. Therefore, if ρ(x) = 1, Lemma 10.2 also yields 1 for its pushforward T∗ρ, and thus equation (10.1) is satisfied.
If instead T is defined as x →τx modulo 1, where τ is any integer other than ±1 or 0, the situation is different, but still not very complicated. We will call these transformations angle multiplications for short. Now each y has |τ| inverse images {x1,···xτ} and T ′(xi) = 1 τ . So if ρ(x) = 1, Lemma 10.2 yields T∗ρ(x) = 1 for the pushforward.
The situation is slightly more complicated for the Gauss map of Defi-nition 6.1.
Proposition 10.3. i) Rotations and angle multiplying transformations on R/Z preserve the Lebesgue measure.
ii) The Gauss map preserves the probability measure dν = 1 ln2 dx 1+x .
Proof. We already proved item (i). For item (ii), notice that ν([0,x]) = 1 ln2 Z x 0 1 1+s ds = 1 ln2 ln(1+x), so ν([0,1]) = 1 and ν is as probability measure. It is easy to check that that the inverse image under T of [0,x] is the union of the intervals 1 a+x, 1 a (see 212 10. Three Maps and the Real Numbers Figure 56), and so ν(T −1([0,x])) = ν S∞ a=1 1 a+x, 1 a = 1 ln2 ∑∞ a=1 ln 1+ 1 a −ln 1+ 1 a+x = 1 ln2 ∑∞ a=1 ln a+1 a −ln a+1+x a+x = 1 ln2 ∑∞ a=1 ln a+x a −ln a+1+x a+1 = 1 ln2 ln(1+x) .
The last equality follows because the sum telescopes.
0 0 1 1 1/2 1/3 x Figure 56. The interval [0,x) (shaded red) and its pre-image under the Gauss map (shaded blue).
This computation shows that the measure on intervals of the form [0,x] or (0,x) is invariant. Taking a difference, we see that the measure an any interval (x,y) is invariant. Therefore, the same is true for any open set (see 9.4). Thus it is true any Borel set (Definition 9.1). Since Lebesgue measurable sets can be approximated by Borel sets (Proposition 9.4), the result follows.
■ At the end of this last proof, we needed to jump through some hoops to get from the invariance of the measure of simple intervals to that of all Borel sets. This can be avoided if we prove the invariance of the density directly via equation (10.1). But to do that, you first need to know a tricky sum, see exercise 10.5.
10.2. The Lebesgue Density Theorem 213 With the invariant measures in hand, we can now turn to proving the ergodicity of the three maps starring in this chapter.
10.2. The Lebesgue Density Theorem Proposition 10.4. Given a measurable set E ⊆[0,1] with µ(E) > 0, then for all ε > 0 there is an interval I such that µ(E ∩I) µ(I) > 1−ε .
We will say that the density of E in I is greater than 1−ε.
Proof. By Proposition 9.4, there are open sets On containing E such that µ(On\E) = δn, where δn tends to 0 as n tends to infinity. Using property (4) of Corollary 9.6, we see that µ(On) = µ(On\E)+ µ(E) = µ(E)+δn .
(10.2) According to exercise 9.4, for each n, there is a collection of disjoint open intervals {In,i} such that On = ∪i In,i .
Now fix an ε > 0 and suppose that µ(E ∩I) ≤(1−ε)µ(I) for all intervals.
In particular this holds for those intervals belonging to the collection of intervals {In,i}. So for any n, we have µ(E ∩On) = µ (E ∩(∪iIn,i)) = ∑ i µ(E ∩In,i) ≤∑ i (1−ε)µ(In,i).
The middle equality follows again from property (4) of Corollary 9.6. No-tice that the left-hand side equals µ(E), since On contains E, and the right-hand side equals (1 −ε)µ(On) by definition of the intervals In,i. Together with equation (10.2), this gives µ(E) ≤(1−ε)µ(On) = (1−ε)(µ(E)+δn).
If n tends to infinity, δn tends to 0, and thus µ(E) must be 0.
■ This is a weak version of a much better theorem. We do not actually need the stronger version, but its statement is so much nicer, it is probably best to remember it and not the proposition. A proof can be found in .
214 10. Three Maps and the Real Numbers Theorem 10.5 (Lebesgue Density Theorem). If E is a measurable set in Rn with µ(E) > 0, then for almost all x ∈E lim ε→0 µ(E ∩Bε(x)) µ(Bε(x)) = 1, where Bε(x) := {y ∈Rn : |y−x| < ε}, the open ε ball centered at x. That is: this holds for all x in E, except possibly for a set of µ measure 0.
10.3. Rotations and Multiplications on R/Z In this section, we will invoke the Lebesgue density theorem, to prove the ergodicity of multiplications by τ ∈{±2,±3,···} modulo 1 and transla-tions by an irrational number ω modulo 1 on R/Z. In each case, however, Proposition 10.4 is sufficient. We denote the Lebesgue measure by µ.
Lemma 10.6. Every orbit of an irrational rotation Rr is dense in R/Z.
Proof. We want to show that for all x and y, the interval [y−δ,y+δ] con-tains a point of the orbit starting at x. Denote by pn qn the continued fraction convergents of r (of Definition 6.4). By Lemma 6.13 lim n→∞x+qnr −pn = lim n→∞x+dn = x.
Fix n large enough enough, so that the distance (on the circle) between x and x + qnr −pn is less than δ. Then the points xi := x + iqnr modulo 1 advance (or recede) by less than δ. And thus at least one must land in the stipulated interval (see Figure 57).
■ x 0 1 x+qr x+2qr x+r x+r+qr Figure 57. r is irrational and p q is a convergent of r. Then x+qr modulo 1 is close to x. Thus adding qr modulo 1 amounts to a translation by a small distance.
Theorem 10.7. Irrational rotations modulo 1 are ergodic with respect to the Lebesgue measure.
10.3. Rotations and Multiplications on R/Z 215 Proof. By Proposition 10.3, the Lebesgue measure is invariant.
Suppose the conclusion of the theorem is false. Then there is an invari-ant set A such that both it and its complement Ac — which is also invariant — have strictly positive measure. By Proposition 10.4, for every ε there are intervals I and J where A, respectively Ac, have density greater than 1−ε.
Suppose that the length ℓ(I) of I is less than ℓ(J). Then there is an n ≥1 so that nℓ(I) ≤ℓ(J) < (n+1)ℓ(I).
By Lemma 10.6, there is i such that R−i ω (I) falls in the first 1 n-fraction of J, another one in the second, and so forth (see Figure 58). In all cases, this 0 1 I J Figure 58. ℓ(I) is between 1 3 and 1 2 of ℓ(J). So there are two disjoint images of I under R−1 ω that fall in J.
means that at least half of J is covered by images of I. By invariance, the images of I have A density greater than 1−ε. That means that A has density at least 1 2(1−ε) in J, which is a contradiction. The case where ℓ(I) = ℓ(J) is easy: split I into 2 equal intervals; at least one of these must have density greater than 1−ε.
■ In the proof of the next theorem, we employ the same strategy as in the proof of Proposition 9.16 and Corollary 9.17. But this time, the Lebesgue density theorem helps us to get a much stronger result. We wish to prove that angle multiplications are ergodic. But it turns out that with the same effort we can prove the result for a larger class of maps. An example of such a map is given in Figure 59.
Definition 10.8. Let {Ii} be a finite or countable partition of [0,1] of in-tervals of positive length ℓi so that ∑i ℓi = 1. On each interval Ii, define fi : Ii →[0,1] to be an affine map onto [0,1]. T = ∪i fi is called a complete affine interval map .
To make the exposition a little clearer, let us first define nth level in-tervals. These are the domains I0 where T n : I0 →[0,1) is a bijection. For 216 10. Three Maps and the Real Numbers 0 0 1 1 Figure 59. An example of the transformation T : [0,1) →[0,1) de-scribed in Theorem 10.9.
example, the 1st level intervals of the Gauss map are (1/(a+1),1/a] for a ∈ N; the third level intervals for the angle doubling map are [k/8,(k + 1)/8) for k in {0,···7}.
Theorem 10.9. Complete, affine interval maps preserve the Lebesgue mea-sure and are ergodic with respect to that measure.
Proof. By hypothesis we have |f ′ i | = 1/ℓi and so ∑i |f ′ i |−1 i = ∑i ℓi = 1 and so the Perron-Frobenius equation (10.1) immediately implies that the Lebesgue measure is preserved.
Suppose that the set A is T-invariant and has positive µ measure and let J be an arbitrary interval. We will prove that µ(A∩J) = µ(a)· µ(J) or µ(Ac ∩J) = (1−µ(A))µ(J).
(10.3) This implies that Ac has measure 0, otherwise a contradiction with the Lebesgue density would result. And so µ(A) = 1, and thus T is ergodic.
Let b : I0 →[0,1) be an arbitrary branch of T n. Since b is affine b−1(A) occupies the same proportion in I0 as A does in [0,1). See Figure 60 All of the |f ′ i | are bounded away from 1, say, larger than d > 1. It follows that the length of any nth level interval is at most d−n. Thus we can approximate J arbitrarily well with a collection of nth level intervals, provided n is large enough. Since in each nth level interval, the inverse image of A occupies a fraction of µ(A) of that interval, we have lim n→∞µ(T −n(A)∩J) = µ(A)· µ(J).
(10.4) But since A is invariant, this proves (10.3), and thus the theorem.
■ 10.3. Rotations and Multiplications on R/Z 217 0 0 1 1 I0 Figure 60. The pre-image in the nth level interval of A (in red) is the blue interval. This branch of T n is affine and so there is no distortion.
As a consequence, the proportions that A and its pre-image occupy are the same.
For completeness, we record the result in the important special case of angle multiplications.
Corollary 10.10. Multiplication by τ ∈Z with |τ| > 1 modulo 1 is ergodic.
The results in this section have interesting consequences. One of the most important ones is the following.
Corollary 10.11. Suppose T : [0,1) →[0,1) is ergodic with respect to Lebesgue measure, then T i(x) is equidistributed (see Definition 9.22) mod-ulo 1 for almost every x.
Proof. Define f as f(x) = 1 if x is in the interval [a,b] and 0 elsewhere. By Corollary 9.13 to f, for almost all x lim n→∞ 1 n n−1 ∑ i=0 f(T i(x)) = Z X f(x)dx = b−a.
■ In particular, this applies to both rotations by irrational angles and an-gle multiplications, as well as the maps defined in Theorem 10.9. In the case where T is an irrational rotation, the sequences { f(T i(x))}∞ i=1 and { f(T i(x′))}∞ i=1 differ only by a translation on the circle. So if one is equidis-tributed, then the other must be too. So in this case, the result holds for all x.
218 10. Three Maps and the Real Numbers The principal consequence of the ergodicity of angle multiplication is the absolute normality (see Definition 9.21) of almost all numbers. This was discussed at length in the exercises of Chapter 9.
10.4. The Return of the Gauss Map Our next aim is to show that the Gauss map T of Definition 6.1 is ergodic.
Thanks to Proposition 10.3, we know the invariant measure. It might seem that Theorem 10.9 proves the rest. It almost does! The only problem is that the that the branches of the Gauss map are not affine. Here is what the problem with that is.
We suppose again that A ⊆[0,1) is an invariant set of positive µ mea-sure. Just as before, we are given an arbitrary interval J and we want to prove that the density of Ac (A’s complement) cannot be very close to 1.
Then, by the Lebesgue density theorem, we can conclude that µ(Ac) = 0.
This proves that T is ergodic.
0 0 1 1 I0 Figure 61. The pre-image in the nth level interval of A (in red) is the blue interval. This branch of T n is far from affine and so the distortion is large. As a consequence, the proportions that A and its pre-image occupy are very different.
We need to make sure of two things. The first is that the lengths of the nth level intervals tend to zero, so that we can cover J efficiently with them. This is relatively easy. The second task is more subtle. The inverse image of A in any nth level interval have density bounded away from 0 to prevent its complement Ac from having density close to 1 in I0. What is the problem? That branch is not affine. Especially for large n, it might have a much bigger derivative in A ∩I0 than it does in Ac ∩I (see Figure 10.4. The Return of the Gauss Map 219 61). This would distort the image under T n in such a way that µ(A∩I0) µ(I0) is much smaller than µ(A), and could even tend to zero as n grows. The solution to this dilemma lies in controlling that distortion. If we can prove that for that particular branch ∂T n(x0) ∂T n(y0) for x0 and y0 in I0, is contained in, say, the interval [1/K,K], K > 0, independent of n, then the argument of the proof of Theorem 10.9 gives that the densities of A on [0,1) and of A in I0 cannot differ too much. This ensures that µ(A∩I0) µ(I0) is bounded away from 0 independently of n (the level).
The exposition in the remainder of this section and the next closely follows .
Definition 10.12. Let I0 be an interval. The distortion D of T n on that interval is defined as D := sup x0,y0∈I0 ln ∂T n(x0) ∂T n(y0) .
Here, ∂stands for the derivative with respect to x.
Proposition 10.13. Let T be the Gauss map. The distortion of T n on any nth level interval I0 is uniformly bounded in n.
Proof. Denote the forward images of I0 by I1, I2, et cetera. Similarly for x0 and y0. Set In = [0,1]. The chain rule gives ∂T n(x0) = ∂T(x0)·∂T(x1)···∂T(xn−1).
Substitute this into the definition of the distortion to get D ≤ n−1 ∑ i=0 sup xi,yi∈Ii |ln|∂T(xi)|−ln|∂T(yi)|| .
(This is an inequality due to the fact that we take the supremum over all xi and yi instead of just the initial points x0 and y0 in the definition.) By the mean value theorem, there is zi ∈Ii such that the right-hand side of this expression equals n−1 ∑ i=0 |∂ln|∂T(zi)||·|yi −xi| ≤ n−1 ∑ i=0 sup zi∈Ii |∂ln|∂T(zi)||·|Ii|.
Now we note that ∂ln|∂T| equals ∂2T ∂T . Furthermore, the mean value the-orem (once again) gives |Ii| = |Ii+1| |∂T(ui)| for some ui ∈Ii. Substituting this into 220 10. Three Maps and the Real Numbers the last equation, we get D ≤ n−1 ∑ i=0 sup zi,ui∈Ii ∂2T(zi) ∂T(zi) ∂T(ui) ·|Ii+1|.
(10.5) We need to estimate the expression in the right-hand side. Recall that we are analyzing a single branch of T n. That implies that each interval Ii lies in one of the basic — or first level — intervals ( 1 ai+1, 1 ai ] depicted in figure 25, where ai is the continued fraction coefficient associated with that particular branch. Since ∂T = −x−2 and ∂2T = 1 2x−3, for that particular branch, we have for x ∈( 1 ai+1, 1 ai ] ∂2T(zi) ∂T(zi) ≤2(ai +1) and 1 ∂T(ui) ≤1 a2 i .
Next we estimate the length on nth level interval |I|. In figure 25, one can see that the only place where |∂T(x)| is small is when x is close to 1. These points are then mapped by T to a neighborhood of zero where they pick up a large derivative. It follows that the derivative of T 2 is positive and bounded by some d > 1 and thus the length of the intervals In−i decays as Kd−i/2.
Putting this together, we see that (10.5) gives D ≤ n−1 ∑ i=0 2(ai +1) a2 i Kd(i+1−n)/2 .
Since ai ∈N, this tells us that the expression in (10.5) is uniformly bounded in n.
■ As explained in the introduction to this section, our main result follows immediately.
Corollary 10.14. The Gauss map is ergodic with respect to dν = 1 ln2 dx 1+x.
10.5. Number Theoretic Implications Finally, it is pay-back time! We have seen some rewards for our efforts to understand ergodic theory in terms of understanding normality in the exercises of Chapter 9 (Definition 9.21). But the biggest pay-off is in un-derstanding some basic properties of the continued fraction expansion of “typical” real numbers. That is what we do in this section.
10.5. Number Theoretic Implications 221 In this section, T denotes the the Gauss transformation and ν its in-variant measure (see Proposition 10.3) while µ will denote the Lebesgue measure. Note that a set has µ measure zero if and only if it has ν mea-sure zero (exercise 10.1). For the continued fraction coefficients an and the continued fraction convergents pn qn , see Definition 6.4.
We start with a remarkable result that says that the arithmetic (usual) mean of the continued fraction coefficients diverges (item (i)) for almost all numbers, but their geometric mean is almost always converges (item (ii)).
Theorem 10.15. For almost all numbers x, the continued fraction coeffi-cients an = an(x) satisfy: i) limn→∞ a1 +...+an n = ∞and ii) limn→∞(a1 ·...·an)1/n = ∏∞ a=1 1− 1 (a+1)2 −log2 a < ∞.
This last constant is approximately equal to 2.86542··· is called Khinchin’s constant.
Proof. i) Define fk : [0,1] →N by For a ∈{1,···k} : fk(x) = a if x ∈ 1 a+1, 1 a fk(x) = 0 elsewhere .
Denote the pointwise limit by f∞. We really want to use Corollary 9.13 to show that the “time average” lim n→∞ a1 +...+an n = lim n→∞ 1 n n−1 ∑ i=0 f∞(T i(x)) is unbounded. But f∞is not integrable and so cannot be used. However the sum on the left hand side is bounded from below by the right-hand side if we replace f∞by fk (which is integrable). Proposition 10.3 and Corollary 9.13 say that the time average of fk equals 1 ln2 Z 1 0 fk(x) 1+x dx = 1 ln2 k ∑ a=1 Z 1 a 1 a+1 a 1+x dx (10.6) The integral of 1/(1+x) is of course ln(1+x) and so the above gives 1 ln2 k ∑ a=1 a ln a+1 a −ln a+2 a+1 .
(10.7) 222 10. Three Maps and the Real Numbers This sum telescopes and the student should verify (see exercise 10.12) that this gives 1 ln2 ln(k +1)−kln 1+ 1 k +1 , (10.8) which diverges as k →∞and proves the first statement.
ii) This proof is very similar to that of (i), except that now we want to compute the “time average” lim n→∞ lna1 +...+lnan n .
The exponential of this will give us the result we need. So this time, we define For a ∈N : g∞(x) = lna if x ∈ 1 a+1, 1 a .
(10.9) This time around, g∞is ν-integrable (as we will see below) and we get 1 ln2 Z 1 0 g∞(x) 1+x dx = ∞ ∑ a=1 lna ln2 ln a+1 a −ln a+2 a+1 .
(10.10) (Note that lna ln2 = log2 a.) Since we can write ln a+1 a −ln a+2 a+1 = −ln 1− 1 (a+1)2 , (10.11) we finally get the result (as well as the assertion that g∞is ν-integrable) by taking the exponential of the sum in (10.10). See exercise 10.13.
■ An example of a sequence {an}∞ n=1 that has a diverging running average but whose running geometric average converges, is given by an = 1, except when n = 22k we set a22k = 22k. For n = 22k, we have a1 +...+an n > an n = 22k−2k , which clearly diverges as k →∞. Meanwhile, the geometric average at that point is (after taking the logarithm and noting that ln1 = 0): lna1 +...+lnan n = ∑n j=1 ln22k 22k = 2k ln2 22k .
The latter converges to 0, which makes the geometric average 1.
10.5. Number Theoretic Implications 223 Theorem 10.16. For almost all numbers x, the convergents pn(x)/qn(x) satisfy i) limn→∞ lnqn n = π2 12 ln2, and ii) limn→∞1 n ln x−pn qn = −π2 6 ln2.
Remark 10.17. The constant π2 12 ln2 ≈1.1866··· is called L´ evy’s constant.
Proof. Item (ii) follows very easily from (i), see exercise 10.17. So here we will prove only (i).
To simplify notation in this proof, we will write xi := T i(x0) where T is the Gauss map. For the nth approximant of x0 ∈(0,1), see Definition 6.4, we will write pn(x) qn(x). From that same definition, we conclude pn(x0) qn(x0) = 1 a1(x0)+ pn−1(x1)/qn−1(x1) = qn−1(x1) a1(x0)qn−1(x1)+ pn−1(x1) .
See also exercise 10.2 (a). By Corollary 6.8 (ii)), gcd(pn,qn) = 1, and so using exercise 10.2 (b), we see that pn(x0) equals qn−1(x1). More generally, we have by the same reasoning pn(xj) = qn−1(xj+1).
(10.12) This implies that pn(x0) qn(x0) · pn−1(x1) qn−1(x1) · pn−2(x2) qn−2(x2) ··· p1(xn−1) q1(xn−1) = 1 qn(x0) , since p1 = 1 by Theorem 6.6. Now we take the logarithm of the last equa-tion. This yields −1 n lnqn(x0) = 1 n n−1 ∑ i=0 lnxi −1 n n−1 ∑ i=0 lnxi −ln pn−i(xi) qn−i(xi) .
(10.13) Two more steps are required. The first is showing that the last average on the right side of (10.13) tends to zero. This not difficult, because n−1 ∑ i=0 ln qn−i(xi)xi pn−i(xi) = n−1 ∑ i=0 ln 1+ qn−i(xi)xi −pn−i(xi) pn−i(xi) .
Corollary 6.7 or, more precisely, exercise 6.12 yields that |qn−i(xi)xi −pn−i(xi)| pn−i(xi) < 1 pn−i(xi)qn−i+1(xi) < 2−(n−i) √ 2 , 224 10. Three Maps and the Real Numbers where the last inequality follows from Corollary 6.8 (i). The fact that for small x, ln(1+x) ≈x concludes the first step (see also exercise 10.10).
Since the second term on the right side of (10.13) tends to zero and xi = T i(x0), we take a limit to get lim n→∞−1 n lnqn(x0) = lim n→∞ 1 n n−1 ∑ i=0 lnT i(x0).
The second step is then to compute the right-hand side of this expression.
Naturally, the ergodicity of the Gauss map invites us to employ Birkhoff’s theorem in the guise of Corollary 9.13 with f(x) set equal to ln(x).
1 n n−1 ∑ i=0 lnT i(x0) = Z 1 0 lnx (1+x)ln2 dx.
The integral is evaluated in exercise 10.16.
■ 10.6. Exercises Exercise 10.1. a) Show that for a measurable set A: µ(A) = 0 (Lebesgue measure) if and only if ν(A) = 0 (invariant measure of the Gauss map).
(Hint: using Lebesgue integrals, write ν(A) = 1 ln2 R A (1+x)−1 dx.) b) Let ν be absolutely continuous with respect to the Lebesgue measure µ.
Show that if a set has full µ measure then it has full ν measure. (Hint: see comments after Definition 9.14.) Exercise 10.2. To reacquaint ourselves with continued fractions, consider x0 = 1 a1 + 1 a2 + 1 a3 +··· def ≡[a1,a2,a3,···] .
(10.14) a) Show that ⌊x−1 0 ⌋= a1 and that T(x0) = x−1 0 −a1 = 1 a2 + 1 a3 + 1 a4 +··· def ≡[a2,a3,···] .
b) Show that if gcd(p,q) = 1, then gcd(p+aq,q) = 1. (Hint: use Lemma 2.5.) c) Prove equation (10.12).
10.6. Exercises 225 Exercise 10.3. a) Show that every probability density ρ on R/Z gives rise to an invariant measure under the identity.
b) What are the absolutely continuous measures — i.e. with a density, see Section 9.5 — that are invariant under rotation by 1/2? (Hint: consider densities with period 1/2.) c) The same for rotation by p/q for p and q in N.
d) Show that the uniform density — with density ρ(x) = 1 — is invariant under x →nx modulo 1 (where n ∈N).
Exercise 10.4. Let R0 be identity on R/Z.
a) Show that for any x, the delta measure δx is an invariant measure for R0, and that R0 is ergodic with respect to that measure.
b) Show that for any of the invariant measures in exercise 10.3 (a), R0 is not ergodic.
c) Show that R0 is not ergodic with respect to any of the measures of exer-cise 10.3 (c).
Exercise 10.5. a) Show that ∑∞ i=1 1 (y+i)(y+i+1) = 1 y+1 for all y ∈R except the negative integers. (Hint: use partial fractions, then note that the result-ing sum telescopes.) b) Show that the inverse images of y ∈[0,1) under the Gauss map T are ∪i∈N{1/(y+i).
c) Let ρ(y) = c/(1+y) a density and compute ρ and |T ′| at these inverse images.
d) Use (a), (b), and (c) to prove directly via equation (10.1) that the Gauss map preserves the measure of Proposition 10.3. Do not use the computa-tion in the proof of that proposition.
Exercise 10.6. Show that ρ(x) = 1 is the only continuous invariant prob-ability density of an irrational rotation R. (Hint: if ρ is invariant under R, it must be invariant under Ri for all positive i. Use equation (10.1) and Lemma 10.6.) Exercise 10.7. a) Show that ρ(x) = 1 is the only continuous invariant den-sity for the angle doubling map. (Hint: as in exercise 10.6: use that inverse images of a point are dense, see Proposition 9.16.) b) Check that the same is true for the map x →τx modulo 1 where τ ∈Z and |τ| > 1.
226 10. Three Maps and the Real Numbers Exercise 10.8. The orbit of any irrational rotation is uniformly distributed.
So why do we encounter specifically the golden mean in phyllotaxis — the placement of leaves? Research this and try to include illustrations. (Hint: the wikipedia page on “golden ratio” is a good start. You can find even more information in . The slightly tongue-in-cheek paper shows that the golden mean does indeed show up in the most unexpected places.
Words of caution on the ‘cult of the golden ratio’ can be found in .) Exercises 10.9 and 10.10 discuss some very useful properties of the loga-rithm for later reference. In fact, they are useful in a much wider context than discussed here. For instance, exercise 10.9 comes up in any discus-sion of entropy or in deciding the stability of Lotka-Volterra dynamical systems . Exercise 10.10 is important for deciding the convergence of products of the form ∏(1+xi).
Exercise 10.9. a) Show that x > −1 = ⇒ ln(1+x) ≤x.
with equality iff x = 0. (Hint: draw the graphs of ln(1+x) and x.) b) Let pi and qi positive and ∑i pi = ∑i qi < ∞. Use (a) to show that −∑ i pi ln pi ≤−∑ i pi lnqi .
(Hint: −∑i pi(ln pi −lnqi) = ∑i pi ln qi pi ≤∑i(qi −pi) by (a).) c) Let Sn be the open n-dimensional simplex pi > 0 and ∑n i=1 pi = 1. Use (b) to show that −∑ i pi ln pi ≤lnn.
(Hint: set qi = 1/n in (b).) d) Show h : Sn →R given by h(p) = −∑i pi ln pi has a unique global max-imum at pi = 1 n. (Hint: The constraint is C(p) := ∑i pi = 1. Deduce that at the maximum, the gradients of h and C with respect to p = (p1,··· , pn) must be parallel. If in doubt, look up Lagrange multiplier in, for example, .) In the next exercise, we prove this lemma.
Lemma 10.18. Suppose that xn > −1 and limn→∞xn = 0. Then ∑n ln(1+ xn) converges absolutely if and only if ∑n xn converges absolutely. Also ∑n ln(1+xn) diverges absolutely if and only if ∑n xn diverges absolutely.
10.6. Exercises 227 Exercise 10.10. a) Show that limx→0 ln(1+x)−x x2 = −1 2. (Hint: use L’Hˆ opital twice.) b) From (a), conclude that if xn > −1 and limn→0 xn = 0, then ∃b1 > 0 such that for all n large enough |ln(1+xn)| ≤b1|xn|. (Hint: use the direct comparison test.) c) From (a), conclude that if xn > −1 and limn→0 xn = 0, then ∃b2 > 0 such that for all n large enough |xn| ≤b2 |ln(1+xn)|.
d) Show that (b) and (c) imply Lemma 10.18.
The next four exercises provide some computational details of the proof of Theorem 10.15.
Exercise 10.11. Compute the frequency with which the coefficient an(x) = a occurs in the continued fraction expansion of almost all x. (Hint: set f(x) = 1 on (1/(1+a),1/a]. Then use Birkhoff.) Exercise 10.12. a) Show that the right-hand side of (10.6) gives (10.7).
b) Show that (10.7) gives (10.8). (Hint: write out the first few terms explic-itly.) c) Use exercise 10.9 (a) to bound the second term of (10.8).
d) Conclude that (10.8) is unbounded. (Hint: (1+1/k)k →e.) Exercise 10.13. a) Show the equality in (10.10) holds.
b) Show the equality in (10.11) holds.
c) Show that (10.11) implies part (ii) of Theorem 10.15.
Exercise 10.14. a) Show that instead of (10.11), we also have ln a+1 a −ln a+2 a+1 = ln 1+ 1 a2 +2a .
b) Use exercise 10.9 (a) to show that ln 1+ 1 a2 +2a ≤1 a2 .
c) Use (a) and (b) and equation (10.10) to show that 1 ln2 Z 1 0 g∞(x) 1+x dx ≤ 1 ln2 ∞ ∑ a=1 lna a2 .
(Hint: indeed, this is equivalent to the fact that g∞is integrable. Can you explain that?) d) Show that (c) implies that Khinchin’s constant is bounded. (Hint: find the maximum of lna−2√a. Then use Figure 11.) Exercise 10.15. Use exercise 10.14 (a) to show that Khinchin’s constant equals ∏∞ a=1 1+ 1 (a2+2a)2 log2 a .
228 10. Three Maps and the Real Numbers Figure 62. Plot of the function ln(x)ln(1+x) .
Exercise 10.16. a) Show that limx→0 ln(x)ln(1+x) = 0 (Figure 62). (Hint: for the limit as x →0, first use exercise 10.10 (b) and then substitute x = ey.) b) Use (a) to show that I := R 1 0 lnx (1+x) dx = − R 1 0 ln(1+x) x dx. (Hint: integra-tion by parts.) c) Show that for |x| < 1, ln(1+x) = ∑∞ i=1 (−1)n+1xn n .
d) Substitute (c) into I and integrate term by term to get I = ∑∞ n=1 (−1)nn−2.
e) The sum in (d) equals −π2 12 . Show that that gives the result advertised in Theorem 10.16. (Observation: we sure took the cowardly way out in this last step; to really work out that last sum from first principles is elementary but very laborious. The interested student should look this up on the web.) In exercise 10.16, note the curious fact that ∑∞ n=1 (−1)nn−2 = −π2 12 while from exercise 2.26 we have that ζ(2) = ∑∞ n=1 n−2 = π2 6 .
Exercise 10.17. a) Show that lim n→∞ lnqn+1 n = lim n→∞ lnqn+1 n+1 n+1 n = lim n→∞ lnqn+1 n+1 = lim n→∞ lnqn n .
b) Use (a), exercise 6.12, and Theorem 10.16 (i) to show that for almost all x ∈[0,1] lim n→∞ 1 n ln x−pn qn = −π2 6ln2 .
c) What do you in (b) get if x is rational? Is that a problem?
10.6. Exercises 229 Exercise 10.18. a) Use Corollary 6.7 to show that ln x−pn qn < lnq−2 n b) Use Theorem 10.16 to show that lim n→∞ 1 n ln x−pn qn = lim n→∞ 1 n lnq−2 n .
Exercise 10.19. a) What the equivalent of Theorem 10.16 for decimal ex-pansions of irrational numbers? (Note: I haven’t worked this out myself yet.) b) Use (a) to compare decimal approximation with continued fraction ap-proximation.
Definition 10.19. Given a one dimensional smooth map T : [0,1] →[0,1], the Lyapunov exponent λ(x) at a point x is given by λ(x) := lim n→∞ 1 n ln|∂T n(x)|, assuming that the limit exists. Here ∂T n(x) stands for d dx (T n)| evaluated at x.
Remark 10.20. Although we do not need it here, we remark that the gen-eralization to higher dimension of this notion is not completely straightfor-ward .
Exercise 10.20. Suppose that λ(x) = λ0, a constant. What does Definition 10.19 tell you about how fast |T ny−T nx| if y is very close to x?
Exercise 10.21. a) Let T be the Gauss map and µ its invariant measure.
Show that the Lyapunov exponent at x satisfies λ(x) = lim n→∞ 1 n n−1 ∑ j=0 ln ∂T(T j(x)) .
(Hint: think chain rule.) b) Show that Birkhoff’s theorem (Corollary 9.13) implies that for almost all x ∈[0,1] λ(x) = Z 1 0 −2lnx ln2(1+x) dx.
c) Use the last part of the proof of Theorem 10.16 and exercise 10.16 to show that for almost all x, the Lyapunov exponent equals π2 6ln2 ≈2.3731.
230 10. Three Maps and the Real Numbers Exercise 10.22. a) See exercise 10.21.
Let T be the Gauss map and x = [n,n,···]. Determine the Lyapunov exponent at x. (Hint: see also exercise 6.2.) b) Why does it not contradict Birkhoff’s theorem that these exponents dif-ferent from the one computed in exercise 10.21?
Exercise 10.23. Let T be the map given in Theorem 10.9. We will assume that T has n branches.
a) Show that for almost all points x, the Lyapunov exponent is given by λ(x) = −∑i ℓi lnℓi. (Hint: see also exercise 10.21.) b) Show that λ(x) > 0.
c) Show that λ(x) ≤lnn (Hint: exercise 10.9 (c).) d) Show that λ(x) = lnn if and only if the slopes of all branches have the same absolute value. (Hint: exercise 10.9 (d).) Exercise 10.24. a) Show that if k ∈N is such that log10 k is rational, then k = 10r, r ∈N. (Hint: use prime factorization on kq = 10p.) b) From now on, suppose that log10 k is irrational. Show that T : x → x+log10 k modulo 1 is ergodic with respect to the Lebesgue measure.
c) Let f(x) = 1 when x ∈[log10 7,log10 8] and 0 elsewhere.
Compute limn→∞1 n ∑n i=0 f(T i(x)).
d) Explain how often, for almost every x, 1 through 9 occur in {kix}∞ i=0 as first digits.
e) How often does any combination of any 2 successive digits, say 36, occur as first digits?
Stock prices undergo multiplicative corrections, that is: each day their price is multiplied by a factor like 0.99 or 1.01. On the basis of the previous prob-lem, it seems reasonable that the distribution of their first digits satisfies the logarithmic distribution of exercise 10.24. In fact, a much wider range of real world data satisfies this distribution than this “multiplicative” explana-tion would suggest. This phenomenon is called Benford’s law and appears to be only partially understood .
Definition 10.21. A map T that preserves the measure µ is called mixing if lim n→∞µ(T −n(A)∩B) = µ(A)· µ(B) for any two measurable sets A and B.
We prove the following result in exercise 10.25 Proposition 10.22. If T is mixing with an invariant probability measure, then it is ergodic.
10.6. Exercises 231 Exercise 10.25. a) Prove Proposition 10.22. (Hint: let A be the invariant set and choose B = A in Definition 10.21; get a quadratic equation for µ(A)) b) Show that ergodic does not imply mixing. (Hint: irrational rotation.) c) Show that the transformation T in Theorem 10.9 is mixing. (Hint: a measurable set in [0,1) can be approximated by finitely many open inter-vals, then see equation (10.4).) Part 3 Topics in Number Theory Chapter 11 The Cauchy Integral Formula Overview. Again, we need to venture very far, apparently, from number theory to make progress. In the mid 19th century, the main insight in num-ber theory came from Riemann, who realized that the distribution of the primes was intimately connected to the properties of the (analytic continu-ation of the) Riemann zeta function to the complex plane. In this chapter, we develop the necessary complex analysis tools — essentially the Cauchy integral formula — to study the convergence of a certain improper integral (Theorem 11.18), which is the key to the proof of the prime number theo-rem in the next chapter (Theorem 12.15). For more detailed introductions to complex analysis, we refer to [3,26,45].
11.1. Analyticity versus Isolated Singularities Definition 11.1. A set is open if it contains no points lying on its boundary and connected if it is not the disjoint union of two non-empty open sets. A domain or region is an open, connected set in C.
An excellent source for information on topological notions such as connect-edness is .
235 236 11. The Cauchy Integral Formula Definition 11.2. Let A be a domain. A function f : A →C is analytic at z0 if f ′(z) := lim δ→0 f(z0 +δ)−f(z0) δ exists in a neighborhood of z0. The word holomorphic is synonymous with analytic. If f is analytic on all of C, it is also called an entire function.
We will use the fact that this says that analyticity is an open condition.
Corollary 11.3. If f is analytic at z0, then it is analytic in an open neigh-borhood of z0.
This creates, as it were, a loophole which will be crucial in the proof of Theorem 11.18. Suppose we know f is analytic some closed set S. Then in fact, f must be analytic in some open set containing S. Because if not, there must be a sequence of points zi converging to z ∈S where f is not analytic.
Then at z, f would not be analytic! See Figure 63.
z {z } i S Figure 63. If f is analytic on the closed set S, then f must be analytic on some open set containing S.
Naturally, not all functions are analytic everywhere. What happens at or near a point z where f is not analytic? We say that f is singular at such a point z0. If there is a neighborhood1 in which it is the only singularity, we call it an isolated singularity. One can prove that every isolated singularity is one of the ones defined below.
Definition 11.4. These are the types of isolated singularities at z0.
i) f may have a removable singularity. In this case, f(z0) can be given a value in such a way that f becomes analytic at z0. An example is sin(z−z0) z−z0 .
1An open neighborhood of z0 minus the point z0 itself is often called a punctured neighborhood of z0.
11.1. Analyticity versus Isolated Singularities 237 ii) f may have a pole of order k ∈N . An example of this is ak(z−z0)−k + ak−1(z−z0)−(k−1) with ak ̸= 0. A pole of order 1 is called a simple pole.
iii) f may have an essential singularity. This is a pole of “infinite order”.
An example is e1/(z−z0). (Expand as eu and substitute u = (z−z0)−1.) One might be tempted to say that the example in item (ii) above consists of two singularities, one of order k and one of order k−1. However, we have ak (z−z0)k + ak−1 (z−z0)k−1 = (ak +ak−1)z−ak−1z0 (z−z0)k .
The numerator does not vanish at z0, and so we have one singularity of order k. A pole of “infinite order” in item (iii) means that the expansion contains infinitely many non-zero terms ak(z−z0)−k with k ∈N.
Remark 11.5. A subtle — but sometimes important — point that is the observation that branch points like the origin for z →(z −z0)1/2 or z → ln(z−z0) are not isolated singularities. The reason is that in any punctured neighborhood of the origin these “functions” are not one-valued. In other words, they are not functions, and therefore a fortiori they are not analytic functions. Even if you redefine the function in this neighborhood so that it describes a single branch of that function, then still there is a line of discontinuities (the branch cut) with the branch point as its endpoint.
For completeness, we mention the only other types of singularities: cluster points , these are limit points of other singularities; and natural boundaries , entire sets where singularities are dense. An example of the latter is the unit circle for the function ∑∞ n=1 zn!. Needless to say, these singularities are not isolated.
All singularities mentioned in this remark are non-isolated, and if z0 is the locus of such a singularity, it is not possible to approximate its behavior in terms of integral powers of (z−z0).
Definition 11.6. A function is meromorphic in a domain if it has only iso-lated poles in the domain. It is meromorphic if this holds on all of C.
We need a criterion for uniform convergence.
Lemma 11.7 (Weierstrass M test). Let A ⊆C and gn : A →C a sequence of functions. Suppose that |gn(z)| ≤mn on A and that ∑n mn converges (uniform absolute convergence ). Then for all z in A: 238 11. The Cauchy Integral Formula i) ∑∞ n=1 |gn(z)| converges (absolute convergence), and ii) For all ε > 0, there is n0 so that for all n > n0: ∑∞ n+1 gi(z) < ε (uniform convergence).
Proof. Item (i) follows immediately from the hypotheses. Item (ii) follows from the fact that ∑∞ n+1 gi(z) ≤∑∞ n+1 |gi(z)| ≤∑∞ n+1 mi and the conver-gence of ∑n mn (so the partial sums of {mn} form a Cauchy sequence).
■ 11.2. The Cauchy Integral Formula First we set the scene with some notation. Let [a,b] be an interval in R of positive length. A curve is a piecewise differentiable function γ : [a,b] →C.
Its orientation is the direction of increasing t ∈[a,b]. A simple, closed curve is a curve without self-intersections and whose endpoints are identical (or γ(a) = γ(b). It follows that the complement of γ consists a well-defined “inside” component and a “outside” component (see Figure 64). A line integral evaluated along the curve γ is denoted by R γ. If the curve is simple and closed, one often writes H γ or simply H .
Figure 64. Left, a curve. Then two simple, closed curves with opposite orientation. The curve on the right is a union of two simple, closed curves.
Proposition 11.8 (Cauchy’s Theorem). Let γ be a simple, closed curve and assume f is analytic on γ and in its interior with at most finitely remov-able singularities. Then we have H γ f(z)dz = 0 For students familiar with differential forms and Stokes’ theorem, we give a very simple proof. Students unfamiliar with that material can skip the first paragraph of the proof. A full proof without assuming Stokes is more laborious and can be found in and in . A proof of Stokes’ theorem can be found in Chapter 5, Section 9.
11.2. The Cauchy Integral Formula 239 Proof. By assumption γ bounds an ‘inside’ region D: γ = ∂D. First assume f is analytic in D (including boundary). As usual, we write f = u+iv and z = x+iy to relate the complex notation to calculus in Rn.
I ∂D f dz = Z ∂D u(dx+idy)+ Z ∂D iv(dx+idy) Z ∂D udx−vdy+i Z ∂D udy+vdx.
By Stokes’ theorem, for any differential form ω on a region D with a piecewise differentiable boundary as specified by the proposition, we have R ∂D ω = R D dω, where d stands for the exterior derivative. Now d(udx−vdy) = −(∂yu+∂xv)dxdy and d(udy+vdx) = (∂xu−∂yv)dxdy.
both of which are zero by the Cauchy-Riemann equations of Proposition 11.23 (exercises 11.11, 11.12, and 11.13). in the exercises. Hence, I ∂D f dz = I D d(f dz) = 0 if f is analytic.
Since f has only finitely many singularities, they cannot accumulate.
Now suppose that f has an isolated singular point z0 at which it is, however, continuous. Let c be a circular path of small radius ε around z0 so that the ε-disk around z0 does not contain any other singular points or points of γ (see Figure 65). Let p be a path that connects γ to c. Now the curve Γ D c p gamma z0 Figure 65. In the interior of the curve obtained by concatenating γ, p, c, and −p, f is analytic. Therefore H γ f dz − H c f dz = 0. If f is also bounded inside c, we also have H c f dz = 0.
obtained by concatenating γ, p, c, and −p is a simple closed curve and thus H Γ f = 0. The integrals along p and −p cancel one another. By continuity, |f| is bounded by some M and so | H c f| is bounded by 2πεM. We can 240 11. The Cauchy Integral Formula choose ε as small as we want, and so | H c f| must be 0. Therefore, H γ f = H Γ f − H c f = 0.
■ It is also instructive to compare this with calculus on the real line. If f : R →R is piecewise differentiable and continuous, then from calculus, we know that Z b a f dx = F(b)−F(a).
This does not depend on the path we choose to get from a to b. Let yi : [0,1] →[a,b] be different parametrizations of the segment [a,b]. Then I γ1−γ2 f = Z γ1 − Z γ1 f = Z 1 0 f(y1(t))y′ 1(t)dt − Z 1 0 f(y2(t))y′ 2(t)dt = 0.
It is this statement that Cauchy’s theorem generalizes.
Theorem 11.9 (Cauchy’s Integral Formula). Let γ be a simple, closed curve going around z once in counter-clockwise direction and suppose that f is analytic on and inside γ. Then f(z) = 1 2πi Z γ f(w) w−z dw.
Proof. Define the function g g(w) = f(w)−f(z) w−z w ̸= z f ′(z) w = z The function g is continuous and therefore analytic (also at z). So H γ g = 0.
By linearity, Z γ f(w) w−z dw = Z γ g(w)dw+ f(z) Z γ 1 w−z dw.
The first integral in the right-hand side is zero by Cauchy’s theorem (Propo-sition 11.8). Now let c be the curve w = z+reit with t ∈[0,2π]. The same construction as in the second part of the proof of Proposition 11.8 shows that H γ − H c = 0 (see Figure 65) and thus H γ = H c. So Z γ 1 w−z dw = Z 2π 0 ireit reit dt = 2πi.
Substituting this into the earlier equation yields the statement.
■ 11.3. Corollaries of the Cauchy Integral Formula 241 Remark. The surprising aspect of this formula is that the value of an ana-lytic function at z0 is determined by the values of that function on a simple, closed curve that encircles z0.
11.3. Corollaries of the Cauchy Integral Formula Cauchy integral formula can be used to show the remarkable result that a function that is analytic at z0 has derivatives of all orders at that point. These derivatives are denoted by f (k)(z0). The simplest way of proving this is by actually calculating an expression for these derivatives.
Lemma 11.10. Suppose that w−z ̸= 0, then for |d| small enough w−z− d ̸= 0 and for some K we have 1 d 1 (w−z−d)k − 1 (w−z)k = k(w−z)k−1 +R(d)d (w−z−d)k(w−z)k , with |R(d)| ≤K.
Proof. First set 1 (w−z−d)k − 1 (w−z)k = (w−z)k −[(w−z)−d]k (w−z−d)k(w−z)k .
According to the binomial theorem (Theorem 5.30), there is a K such that −[(w−z)−d]k = −(w−z)k +k(w−z)k−1d +R(d)d2 .
with |R(d)| ≤K. Inserting this and canceling d in the left-hand side yields the lemma.
■ Theorem 11.11. Let γ be a simple, closed curve going around z once in counter-clockwise direction and suppose that f analytic on and inside γ (see Figure 66). Then for the kth derivative of f at z0, or f (k)(z0), we have i) f (k)(z) k!
= 1 2πi I γ f(w) (w−z)k+1 dw.
ii) f (k)(z) k!
≤Mℓ(γ) rk+1 , where M = maxw∈γ(|f(w)|), ℓ(γ) is the length of γ, and r is a lower bound for the distance of z to γ.
242 11. The Cauchy Integral Formula Proof. Cauchy’s integral formula establishes the result for k = 0. The in-duction step proceeds as follows. Suppose we are given f (k−1)(z) = (k −1)!
2πi I γ f(w) (w−z)k dw.
Since z lies inside γ, so does z+d if d is small enough (Figure 66). We use z z+d r Figure 66. The curve γ goes around z exactly once in counter-clockwise direction. If d is small enough, z+d also lies inside γ.
the induction hypothesis to compute the next derivative as limd→0 f (k−1)(z+d)−f (k−1)(z) d .
This equals ··· = lim d→0 (k −1)!
2πid I γ f(w) (w−z−d)k dw− I γ f(w) (w−z)k dw = lim d→0 (k −1)!
2πid I γ f(w) 1 (w−z−d)k − 1 (w−z)k dw = lim d→0 (k −1)!
2πi I γ f(w) k(w−z)k−1 +R(d)d (w−z−d)k(w−z)k dw.
The first and second equalities above follow by linearity of integration. The final equality uses Lemma 11.10. The limit can now be taken safely, be-cause the denominator is never zero, and so everything is nice and continu-ous.
··· = k!
2πi I γ f(w) (w−z)k+1 dw.
This establishes (i). Item (ii) follows immediately.
■ This has the remarkable implication that an analytic function — defined as having one derivative, Definition 11.2 — has derivatives of all orders. In particular, we have the following result.
11.3. Corollaries of the Cauchy Integral Formula 243 Corollary 11.12. The derivative of an analytic function is again analytic.
Proposition 11.13 (Morera’s Theorem). If f is continuous and if always H f dz = 0 in some region A, then f is analytic in A.
z z z+d 0 c Figure 67. F(z) does not depend on the path. So F(z + d) −F(z) = R c f ≈f(z)d Proof. Pick a point z0 and set F(z) := R z z0 f(w)dw. Because H f(w)dw = 0, F(z) does not depend on the path from z0 to z and so is uniquely defined.
Thus F(z+d)−F(z) = R c f ≈f(z)d, where c is a short, linear path from z to z+d (see Figure 67). Then F′(z) = f(z) and so f is the derivative of an analytic function and therefore is itself analytic.
■ Proposition 11.14. Let {gi} be a sequence of functions that are analytic in a region A and suppose that ∑∞ i=1 gi(z) converges uniformly on every closed disk contained in A. Then i) For any curve γ in A: R γ limn ∑n i=1 gi = limn R γ ∑n i=1 gi.
ii) limn ∑n i=1 gi is analytic in A.
iii) d dz limn ∑n i=1 gi(z) = limn d dz ∑n i=1 gi(z).
Proof. Write fn = ∑n i=1 gi and call the limit f. Then for all n > N Z γ fn − Z γ f = Z γ fn −f ≤ Z γ |fn −f| ≤εℓ(γ).
where ℓ(γ) is the length of γ (a curve whose image is a compact set). The fact that |fn(z)−f(z)| ≤ε for all z ∈γ is due to uniform convergence. This proves (i).
Next, we prove (ii). Pick z0 ∈A and let B = Br(z0) be an open disk whose closure ¯ B is contained in A. By assumption, fn →f uniformly on ¯ B and thus f is continuous on ¯ B (see exercise 11.17). Now let γ be any simple, closed curve in ¯ B. Then by Cauchy’s theorem, H γ fn = 0. Item (i) implies that H γ f = 0. Finally, Morera’s theorem implies that f is analytic at z0.
244 11. The Cauchy Integral Formula For part (iii), we have to show that |f ′ n(z) −f ′(z)| tends to zero as n tends to infinity. We use Theorem 11.11 to do that. Fix some small r and so that γ(t) := z0 +reit is contained in A. Then |f ′ n(z0)−f ′(z0)| ≤1 2π I γ fn(z)−f(z) (z−z0)2 |dz|.
By uniform convergence, for large n, |fn(z)−f(z)| is less than ε on γ while |z−z0| = r and the length of γ is 2πr.
■ Lemma 11.15. If |z−z0| < |w−z0|, then ∞ ∑ k=0 (z−z0)k (w−z0)k+1 = 1 w−z .
Proof. ∑∞ k=0 h z−z0 w−z0 ik is a geometric series that can be written as ∑∞ k=0 xk, where |x| < 1. This equals 1 1−x. Substituting this in the right-hand side of the lemma gives the result.
■ Theorem 11.16 (Taylor’s Theorem). Suppose f is analytic in a region A and let D be any open disk centered on z0 whose closure is contained in A.
Then for all z ∈D we have f(z) = ∞ ∑ n=0 f (n)(z0) n!
(z−z0)n , which converges on D. This is called the Taylor series of f at z0.
Proof. Let D be the disk bounded by the curve γ given by w(t) = z0 +reit.
Take z inside D (see Figure 68) so that |z−z0| < |w−z0|. By Theorem 11.9 and Lemma 11.15, we have f(z) = 1 2πi I γ f(w) (w−z) dw = 1 2πi I γ ∞ ∑ k=0 f(w) (z−z0)k (w−z0)k+1 dw.
Again because |z −z0| < |w −z0|, the sum converges uniformly, and so Proposition 11.14 (i) implies that the sum and integral can be interchanged.
To the expression that then results, we apply Theorem 11.11 to get ··· = 1 2πi ∞ ∑ k=0 I γ f(w) (z−z0)k (w−z0)k+1 dw = ∞ ∑ k=0 f (k)(z0) k!
(z−z0)k .
11.4. A Tauberian Theorem 245 A z z 0 D r w Figure 68. The curve w goes around z0 exactly once in counter-clockwise direction. .
By Theorem 11.11 (ii), the last expression is bounded by M ∑∞ k=0 |z−z0|k rk .
Uniform convergence on compact sets contained in the open disk of radius r follows from Lemma 11.7. The series is analytic by Proposition 11.14.
■ Remark 11.17. Note that it follows that the Taylor series of an entire func-tion (Definition 11.2) converges in all of C.
11.4. A Tauberian Theorem There is no formal definition of what a Tauberian theorem is, but generally it is something along the lines of the theorem below: we know that some transform TF(z) of f(t) converges for Rez > 0, but we want to know that it converges for z = 0. The price we pay is some extra information on f as in the case below, where we stipulate a bound on |f(t)|. The reader is strongly encouraged to first have a look at the examples in exercise 11.23.
Theorem 11.18. Let f : [0,∞) →R be integrable on compact intervals in [0,∞) and bounded by |f| ≤F for some F > 0 and define g(z) := Z ∞ 0 f(t)e−zt dt .
If g(z) has an analytic continuation defined on Rez ≥0, then R ∞ 0 f(t)dt exists and equals g(0).
Remark 11.19. The function g in Theorem 11.18 is called the Laplace transform of f.
Proof. First define gT(z) := Z T 0 f(t)e−zt dt .
246 11. The Cauchy Integral Formula Note that g′ T exists (exercise 11.24) and so gT is entire. Pick any ε > 0, we will prove that for any ε > 0, we can choose T such that lim T→∞|gT(0)−g(0)| < ε .
(11.1) Since gT(0) is finite, this implies that g(0) also exists. So, fix ε > 0.
R C C d + − − L R d DR Figure 69. g is analytic in DR := {Rez ≥−dR} ∩{|z| ≤R} (shaded).
The red curve is given by C+(s) = Reis with s ∈(−π 2 , π 2 ). The green curve is given by C+(s) = Reis with s ∈( π 2 , 3π 2 ). The blue L−con-sists of 2 small circular segments plus the segment connecting their left endpoints at a distance 0 < d < dR to the left of the the imaginary axis.
For the definition of the region DR and the curves C+, C−, and L−, we refer to Figure 69. Because g is analytic on Rez ≥0, Corollary 11.3 says that for any R, there is a dR so that g is analytic in the compact region DR. Since gT is analytic on all of C, the Cauchy integral formula (Theorem 11.9) tells us that2 g(0)−gT(0) = 1 2πi I C+∪L− (g(z)−gT(z)) 1+ z2 R2 ezT z dz.
(11.2) We will show that |g(0)−gT(0)| < ε by cleverly splitting up this integral.
First compute the full integral along C+ where z = Reis = R(coss + isins). We will abbreviate coss by c. For c > 0, we first estimate the three factors in the integrand of (11.2).
|g(z)−gT(z)| = Z ∞ T f(t)e−zt dt ≤F Z ∞ T e−Rct dt = Fe−RcT Rc .
(11.3) 2The factor (1+ z2 R2 ) in the integrand, introduced by Newman , may seem artificial and unnecessary at this point, but is in fact essential, see exercise 11.25.
11.4. A Tauberian Theorem 247 Furthermore, 1 z 1+ z2 R2 = 1 R 1+e2is = 1 R e−is +eis = 2|c| R .
(11.4) And finally ezT = eRTc+iRT sins = eRTc .
(11.5) Since the length of C+ is πR, we thus obtain from (11.2) that 1 2πi Z C+ ≤1 2π · Fe−RcT Rc · 2c R ·eRcT ·πR = F R .
(11.6) For the second step, analyticity of gT and Theorem 11.9 imply that 1 2πi Z C− gT(z) 1+ z2 R2 ezT z dz = 1 2πi Z L− gT(z) 1+ z2 R2 ezT z dz, allowing us to evaluate the integral along C−. We have, now for c < 0, |gT(z)| = Z T 0 f(t)e−zt dt ≤F Z T 0 e−Rct dt = Fe−RcT R|c| .
Substituting this into the integral over C−and using (11.4) and (11.5) gives 1 2πi Z C− gT(z) 1+ z2 R2 ezT z dz ≤1 2π Fe−RcT R|c| 2|c| R eRcT πR = F R . (11.7) The third (most painful) step is the evaluation of the remaining integral, Z L− G(z)ezt dz, (see again Figure 69) where G(z) := g(z)(1+ z2 R2 )/(2πiz). On the two (com-pact) circular segments z = Reis with Rez ∈[−dR,0], |G| is maximized by the constant Mh(R,dR). The combined length of these segments is less than 4d. Thus the integral over these pieces contributes at most Mh(R,dR)4d.
On the vertical segment, |G| is bounded by another constant, Mv(R,d). This may very well increase as d decreases, since, with decreasing d, the path passes very close to the origin. We have that ezT = e−dT and the path length is less than 2R. So the contribution of the vertical segment is at most Mv(R,d)e−dT 2R. Summarizing, this gives Z L− G(z)ezt dz ≤4d Mh(R,dR)+2RMv(R,d)e−dT .
(11.8) 248 11. The Cauchy Integral Formula Now we add up the contributions of equations (11.6), (11.7), and (11.8).
1 2π I C+∪L− ≤2F R +4d Mh(R,dR)+2RMv(R,d)e−dT .
There are now three parameters, R, d, and T, whose values have not been fixed yet. We use these to “talk” the right hand side into being less than ε. Start by choosing R so that the first term is less than ε/3. Then choose d ∈[0,dR] so that 4d Mh(R,dR) < ε/3. Finally, we choose T so that the last term is also less than ε/3.
■ 11.5. A Polynomial Must Have a Root While we are on the topic of complex analysis, we take advantage of the opportunity to fill a gap in our proof of the fundamental theorem of algebra (Theorem 3.19).
Proposition 11.20. Every polynomial of degree d ≥1 has a root in C.
Proof. Let p(z) = ∑d i=0 aizi be a non-constant polynomial (with non-zero leading coefficient ad). The proof consists of showing that |p(z)| has a minimum and that that minimum equals zero.
We write z = reiϕ (polar coordinates) and immediately obtain p(reiϕ) = adrdediϕ 1+ ad−1 ad r−1e−iϕ +···+ a0 ad r−de−diϕ .
The term in parentheses can be written as 1 + r−1A(r), where A(r) can be bounded from above by a geometric series in 1/r. Thus for r greater than some R, A(r) is bounded by A0 ≥0. We then get for r > R |p(reiϕ)| = |ad|rd 1+A(r)r−1 where |A(r)| ≤A0 .
Thus for R large enough, |p(2Reiϕ)| is larger than |p(Reiϕ)|. The closed disk D of radius 2R is compact and p is continuous, so it follows that |p(z)| must have a minimum in in the interior of that disk (see Figure 70).
Let z0 be this minimum. Take δ in the ball |δ| < ε, and ε small so that the ε-disk around z0 is in the interior of D (see Figure 70). Now expand p(z0 +δ) = ∑d i=0 ai(z0 +δ)i. The expansion must contain non-trivial terms, because otherwise p would be constant. So for some 0 < k ≤d, p(z0 +δ) = p(z0)+bkδ k +bk+1δ k+1 +···+bdδ d , 11.6. Exercises 249 0 2R z0 D Figure 70. In the proof of Proposition 11.20, |p(z)| must have a mini-mum z0 in the interior of the disk |z| < 2R and it cannot have a minimum unless at z0 unless it is zero.
where bk ̸= 0. Thus p(z0 +δ) = p(z0)+bkδ k (1+δB(δ)) , where again for ε small enough |B(δ)| is bounded and so p(z0 + δ) ≈ p(z0) + bkδ k.
By choosing the phase of δ appropriately and |δ| small enough, one make sure that if |p(z0)| > 0, then |p(z0)+bkδ k| < |p(z0)|.
■ Lest one might think that every complex function must have a zero, we warn the reader that ez has no zero (see also exercise 11.16).
Together with exercise 3.24, the last result establishes the fundamental theorem of algebra (Theorem 3.19), which we repeat verbatim here.
Theorem 11.21 (Fundamental Theorem of Algebra). A polynomial in C[x] (the set of polynomials with complex coefficients) of degree d ≥1 has exactly d roots, counting multiplicity.
11.6. Exercises Exercise 11.1. Which of the following sets are regions or domains in C?
a) C{0}.
b) C\N.
c) C minus the negative real axis.
d) C minus the real axis.
e) The union of the closed unit disks with centers at 1 and -1.
f) The same as (d), but minus the boundary.
g) The same as (e), but now add the imaginary axis.
250 11. The Cauchy Integral Formula In exercise 11.2 briefly discuss two “bad” (non-isolated) singularities. Around such a singularity no power series expansions can be used to approximate the functions. Pictures of the two singularities can be found in [Sections 2.4 and 3.1].
Exercise 11.2. On D = {z : |z| < 1}, define f(z) = ∞ ∑ n=1 zn! and g(z) = 1 sin(1/z) .
a) Let p and q be co-prime integers and set z = re2πip/q. Show that |f(z)| ≥−q+ ∑ n≥q rn! , and that this diverges as r ↗1. (Note: the unit circle is a natural bound-ary.) b) Conclude that the singularities of f are dense on the unit circle.
c) Show that g has a cluster point at the origin.
Exercise 11.3. a) Show that on (0,1), ∑∞ n=1 xn is absolutely convergent but not uniformly convergent.
b) Show that on (0,1), ∑∞ n=1 (−1)nx n is uniformly convergent but not abso-lutely convergent. (Hint: the sum is −xln2.) Exercise 11.4. a) Let z = x+iy and show that for n ∈N, n−z = n−xe−iylnn.
b) From (a), show that n−z = n−x.
c) Use (b) to show that ζ(z) = ∑∞ n=1 n−z is uniformly convergent on com-pact disks in Rez > 1. (Hint: use Lemma 11.7 and exercise 2.25 (e).) Exercise 11.5. Let f analytic at z0 and suppose furthermore that there is a sequence {zn} converging to z0 such that f(zn) = 0.
a) Show that f has all derivatives at z0. (Hint: Theorem 11.11.) b) Show that if at least one of f (n)(z0) ̸= 0, then for z close enough to z0, f(z) ̸= 0. (Hint: the first non-zero term in the Taylor expansion dominates as in Section 11.5.) c) Use (a) and (b) to show that for all n ≥0, f (n)(z0) = 0.
d) Use Taylor’s theorem to show that f is zero in an open disk containing z0.
11.6. Exercises 251 Exercise 11.6. Let A be a region (that is: an open, connected set) contain-ing a sequence {zn} converging to z0. Let f and g be analytic functions on A such that f(zn) = g(zn) for all n.
a) Show that h := f −g is analytic of A and satisfies h(zn) = 0.
b) Use exercise 11.5 to show that h = 0 in an open disk containing z0.
c) Write A as the disjoint union of A0 := {z0 ∈A : h(z) = 0 on an open neighborhood of z0} and A1 := A\A0 .
Show that A0 is open in A. (Hint: by definition of A0.) d) Show A1 is open in A. (Hint: consider z ∈A1, if h(z) ̸= 0, use continuity of h; if h(z) = 0, use exercise 11.5 that h is not zero in a neighborhood of z.) e) Show that one of A0 or A1 must be empty. (Hint: use Definition 11.1.) f) Conclude that the analytic continuations of f and g in A coincide. (Note that this was remarked more informally in Section 2.5.) The last result of exercise 11.6 will be relevant when we discuss the analytic continuation of the zeta function. We isolate the result here.
Theorem 11.22 (Uniqueness of Analytic Continuation). Suppose f and g are analytic in a region or domain A ∈C. Let Z be the set of points such that f(z) −g(z) = 0 and suppose that Z has a limit point in A. Then the analytic continuation of f and g coincide on A.
Exercise 11.7. For z ∈C, define ez = ∞ ∑ n=0 zn n! .
a) Assume or provea that the sum converges uniformly on every closed disk. Conclude that ez is entire. (Hint: Proposition 11.14 (ii).) b) Use exercise 11.6 to show that it is the unique analytic continuation of the real function ex.
c) Compare the expansion of eiy with those of cosy and siny and conclude that eiy = cosy+isiny.
d) Use ea+b = ea eb to establish that ex+iy = ex(cosy+isiny).
e) Use (a) and (d) to show that ez is entire but never equal to 0.
aThe factorial always wins out.
252 11. The Cauchy Integral Formula Exercise 11.8. a) Use exercise 11.7 (c) to show that (Figure 71) for y ∈R cosy = 1 2 eiy +e−iy and siny = 1 2i eiy −e−iy .
b) Use exercise 11.6 to show that cosz = 1 2 eiz +e−iz and sinz = 1 2i eiz −e−iz are the unique extensions of the sine and cosine functions to the complex plane.
c) Find a formula with only exponentials for tanz. (Hint: tanx = sinx cosx.) exp(it) exp(−it) −exp(−it) i sin(t) cos(t) Figure 71. The complex plane with eit, −e−it and e−it on the unit cir-cle. cost is the average of eit and e−it and isint as the average of eit and −e−it.
Exercise 11.9. Use eit e−it +eit = 1+e2it to show that a) 2cos2(t) = 1+cos(2t), and b) 2sint cost = sin2t.
L r ei phi 0 r Figure 72. Moving around the origin once in the positive direction in-creases ϕ, and thus lnz, by 2π. Discontinuities can be avoided if we agree never to cross the half line or branch cut L.
11.6. Exercises 253 Exercise 11.10. The complex logarithm lnz is the (local) inverse of ez. See Figure 72.
a) Use “polar” coordinates, i.e. z = reiϕ, to show that lnz = lnr +iϕ.
b) Fix r and increase ϕ from 0 to 2π. Assuming that you do not encounter discontinuities, show that lnz has increased by 2πi while its real part re-mained constant.
c) Conclude that lnz is multivalued .
d) Let L be any half line from the origin to infinity. Show that lnz is analytic of C minus L. L is called a branch cut .
For any function f : C →C, we can always write z = x + iy and f(z) = u(x + iy) + iv(x + iy). In the next three exercises, we prove the following result.
Proposition 11.23. f : C →C is analytic (see Definition 11.2) at z0 if and only if in a neighborhood of z0, f is differentiable3 as a function from R2 to itself and the Cauchy-Riemann equations hold: ∂xu = ∂yv and ∂xv = −∂yu.
Exercise 11.11. a) Show that if f is analytic at z0, then in a neighborhood of z0, f ′(z) = limδ→0 f(z+δ)−f(z) δ does not depend on δ (as long as it tends to 0).
b) Compute the derivative in (a) for δ real and δ imaginary.
c) Use (a) to show these two are equal.
d) Use (c) to prove that analyticity implies that u and v satisfy the Cauchy-Riemann equations.
Exercise 11.12. For real a and b, let A = a −b b a and z = x y .
a) Show that multiplication by A of z in R2 acts exactly like multiplication by a+ib of x+iy in C.
b) Write the matrix A as Reiθ. (Hint: R = √ a2 +b2. What is θ?).
c) Use (b) to show that a non-zero derivative at a point z0 of an analytic function is a dilatation composed with a rotation.
d) Explain that if f ′(z0) is non-zero, f “locally” preserves angles.
Definition 11.24. A map f from a region A ⊂C to C is conformal at z0 if its derivative at z0 exists and is non-zero.
3This means that the partial derivatives exist and are continuous.
254 11. The Cauchy Integral Formula Exercise 11.13. Write z = x+iy and f(z) = u(x+iy)+iv(x+iy), where u and v are real functions. In a neighborhood of (x0 +iy0), suppose that the matrix of (continuous) derivatives D f(x,y) satisfies Cauchy-Riemann.
a) Use exercise 11.12 to show that this implies that D f(x,y) acts like a complex number.
b) Use (a) to imply that f is analytic.
Exercise 11.14. Write z = x+iy and f(z) = u(x+iy)+iv(x+iy), where u and v are real functions.
a) Given that u(x + iy) = e−y cosx, compute v and f(z). (Hint: use the Cauchy-Riemann equations to compute ∂xv and ∂yv. Integrate both to get v. Finally, express u+iv as f(z).) b) Given that v(x+iy) = −y3 +3x2y−y, compute u and f.
c) Given that f(z) = tanz, compute u and v. (Hint: use exercise 11.8 (c).) An interesting result — though we will not prove it — is the following. A weaker version of this is called the Casorati-Weierstrass Theorem and has an easy proof [chapter 4] [chapter 3].
Theorem 11.25 (Picard Theorem). Let f have an isolated essential singu-larity at z0. Then the image of any punctured neighborhood of z0 contains all values infinitely often with at most one exception.
The next results are important corollaries (proof in exercise 11.15).
Corollary 11.26 (Little Picard). Let f be entire and not constant. Then the image of f contains all values with at most one exception.
Corollary 11.27 (Liouville’s theorem). A bounded entire function must be constant.
Exercise 11.15. Assume f is entire and not constant.
a) Show that f has an expansion ∑∞ i=0 anxn that converges in all of C. (Hint: see Taylor’s theorem.) b) Show that if f is a polynomial (only finitely many non-zero an), then it has a pole at infinity. (Hint: effect a coordinate change that moves ∞to 0, i.e. set w = 1/z. What does f look like in terms of the new coordinate?) c) Show that in case (b), for all z0 ∈C, f(z) −z0 has a zero. (Hint: the fundamental theorem of algebra (Theorems 3.19 and 11.21).) d) Show that if f is a not a polynomial, then it has an essential singularity at infinity.
e) Show that (c) and (d) and the Picard Theorem imply little Picard.
f) Show that Little Picard implies Liouville’s theorem.
The function ez is a good illustration of little Picard (see exercise 11.7 (e)).
The next problem illustrates the Picard Theorem (Theorem 11.25).
11.6. Exercises 255 Exercise 11.16. a) Show that if z = x+iy, then 1 z = x x2 +y2 −i y x2 +y2 .
b) Show that f(z) := e 1 z = e x x2+y2 cos y x2 +y2 +isin y x2 +y2 .
c) Show that if y = 0 and x ↘0, then f(z) is real and tends to infinity.
d) Show that if y = 0 and x ↗0, then f(z) is real and tends to zero.
e) Show that if you approach 0 in any other direction, f has arbitrarily large oscillations. (Hint: fix t and set y = tx and let x ↘0.) f) Show that f(z) ̸= 0 for all z.
Exercise 11.17. Let {fn} be a sequence of continuous functions on a com-pact set S in Rn or C. Suppose fn →f uniformly on S and let x, y ∈S.
a) Show that there is an n such that | fn(x)−f(x)| < ε/3.
b) Given n as in (a), show that there is a δ such that for all x with |y−x| < δ, we have |fn(y)−fn(x)| < ε/3.
c) Show that (a) and (b) imply that |f(y)−f(x)| < ε. (Hint: this is called the “ε/3 trick”.) d) Show that (c) implies that f is continuous.
Exercise 11.18. We give an easy informal “proof” of Theorem 11.11 by interchanging differentiation and integration without justification.
a) Let k a non-negative integer. Suppose that f (k)(z0) = k!
2πi I γ f(z) (z−z0)k+1 dz.
Change the order of integration and differentiation to show that d dz0 f (k)(z0) = k!
2πi I γ d dz0 f(z) (z−z0)k+1 dz = (k +1)!
2πi I γ f(z) (z−z0)k+2 dz.
b) Use (a) to give a proof by induction of Theorem 11.11.
Integrals and limits cannot always be exchanged, and the same holds for derivatives. The following exercise provides examples (see Figure 73). For uniformly converging analytic functions, the changes can be made (Propo-sition 11.14).
256 11. The Cauchy Integral Formula Exercise 11.19. On [0,1], consider the functions gk(x) = k2xk(1−x) and hk(x) = sin(kπx) k .
a) Show that limk→∞gk(x) = 0.
b) Show that R 1 0 limk→∞gk(x)dx = 0 while limk→∞ R 1 0 gk(x)dx = limk→∞ k2 (k+1)(k+2) = 1.
c) Show that limk→∞hk(x) = 0.
d) Show that d dx limk→∞hk(x) = 0 while limk→∞d dxhk(x) = 0 does not exist at x = 1/2 (for example).
Figure 73. The functions gk and hk of exercise 11.19 for k ∈{2,8,15,30}.
Exercise 11.20. Set α = a + ib where a and b real and greater than zero and let f(z) = (z−α)−1.
a) Show that f is analytic inside and on the contour C given in Figure 74.
b) Show H C f = 0.
c) Show that R bi f tends to 0 as R tends to infinity. (Hint: |f| →0 while the path length remains finite.) d) Show that R r f tends to πi as R tends to infinity. (Hint: set z(t) = ib+Reit with t ∈[0,π].) e) Show that H p f tends to −2πi as R tends to infinity. (Hint: set z(t) = α +re−it with t ∈[0,2π].) f) Conclude that limR→∞ R +R −R f(z)dz = πi. (Hint: use (a).) 11.6. Exercises 257 b1 b2 c g +R −R a ib r1 p r2 Figure 74. The contour C is the concatenation of c (celeste), b1 (blue), r1 (red), g (green), p (purple), −g, r2, and b2. The path r is a semi-circle of radius R. The path p is a small circle of radius r. See exercise 11.20.
Exercise 11.21. We check the outcome of exercise 11.20 by direct integra-tion. We use the notation of that problem.
a) Show that Z +R −R f(z)dz = Z +R −R x−a+ib (x−a)2 +b2 dx.
b) Sustitute s = x−a and show that Z +R −R f(z)dz = Z +R−a −R−a s+ib s2 +b2 ds.
c) Show that the real part of this integral tends to zero as R →∞. (Hint: it is odd plus something that tends to zero.) d) Show that limR→∞ R +R−a −R−a ib s2+b2 ds = πi. (Hint: substitute bt = s and use that the derivative of arctanx equals 1/(x2 +1).) Exercise 11.22. Let f(z) = ∑n≥−k ak(z−z0)k with k > 0.
a) Compute Res( f,z0) := 1 2πi I f(z)dz along the path γ(t) = z0 + εeit, t ∈[0,2π] for small ε > 0. This is called the residue of f at z0.
b) Let Γ be any piecewise smooth contour winding exactly once around z0 in the anti-clockwise direction. Show that I Γ f(z)dz = 2πiRes( f,z0).
(Hint: consider a contour that narrowly avoids the singularity such as the contour C in Figure 74.) 258 11. The Cauchy Integral Formula Exercise 11.23. a) Let f(t) = 1. Show that its Laplace transform as defined in Theorem 11.18 does not have an analytic continuation to the imaginary axis.
b) In (a), show that the conclusion of Theorem 11.18 does not hold.
c) Repeat (a) and (b), but now for f(t) = eiωt.
Exercise 11.24. Consider gT (z) as in the proof of Theorem 11.18.
a) Write out Hε := 1 ε (gT (z+ε)−gT (z)).
b) Use linearity of integration to show that limε→0 Hε = R T 0 −t f(t)e−zt dt.
c) Show that the integral in (b) exists.
d) Conclude that gT is entire.
Exercise 11.25. a) Explain why it is crucial in the proof of Theorem 11.18 that g(z) is analytic on the imaginary axis.
b) Explain why the factor (1 + z2 R2 ) is essential to the proof of Theorem 11.18.
Chapter 12 The Prime Number Theorem Overview. In 1850, it seemed that Chebyshev was awfully close to proving the prime number theorem (Theorem 2.21). But to bridge that last brook, a whole new approach to the problem was needed. That approach was the connection with analytic functions in the complex domain pioneered by Riemann in 1859 . A very weak version of the Riemann hypothesis (Conjecture 2.22), namely the absence of zeroes of ζ(z) in Rez ≥1 turns out to be an essential step. We look at this in Section 12.3 and in particular Lemma 12.12. It would take another 37 years after Riemann’s monumental contribution before the result was finally proved by De La Vall´ ee Poussin and Hadamard in 1896. The version we prove is a highly streamlined de-rivative of that proof, the last stage of which was achieved by Newman in 1982 . We made heavy use of Zagier’s rendition of this proof and of .
12.1. Preliminaries Recall that π(x) denotes the number of primes in the interval [2,x]. So π(2) = 1, π(3.2) = 2, and so on. The reason that the variable x is real is that it simplifies the formulas to come. The Riemann zeta function is denoted by ζ(s), see Definition 2.19 and Proposition 2.20. In this chapter, we will frequently encounter sums of the form ∑p. For example see Definition 12.1 259 260 12. The Prime Number Theorem below. Such sums will always be understood to be over all positive primes.
On the other hand, ∑p≤x indicates a sum over all positive primes p less than or equal to x. A similar convention holds for products ∏p and ∏p≤x. The letter z will always denote a complex variable.
We now define a couple of new functions.
Definition 12.1. The first Chebyshev function is defined as θ(x) := ∑ p≤x ln p.
The function Φ : {z ∈C : Rez > 1} →C is defined as Φ(z) := ∑ p ln p pz .
It is analytic in Rez > 1.
x x i i+1 2 3 5 ln 2 + ln 3 ln 2 + ln 3 + ln 5 theta(x) f(x) Figure 75. The Riemann-Stieltjes integral (12.1) near x = 5 picks up the value f(5)(θ(xi+1)−θ(xi)).
In what follows, we will need to integrate expressions like I(x) := Z x 1 f(t)dθ(t), (12.1) where f is differentiable. If we partition the interval [1,x] by 1 = x0 < x1 ···xn = x, then I(x) can be approximated as I(x) ≈ n ∑ i=1 f(ci)(θ(xi+1)−θ(xi)), where ci ∈(xi,xi+1) and then the appropriate limit (assuming it exists) can be taken. This is a Riemann-Stieltjes integral. It is very similar to the Riemann integral from calculus, except that instead of the increments xi+1− xi, we look at increments of a function: θ(xi+1) −θ(xi) (see ), see 12.1. Preliminaries 261 Figure 75. Now, θ(t) is constant except at the values t = p (a prime) where it has a jump discontinuity of size ln p. Thus, in this case, I(x) simplifies to I(x) = Z x 1 f(t)dθ(t) = ∑ p≤x f(p)ln(p).
(12.2) On the other hand, we can find a different expression for I(x) by integration by parts (sometimes called partial integration) I(x) = Z x 1 d f(t)θ(t) − Z x 1 θ(t)d f(t) = f(t)θ(t) x 1 − Z x 1 f ′(t)θ(t)dt .
(12.3) The point of this operation is usually that now we have expressed the in-tegral in (12.2) as fixed expression plus another integral which has better convergence properties than the original integral. For instance if f(t) = t−k, then f ′(t) ∝t−k−1 and so the integral converges faster.
Lemma 12.2. We have for x ≥2 π(x) = θ(x) lnx + Z x 2 θ(t) t (lnt)2 dt .
Proof. First note that since 2 is the smallest prime, equation (12.2) gives π(x) = Z x 2−ε d θ(t) lnt .
Apply integration by parts (12.3) to obtain π(x) = θ(x) lnx − Z x 2−ε θ(t)d 1 lnt .
Using d 1 lnt = − dt t(lnt)2 to work out the last term yields the lemma with lower limit 2−ε in the integral. But since θ(t) = 0 for t < 2, we may replace that limit by 2.
■ Lemma 12.3. For Rez > 1, we have Φ(z) z − 1 z−1 = Z ∞ 1 θ(x) x −1 x−z dx = Z ∞ 0 θ(et)e−t −1 e−zt+t dt .
262 12. The Prime Number Theorem Proof. Using (12.2), we can write Φ(z) as R ∞ 1 x−zdθ(x). Then apply (12.3) (partial integration) to obtain Φ(z) = x−zθ(x) ∞ 1 +z Z ∞ 1 x−z−1θ(x)dx.
We will see in equation (12.6) that for Rez > 1, the boundary term x−zθ(x) ∞ 1 vanishes. This gives Φ(z) z = Z ∞ 1 θ(x) x x−z dx.
Noting that 1/(z −1) = R ∞ 1 x−z dx, the first equality follows. The second equality follows by substitution of x by t where et = x.
■ 12.2. Chebyshev’s Theorem We prove Theorem 12.7, an approximate version of the prime number theo-rem (Theorem 2.21). Recall that ⌊x⌋is the greatest integer less than or equal to x (see Definition 2.1), whereas a b indicates the binomial factor a!
b!(a−b)!
(see Theorem 5.30).
We start with a remarkable lemma. Let a, b, and k > 0 be integers. We introduce the notation ak ∥b to mean that ak | b but not ak+1 | b. In words, this is expressed by saying that ak divides b exactly.
Lemma 12.4. Let p prime and suppose that pk ∥ n m with n > m > 0. Then we have pk ≤n.
Proof. Let p prime and suppose that pk ∥(1 · 2···n). We want to find k.
Any multiple ap ≤n in the product 1 · 2···n contributes one factor p to pk. The number of multiples ap less than or equal to n equals j n p k . So these contribute j n p k to k. If ap is also a multiple of p2 then it contributes two factors to k. Thus we need to add another factor in the form of j n p2 k .
Continuing like that, we find pk ∥n!
= ⇒ k = ∞ ∑ j=1 n p j .
(12.4) As a consequence, we obtain for the binomial factor pk ∥ n m = ⇒ k = ∞ ∑ j=1 n pj − m pj − n−m pj .
(12.5) 12.2. Chebyshev’s Theorem 263 Consider the expression E = ⌊x1 + x2⌋−⌊x1⌋−⌊x2⌋. By substituting x1 = a1 + ω1 and x2 = a2 + ω2, where ai are integers and ωi ∈(0,1), one sees that E ∈{0,1}. Going back to the expression in equation (12.5), we see that if pj > n, then the contribution is always zero. Thus if n > m > 0, the last positive contribution occurred for j = k such that pk ≤n.
■ The crux of the proof of Chebyshev’s theorem is contained in two sim-ple, yet very clever, lemmas.
Lemma 12.5. For n ≥2, we have 2n n+1 < n ⌊n/2⌋ ≤2n−1.
Proof. We prove the right-hand side first. Since 2 1 = 21 and for n > 0 2n+2 n+1 = (2n+1)(2n+2) (n+1)2 2n n < 4 2n n , The result follows in the even case. The odd case is similar. For the left-hand side, we note that n ⌊n/2⌋ is the largest of the n + 1 numbers n i and so (n+1) n ⌊n/2⌋ > n ∑ i=0 n i = 2n .
■ Lemma 12.6. i) For all n ≥2, we have n ⌊n/2⌋ ≤nπ(n).
ii) For n ≥2 a power of 2, we have eθ(n)−θ(n/2) ≤ n n/2 .
Proof. For the first inequality, use unique factorization (Theorem 2.11) and the definition of π(n) to write n ⌊n/2⌋ = π(n) ∏ i=1 pki i .
By Lemma 12.4, pki i ≤n. Thus ∏ π(n) i=1 pki i ≤nπ(n), which yields the inequal-ity.
For the second inequality, we start by noticing that n is even and so any prime p in the interval n 2,n is a divisor of n! but not of the denominator of n n/2 . Therefore any such p divides n n/2 . This implies that ∏ n 2 <p≤n p ≤ n n/2 .
264 12. The Prime Number Theorem Noting that p = eln p and inserting the definition of θ(x) (Definition 12.1) yields the last inequality.
■ Theorem 12.7 (Chebyshev’s Theorem). For all a < ln2 and b > ln4, there is a large enough K such that ∀x ≥K : π(x) x/lnx ∈[a,b].
Proof. Putting Lemmas 12.5 and 12.6 together gives 2n n+1 ≤nπ(n) and eθ(n)−θ( n 2 ) ≤1 22n (if n a power of 2).
Taking the logarithm of the first of these inequalities gives ln2−ln(n+1) n n lnn < π(n), which yields an estimate for a.
For n a power of 2, we get from the second inequality θ(n)−θ n 2 ≤n 2 ln2 and θ n 2 −θ n 4 ≤n 4 ln2 and ··· and so on. Thus θ(n) ≤nln2. For x ≥2, there is an n′ that is a power of 2 in the interval [x,2x). Thus θ(x) ≤θ(n′) ≤2xln2. Therefore, θ(x) ≤x ln4.
(12.6) Substituting this into Lemma 12.2 gives that π(x) ≤ln4 x lnx +ln4 Z x 2 (lnt)−2 dt .
(12.7) L’Hˆ opital’s rule implies that lim x→∞ R x 2 (lnt)−2 dt x(lnx)−2 = 1.
(12.8) Thus the integral in (12.7) can be replaced by x(lnx)−2. The dominant term of the right-hand side of that equation is the first one. Thus for any b > ln4, we have for x large enough that π(x) < b x lnx.
■ Remark 12.8. In fact, Chebyshev was able to prove the sharper result that if limx→∞ π(x) x/lnx exists, it has to be 1.
Equations (12.6) and (12.8) will also play an important role in the proof of the (full) prime number theorem.
12.3. Zeroes and Poles of the Zeta Function 265 12.3. Zeroes and Poles of the Zeta Function The proof of the prime number theorem relies in part on a careful study of the analytic extensions of some functions related to the zeta function. We do that in this section and the next.
Remark 12.9. From Chapter 4, equation (4.8) (see also exercise 4.27), we know that ζ(z) and 1/ζ(z) both converge for Rez > 1. Therefore neither has zeroes or poles in that region.
Here we prove a stronger statement.
Lemma 12.10. For Rez > 1, we have that lnζ(z) = −∑ p ln 1−e−zln p = ∑ p ∞ ∑ n=1 e−znln p n = ∑ p ∞ ∑ n=1 1 npnz .
is analytic.
Proof. First, set w := p−z = e−zln p. Using that the Taylor series at 0 −ln(1−w) = ∑ n≥1 wn n , converges uniformly on |w| < 1 on compact subsets, we see from Proposi-tion 11.14 (ii) that −ln 1−p−z = −ln 1−e−zln p = ∞ ∑ n=1 e−znln p n = 1 npn (12.9) is analytic on Rez > 0 (and thus also on Rez > 1).
Next, from Proposition 2.20, we conclude that lnζ(z) = −∑p ln(1−p−z).
By Lemma 10.18, this converges absolutely iff ∑p p−z converges abso-lutely. But if we set z = x+iy, then ∑ p p−z ≤∑ n n−z = ∑ n n−x , which converges absolutely by exercise 2.25 (e), and thus uniformly on closed disks in Rez > 1. Therefore, by Proposition 11.14, −∑p ln(1−p−z) is analytic on Rez > 1.
■ 266 12. The Prime Number Theorem We saw in see exercise 2.24 (c) that ζ(z) diverges as z ↘1+. Here is a more precise statement. Recall that analytic continuations are well-defined (i.e. unique) in domains with only isolated singularities (see Theo-rem 11.22).
Proposition 12.11. i) The functions (z −1)ζ(z) and (z −1)ζ ′(z) + zζ(z) have well-defined analytic continuations on Rez > 0.
ii) (The analytic continuation of) (z−1)ζ(z) evaluated at z = 1 equals 1.
Remark: The factor (z−1) precisely cancels the simple pole in ζ at z = 1.
u x n n+1 n n+1 Figure 76. Integration over the shaded triangle of area 1/2 in equation (12.11).
Proof. We have that R ∞ 1 x−z dx = 1/(z −1) and R n+1 n n−z dx = n−z. Using the definition of the zeta function (Definition 2.19), we define in Rez > 1 h(z) := ζ(z)− 1 z−1 = ∞ ∑ n=1 n−z − Z ∞ 1 x−z dx = ∞ ∑ n=1 Z n+1 n n−z −x−z dx.
(12.10) Next, since n−z −x−z = R x n −zu−z−1 du, we also have ··· = ∞ ∑ n=1 Z n+1 n Z x n −zu−z−1 dudx.
(12.11) Each term of the sum is an integral over a triangular domain of area 1/2 (Figure 76). The maximum of the integrand is zn−z−1 = p σ2 +τ2 n−σ−1 , where z = σ +iτ (with σ, τ real). So, each summand has absolute value less than half that. Thus (12.11) converges uniformly on compact disks in σ > 0 (see also exercise 11.4) and so h has an analytic continuation to Rez > 0.
12.3. Zeroes and Poles of the Zeta Function 267 To prove (i), note that, by the above, (z −1)h(z) + 1 = (z −1)ζ(z) is analytic. Therefore so is its derivative — by Corollary 11.12. The second function of part (i) is the sum of these two. Finally, evaluating (z−1)ζ(z) = (z−1)h(z)+1 at z = 1 establishes part (ii).
■ Now follows a lemma that is brilliant and an essential step in proving the prime number theorem. It will make its appearance in Proposition 12.13.
Lemma 12.12. ζ(z) has no zeroes on the line Rez ≥1.
Proof. By remark 12.9, ζ has no zeroes in Rez > 1. So we only need to prove that ζ has no zeroes on Rez = 1.
Let z = σ + iτ with σ > 1 and τ ̸= 0 real. We start by computing the admittedly strange expression E := ln ζ(σ)3ζ(σ +iτ)4ζ(σ +2iτ) . By Proposition 12.11, ζ has a simple pole at 1 and no poles in Rez > 1. Thus if ζ has a zero at 1+iτ, it cannot be compensated by a pole at 1+4iτ and the pole of order 1 at z = 1. Thus in this case, the expression eE evaluated at σ + iτ where σ is slightly greater than 1, would yield a number that is very close to zero. We now show that this cannot happen.
Combining the fact that ln(ab) = lna+lnb and Lemma 12.10, we get E = 3lnζ(σ)+4lnζ(σ +iτ)+lnζ(σ +2iτ) = ∑ p ∑ n≥1 e−σnln p n 3+4e−iτnln p +e−2iτnln p .
Now consider the real part of this expression: ReE = ∑ p ∑ n≥1 3+4cos(τnln p)+cos(2τnln p) npnσ .
Noting that 1+cos2x = 2cos2 x (exercise 11.9), we obtain ReE = ∑ p ∑ n≥1 2+4cos(τnln p)+2cos2(τnln p) npnσ = ∑ p ∑ n≥1 2(1+cos(τnln p))2 npnσ > 0.
But ReE > 0 yields |eE| > 1, which implies the lemma.
■ 268 12. The Prime Number Theorem 12.4. The Function Φ(z) The proof we will give of the prime number theorem (Theorem 12.15) really consists of inserting an analyticity property of the function Φ into Theorem 11.18 to prove the convergence of an improper integral. Here is the analyt-icity property we need.
Proposition 12.13.
Φ(z) z − 1 z−1 has an analytic continuation in the closed half plane Rez ≥1.
Proof. Taking a derivative with respect to z on both sides of the first equal-ity of Lemma 12.10, we obtain −ζ ′(z) ζ(z) = ∑ p ln p e−zln p 1−e−zln p = ∑ p ln p pz −1 .
To express this in terms of the function Φ, we use 1 x−1 = 1 x + 1 x(x−1) to get −ζ ′(z) ζ(z) = ∑ p ln p pz +∑ p ln p pz (pz −1) .
The first term on the right, of course, is Φ(z) (Definition 12.1). Subtracting the second term on the right, we see that Φ(z) z − 1 z−1 = −ζ ′(z) zζ(z) − 1 z−1 −1 z ∑ p ln p pz (pz −1) = −(z−1)ζ ′(z)+zζ(z) z(z−1)ζ(z) −1 z ∑ p ln p pz (pz −1) .
We tackle the first term on the right-hand side. From Proposition 12.11 (i), we obtain that both the numerator and the denominator are analytic on Rez > 0. We only need to make sure the denominator does not have zeros in Rez ≥1. By Proposition 12.11 (ii), we know that it does not have a zero at z = 1. Lemma 12.12 says that it has no zeroes if Rez ≥1.
Next we look at the second term on the right-hand side. Since ln p is smaller than any positive power of p, the last term on the right-hand side is comparable to p−2z. Since p−2z ≤p−2 whenever Rez ≥1, it converges uniformly in that region and is thus analytic in the desired region (Proposi-tion 11.14 (ii)).
■ 12.5. The Prime Number Theorem 269 12.5. The Prime Number Theorem Here we first prove that the prime number theorem is equivalent to the exis-tence of a certain improper integral. Then we use the Tauberian theorem to prove that that integral exists. That will finally establish the prime number theorem.
Lemma 12.14. We have i) Z ∞ 1 θ(y)−y y2 dy exists = ⇒lim x→∞ θ(x) x = 1.
ii) lim x→∞ θ(x) x = 1 ⇐ ⇒ lim x→∞ π(x) x/lnx = 1 .
Proof. We first prove (i). Suppose that the conclusion of the lemma does not hold. Then for some ε > 0 either there is a sequence of xi such that limi→∞xi = ∞with θ(xi) > (1 + ε)xi or the same holds with θ(xi) < (1 − ε)xi.
Let us assume the former. Since θ is monotone, we have for all i Z (1+ε)xi xi θ(y)−y y2 dy > Z (1+ε)xi xi (1+ε)xi −y y2 dy = −(1+ε)xiy−1−lny (1+ε)xi xi .
The latter can easily be worked out and yields ε −ln(1+ε) for each i. Since this is strictly greater than 0 by exercise 10.9, I(s) = R s 1 θ(y)−y y2 dy cannot converge to a fixed value as s tends to infinity.
The proof of non-convergence if θ(xi) < (1 −ε)xi is almost identical (exercise 12.17).
To prove (ii), we use Lemma 12.2 to establish that π(x)−θ(x) lnx = Z x 2 θ(t) t (lnt)2 dt .
Next we use (12.6) to get rid of the θ(x) in the integrand, and subsequently (12.8) to estimate the remaining integral. For large x, this gives π(x)−θ(x) lnx ≤ln4 x (lnx)2 (1+ε), for any ε > 0. Now we multiply both sides by lnx/x to obtain the result.
■ 270 12. The Prime Number Theorem So this lemma implies that to prove the prime number theorem at this point, we need to show that R ∞ 1 θ(x)−x x2 dx = R ∞ 0 (θ(et)e−t −1) dt exists. We restate Theorem 2.21 in its full glory.
Theorem 12.15 (Prime Number Theorem). We have 1) lim x→∞ π(x) (x/lnx) = 1 and 2) lim x→∞ π(x) R x 2 lnt dt = 1.
Proof. The equivalence of parts (1) and (2) is due to the fact that L’Hopital’s rule implies that limx→∞ x(lnx)−1 R x 2 (lnt)−1 dt = 1. Thus, for example, lim x→∞ π(x) R x 2 lnt dt = lim x→∞ π(x) x/lnx x/lnx R x 2 lnt dt .
The same reasoning works vice versa (exercise 12.10).
So we only need to prove part (1). Lemma 12.3 gives Φ(z+1) z+1 −1 z = Z ∞ 0 θ(et)e−t −1 e−zt dt .
Proposition 12.13 says that the left-hand side has an analytic continuation in Rez ≥0 while equation (12.6) says that θ(et)e−t −1 is bounded. But then, by Theorem 11.18, R ∞ 0 (θ(et)e−t −1) dt exists. Finally, Lemma 12.14 implies that then (1) holds.
■ 12.6. Exercises Exercise 12.1. Write out in full the computations referred to in the proofs of Lemmas 12.2 and 12.3.
Proposition 12.16 (Abel Summation). For the sequence {an}∞ n=1, denote A(x) = ∑n≤x an. Then for any differentiable f, we have ∑ n≤x an f(n) = A(x) f(x)− Z x 1 A(t) f ′(t)dt .
12.6. Exercises 271 Exercise 12.2. a) Show that for any small ε > 0, ∑ n≤x an f(n) = Z x 1−ε f(t)dA(t).
b) Apply integration by parts to (a), to get ∑ n≤x an f(n) = A(x)f(x)− Z x 1−ε A(t) f ′(t)dt .
(Hint: you need that A(1−ε) = 0.) c) Prove Proposition 12.16. (Hint: you need that A(t)f ′(t) is finite and continuous at t = 1.) Exercise 12.3. Recall the notation ⌊x⌋(floor) and {x} (fractional part) from Definition 2.1.
a) Use Abel summation to show that ∑ n≤x 1 n = x−{x} x + Z x 1 t −{t} t2 dt .
(Hint: set an = 1 and f(x) = 1 x .) b) Use (a) to show that ∑ n≤x 1 n −1−lnx = −{x} x − Z x 1 {t} t2 dt .
c) Use (a) to show that lim x→∞ ∑ n≤x 1 n −1−lnx+ Z x 1 {t} t2 dt = 0.
d) Show that the Euler-Mascheroni constant γ := 1−limx→∞ R x 1 {t} t2 dt sat-isfies 1−π2 12 < γ < 1. (Hint: show that R n+1 n t−n−1/2 t2 dt is negative. Then use exercise 2.24 (c) and the fact that ζ(2) = π2 6 . Note: in fact γ ≈0.577··· . At the time of this writing (2021), it is unknown whether γ is irrational.) Exercise 12.4. a) Follow exercise 12.3 to show that ∑ n≤x lnn = ⌊x⌋lnx−(x−1)+ Z x 1 {t} t dt .
b) Show that (a) implies that 1 n nn en < n! < n nn en .
(Hint: use that the absolute value of the integral in (a) is less than lnx.) Exercise 12.4 proves part of what is known as Stirling’s formula , namely: n! = √ 2πn en nn 1+ 1 12n +··· .
272 12. The Prime Number Theorem Exercise 12.5. a) Use Proposition 12.16 to show that for Rez > 1 ζ(z) = z Z ∞ 1 ⌊t⌋ tz+1 dt .
(Hint: write an = 1 and f(x) = x−z.) b) Use that ⌊t⌋= t −{t} to show that ζ(z) = z z−1 −z Z ∞ 1 {t} tz+1 dt = 1 z−1 +1−z Z ∞ 1 {t} tz+1 dt .
c) Use (b) to reprove Proposition 12.11. (Hint: you need to prove analyt-icity of h in Rez > 1.) Exercise 12.6. a) How many trailing zeros does 400! (in decimal notation) have? (Hint: use the proof of Lemma 12.4 with p = 5 and p = 2.) b) How about 400 200 ?
Exercise 12.7. Consider E(x1,x2) := ⌊x1+x2⌋−⌊x1⌋−⌊x2⌋as in the proof of Lemma 12.4 and show that E ∈{0,1}.
Exercise 12.8. a) In Theorem 12.7, show that we can take a = ln2−1 2 ln3 ≈0.14, for all x ≥2.
b) Establish numerically that lnx x Z x 2 (lnt)−2 dt < 1.
(Note: an analytic estimate of this expression is tricky and the reward is modest.
But enthusiastic students can try the following.
Show that R x 2 (lnt)−2 dt −lnx x has a maximum at x = e2. Then give a rough estimate of the expression in (b) for that value of x. You will likely get a much worse estimate than 1.) c) Use (b) and equation 12.7 to show that b = 5ln2 works for all x ≥2.
Exercise 12.9. Suppose we had an “perfect” estimate for Lemma 12.5 of the form n n/2 = c 2n √n for some c > 0. Can you improve Theorem 12.7? (Hint: no. Conclusion: we need a different method to make further progress.) In the next exercise, we prove the equivalence of Theorem 12.15 (a) and (b).
12.6. Exercises 273 Exercise 12.10. a) Compute the derivative of x(lnx)−r for r > 0.
b) Use (a) and L’Hopital to prove that for r > 0 lim x→∞ R x 1 (lnt)−r dt x(lnx)−r = 1.
c) Use (b) to show that parts (a) and (b) of Theorem 12.15 are equivalent.
d) Compare (b) to (12.8).
In the next two problems we prove the following result.
Proposition 12.17. Let pn denote the nth prime. The prime number theorem is equivalent to lim n→∞ pn n lnn = 1.
Exercise 12.11. For this exercise, assume that limx→∞ y x/lnx = 1 and that x →∞if and only if y →∞. (In fact, y stands for π(x), and we know that x →∞if and only if π(x) →∞, see Theorem 2.17.) a) Suppose limx→∞fi(x) = ∞and limx→∞ f1(x) f2(x) = 1. Show that lim x→∞ ln f1(x) ln f2(x) = 1.
(Hint: for x large, (1 −ε) < f1(x) f2(x) < (1 + ε), multiply by f2(x), and take logarithms.) b) Show that limx→∞lnlnx lnx = 0. (Hint: substitute x = eet.) c) Use the hypotheses and (a) to show that x ylny = x ylny y x/lnx lny lnx−lnlnx = 1 1−lnlnx lnx .
d) Use (b) to show that the limit in (c) as x →∞tends to 1. Use the hypotheses to change to the limit as y →∞.
e) Show that (d) implies one way of Proposition 12.17.
Exercise 12.12. For this exercise, assume that limy→∞ x ylny = 1 and that x →∞if and only if y →∞. See exercise 12.11.
a) Follow exercise 12.11 in reverse to show that lim x→∞ y x/lnx = lim x→∞ x ylny y x/lnx lny lnx−lnlnx = lim x→∞ 1 1−lnlnx lnx = 1.
b) Show that (b) implies the other direction of Proposition 12.17.
c) Whereabouts is the nth prime located?
274 12. The Prime Number Theorem Exercise 12.13. In this exercise, we fix any K > 1 and {xi}∞ i=1 is a sequence such that limi→∞xi = ∞. We also set x′ = Kx for notational ease.
a) Show that if π(x′ i) = π(xi) and limi→∞ π(xi) xi/lnxi exists, then lim i→∞ π(x′ i) x′ i/lnx′ i = 1 K lim i→∞ π(xi) xi/lnxi .
b) Show that (a) and the prime number theorem imply that for large enough x, there are primes in (x,x′]. (Hint: if (a) holds, then there are no primes in [xi,Kxi].) c) Show that in fact, the prime number theorem implies lim i→∞ π(x′ i) π(xi) = K .
d) Show that (c) implies that for large enough x, there are approximately (K −1)π(x) primes in (x,x′].
In fact, the following holds for all n. We omit the proof, which involves some careful computations. It can be found in .
Proposition 12.18 (Bertrand’s Postulate). For all n ≥2 there is a prime in the interval [n,2n).
The same reference also mentions an open (in 2018) problem in this direction: Is there always a prime between n2 and (n+1)2?
Exercise 12.14. a) Show for every m ∈N, the set {m! + 2,··· ,m! + m} contains no primes. (Hint: for 2 ≤j ≤m we have j | (m!+ j).) b) Show that from Proposition 12.17, we might reasonably expect the “ex-pected” prime gap pn+1 −pn to be equal to Gn := (n+1)ln(n+1)−nlnn ≈ln((n+1)e), if n large.
c) Use the prime number theorem to show that Gn ≈ln pn+1 −lnln pn+1 +1 ≈ln pn+1 .
d) Assume the twin prime conjecture to show that pn+1−pn ln pn+1 does not con-verge to a limit. See also Figure 77.
d) Use lemma 12.14 to show that the prime number theorem is equivalent to saying that the sum of the first n “expected” prime gaps equals pn+1.
Exercise 12.15. a) Show for every m ∈N, the set {m! + 2,··· ,m! + m} contains no primes. (Hint: for 2 ≤j ≤m we have j | (m!+ j).) b) Compare that prime gap at pn ∼m! with the gap you expect from exer-cise 12.14. (Hint: use exercise 12.4 (b).) ( 12.6. Exercises 275 Figure 77. The prime gaps pn+1 −pn divided by ln pn+1 for n in {1,··· ,1000}.
Exercise 12.16. We give a different proof of Lemma 12.14 (ii) (following ).
a) Show that θ(x) ≤π(x)lnx.
b) Show that (1−ε)lnx ∑ x1−ε≤p≤x 1 ≤ ∑ x1−ε≤p≤x ln p ≤θ(x).
c) Show that for all ε > 0 π(x)−x1−ε ≤ ∑ x1−ε≤p≤x 1.
d) Use (a), (b), and (c) to show that for all ε > 0 (1−ε)(π(x)−x1−ε)lnx x ≤θ(x) x ≤π(x)lnx x .
e) Use (d) to prove Lemma 12.14 (ii). (Hint: show that limx→∞x−ε lnx = 0 by substituting x = et.) Exercise 12.17. a) Suppose that in the proof of Lemma 12.14 there is a sequence of xi such that limi→∞xi = ∞with θ(xi) < (1 −ε)xi for some ε > 0. Show that the integral in the lemma cannot converge.
b) How about if both occur and alternate?
We define two new functions. This definition usually accompanies Defini-tion 12.1.
Definition 12.19. The von Mangoldt function is given by Λ(n) := ln p if n = pk where p is prime and k ≥1 0 otherwise 276 12. The Prime Number Theorem The second Chebyshev function is given by ψ(x) := ∑ n≤x Λ(n).
Just like the first Chebyshev function θ(x), the second Chebyshev function ψ(x) is often used as a more tractable version of the prime counting function π(x). In particular, in exercises 12.18 and 12.19, we will prove a lemma similar to Lemma 12.14, namely Lemma 12.20. We have lim x→∞ ψ(x) x = 1 ⇐ ⇒lim x→∞ π(x) x/lnx = 1.
Exercise 12.18. a) Show that ψ(x) = ∑pk≤x ln p. (Hint: from Definition 12.19. Note that this means that ψ counts all prime powers no greater than x.) b) Show that ψ(x) = ∑p≤x ln p j lnx ln p k . (Hint: this expression only increases at x a power of a prime.) c) Show that ψ(x) ≤∑p≤x lnx. (Hint: ⌊a⌋≤a.) d) Show that (c) implies that ψ(x) ≤π(x)lnx.
Exercise 12.19. a) Show that Definitions 12.1 and 12.19 imply that θ(x) ≤ ψ(x).
b) Use (a) and exercises 12.16 (d) and 12.18 (d) to show that (1−ε)(π(x)−x1−ε)lnx x ≤θ(x) x ≤ψ(x) x ≤π(x)lnx x .
c) Use (b) and Lemma 12.14 (ii) to prove Lemma 12.20.
Exercise 12.20. Plot θ(x)/x, ψ(x)/x, and π(x)lnx/x in one figure. (See for example, Figure 78). Compare with exercise 12.18. b) Show that all three tend to 1 as x tends to infinity.
12.6. Exercises 277 Figure 78. The functions θ(x)/x (green), ψ(x)/x (red), and π(x)lnx/x (blue) for x ∈[1,1000]. All converge to 1 as x tends to infinity. The x-axis is horizontal.
Exercise 12.21. The analysis of this exercise should be compared to the proof of Proposition 12.13.
a) Use Lemma 12.10 and Corollary 11.12 to show that −ζ ′(z) ζ(z) = ∑ p ∞ ∑ n=1 ln p pnz is analytic for Rez > 1.
b) Use Definition 12.19 to show that for Rez > 1 −ζ ′(z) ζ(z) = ∞ ∑ n=1 Λ(n) nz .
c) Use Abel summation (Proposition 12.16) and Definition 12.19 to show that for Rez > 1 −ζ ′(z) ζ(z) = z Z ∞ 1 ψ(x)x−z−1 dx.
(Hint: in the proposition, set f(x) = x−z and A(x) = ψ(x). Then use that in the boundary term, ψ(x)/x converges to 1.) d) Subtract z/(z−1) from (c) and divide by z to conclude that for Rez > 1 −ζ ′(z) zζ(z) − 1 z−1 = Z ∞ 1 ψ(x)−x xz+1 dx.
Exercise 12.22. Show that limx→∞a(x)/x = 1 is equivalent to the follow-ing. For all ε > 0, we have |a(x)−x| < εx for x large enough.
278 12. The Prime Number Theorem Exercise 12.23. For this problem, we assume that there is a θ ∈(1/2,1) so that |ψ(x)−x| ≤Kxθ.
a) Note (exercise 12.22) that this is stronger than limx→∞ψ(x)/x = 1.
b) Use exercise 12.21 to show that −ζ ′(z) zζ(z) − 1 z−1 = ∑ n≥1 Z n+1 n ψ(x)−x xz+1 dx.
c) Show that our hypothesis for this exercise implies that Z n+1 n ψ(x)−x xz+1 dx ≤2Knθ−Rez−1 .
d) Use Proposition 11.14 to show that the right hand side of (b) is analytic for Rez > θ.
e) Show that (d) implies that ζ(z) has no zeros in Rez > θ.
In the next two problems, we prove a second version of the Tauberian theorem in Chapter 11. This is essentially just a reformulation of Theorem 12.15 (2), but with θ(n) replaced by an arbitrary sequence an satisfying certain condi-tions. The proof is also essentially the same.
Theorem 12.21. Suppose an ≥0 so that there is a K > 0 with A(x) := ∑n≤x an ≤Kx. Define G(z) := ∞ ∑ n=1 an nz .
G is analytic on Rez > 1. Assume also that G admits an analytic continua-tion to Rez ≥1 except for a simple pole at 1 with residue 1. Then lim x→∞ A(x) x = 1.
Exercise 12.24. a) Show that G is analytic on Rez > 1. (Hint: use the condition on A(x) and Proposition 11.14 (ii).) b) Use (a) and Abel summation to show that G(z) = z Z ∞ 1 A(x)x−1−z dx.
c) Show that G(z)− z z−1 = z Z ∞ 1 A(x)−x x1+z dx = z Z ∞ 0 A(et)−et e−zt dt .
d) In (c), set z′ +1 = z and then drop the prime to show that H(z) := G(1+z) z+1 −1 z = Z ∞ 0 A((x)−x x2+z dx = Z ∞ 0 A(et)e−t −1 e−zt dt .
12.6. Exercises 279 Exercise 12.25. a) Show that the function H(z) of exercise 12.24 (d) has an analytic continuation to Rez ≥0. (Hint: the pole at z = 0 has been canceled by the subtraction of 1/z.) b) Use Theorem 11.18 to show that R ∞ 0 A(x)−x x2 dx converges.
c) Use Lemma 12.14 (i) to show that limx→∞ A(x) x = 1.
Exercise 12.26. a) Show that lim n→∞ ∏ p≤n p ! 1 n = e if and only the prime number theorem holds. (Hint: see Lemma 12.14 (ii).) b) See Figure 79). Show that lim n→∞(lcm(1,2,··· ,n)) 1 n = e if and only the prime number theorem holds. (Hint: see Lemma 12.20.) Figure 79. Plot of the function f(n) := (lcm(1,2,··· ,n)) 1 n for n in {1,··· ,100} (left) and in {104,··· ,105} (right). The function converges to e indicated in the plots by a line.
Chapter 13 Primes in Arithmetic Progressions Overview. An arithmetic progression is a set S of the form S := {a + kq | k ∈N}. If gcd(a,q) = d > 1, then any two distinct numbers in S are not co-prime. Thus, in that case, S can contain at most one prime. We will see that asymptotically the primes are distributed equally over the remaining arith-metic progressions, namely the sets {a+kq | k ∈N} such that gcd(a,q) = 1.
One of the more accessible introductions to the material in this chapter is . For section 13.6, we used and .
13.1. Finite Abelian Groups Definition 13.1. Two groups g and H are isomorphic if there exists a bijec-tive homomorphism f : G →H (see also exercise 13.3).
Definition 13.2. A cyclic group is a group generated by a single element.
The proof of the following proposition is loosely based on the analo-gous proof in [appendix 3C]. It uses the simple observation that every element g of a finite Abelian group generates a cyclic group. This is evi-dent, because the sequence {gi} can have finitely many distinct elements, and so the smallest value of i ≥0 where a repeated value occurs must be the order o of the element g.
281 282 13. Primes in Arithmetic Progressions Proposition 13.3 (Fundamental Theorem of Finite Abelian Groups).
Any finite Abelian group G of order n is isomorphic to a cartesian prod-uct of finite cyclic groups Z+ o1 ×···×Z+ or. Furthermore, ∏r i=1 oi = n.
Proof. There are finitely many ways of choosing a non-empty subset S of elements of G. Since each element has order at most n, for each of these subsets, we can find out whether it generates G. Let r be the minimal car-dinality of the subsets that generate G and denote by Sr the (non-empty) collection of all such sets of generators of cardinality r.
Pick S in Sr, denote its elements by gi, and the order of gi ∈S by oi(S). By construction, there is a map σ(S) from ∏r i=1 {0,1,··· ,oi(S) − 1} = Z+ o1 ×···×Z+ or onto G given by σ : (a1,··· ,ar) → r ∏ i=1 gai i .
Now let us assume that there is a non-empty set S r ⊆Sr so that for S in S r, σ(S) is not a bijection. We will show that this leads to a contradiction.
For S in S r, there are i and 0 ≤ai,a′ i < oi(S) such that r ∏ i=1 gai i = r ∏ i=1 ga′ i i ⇐ ⇒ r ∏ i=1 g(ai−a′ i) mod oi(S) i = r ∏ i=1 gci i = 1, where we have set ci equal to the least residue of (ai −a′ i) modulo oi(S).
Note that in this expression at least two of the coefficients ci are greater than 0. Now let s(S) := min ci∈{0,···,oi(S)−1} ( r ∑ i=1 ci : r ∏ i=1 gci i = 1 ) ≥2.
Finally, minimize s(S) over S r s−:= min S∈S r s(S) ≥2.
(13.1) Let {gi}r i=1 be the collection of generators at which this minimum is assumed. At least two of the ci’s are greater than 0, say, c2 ≥c1 > 0. Define f1 = g1g2 and fi = gi for all i > 1. This change of variables is invertible, so {hi}r i=1 still generate G. A simple calculation gives 1 = r ∏ i=1 gci i = f c1 1 f c2−c1 2 f c3 3 ··· f cr r .
13.2. The Hermitian Inner Product 283 Thus s−has decreased, which contradicts (13.1). This shows that S r is empty and thus for all S in Sr, σ(S) is a bijection.
It is also a homomorphism, because for a and a′ in Z+ o1 ×···×Z+ or σ(a+a′) = r ∏ i=1 gai i · r ∏ i=1 ga′ i i .
Thus σ is an isomorphism (see exercise 13.3). Clearly, the number of ele-ments in G equals ∏r i=1 oi(S) which must therefore be equal to n.
■ 13.2. The Hermitian Inner Product Later on, we will briefly need to consider V = Cn as an inner product space.
The Hermitian inner product generalizes the dot product of Rn.
Definition 13.4. The (standard) Hermitian inner product on Cn is given by (x,y) = x1y1 +···+xnyn , where y indicates the complex conjugate of y.
One easily checks that this binary operation satisfies the requirements that for all x, y, and z in V and α in C 1) (x,x) ≥0 positivity 2) (x,x) = 0 ⇐ ⇒x = 0 definiteness 3) (x,αu+v) = α(x,u)+(x,v) linearity 4) (x,y) = (y,x) conjugate symmetry More generally, any function V × V that satisfies these requirements is called an inner product, but we will not be needing that generality here.
Definition 13.5. A set {ei}n i=1 of vectors in V is an orthonormal basis if for all i ̸= j and all x in V 1) (ei,ei) = 1 unit vectors 2) (ei,e j) = 0 orthogonality 3) ∃αi such that x = ∑n i=1 αiei basis 284 13. Primes in Arithmetic Progressions The property that is crucial for us is that the αi in item (3) of this defi-nition can be computed easily, namely x = n ∑ i=1 (ei,x)ei .
(13.2) For more details and a good general introduction, see [Chapter 6].
13.3. Characters of Finite Abelian Groups Definition 13.6. A character of a group G is a complex-valued homomor-phism f : G →C×.
For example, the identity e of a group G satisfies e2 = e and since f is a multiplicative homomorphism, we have that f(e)2 = f(e) and so f(e) = 0 or f(e) = 1. The former is excluded because 0 is not in the domain of C×.
Thus f(e) = 1 for any character. An example of a character of G is the constant function, f(g) = 1, also called the principal character. We indicate it by f0.
Remark 13.7. From now on, we will denote the field of units in Zb by Z× b .
Sometimes we will consider it as multiplicative groups (as directly below), and other times we may consider them a field. See Section 5.4.
Before continuing, let us look at a few examples of characters, namely G = Z× 5 and G = Z× 8 .
mod 5 f0 f1 f2 f3 1 1 1 1 1 2 1 i -1 -i 3 1 -i -1 i 4 1 -1 1 -1 mod 8 f(0,0) f(0,1) f(1,0) f(1,1) 1 1 1 1 1 3 1 1 -1 -1 5 1 -1 1 -1 7 1 -1 -1 1 (13.3) The table on the left lists the characters of Z× 5 . Each column corresponds to a different character. The table on the right lists the characters of Z× 8 . Note that each of these groups has four characters, but they are not the same.
How do we determine these characters? The short answer is: exploit multiplicativity. First look at Z× 5 . We note it is a cyclic group generated by the element 2, namely 2k mod 5 cycles through the values 2, 4, 3, and 1 for 13.3. Characters of Finite Abelian Groups 285 k ∈{1,2,3,4}. Thus f(24) = ( f(2))4 = 1, and so for any character f, the value of f(2) must be a 4th root of unity. So choose (as in the left table of (13.3)) fm(2) = e2πi m 4 .
For any choice of m, we can obtain a multiplicative function as follows e2πi(k+ℓ) m 4 = e2πik m 4 e2πiℓm 4 = ⇒ fm(2k+ℓ) = fm(2k)fm(2ℓ).
(13.4) This example is wonderful, because it turned out that Z× 5 is isomorphic to Z+ 4 which simplifies things: we get something very reminiscent of a discrete Fourier transform (see Definition 13.27).
The group Z× 8 also has 4 elements, namely {1,3,5,7}. But none of these elements has order 4, for 32 =8 52 =8 72 =8 1. Thus for any character f, each of f(3), f(5), and f(7) must be square roots of unity. This group is therefore not isomorphic to Z+ 4 . However, consider (a1,a2) (0,0) (0,1) (1,0) (1,1) 3a1 ·5a2 mod 8 1 5 3 7 This gives a bijection h : Z× 8 →Z+ 2 ×Z+ 2 . But in Z+ 2 ×Z+ 2 , h(3a15a2)h(3b15b2) = (a1,a2)(b1,b2) = (a1 +b1,a2 +b2) h(3a15a2 ·3b15b2) = (a1 +b1,a2 +b2) .
It also shows that h is a homomorphism, and thus Z× 8 is isomorphic to Z+ 2 × Z+ 2 . So let m = (m1,m2) where mi ∈{0,1} and set fm(3) = e2πi m1 2 and fm(5) = e2πi m2 2 .
So that (as illustrated in the right table of (13.3)) fm(3a15a2) = e2πi a1m1 2 e2πi a2m2 2 = e2πi( a1m1 2 + a2m2 2 )) .
(13.5) fm is multiplicative by the same calculation as done in (13.4), but now sep-arated out in ‘components’ to prove that fm(3k1+k25ℓ1+ℓ2) = fm(3k15ℓ1)fm(3k25ℓ2).
(13.6) The student is asked to provide a few more details in exercise 13.1.
These computations tell us what is going on. We first simplify the notation, and then formulate the relevant theorem.
286 13. Primes in Arithmetic Progressions Definition 13.8. For the remainder of this chapter, we abbreviate: Z+ o := Z+ o1 ×···×Z+ or ; n := r ∏ i=1 oi ; and for a and m in Z+ o , we set m/o := ( m1 o1 ,··· , mr or ) and ga := r ∏ j=1 g a j j ; a·(m/o) := r ∑ i=1 ajm j mod o j oj .
Theorem 13.9. Let G be an n element Abelian group. With the notation of Definition 13.8, we have: i) The characters fm of G are given by fm(ga) = e2πia·(m/o) .
ii) The characters fm are all orthogonal to one another in the sense that: ∀m, ℓ∈Z+ o : ∑ a∈Z+ o fm(ga) fℓ(ga) = n if m = ℓ 0 if m ̸= ℓ (see Figure 80). Thus the n characters are all distinct.
(1,1) (1,−1) Figure 80. The two characters modulo 3 illustrate the orthogonality of the Dirichlet characters.
Proof. By complete multiplicativity (because f is a homomorphism), any character f is completely determined by its value f(gi) on the r generators 13.3. Characters of Finite Abelian Groups 287 of G. So f r ∏ j=1 g a j j !
= r ∏ j=1 f(gj)aj .
f(gj) must be an ojth root of unity, and is thus equal to exp(2πimj/oj).
The second statement follows using Definition 13.8.
∑ a∈Z+ o fℓ(ga) fm(ga) = ∑ a∈Z+ o e2πia·((m−ℓ)/o) .
(13.7) If m = ℓ, the exponent is zero and so ∑a∈Z+ o 1 = n. The above sum is really a sum of products (Definition 13.8) which can be converted into a product of sums (exercise 4.3) of the form ∑ a j∈Z+ oj e 2πi a j(m j−ℓj)k j oj .
So if m ̸= ℓ, then there is an j such that m j −ℓj ̸= 0. The above sum then has the form o−1 ∑ a=0 e2πi aK o = e2πi oK o −1 e2πi K o −1 = 0, and so the product of the sums also reduces to zero.
■ Theorem 13.9 implies that there is an injection from Z+ o to the char-acters given by F : m →fm. It is actually a bijection, because an injection between sets of the same size — namely {m} and {fm} — must be a bijec-tion. A slight variation on equation (13.7) allows us to go a little further, namely fm+ℓ(ga) = fm(ga) fℓ(ga).
Thus the bijection becomes a group homomorphism. Using Proposition 13.3, we obtain the following corollary.
Corollary 13.10. The characters of a finite Abelian group G together with the multiplication ( fm fn)(ga) = fm(ga) fn(ga) form a group that is isomor-phic to G which in turn is isomorphic to Z+ o .
There is another interesting way to look at these characters. Order the elements of Z+ o by defining some bijection, or counter, ϕ from Z+ o to {1,··· ,n}. We can then think of fm(ga) as the ϕ(a)th component of the vector fm in Cn. This is what we did in the tables (13.3). Theorem 13.9 implies that the vectors fm now form an orthogonal basis of Cn equipped 288 13. Primes in Arithmetic Progressions with the Hermitian inner product (Definition 13.4). Reformulating the theo-rem gives yet another corollary. See Definition 13.27 and the exercises that follow it for more details.
Corollary 13.11. If we define the vectors em as n−1/2 fm, then the set {em}m∈Z+ o is an orthonormal basis (Definition 13.5) of Cn.
13.4. Dirichlet Characters and L-functions The Dirichlet characters are essentially the characters of Z× q , the multiplica-tive group of the reduced residues of Zq (Z modulo q) with identity element 1. Since we will use Dirichlet characters as the coefficients in Dirichlet series, we need to convert them into arithmetic functions.
Definition 13.12. Corresponding to each character f : Z× q →C×, we define a q-periodic arithmetic function χ f , the Dirichlet character modulo q, as follows: χf (n) = f(Resq (n)) if gcd(n,q) = 1 χf (n) = 0 if gcd(n,q) > 1 By Corollary 13.10, these characters form a multiplicative subgroup of C that we will denote by Xq.
Recall that the principal Dirichlet character evaluates to 1 on numbers relatively prime to q and equals 0 elsewhere. It will be denoted by χf0 or χ1.
More generally, it is easy to see that the Dirichlet characters are com-pletely multiplicative (Definition 4.2) arithmetic (Definition 4.1) functions.
For if gcd(ab,q) > 1, then gcd(a,q) > 1 or gcd(b,q) > 1 (or both). And so from Definition 13.12, we see that then χ(ab) = χ(a)χ(b) = 0. On the other hand, if both gcd(a,q) = 1 and gcd(b,q) = 1, then since f is a homo-morphism, χ(ab) = χ(a)χ(b). That means that for any Dirichlet character χ, we get χ(1) = 1.
Remark 13.13. Since from now on, we will only deal with Dirichlet char-acters modulo q ∈N, we will, in the interest of brevity, refer to these simply as characters from now on.
13.4. Dirichlet Characters and L-functions 289 Definition 13.14. The Dirichlet L-series associated to a Dirichlet character χ is defined as L(χ,z) := ∞ ∑ n=1 χ(n) nz .
The Dirichlet L-function associated to a Dirichlet character χ is the func-tion defined by the analytic continuation of the Dirichlet L-series.
Often these are abbreviated to L-series and L-function , though some authors reserve those names for generalizations of those notions.
These L-function have the “feel” of a zeta function as the next result indicates. We will use a complicated combination of L-functions as a “new” zeta function to prove our main theorem. In the remainder of this chapter, we abbreviate the function f has a well-defined analytic continuation in the region S by f is analytic in S.
Proposition 13.15. If ψ is bounded and completely multiplicative, then L(ψ,z) is analytic in Rez > 1 and ln ∞ ∑ n=1 ψ(n)n−z !
= lnL(ψ,z) = −∑ p prime ln(1−ψ(p)p−z) =∑ p ∞ ∑ n=1 ψ(pn) npnz .
If ψ is periodic and has average zero, then L(ψ,z) is analytic in Rez > 0.
Proof. The first equality follows from the definition of L. We paraphrase the second proof of Proposition 2.20. Using the complete multiplicativity of ψ, we obtain ψ(2)2−zL(ψ,z) = ∞ ∑ n=1 ψ(2)ψ(n)2−zn−z = ∞ ∑ n=1 ψ(2n)(2n)−z .
Thus 1−ψ(2)2−z L(ψ,z) = ∑ 2∤n ψ(n)n−z .
Subsequently we multiply this expression by (1 −ψ(3)3−z). This has the effect of removing multiples of 3 from the remaining terms. Continuing like this, it follows that eventually1 ∏ p prime (1−ψ(p)p−z) !
L(ψ,z) = 1.
1Note that we use factorization in terms of primes here 290 13. Primes in Arithmetic Progressions Upon taking the logarithm, we arrive at the second equality. The third one — and analyticity — follows from Lemma 12.10.
To prove the last part, we use Proposition 12.16 and compute L(ψ,z) = ∑ n≤x ψ(n)n−z = Ψ(x)x−z +z Z x 1 Ψ(t)t−z−1 dt , where Ψ(x) = ∑n≤x ψ(n). Since ψ has period, say, q with average 0, we have Ψ(x+q) = Ψ(x), and so Ψ is bounded. Thus both terms in the above equation converge for Rez > 0.
■ 13.5. Preliminary Steps The way we want to prove the prime number theorem for arithmetic pro-gressions is by defining an arithmetic function hq,a : N →C — the so-called indicator function — that equals 1 when n is equal to a modulo q and 0 else-where. With that function in hand, we then define ∑p hq,a(n)n−z and use the machinery in chapter 12 to compute the density of the primes in the arith-metic progression (a,a + q,a + 2q,···). But there is a problem here. The function h is not multiplicative: hq,a(a2) is not generally equal to (hq,a(a))2 — by way of example, h3,2(2) = 1 while h3,2(22) = 0. So we have to be more careful.
Lemma 13.16. Let gcd(a,q) = 1. We have ∑ χ∈Xq χ(a)χ(n) = ϕ(q) if n =q a 0 else .
Thus the indicator function hq,a equals (ϕ(q))−1 ∑χ∈Xq χ(a−1)χ(n).
Proof. Since χ(a) has unit modulus, we have that χ(a)χ(a) = 1. Because there are ϕ(q) characters, the first equality follows.
The second equality is automatic if either a or n is not co-prime to q. If a and n are distinct co-primes, then recall that the characters form an orthogonal basis. Thus there must be another character χ∗∈Xq so that χ∗(a−1n) ̸= 1. Since the reduced residues mod q form a field, from the above we must have that χ(a) = χ(a−1). Using multiplicativity, we obtain 13.5. Preliminary Steps 291 that ∑χ∈Xq χ(a)χ(n) equals ∑ χ∈Xq χ(a−1n) = ∑ χ∈Xq (χ∗χ)(a−1n) = χ∗(a−1n) ∑ χ∈Xq χ(a−1n) = 0.
The first equality holds because χ∗χ runs through the entire group (es-sentially the same argument as Lemma 5.3). The second by multiplica-tivity.
■ We will define quantities that allow us to mimic the proof of the prime number theorem. To facilitate this, we use uppercase letters of the cor-responding notation we used earlier. So ζ becomes Z, π becomes Π, θ becomes Θ, and Φ stays the same. We will then proceed to give a proof of the prime number theorem for arithmetic progressions that follows the proof of Theorem 12.15 as closely as possible. As in Chapter 12, ∑p and ∏p mean sum or product over the (positive) primes.
The following definition should be compared with the definition of the Riemann zeta function (Definition 2.19), of the prime counting function (in Theorem 2.21), and Definition 12.1.
Definition 13.17. We introduce a new zeta function Zq,a, a function Πq,a that counts the primes congruent to a mod q, and two auxiliary functions.
Zq,a(z) := ∏ χ∈Xq L(χ,z)χ(a) = exp ∑ χ∈Xq χ(a)ln(L(χ,z)) !
.
Πq,a(x) := ∑ p≤x p=qa 1.
Θq,a(x) := ϕ(q) ∑ p≤x p=qa ln p and Φq,a(z) := ϕ(q) ∑ p=qa ln p pz .
From now on, we restrict a to Z× q , the reduced residues modulo a.
Remark 13.18. Recall that there is at most 1 prime in each congruence class that is not co-prime with q.
Note that Θq,a(x) ≤ϕ(q)θ(x). Our first inequality follows from (12.6).
∃C > 0 such that Θq,a(x) ≤Cx.
(13.8) The factor 1/ϕ(q) that figures so prominently in our main result, The-orem 13.26, shows up in the following lemma.
292 13. Primes in Arithmetic Progressions Lemma 13.19. We have for x ≥2 Πq,a(x) = Θq,a(x) ϕ(q)lnx + 1 ϕ(q) Z x 2 Θq,a(t) t (lnt)2 dt .
Proof. First note that since 2 is the smallest prime, equation (12.2) gives Πq,a(x) = 1 ϕ(q) Z x 2−ε d Θq,a(t) lnt .
The rest follows as in Lemma 12.2 ■ Lemma 13.20. For Rez > 1, we have Φq,a(z) z − 1 z−1 = Z ∞ 1 Θq,a(x) x −1 x−z dx = Z ∞ 0 Θq,a(et)e−t −1 e−zt+t dt .
Proof. Using (12.2), we can write Φq,a(z) as R ∞ 1 x−zdΘq,a(x). Then apply (12.3) (partial integration). The proof follows that of Lemma 12.3, except that (12.6) is replaced by (13.8) ■ 13.6. Primes in Arithmetic Progressions Now we follow the reasoning of Sections 12.3 to 12.5 as closely as possible.
Lemma 13.21. For Rez > 1, we have that lnZq,a(z) = −∑ χ∈Xq∑ p χ(a)ln 1−χ(p)e−zln p = ϕ(q) ∑ p∤q ∞ ∑ n=1 pn=qa 1 npnz .
and is analytic in that region.
Proof. The first equality follows from Proposition 13.15. Then we follow the reasoning of Lemma 12.10 to get −ln 1−χ(p)e−zln p = ∞ ∑ n=1 χ(pn) npnz , where we used complete multiplicativity of χ. Since |χ| = 1, this is analytic on Rez > 1. Substitute this back into the lemma. Analyticity then allows us to perform the finite sum over χ first. By Lemma 13.16, this gives a contribution ϕ(q) if both pn =q a and gcd(pn,q) = 1, and else zero. This 13.6. Primes in Arithmetic Progressions 293 proves the second equality of the lemma. Now the proof follows verbatim the second paragraph of the proof of Lemma 12.10.
■ Proposition 13.22. i) The functions (z −1)Zq,a(z) and (z −1)Z′ q,a(z) + zZq,a(z) have well-defined analytic continuations on Rez > 0.
ii) (The analytic continuation of) (z−1)Zq,a(z) evaluated at z = 1 does not equal 0.
Proof. Since χ1(a) = 1, Definition 13.17 gives (z−1)Zq,a(z) = (z−1)L(χ1,z)·exp ∑ χ∈Xq χ̸=χ1 χ(a)ln(L(χ,z)) .
We need to show that (z −1)L(χ1,z) and L(χ,z), χ ̸= χ1, are analytic in Rez > 0, and therefore so is (z −1)Zq,a(z). Adding this function to its derivative gives (z−1)Z′ q,a(z)+zZq,a(z).
Since χ1(n) equals 0 or 1, we can define h(z) := L(χ1,z)− 1 z−1 .
The same argument presented in Proposition 12.11, shows that also here, h is analytic in Rez > 0. Therefore, the same holds for (z−1)L(χ1,z) = (z−1)h(z)+1.
(13.9) Recall that any non-principal χ is orthogonal to the principal character χ1.
Since χ1 is always 1 (on co-primes), χ must have average zero. All charac-ters are periodic by construction, so Proposition 13.15 implies that lnL(χ,z) is analytic in Rez > 0. This proves part (i).
Part (ii) is implied by the fact that (13.9) implies that (z −1)L(χ1,z) evaluated at z = 1 gives 1 and that the exponential in the above expression for (z−1)Zq,a(z) cannot give zero.
■ Lemma 13.23. Zq,a(z) has no zeroes on the line Rez ≥1.
Proof. By Lemma 13.21, we only need to check at z = 1 + iτ for τ real.
Define E := ln(Zq,a(σ)3Zq,a(σ +iτ)4Zq,a(σ +2iτ)). By Proposition 13.22, Zq,a has a simple pole at 1 and no poles in Rez > 1. Thus if Zq,a has a zero at 1+iτ, then the expression eE evaluated at σ +iτ where σ is slightly greater 294 13. Primes in Arithmetic Progressions than 1, would yield a number that is very close to zero. The rest of the proof follows that of Lemma 12.12 verbatim.
■ Proposition 13.24.
Φq,a(z) z − 1 z−1 has an analytic continuation in the closed half plane Rez ≥1.
Proof. By Lemma 13.21, −Z′ q,a(z) Zq,a(z) = ∑ χ∈Xq∑ p χ(a)χ(p)p−z ln p 1−χ(p)p−z = ∑ χ∈Xq∑ p χ(a)χ(p)ln p pz −χ(p) .
To express this in terms of the function Φq,a, we use 1 x−k = 1 x + k x(x−k) to get −Z′ q,a(z) Zq,a(z) = ∑ χ ∑ p χ(a)χ(p)ln p pz +∑ p χ(a)χ(p)2 ln p pz (pz −χ(p)) .
Now we note that by Lemma 13.21, in the region z > 1, we may do the summation over χ first. We then see that by Lemma 13.16, the first term on the right hand side equals Φq,a(z). The rest of the proof follows that of Lemma 12.13 ■ Lemma 13.25. For all q ≥2 and a such that gcd(a,q) = 1: i) Z ∞ 1 Θq,a(y)−y y2 dy exists = ⇒lim x→∞ Θq,a(x) x = 1.
ii) lim x→∞ Θq,a(x) x = 1 ⇐ ⇒lim x→∞ Πq,a(x) x/lnx = 1 ϕ(q) .
(If gcd(a,q) > 1, the density of primes is 0.) Proof. The proof of (i) is entirely parallel to that of Lemma 12.14. For the proof of (ii), we use Lemma 13.19 and (13.8) instead of Lemma 12.2 and (12.6). So, Πq,a(x)−Θq,a(x) ϕ(q)lnx = 1 ϕ(q) Z x 2 Θq,a(t) t (lnt)2 dt ≤ 1 ϕ(q) Cx (lnx)2 (1+ε).
for any ε > 0. Multiply both sides by lnx/x to obtain the result.
■ Theorem 13.26 (Prime Number Theorem for Arithmetic Progressions).
We have 1) lim x→∞ Πq,a(x) (x/lnx) = 1 ϕ(q) and 2) lim x→∞ Πq,a(x) R x 2 lnt dt = 1 ϕ(q) .
13.7. Exercises 295 Proof. The equivalence of (1) and (2) is the same as in Theorem 12.15.
So we only need to prove part (1). Lemma 13.20 gives Φq,a(z+1) z+1 −1 z = Z ∞ 0 Θq,a(et)e−t −1 e−zt dt .
Proposition 13.24 says that the left-hand side has an analytic continuation in Rez ≥0 while equation (13.8) says that Θq,a(e−t)e−t −1 is bounded. But then, by Theorem 11.18, R ∞ 0 (θ(et)e−t −1) dt exists. Finally, Lemma 13.25 implies that then (1) holds.
■ 13.7. Exercises Exercise 13.1. a) Finish the computation of (13.6) to show that fm is mul-tiplicative. (Hint: see equation (13.4).) b) Check that the entries table on the right in (13.3) correspond to (13.5).
Exercise 13.2. a) Show that Z× 5 as a group is isomorphic to Z+ 4 . In other words, find a bijection f : Z× 5 →Z+ 4 such that for all a, b in Z× 5 , f(ab) = f(a)+ f(b).
b) Show that Z× 7 is isomorphic to Z+ 6 .
c) Show that Z+ 6 is isomorphic to Z+ 2 ×Z+ 3 .
Exercise 13.3. Let f : G →H a bijective homomorphism between groups.
Use multiplicative notation.
a) Show that for every a and b in H, there are unique x and y in G such that x = f −1(a) and y = f −1(b), where f −1 is the inverse of f.
b) Show that (a) implies that xy = f −1(a)f −1(b).
c) Show that (b) implies that f(xy) = ab and thus xy = f −1(ab).
d) Conclude that f −1 is also a homomorphism.
Exercise 13.4. a) Show that Z× 16 is isomorphic to Z+ 2 ×Z+ 4 .
b) Show that Z× 16 is not isomorphic to Z+ 8 . (Hint: find the elements of order 8.) c) Consider the residues modulo 16 with addition and multiplication and verify that it is a ring.
d) Find the units (Definition 5.25) of this ring.
e) Show that the units of a (commutative) ring form a multiplicative Abelian group.
296 13. Primes in Arithmetic Progressions Exercise 13.5. a) Find a primitive root a modulo 26 (see Definition 5.5).
b) Find a primitive root b modulo 13.
c) Show that Z× 26 is isomorphic to Z× 13. (Hint: let h map ai to bi and show that h is a bijective homomorphism.) d) Use Theorem 5.7 to prove that for odd primes, Z× pk is isomorphic to Z× 2pk.
Definition 13.27. Given x = (x0,x1,··· ,xn−1)T ∈Cn. The discrete Fourier transform is defined as b xm = n−1 ∑ k=0 xke−2πi km n , for m ∈{0,··· ,n−1}. The inverse discrete Fourier transform is given by xk = 1 n n−1 ∑ m=0 b xme2πi km n .
Exercise 13.6. a) What are the characters of the group Z+ n ?
b) Show that the composition of the discrete Fourier transform and the inverse discrete Fourier transform of Definition 13.27 is the identity (i.e.
they are inverses of one another). (Hint: use Theorem 13.9 and equation (13.2).) c) Set α := e2πi 1 n .
Let F be the n by n matrix whose (m,k) entry is α−(k−1)(m−1). Show that the discrete Fourier transform is: b x = Fx.
d) From Definition 13.27, deduce the inverse F−1 of the matrix F.
Exercise 13.7. a) What are the characters of the group Z+ n ×Z+ m?
b) What are the formulas in this case for the discrete Fourier transform and its inverse? (Hint; think of this as a two-dimensional version of the Fourier transform.) Exercise 13.8. a) Use Theorem 13.9 and exercise 13.5 to construct the characters of Z× 13 and Z× 26.
b) Show that these characters basically correspond to the Fourier transform of Definition 13.27, except that the xk are re-ordered (see also exercise 13.10).
Exercise 13.9. Proposition 2.20 is very similar to Proposition 13.15, but the former was proved in two different ways. Give the “other” proof of Proposition 13.15.
b) Is it sufficient for χ to be multiplicative (i.e. not completely multiplica-tive)?
13.7. Exercises 297 Exercise 13.10. a) For any odd prime p denote by g its smallest primitive root. Show that there is a bijection ind p : Z× p →Z+ p−1 given by ind p(ga) = a.
The value ind p(x) is called the index of x relative top p. The prime root g is called the base.
b) For every odd prime less than 20, choose the smallest primitive root as base, and determine the indices of {1,2,··· , p−1}. Hint: as an example, for p = 17 with base 3, we obtain the following table 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 16 14 1 12 5 15 11 10 2 3 7 13 4 9 6 8 c) Prove that the indices behave like logarithms, that is: indp(ab) =ϕ(p) ind p(a)+ ind p(b) and ind p(ak) =ϕ(p) kind p(a).
Exercise 13.11. In this exercise, we use indices (exercise 13.10) to solve 9x8 =17 8.
a) Use exercise 13.10 (c) to show that the equation above is equivalent to ind17(9)+8ind17(x) =16 ind17(8).
b) Use exercise 13.10 (b) to show that (a) is equivalent to 8ind17(x) =16 8.
c) Use Corollary 3.8 to find the solutions to this equation. (Hint: there are 8 solutions.) Exercise 13.12. Show that for any k > 0 there are infinitely many primes ending in k consecutive 9’s.
There are useful relations between the newly minted functions in this chap-ter and their counterparts in Chapter 12. We prove the following lemma in exercise 13.13 Lemma 13.28. Let q = ∏r i=1 pki i . We have the following equalities: i) L(χ1,z) = ζ(z) ∏p|q 1−p−z i , ii) ∏a∈Z× q Zq,a(z) = ζ(z)ϕ(q) ∏p|q 1−p−z i ϕ(q) .
298 13. Primes in Arithmetic Progressions Exercise 13.13. a) Using the Euler product of Proposition 13.15, show that lnL(χ1,z) = −∑ p ln(1−p−z)+∑ p|q ln(1−p−z), b) Show that (a) implies item (i) of Lemma 13.28.
c) Show that ∏ a Zq,a(z) = L(χ1,z)ϕ(q) ∏ χ̸=χ1 L(χ,z)∑a χ(a) .
d) Show that (c) implies item (ii).
Many special cases of Dirichlet’s theorem can be proved without using the machinery we have developed chapters 11 and 12 and applied in the current chapter. We discuss these cases in the next three problems.
Exercise 13.14. Define S := {3 + 4k | k ∈N}. Assume there are finitely many primes in S, namely {p1,··· , pk} and derive a contradiction. Denote P = 4 k ∏ i=1 pi and D = P−1.
a) Show that D is not prime. (Hint: D =4 3.) b) Use (a) to show that D must have a prime divisor pi in S. (Hint: xy =4 3 iff one of x or y is congruent to 3.) c) Use (a) and (b) to show that there is a k such that D = kpi = −1+4pi k ∏ j̸=i p j .
d) Use (c) to derive that pi | 1, a contradiction.
Exercise 13.15. Define S := {1 + 3k | k ∈N}. Assume there are finitely many primes in S, namely {p1,··· , pk} and derive a contradiction. Denote P = 3 k ∏ i=1 pi and D = P2 +P+1.
a) Show that D must have a non-trivial prime divisor r ̸= 3 and r ̸∈S. (Hint: D =3 1 and D =pi 1.) b) Show that P3 =r 1. (Hint: P3 −1 = (P−1)D.) c) Show that Ord× r (P) = 3. (Hint: if P2 =r 1, then P =r 1 by (b) and so D =r 3; the latter is impossible, because by (a), D =r 0 and r ̸= 3.) d) Use (a) to show that gcd(P,r) = 1 and so Pr−1 =r 1. (Hint: Fermat’s little theorem.) e) Use (c) and (d) to show that 3 | (r −1).
f) Point out the contradiction. (Hint: if r ̸∈S, then r ̸=3 1.) 13.7. Exercises 299 Exercise 13.16. For any q > 1, define S := {1 + qk | k ∈N}. Assume there are finitely many primes in S, namely {p1,··· , pk} and derive a contradiction. Denote P = q k ∏ i=1 pi and D = q−1 ∑ i=0 Pi .
a) Show that D must have a prime divisor r ∤q and r ̸∈S. (Hint: for any divisor e > 1 of q, D =e 1 and similarly D =pi 1. Recall B´ ezout.) b) For r as in (a), show that Pq =r 1. (Hint: Pq −1 = (P−1)D and D = xr.) c) For r as in (a), show that Ord× r (P) = q. (Hint: by (b), d := Ord× r (P) | q; if q = de with d,e > 1, then D = ∑d−1 i=0 Pi ∑e−1 i=0 Pid and so D =e 0; the latter is impossible by (a).) d) Use (a) to show that gcd(P,r) = 1 and so Pr−1 =r 1. (Hint: unique factorization and Fermat’s little theorem.) e) Use (c) and (d) to show that q | (r −1).
f) Point out the contradiction.
Dirichlet proved a weaker version of Theorem 13.26 that does not use the Tauberian convergence argument of Theorem 11.18. We discuss the proof in exercise 13.17 below.
Theorem 13.29 (Dirichlet’s Theorem). Define S := {n ∈N : n =q a}.
Then lim z→1+ ∑p∈S p−z ∑p p−z = 1 ϕ(q) .
Exercise 13.17. a) Use Proposition 13.15 to show that for real z ≥1 and χ ̸= χ1, ∑p χ(a)χ(p) pz is bounded.
b) Use Lemma 13.16 to show that for Rez > 1 ∑ p=qa 1 pz = 1 ϕ(q) ∑ χ∈Xq∑ p χ(a)χ(p) pz = 1 ϕ(q) L(χ1,z)+ ∑ χ̸=χ1∑ p χ(a)χ(p) pz !
.
c) Use Proposition 12.11 (ii) and (13.9) to show that limz↘1+ L(χ1,z) ζ(z) = 1.
d) Show that (a), (b), and (c) imply Dirichlet’s theorem.
Definition 13.30. The natural density of a set S ⊆T relative to T is lim x→∞ S(x) T(x) , 300 13. Primes in Arithmetic Progressions where S(x) = card(S ∩[1,x]) and T(x) = card(T ∩[1,x]). The Dirichlet density of a set S ⊆T relative to T is lim z↘1+ ∑n∈S n−z ∑n∈T n−z .
Usually, the set T is understood to be the set of primes in N or N itself. The function ∑p p−z is sometimes called the prime zeta function.
Exercise 13.18. a) Show that for n ≥2 ∑ p 1 npnz < 1 n Z ∞ 1 x−nz dx = 1 n(nz−1) .
b) Use (a) and Lemma 12.10 to show that as z ↘1+ lnζ(z) = ∑ p p−z +bounded.
c) Use (12.10) to show that as z ↘1+ lnζ(z) = −ln(z−1)+bounded.
(Hint: h is analytic near z = 1 and from (12.10), one easily sees that it is negative for z near 1 and real.) d) Use (b) and (c) to show that as z ↘1+ ∑ p p−z = −ln(z−1)+bounded.
e) Therefore lim z↘1+ ∑p f(p)p−z ∑p p−z = lim z↘1+ ∑p f(p)p−z −ln(z−1) .
The relation between natural density and Dirichlet density (Definition 13.30) is somewhat subtle. If the natural density exists then so does the Dirichlet density, but not vice versa. To establish the former, we prove Lemma 13.31 below in exercise 13.19. The other direction of this statement is not so easy; it is established by way of a counter-example in exercises 13.20 and 13.21.
Lemma 13.31. Let A and B be non-empty subsets of N and an and bn are their indicator functions. That is: an equals 1 if n ∈A and 0 elsewhere, and similar for bn. Furthermore, A(x) = ∑n≤x an and similar for B(x). Now we have for Rez > 1: lim x→∞ A(x) B(x) = µ = ⇒ lim z↘1+ ∑∞ n=1 ann−z ∑∞ n=1 bnn−z = µ .
13.7. Exercises 301 Exercise 13.19. a) Use Abel summation to show that for Rez > 1 ∞ ∑ n=1 ann−z = z Z ∞ 1 A(t)t−z−1 dt .
(Hint: also use that A(x) ≤x.) b) Show that the hypothesis of Lemma 13.31 implies that for all ε > 0, we have |A(x)−µB(x)| < εB.
c) Show that under the hypothesis of that lemma, we have that for all ε > 0, R ∞ 1 A(t)t−z−1 dt R ∞ 1 B(t)t−z−1 dt −µ < ε .
(Hint: write µ as R ∞ 1 µB(t)t−z−1 dt R ∞ 1 B(t)t−z−1 dt and use (b).) Definition 13.32. The logarithmic density of a set S ⊆T relative to T is lim x→∞ ∑k∈S,k≤x k−1 ∑k∈T,k≤x k−1 .
Usually, the set T is understood to be the set of primes in N or N itself.
2 4 8 16 32 64 0 2 4 8 16 32 64 1 128 256 Figure 81. The set S consists of the natural numbers contained in in-tervals shaded in the top figure of the form [22n−1,22n). The bottom picture is the same but with a logarithmic horizontal scale.
302 13. Primes in Arithmetic Progressions Exercise 13.20. We show that the set S depicted in top of Figure 81 does not have a natural density (relative to N), but that it does have a logarithmic density.
a) Show that the limsup of the natural density is at least 5/8 while the liminf of the density is at most 3/8. (Hint: first, take the average up to the green points in the figure, and then up to the blue points) b) Use Figure 82 to show that 2m−1 ∑ j=0 1 2m + j = 2m−1 ∑ j=0 2−m 1+ j2−m = Z 1 0 1 1+x dx+rm = ln2+rm , where rm ∈[0,2−m+1]. (Note: for Riemann sum, see chapter 6.) c) Use (a) to show that ∑ k∈S,k≤n k−1 = 1 2 log2 n ln2+R(n) = 1 2 lnn+R(n), where |R(n)| ≤2(1+ln2).
d) Use (b) and exercise 12.3 (c) to show that the logarithmic density of S is 1/2. (Note: a much simpler heuristic argument gives that according to exercise 12.3 the logarithmic density corresponds to redrawing S with the horizontal coordinate logarithmic as in the bottom picture of Figure 81, and then computing the density.) 0 1 f(0) f(1) dx Figure 82. Proof that R 1 0 f(x)dx is between ∑k j=1 f( jdx)dx and ∑k−1 j=0 f( jdx)dx if f is strictly decreasing.
13.7. Exercises 303 Exercise 13.21. We show that if the logarithmic density of a set S (Defini-tion 13.32) exists, then its Dirichlet density equals the logarithmic density.
a) Denote the elements of S by {n1,n2,···} and show that for Rez > 1 ∑∞ n=1 ∑k∈S,k≤n k−1 n1−z −(n+1)1−z = n−1 1 n1−z 1 −(n1 +1)1−z +···+(n2 −1)1−z −n1−z 2 + n−1 1 +n−1 2 n1−z 2 −(n2 +1)1−z +···+(n3 −1)1−z −n1−z 3 + n−1 1 +n−1 2 +n−1 3 n1−z 3 −(n3 +1)1−z +···+(n4 −1)1−z −n1−z 4 +··· = ∑n∈S n−z .
(Hint: n−1 1 gets multiplied by (n1−z −(n + 1)1−z) for n ≥n1, n−1 2 by (n1−z −(n + 1)1−z) for n ≥n2, and so on. The sums as given telescope to n−1 1 (n1−z 1 −n1−z 2 ), (n−1 1 +n−1 2 )(n1−z 2 −n1−z 3 ), and so forth.) b) Show that if the logarithmic density of S (with respect to N) equals µ, then, by (a), we have ∑ n∈S n−z = ∞ ∑ n=1 ∑ k∈S,k≤n k−1 ! n1−z −(n+1)1−z = = ∞ ∑ n=1 µ ∑ k≤n k−1 ! n1−z −(n+1)1−z = = µ ∑ n∈N n−z .
c) Use (b) to demonstrate the statement heading this exercise.
To emphasize once again the similarity between our generalized zeta func-tions and ζ of Chapter 12, we show that Zq,a has no zeroes in Rez > 1. The proof can be copied from exercise 4.24, provided you make the requisite substitutions.
Definition 13.33. The function Mq,a : N →Z is given by: Mq,a(n) = 1 if n = 1 0 if ∃p prime with p ̸=q a and p | n 0 if ∃p prime with p2 | n (−1)r if n = p1 ··· pr and pi =q a .
This the counterpart of the M¨ obius function of Definition 4.6.
304 13. Primes in Arithmetic Progressions Exercise 13.22. a) Show that Mq,a is a multiplicative function. (Hint: com-pare with the M¨ obius function in Chapter 4.) b) Use Euler’s product formula and Definition 13.33 to show that in Rez > 1 1 Zq,a(z) = ∏ p=qa 1−p−z = ∏ p prime ∑ i≥0 Mq,a(pi)p−iz !
.
c) Without using equation (4.7), prove that the expression in (b) equals ∑n≥1 Mq,a(n)n−z. (Hint: since Mq,a is multiplicative, you can write a proof re-arranging terms as in the first proof of Euler’s product formula.) Exercise 13.23. Show that for q > 1 in N: lim n→∞ ∏ p≤n, p=qa p !1/n = e1/ϕ(q) if and only the prime number theorem for arithmetic progressions holds.
(Hint: see Lemma 13.25 (ii). See also exercise 12.26.) In exercises 13.24 and 13.25, we prove partial versions of some remarkable results knowns as Mertens’ theorems. These were proved 22 years before the prime number theorem . More details can be found in [Section 22]. The version we give summarizes the statements given in .
Theorem 13.34 (Mertens’ Theorems).
i) limx→∞ ∑p≤x ln p p −lnx = −B3 ≈−1.3326.
ii) limx→∞ ∑p≤x 1 p −lnlnx = B1 ≈0.2615.
iii) limx→∞ ∑p≤x ln 1−p−1 −lnlnx = −γ .
B1 and B3 are sometimes called Mertens constants , but also go by other names. γ is the Euler-Mascheroni constant (see exercise 12.3).
13.7. Exercises 305 Exercise 13.24. a) Deduce from (12.4) and unique factorization that 1 n ln(n!) = ∑ pk≤n 1 n n pk ln p.
(Note: we sum over both the relevant integers k and primes p.) b) Show that 1 n n pk ≤1 pk .
c) Show that (a) and (b) imply that for some K1 > 0 1 n ln(n!)−∑ p≤n ln p p < K1 .
(Hint: ∑k≥2 p−k = 1/(p(p−1)).) d) Use exercise 12.4 (a) to show that there is a K2 so that 1 n ln(n!)−lnn < K2 .
e) Conclude that R(x) is bounded where R(x) := ∑ p≤n ln p p −lnn.
Figure 83. The function ln(ln(x)) for x ∈[1,1040].
306 13. Primes in Arithmetic Progressions Exercise 13.25. a) Let an = ln p p is n = p a prime and 0 else and set f(t) = 1/lnt. Now use Abel summation (Proposition 12.16) to show that ∑ n≤x 1 p = 1 lnx ∑ p≤x ln p p + Z x 2 1 t(lnt)2 ∑ p≤t ln p p dt .
b) Use exercise 13.24 (d) applied to the previous item to show that ∑ n≤x 1 p = 1+ R(x) lnx + Z x 2 1 t lnt dt + Z x 2 R(t) t(lnt)2 dt c) Conclude that ∑ p≤x 1 p = lnlnx+o(lnlnx).
d) Compare (a) with exercise 13.18(d).
e) To appreciate how agonizingly slow the approach of lnlnx to infinity is, approximate lnln101010. (Hint: about 25).
f) To write that number — 101010 — in full decimal notation in a series of books, how many books would you fill? Assume that you write 2000 characters on a page and that 500 pages make one book.
Chapter 14 The Birkhoff Ergodic Theorem Overview. To fully understand and appreciate the proof of the Birkhoff ergodic theorem, we have to dig a little deeper in analysis. We give the necessary background in this chapter and then prove the theorem. It is rec-ommended that you carefully read Sections 9.1 and 9.2 again before starting Sections 14.1, 14.2, and 14.3 below.
14.1. Measurable Sets We recall from Section 9.2 that if we have a space X and a collection Σ of measurable sets, then the pair (X,Σ) is called a measurable space.
Definition 14.1. A function F : X →Y is called measurable function if the inverse image under F of any measurable set in Y is measurable in X.
A measure µ is a non-negative function from Σ to [0,∞] that is count-ably additive on disjoint measurable sets (Definition 9.5). A triple (X,Σ,µ) is called a measure space. A probability measure is a measure that assigns a measure 1 to the entire space. It is time to refine our understanding of those concepts.
Definition 14.2. A sigma algebra or σ-algebra is a collection Σ of sets with the following properties: 307 308 14. The Birkhoff Ergodic Theorem / 0 ∈Σ and Σ is closed under complementation and under countable union.
In any topological space, the smallest σ-algebra that contains the open sets is called the Borel sets (Definition 9.1).
Remark 14.3. Since (∪i∈N Ai)c = ∩i∈N Ac i (exercise 9.1), we see that a σ-algebra is also closed under countable intersection.
We are now in a position to give a more formal definition of a measure (see Definition 9.5).
Definition 14.4. Let (X,Σ) be a measure space. A measure is a function µ : Σ →[0,∞] such that µ(/ 0) = 0 and for every countable sequence of disjoint (measurable) sets Si: µ(∪∞ i=1 Si) = ∞ ∑ i=1 µ(Si).
If Σ contains the open sets, then µ is called a Borel measure .
Definition 14.5. Let X be a set and µ : Σ →[0,∞] a measure. A set S ⊆X is measurable or, more accurately, µ-measurable if S ∈Σ. A function f : X →R is measurable if for every y ∈R, f −1((y,∞)) is in Σ . (Here, f −1(S) means the inverse image of the set S) Why is this so complicated? Recall from Section 9.1 that if we try to assign a measure µ on the real line consistent with our intuitive notion of length, that is: intervals (a,b) must have “measure” b−a and the measure is translation invariant1, we may run into trouble. In that section we showed, using the axiom of choice, that we can construct sets (Vitali sets) that cannot be assigned a measure. Thus we cannot assume that any arbitrary set is measurable. So if we define certain abstract sets and want to talk about their measure, we have to always be very careful that we didn’t leave the sigma algebra of measurable sets.
In fact, the determination that certain combinations of measurable func-tions are still measurable plays a significant role in the proof of Birkhoff’s theorem. So let us have a closer look at this.
1This is the Lebesgue measure.
14.2. Measurable Functions 309 14.2. Measurable Functions Not all functions are measurable. For instance, the function from R to itself that is 1 on the Vitali set V (see Section 9.1) and zero elsewhere, is not Lebesgue measurable. For a more interesting example, see exercise 14.3.
Suppose f is measurable. Recall that the collection of measurable sets must be closed under complementation, countable intersection, and count-able union. Since f −1((−∞,y]) is the complement of the measurable set f −1((y,∞)), it is also measurable. f −1([y,∞)) can be written as the (count-able) intersection ∩n∈N f −1((y −1 n,∞)), it, too, is measurable. Again, by complementation, f −1((−∞,y)) is measurable.
f g y z Figure 84. Illustration of the fact that if h+(x) = sup{ f(x),g(x)}, then h−1 + ((y,∞)) = f −1((y,∞)) ∪g−1((y,∞)).
Similarly, if h−(x) = inf{f(x),g(x)}, then h−1 −((z,∞)) = f −1((z,∞))∩g−1((z,∞)).
It is easy to see that if f and g are measurable, then h(x) = sup{ f(x),g(x)} is measurable because h−1((y,∞)) = f −1((y,∞)) ∪g−1((y,∞)) (see Figure 84). Similar for inf{f(x),g(x)}. Almost as easy is the fact that also f + g and f ·g are measurable. For the set Ar1,r2 := {x | f(x) > r1}∩{x | g(x) > r2} is measurable for all rationals ri, and therefore so is the (countable) union of Ar1,r2 over those rationals such that r1 +r2 > y or such that r1r2 > y.
Lemma 14.6. Let { fn} be a sequence of measurable functions. Then supn fn(x), infn fn(x), limsupn fn(x), and liminfn fn(x) are measurable.
Proof. Set h± equal to supn fn(x) and infn fn(x), respectively. Then h−1 + ((y,∞)) = ∪∞ n=1 f −1 n ((y,∞)), 310 14. The Birkhoff Ergodic Theorem which proves the first case. The proof for h−is same, except that the union must be replaced by an intersection.
Set g± equal to limsupn fn(x) and liminfn fn(x), respectively. Since g+(x) = lim n→∞sup i≥n fi(x) and supi≥n fi(x) is non-increasing (in n), we can replace the above limit by the infimum, and use the above results for supremum and infimum to get g−1 + ((y,∞)) = ∩n≥1 ∪i≥n f −1 i ((y,∞)).
And so g+ is measurable. A similar reasoning works for g−.
■ Remark 14.7. As a result, the pointwise limit (if it exists) of a sequence of measurable functions is also measurable.
14.3. Dominated Convergence In this section, we prove — largely inspired by — Lebesgue’s dominated convergence theorem. This is a result of fundamental importance in its own right. It is widely used not only in analysis but also in applications of analysis to the study of partial differential equations and probability theory among others. Here, we will need it to prove the ergodic theorem.
The following theorem says that almost everywhere pointwise2 conver-gence implies nearly 3 uniform convergence, that is: convergence is uniform, except on a set of small measure. See Figure 85.
Theorem 14.8 (Egorov’s Theorem). Let (X,Σ,µ) a finite measure4 space.
Suppose that {fi} is a sequence of measurable functions on X, so that µ almost everywhere, fi(x) converges pointwise to f(x). Then for every ε > 0, there is a set U ∈Σ on which the convergence of fi →f is uniform, and so that the exceptional set µ(X\U) < ε.
Proof. Let Am,n := x ∈X : ∀i ≥m, |fi(x)−f(x)| < 1 n .
2Pointwise convergence means that for x fixed, limi→∞fi(x) = f(x).
3For the usage of the word ‘nearly’ here, see Section 14.4.
4A space with µ(X) < ∞ 14.3. Dominated Convergence 311 f f f 1 2 3 A A A 1,n 2,n 3,n 1/n 0 0 1 1 Figure 85. A sequence of functions fn(x) = x1/n that converge almost everywhere pointwise to f(x) = 1 on [0,1]. The convergence is uniform on U = [ε,1] for any ε ∈(0,1).
We have Am,n ⊆Am+1,n and ∪m Am,n covers all of X, except for a measure zero set Zn (see Figure 85). Thus we can choose mn such that µ(X\Amn,n) = µ(Ac mn,n) < ε 2n , (14.1) where the superscript indicates the complement in X. For any x in the inter-section of all Amn,n, we have that for i ≥mn, |fi(x)−f(x)| < 1/n.
And thus on U := ∩n≥1 Amn,n , we have uniform convergence. Within X, we have X\U = (∩n≥1 Amn,n)c = ∪n≥1 Ac mn,n , And so, by equation (14.1) and subadditivity (9.1), µ(X\U) < ε.
■ Next we prove first that integrable functions nearly live on sets of finite measure and that integrals of measurable functions over small sets are small.
Lemma 14.9. Suppose g : X →[0,∞] is measurable and integrable. Then: i) for every ε > 0, there is a set F of finite measure such R X\F gdµ < ε.
ii) for every ε > 0, there is a δ > 0 such that for all small sets S with µ(S) < δ, we have R S gdµ < ε.
Proof. Let {yi} be a countable partition of the range of an integrable func-tion g. Denote Ai = g−1 ({y : y ≥yi+1}) and ∆i = yi+1 −yi. From the 312 14. The Birkhoff Ergodic Theorem definition of the Lebesgue integral (Section 9.2 and Figure 86), we see that for every ε > 0 we can choose a partition so that ∞ ∑ i=1 µ(Ai)∆i < Z gdµ < ε 2 + ∞ ∑ i=1 µ(Ai)∆i .
Since the sum must converge, we can truncate at some n to get n ∑ i=1 µ(Ai)∆i < Z gdµ < ε + n ∑ i=1 µ(Ai)∆i .
(14.2) We may assume ∆i > 0 in this sum (the ∆i = 0 terms do not contribute).
The union F := ∪n i=1 Ai must have finite measure (otherwise ∑n i=1 µ(Ai)∆i would diverge). Now we compute Z X\F gdµ = Z X gdµ − Z F gdµ < ε + n ∑ i=1 µ(Ai)∆i !
− n ∑ i=1 µ(Ai)∆i = ε .
We used both inequalities in (14.2) to derive the last inequality.
y y y y y 0 1 2 3 4 5 y g(x) Figure 86. The definition of the Lebesgue integral. Let {yi} be a count-able partition of the range of g. We approximate R gdµ from below by ∑i µ g−1 ({y : y ≥yi+1}) (yi+1 −yi). Then g is integrable if the limit converges as the mesh of the partition goes to zero. The function y in the proof of Lemma 14.9 (ii) is indicated in red. (Here µ is the Lebesgue measure.) To prove (ii), denote by y the function whose value equals yi on Ai (see Figure 86). Let y+ be the maximum of the yi (in the definition of Ai) for which the ∆i are positive. Then for any measurable set B with µ(B) < δ Z B gdµ = Z B (g−y)dµ + Z B ydµ < Z X (g−y)dµ + Z B ydµ .
14.3. Dominated Convergence 313 Since R X ydµ is simply ∑n i=1 µ(Ai)∆i, the first integral is less than ε by (14.2). The same holds for the second integral if we choose δ so that y+δ < ε.
■ Theorem 14.10 (Lebesgue’s Dominated Convergence Theorem). Let { fk} be a sequence of real valued measurable functions on (X,Σ,µ). Suppose that the sequence converges µ almost everywhere to f and that it is domi-nated by an integrable function g so that for all k, |fk(x)| ≤g(x). Then lim k→∞ Z fk dµ = Z lim k→∞fk dµ = Z f dµ .
Proof. For any set U ∈Σ, we have (using linearity of the integral) Z X fk dµ − Z X f dµ = Z X\U fk dµ − Z X\U fk dµ + Z U fk dµ − Z U f dµ ≤ 2 Z X\U gdµ + Z U ( fk −f)dµ .
We consider the finite measure case (where µ(X) < ∞) and the infinite mea-sure case separately.
When µ(X) < ∞, we use Egorov’s theorem to choose the set U on which we have uniform convergence so that µ(X\U) < δ, where δ is as in Lemma 14.9 (ii). So for any η > 0, we can choose k large enough so that the above inequality becomes Z X fk dµ − Z X f dµ < 2ε +ηµ(U) Upon choosing η small enough, the result follows because µ(U) < ∞.
In the case where µ(U) is infinite, we need to do one step extra. Use Lemma 14.9 (i) to find a set F ⊆U of finite measure so that R U\F gdµ < ε.
The first inequality is now followed by Z U ( fk −f)dµ = Z U\F ( fk −f)dµ + Z F ( fk −f)dµ ≤ 2 Z U\F gdµ + Z F ( fk −f)dµ .
The second integral can now be estimated in the same way as before.
■ 314 14. The Birkhoff Ergodic Theorem Remark 14.11. While we proved the theorem here for real valued func-tions, it also holds for complex valued functions. One simply proves the result for the real and imaginary parts separately.
14.4. Littlewood’s Three Principles The subject of real analysis, and measure theory, and Lebesgue integration in particular, overtook the older, more informal notions of length and Rie-mann integration in part because extremely useful theorems like the domi-nated convergence theorem simply do not hold in the older setting. Here is a simple example to illustrate that.
Let fn : [0,1] →R be given by fn(x) = 1 if x ∈Sn where Sn := i k : 0 < k ≤n and 0 ≤i ≤k , and fn(x) = 0 elsewhere. Clearly, each fn is Riemann integrable (having only finitely many discontinuities). Also the fn are dominated by g(x) = 1.
See Figure 87. However, limn→∞ R fn dx = 0 while R limn→∞fn dx is not 1 1 1/2 1/3 1/4 2/3 3/4 0 0 Figure 87. The non-zero values of the function f4 in red.
defined since limn→∞fn is not Riemann integrable (as it has a dense set of discontinuities). In exercises 14.5-14.8, we give other interesting “counter-examples”. For now, note that if we switch to Lebesgue integration, there is no problem because then limn→∞fn is integrable, and non-zero only on the rationals (measure zero) and and so its integral is zero.
Nonetheless, this more powerful mode of reasoning seems very ab-stract and for that reason it is difficult to develop an intuition in the subject.
It is perhaps comforting to know that at least some of the masters of the sub-ject themselves recognized this. The most famous instance of this is formed by Littlewood’s three principles .
14.4. Littlewood’s Three Principles 315 Each of these principles describes a desirable behavior that indeed holds if only one excludes sets of arbitrarily small measure. This is ex-pressed by the word “nearly”, which we encountered a few times in Section 14.3: we say that in these cases the behavior nearly holds.
• Every measurable set is nearly a finite union of disjoint open in-tervals.
• Every measurable function is nearly continuous.
• Every pointwise convergent sequence of functions is nearly uni-formly convergent.
The first principle is in fact Proposition 9.4 (ii). The third principle is of course Egorov’s theorem (Theorem 14.8). For the second principle is vir-tually all texts refer to Luzin’s Theorem (see below). However, Littlewood himself mentions a slightly different theorem in this context, [Section 4.1]. For completeness, we state Luzin’s theorem with only a sketch of the proof.
Theorem 14.12 (Luzin’s Theorem). Let f be measurable in (R,Σ,µ) where Σ are the Borel sets and µ is the Lebesgue measure. For every ε > 0, there is a small open set O of (Lebesgue) measure less than ε so that f is continuous when restricted to R\O.
Sketch of proof. We approximate f by stepfunctions fnas in Figure 86, so that almost everywhere fn →f. The fn are constant except on a exceptional set En of measure zero where the discontinuities are located. By Egorov, we now have that fn →f is uniform except on a arbitrarily small set F. So we set S := R−F −∪j≥nEn .
Now on S, the continuous fn converge uniformly to f and therefore f is continuous5 on S. Now all we need to do, is to approximate S with a closed set C as in Proposition 9.4 (i) and let O be the complement of C.
■ One must be careful in the interpretation of this last result: it does not mean that the points of the R\S are points of continuity of f. As an exam-ple, consider the function that is 1 on the rational numbers and 0 everywhere else. As a function R →R, it is nowhere continuous, but it’s restriction to 5This is proved in any introductory analysis course.
316 14. The Birkhoff Ergodic Theorem the irrational numbers is continuous. Luzin’s theorem still goes a little fur-ther, and asserts that we can contain the rationals in an open sets of arbitrary small measure (exercise 9.3).
14.5. Weyl’s Criterion To get us in the mood for the ergodic theorem, we first look at another which is very often used in number theory. First, we need a basic result from analysis.
Theorem 14.13 (Fej´ er’s Theorem, informal version). Let S be the unit cir-cle parametrized by z = e2πix in the complex plane. Let f : S →R be contin-uous. Then for any ε > 0 there is pM(e2πix) := ∑M m=−M ame2πimx such that |f(z)−pM(z)| < ε.
The full version of this theorem explicitly constructs the approximating polynomials, see [Chapter 12] or [Chapter 11]. We will not prove this result here, as that would take us too far afield.
In what follows, we let 1a,b denote the function that is 1 if x is in [a,b] and 0 elsewhere.
Theorem 14.14 (Weyl’s Criterion). The following are equivalent.
i) The real sequence {xn} is equidistributed modulo 1, i. e. for all a and b lim n→∞ 1 K K ∑ k=1 1a,b = (b−a).
ii) For every continuous function f : R/Z →C, lim n→∞ 1 K K ∑ k=1 f(xk) = Z 1 0 f dx.
iii) For all m ̸= 0 in Z lim n→∞ 1 K K ∑ k=1 e2πimxk = 0.
Proof. We first prove the equivalence of (i) and (ii). Since real and imagi-nary of f can be dealt with in the same way, it is sufficient to consider only the real case.
14.5. Weyl’s Criterion 317 Since f is continuous on a compact domain, it is uniformly continuous (see ). So let {xi}m i=0 is a partition of the circle R/Z and ci ∈[xi,xi+1], then for any ε > 0, we can choose a fine enough partition so that fm(x) = m−1 ∑ i=0 f(ci)1xi,xi+1 and |f(x)−fm(x)| < ε .
(14.3) Now (i) implies that lim K→∞ 1 K K ∑ k=1 f(xk) = m−1 ∑ i=0 f(ci)(xi+1 −xi) = Z fm dx.
With (14.3), this implies that lim K→∞ 1 K K ∑ k=1 fm(xk)− Z f dx < 2ε , and (ii) follows. The reverse follows by approximating 1[a,b] by the con-tinuous function gε as indicated Figure 88. It is easy to see that (ii) implies 0 1/2 1 Figure 88. The functions 1[a,b] in black and gε in red. ε is the width (indicated by two-sided arrows) of the regions where gε and the step function do not agree.
lim K→∞ 1 K K ∑ k=1 gε(xk) = Z gε dx = b−a.
Taking the limit as ε tends zero establishes (i).
It is clear that (ii) immediately implies (iii). Thus we only need to prove that (iii) implies (ii). Let f : S →R be continuous. Theorem 14.13 implies 318 14. The Birkhoff Ergodic Theorem that for any ε > 0 there is a pM(e2πix) := ∑M m=−M ame2πimx so that Z 1 0 f(e2πix)dx− Z 1 0 pM(e2πix)dx = Z 1 0 f(e2πix)dx−a0 < ε .
If we also appeal to item (iii), we see that lim K→∞ 1 K K ∑ k=1 f(e2πixk)−pM(e2πixk) = lim K→∞ 1 K K ∑ k=1 f(e2πixk)−a0 < ε .
Comparison of the last two inequalities yields the implication.
■ If T is ergodic with respect to Lebesgue, and if we set xi = T i(x0) and f(x) := e2πimx. Now, of course, item (ii) is exactly Corollary 9.13, which says that time averages equal space averages. The standard example of this is T(xk+1) = xk +ρ where ρ is irrational, as we discussed at length in Chapter 10. However, it is still amusing to give a very simple and direct proof of this, based on Weyl’s criterion.
Indeed, it requires no more than than summing a geometric series to see that 1 n n−1 ∑ k=0 e2πimxk = e2πimx0 n n−1 ∑ k=0 e2πimkρ = e2πimx0 n · e2πimnρ −1 e2πimρ −1 .
Since ρ is irrational and m ̸= 0, we have that e2πimρ −1 ̸= 0, and so the factor 1/n drives the limit to zero. (If m = 0 the left hand side immediately yields one).
A final remark is that in the proof of Weyl’s criterion, it might seem more reasonable to prove (iii) implies (i). But there is a subtle obstruction to an easy proof. Such a proof would express 1[a,b] as a trigonometric sum that converges uniformly. However, this is not possible. It turns out that if we try that the trigonometric sum s(x) approximating functions with a jump discontinuity always “overshoots” by almost 9%. This is called the Gibbs phenomenon [3,26], and also goes by the name of ringing .
14.6. Proof of Birkhoff’s Ergodic Theorem Our proof is based on [section 9]. As before, we will denote iterates under T by subscripts.
T(x0) = x1 , T(T(x0)) = T 2(x0) = T(x1) = x2 , ··· 14.6. Proof of Birkhoff’s Ergodic Theorem 319 and so on. We also define the sums Sn f (x0) = n ∑ i=1 f(T i(x0)).
Remark 14.15. In this section, we work in a measure space (X,Σ,µ). We stipulate that T : X →X is a measurable transformation that preserves a probability measure µ and that f : X →R (or C) is an arbitrary µ-integrable function.
Proposition 14.16 (Maximal Ergodic Theorem). If for µ-almost every x, there is an n(x) such that Sn(x) f (x) ≥0 (≤0), then R f dµ ≥0 (≤0).
k N 0 n p N−k S (x ) 0 f Figure 89. A plot of Sn f (x0) for some fixed x0 for n ∈{0,··· ,N}.
Proof. Note that this statement holds for f with “≥” if and only if it holds for g = −f with “≤”. So it is sufficient to prove only the ≥version.
First assume that n(x) is bounded (for almost all x) by some k > 0.
Then no matter how large we take N, there is some p(x) in {N −k,··· ,N} such that Sp(x) f (x) ≥0 (see Figure 89). We then have for µ-almost all x0 SN f (x0) = Sp(x0) f (x0)+SN−p(x0) f (xp) ≥−SN−p(x0) | f| (xp) ≥−Sk |f|(xN−k).
Therefore, for µ-almost all x, N ∑ i=1 f(T i(x)) ≥− N ∑ i=N−k+1 f(T i(x)) .
Bearing in mind that µ is invariant, we integrate this inequality. So by Lemma 10.1, R f(T i(x))dµ = R f(x)dµ(x) and similarly for |f|. In this way we obtain, after integrating, that N R f dµ ≥−k R |f|dµ. But since we may take N arbitrarily large, it follows that R f dµ ≥0.
320 14. The Birkhoff Ergodic Theorem Now, let k be arbitrary positive integers and define fk(x) = f(x) if n(x) ≤k 0 else We have |fk| ≤|f| and so the fk are dominated by |f| and since f is µ-integrable, so are the fk. Since the fk converge pointwise to f, we have Z f dµ = lim k→∞ Z fk dµ ≥0, by dominated convergence (Theorem 14.10).
■ We will need the contra-positive of this result. Here it is explicitly.
Corollary 14.17. Suppose R f dµ < 0 (> 0), then there is a set P of positive µ-measure such that for all x in P, Sn f (x) < 0 (> 0) for all n.
Under the hypotheses of remark 14.15, the statement of Theorem 9.10 is as follows.
Theorem 14.18 (Birkhoff or Pointwise Ergodic Theorem). The limit of the time average ⟨f⟩(x) := lim n→∞ 1 n n−1 ∑ i=0 f(T i(x)) is defined on a set of full measure. It is an integrable function and satisfies (wherever defined) Z X ⟨f⟩(x)dµ = Z X f(x)dµ .
Proof. We want to compute the limit of the time average of f. So let ⟨f⟩+(x) = limsup n→∞ 1 n Sn f (x) and ⟨f⟩−(x) = liminf n→∞ 1 n Sn f (x).
By Lemma 14.6 and the comments immediately prior to it, ⟨f⟩± are mea-surable functions. First suppose they are bounded. Then they are also inte-grable, because µ(X) = 1.
Suppose that the following statement is false: Z X ⟨f⟩−dµ ≥ Z X f dµ .
14.6. Proof of Birkhoff’s Ergodic Theorem 321 Then, since ⟨f⟩± and µ(X) are bounded, there must be an ε > 0 so that Z X ⟨f⟩−dµ < Z X ( f −ε)dµ = ⇒ Z X (⟨f⟩−−f +ε)dµ < 0.
By the contrapositive of the maximal ergodic theorem, this gives that there are (a positive measure of) x so that Sn ⟨f⟩−−f+ε(x) < 0 for all n. Now, it is easy to see that Sn f+g = Sn f +Sn g and that ⟨f⟩−is invariant along orbits. Thus for any such x, we obtain that n⟨f⟩−(x)−Sn f (x)+nε < 0 or ⟨f⟩−(x) < 1 n Sn f (x)−ε .
If we take the liminf as n →∞on both sides, we arrive at a contradiction.
This establishes that Z X ⟨f⟩−dµ ≥ Z X f dµ .
(14.4) In a similar way (exercise 14.12), one derives that Z X f dµ ≥ Z X ⟨f⟩+ dµ .
(14.5) Putting (14.4) and (14.5) together shows that if ⟨f⟩± are bounded, then Z X ⟨f⟩+ dµ ≤ Z X f dµ ≤ Z X ⟨f⟩−dµ .
and thus all three are equal. Since ⟨f⟩+(x) ≥⟨f⟩−(x), we also have that these two quantities must be equal except on a set of measure zero. This proves that for almost all x, the limit of 1 n Sn f (x) exists.
Now we drop the assumption that ⟨f⟩± are finite. So let Xn := {x ∈X : −n ≤⟨f⟩−(x) ≤⟨f⟩+(x) ≤n}.
T maps Xn to itself and so all hypotheses hold for Xn and therefore the above conclusion holds for Xn instead of X, and thus for X∞= ∪nXn. We are done if X\X∞has µ-measure zero. Now, Xn is measurable because ⟨f⟩± are, and so X∞and its complement are also measurable. Suppose the complement has positive measure, then since f is integrable, there must be a c > 0 so that Z X\X∞ −cdµ < Z X\X∞ f dµ < Z X\X∞ cdµ .
We apply again the contrapositive of the maximal ergodic theorem, to get that then there must be a (positive measure of) x in X\X∞so that for all n Sn (f+c)(x) > 0 and Sn ( f−c)(x) < 0 = ⇒ −nc < Sn f (x) < nc.
322 14. The Birkhoff Ergodic Theorem But this contradicts the definition of X\X∞.
■ As mentioned in Chapter 9, it is frequently the following Corollary which one has in mind when referring to Birkhoff’s ergodic theorem.
Corollary 14.19. A transformation T : X →X that preserves a probability measure µ has the property that every T-invariant set has measure 0 or 1 if and only if for every integrable function f lim n→∞ 1 n n−1 ∑ i=0 f(T i(x)) = Z X f(x)dµ for all x except possibly on a set of measure 0.
c 0 1 c+ − Figure 90. The functions µ(X− c ) and µ(X+ c ).
Proof. By Theorem 14.18, ⟨f⟩(x) is defined on a set of full measure. So let X− c := {x ∈X : ⟨f⟩(x) < c} and X+ c := {x ∈X : ⟨f⟩(x) > c} .
Replacing x by an inverse image (under T) of x does not change the value of ⟨f⟩(x), and so X± c are invariant sets. By the ergodic assumption, µ(X− c ) (as a function of c) must have measure 0 or 1, and is therefore an increasing step function with the step of height 1 occurring, say, at c = c−. Similarly, µ(X+ c ) is a step function, with an decreasing step of height 1 occurring at c = c+. See Figure 90.
If c−< c+, then for any interval [c1,c2] ∈(c−,c+), we obtain that µ(X− c1) = µ(X+ c2) = 1, which is impossible, since these sets do not inter-sect. In the same way, if c+ < c−, then for any interval [c1,c2] ∈(c+,c−) µ(X+ c1) = µ(X− c2) = 0, which contradicts the fact that the union of X+ c1 and X− c2 is the entire space and so must have measure 1. So c−= c+ = c0. Thus ⟨f⟩(x) = c0 on a set of full measure. And therefore Theorem 14.18 implies 14.7. Exercises 323 that R X f(x)dµ = R X ⟨f⟩(x)dµ = c0, which implies that time average equals space average.
Vice versa, if T is not ergodic, then there are invariant sets X1 and its complement X2 both of positive measure. Let 1X1 be the function that is 1 on X1 and 0 elsewhere. The time average ⟨1X1⟩(x) is 1 or 0, depending on where the starting point x is. In either case, it is not equal to the spatial average R X 1X1(x)dµ ∈(0,1).
■ In Section 9.3, we observed that ergodic measures are the building blocks of chaotic dynamics. Thus transformations where there is a unique ergodic measure are especially interesting.
Definition 14.20. A transformation T of a measure space is uniquely ergodic if there is a unique Borel probability measure with respect to which T is er-godic.
14.7. Exercises Exercise 14.1. a) Show that gn(X) = supi≥n fi(x) is non-increasing (in n).
b) Let fn(x) = sinnx. Determine limsupn fn(1). (Hint: use Lemma 10.6).
c) Show that the twin prime conjecture (Conjecture 1.29) is equivalent to liminfn pn+1 −pn = 2 (pn is the nth prime).
Exercise 14.2. a) Give a definition of a measurable function f from a topo-logical space to C. (Hint: split up the real and imaginary parts and use the Borel sigma algebra as measurable sets, then follow Section 14.1.) b) Show that if c is a constant and a real function f measurable, then c f is measurable.
c) Consider the set V in Section 9.1 and show that it is not measurable.
d) Consider the function 1V which is 1 on points in V and 0 elsewhere.
Show that 1V is not measurable.
324 14. The Birkhoff Ergodic Theorem 1/3 2/3 1/9 2/9 7/9 8/9 0 1 0 2 1/3 2/3 1/9 2/9 7/9 8/9 0 1 0 2 Figure 91. Left: an impression of the function g(x) = c(x) + x where c(x) is taken from figure 54. Left: its inverse h := g−1 is well-defined but not Lebesgue measurable. Linear interpolated segments are colored red.
Exercise 14.3. In this exercise, we exhibit a non-trivial function g that is not Lebesgue measurable. The student should review exercises 9.8–9.13 for the properties of the Cantor function c(x). Define g(x) = c(x) + x so that g : [0,1] →[0,2]. Its inverse is denoted by h (see Figure91). The Lebesgue measure is denoted by µ.
a) Show that g is invertible and call its inverse h. (Hint: c is strictly in-creasing on the Cantor set C, so g is strictly increasing on [0,1]).
b) Show that the complement O of g(C) is Lebesgue measurable, and that µ(O) = 1. (Hint: O is a countable collection of open intervals whose lengths sum to 1.) c) Use (b) to show that µ(g(C)) = 1. (Hint: Corollary 9.6.) d) Show that there is a non-measurable set W ⊂g(C). (Hint: recall that V ⊆[0,1]; insert the pieces corresponding to O into [0,1]. Bits of V will be translated, but this does not affect their measure. When finished, you have a non-measurable set W ⊂g(C) ⊂[0,2].) e) Show that µ(h(W)) = 0, and so h(W) is measurable. (Hint: h(W) is contained in C. Use Lemma 9.20.) f) Use (a) and (e) to show that there is a Lebesgue measurable set Z such that h−1(Z) = W and W is not measurable.
14.7. Exercises 325 Exercise 14.4. Explain why Henri Lebesgue wrote the following about his method of integration (as cited by [ page 796]): “I have to pay a certain sum, which I have collected in my pocket. I take the bills and coins out of my pocket and give them to the creditor in the order I find them until I have reached the total sum. This is the Riemann integral. But I can proceed differently. After I have taken all the money out of my pocket I order the bills and coins according to identical values and then I pay the several heaps one after the other to the creditor. This is my integral.” Exercise 14.5. In this exercise, we show that the dominated convergence with Riemann integration cannot be saved even by restricting to continuous functions f : [0,1] →[0,1]. Let fn be given as follows, see Figure 92. For every pair ( j,k) with gcd( j,k) = 1 and so that j/k ∈[0,1], define hn(j,k,x) = max 0,1−n3 x−j k and fn(x) := ∑ j/k∈[0,1] gcd( j,k)=1 k≤n h(j,k,x).
a) Show that fn is continuous and dominated by g(x) = 1. (Hint: show that the minimal distance between the centers of any two “triangles” hn( j,k,x) defined in (a) is at least 1/n2.) b) Show that limn→∞ R fn dx = 0. (Hint: first give a rough estimate how many rationals with denominator less than n+1 there are in the unit inter-val.) c) Let r ∈[0,1] be an algebraic number of degree at least two. Show that Roth’s theorem (Theorem 1.21) implies that for all ε > 0 ∃c(r,ε) > 0 such that ∀j k ∈Q : r −j k > c(r,ε)k1−ε n3 .
d) Show that (c) implies that for every algebraic number r of degree at least two, limn fn(r) = 0.
e) Use exercise 7.14 (e) to show that for this example dominated conver-gence does not hold for Riemann integration. (Hint: show that the Rie-mann integral of limn fn is not defined.) f) Explain that if you use Lebesgue integration, there is no problem.
Exercise 14.6. Let fn(x) = n for x ∈[0,1/n] and 0 elsewhere and set g(x) = 1/x.
a) Show that g dominates the fk. See Figure 93.
b) Show that limk→∞ R fk dµ ̸= R limk→∞fk dµ.
c) Why do (a) and (b) not contradict Theorem 14.10?
Exercises 14.7 and 14.8 provide a more interesting illustration of the dom-inated convergence theorem. Generalizing exercise 11.19, for fixed r ≥ 326 14. The Birkhoff Ergodic Theorem j/k 0 0 1 1 1/n3 1/n3 Figure 92. The function fn (in red) in exercise 14.5 is a sum of very thin triangles with height 1. Each triangle is given by hn( j,k,x) (in black).
1 2 3 4 1 1/2 1/3 1/4 Figure 93. The functions fn in exercise 14.6 and the function g (in red) that dominates them.
1, consider the functions gk(x) = krxk(1 −x) on [0,1]. Define Gk(x) = supi≤k gi(x) and G(x) = supi gi(x).
14.7. Exercises 327 Exercise 14.7. a) Show that gk(x) is increasing on [0, k k+1] and decreasing on [ k k+1,1].
b) Show that gk has maximum kr−1 k k+1 k+1 ≈kr−1e−1.
(Hint: limk→∞(1+a/k)k = ea.) c) Show that gk−1(x) = gk(x) iff x ∈ n 0, k−1 k r ,1 o and that gk k −1 k r = kr k −1 k rk 1− k −1 k r ≈kr−1re−r .
d) Show that gk(x) = gk+1(x) iff x ∈ n 0, k k+1 r ,1 o and that gk k k +1 r = (k+1)r k k +1 r(k+1) 1− k k +1 r ≈(k+1)r−1re−r .
e) Show that k k+1 r − k−1 k r ≈rk−2. (Hint: compute the first non-zero term in the expansion of (1+x)−r −(1−x)r.) f) Show that R G(x)dx is “sandwiched” between the sum S1 of the areas of the rectangles like the one shaded red in Figure 94 and the sum S2 of the red ones plus the green ones.
g) Use (c) through (f) to show that that there are functions c1 and c2 such that S1 = c1(r)∑kr−1k−2 and S2 = c2(r)∑(k +1)r−1k−2 .
h) Conclude that G is integrable iff r < 2.
k+1 0 1 k e −2 2 2 ((k−1)/k) (k/(k+1)) 0 k−1 g k g g (k+1) e−2 2 2 Figure 94. In this figure r = 2. We show the function gk(x) (red) on [0,1] and its intersections. The sum of the rectangles like the one shaded in red give a lower bound for R Gk dx while the sum of the red and green rectangles give an upper bound.
328 14. The Birkhoff Ergodic Theorem Exercise 14.8. a) Use exercise 14.7 (f) to show that the dominated conver-gence theorem implies that for r < 2, we have Z 1 0 lim k→∞gk(x)dx = lim k→∞ Z 1 0 gk(x)dx.
b) What goes wrong for r ≥2?
c) Show that Z 1 0 lim k→∞gk(x)dx = 0 and Z 1 0 gk(x)dx = kr (k +1)(k +2) .
d) Why is (c) consistent with (a) and (b)?
Exercise 14.9. Let A be a compact collection of irrational numbers and and {ni} a sequence of natural numbers whose partial sums satisfy Sk = k ∑ i=1 ni where lim k→∞ Sk k = ∞.
Create a sequence {xi} of real numbers as follows. Choose an x0 and set n0 = 0. For i ∈{1,··· ,n1}, let xi = xi−1 + α1 where α1 ∈A; for i ∈ {S1 +1,··· ,S2}, let xi = xi−1 +α2 where α2 ∈A; and so on.
a) Show that for any fixed m ̸= 0 in Z, there is a εm > 0 so that min α∈A e2πimα −1 > εm .
(Hint: the compactness of the set of irrational numbers is crucial.) b) Show that 1 Sk Sk−1 ∑ n=0 e2πimxn = 1 Sk ( e2πimx0 n1−1 ∑ n=0 e2πimnα1 +···+e2πimxSk−1 nk−1 ∑ n=0 e2πimnαk ) .
c) Use the geometric series as in Section 14.5 to show that for each sum in (b), we obtain nℓ−1 ∑ n=0 e2πimnαk = e2πimnℓαk −1 e2πimαk −1 .
d) Use (a) and (c) to show that nℓ−1 ∑ n=0 e2πimnαk < 2ε−1 m .
e) Use (d) and the condition on the partial sums to show that lim k→∞ 1 Sk Sk−1 ∑ n=0 e2πimxn = 0.
f) Show that (e) and Weyl’s criterion imply that the sequence {xi} is equidistributed modulo 1.(Hint: you need to pass from limk→∞1 Sk ∑Sk−1 n=0 to limS→∞1 S ∑S n=0; so vary the value of the last nk.) 14.7. Exercises 329 Many number theory textbooks (correctly) state that the fractional parts of f(n) = ln pn, where pn denotes the nth prime, are not equidistributed. This is slightly misleading because an unsuspecting student could be tempted into wondering to what mysterious distribution the numbers the fractional parts of ln pn would deign to converge to? The answer — perhaps somewhat disappointingly — is that the logarithm increases so slowly that in fact those numbers do not converge at all as we show in exercises 14.10 and 14.11.
We denote the fractional part of x by {x}. For a slowly increasing function f : N →R and an interval J of the unit circle [0,1]/{0 = 1}, we define the “hitting frequency” as follows: FJ(0,n) := #{{ f(i)} ∈J for i ∈{1,··· ,n}} n .
Note that if the fractional parts of { f(n)} converge to any distribution what-soever, then there is a c ∈[0,1] so that limn→∞FJ(0,n) = c.
Exercise 14.10. In this exercise, we set f(n) := lnn and let J = [α,α +δ) be an arbitrary interval of length δ in the unit circle. For K ∈N and nK, choose n′ K so that f(nK) ≤K +α < f(nK +1) and f(n′ K) ≤K +α +δ < f(n′ K +1).
a) Show that lim K→∞ nK n′ K = e−δ .
b) Assuming that limn→∞FJ(0,n) = c, show that (see Figure 95) n′ KFJ(0,n′ K) ≈nK ·c+(n′ K −nK)·1.
c) Show that (b) implies that 0 = lim K→∞FJ(0,n′ K)−c = (1−c)(1−e−δ ).
d) Conclude that the fractional parts of f(n) = lnn do not converge to any distribution. (Hint: we can only get convergence if the the hitting frequency converges to 1 for every interval J.) 330 14. The Birkhoff Ergodic Theorem n n ’ K K K+J K K+1 x f(x) Figure 95. A schematic illustration of the quantities defined in exer-cises 14.10 and 14.11.
Exercise 14.11. In this exercise, we set f(n) := ln pn, where pn are the primes. The definitions of J, nK, and n′ K are as in exercise 14.10.
a) Recall the prime counting function (defined in Theorem 2.21) and show that nK = π eK+α and n′ K = π eK+α+δ .
b) Use Chebyshev’s theorem (Theorem 12.7) to show that there are positive a and b (with a < 1 < b) so that for large enough K, qK := nK n′ K ∈ a b e−δ , b a e−δ .
c) Assuming that limn→∞F(0,n) = c, use the reasoning of exercise 14.10 to show that 0 = lim K→∞FJ(0,n′ K)−c = lim K→∞(1−c)(1−qK).
d) Show that c = 1 leads to a contradiction.
e) Show that a and b and δ can be chosen so that limsupK→∞qK < 1.
f) Show that for any interval with the choices as in (e), we must have c = 1.
g) Let {Ii} be a finite set of intervals as in (f). Show that the hitting fre-quency in J := ∩Ii must be be 1.
h) Show that the intervals in (f) can be chosen so that they have empty intersection.
Exercise 14.12. See the proof of Theorem 14.18.
a) Show that Sn f+g = Sn f +Sn g.
b) Show that ⟨f⟩−is invariant along orbits.
c) Use (a) and (b) to show that Sn ⟨f⟩−−f+ε(x) = n⟨f⟩−(x)−Sn f (x)+nε.
d) Use (a), (b), and (c) to deduce a contradiction from ε > 0 and liminf n→∞⟨f⟩−(x) < liminf n→∞ 1 nSn f (x)−ε .
14.7. Exercises 331 Exercise 14.13. See the proof of Theorem 14.18.
a) Show that T maps Xn to itself.
b) Use the results in Section 14.2 to show that X\X∞is measurable.
c) Show that under the hypotheses of the proof, there must be a c > 0 so that R X\X∞( f −c)dµ > 0.
0 0 1 1 1/2 1/3 "2" "3" "1" Figure 96. A few branches of the L¨ uroth map T of exercise 14.14. The names of the branches are as indicated in the figure. The L¨ uroth expan-sion of x ∈)0,1] is [a1,a2,···], where the ith digit ai equals k if T i−1(x) falls in the domain of the kth branch.
Exercise 14.14. The L¨ uroth map T : [0,1) →[0,1) is defined by T(x) = n(n+1)x−n, if x ∈[ 1 n+1, 1 n) 0 if x = 0 where n ≥1. See Figure 96.
a) Show that T preserves the Lebesgue measure and is ergodic. (Hint: see Theorem 10.9.) b) Show that for almost all x, the digit k defined in Figure 96 has a fre-quency of 1 k(k+1) in the expansion of x for k ≥1.
c) Show that almost all x have Lyapunov exponent (Definition 10.19) λ(x) = ∞ ∑ k=1 lnk(k +1) k(k +1) ≈2.05.
(Hint: see exercise 10.23.) 332 14. The Birkhoff Ergodic Theorem Exercise 14.15. This exercise relies on exercise 14.14 and Section 6.6.
Let bk(x) : Ik →[0,1) be the branch of T k−1 such that x ∈Ik, then the kth convergent [a1,··· ,ak] of x is the (unique) endpoint of Ik that maps to zero under T k (see Proposition 6.16). The branches of T are labeled as indicated in Figure 96. For simplicity, we note (without proof) that the kth convergent is always a rational number also denoted by pk/qk. The L¨ uroth expansion of a number x ∈[0,1) is the list [a1,a2,···] where ai is the label of the branch in whose domain T i−1(x) is located. For more details, see .
a) Show that x−pk qk < |Ik|, where |Ik| is the length of Ik.
b) Show that T k : Ik →[0,1) is an affine bijection.
c) Show that |Ik+1| ≤ x−pk qk < |Ik|, (Hint: bk maps Ik affinely onto [0,1) (see Figure 97) and so the sub-intervals of Ik have the same proportions as the sub-intervals of the unit interval in the L¨ uroth map of Figure 96.) d) Use (b) to show that ln 1 |Ik| = k−1 ∑ j=0 ln ∂T(T j(x)) .
(Hint: see exercise 10.21; here we use the same notation.) e) Use (c) and (d) to show that lim k→∞ 1 k ln x−pk qk = −λ(x), where λ(x) is the Lyapunov exponent (Definition 10.19) of T at x.
Exercise 14.16. This exercise is based on exercises 14.14 and 14.15.
a) Compare the almost everywhere convergence of the continued fraction convergents with the Luroth convergents. (Hint: one is alternating and converges faster.) b) Can you venture an intuitive explanation for the faster convergence?
(Hint: look at T 2 in both cases.) 14.7. Exercises 333 0 1 I I k k+1 x p /q k k Figure 97. A few branches of the k + 1st iterate of the L¨ uroth map T restricted to the interval Ik. In red a branch of T k and in black a few branches of T k+1.
Exercise 14.17. Two measures ν and µ are said to be in the same mea-sure class if they have the same sets measure zero sets. Suppose we fix a measure class and are given that T is ergodic with respect to an (unknown) ergodic measure in this class.
a) Given a set S and its characteristic function {1}S. Show that µ(S) = R {1}S dµ.
b) Use (a) and Corollary 14.19 to show that µ(S) = lim n→∞ 1 n n−1 ∑ i=0 f(T i(x)).
c) Show that this determines the measure µ.
d) Show that if there was another ergodic measure ρ, then it would live entirely in the sets of µ-measure zero. (Hint: see Corollary 9.15.) As noted before, in the more common definition of ergodicity (Definition 9.12) “weak invariance” is replaced by the “strict invariance”. In exercise 14.18, we show that these notions are in fact equivalent. We assume that T is ergodic with respect to the (invariant) measure µ.
334 14. The Birkhoff Ergodic Theorem Exercise 14.18. a) Show that ergodicity with weak invariance implies er-godicity with strict invariance. (Hint: this is trivial.) b) Now assume that T is “strict invariance” ergodic with respect to the measure µ, and let S0 be a weakly invariant set of positive measure. Show that S := ∞ \ k=0 [ i≥k T −i(S0), is strictly invariant. (Hint: x ∈S if and only if x ∈T −k(S0) for infinitely many distinct k.) c) Denote the symmetric difference by △(see Figure 98). Show that for k ≥0 : µ(T −k−1(S0)△T −k(S0)) = 0.
(Hint: for k = 0, this follows from the definition of S0; then use the fact that T preserves µ.) d) Use (c) and Figure 98) to show that for k ≥0 : µ(T −k(S0)△S0) = 0.
e) Show that S△S0 ⊆ [ k≥0 T −k(S0)△S0 .
f) Use (c) and (d) to show that µ(S0) = µ(S). (Hint: T preserves measure.) g) Use ergodicity to show that µ(S0) has full measure.
X A1 A2 X A1 A2 A3 Figure 98. Left: the symmetric difference of two sets A1 and A2 is A1△A2 := (A1 ∪A2)(A1 ∩A2). On the right side, we illustrate that A3△A1 is contained in (A3△A2)∪(A2△A1) (green).
Exercise 14.19. Suppose T : X →X where X is a compact, metric space.
If T has a periodic orbit O, exhibit a (discrete) invariant measure µ which is ergodic. (Hint: for any set A not intersecting O, µ(A) = 0. See also Section 9.4.) 14.7. Exercises 335 Exercise 14.20. For this exercise, assume that the linear combinations of the functions e2πimx are dense in the set L1(R/Z) of integrable function on the circle. (Let f be integrable, then for every ε > 0, there is a finite combination ˜ f of multiples of e2πimx so that R |f −˜ f|dx < ε.) a) Show that the Lebesgue measure is ergodic and measure preserving if and only if for all m ̸= 0 in Z lim n→∞ 1 n n−1 ∑ k=0 e2πimT k(x) = 0.
(Hint: use the proof of Theorem 14.14.) b) Show that T in (a) is ergodic if and only if {T k(x)} is equidistributed.
We saw in Section 9.4 that a given transformation T may have uncountably many coexisting invariant measures. The Krylov-Bogoliubov theorem (see ) states that a continuous map T from a compact metric space to itself has an (at least one) invariant Borel probability measure. Exercise 14.21 gives a counterexample if we drop continuity.
Exercise 14.21. T : [0,1] →[0,1] is given by T(x) = x/2 if x > 0 and T(0) = 1. Assume that there exists an invariant probability measure µ sat-isfies µ(T −1(A)) = µ(A) and such that µ is defined on all open sets.
a) Show that if µ((1/2,1)) = p > 0, then µ((0,1)) is unbounded, a con-tradiction. (Hint: use Definition 14.4.) b) Show that (a) implies that all measure must be concentrated on the points {2−i}∞ i=0 and {0}.
c) Show that if any of the points in (b) carry positive measure, then we also get a contradiction. (Hint: similar to (a).) d) Conclude that it is impossible to consistently assign an invariant mea-sure to open sets.
e) Show that there does exist an invariant measure on the trivial sigma alge-bra. (Hint: what is the smallest σ-algebra possible under Definition 14.2?) Exercise 14.22. T : [0,1] →[0,1] is given by T(x) = x/2.
a) Show that T has a unique invariant Borel probability measure, namely µ({0}) = 1. (Hint: use the strategy of exercise 14.21.) b) Show that with respect to the measure in (a), T is ergodic.
c) Show that T is uniquely ergodic (Definition 14.20).
Proposition 14.21. Suppose T : X →X where X is a compact, metric space.
If T has a unique invariant Borel probability measure µ, then that measure is the unique ergodic measure for T.
336 14. The Birkhoff Ergodic Theorem Exercise 14.23. We prove Proposition 14.21 as in [Section 4.1]. For any measurable set A, define the conditional measure µA as µA(B) = µ(A∩B) µ(A) .
Assume that there is an invariant measurable set S with 0 < µ(S) < 1.
a) Show that µS is an invariant measure.
b) Show that µX\S is an invariant measure.
c) Show that the measures in (a) and (b) are distinct. (Hint: what is the measure of S?) d) Show that (c) contradicts the hypothesis of Proposition 14.21.
Chapter 15 The Unsolvability of the Quintic Overview. In 1963, V. I. Arnold proposed a proof with a topological flavor of the insolvability by radicals of the quintic that was (supposed to be) ac-cessible for high school students. The proof is notable in that it uses much less algebra than the traditional proof based on Galois theory. It also has the advantage that it is conceptually much clearer.
To our knowledge this proof made the (English language) press only in 2004 in . It was the subject of a paper in the Monthly as recently as 2022. We will give a proof loosely based on these recent publications.
15.1. Solvable Groups We start by defining the permutation group on n symbols. Recall that a permutation Definition 15.1. The group of permutations of {1,··· ,n} is denoted by Sn.
These groups are called symmetric groups.
The reason for that name is as profound — in this context — as it is obvious. Given a polynomial f of degree n, it has n roots counting multi-plicity. Typically these roots are distinct. They do not come in any specified 337 338 15. The Unsolvability of the Quintic order, so if we permute them, they still are the roots of that exact same poly-nomial f. Thus the group of permutations on n (generally) distinct roots are a groups of symmetries of the polynomial f.
The identity element {e} of a group G is a normal subgroup. Thus if G is Abelian, then the ‘quotient’ G/{e} is obviously also Abelian. In this sense, the following definition can be seen as a generalization of Abelian.
Definition 15.2. A group G is solvable if it has subgroups Gi with {e} = Gm ⊆Gm−1 ··· ⊆G0 = G, such that Gi is normal in Gi+1 and, furthermore, Gi+1/Gi is Abelian for i ∈{0,··· ,m−1}.
The denomination solvable is, again, no coincidence. As we shall see, the polynomials of degree 4 and less can be solved by radicals (defined below) precisely because the group of permutations of the 4 (or less) roots has the above property of being solvable. This is not true for the group of permutations of 5 (or more)elements, and hence, the general quintic is not solvable by such elementary means.
In an alternative definition of a solvable group, the requirement that Gi+1/Gi is Abelian is replaced by the requirement that Gi+1/Gi must be cyclic. However, by the fundamental theorem for finite Abelian groups (Proposition 13.3), this is the same thing.
We recall that a subgroup H of G gives rise to a quotient group G/H if and only the subgroup is normal (Definition 7.30). In this case, the cosets Hx partition the original group G. It follows that the order of H is a divisor of the order of G. Among other things, it also means that cosets can be multiplied and left and right cosets are the same. So for x and y in G: HxHy = Hxy and Hx = xH .
(15.1) Now G/H is Abelian if and only if for all x and y, HxHy = HyHx, or, equivalently, HxHy(Hx)−1(Hy)−1 = H. This always holds if and only if elements of the form xyx−1y−1 are in the subgroup H. See exercise 15.5 and, for more details, Chapter 15 of .
Definition 15.3. Elements of a group G of the form aba−1b−1 are called commutators of G. The subgroup generated by the commutators is called the commutator subgroup or the derived subgroup and indicated by [G,G].
15.1. Solvable Groups 339 Lemma 15.4. The commutator subgroup is normal.
Proof. We use Definition 7.30. It is easy to see that xaba−1b−1x−1 = xax−1 xbx−1 xa−1x−1 xb−1x−1 , which is again a commutator and is thus in the commutator subgroup. The same works for a product of commutators (exercise 15.1).
■ Two important remarks are in order here. The first is that the com-mutator subgroup contains the commutators, of course, but may contain other elements as well (exercises 15.2 and 15.3). The second is that group G/[G,G] is, in a sense, the ‘Abelian part’ of G. So the smaller the derived group is, the ‘more’ Abelian G is.
If a finite group G “lacks” solvability, it must be because at some point, the commutator subgroup Gi+1 of Gi stopped becoming smaller than Gi itself. In other words, the commutator subgroup of Gi is equal to itself.
Thus a natural way to test if a group G := G0 is solvable, is by considering the sequence of commutator subgroup Gi+1 := [Gi,Gi]. If we start with a finite group, this sequence of subgroups of decreasing must eventually end with Gm+1 = Gm for some m. If Gm consists of the identity, then G is solvable. If Gm is a non-trivial group, then G is not solvable.
Definition 15.5. The descending series of normal subgroups, G := G0 ▷G1 ▷···Gm ▷··· , where each group is the commutator subgroup of the previous entry, is called the derived series. Here we used the standard notation G ▷H for H a proper normal subgroup of G.
Definition 15.6. A simple group is a group with no non-trivial normal sub-group. A perfect group is a group that is generated by its commutators.
Thus the commutator subgroup of a simple group must be equal to the identity or to itself. In other words: that group is perfect. The reverse, however, is false as be seen in exercise 15.4, where we show that SL2(R) has a non-trivial normal subgroup, even though its commutator subgroup equals the whole group.
340 15. The Unsolvability of the Quintic At this point, the reader could be excused for fearing that a (finite) group G might admit several very different and non-unique descending se-ries of proper subgroups. But it turns out that if at every stage you quotient by the maximal proper subgroup, the sequence is essentially unique. This general statement is called the Jordan-H¨ older theorem , see . Applied to cyclical groups Zn, this yields the unique factorization theorem (Theorem 2.11). The Jordan-H¨ older theorem can, in fact, be seen as a generalization of the unique factorization theorem.
15.2. The Derived Series of Sn Recall that a permutation on n symbols can be written as a product of a number of transpositions. The parity of the permutation indicates whether that number is even or odd and is well-defined (see exercise 15.7).
In what follows, we assume the reader knows cycle notation for permuta-tions; details can be found . To avoid ambiguities, we only remark that σ = (bdce)(abc) is read as follows: σ maps b first to c and then to e, so σ(b) = e, and so forth. See Figure 99.
a b c d e r g Figure 99. The permutation σ = r◦g consist of first applying the green cycles g and then the red cycle r.
Definition 15.7. A alternating group of degree n is the group of even per-mutations on n symbols. It is denoted by An.
Proposition 15.8. An consists of products of 3-cycles. Furthermore, the commutator subgroup of Sn for n > 2 is An: [Sn,Sn] = An.
15.2. The Derived Series of Sn 341 Proof. Given a permutation in An. Since it is even, we can write it as a composition of pairs of transpositions. There are three possibilities for each pair: (ab)(ab) = I (ab)(bc) = (abc) (ab)(cd) = (abc)(bcd).
Thus every even permutation can be written as a product of 3-cycles.
On the other hand, every 3-cycle can be written as a commutator. This can be most easily seen by realizing that a transposition is its own inverse, and so (ab)(ac) = (acb) = ⇒ (ab)(ac)(ab)(ac) = (abc).
(15.2) Thus every element of An can be written as a product of 3-cycles and there-fore of commutators. Furthermore, the number of transpositions in a com-mutator must be a multiple of four and so their product must be in An.
■ Proposition 15.9. For n ≥5, An is perfect: [An,An] = An.
r g 1 2 3 4 5 Figure 100. Denote the red 3-cycle by r and the green by r. The com-mutator r−1g−1rg equals the 3-cycle (123).
Proof. The crux here is that if we have at least five symbols at our disposal, we can write every 3-cycle as a commutator of even permutations. From (15.2), we see that every 3-cycle is a commutator. So it is sufficient to write an arbitrary 3-cycle as a commutator of 3-cycles. From Figure 100 we conclude that the cycle (1,2,3) can be written as (235)(421)(532)(124) = (532)−1(124)−1(532)(124) = (123).
We need to use two ‘extra’ symbols (4 and 5) to do this.
■ 342 15. The Unsolvability of the Quintic This gives us the derived series for Sn with n ≥5, that leaves the cases n ∈{2,3,4} open. Now S2 has only two permutations, namely the identity and (12). Its derived series is S2 ▷{e}. S3 has 3! = 6 permutations. By Proposition 15.8, A3 is generated by the 3-cycles and so it must consist of (123) plus its iterates. This group has three elements. The order of any normal subgroup must be a divisor of 3, and therefore can only be A3 or the identity I = {e}. A trivial check shows that the latter is the case. That leaves only S4 to be analyzed.
Lemma 15.10. LetV = {I,(12)(34),(13)(24),(14)(23)}. We have [A4,A4] = V and [V,V] = {e}.
2 3 4 1 p p s conjugate by s 2 3 4 1 Figure 101. The effect of conjugating a permutation p by the reflection s in the dotted line.
Proof. We start by noting that the commutator subgroup V is invariant un-der conjugation by Lemma 15.4. Thus if p ∈V then for any s ∈A4, we have that ps := sps−1 is in V. It follows that (see Figure 101) if p : i →j then ps : s(i) →s(j).
Next, we use that A4 is generated by 3-cycles to compute one element of [A4,A4].
(123)(234)(321)(432) = (14)(23).
But by the previous paragraph, relabeling will give another element of [A4,A4]. All relabelings plus the identity give the elements in the state-ment of the proposition. Since we the order of [A4,A4] must be a non-trivial divisor of 12 (the order of A4), we can have at most two more elements in [A4,A4].
If [A4,A4] had any other element, it could not be a single 2-cycle or a 4-cycle (they have the wrong parity). So it would have to be a 3-cycle.
15.3. Solving Cubic and Quartic Equations 343 In view of the first paragraph, this gives more than two elements (see also exercise 15.9), which completes the proof.
■ The main result of this section now follows.
Theorem 15.11. We have the following derived series: S2 ▷{e} S3 ▷A3 ▷{e} S4 ▷A4 ▷V ▷{e} ∀n > 4 : Sn ▷An .
Proof. This follows immediately from putting all results in this section to-gether.
■ Remark 15.12. We can naturally associate a sequence of Abelian groups to each derived series, namely S2/{e} ∼ = Z2 S3/A3 ∼ = Z2 , A3/{e} ∼ = Z3 S4/A4 ∼ = Z2 , A4/V ∼ = Z3 , V/{e} ∼ = Z2 ×Z2 ∀n > 4 : Sn/An ∼ = Z2 .
15.3. Solving Cubic and Quartic Equations We all know how to solve for the roots of linear and quadratic polynomial equations in C and are probably aware of the fact that there are general formulas for the roots of cubic and quartic polynomials. However, these general formulas are sufficiently complicated (see [55, 56]) that they have little or no practical use. What is interesting, though, is that at n = 4 it stops: there exist no general formulas for n ≥5 that beside continuous functions involve finitely many radicals!
To understand this better, we first look at the formulas for the roots of cubic and quartic polynomials. We consider polynomials with coefficients in a field. In fact, we will assume the field to be C. The field operations 344 15. The Unsolvability of the Quintic are, of course, addition, subtraction, multiplication, and division. To these operations we add taking an nth root (specifying which one).
Definition 15.13. An equation is solvable via radicals1 if its roots are func-tions of the coefficients that use only the field operations and (specified) nth roots.
We now consider a general degree n polynomial q(z) := ∑n i=0 cizi. We start by simplifying the problem.
Lemma 15.14. If we can find the roots of the general depressed polynomial p(z) := zn +∑n−2 i=0 aizi, then we can find the roots of the general polynomial q(z) := ∑n i=0 cizi.
Proof. Since the coefficients are in C, solving q(z) = 0 is equivalent to solving zn +∑n−1 i=0 ci cn zi = 0. Next, eliminate z in this polynomial in favor of x, where2 z = x−cn−1 ncn .
(15.3) It is a straightforward computation to check that the term of order n −1 cancels and the polynomial now becomes x−cn−1 ncn n −cn−1 cn x−cn−1 ncn n−1 +··· = xn −cn−1 cn xn−1 + n 2 c2 n−1 n2c2 n xn−2 +··· + cn−1 cn xn−1 −(n−1)cn−1 ncn xn−2 +··· +··· = xn +∑n−2 i=0 aixi , upon suitably defining the ai in terms of the ci. So a root x∗= F(a1,··· ,an−2) of p(x) corresponds to a solution z∗= x∗−cn−1 ncn = F(a1,··· ,an−2)−cn−1 ncn , where the ai are defined in terms of the ci.
■ Lemma 15.15 (Viete’s Formula). Two numbers x± in C satisfy x−+x+ = −b and x−x+ = −c, 1A radical is an nth root; radix being the Latin word for root.
2It is important to note that this might not work if the coefficient are in some other field, such as as a degree p polynomial with coefficients in Fp (see Chapter 5). The number p satisfies p · 1 = 0 in this field.
Such a number the characteristic of the field, or ring. It is zero if there is no natural number p satisfying that equation, as in the case of Z or C.
15.3. Solving Cubic and Quartic Equations 345 if and only if x± = −b 2 ± r b2 4 +c.
Proof. Both x−and x+ can be seen to satisfy x(−b −x) = −c. This has precisely the two indicated solutions. Vice versa, it is easy that the two relations hold.
■ Now we turn to the cubic equation and set p3(x) = x3 +a1x+a0. First, substitute x = x−+ x+. After a slight rearrangement of the terms, we get the following equation for the roots p3(x−+x+) = x3 −+x3 + +(a1 +3x−x+)(x−+x+)+a0 = 0.
We can solve this if x3 −+x3 + = −a0 and x−x+ = −a1 3 and so x3 −+x3 + = −a0 and x3 −x3 + = −a3 1 33 .
We apply Vi ete’s forula (Lemma 15.15) to this last pair of equations.
x3 ± = −a0 2 ± s a2 0 4 + a3 1 33 .
(15.4) So x± are third roots of this expression, and the roots x∗of p3 are x∗= x−ω−+x+ω+ , (15.5) where ω± are third roots of unity and must chosen so that also x−x+ = −a1 3 is satisfied. There are exactly three such third roots, so if we choose ω−, then ω+ is uniquely determined. Thus this gives three solutions of the cubic equation p3(x) = 0.
Moving along to the quartic equation, after reducing the equation as before, we set p4(x) = x4 + a2x2 + a1x + a0. If a1 = 0, then we apply the usual quadratic formula to get two solutions for x2 and hence four solutions for x. So assume that a1 ̸= 0. To solve for the roots, we try to equate p4(x) to a product of quadratic equations, as follows.
p4(x) = (x2 −yx+b)(x2 +yx+c) = x4 +(−y2 +b+c)x2 +(b−c)yx+bc.
(15.6) 346 15. The Unsolvability of the Quintic For this to hold, we need that b+c = y2 +a2 and b−c = a1 y and bc = a0 .
Note that we may assume that y ̸= 0 (because a1 ̸= 0). The first two of these three equations can be used to deduce that b = 1 2 y2 +a2 + a1 y and c = 1 2 y2 +a2 −a1 y (15.7) Substitute these into bc = a0 to get y2 +a2 + a1 y y2 +a2 −a1 y −4a0 = y4+2a2y2+(a2 2−4a0)−a2 1 y2 = 0.
(15.8) This is a cubic equation in y2. We can use the cubic formula to solve and get 6 possible values for y. Pick any non-zero value for y and compute b and c from (15.7) (and check that bc = a0). Finally, substitute these values into (15.6) and solve the quadratic equations. Notice the somewhat mysterious fact that any non-zero value for y must give the same four solutions, as there are only four solutions by the fundamental theorem of algebra (Theorem 11.21).
We will see in exercises 15.10 and 15.11 that the practical implementa-tion of these schemes is far from trivial. In practice one either ‘guesses’ one or more roots of such equations or uses a numerical scheme to approximate solutions to arbitrary precision, such as Newton’s method.
15.4. Monodromy One more piece of information is needed to finalize the proof of the non-solvability of the quintic. This goes by the name of monodromy. This is a rather technical and very important concept. Luckily we only need the basic material as it relates to taking nth roots in the complex plane.
For concreteness, we let p : z →zn for n some integer greater than 1. A loop in C is a continuous path α : [0,1] →C whose endpoints are the same (denoted as the base-point). Choose a (non-zero) base-point in C, say 1.
Consider loops with 1 as its endpoints. Furthermore since under p, every point has exactly n inverses, except 0, we want our loops to avoid 0. So we consider loops in C{0} based at, say, 1. To every loop α avoiding zero, there is a well-defined integer k we can assign to it that counts the number of 15.4. Monodromy 347 1 Figure 102. Two loops in C{0} based at 1 that wind around 0 exactly once. The red arrows indicate how the outer curve can be continuously deformed in C{0} to the inner curve.
times it circles around the origin in the counter-clockwise direction. This is called the winding number of the loop. It turns out that any two curves with the same k can be continuously deformed into one another. We illustrate this in Figure 102; details can be found in , Chapters 9 and 10. This continuous deformation defines an equivalence relation on loops in C{0} based at 1. This is called homotopy equivalence .
Definition 15.16. The set of loops in C{0} based at 1 and up to homotopy equivalence is called the fundamental group of C{0} relative to the point 1. It is denoted by π1(C{0},1).
Since in this example all loops with the same winding number are (homo-topy) equivalent, the fundamental group π1(C{0},1) is just Z.
It is clear how loops map under p, as p is a well-defined continuous function. The problem is when we try to define the image of a loop α around 0 under its multi-valued ‘inverse’, n √z. Informally, here is how we do it. We refer to Figure 103. The complete inverse image of the base-point under p, or p−1(1), is called the fiber of 1 under p. In this case, the fiber of 1 consists of the n points aℓ:= e2πiℓ/n, ℓ∈{0,··· ,n−1}. Suppose the loop α has base-point 1 and runs around zero exactly once. Choose one of the points in the fiber of 1, say aℓ. Take some small neighborhood of 1 so that its inverse images under p around the ai are disjoint. Then locally in each of these inverse images, the map is invertible. In particular, we can draw the inverse image ˜ α for some small t ∈[0,ε] such that ˜ α(0) = aℓ. By chopping up all of α into small pieces, this process can be continued until we have an inverse image ˜ α of α with ˜ α(0) = aℓ. This is called a lift of α and has the 348 15. The Unsolvability of the Quintic p lift 1 1 −1 Figure 103. The curious behavior of lifts loops based at 1 through a 4th root. The fiber of 1 is {1,i,−1,−i}. The green loop on the left has winding number 1 and lifts to curves whose endpoints are not the same. We exhibit one lift based at 1 and one based at -1. The red loop on the left has winding number 0 and lifts to loops: we exhibit the ones based at 1 and -1. The blue neighborhood on the left lifts to disjoint blue neighborhoods on the right. Restricted to these local neighborhoods, p is invertible.
property that p( ˜ α) = α. What is important — and should be clear — is that in a lifted path ˜ α, the endpoints do not necessarily coincide anymore.
What happened here is that the loop α induces a permutation on the fiber of its base-point. In the case of nth roots, it is easy to see3 that if α has winding number k, then its lift ˜ α based at ai is a curve from ai to a(i+k) mod n. We can think of this as the action of the winding number on the points of the fiber p−1(1). In this case, we see that this action is a cyclic permutation of those points and they form a group. Thus the lifts of a curve give rise to a cyclic permutation of the points of the fiber.
Definition 15.17. Let p(z) = zn. The action of the winding number on the roots gives rise to a homomorphism π1(C{0},1) →Sn. The image of that homomorphism in this case is isomorphic to Zn and is called the monodromy group4.
Before we state the central result of this section, we need one more definition. The commutator of two loops α and β with the same base-point 3More general formal proofs tend to get very technical.
4We note here that there is no immediate connection with the similarly named ‘monodromy operator’ for differential equations with periodic coefficients, see , Section 26 15.5. The General Quintic is Not Soluble via Radicals 349 is given by concatenating first α(t), then β(t), then α(1 −t), and finally β(1 −t) and finally rescaling the parameter t so that its domain is [0,1].
The commutator is denoted by [α,β] .
Lemma 15.18. Let α(t) and β(t) be continuous loops in C with the same base-point. Then γ := n p [α,β] is a loop. (In other words: their endpoints coincide.) Proof. Suppose α(t) and β(t) loop around the origin a and b times respec-tively, the their commutator loops around a+b−a−b = 0 times, and so its lift is a loop. The underlying reason, of course, is that Zn is Abelian.
■ The astounding simplicity of this lemma notwithstanding, it represents the essence of the proof of the non-solvability of the quintic by radicals.
15.5. The General Quintic is Not Soluble via Radicals We now address the central question of this chapter: when can we solve for the roots of polynomials by radicals? Here is the definition.
Definition 15.19. We say that a polynomial p(x) over a field F admits solution by radicals if its roots can be obtained from the the coefficients of the polynomial p using the usual field operations (addition, subtrac-tion, multiplication, and division) and application of radicals (nth roots for n = 2,3,4,···).
Consider the reverse situation where we know the roots si and want to compute the coefficients of the above polynomial. This is easy, because p(z) = (z−s1)···(z−sn) = zn −(s1 +···+sn)zn−1 +···+(−1)ns1s2 ···sn .
From this, the coefficients follow as continuous functions of the roots. From now on, we will denote this function by F(s1,··· ,sn). The ith coefficient ci is given by the ith component of F : Cn →Cn.
There is one overwhelmingly important observation that will dominate this conversation. It is this: The polynomial p is unchanged by permuta-tions of its roots! Note that this is very much not true for its coefficients.
The way we are going to take advantage of this is as follows. We are go-ing to look at continuous paths for the roots of polynomials with the prop-erty that at t = 0 we start at roots (s1,s2,··· ,sn) and at t = 1, we end at 350 15. The Unsolvability of the Quintic (π(s1),π(s2),··· ,π(sn)), where π is some permutation of the n roots. The point of this is that it gives us a continuous way to effect a permutation.
This enables us to use topological arguments to analyze what expressions can be used for the zeros of polynomials.
c c s s 1 2 1 0 Q F Figure 104. Left, the coefficients c1(t) = 0 and c0(t) = e2πit of a qua-dratic polynomial. Right, its solutions s1(t) and s2(t). Between t = 0 and t = 1, the solutions are permuted.
To make this more concrete, we look at a quadratic case. Consider the polynomial z2 +2e2πit where t ∈[0,1], see Figure 104. On the left, we have the coefficient space. The coefficients describe a loop in C2.
c(t) = (c0(t),c1(t)) = (2e2πit,0).
The corresponding lifts γ(t) in the solution space on the right give γ(t) = (s1(t),s2(t)) = (i √ 2eπit,−i √ 2eπit).
Note that this is not a loop, for γ(0) = (i √ 2,−i √ 2) and γ(1) = (−i √ 2,i √ 2).
This gives us one way of verifying that Q must be multi-valued, because un-der F both these points map to (0,2). See also exercise 15.13. Below we will give a different proof.
The seemingly rather trivial observation that the roots are exchangeable immediately leads to a remarkable result.
Proposition 15.20. For n ≥2, let p(z) = zn + ∑n−1 i=0 cizi have roots {si}.
There is no general solution in the form of a continuous function. That is: the solution must be in the form of an n-valued function q : Cn →C.
Proof. Suppose there exists a continuous function Q : Cn →Cn with com-ponents qi that takes the coefficients {ci} to the roots {si}.
si = qi(c1,c2,··· ,cn).
15.5. The General Quintic is Not Soluble via Radicals 351 This solution is general and therefore must remain the same if we arbi-trarily permute the si. It follows that all qi are the same: qi(c1,··· ,cn) = q(c1,··· ,cn). But we still have n distinct solutions, so q must be n-valued and therefore is not a function.
■ Remark 15.21. This result says that there is no general single-valued for-mula that gives the roots. However, in special cases, there may very well be such a formula. A case in point is p(z) = (z−1)5. The roots are x = 1 (with multiplicity 5).
We will from now on concentrate on describing q. It is a multi-valued ‘function’ Cn →C and so, according to Definition 15.19, it must contain at least one radical. But roots act on complex numbers, not vectors. So at some point in the evaluation of q (see Figure 105), we must take a multi-valued nth root √.. : C →C. Then further continuous, single-valued operations may follow to eventually yield the zeros of p. The final form of q may be a sum, product, et cetera of such expressions. However, the interesting ‘action’ is where we take the nth roots, everything consists of continuous and single-valued functions. Accordingly, we will visualize Q only in terms of the nth root operation.
c c c 1 2 0 s3 s s 1 2 function nth root Figure 105. At some point in the evaluation of q, we must take a multi-valued nth radical n √.. : C →C.
Before stating our main results, we recall from the proof of Lemma 15.14 that it is sufficient to study the zeros of monic polynomials p(z) = zn +∑n−1 i=0 cizi. The reader should compare these results with Theorem 15.11 and the remark following it.
Theorem 15.22. Let p(z) = z3 +∑2 i=0 cizi have roots {si}3 i=1. The general solution by radicals Q must at least contain nested radicals (i.e. p..√..).
352 15. The Unsolvability of the Quintic s3 s1 c Q F s2 Figure 106. Right, paths s1(t) and s2(t) that swap the first two roots in green. This results in a green loop g(t) on the left, upon taking the 3rd power; Similarly, in red, paths that swap the second and third loop and their 3rd power.
Proof. Suppose that the general solution is an expression that contains rad-icals but does not contain nested radicals. The radical in the description of Q is depicted in Figure 106. If we swap two of the three roots as depicted in green, then the image under z →z3 is a green loop g. Take a different pair and swap those in red. This will give a red loop r.
Now consider the commutator r−1g−1rg. If we just compute that com-mutator on the right directly, we obtain (23)(12)(23)(12) = (123).
However, according to Lemma 15.18, the action of a commutator on the permutation is trivial, a direct contradiction.
■ Theorem 15.23. Let p(z) = z4 + ∑3 i=0 cizi have roots {si}4 i=1. The gen-eral solution by radicals Q must at least contain triply nested radicals (i.e.
q ..p..√..).
Proof. Suppose that the general solution is an expression that contains nested radicals but nothing ‘worse’ than that. Consider 4 loops numbered from 1 to 4, as in Figure 107. Apply Lemma 15.18. Under the first root, the commutator [1,2] goes to a loop, say α. Similarly the [3,4] goes to β. The commutator of these commutators, is mapped under the root to a commutator. Upon taking the second root, the same lemma ensures that the final image is a loop again. This is in direct contradiction with Lemma 15.10 which says that the commutator subgroup of the commutator subgroup is, in fact, a non-trivial group.
■ 15.5. The General Quintic is Not Soluble via Radicals 353 c 1 2 3 4 Figure 107. Four loops numbered from 1 to 4 make a commutator of commutators, namely .
Theorem 15.24. Let n ≥5 and p(z) = zn + ∑n−1 i=0 cizi have roots {si}n i=1.
There is no general solution by containing only finite nests of radicals.
Proof. Now we consider paths that are commutators of commutators of ··· of commutators, as long as the number is finite, say M. These are con-structed in a similar way as illustrated in Figure 107, but with 2M loops instead of just 2. After taking the M nested roots, though, this commutator must be trivial by Lemma 15.18 (again). However, this contradicts Propo-sition 15.9.
■ Remark 15.25. This theorem says that there is no general single-valued formula that gives the roots. However, in special cases, there may very well be such a formula. A case in point is p(z) = (z −1)5. The roots are x = 1 (with multiplicity 5).
Remark 15.26. There are important differences with the more traditional treatment of this topic, which uses Galois theory to address the question of solvability by radicals [54, 68]. In the Galois theory approach, solubil-ity of individual polynomials is considered, not just ‘general’ solubility. In that sense, the Galois theory approach is stronger than the theory described here. However, in Galois theory, we are only allowed to apply the field oper-ations to the coefficients— addition and multiplication and their inverses — whereas here, any continuous function — such lnci or |ci| — is fair game.
So in that sense the theory described here is stronger.
354 15. The Unsolvability of the Quintic 15.6. Exercises Exercise 15.1. Complete the proof of Lemma 15.4, that is: a) Check that (xax−1)−1 = xa−1x−1, and b) Check the formula in the proof for a product of commutators.
The commutator subgroup [G,G] of a group G is generated by and therefore contains, of course, its commutators. What is less clear is that it may contain more. Examples of this are surprisingly hard to exhibit. In the next few exercises, we show that the commutator subgroup of SL(2,R), the 2 × 2 real matrices with determinant 1 with matrix multiplication as its operation, has this property. To do so, we will show that −I is not a commutator itself, but is a product of commutators.
Exercise 15.2. Consider the group G := SL(2,R) and assume that −I is a commutator.
a) Let Ai ∈G and show that A1A2A−1 1 A−1 2 = −I implies that A1A2A−1 1 = −A2 and A2A1A−1 2 = −A1 .
b) Conclude that TrA1 = TrA2 = 0. (Hint: the trace is invariant under conjugation.) c) Show that Ai = ai bi −1+a2 i bi −ai .
d) Compute that A1A2 = −a1a2 −b1 b2 (1+a2 2) a1b2 −a2b1 −a2 b1 (1+a2 1)−a1 b2 (1+a2 1) −a1a2 −b2 b1 (1+a2 1) .
e) Now set A1A2 = −A2A1 and conclude from equating the 12 and 22 entries that 2a1a2 = b1 b2 (1+a2 2)+ b2 b1 (1+a2 1) 0 = a2 b1 (1+a2 1)+ a1 b2 (1+a2 1).
f) Multiply the first equation by b1b2 to show that a1a2b1b2 > 0 and rework the second equation to get that a1b1 a2b2 < 0.
g) Show that the assumption that −I is a commutator, is false.
15.6. Exercises 355 Exercise 15.3. a) For µ2 ̸= 1, compute µ 0 0 µ−1 1 λ µ2−1 0 1 µ−1 0 0 µ 1 0 − λ µ2−10 1 = 1 λ 0 1 , and use that to show that 1 λ 0 1 is a commutator.
b) Similarly, show that 1 0 λ 1 is a commutator.
c) Compute that 1 a−1 0 1 1 0 −a 1 1 b 0 1 = 0 a−1 −a 1−ab .
d) Choose a and b so that you can conclude that 0 1 −1 0 is product of commutators.
e) Conclude that −I is a product of commutators. (Hint: the matrix in (d) represents a rotation by π/2. Apply that rotation twice.) Remark: In fact, with a little more patience, one can show that the matri-ces 1 λ 0 1 and 1 0 κ 1 with λ and κ in R, generate the entire group SL(2,R). Thus this group is a perfect group (Definition 15.6).
Remark: For finite groups, it is known that the smallest group whose com-mutator subgroup contains more elements than just its commutators has 96 elements.
Exercise 15.4. a) Show that the commutator subgroup of a finite group G is not necessarily the largest normal subgroup. (Hint: for primes p and q, consider the cyclic group of order pq.) b) Use the above remarks to show SL2(R) is perfect but not simple. (Hint: consider the set {I,−I}.) Exercise 15.5. Let G be a group and H a normal subgroup.
a) Use (15.1) to show that (Hx)−1 = Hx−1.
b) Use (a) and (15.1) to show that HxHy(Hx)−1(Hy)−1 = H if and only if xyx−1y−1 ∈H.
c) Conclude that G/H is Abelian iff H contains all commutators.
356 15. The Unsolvability of the Quintic Exercise 15.6. In this exercise, we prove that if you write the identity of the permutation group Sn in terms of transpositions, the number of trans-positions is always even. To do so, suppose that I = τ1 ···τm where the τi are transpositions. Fix any x ∈{1,··· ,n}, and let τk+1 be the rightmost transposition that contains the symbol x.
a) Suppose τk does not contain the same two symbols as τk+1. Show that I can be written as a product of m−2 transpositions.
b) Suppose τk contains at most one symbol that also occurs in τk+1. Show that I can be written as the following product of transpositions I = τ1 ···τk−1σkσk+1τk+2 ···τm , where now σk is the rightmost transposition containing x. (Hint: use that (xa)(xb) = (xa)(ab), et cetera.) c) Show that if the symbol x never gets cancelled as in (a), then I ends up being written as a product of transposition where only the leftmost trans-position contains x.
d) Show that (c) is impossible.
e) Conclude that every x must get cancelled, and thus I must have had an even number of transpositions.
Exercise 15.7. a) Use exercise 15.6 to show that there is a well-defined surjective homomorphism P : Sn →{±1}. The image under P of a permu-tation σ is called the parity of σ.
b) Use Corollary 7.31 to show that An is a normal subgroup of Sn.
1 2 3 4 1 2 3 4 Figure 108. A permutation on n symbols can be decomposed into transpositions graphically, and thus its parity can be determined.
15.6. Exercises 357 Exercise 15.8. A convenient way to decompose a permutation into trans-positions, and indeed to find the parity defined in exercise 15.7, is to count the number of crossings of the diagram of the permutation σ as depicted in Figure 108. In this drawing only two lines can cross at the same point.
Such a figure in also called a matching diagram .
a) Use Figure 108 to show that σ(1) = 4, σ(2) = 3, and so on.
b) Show that σ = 1234 4312 !
.
c) Use Figure 108 to show that σ can be decomposed into transpositions as follows σ = (13)(14)(34)(24)(23).
d) Verify in this example that deforming the lines may lead to a different decomposition into transpositions, but does not affect the parity.
Exercise 15.9. a) Show that the group V in Lemma 15.10 is isomorphic to Z2 ×Z2.
b) Show that the parity of a single 2-cycle or a 4-cycle is odd (and so they are not in An).
c) Show that there are four distinct 3-cycles in A4, each of which has two distinct non-identity iterates.
The groupV in exercise 15.9 and Lemma 15.10 is called the Klein four-group or simply the Klein group . It is usually denoted by V or K4. It is the small-est non-cyclic group. It is not isomorphic to Z4 (which is cyclic).
In the following two exercises, we follow the method of Section 15.3 to solve for the roots of of a cubic and a quartic polynomial. To get a feel for the considerable subtleties in this process, the reader is encouraged to do all the steps by hand, except where otherwise indicated. Allow ample time for these exercises!
358 15. The Unsolvability of the Quintic Exercise 15.10. We solve for the roots of p3(z) = z3 −34z2 +313z−400 (see Figure 109).
a) Show that it is sufficient to solve q(x) = x3 −217 3 x + 6370 33 = 0. (Hint: set z = x+ 34 3 .) b) Use (15.4) and 31852 −2173 = −74088, to show that x3 ± = 3−3 −3185± p 31852 −2173 = 3−3 −3185± √ −74088 .
c) Take x−and x+ to be (standard) third roots in the right half plane, and show that x−x+ = 3−2 31852 +74088 1/3 = 217 9 = −a1 3 .
(Hint: 31852 +74088 = 2173.) d) So let ω := e2πi/3 and use (15.5) to show that the roots of q are given by x−+x+ , x−ω +x+ω−1 , x−ω−1 +x+ω .
e) Deduce that the roots of q are equal to { 14 3 ,−7 3 ±2 √ 14}. (Hint: approx-imate numerically or divide q by (x−14/3).) f) Conclude that the roots of p3 are equal to {16,9±2 √ 14} ≈{1.52,16,16.48}.
Figure 109. The graph of p3(z) of exercise 15.10 showing its three roots at approximately {1.52,16,16.48}.
15.6. Exercises 359 Exercise 15.11. We solve for the roots of p4(z) = z4 −17z2 +20z−6 (see Figure 110).
a) Show that the associated ‘cubic’ polynomial (15.8) is q(y) = y6 −34y2 +313x−400.
b) Conclude from exercise 15.10 that this polynomial has 4 as a root.
c) Use (15.6) and (15.7) to show that p4 satisfies p4(x) = x2 −2x−3 2 x2 −2x−3 2 d) Conclude that the roots of p4 are equal to {−2± √ 2,2± √ 7} ≈{−3.41,−0.59,−0.65,4.65}.
e) Take another root of the ‘cubic’ and check that you get the same roots for p4.
Figure 110. The graph of p4(z) of exercise 15.11 showing its roots are approximately {−3.41,−0.59,−0.65,4.65}.
Exercise 15.12. Let p(z) = z4.
a) What is the fiber of 16 under p?
b) Let α(t) = 16e2πit. Define 4 different lifts of α.
c) For each of the four lifts ˜ α in (b), give the start and end point of ˜ α.
360 15. The Unsolvability of the Quintic Exercise 15.13. Let p(z) = (z−s1)(z−s2)(z−s3).
a) Show that F in Section 15.5 is given by F(s1,s2,s3) = (−s1s2s3,s1s2 +s1s3 +s2s3,s1 +s2 +s3).
b) Compute p(z) if the solutions si lie on a curve γ in Cn given by γ(t) = ( 3 √ 2e 1 3 πi(2t+1), 3 √ 2e 1 3 πi(2t−1), 3 √ 2e 1 3 πi(2t+3)).
c) Draw a picture of the coefficient space and the solution space similar to Figure 104.
d) Show that γ(0) ̸= γ(1), but F ◦γ (0) = F ◦γ (1).
e) Conclude from (d) that Q, the ‘reverse’ of F, is not single-valued.
Exercise 15.14. Newton’s method if another problem is needed.
There are various techniques that address roots of equations that are insolu-ble via radicals. In exercise 15.15, we briefly discuss an example of one of these.
Definition 15.27. The Bring radical of a real number a is the unique real root of f(x) := x5 +x−a. It is denoted by Br(a). See Figure 111.
Figure 111. The graph of the Bring radical as function of (real) a.
Exercise 15.15. a) Show that Br(a) is a uniquely determined real if a is real.
b) Show that ∂aBr(a) > 0.
c) Show that Br : R →R has a well-defined, differentiable inverse.
d) Show that a root of x5 + px + q is given by p1/4Br p−5/4q . (Note: once we have one root, formulas for the other roots can be derived via the standard formulas for a quartic polynomial.) 15.6. Exercises 361 We remark here that to actually bring an arbitrary quintic into the form x5 + px+q of exercise 15.15 so that it can be solved using the Bring radical is very complicated and requires a lot of computation (see ).
Bibliography M. Aigner and G. M. Ziegler, Proofs from the book, 6th edition, Springer Verlag, Providence, RI, 2018.
V. B. Alekseev, Introduction to analytic number theory, Kluwer Academic Publishing, New York, 2004.
T. M. Apostol, Mathematical analysis, 2nd edn, Addison-Wesley, Philippines, 1974.
, Introduction to analytic number theory, Springer Verlag, New York, 1989.
V. I. Arnold, Geometrical methods in the theory of ordinary differential equations, Springer Verlag, New York, 1983.
V. I. Arnold and A. Avez, Ergodic problems in classical mechanics, Benjamin, New York, 1968.
S. Axler, Linear algebra done right, 3rd edition, Springer International Publishing, Cham, Switzerland, 2015.
, Measure, integration & real analysis, Springer Nature Switzerland, Cham, Switzerland, 2020.
L. Barreira and G. Iommi, Frequency of digits in the l¨ uroth expansion, Journal of Number Theory (2009), 1479–1490.
B. Baumslag, A simple way of proving the jordan-holder-schreier theorem, The American Mathemat-ical Monthly 113 (2006), 933–935.
A. F. Beardon, Continued fractions, m¨ obius transformations and clifford algebras, Bull. London Math.
Soc. (2003), 302–308.
C. M. Bender, D. C. Brody, and M. P. M¨ uller, Hamiltonian for the zeros of the riemann zeta function, Phys. Rev. Lett. 118 (2017Mar), 130–201.
A. Berger and T. P. Hill, Benford’s law strikes back: No simple explanation in sight for mathematical gem, Mathematical Intelligencer (2011), 85–91.
P. Berrizbeitia and B. Iskra, Gaussian mersenne and eisenstein mersenne primes, Mathematics of Com-putation 79 (2010), 1779–1791.
M. V. Berry and J. P. Keating, The riemann zeros and eigenvalue asymptotics, SIAM Review 41 (1999), 236–266.
I. Boreico, My favorite problem: Linear independence of radicals, The Harvard College Mathematics Review (2007), 83–87.
Y. Bugeaud, Distribution modulo one and diophantine approximation, Cambridge University Press, Cambridge, UK, 2012.
D. M. Burton, Elementary number theory, 7th edition, McGraw-Hill, New York, NY, 2011.
363 364 Bibliography G. Cantor, ¨ Uber eine elementare frage der mannigfaltigkeitslehere 1 (1890), 75–78.
J. S. Caughman, 2018. Personal communication.
D. A. Clark, A quadratic field which is euclidean but not norm-euclidean, Manuscripta Mathematica 83 (1994), 327–330.
I. P. Cornfeld, S. V. Fomin, and Ya. S. Sinai, Ergodic theory, Springer-Verlag, New York, NY, 1982.
R. J. Drociuk, On the complete solution to the most general fifth degree polynomial (2000).
J. C. Lagarias (editor), The ultimate challenge: The 3x+1 problem, AMS, Providence, RI, 2010.
J. Esmonde and M. Ram Murty, Problems in algebraic number theory, Springer, New York, NY, 1999.
B. Fornberg and C. Piret, Complex variables and analytic functions, an illustrated introduction, SIAM, Philadelphia, 2020.
M. Gardner, The cult of the golden ratio, Skeptical Inquirer 18 (1994), 243–247.
W. J. Gilbert, Modern algebra with applications, John Wiley & Sons, New York, 1976.
T. Gowers, The princeton companion to mathematics, Princeton University Press, Princeton, NJ, 2008.
A. Granville, Number theory revealed: a masterclass, AMS, Providence, RI, 2019.
Dumas H. S., The kam story: A friendly introduction to the content, history, and significance of classi-cal kolmogorov-arnold-moser theory, World Scientific, Singapore, 2014.
G. H. Hardy and E. M. Wright, An introduction to the theory of numbers, sixth edition, Oxford Uni-versity Press, London, UK, 2008.
D. Harel, R. Unger, and J. L. Sussman, Beauty is in the genes of the beholder, Trends in Biochemical Sciences 11 (1986), 155–156.
Th. W. Hungerford, Algebra, Springer Verlag, New York, 1974.
H. S. Zuckerman I. Niven and H. L. Montgomery, An introduction to the theory of numbers, Wiley & Sons, New York, NY, 1991.
I. Kaplansky, Set theory and metric spaces, 2nd edition, AMS Chelsea Publishing, Providence, RI, 2001.
D. Katahdin, 2018. Personal communication.
A. Katok and B. Hasselblatt, Cambridge, UK.
A. Ya. Khintchine, Continued fractions, P. Noordhoff, Ltd, Groningen, 1963.
A. N. Kolmogorov and S. V. Fomin, Introductory real analysis, Dover, New York, NY, 1970.
S. Lang, Undergraduate analysis, 2nd edition, Springer Verlag, New York, NY, 1997.
J. E. Littlewood, Lectures on the theory of functions, Oxford University Press, Oxford, UK, 1944.
M. Livio, The golden ratio: The story of phi, the world’s most astonishing number, Broadway Books, New york, NY, 2010.
A. Lopez-Ortiz, Frequently asked questions in mathematics. Available online at:
uwaterloo.ca/˜alopez-o/math-faq/mathtext/math-faq.html.
J. E. Marsden and M. J. Hoffman, Basic complex analysis, 3rd edn, W. H. Freeman, New York, NY, 1999.
Wolfram MathWorld, Mertens constant. Available online at:
com/MertensConstant.html.
F. Mertens, Ein beitrag zur analytischen zahlentheorie, J. reine angew. Math. 78 (1874), 46–62.
J. W. Milnor, Dynamics: Introductory lectures, University of Stony Brook, 2001.
C. M. Moore, Ergodic theorem, ergodic theory, and statistical mechanics, PNAS 112 (2015), 1907– 1911.
J. R. Munkres, Topology, 2nd edition, Prentice-Hall, Hoboken, NJ, 2000.
D. J. Newman, Simple analytic proof of the prime number theorem, The American Mathematical Monthly 87 (1980), 693–696.
Bibliography 365 A. M. Odlyzko, On the distribution of spacings between zeros of the zeta function, Mathematics of Computation 48 (1987), 1003–1026.
Fitzpatrick P. M., Advanced calculus. a course in mathematical analysis, PWS Publishing Company, Boston, 1996.
C. C. Pinter, A book of abstract algebra, 2nd edition, Dover, New York, 1990.
PlanetMath, Cubicformula. Available online at: February 2023.
, Quarticformula. Available online at: February 2023.
C. Pomerance, J. L. Selfridge, and S. S. Wagstaff, The pseudoprimes to 25·109, Mathematics of Com-putation 35 (1980), 1003–1026.
C. C. Pugh, Real mathematical analysis, 2nd edn, Springer, Cham, Switzerland, 2015.
P. Ramond, The abel-ruffini theorem: Complex but not complicated, The American Mathematical Monthly 129 (2022), 231–245.
B. Riemann, Ueber die anzahl der primzahlen unter einer gegebenen gr¨ osse, Monatsberichte der Berliner Akademie (1859).
S. Roman, An introduction to discrete mathematics, Harcourt Brace Jovanovich, Orlando, FL, 1989.
W. Rudin, Real and complex analysis, 3rd edn, McGraw-Hill International, New York, NY, 1987.
J. H. Shapiro, 2020. Informal Lecture Notes.
C. L. Siegel, Algebraische abh¨ angigkeit von wurzeln. (german), Acta Arith. 21 (1972), 59–64.
I. Soprounov, A short proof of the prime number theorem for arithmetic progressions (2010). Available online at:
H. M. Stark, On the gap in the theorem of heegner, Journal of Number Theory 1 (1) (1969), 16–27.
S. Sternberg, Dynamical systems, revised in 2013, Dover Publications, United States, 2013.
I. Stewart and D. Tall, Algebraic number theory and fermat’s last theorem, third edition, A. K. Peters, Natick, MA, 2002.
J. Stillwell, Mathematics and its history, third edition, Springer, New York, NY, 2010.
S. Sutherland, V’ir Tbg n Frperg. Available online at: ˜scott/papers/MSTP/crypto.pdf.
J. J. P. Veerman, Symbolic dynamics of order-preserving sets, Physica D 29 (1986), 191–201.
, Symbolic dynamics and rotation numbers, Physica A 134 (1987), 543–576.
, The dynamics of well-ordered orbits, Autonomous University of Barcelona, Barcelona, Spain, 1995.
R. A. Wilson, An example of a pid which is not a euclidean domain. Robert A. Wilson’s website, accessed in December 2021.
D. Zagier, Newman’s short proof of the prime number theorem, The American Mathematical Monthly 104 (1997), 705–708.
Index Bε(x), 214 F((x)), 135 F/E, 142 G▷H, 339 I(n), 68 L-function, 289 L-series, 289 L(χ,z), 289 Mq,a(n), 303 R(x), 134 R, 134 R[x], 134 Sn, 337 Zq,a(z), 291 [F : E], 142 [G,G], 338 [α,β], 349 C[x], 8 Fp, 94 Λ(n), 275 Ω(n), 64 Φ(z), 260 Φq,a(z), 291 Πq,a(x), 291 Q[x], 8 R[x], 8 Θq,a(z), 291 Z(γ), 145 Z[x], 8 Z× b , 284 χ1, 288 ε(n), 64, 68 γ, 271 ind p(x), 297 λ(n), 78 ⌈θ⌉, 24 ⌊θ⌋, 24 µ(n), 63 ω(n), 64, 78 ϕ(n), 66 π(x), 33 π1(C{0},1), 347 ψ(x), 276 Resm (a), 5 σ-algebra, 307 σk(n), 62 τ(n), 63 Frac(R), 178 θ(x), 260 ζ(z), 32 {1,2,···b−1}N, 12 {θ}, 24 a | b, 4 ak ∥b, 262 deg( f), 8 1(n), 68 Abel summation, 270 Abelian group, 92 absolute convergence, 238 367 368 Index absolutely continuous measure, 200 absolutely normal, 206 absorbs products, 135 additive function, 64 additive order, 82 affine cipher, 99 agonizingly slow, 306 algebraic extension, 139 algebraic integer, 8 algebraic number, 8 algebraic number theory, 129 aliquot sequence, 75 alternating group, 340 amicable numbers, 74, 75 analytic at z0, 236 analytic continuation, 251 angle multiplications, 211 arithmetic function, 61 arithmetic progression, 281 ascending chain of ideals, 173 aspiring numbers, 75 associative, 77, 92 axiom of choice, 189 B´ ezout’s Lemma, 25 Baker-Heegner-Stark theorem, 177 base, 297 base-point, 346 Benford’s law, 230 Bertrand’s Postulate, 274 binomial theorem, 99 Birkhoff Ergodic Theorem, 194, 320 Borel measure, 308 Borel sets, 190 branch cut, 237, 253 branch point, 237 Bring radical, 360 can of worms, 95 Cancellation Theorem, 26, 162 Cantor function, 204 cardinality, 14 cardinality of a finite set, 14 Carmichael number, 87 Casorati-Weierstrass Theorem, 254 Catalan number, 101 Catalan-Dickson conjecture, 75 Cauchy’s Integral Formula, 240 Cauchy’s Theorem, 238 Cauchy-Riemann equations, 253 ceiling function, 24 Champernowne number, 207 character of a group, 284 characteristic of ring or field, 344 Chebyshev function, 276 Chebyshev’s Theorem, 264 Chinese Remainder Theorem, 52 Chinese remainder theorem, generalized, 58 closed, 92 closed set, 190 closest return, 116 cluster points, 237 Collatz conjecture, 17 commutative, 77, 92 commutative ring, 93 commutator, 338, 348 commutator subgroup, 338 companion matrix, 144 complete affine interval map, 215 complete set of least residues, 96 complete set of residues, 82 completely additive function, 64 completely multiplicative function, 61 composite, 4, 5 conditional measure, 336 conformal, 253 congruent, 5 conjugate, 147, 149 continued fraction approximants, 108 continued fraction coefficients, 108 continued fraction convergents, 108 continued fraction expansion, 108 contrapositive, 84 cool, 69 countable, 11 countably infinite, 10 cyclic group, 281 De Morgan laws, 202 degree, 8 degree of an extension, 142 dense, 6 density, 299, 301 density of a measure, 200 density of a set, 213 Index 369 depressed polynomial, 344 derived series, 339 derived subgroup, 338 Devil’s staircase, 204 diagonal argument, 11 dimension, 142 Diophantine equation, 43 Dirichlet L-function, 289 Dirichlet L-series, 289 Dirichlet character modulo q, 288 Dirichlet convolution, 68 Dirichlet density, 300 Dirichlet inverse, 102 Dirichlet ring, 77 Dirichlet series, 69 Dirichlet’s Theorem, 299 discrete Fourier transform, 296 discrete Fourier Transform, inverse, 296 discrete measure, 201 distortion, 219 distributive, 77, 93 division algorithm, 25 divisor, 4, 94 domain, 162, 235 dominated convergence theorem, 313 dynamical system, 107 Egorov’s theorem, 310 eigenpair, 143 Eisenstein integers, 171 Eisenstein Mersenne primes, 181 Eisenstein’s criterion, 134 elementary, 36 emphasis, 48 empty product, 165 encryption, 98 entire function, 236 equidistribution, 207 equivalence relation, 14 Eratosthenes’ sieve, 5 ergodic transformation, 195 essential singularity, 237 Euclid numbers, 38 Euclid’s division lemma, 25 Euclid’s Lemma, 26 Euclidean algorithm, 44 Euclidean domain, 164 Euclidean function, 164 Euclidean ring, 164 Euler’s phi function, 66 Euler’s Product Formula, 32 Euler’s Theorem, 89 Euler-Mascheroni constant, 271 evaluation map, 140 extension field, 139 fat Cantor set, 204 Fej´ er’s theorem, 316 Fermat number, 88 Fermat prime, 88 Fermat’s little theorem, 85 fiber, 347 Fibonacci numbers, 56–58, 108 field, 94 field extension, 139 field norm, 147 field of fractions, 178 field of quotients, 178 finite, 10 finite extension, 139 first Chebyshev function, 260 floor function, 24 formal expression, 107 Fourier transform, 296 Fourier transform, inverse, 296 fractional part, 24 fundamental domain, 168 fundamental group, 347 fundamental homomorphism theorem, 137 fundamental homomorphism theorem for groups, 149 fundamental theorem of algebra, 60, 248, 249 fundamental theorem of arithmetic, 27 fundamental theorem of finite Abelian groups, 281 Gauss map, 105 Gauss’ lemma, 133 Gauss’ Theorem, 66 Gaussian integers, 166 Gaussian Mersenne primes, 181 Gaussian Primes, 179 general solution, 48 generalized continued fraction, 128 370 Index Gibbs phenomenon, 318 Goldbach conjecture, 15 golden mean, 56, 108 golden spiral, 57 greatest common divisor, 4, 39, 51, 160, 169 greatest common divisor, polynomial, 53, 130 greedy algorithm, 54 group, 92 Hermitian inner product, 283 holomorphic, 236 homogeneous, 48 homotopy equivalence, 347 hush-hush, 77 ideal, 135 ideals, addition, 135 ideals, multiplication, 136 identity, 92 inclusion-exclusion, 76 index, 297 indicator function, 290 Infinitude of Primes, 30 inhomogeneous, 48 inner product, 283 integer, 4, 6 integers of Q(γ), 145 integrable function, 193 integral domain, 162 integration by parts, 261 invariant measure, 194 invariant set, 195 inverse, 92 invertible element, 94 irrational number, 6 irreducible, 5, 159 irreducible polynomial, 52, 130 isolated singularity, 236 iterates, 318 Jordan-H¨ older theorem, 340 kernel of a ring homomorphism, 136 key, 98 Khinchin’s constant, 221 Klein four-group, 357 Klein group, 357 Kronecker product, 143 Krylov-Bogoliubov theorem, 335 L¨ uroth expansion, 332 L¨ uroth map, 331 L´ evy’s constant, 223 Lagrange’s Theorem, 184 Lam´ e’s theorem, 56 Lambert series, 69 laplace transform, 245 Laurent series, 135 least common multiple, 4, 39, 51, 160, 169 least common multiple, polynomial, 53, 130 least residue, 96 Lebesgue Decomposition, 200 Lebesgue density theorem, 214 Lebesgue integral, 193 Lebesgue measurable sets, 189, 191 Lebesgue’s dominated convergence theorem, 313 left module, 146 Liouville λ-function, 78 Liouville number, 9 Liouville’s theorem, 9, 254 Little Picard, 254 logarithm, useful properties of, 226 loop, 346 Luzin’s theorem, 315 Lyapunov exponent, 229 M¨ obius inversion, 65 M¨ obius function, 63 matching diagram, 357 Maximal Ergodic Theorem, 319 maximal ideal, 135 measurable function, 308 measurable sets, 189 measurable space, 193 measurable transformation, 193 measure, 192, 308 measure preserving, 194 measure space, 193 Meissel-Mertens constant, 304 meromorphic, 237 Mersenne number, 88 Mersenne prime, 88 Index 371 Mertens constant, 304 Mertens’ Theorems, 304 middle third Cantor set, 16, 203 minimal polynomial, 8, 52, 130 mixing, 230 modular arithmetic, 5, 81 modular exponentiation, 86 module, 146 modulo, 5 monodromy group, 348 Morera’s Theorem, 243 multiple, 4 multiplicative cipher, 98 multiplicative function, 61 multiplicative order, 82 multiplicative unit, 5 multivalued, 253 mutually singular measures, 196 name-calling, 129 natural boundaries, 237 natural density, 299, 301 nearly, Littlewood’s principles, 315 non-unit, 5 norm, 147 normal in base b, 206 normal subgroup, 149 number theoretic function, 61 open set, 190 order, 81 order of a modulo b, 82 orthonormal basis, 283 outer measure, 190 parity, 356 partial integration, 261 particular solution, 48 Pell’s equation, 183 perfect group, 339 perfect number, 74, 88 Perron-Frobenius operator, 210 Picard Theorem, 254 PID, 173 plain text, 98 pointwise convergence, 310 Pointwise Ergodic Theorem, 194, 320 pole of order k, 237 power set, 18 prime, 4, 5, 159 prime counting function, 33 prime gap, 274 Prime Number Theorem, 33, 270 Prime Number Theorem for Arithmetic Progressions, 126 Prime Number Theorem for arithmetic progressions, 294 prime omega functions, 64 prime polynomial, 130 prime zeta function, 300 primitive element, 141 primitive element theorem, 141 primitive root, 83 principal character, 284 principal ideal, 135 principal ideal domain, 173 probabilistic number theory, 187 probability measure, 193 pseudoprime (to the base b), 87 punctured neighborhood, 236 pushforward of a measure, 194 quadratic congruences, 100 radical, 344, 349 Radon-Nikodym, 200 rational number, 6 reduced set of residues, 82 reducible, 4, 5, 159 region, 235 relatively prime, 5 removable singularity, 236 residue, 5, 257 Riemann Hypothesis, 34 Riemann zeta function, 32 Riemann-Stieltjes integral, 260 right module, 146 ring, 93 ring homomorphism, 136 ring isomorphism, 136 ring of polynomials, 130 ringing, 318 rng, 94 Roth’s Theorem, 10 second Chebyshev function, 276 sequences, 61 sigma algebra, 307 372 Index simple closed curve, 238 simple group, 339 simple pole, 237 singular continuous measure, 201 Skewes number, 34 sociable numbers, 75 soluble group, 338 solution by radicals, 349 solvable group, 338 solvable via radicals, 344 sophistry, 29 square free, 38, 63 square full, 38 Stirling’s formula, 271 Stokes’ theorem, 239 strictly invariant set, 195 sub-additivity, 192 sum of divisors, 62 symmetric group, 337 Tauberian theorem, first version, 245 Tauberian theorem, second version, 278 Taylor series, 244 Taylor’s Theorem, 244 time average equals spatial average, 195 Titchmarsh’s convolution theorem, 153 totient function, 66 tower of fields, 157 trace, 147 transcendental extension, 139 transcendental number, 8 twin prime conjecture, 15 twin primes, 14 UFD, 163 uncountable, 11 underlined, 3 uniform absolute convergence, 237 unique factorization domain, 163 unique factorization theorem, 28 uniquely ergodic, 323 unit, 5, 94 untouchable numbers, 75 vector space, 141 Veerman, 3 Vi` ete’s formula, 344 Vitali set, 189 von Mangoldt function, 275 weakly invariant set, 195 Weierstrass M test, 237 well-ordering principle, 6 Weyl’s criterion, 316 Wilson’s theorem, 95 winding number, 347 word, 12 zeta function, 32 |
189241 | https://phys.libretexts.org/Courses/Kettering_University/Electricity_and_Magnetism_with_Applications_to_Amateur_Radio_and_Wireless_Technology/04%3A_Potential_and_Field_Relationships/4.05%3A_Applications_of_Electric_Potential_and_Conductors_in_Electrostatic_Equilibrium | 4.5: Applications of Electric Potential and Conductors in Electrostatic Equilibrium - Physics LibreTexts
Skip to main content
Table of Contents menu
search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode
school Campus Bookshelves
menu_book Bookshelves
perm_media Learning Objects
login Login
how_to_reg Request Instructor Account
hub Instructor Commons
Search
Search this book
Submit Search
x
Text Color
Reset
Bright Blues Gray Inverted
Text Size
Reset
+-
Margin Size
Reset
+-
Font Type
Enable Dyslexic Font - [x]
Downloads expand_more
Download Page (PDF)
Download Full Book (PDF)
Resources expand_more
Periodic Table
Physics Constants
Scientific Calculator
Reference expand_more
Reference & Cite
Tools expand_more
Help expand_more
Get Help
Feedback
Readability
x
selected template will load here
Error
This action is not available.
chrome_reader_mode Enter Reader Mode
4: Potential and Field Relationships
Electricity and Magnetism with Applications to Amateur Radio and Wireless Technology
{ }
{ "4.01:Electric_Potential_from_Electric_Field" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.02:_Electric_Field_from_Electric_Potential" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.03:_Equipotential_Curves_and_Surfaces" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.04:_Conductors_in_Electrostatic_Equilibrium" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.05:_Applications_of_Electric_Potential_and_Conductors_in_Electrostatic_Equilibrium" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.06:_Potential_and_Field_Relationships(Summary)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.07:Potential_and_Field_Relationships(Exercises)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.08:Potential_and_Field_Relationships(Answers)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
{ "00:Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Preliminary_Concepts" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_The_Electric_Field" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_The_Electric_Potential" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Potential_and_Field_Relationships" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Electric_Current_and_Resistance" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Direct-Current(DC)Resistor_Circuits" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Capacitance" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_The_Magnetic_Field" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Electromagnetic_Induction" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Inductance" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Electromagnetic_Waves" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Antenna_Systems" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Propagation_of_Electromagnetic_Waves" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "14:_Introduction_to_Semiconductor_Devices" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15:_Part_2-Detailed_and_or_Advanced_Content" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "16:_Direct_Calculation_of_Electrical_Quantities_from_Charge_Distributions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "17:_Gauss\'s_Law_for_Calculation_of_Electrical_Field_from_Charge_Distributions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "18:_Calculation_of_Magnetic_Quantities_from_Currents" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "19:_Alternating-Current(AC)_Circuits" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "20:_Maxwell\'s_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "21:_Electrical_Transmission_Lines" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "22:_Generation_and_Detection_of_Electromagnetic_Waves" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "23:_Signal_Modulation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
Tue, 15 Jul 2025 19:43:46 GMT
4.5: Applications of Electric Potential and Conductors in Electrostatic Equilibrium
102688
102688
Ron Kumon
{ }
Anonymous
Anonymous
2
false
false
[ "article:topic", "conductor", "polarized", "free charge", "electrostatic equilibrium", "ionosphere", "Faraday cage", "license:ccby", "showtoc:yes", "program:openstax", "licenseversion:40", "source-phys-2566", "source@ "author@Ronald Kumon", "author@OpenStax", "source-phys-102404" ]
[ "article:topic", "conductor", "polarized", "free charge", "electrostatic equilibrium", "ionosphere", "Faraday cage", "license:ccby", "showtoc:yes", "program:openstax", "licenseversion:40", "source-phys-2566", "source@ "author@Ronald Kumon", "author@OpenStax", "source-phys-102404" ]
Search site Search Search Go back to previous article
Sign in
Username Password Sign in
Sign in
Sign in
Forgot password
Contents
1. Home
2. Campus Bookshelves
3. Kettering University
4. Electricity and Magnetism with Applications to Amateur Radio and Wireless Technology
5. 4: Potential and Field Relationships
6. 4.5: Applications of Electric Potential and Conductors in Electrostatic Equilibrium
Expand/collapse global location
Electricity and Magnetism with Applications to Amateur Radio and Wireless Technology
Front Matter
1: Preliminary Concepts
2: The Electric Field
3: The Electric Potential
4: Potential and Field Relationships
5: Electric Current and Resistance
6: Direct-Current (DC) Resistor Circuits
7: Capacitance
8: The Magnetic Field
9: Electromagnetic Induction
10: Inductance
11: Electromagnetic Waves
12: Antenna Systems
13: Propagation of Electromagnetic Waves
14: Introduction to Semiconductor Devices
15: Part 2 - Detailed and/or Advanced Content
16: Direct Calculation of Electrical Quantities from Charge Distributions
17: Gauss's Law for Calculation of Electrical Field from Charge Distributions
18: Calculation of Magnetic Quantities from Currents
19: Alternating-Current (AC) Circuits
20: Maxwell's Equations
21: Electrical Transmission Lines
22: Generation and Detection of Electromagnetic Waves
23: Signal Modulation
Back Matter
4.5: Applications of Electric Potential and Conductors in Electrostatic Equilibrium
Last updated Jul 15, 2025
Save as PDF
4.4: Conductors in Electrostatic Equilibrium
4.6: Potential and Field Relationships (Summary)
picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report View on CommonsDonate
Page ID 102688
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Applications of Electric Potential and Field
1. Grounding and Bonding
2. Electrophysiology
3. Earth’s Electric Field
Applications of Conductors
Lightning Rods
Lightning Mapping
Screening
References
Technician Exam Questions
Learning Objectives
By the end of this section, you will be able to:
Define grounding and bonding.
Identify some effects of electricity on the human body and their implications for electrical safety.
Describe the electric field surrounding Earth.
Explain how a lightning rod works.
Define screening, and explain how a Faraday cage works.
Applications of Electric Potential and Field
There are numerous practical applications of electric potential and electric field in the practice of Amateur Radio and other wireless technology. We highlight a few important examples in this section.
Grounding and Bonding
With respect to electricity, a ground is considered a reference point in an electrical circuit from which voltages are measured . The term "ground" is used because, it is common to make physical connections to the Earth, often by a metal stake placed in ground (Fig. 4.5.1), via a conductor. From Conductors in Electrostatic Equilibrium, we know that a conductor will have a constant potential. If multiple connections are made to ground via conductors, then they will all share the same potential. It is conventional to set the potential to zero volts for convenience. The process of connecting a circuit to ground is called grounding. Bondingis the process of connecting together multiple electrical devices with a conductor, usually a thick wire, to maintain a constant potential among the devices, and is often done before connecting the devices to ground.
Figure 4.5.1: Example of grounding. The electrical circuits from a building are connected by a conductor (green and yellow wire) into a metal stake into the ground.
Proper grounding and bonding are important for electrical safety. For example, grounding the metal case of an electrical appliance ensures that it is at zero volts relative to Earth. This practice can minimize the chance of electric shock hazards when a fault occurs in the wiring. Grounding can also help to minimize potential differences between devices during sudden changes in voltage in the electrical distribution system. During lighting strikes, it can also help to dissipate charge, routing away from valuable equipment and into the Earth. While these processes sound simple enough, there can be subtleties when implementing them in practice. You should follow the grounding requirements in your local electrical codes, particularly for Amateur Radio towers and antennas. (See Ref for a comprehensive discussion of best practices for grounding and bonding for Amateur Radio.)
Electrophysiology
Electrophysiology is a branch of physiology that studies the electrical properties of cells and tissue. The human body contains several systems that use electrical signals. For example, the heart relies on electrical signals to maintain its rhythm . The movement of electrical signals causes the chambers of the heart to contract and relax. The equipotential lines around the heart and surrounding areas are useful ways of monitoring the structure and functions of the heart. An electrocardiogram (ECG) measures the small electric signals being generated during the activity of the heart. When a person has a heart attack, the movement of these electrical signals may be disturbed. When a person has irregular heartbeat, an artificial pacemaker and a defibrillator can be used to initiate the rhythm of electrical signals. Electrical signals also play an important role in the nervous system. Like cardiac tissue, nerves are also capable of transmitting electrochemical nerve impulses called action potentials (voltages) along the body of the nerve cells (Fig. 4.5.2).
Figure 4.5.2: Electrical impulses in a nerve. As a nerve impulse travels down the axon, there is a change in polarity across the membrane. The Na+ and K+ gated ion channels open and close in response to a signal from another neuron. At the beginning of action potential, the Na+ gates open and Na+ moves into the axon. This is depolarization. Repolarization occurs when the K+ gates open and K+ moves outside the axon. This creates a change in polarity between the outside and inside of the cell . The impulse continuously travels down the axon in one direction only, through the axon terminal and to other neurons.
Human electrophysiology has implications for electrical safety. As we will see, electrical charges will move from regions of high electrical potential to regions of low potential (ground). When touching electrical circuits, you want to ensure you do not become the path to ground! One good practice is only to touch circuits at one point. (Keep your other hand in a pocket!) In particular, avoid allowing charge to move in a path across the heart, for example, from one hand to the other. Electric shocks can also cause nervous system responses, including involuntary muscle contractions. In the worst cases, these contractions are so strong that they may prevent you from releasing from the charged object, making it difficult to remove yourself from the hazard! As will be discussed later, electrical shocks may also cause heating or burning of tissue. Other common safety practices include removing metal jewelry, working in dry conditions, and being aware of overhead wires.
Earth’s Electric Field
A near uniform electric field of approximately 100 to 150 N/C, directed downward, surrounds Earth, with the magnitude increasing slightly as we get closer to the surface. What causes the electric field? At around 100 km above the surface of Earth, we have a layer of charged particles, called the ionosphere. The ionosphere is predominantly caused by ultraviolet radiation from the sun. The ionosphere is responsible for a range of interesting phenomena, including the electric field surrounding Earth and the bending of radio waves. In fair weather, the ionosphere is positive and the Earth largely negative, maintaining the electric field (Figure 4.5.3a) .
In storm conditions, clouds form, and localized electric fields can be larger and even reversed in direction (Figure 4.5.3b). The exact charge distributions depend on the local conditions, and variations of Figure 4.5.2b are possible. If the electric field is sufficiently large, the insulating properties of the surrounding material experience electrical breakdown, and it becomes conducting. For air, breakdown starts to occur at around 3×10 6 N/C. The field ionizes the air, and electrons can move through the air, resulting in effects like lightning [8, 9] and corona discharge.
Figure 4.5.3: Earth’s electric field. (a) Fair weather field. Earth and the ionosphere (a layer of charged particles) are both conductors. They produce a uniform electric field of about 100 to 150 N/C. (credit: D. H. Parks) (b) Storm fields. In the presence of storm clouds, the local electric fields can be larger. At very high fields, the insulating properties of the air break down, and lightning can occur. (credit: Jan-Joost Verhoef)
Applications of Conductors
Lightning Rods
As discussed in Conductors in Electrostatic Equilibrium, a charged conductor will have a higher electric field and surface charge density where it has higher curvature. A practical application of this phenomenon is the lightning rod, which is simply a grounded metal rod with a sharp end pointing upward (Fig. 4.5.4, 4.5.5a). As positive charge accumulates in the ground due to a negatively charged cloud overhead, the electric field around the sharp point gets very large. When the field reaches a value of approximately 3.0×10 6N/C (the dielectric strength of the air), the free ions in the air are accelerated to such high energies that their collisions with air molecules actually ionize the molecules. The resulting free electrons in the air then flow through the rod to Earth, thereby neutralizing some of the positive charge. This keeps the electric field between the cloud and the ground from getting large enough to produce a lightning bolt in the region around the rod.
Lightning protection is an important part of the safety considerations for Amateur Radio operators, who often have tall antennas or towers to improve their ability to transmit and receive radio signals. As the height of an antenna or tower increases, so does its probability of being struck. The risk level also depends on the prevalence of thunderstorms in the vicinity. This property of conductors must also be considered when performing the grounding of a lightning rod. The connections from the rod should be as short and direct as possible and should not connect sharp angles that could result in the electric charges arcing through the air prior to reaching ground.
Figure 4.5.4: A very pointed conductor has a large charge concentration at the point. The electric field is very strong at the point and can exert a force large enough to transfer charge on or off the conductor. Lightning rods are used to prevent the buildup of large excess charges on structures and, thus, are pointed.
Lightning Mapping
It is possible to create maps of lightning activity in near real-time. Each time a lightning strike occurs, very low frequency (VLF) radio waves (3 to 30 kHz) are emitted. By detecting these signals at multiple receiving stations at known locations, it is possible to triangulate the location of the lightning strike . See the maps at Blitzortung.org for examples in your region.
Of course, we sometimes wish to prevent the transfer of charge rather than to facilitate it. In that case, the conductor should be very smooth and have as large a radius of curvature as possible (Figure 4.5.5b). For example, smooth surfaces are used on high-voltage transmission lines to avoid leakage of charge into the air.
Figure 4.5.5: (a) A lightning rod is pointed to facilitate the transfer of charge. (credit: Romaine, Wikimedia Commons) (b) This Van de Graaff generator has a smooth surface with a large radius of curvature to prevent the transfer of charge and allow a large voltage to be generated. The mutual repulsion of like charges is evident in the person’s hair while touching the metal sphere. (credit: Jon ‘ShakataGaNai’ Davis/Wikimedia Commons).
Screening
We also learned in Conductors in ElectrostaticEquilibrium that no electric field exists inside a cavity in a conductor. This effect can be used practically to perform screening or shielding of objects from electric fields. For example, a Faraday cage is a metal shield that encloses a volume (Fig. 4.5.6) . The Faraday cage is commonly used to prohibit stray electrical fields in the environment from interfering with sensitive measurements, such as the electrical signals inside a nerve cell. Perhaps the most common example of a (partial) Faraday cage is a typical microwave oven, which will have 5 complete metal sides and one metal screen on the oven's door to contain the microwaves in the oven and prevent exposure to nearby users.
Figure 4.5.6: Animated illustration of the screening effect of the Faraday cage. In the presence of an external electric field, the electrical charges will reside on the outside surface of this shield and arrange themselves such that is no electrical field inside.
By similar reasoning, the body of your car can serve as a partial Faraday cage. As a result, if you are driving a car during an electrical storm, it is best to stay inside the car. Even if the car is in the vicinity of a lightning strike, the effect of the strike will be felt on the outside of the car. You will be unaffected, provided you remain totally inside and not in contact with the body of vehicle. This protective effect is also true if an active (“hot”) electrical wire breaks in a storm or an accident and falls on the car.
Amateur radio operators will use screening to try to protect their radio equipment from interference from surrounding devices. Some cables, like the coaxial cables used in radio and cable television, also use screening to minimize the effects of surrounding electrical noise. The American Radio Relay League headquarters contains a Faraday cage used to test radio equipment without outside interference. However, the screening effects can also be detrimental to radio reception. For example, it can sometimes be difficult to get good reception inside a vehicle using a handheld VHF transceiver with a directly-mounted antenna (see photo in Amateur Radio Equipment Basics). Instead, the antenna should be placed on the exterior of the vehicle and then connected to the radio via a transmission cable.
References
Wikipedia contributors. Ground (electricity) [Internet]. Wikipedia, The Free Encyclopedia.
Silver, HW (N0AX). Grounding and bonding for the radio amateur: Good practices for electrical safety, lightning protection, and RF management, 2nd edition. Newington, CT: American Radio Relay League; 2021.
Wikimedia Commons contributors. File:HomeEarthRodAustralia1.jpg [Internet]. Wikimedia Commons. (Ali K., CC-BY-SA 3.0)
Wikipedia contributors. Cardiac conduction system [Internet]. Wikipedia, The Free Encyclopedia.
Wikipedia contributors. Nerve [Internet]. Wikipedia, The Free Encyclopedia.
Wikimedia Commons contributors. File:Action Potential.gif [Internet]. Wikimedia Commons. (Laurentaylorj, CC-BY-SA 3.0)
Wikipedia contributors. Atmospheric electricity [Internet]. Wikipedia, The Free Encyclopedia.
Wikipedia contributors. Lightning [Internet]. Wikipedia, The Free Encyclopedia.
JetStream - An Online School for Weather.Lightning: introduction to lightning. [Internet] National Oceanic and Atmospheric Administration.
Wikipedia contributors. Corona discharge [Internet]. Wikipedia, The Free Encyclopedia.
Lightningmaps.org. Lightning detection [Internet]. Lightningmaps.org.
Blitzortung.org. Overview map [Internet]. Blitzortung.org.
Wikipedia contributors. Faraday cage [Internet]. Wikipedia, The Free Encyclopedia.
Wikimedia Commons contributors. File:Faraday cage.gif [Internet]. Wikimedia Commons. (Stanisław Skowron, public domain)
Technician Exam Questions
Relevant exam questions include: T9A07, T0A02, T0A09, T0B01, T0B10, T0B11
4.5: Applications of Electric Potential and Conductors in Electrostatic Equilibrium is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ronald Kumon & OpenStax.
7.6: Equipotential Surfaces and Conductors by OpenStax is licensed CC BY 4.0. Original source:
18.7: Conductors and Electric Fields in Static Equilibrium by OpenStax is licensed CC BY 4.0. Original source:
Toggle block-level attributions
Back to top
4.4: Conductors in Electrostatic Equilibrium
4.6: Potential and Field Relationships (Summary)
Was this article helpful?
Yes
No
Recommended articles
3.8: Conductors and Electric Fields in Static Equilibrium
18.7: Conductors and Electric Fields in Static EquilibriumConductors contain free charges that move easily. When excess charge is placed on a conductor or the conductor is put into a static electric field, ch...
4.4: Conductors in Electrostatic EquilibriumWhen charges are stationary in a conductor, it is in a state of electrostatic equilbrium. This section describes the properties of conductors in elec...
2.3: Conduction and ChargingIn the preceding section, we said that scientists were able to create electric charge only on nonmetallic materials and never on metals. To understand...
4.1: Electric Potential from Electric FieldThe electric potential is different from the electric field, but the two quantities are related. In this section, we learn how to calculate the elect...
Article typeSection or PageLicenseCC BYLicense Version4.0OER program or PublisherOpenStaxShow TOCyes
Tags
author@OpenStax
author@Ronald Kumon
conductor
electrostatic equilibrium
Faraday cage
free charge
ionosphere
polarized
source-phys-102404
source-phys-2566
source@
© Copyright 2025 Physics LibreTexts
Powered by CXone Expert ®
?
The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org.
Support Center
How can we help?
Contact Support Search the Insight Knowledge Base Check System Status×
contents readability resources tools
☰
4.4: Conductors in Electrostatic Equilibrium
4.6: Potential and Field Relationships (Summary)
Complete your gift to make an impact |
189242 | https://www.shaalaa.com/question-bank-solutions/the-line-y-mx-1-tangent-curve-y2-4x-if-value-m_45987 | The Line Y = Mx + 1 is a Tangent to the Curve Y2 = 4x, If the Value Of M Is - Mathematics | Shaalaa.com
English
0
CBSECommerce (English Medium) Class 12
Question Papers
Question Papers 2491
Textbook Solutions 20222
MCQ Online Mock Tests 42
Important Solutions 18952
Concept Notes & Videos 235
Time Tables 24
Syllabus
The Line Y = Mx + 1 is a Tangent to the Curve Y2 = 4x, If the Value Of M Is - Mathematics
Advertisements
Advertisements
Advertisement
Powered by
00:01
00:00
Question
The line y=mx+ 1 is a tangent to the curve y 2 = 4x, if the value of m is ____ .
Options
1
2
3
1 2
MCQ
Advertisements
Solution Show Solution
1
Let (x 1, y 1) be the required point.
The slope of the given line is m.
We have
𝑦 2=4𝑥
⇒2𝑦𝑑𝑦 𝑑𝑥=4
⇒𝑑𝑦 𝑑𝑥=4 2𝑦=2 𝑦
Slope of the tangent=(𝑑𝑦 𝑑𝑥)(𝑥 1,𝑦 1)=2 𝑦 1
Given:
Slope of the tangent=𝑚
Now,
2 𝑦 1=𝑚...(1)
Because the given line is a tangent to the given curve at point (x 1, y 1), this point lies on both the line and the curve.
∴𝑦 1=𝑚𝑥 1+1 and 𝑦 1 2=4𝑥 1
⇒𝑥 1=𝑦 1−1 𝑚 and 𝑥 1=𝑦 1 2 4
𝑆𝑜,
𝑦 1−1 𝑚=𝑦 1 2 4
⇒𝑦 1−1(2 𝑦 1)=𝑦 1 2 4[From(1)]
⇒𝑦 1(𝑦 1−1)2=𝑦 1 2 4
⇒2𝑦 1 2−2𝑦 1=𝑦 1 2
⇒𝑦 1 2−2𝑦 1=0
⇒𝑦 1 2−2𝑦 1=0
⇒𝑦 1(𝑦 1−2)=0
⇒𝑦 1=0,2
So, For 𝑦 1=0,𝑚=2 0=∞
For 𝑦 1=2,𝑚=2 2=1
shaalaa.com
Tangents and Normals
Report Error Is there an error in this question or solution?
Q 27Q 26Q 28
Chapter 16: Tangents and Normals - Exercise 16.5 [Page 43]
APPEARS IN
RD Sharma Mathematics [English] Class 12
Chapter 16 Tangents and Normals
Exercise 16.5 | Q 27 | Page 43
Video Tutorials VIEW ALL
view
Video Tutorials For All Subjects
Tangents and Normals video tutorial 01:19:39
Tangents and Normals video tutorial 00:51:08
Tangents and Normals video tutorial 01:01:06
RELATED QUESTIONS
Show that the equation of normal at any point t on the curve x = 3 cos t – cos 3 t and y = 3 sin t – sin 3 t is 4 (y cos 3 t – sin 3 t) = 3 sin 4t
Find the slope of the normal to the curve x=a cos 3 θ,y=a sin 3 θ at 𝜃=𝜋 4
Find the equation of the normals to the curve y=x 3+ 2 x+ 6 which are parallel to the line x+ 14 y+ 4 = 0.
Find the equations of the tangent and normal to the parabola y 2= 4 ax at the point (at 2, 2 at).
The slope of the tangent to the curve x = t 2 + 3t – 8, y = 2t 2 – 2t – 5 at the point (2,– 1) is (A) 2 2 7 (B) 6 7 (C) 7 6 (D) −6 7
Find the points on the curve y = 4𝑥 3−3𝑥+5 at which the equation of the tangent is parallel to the x-axis.
Find the slope of the tangent and the normal to the following curve at the indicted point 𝑦=√𝑥 at 𝑥=9 ?
Find the slope of the tangent and the normal to the following curve at the indicted point x 2+ 3 y+y 2= 5 at (1, 1) ?
Find the points on the curve y 2= 2 x 3 at which the slope of the tangent is 3 ?
Find the points on the curve 2a 2 y=x 3− 3ax 2 where the tangent is parallel to x-axis ?
Find the points on the curve 𝑥 2 4+𝑦 2 2 5=1 at which the tangent is parallel to the x-axis ?
Find the equation of the tangent and the normal to the following curve at the indicated point y=x 4−bx 3+ 13 x 2− 10 x+ 5 at (0, 5) ?
Find the equation of the tangent and the normal to the following curve at the indicated point 𝑥 2 𝑎 2−𝑦 2 𝑏 2=1 at(𝑎s e c𝜃,𝑏t a n𝜃) ?
Find the equation of the tangent and the normal to the following curve at the indicated point 𝑥 2 𝑎 2−𝑦 2 𝑏 2=1 at(𝑥 0,𝑦 0) ?
Find the equation of the tangent and the normal to the following curve at the indicated point 𝑥 2 𝑎 2−𝑦 2 𝑏 2=1 at(√2𝑎,𝑏) ?
Find the equation of the tangent line to the curve y=x 2+ 4 x− 16 which is parallel to the line 3 x−y+ 1 = 0 ?
Find the equation of the tangent line to the curve y=x 2− 2 x+ 7 which perpendicular to the line 5 y− 15 x= 13. ?
Find the equations of all lines having slope 2 and that are tangent to the curve 𝑦=1 𝑥−3,𝑥≠3 ?
Show that the curves 2x=y 2 and 2xy=k cut at right angles, if k 2= 8 ?
Find the condition for the following set of curve to intersect orthogonally 𝑥 2 𝑎 2−𝑦 2 𝑏 2=1 and 𝑥𝑦=𝑐 2 ?
If the tangent line at a point (x,y) on the curve y=f(x) is parallel to x-axis, then write the value of 𝑑𝑦 𝑑𝑥 ?
Write the value of 𝑑𝑦 𝑑𝑥 , if the normal to the curve y=f(x) at (x,y) is parallel to y-axis ?
Write the equation of the normal to the curve y=x+ sin x cos x at 𝑥=𝜋 2 ?
Write the equation on the tangent to the curve y=x 2−x+ 2 at the point where it crosses the y-axis ?
The point on the curve y 2 = x where tangent makes 45° angle with x-axis is __ .
If the line y=x touches the curve y=x 2 + bx + c at a point (1, 1) then _______ .
Find the angle of intersection of the curves y 2 = x and x 2 = y.
Find the condition for the curves 𝑥 2 a 2−𝑦 2 b 2 = 1; xy = c 2 to interest orthogonally.
Find the angle of intersection of the curves y 2 = 4ax and x 2 = 4by.
Show that the line 𝑥 a+𝑦 b = 1, touches the curve y = b · e– x/a at the point where the curve intersects the axis of y
The curve y = 𝑥 1 5 has at (0, 0) ______.
If the curve ay + x 2 = 7 and x 3 = y, cut orthogonally at (1, 1), then the value of a is ______.
The slope of tangent to the curve x = t 2 + 3t – 8, y = 2t 2 – 2t – 5 at the point (2, –1) is ______.
The distance between the point (1, 1) and the tangent to the curve y = e 2x + x 2 drawn at the point x = 0
Find a point on the curve y = (x – 2)2. at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
The line y = x + 1 is a tangent to the curve y 2 = 4x at the point
The slope of the tangentto the curve 𝑥=𝑡 2+3𝑡−8,𝑦=2𝑡 2−2𝑡−5 at the point (2,−1) is
If m be the slope of a tangent to the curve e 2y = 1 + 4x 2, then ______.
Question Bank with Solutions
Maharashtra State Board Question Bank with Solutions (Official)
Textbook Solutions
Balbharati Solutions (Maharashtra)
Samacheer Kalvi Solutions (Tamil Nadu)
NCERT Solutions
RD Sharma Solutions
RD Sharma Class 10 Solutions
RD Sharma Class 9 Solutions
Lakhmir Singh Solutions
TS Grewal Solutions
ICSE Class 10 Solutions
Selina ICSE Concise Solutions
Frank ICSE Solutions
ML Aggarwal Solutions
NCERT Solutions
NCERT Solutions for Class 12 Maths
NCERT Solutions for Class 12 Physics
NCERT Solutions for Class 12 Chemistry
NCERT Solutions for Class 12 Biology
NCERT Solutions for Class 11 Maths
NCERT Solutions for Class 11 Physics
NCERT Solutions for Class 11 Chemistry
NCERT Solutions for Class 11 Biology
NCERT Solutions for Class 10 Maths
NCERT Solutions for Class 10 Science
NCERT Solutions for Class 9 Maths
NCERT Solutions for Class 9 Science
Board / University Study Material
CBSE Study Material
Maharashtra State Board Study Material
Tamil Nadu State Board Study Material
CISCE ICSE / ISC Study Material
Mumbai University Engineering Study Material
Question Paper Solutions
CBSE Previous Year Question Papers With Solutions for Class 12 Arts
CBSE Previous Year Question Papers With Solutions for Class 12 Commerce
CBSE Previous Year Question Papers With Solutions for Class 12 Science
CBSE Previous Year Question Papers With Solutions for Class 10
Maharashtra State Board Previous Year Question Papers With Solutions for Class 12 Arts
Maharashtra State Board Previous Year Question Papers With Solutions for Class 12 Commerce
Maharashtra State Board Previous Year Question Papers With Solutions for Class 12 Science
Maharashtra State Board Previous Year Question Papers With Solutions for Class 10
CISCE ICSE / ISC Board Previous Year Question Papers With Solutions for Class 12 Arts
CISCE ICSE / ISC Board Previous Year Question Papers With Solutions for Class 12 Commerce
CISCE ICSE / ISC Board Previous Year Question Papers With Solutions for Class 12 Science
CISCE ICSE / ISC Board Previous Year Question Papers With Solutions for Class 10
Other Resources
Entrance Exams
Video Tutorials
Question Papers
Question Bank Solutions
Question Search (beta)
Privacy Policy
Terms and Conditions
Contact Us
About Us
Shaalaa App
Ad-free Subscriptions
© 2025 Shaalaa.com |Contact Us|Privacy Policy
Share
0
0
0
0
0
Notifications
Select a course
CANCEL
Englishहिंदीमराठी
user
Login
Create free account
email:
password:
Log inForgot password?
Course
Commerce (English Medium) Class 12 CBSE
PUC Science 2nd PUC Class 12 Karnataka Board PUC
Arts (English Medium) Class 12 CBSE
Science (English Medium) Class 12 CBSE
change
Home
Class 1 - 4
Class 5 - 8
Class 9 - 10
Class 11 - 12
Entrance Exams
Search by Text or Image
Textbook Solutions
Study Material
Remove All Ads
Change mode
Log out
Use app×
x |
189243 | https://academics.hamilton.edu/economics/cgeorges/macro-topics-files/growth.pdf | Economics 504 Chris Georges Handout on Growth Rates Discrete Time Analysis: All macroeconomic data are recorded for discrete periods of time (e.g., quarters, years). Consequently, it is often useful to model economic dynamics in discrete periods of time.
Let’s consider annual GDP per capita (gross domestic output for a year divided by the number of people in the population in that year) of a country. Call this Y .
Define the annual growth rate g of Y in any year t as the annual percentage change in Y from the previous year.
gt = b Yt = Yt −Yt−1 Yt−1 Defined in this way, growth rates are compounding over time. Starting at time 0, we have Y1 −Y0 Y0 = g1 and solving this for Y1 Y1 = (1 + g1) · Y0 Similarly, Y2 = (1 + g2) · Y1 = (1 + g2) · (1 + g1) · Y0 And for any future time t Yt = (1 + g1) · (1 + g2) · · · (1 + gt) · Y0 If gt happened to be constant over the t years following year 0, then Yt = (1 + g)t · Y0 and the path of Y over time would look like that in figure A below.
Starting from the same initial level Y0, a larger growth rate g rotates the curve upward (figure B above).
The gap between two economies which start at the same level but grow at different rates grows over time.
In other words, due to compounding, small differences in permanent growth rates have large effects in the future.
1 Consider also two economies (1 and 2) which start at different initial levels Y0 but grow at the same rate (figure C above). The absolute gap (Y1t −Y2t) between the levels of GDP per capita of the two countries grows over time, but the ratio Y1t/Y2t remains constant.
The following approximations for percentage growth rates are useful. For small changes in any two variables x and y: d x/y ≈b x −b y d x · y ≈b x + b y For the really interested reader, a proof of first proposition appears in the appendix at the end of this handout.
Application: Post WWII growth in the U.S. and Japan Here are some measures of per capita real GDP for the US and Japan in 1950 and 1989: 1950 1989 US 8,611 18,317 Japan 1,563 15,101 What are the annual average growth rates over this period for the US and Japan?
Here is one way to answer this question: Y1989 = (1 + g)39 · Y1950 Consequently, g can be calculated (1 + g) = Y1989 Y1950 1 39 Yielding g = 0.0195 for the US and g = 0.0597 for Japan. The US grew at an average growth rate of about 2% annually over the period while Japan grew at about 6% annually.1 Log Growth Rates: The following method gives a close approximation to the answer above, and will be useful in other contexts. A useful approximation is that for any small number x: ln(1 + x) ≈x Now we can take the natural log of both sides of Y1989 = (1 + g)39 · Y1950 to get ln(Y1989) = 39 · ln(1 + g) + ln(Y1950) which rearranges to ln(1 + g) = ln(Y1989) −ln(Y1950) 39 and using our approximation g ≈ln(Y1989) −ln(Y1950) 39 1 It is worth noting that these numbers are not the same as the averages of the actual annual growth rates taken year by year — which just goes to show that there is rarely a single correct method for measuring things.
2 In other words, log growth rates are good approximations for percentage growth rates. Calculating log growth rates for the data above, we get g ≈0.0194 for the U.S. and g ≈0.0582 for Japan. The approximation is close for both, but closer for the U.S. than Japan as the log approximation will be closer, the closer g is to zero. Log growth rates are often used in economic modeling and empirical work. For example, for year to year growth, researchers will often just use the change in the log: ∆ln(Yt).
Log Plots: Recall that, with a constant growth rate g and starting from time 0, output in time t is Yt = (1 + g)t · Y0 Taking logs of both sides, ln Yt = lnY0 + ln(1 + g) · t we see that log output is linear in time. Thus, if the growth rate is constant, a plot of log output against time will yield a straight line. Consequently, plotting log output against time is a quick way to eyeball whether growth rates have changed over time.
Time to Doubling: How long will it take for standards of living to double? If we measure the standard of living by GDP per capita, for example, then this reduces to the question, in what year t will GDP per capita be twice that of year 0.
To answer this question, we want to solve Yt = 2Y0 for t.
(1 + g)t · Y0 = 2Y0 (1 + g)t = 2 Use logs to get t out of the exponent: t · ln(1 + g) = ln(2) t = ln(2) ln(1 + g) We can get a good approximation to this by calculating ln(2) ≈0.7 and using our approximation ln(1 + g) ≈g. Thus: t ≈.7/g.
3 Notice that this is smaller than 1/g due to compounding. Again, small differences in growth rates have increasingly large effects on future standards of living.
Then in the US for 1950–1989, with g ≈0.02, GDP per capita doubled roughly every 35 years over the period. In Japan with g ≈0.06, GDP per capita doubled roughly every 12 years over the period.
At these kinds of growth rates, successive generations are substantially better offthan their predecessors.
Notice that if per capita growth falls to 1%, years to doubling rises to about 70 years. A few percentage points in growth rates makes a big difference.
Continuous Time: For modeling purposes it is sometimes useful (and less clumsy) to work in continuous time. Suppose that we are interested in annual growth patterns, but also want to consider periods of time shorter than a year. In the extreme case we can think of there being a growth rate (measured as an annual rate) at each instant. I.e, the annual rate can be constantly changing, and the actual increase in output over the course of any year depends on all the growth rates during the year (i.e., on average growth during the year).
In this case, rather than defining the growth rate g as the percentage change in GDP from one year to the next, we define it as the instantaneous rate of growth of GDP.
gt = b Yt = ˙ Yt Yt = dYt dt Yt where ˙ Yt is shorthand for the derivative of output with respect to time, dYt dt .
As in the discrete time case, we need to add up the changes in output over time to calculate future levels of output. However, in continuous time we would do this by integrating over time, which in the case of a constant growth rate g would yield Yt = Y0 · eg·t This looks very similar to our formula under discrete time (which was Yt = Y0 · (1 + g)t), and is close numerically as well.2 The analytical convenience of continuous time analysis stems from the fact that the approximations that I discussed above under discrete time are exact equalities under continuous time. The following equalities hold exactly in continuous time: d x/y = b x −b y d x · y = b x + b y ln xt −ln x0 t = g b xt = d dt lnxt tdoubling = ln 2 g The third and sfourth equation say that log diferrences are exactly equal to the growth rate.
Again, the curious reader can see the appendix for proofs of some of these propositions.
2 In continuous time, output is slightly greater in the future than it is in discrete time (i.e., egt > (1 + g)t), because growth is compounding continuously rather than annually.
4 The Solow Residual: The Solow residual is an empirical measure of total factor productivity (TFP) growth and is often used as a rough measure of the contribution of technological progress to economic growth.
Consider the Cobb Douglas production function with constant returns to scale: Y = A · Kα · L1−α The parameter A is total factor productivity TF P . The parameter α is the elasticity of output with respect to capital and also reflects the relative productivities of capital and labor. Empirical estimates of α often put it at around 1/3 (more on that at a later date).
If we take growth rates of each side of this equation and rearrange, we have:3 b Y = b A + α · b K + (1 −α) · b L b A = b Y −α · b K −(1 −α) · b L Consider the second equation abovve. If we use an independent estimate of α (like 1/3), then we can take the right hand side of the second equation above (the Solow residual) as an observable measure of the unobserved left hand side (TFP growth).
Appendix A: Proof of proposition that d x/y ≈b x −b y in discrete time.
Exact method: d x/y = x1/y1 −x0/y0 x0/y0 = x1 x0 /y1 y0 −1 = 1 + b x 1 + b y −1 = b x −b y 1 + b y ≈b x −b y The approximation makes use of the fact that, for small b y, 1 + b y is close to 1.
Alternative method: use the fact introduced in the handout that log differences are close approximations to percentage growth rates.
d x/y ≈∆lnx/y = lnx1/y1 −lnx0/y0 = lnx1 −ln x0 −lny1 + lny0 ≈b x −b y B: Proof of proposition that d x/y = b x −b y in continuous time.
The proof makes use of the fact that in continuous time, the time derivative of the log of a variable is the growth rate of that variable. To see this, recall that the derivative of lnx is 1/x. Thus, d dt ln xt = 1 xt · d dtxt = ˙ xt/xt = b xt.
3 Note that, from what we have learned above, the following equations are exact equalities if we are working in continuous time and close approximations if we are working with discrete time data.
5 The proof is then: d x/y = d dt ln(x/y) = d dt(ln x −ln y) = d dt ln x −d dt lny = b x −b y 6 |
189244 | https://www.quora.com/Are-there-infinitely-many-pairs-of-positive-integers-a-b-such-that-a-b-2-1-and-b-a-2-1 | Something went wrong. Wait a moment and try again.
Infinite Set
Integer Sequences
Proofs (mathematics)
Prime Number Theory
Finite Sets & Infinite Se...
Studying Number Theory
Positive Integers
Mathematical Proof
5
Are there infinitely many pairs of positive integers
(
a
,
b
)
such that
a
|
b
2
+
1
and
b
|
a
2
+
1
?
Kevin Gomez
Number theorist in training
·
Upvoted by
Siva Tej
, PhD Mathematics, University of Southern Denmark (2025) and
Horst H. von Brand
, PhD Computer Science & Mathematics, Louisiana State University (1987) · Author has 520 answers and 2M answer views
·
7y
Let’s start off with a trivial pair:
(a,b)=(1,2)
This pair quite clearly satisfies our relation, and now we’re going to use it to demonstrate that
(a,b)=(F2n−1,F2n+1)
where Fn is a Fibonacci number
is always a valid solution, and thus the number of positive integer solutions is infinite.
Firstly, how might you (or anyone else) come across the Fibonacci family of solutions? Well, it turns there is a very simple way to generate them, even if you aren’t sure exactly how to prove the method.
We can consider the relation (a,b)→(b2+1,a2+1). Start with (1,2):
(1,2)→(5,2
Fibonacci number - Wikipedia
Let’s start off with a trivial pair:
(a,b)=(1,2)
This pair quite clearly satisfies our relation, and now we’re going to use it to demonstrate that
(a,b)=(F2n−1,F2n+1)
where Fn is a Fibonacci number
is always a valid solution, and thus the number of positive integer solutions is infinite.
Firstly, how might you (or anyone else) come across the Fibonacci family of solutions? Well, it turns there is a very simple way to generate them, even if you aren’t sure exactly how to prove the method.
We can consider the relation (a,b)→(b2+1,a2+1). Start with (1,2):
(1,2)→(5,2)
Take (a2+1,b2+1a) for our next pair:
(2,5)→(26,5)
And again with (b2+1)/a=26/2=13 and a2+1=5
(5,13)→(170,26)
Performing this process indefinitely will produce an infinite family of solutions. Neat, right? But why?
We must first establish the connect between this family and the Fibonacci numbers. The sequence
1,2,5,13,34,…
represents the odd-index Fibonacci numbers (with F0=0)
0,1,1,2,3,5,8,…
We must prove that
F22n−1+1F2n−3=F2n+1
Here we go (this is just one possible method):
F22n−1+1F2n−3=F2n+1F22n−1+1=F2n+1F2n−3F22n−1−F2n+1F2n−3=−1
We now apply Catalan’s identity
:
F2m−Fm−rFm+r=(−1)m−rF2r
with m=2n−1 and r=2:
F22n−1−F2n−3F2n+1=(−1)2n−3F22=−1
Now we must prove that the pair (F2n−1,F2n+1) is always a solution. Specifically:
F2n−1|F22n+1+1F2n+1|F22n−1+1
(2) is easily shown by recalling that we have already proven that
F22n−1+1F2n−3=F2n+1⟹F22n−1+1F2n+1=F2n−3
(1) can be proven by a simple change of variable n→n+1:
F22(n+1)−1+1F2(n+1)−3=F2(n+1)+1→F22n+1+1F2n−1=F2n+3
And there we have it! Not only do we have infinite solutions to a|b2+1,b|a2+1, we have an easy way to generate them!
During my pursuit of this problem (i.e. toying with my calculator over the last hour), I discovered a similar result for the even-indexed Fibonacci numbers. See if you can prove that (F2n,F2n+2) is always a solution to a|b2−1 and b|a2−1.
Footnotes
Fibonacci number - Wikipedia
Cassini and Catalan identities - Wikipedia
Related questions
Are there infinitely many integers
n
such that
n
2
+
1
divides
n
!
?
How do you prove that there are infinitely many integers
x
,
y
such that
x
+
y
=
100
and
?
What can I do with a BS in Mathematics?
Does being good at mathematics make you intelligent?
How do I find all pairs of positive integers
such that
?
Hyunbok Wi
7y
This is a way to find ALL solutions to a|b^2+1, b|a^2+1..
Take the product: ab|(a^2+1)(b^2+1), and so
ab|a^2+b^2+1.
Let k=(a^2+b^2+1)/ab , and for some fixed k,
Let (a,b) be the pair such that a+b is smallest,
And WLOG a>=b.
a satisfies X^2-(kb)X+(b^2+1)=0, but this equation has another root, let’s call it a_1.
Then a_1=kb-a=(b^2+1)/a, and so a_1 is a natural number. a_1 must be greater than a(since (a_1,b)is a new solution, and we assumed the smallest solution) , and so b^2+1>=a^2.
Here, I assumed a>=b, so the only way that b^2+1>=a^2 is a=b. That gives a=1, and k=3.
So, for any (x,y) satisfying the g
This is a way to find ALL solutions to a|b^2+1, b|a^2+1..
Take the product: ab|(a^2+1)(b^2+1), and so
ab|a^2+b^2+1.
Let k=(a^2+b^2+1)/ab , and for some fixed k,
Let (a,b) be the pair such that a+b is smallest,
And WLOG a>=b.
a satisfies X^2-(kb)X+(b^2+1)=0, but this equation has another root, let’s call it a_1.
Then a_1=kb-a=(b^2+1)/a, and so a_1 is a natural number. a_1 must be greater than a(since (a_1,b)is a new solution, and we assumed the smallest solution) , and so b^2+1>=a^2.
Here, I assumed a>=b, so the only way that b^2+1>=a^2 is a=b. That gives a=1, and k=3.
So, for any (x,y) satisfying the given relation, x^2–3xy+y^2=1. And for any (x,y) that satisfies that, it also satisfies the given relation.
And now, it’s possible to solve.
4x^2–12xy+4y^2=4
(2x-3y)^2-5y^2=4.
Let 2x-3y=u, y=v. We have to find all solutions which satisfy u^2–5v^2=4. This is a Pell equation, and you can solve it pretty easily.(Well, you can just look up Pell’s equation, if you dont know what I’m doing.)
The smallest solution to a^2–5b^2=1 is (9,4),
And the solutions of a^2–5b^2=4 satisfying a<=18 is (3,1), (7,3),(18,8).
So all solutions of u^2–5v^2=4 Look like:
u+v(sqrt 5)=(a+bsqrt5)•(9+4sqrt5)^n
Where (a,b) is the three solutions .
y=v, and x=(u+3v)/2, so you can find all solutions to our equations now.
Here, (9+4sqrt5)=[(1+sqrt5)/2]^6, and from that you can prove, with some algebraic manipulations, that indeed all solutions are of the fibonacci form mentioned in the other answer.
David Smith
BSc (Hons) in Mathematics & Computer Science, University of Bristol (Graduated 1986)
·
Author has 3.6K answers and 4M answer views
·
7y
Yes there are.
Take alternate members of the Fibonacci series (highlighted with bold):
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, …
Create pairs from adjacent highlighted numbers:
Each such pair has the desired property.
Mehdi Khayeche
IMO Silver Medalist
·
Upvoted by
Michael Jørgensen
, PhD in mathematics and
Will Cadegan-Schlieper
, PhD Mathematics, University of California, Los Angeles (2018) · Author has 75 answers and 1.4M answer views
·
6y
Related
Are there infinitely many integers
such that
and
? Can this be proved without assuming knowledge of maths beyond high school?
First of all, notice that and have to be relatively prime.
divides implies divides and similarly divides . That’s equivalent to: divides
We want to find infinitely many solutions, so let’s start by finding one:
Obviously works, and so does , with bit more trials we get that is also a solution. Now here’s the subtle observation, in all of our examples we have:
(in fact, you can show that’s always the case, but we don’t need that for this question)
Let’s rewrite as :
If w
First of all, notice that and have to be relatively prime.
divides implies divides and similarly divides . That’s equivalent to: divides
We want to find infinitely many solutions, so let’s start by finding one:
Obviously works, and so does , with bit more trials we get that is also a solution. Now here’s the subtle observation, in all of our examples we have:
(in fact, you can show that’s always the case, but we don’t need that for this question)
Let’s rewrite as :
If we fix , then this becomes a quadratic in . So I had this idea to prove that we have infinitely many solutions:
Start with a solution , the quadratic has a second root so we get a new solution and then we can iterate the process to get a solution etc.. This seems like it could work so let’s formalize it as follows:
Assume, for a sake of contradiction, that we have finitely many solutions to equation :
Let be the solution that maximizes among all the solutions (we can do this since there’s a finite number of them). WLOG assume .
We get that is a root of and the other root is (since the constant coefficient is the product of the roots) so is also a solution. But then by the maximality of we must have:
That’s impossible since we assumed .
Hence we have infinitely many solutions.
Sponsored by Grammarly
Is your writing working as hard as your ideas?
Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step.
Related questions
If
[math]7^x=49[/math]
, what is
[math]x[/math]
?
How should I study math?
How would you show that for any fixed integer
[math]k>1[/math]
there exist infinitely many positive integers
[math]n[/math]
such that
[math]n|1^n+2^n+3^n+…+k^n[/math]
?
How should I improve my maths?
What's a good math book for somebody who hates math?
Mehdi Khayeche
IMO Silver Medalist
·
Upvoted by
Alex Eustis
, Ph.D. Mathematics, University of California, San Diego (2013) and
Alexey Godin
, Ph.D. Mathematics & Economics, Moscow State University (1998) · Author has 75 answers and 1.4M answer views
·
6y
Related
Can you determine all of the pairs of positive integers (a,b) such that:
[math]\quad\frac{a^2}{2ab^2-b^3+1}\quad[/math]
is a positive integer?
ahh I remember this problem. If my memory serves me right, this was Problem [math]X[/math] of IMO [math]Y[/math] with [math]X \in {2,5}[/math] and [math]Y \in [1989,2006][/math]. I’m too lazy to check.
Let’s get to work, and as usual, we start by simple observations.
Usually Olympiad problems ask to find integers such that :
[math]\text{(Some Expression) divides (Another Expression)}\tag{}[/math]
However, this question also gives the information that the quotient of the two expressions is positive, so I’m guessing that additional fact has to be one of the keys of the solution. Let’s try to exploit that:
[math]a^2[/math] is always positive, so we must have :
[math]2ab^2-b^3+1 \g[/math]
ahh I remember this problem. If my memory serves me right, this was Problem [math]X[/math] of IMO [math]Y[/math] with [math]X \in {2,5}[/math] and [math]Y \in [1989,2006][/math]. I’m too lazy to check.
Let’s get to work, and as usual, we start by simple observations.
Usually Olympiad problems ask to find integers such that :
[math]\text{(Some Expression) divides (Another Expression)}\tag{}[/math]
However, this question also gives the information that the quotient of the two expressions is positive, so I’m guessing that additional fact has to be one of the keys of the solution. Let’s try to exploit that:
[math]a^2[/math] is always positive, so we must have :
[math]2ab^2-b^3+1 \geq 1 \Leftrightarrow 2ab^2 \geq b^3 \Leftrightarrow 2a \geq b \tag{1}[/math]
There’s another inequality that we can establish:
[math]2ab^2-b^3+1\leq a^2 \Leftrightarrow b^2(2a-b) +1 \leq a^2\tag{}[/math]
If [math]2a=b[/math] we can verify that we indeed have a solution, otherwise:
[math]2a-b \geq 1 \tag{}[/math]
Which implies:
[math]b^2 < b^2(2a-b) +1 \leq a^2 \Leftrightarrow b < a \tag{2}[/math]
Let’s hold on to that for now.
The annoying thing about this problem is the “[math]+1[/math]” in the denominator, which makes it infeasible to prove relations of the form:
[math]a \text{ divides (something) or }b\text{ divides (something)}\tag{} [/math]
It also discourages us from considering prime factors of [math]a[/math] and [math]b[/math]. After some fruitless efforts, I remembered my good ol’ friend Vieta. So let’s rewrite the equation as follows:
[math]\displaystyle \frac{a^2}{2ab^2-b^3+1}=n \in \N \tag{}[/math]
[math]\Leftrightarrow a^2=n\left(2ab^2-b^3+1\right)\tag{}[/math]
[math] \Leftrightarrow a^2+a(-2nb^2)+n(b^3–1)=0 \tag{} [/math]
If we fix [math]b[/math], then this is a quadratic in [math]a[/math]. Assuming it has a positive solution [math]a_{1}[/math], then the other solution [math]a_{2}[/math] would satisfy:
[math]\begin{cases} a_{1}+a_{2}= 2nb^2 \ a_{1}a_{2}=n(b^3–1) \end{cases} \tag{}[/math]
What can we say about [math]a_{2}[/math]?
Well, [math]a_{2}=2nb^2 -a_{1}[/math], so it has to be an integer. Additionally [math]a_{2}=\frac{n(b^3–1)}{a_{1}} \geq 0[/math].
We can also WLOG assume that [math]a_{2} \leq a_{1}[/math], which would give:
[math]\displaystyle a_{2}=\frac{n(b^3–1)}{a_{1}}\leq \frac{n(b^3-1)}{nb^2} = \frac{b^3-1}{b^2} < b\tag{3}[/math]
If [math]a_{2}=0[/math] we must have [math]n(b^3–1)=0[/math] i.e. [math]b=1[/math] and [math]a_{1}=2n[/math].
If [math]a_{2} > 0[/math], then math[/math] would contradict math[/math] unless we have [math]2a_{2}=b[/math] (in particular [math]b[/math] is even). From here we can solve for [math]a_{1}[/math]:
[math]\displaystyle \frac{1}{a_{1}} + \frac{1}{a_{2}}=\frac{a_{1}+a_{2}}{a_{1}a_{2}}=\frac{2nb^2}{n(b^3–1)}=\frac{2b^2}{b^3–1}\tag{}[/math]
Plugging in [math]a_{2}=\frac{b}{2}[/math] we get:
[math]\displaystyle \begin{align} a_{1} &=\frac{1}{ \frac{2b^2}{b^3–1} - \frac{2}{b}} \ &= \dots\&= \frac{b^4-b}{2} \end{align} \tag{}[/math]
Since [math]b[/math] has to be even, we can let [math]b=2k[/math] so that [math]a_{1}=8k^4-k[/math] and [math]a_{2}=k[/math].
To summarize, the solutions to the problem are :
[math]\boxed{ (a,b) \in {(k,2k),(2k,1),(8k^4-k,2k) | k \geq 1 }}\tag{}[/math]
Rik Bos
Ph.D. Mathematics from Utrecht University (Graduated 1979)
·
Upvoted by
Michael Jørgensen
, PhD in mathematics and
Alon Amit
, Lover of math. Also, Ph.D. · Author has 1.4K answers and 1.3M answer views
·
Aug 29
Related
How do I find all pairs of positive integers
math,[/math]
such that
[math]a^2b-1 \mid ab^3-1[/math]
?
Actually, I came to dislike this problem, but eventually I found a solution. My first approach was to assume [math]a[/math] and [math]b[/math] are both powers of the same number. That gave a bunch of solutions and there seemed to be no others. To actually prove that turned out to be not difficult, once you move in the right direction.
So first suppose [math]a=c^m,b=c^n[/math] where [math]c,m,n[/math] are nonnegative integers and [math]c>0[/math]. Then we need [math]c^{2m+n}-1\mid c^{m+3n}-1[/math]. This holds iff [math]m+3n=k(2m+n)[/math] for some positive integer [math]k[/math]. Then [math]m(2k-1)+n(k-3)=0[/math]. Clearly [math]k<4[/math] and we obtain [math]k=3,m=0[/math], so [math]a=1[/math] and [math]b[/math] can be any positive integer. Or [math]k=2[/math] and [math]n=3m[/math], i
Actually, I came to dislike this problem, but eventually I found a solution. My first approach was to assume [math]a[/math] and [math]b[/math] are both powers of the same number. That gave a bunch of solutions and there seemed to be no others. To actually prove that turned out to be not difficult, once you move in the right direction.
So first suppose [math]a=c^m,b=c^n[/math] where [math]c,m,n[/math] are nonnegative integers and [math]c>0[/math]. Then we need [math]c^{2m+n}-1\mid c^{m+3n}-1[/math]. This holds iff [math]m+3n=k(2m+n)[/math] for some positive integer [math]k[/math]. Then [math]m(2k-1)+n(k-3)=0[/math]. Clearly [math]k<4[/math] and we obtain [math]k=3,m=0[/math], so [math]a=1[/math] and [math]b[/math] can be any positive integer. Or [math]k=2[/math] and [math]n=3m[/math], i.e. [math]b=a^3[/math]. Or [math]k=1[/math] and [math]m=2n[/math], i.e. [math]a=b^2[/math].
Now assuming [math]a^2b\ne 1[/math], there exists some [math]r\ge 1[/math] such that [math]ab^3-1=r(a^2b-1)[/math]. Then subtracting [math]a^2b-1[/math] from both sides gives [math]ab(b^2-a)=(r-1)(a^2b-1)[/math] (which is nonnegative) and since [math]ab[/math] and [math]a^2b-1[/math] are coprime we see that [math]a^2b-1\mid b^2-a\ge 0[/math]. One solution is [math]a=b^2[/math] (and [math]r=1[/math]) which we already found above.
Now we are going to find yet another divisibility relation. First, [math]a^2b-1[/math] obviously divides [math]a^2b^2-b[/math] and since [math]b^2-a[/math] obviously divides [math]a^2b^2-a^3[/math], we see from [math]a^2b-1\mid b^2-a[/math] that [math]a^2b-1\mid (a^2b^2-b)-(a^2b^2-a^3)=a^3-b[/math]. If [math]a^3-b=0[/math] we recover the solution [math]b=a^3[/math].
So the last two paragraphs showed that [math]a^2b-1\mid b^2-a[/math] and [math]a^2b-1\mid a^3-b[/math]. Apart from the solutions [math]a=b^2[/math] and [math]b=a^3[/math] we can now prove there are no others (assuming as we stated above [math]a,b>1[/math]).
First, using the second divisibility relation, if [math]b>a^3[/math] then [math]b-a^3\ge a^2b-1[/math], which implies [math]b(1-a^2)\ge a^3-1[/math], a contradiction since [math]a>1[/math].
Now suppose [math]ba^3+1\ge b(a^2+1)[/math] which implies [math]b<a[/math]. But now we derive a contradiction using the first divisibility relation above: [math]a^2b-1\mid b^2-a[/math], so since [math]2\le b<a[/math] we have [math]2a^2-1\le a^2b-1\le b^2-a<a^2-a[/math], so [math]a^2<1-a<0[/math] which is impossible.
So the only solutions are
[math]a=1,b\ge 1[/math];
[math]a=b^2[/math] and [math]a,b\ge 1[/math]
[math]b=a^3[/math] and [math]a,b\ge 1[/math]
Promoted by The Penny Hoarder
Lisa Dawson
Finance Writer at The Penny Hoarder
·
Updated Sep 16
What's some brutally honest advice that everyone should know?
Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included.
And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did.
Cancel Your Car Insurance
You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix.
Don’t waste your time
Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included.
And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did.
Cancel Your Car Insurance
You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix.
Don’t waste your time browsing insurance sites for a better deal. A company calledInsurify shows you all your options at once — people who do this save up to $996 per year.
If you tell them a bit about yourself and your vehicle, they’ll send you personalized quotes so you can compare them and find the best one for you.
Tired of overpaying for car insurance? It takes just five minutes to compare your options with Insurify andsee how much you could save on car insurance.
Ask This Company to Get a Big Chunk of Your Debt Forgiven
A company calledNational Debt Relief could convince your lenders to simply get rid of a big chunk of what you owe. No bankruptcy, no loans — you don’t even need to have good credit.
If you owe at least $10,000 in unsecured debt (credit card debt, personal loans, medical bills, etc.), National Debt Relief’s experts will build you a monthly payment plan. As your payments add up, they negotiate with your creditors to reduce the amount you owe. You then pay off the rest in a lump sum.
On average, you could become debt-free within 24 to 48 months. It takes less than a minute to sign up and see how much debt you could get rid of.
Set Up Direct Deposit — Pocket $300
When you set up direct deposit withSoFi Checking and Savings (Member FDIC), they’ll put up to $300 straight into your account. No… really. Just a nice little bonus for making a smart switch.
Why switch? With SoFi, you can earn up to 3.80% APY on savings and 0.50% on checking, plus a 0.20% APY boost for your first 6 months when you set up direct deposit or keep $5K in your account. That’s up to 4.00% APY total. Way better than letting your balance chill at 0.40% APY.
There’s no fees. No gotchas.Make the move to SoFi and get paid to upgrade your finances.
You Can Become a Real Estate Investor for as Little as $10
Take a look at some of the world’s wealthiest people. What do they have in common? Many invest in large private real estate deals. And here’s the thing: There’s no reason you can’t, too — for as little as $10.
An investment called the Fundrise Flagship Fund lets you get started in the world of real estate by giving you access to a low-cost, diversified portfolio of private real estate. The best part? You don’t have to be the landlord. The Flagship Fund does all the heavy lifting.
With an initial investment as low as $10, your money will be invested in the Fund, which already owns more than $1 billion worth of real estate around the country, from apartment complexes to the thriving housing rental market to larger last-mile e-commerce logistics centers.
Want to invest more? Many investors choose to invest $1,000 or more. This is a Fund that can fit any type of investor’s needs. Once invested, you can track your performance from your phone and watch as properties are acquired, improved, and operated. As properties generate cash flow, you could earn money through quarterly dividend payments. And over time, you could earn money off the potential appreciation of the properties.
So if you want to get started in the world of real-estate investing, it takes just a few minutes tosign up and create an account with the Fundrise Flagship Fund.
This is a paid advertisement. Carefully consider the investment objectives, risks, charges and expenses of the Fundrise Real Estate Fund before investing. This and other information can be found in the Fund’s prospectus. Read them carefully before investing.
Cut Your Phone Bill to $15/Month
Want a full year of doomscrolling, streaming, and “you still there?” texts, without the bloated price tag? Right now, Mint Mobile is offering unlimited talk, text, and data for just $15/month when you sign up for a 12-month plan.
Not ready for a whole year-long thing? Mint’s 3-month plans (including unlimited) are also just $15/month, so you can test the waters commitment-free.
It’s BYOE (bring your own everything), which means you keep your phone, your number, and your dignity. Plus, you’ll get perks like free mobile hotspot, scam call screening, and coverage on the nation’s largest 5G network.
Snag Mint Mobile’s $15 unlimited deal before it’s gone.
Get Up to $50,000 From This Company
Need a little extra cash to pay off credit card debt, remodel your house or to buy a big purchase?
We found a company willing to help.
Here’s how it works: If your credit score is at least 620, AmONE can help you borrow up to $50,000 (no collateral needed) with fixed rates starting at 6.40% and terms from 6 to 144 months.
AmONE won’t make you stand in line or call a bank. And if you’re worried you won’t qualify, it’s free tocheck online. It takes just two minutes, and it could save you thousands of dollars.
Totally worth it.
Get Paid $225/Month While Watching Movie Previews
If we told you that you could get paid while watching videos on your computer, you’d probably laugh.
It’s too good to be true, right?
But we’re serious. By signing up for a free account with InboxDollars, you could add up to $225 a month to your pocket. They’ll send you short surveys every day, which you can fill out while you watch someone bake brownies or catch up on the latest Kardashian drama.
No, InboxDollars won’t replace your full-time job, but it’s something easy you can do while you’re already on the couch tonight, wasting time on your phone.
Unlike other sites, InboxDollars pays you in cash — no points or gift cards. It’s already paid its users more than $56 million.
Signing up takes about one minute, and you’ll immediately receive a $5 bonus to get you started.
Earn $1000/Month by Reviewing Games and Products You Love
Okay, real talk—everything is crazy expensive right now, and let’s be honest, we could all use a little extra cash. But who has time for a second job?
Here’s the good news. You’re already playing games on your phone to kill time, relax, or just zone out. So why not make some extra cash while you’re at it?
WithKashKick, you can actually get paid to play. No weird surveys, no endless ads, just real money for playing games you’d probably be playing anyway. Some people are even making over $1,000 a month just doing this!
Oh, and here’s a little pro tip: If you wanna cash out even faster, spending $2 on an in-app purchase to skip levels can help you hit your first $50+ payout way quicker.
Once you’ve got $10, you can cash out instantly through PayPal—no waiting around, just straight-up money in your account.
Seriously, you’re already playing—might as well make some money while you’re at it.Sign up for KashKick and start earning now!
Sri Kanth
Did engineering Ph,D, Will do Physics Ph.D.
·
Author has 86 answers and 566.5K answer views
·
Updated 9y
Related
Given two positive integers
[math]a,b[/math]
can there be two other positive integers
[math]x,y[/math]
such that
[math]a^2+b^2 = x^2+y^2[/math]
?
You don't need an 'algorithm' for this answer. This problem was considered by Fermat and it was investigated long after he was gone. You just need a fact from number theory and a little bit of complex numbers.
Before we delve into it, observe the magic for [math]65[/math]:
[math]65 = 13 \cdot 5 = (3^2 + 2^2)(2^2 + 1^2).[/math]
Now, this is where we introduce complex numbers. One way to write the above expression is:
[math]65 = |(3+ 2i)(2 + i)|^2 = |4+7i|^2 = 4^2 + 7^2.[/math]
Another way is two group one piece with the others conjugate:
[math]65 = |(3+ 2i)(2 - i)|^2 = |8+i|^2 = 8^2 + 1^2.[/math]
Therefore [math]8^2 + 1^2 = 4^2 + 7^2.[/math]
Well, this seems like
You don't need an 'algorithm' for this answer. This problem was considered by Fermat and it was investigated long after he was gone. You just need a fact from number theory and a little bit of complex numbers.
Before we delve into it, observe the magic for [math]65[/math]:
[math]65 = 13 \cdot 5 = (3^2 + 2^2)(2^2 + 1^2).[/math]
Now, this is where we introduce complex numbers. One way to write the above expression is:
[math]65 = |(3+ 2i)(2 + i)|^2 = |4+7i|^2 = 4^2 + 7^2.[/math]
Another way is two group one piece with the others conjugate:
[math]65 = |(3+ 2i)(2 - i)|^2 = |8+i|^2 = 8^2 + 1^2.[/math]
Therefore [math]8^2 + 1^2 = 4^2 + 7^2.[/math]
Well, this seems like a cool trick but does it solve the problem of finding ALL numbers that can be written as sum of two squares in two different ways? The answer is yes and we need a little bit of help from Fermat's following observation:
Fact: Every prime of the form [math]4k+1[/math] can be written as sum of two squares in a unique way.
With any number that has two prime factors of the form [math]4k+1[/math], we can pull the trick we did with [math]65[/math] above. The trick is write the number as a product of its prime factors and then expressing the prime factors as sum of squares. Now introduce complex numbers and multiply two non-conjugate complex numbers to get a new complex number. This can be done in two ways. The modulus of each of these multiplied complex numbers is the same and therefore we found a way to write a number as sum of two squares in two different ways.
The prime [math]2[/math] is a little weird. The factors of two are [math]1+i[/math] and [math]1-i[/math]. These factors when grouped with other complex numbers give almost the same complex numbers. But a number of the form [math]2m[/math] where m has at least two prime factors of the form [math]4k+1[/math] still works.
Let me show you the grouping trick for [math]50[/math].
[math]50 = 2 \cdot 25 = (1^2 + 1^2)(1^2 + 2^2)(1^2 + 2^2)[/math]
Using complex numbers:
[math] 50 = |(1 + i)(1- 2i)(1+2i)|^2 [/math]
[math] 50 = |(3-i)(1+2i)|^2 [/math]
[math] 50 = |(3-i)(1+2i)|^2 = |5+5i|^2. [/math]
Here is another way to write it:
[math] 50 = |(1 + i)(1+ 2i)(1+2i)|^2 [/math]
[math] 50 = |(1+i)(-3+4i)|^2 [/math]
[math] 50 = |(1+i)(-3+4i)|^2 = |-7+i|^2. [/math]
Therefore [math] 50=7^2 + 1^2 = 5^2 + 5^2.[/math]
1) Our method explains Tikhon Jelvis's list:
Our method asks us to list all numbers that have two or more prime factors of the form [math] 4k+1 [/math]. The first few primes of this type are [math]5,13,17,29,37,41.[/math] Because of the weird behaviour of 2, we need an odd power of 2 to divide the number. Multiplying them and listing them in increasing order gives us:
[math]50=2.5.5[/math]
[math]65=5.13[/math]
[math]85=5.17[/math]
[math]125=5.5.5[/math]
[math]130=2.5.13[/math]
[math]145=5.29[/math]
[math]170=2.5.17[/math]
[math]185=5.37[/math]
[math]200=2.2.2.5.5[/math]
[math]205=5.41[/math]
2) It explains the weird behavior observed in the article shared by Justin Rising. The authors of that article observe a list similar to Tikhon's list and wonder if all numbers with this nice property are divisible by [math]5[/math]. The above list seems to conform this pattern.
But we know better!! This just happens because 13 is way larger than 5. In fact, if we multiply 13 and 17, we get the first counter example to the claim that such numbers are divisible by 5.
[math]221=13.17 = 10^2 + 11^2 = 14^2 + 5^2[/math]
In fact, any number with two or more prime factors of the form [math]4k+1[/math] which is not divisible by 5 will do!
3) As an exercise try to figure out the decomposition as sum of two squares for Sundaram's example: 85.
Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006)
·
Upvoted by
Michael Jørgensen
, PhD in mathematics · Author has 8.5K answers and 21.1M answer views
·
1y
Related
How do I find all non-negative integers
[math]x, y[/math]
such that
[math]x^3y+x+y=xy+2xy^2[/math]
?
If [math]y = 0[/math], then this leads to the non-negative integer solution math = (0, 0)[/math]. Conversely, [math]x = 0[/math] implies that [math]y = 0[/math] as well.
Hereafter, assume that [math]x, y \geq 1[/math]. Then, we can rewrite the Diophantine equation as
[math]x = y(x + 2xy - x^3 - 1). \tag{}[/math]
Since [math]x + 2xy - x^3 - 1[/math] is an integer, this implies that [math]y \mid x[/math]. Writing [math]x = yz[/math] for some positive integer [math]z[/math], substituting this into our equation and rearranging terms yields
[math]z(y + 2y^2 - 1 - y^3 z^2) = 1. \tag{}[/math]
Since [math]y, z \in \mathbb{N}[/math] and this is an equation in integers, this latter equation forces [math]z = 1[/math] in order for the equation to have positive integ
If [math]y = 0[/math], then this leads to the non-negative integer solution math = (0, 0)[/math]. Conversely, [math]x = 0[/math] implies that [math]y = 0[/math] as well.
Hereafter, assume that [math]x, y \geq 1[/math]. Then, we can rewrite the Diophantine equation as
[math]x = y(x + 2xy - x^3 - 1). \tag{}[/math]
Since [math]x + 2xy - x^3 - 1[/math] is an integer, this implies that [math]y \mid x[/math]. Writing [math]x = yz[/math] for some positive integer [math]z[/math], substituting this into our equation and rearranging terms yields
[math]z(y + 2y^2 - 1 - y^3 z^2) = 1. \tag{}[/math]
Since [math]y, z \in \mathbb{N}[/math] and this is an equation in integers, this latter equation forces [math]z = 1[/math] in order for the equation to have positive integer solutions. Hence, [math]x = y[/math] and letting [math]z = 1[/math] in th last version of our Diophantine equation yields [math]y + 2y^2 - 1 - y^3 = 1[/math], or
[math]y^3 - 2y^2 - y + 2 = (y^2 - 1)(y - 2) = 0. \tag{}[/math]
Solving for positive integer solutions gives us [math]y = 1, 2[/math]. Hence, we also have the solutions math = (1, 1), (2, 2)[/math].
Summarizing our work, the solutions to the given Diophantine equation are
[math]\boxed{(x, y) = (0, 0), \, (1, 1), \, (2, 2)}. \tag{}[/math]
Sponsored by CDW Corporation
How well do your distributed teams work together?
Enable seamless and secure communication for everyone with Cisco collaboration solutions from CDW.
Alon Amit
PhD in Mathematics; Mathcircler.
·
Upvoted by
Jeremy Stone
, PhD Mathematics & Computational Fluid Dynamics, University of Birmingham and
Vladimir Novakovski
, silver medals, IOI 2001 and IPhO 2001 · Author has 8.8K answers and 173.8M answer views
·
Updated 5y
Related
How would you show that for any fixed integer
[math]k>1[/math]
there exist infinitely many positive integers
[math]n[/math]
such that
[math]n|1^n+2^n+3^n+…+k^n[/math]
?
Fun question! Where is it from?
Here’s how I did it, including the false turns and errors of judgement.
My first reaction was that the problem seems wrong. If [math]k[/math] is fixed and we’re looking for lots of exponents [math]n[/math], we are quickly going to have [math]n[/math] much larger than [math]k[/math], and the behavior of [math]k^n[/math] modulo [math]n[/math] for small [math]k[/math] and large [math]n[/math] is very rigid. For example, if [math]n[/math] happens to be prime, then [math]k^n \equiv k \pmod{n}[/math], which doesn’t bode well for the sum [math]1^n+2^n+\ldots+k^n[/math] to become zero.
Of course, that intuition is wrong, and the question is fine, but there’s a helpful clue here: we don’t want to have [math]n[/math] grow uncon
Fun question! Where is it from?
Here’s how I did it, including the false turns and errors of judgement.
My first reaction was that the problem seems wrong. If [math]k[/math] is fixed and we’re looking for lots of exponents [math]n[/math], we are quickly going to have [math]n[/math] much larger than [math]k[/math], and the behavior of [math]k^n[/math] modulo [math]n[/math] for small [math]k[/math] and large [math]n[/math] is very rigid. For example, if [math]n[/math] happens to be prime, then [math]k^n \equiv k \pmod{n}[/math], which doesn’t bode well for the sum [math]1^n+2^n+\ldots+k^n[/math] to become zero.
Of course, that intuition is wrong, and the question is fine, but there’s a helpful clue here: we don’t want to have [math]n[/math] grow uncontrollably in terms of its prime divisors. In fact, it will be best to have [math]n=a^m[/math] for some fixed base [math]a[/math] and growing [math]m[/math], which ensures that the prime divisors of [math]n[/math] are unchanging. But this is all good in hindsight: I didn’t understand it initially.
To prove that the problem really is wrong, I wanted to convince myself that even something as silly as [math]k=2[/math] can’t work. In this case, we are looking for infinitely many [math]n[/math] such that [math]2^n \equiv -1 \pmod{n}[/math]. Because I’m a lazy bum, I used Python to quickly demonstrate that I’m right and there aren’t such [math]n[/math] except for a few random exceptions.
In : [n for n in range(1,100) if (2n+1)%n==0]Out: [1, 3, 9, 27, 81]
Huh. Ok, I’m an idiot. Powers of [math]3[/math] seem to work. Really? Does [math]2^{3^m} \equiv -1 \pmod{3^m}[/math] for all [math]m \geq 1[/math]?
Shifting perspective to prove that this works (rather than show that it can’t), I’ve noticed that this is reminiscent of Hensel’s Lemma (you don’t need to know what it is, but if you do, then you’ll notice the technique used in the proof is very common).
We start with the obvious
[math]2^3 \equiv -1 \pmod{3}[/math]
and now we want to show that this stays true when we switch both [math]3[/math]'s to [math]9[/math]'s. Anticipating an inductive process, we write [math]2^3 = 3q-1[/math] for some unimportant [math]q[/math], and then
[math]2^{3^2} = \left(2^3\right)^3 = (3q-1)^3[/math]
We need to control this mod [math]9[/math], which means (this is where the Henselian intuition comes into play) we can ignore the first few terms in the binomial expansion. In fact, we find that we can ignore all of them except the last one: when you take a high power of [math]3q[/math], you get nothing mod [math]9[/math], and even the linear term [math]3q[/math] itself gets multiplied by the next-to-last binomial coefficient, which is [math]3[/math], so it also disappears. So, indeed,
[math]2^{3^2} \equiv -1 \pmod{3^2}[/math]
and similarly, if we already know that [math]2^{3^m}[/math] is [math]-1[/math] mod [math]3^m[/math], so again it is some [math]3^m q-1[/math], and then
[math]2^{3^{m+1}}=\left(2^{3^m}\right)^3 = (3^m q -1)^3 \equiv -1 \pmod{3^{m+1}}[/math]
Ok. This settles the problem for [math]k=2[/math]: for the infinitely many [math]n[/math] you need, just take [math]n=3^m[/math].
I now got into a pretty lengthy rathole, where I was trying to understand what to do with larger [math]k[/math] and failing to notice that [math]3[/math] is just one more than [math]2[/math]. I had an intuition that the Henselian argument would let me push the powers higher and higher once I have a starting point, but I got lost looking for the starting point, which was staring me right in the face: just take [math]n=k+1[/math], at least when [math]k[/math] is even.
The breakthrough came only when I tried to find a good [math]n[/math] for [math]1^n+2^n+3^n+4^n[/math]. Luckily, I was biking at this point, so I couldn’t shortcut with Python. Since [math]1+2+3+4=10[/math], I wanted to use a prime factor of that, so I took [math]n=5[/math], which worked. But then I was like, why would that continue to work for [math]n=25[/math]? I got hung up on every [math]i^n[/math] being [math]i[/math] modulo [math]5[/math], so when moving on to [math]5^2[/math] I expected every [math]i[/math] to become [math]i^5[/math] mod [math]5^2[/math]. Why would those continue to add up to [math]0[/math]?
I should have seen this immediately, but I didn’t. I had to actually calculate [math]2^5[/math] mod [math]25[/math] which is [math]7[/math], and then [math]3^5[/math] mod [math]25[/math] which is [math]-7[/math], and it hit me that this can’t be a coincidence. Ugh, dumbass, [math]3[/math] is just [math]-2[/math] mod [math]5[/math], so of course the fifth powers come to be the negatives of each other.
I now realized that the way to force [math]1^n+2^n+\ldots+k^n[/math] to be zero mod [math]n[/math] wasn’t using some clever manipulation of sums, but simply by pairing them off in the most natural way. You want [math]k[/math] to be [math]-1[/math] mod [math]n[/math], and [math]k-1[/math] to be [math]-2[/math], and everything will cancel in pairs as long as [math]n[/math] is odd. And, hooray, for this to work you just need [math]n=k+1[/math], and precisely when [math]k[/math] is even two good things happen: everybody has a buddy, and math^n[/math] stays [math]-1[/math].
With that, I was able to quickly repeat the Henselian argument from before. For any even [math]k[/math], choose [math]n=k+1[/math] to begin with. Then [math]i^n[/math] and math^n[/math] cancel each other out mod [math]n[/math], and therefore you have your first good [math]n[/math]. Then, switch [math]n[/math] to [math]n^2[/math], and then to [math]n^3[/math] and so on, and repeat the exact same argument as with [math]k=2[/math] earlier. Whatever [math]i^{n^m}[/math] is modulo [math]n^m[/math], it gets raised to the [math]n^\text{th}[/math] power when you move to [math]i^{n^{m+1}}[/math], and [math]n-i[/math] does the same thing, so they still cancel each other out (since [math]n[/math] is odd!).
At this point I had a full solution for the case of even [math]k[/math], and I took another set of false paths in an attempt to do something similar when [math]k[/math] is odd. I still tried to pair up the numbers symmetrically, and to handle the lonely midpoint I tried to start with it as my [math]n[/math], so that its powers won’t matter. It then felt like I was using some kind of doubling, moving from [math]1[/math] through [math]4[/math] to [math]1[/math] through [math]9[/math] with a [math]5[/math] in the middle, and that seemed to work except I got confused with things like [math]k=7[/math].
Of course, this was all nonsense. All you have to notice is that the exact same choice of [math]n=(k+1)^m[/math] work just great for
[math]1^n+2^n+\ldots+k^n+(k+1)^n[/math]
since the last term doesn’t change anything. So for example, once you know that
[math]1^{5^m}+2^{5^m}+3^{5^m}+4^{5^m} \equiv 0 \pmod{5^m}[/math],
you can immediately conclude that also
[math]1^{5^m}+2^{5^m}+3^{5^m}+4^{5^m}+5^{5^m} \equiv 0 \pmod{5^m}[/math]
since the additional term doesn’t contribute anything.
So, wrapping it all up, we have proven: Given any integer [math]k>1[/math], for any [math]m>0[/math] set [math]n=(k+1)^m[/math] if [math]k[/math] is even or [math]n=k^m[/math] if [math]k[/math] is odd. Then
[math]1^n+2^n+\ldots+k^n \equiv 0 \pmod{n}[/math]
and this gives us infinitely many exponents satisfying the divisibility condition, as required by the question.
To make sure you understand all this, write down the full proof following the outline above – it’s actually really short, once you remove all of my meandering blunders. I thought it would be useful, though, to show those meandering blunders. I always feel bad about taking forever to solve a problem and then writing down a slick solution that makes it look like it was achieved oh so elegantly.
Alon Amit
Math Circle educator, Proof School trustee
·
Author has 8.8K answers and 173.8M answer views
·
6y
Related
Are there infinitely many integers
[math]p,q[/math]
such that
[math]p|q^2+1[/math]
and
[math]q|p^2+1[/math]
? Can this be proved without assuming knowledge of maths beyond high school?
Yes, and yes.
First off: the variable names [math]p[/math] and [math]q[/math] are usually reserved for primes (or prime powers) in number theory. This isn’t the case here: you’re asking for integers with this property. So let’s please change the letters to [math]a[/math] and [math]b[/math]. We want [math]a|b^2+1[/math] and [math]b|a^2+1[/math].
Next: we need examples. Let’s look for small numbers which work. Without too much trouble, we should be able to come up with these pairs math[/math]:
math[/math]
math[/math]
Is there something we can do with [math]3^2+1[/math]? Not quite: the divisors of [math]10[/math] are [math]2[/math] and [math]5[/math] but [math]3[/math] divides neither [math]2^2+1[/math] nor [math]5^2+1[/math]. The next number, [math]4^2+1[/math], isn’t helpful either. But then we
Yes, and yes.
First off: the variable names [math]p[/math] and [math]q[/math] are usually reserved for primes (or prime powers) in number theory. This isn’t the case here: you’re asking for integers with this property. So let’s please change the letters to [math]a[/math] and [math]b[/math]. We want [math]a|b^2+1[/math] and [math]b|a^2+1[/math].
Next: we need examples. Let’s look for small numbers which work. Without too much trouble, we should be able to come up with these pairs math[/math]:
math[/math]
math[/math]
Is there something we can do with [math]3^2+1[/math]? Not quite: the divisors of [math]10[/math] are [math]2[/math] and [math]5[/math] but [math]3[/math] divides neither [math]2^2+1[/math] nor [math]5^2+1[/math]. The next number, [math]4^2+1[/math], isn’t helpful either. But then we come to [math]5^2+1[/math] which is divisible by [math]2[/math] (we’ve already listed that) and [math]13[/math], and hooray, [math]5[/math] indeed divides [math]13^2+1[/math]. So we’ve found our next example:
math[/math]
We notice something funny: the pairs we have so far can be concatenated, meaning the second number in each pair is the first number in the next. Does this trend continue? What else is [math]13^2+1[/math] divisible by? It’s [math]2 \times 5 \times 17[/math]. Trying [math]2,5,17,10,34,85[/math] we hit upon [math]34[/math]: [math]13[/math] divides [math]34^2+1[/math]. We have another pair:
math[/math]
and those are just consecutive numbers in the sequence [math]1,2,5,13,34[/math]. What is this sequence?
There are several ways to answer this question, but the one most likely to be recognizable to a high-schooler is that those are every-other-Fibonacci-number. The Fibonacci sequence is familiar to many high schoolers:
[math]F_0=0[/math]
[math]F_1=1[/math]
[math]F_{n+1} = F_n+F_{n-1}[/math] for [math] n \ge 1[/math]
In words, we start with [math]0,1[/math] and then each number is the sum of the two previous ones.
[math]0,1,1,2,3,5,8,13,21,34,55,89,\ldots[/math]
The examples we’ve found are all here, with skips:
[math]0,\mathbf{1},1,\mathbf{2},3,\mathbf{5},8,\mathbf{13},21,\mathbf{34},55,\mathbf{89},\ldots[/math]
So what we seem to be saying is that each [math]F_{n-1}[/math] divides [math]F_{n+1}^2+1[/math] while at the same time [math]F_{n+1}[/math] divides [math]F_{n-1}^2+1[/math], when [math]n[/math] is odd (but not when [math]n[/math] is even – can you see what happens then?) In fact, even better – we can see what the quotients are! [math]13^2+1[/math] is divisible by [math]5[/math] with a quotient of [math]34[/math], and [math]5^2+1[/math] is divisible by [math]13[/math] with a quotient of [math]2[/math]. So what we really think is that
[math]F_{n+1}^2+1 = F_{n-1}F_{n+3}[/math] (when [math]n[/math] is odd!)
[math]F_{n-1}^2+1 = F_{n+1}F_{n-3}[/math] (when [math]n[/math] is odd!)
happily, these two are the same formula applied at different indices.
Depending on how much Fibonacci-foo our high-schooler has, these formulas may be already familiar (they even have a name; what we need here is Catalan with [math]r=2[/math]), or if they’re not, at least they give you something concrete to prove. Prove it. It’s easy with matrices, but not super hard directly with induction as well.
Once you’ve proven this, you’re done: you’ve produced an infinite sequence of pairs of integers math[/math] which satisfy the requirement. [math]\blacksquare[/math]
Trevor
B.A. with MMath in Mathematics, University of Cambridge (Graduated 2023)
·
Author has 1.3K answers and 5.5M answer views
·
8y
Related
How would you show that for any fixed integer
[math]k>1[/math]
there exist infinitely many positive integers
[math]n[/math]
such that
[math]n|1^n+2^n+3^n+…+k^n[/math]
?
We can find the most basic value that works for all [math]k[/math],
If [math]k[/math] is odd, then we choose [math]n=k[/math]
If [math]k[/math] is even, then we choose [math]n=k+1[/math]
The above can be proved using simple modular arithmetic:
When [math]k[/math] is odd, we pair up this way:
[math]m^k+(k-m)^k \equiv m^k+(-m)^k\equiv 0\mod k[/math]
The remaining [math]k^k[/math] must be divisible by [math]k[/math], so we have proved for the odd case.
The even case:
[math]m^{k+1}+(k+1-m)^{k+1}\equiv m^{k+1}+(-m)^{k+1}\equiv 0 \mod (k+1)[/math]
This relies on the fact that the exponent is odd.
When we have proven the base case, let’s consider when [math]n=k_0^m[/math], where [math]k_0[/math] is the base case, and [math]m[/math] is a postive integer.
I intuitively think tha
We can find the most basic value that works for all [math]k[/math],
If [math]k[/math] is odd, then we choose [math]n=k[/math]
If [math]k[/math] is even, then we choose [math]n=k+1[/math]
The above can be proved using simple modular arithmetic:
When [math]k[/math] is odd, we pair up this way:
[math]m^k+(k-m)^k \equiv m^k+(-m)^k\equiv 0\mod k[/math]
The remaining [math]k^k[/math] must be divisible by [math]k[/math], so we have proved for the odd case.
The even case:
[math]m^{k+1}+(k+1-m)^{k+1}\equiv m^{k+1}+(-m)^{k+1}\equiv 0 \mod (k+1)[/math]
This relies on the fact that the exponent is odd.
When we have proven the base case, let’s consider when [math]n=k_0^m[/math], where [math]k_0[/math] is the base case, and [math]m[/math] is a postive integer.
I intuitively think that this is the case, but don’t have the time to prove so. I think you can use bionimial theorem to do this but I am not sure. If that is proven to be true, since [math]m[/math] can take any value, there are infinitely many cases of [math]n[/math].
Fernando Cagarrinho
college degree in Physics, IST - Instituto Superior Técnico (Graduated 1997)
·
Author has 185 answers and 36.4K answer views
·
May 21
Related
Do there exist infinitely many positive integers
[math]a, b,[/math]
so that
math(b^2+1)((a+b)^2+1)[/math]
is a perfect square?
Each factor is obviously not a square, so they have to share divisors. This reminded me of the problem,
[math]x|y^2+1[/math] and [math]y|x^2+1[/math]
solved by Rik Bos in this answer.
There it’s proven that [math]x,y[/math] are some Fibonacci Numbers . This looks promising, because if [math]a,b[/math] are consecutive members of this sequence, [math]F_n,F_{n+1},[/math] then [math]a+b=F_n+F_{n+1}=F_{n+2},[/math] is the next one.
An identity of interest to our problem is the well known Cassini’s identity,
For even [math]n=2k[/math] this is,
resembling the factors of the problem. This is true only for even indexed FN and if we stay with th
Each factor is obviously not a square, so they have to share divisors. This reminded me of the problem,
[math]x|y^2+1[/math] and [math]y|x^2+1[/math]
solved by Rik Bos in this answer.
There it’s proven that [math]x,y[/math] are some Fibonacci Numbers . This looks promising, because if [math]a,b[/math] are consecutive members of this sequence, [math]F_n,F_{n+1},[/math] then [math]a+b=F_n+F_{n+1}=F_{n+2},[/math] is the next one.
An identity of interest to our problem is the well known Cassini’s identity,
[math]F_n^2+(-1)^n=F_{n-1}F_{n+1}[/math]
For even [math]n=2k[/math] this is,
[math]F_{2k}^2+1=F_{2k-1}F_{2k+1}[/math]
resembling the factors of the problem. This is true only for even indexed FN and if we stay with the hypothesis of [math]a,b[/math] being consecutive FN, it can’t be applied to all factors.
Maybe it’s time to do some experimentation. Starting at index [math]0[/math] ,the first terms of the Fibonacci sequence are:
math(1^2+1)(1^2+1)=4[/math] (square)
math(1^2+1)(2^2+1)=20[/math] (no)
math(2^2+1)(3^2+1)=(2)(5)(10)[/math] (square)
math(3^2+1)(5^2+1)=(5)(10)(26)[/math] (no)
We’re getting squares when [math]a=F_{2n}.[/math] Looking closely at the last example we see,
[math]3^2+1=2\cdot 5[/math]
[math]5^2+1=2\cdot 13[/math]
[math]8^2+1=5\cdot 13[/math]
The first and last equalities come from the Cassini’s identity, but the second sugests the conjecture,
[math]F_{2n+1}^2+1=F_{2n-1}F_{2n+3}[/math] (1)
The first examples follow this pattern and if it’s true we are done. Precisely this equality was proven in the above mencioned answer, by using the FN representation with powers of the golden ratio. This can also be done using the Cassini's identity and the recurrence relation of FN.
To finish, from Cassini's identity we have,
Multipliing these two with the new relation (1),
This proves that the sequence math=(F_{2n},F_{2n+1})[/math] provides an infinite family of solutions.
Carlos Eŭ Th
Triple IMO bronze medalist
·
Author has 5.9K answers and 4.4M answer views
·
Updated May 8
Related
Are there infinitely many integers
[math]p,q[/math]
such that
[math]p|q^2+1[/math]
and
[math]q|p^2+1[/math]
? Can this be proved without assuming knowledge of maths beyond high school?
Let's suppose there are two coprime numbers, [math]p,q[/math] such that [math]q\ge p[/math], and [math]q\mathrel|p^2+1[/math] and [math]p\mathrel|q^2+1[/math]. This means that there are numbers [math]r,s[/math] such as [math]p^2+1=sq[/math], and [math]q^2+1=rp[/math].
So [math]r=\frac{q^2+1}p[/math].
Then [math]r^2+1=\dfrac{q^4+2q^2+1}{p^2}+1=\dfrac{q^4+2q^2+p^2+1}{p^2}=\dfrac{q^4+2q^2+qs}{p^2}[/math].
So [math]r^2+1=q\frac{q^3+2q+s}{p^2}[/math].
We just show that [math]\frac{q^3+2q+s}{p^2}[/math] is integer.
But [math]r[/math] is integer, so [math]r^2+1[/math] is integer, so [math]p^2[/math] divides [math]q(q^3+2q+s)[/math], but [math]p,q[/math] are coprime, so [math]p^2[/math] must divide [math]q^3+2q+s[/math].
We also had [math]p\le q[/math], this means [math]pq\le q^2<q^2+1[/math], so [math]pqq[/math].
This means that if there is one coprime pair [math]p,q;p[/math]
Let's suppose there are two coprime numbers, [math]p,q[/math] such that [math]q\ge p[/math], and [math]q\mathrel|p^2+1[/math] and [math]p\mathrel|q^2+1[/math]. This means that there are numbers [math]r,s[/math] such as [math]p^2+1=sq[/math], and [math]q^2+1=rp[/math].
So [math]r=\frac{q^2+1}p[/math].
Then [math]r^2+1=\dfrac{q^4+2q^2+1}{p^2}+1=\dfrac{q^4+2q^2+p^2+1}{p^2}=\dfrac{q^4+2q^2+qs}{p^2}[/math].
So [math]r^2+1=q\frac{q^3+2q+s}{p^2}[/math].
We just show that [math]\frac{q^3+2q+s}{p^2}[/math] is integer.
But [math]r[/math] is integer, so [math]r^2+1[/math] is integer, so [math]p^2[/math] divides [math]q(q^3+2q+s)[/math], but [math]p,q[/math] are coprime, so [math]p^2[/math] must divide [math]q^3+2q+s[/math].
We also had [math]p\le q[/math], this means [math]pq\le q^2<q^2+1[/math], so [math]pqq[/math].
This means that if there is one coprime pair [math]p,q;p\le q[/math], that fulfill the condition, then there is a greater pair [math]q,r[/math] that fulfill the condition.
We only need the initial pair.
And the initial pair is the trivial coprime pair [math]1,1[/math].
Square the greatest, add one, divide by the previous one, and we get [math]1,2[/math].
Square the greatest, add one, divide by the previous one, and we get [math]2,5[/math].
Square the greatest, add one, divide by the previous one, and we get [math]5,13[/math].
Square the greatest, add one, divide by the previous one, and we get [math]13,34[/math].
We can build the succession 1, 1, 2, 5, 13, 34, 89, 233, … any consecutive numbers in this succession fulfill the condition.
If those numbers seem familiar, let's add a few numbers between:
1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, … step every two numbers.
That is the Fibonacci progression.
Related questions
Are there infinitely many integers
such that
divides
?
How do you prove that there are infinitely many integers
such that
and
?
What can I do with a BS in Mathematics?
Does being good at mathematics make you intelligent?
How do I find all pairs of positive integers
such that
?
If
, what is
?
How should I study math?
How would you show that for any fixed integer
there exist infinitely many positive integers
such that
?
How should I improve my maths?
What's a good math book for somebody who hates math?
What is 6/2(1+2) =?
Why is mathematics so hard?
What are the all integers
such that
?
Do math geniuses struggle with math problems?
What are all pairs of integers
such that
?
About
·
Careers
·
Privacy
·
Terms
·
Contact
·
Languages
·
Your Ad Choices
·
Press
·
© Quora, Inc. 2025 |
189245 | https://www.scribd.com/document/325721170/ANSI-ASME-B106-1M-1985-Design-of-Transmission-Shafting | ANSI/ASME B106.1M-1985, Design of Transmission Shafting | PDF | Bending | Stress (Mechanics)
Opens in a new window Opens an external website Opens an external website in a new window
This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy
Open navigation menu
Close suggestions Search Search
en Change Language
Upload
Sign in
Sign in
Download free for 30 days
100%(1)100% found this document useful (1 vote)
4K views 2 pages
ANSI/ASME B106.1M-1985, Design of Transmission Shafting
The document discusses the design and analysis of shafts. Shafts are designed to transmit power durably by conveying torque while limiting deflections and stresses. Key features of shafts in…
Full description
Uploaded by
Fernando Aguirre
AI-enhanced description
Go to previous items Go to next items
Download
Save Save ANSI/ASME B106.1M-1985, Design of Transmission Sha... For Later
Share
100%100% found this document useful, undefined
0%, undefined
Print
Embed
Translate
Ask AI
Report
Download
Save ANSI/ASME B106.1M-1985, Design of Transmission Sha... For Later
You are on page 1/ 2
Search
Fullscreen
C:\Bill\BEI\ME312\2005\ShaftsHam.doc
Shafting
Mission: Transmit power durably. Requirements: 1. Convey torque 2. Limit deflections (for gears, bearings, sprockets) 3. Limit stresses Features: 1. Steps or shoulders to locate bearings, gears, sheaves 2. Keyways or splines to transfer torque 3. Typically steady torque and alternating bending loads Stress and Deflection:
•
Stress analysis is local 1. Bending moments and torsional shears are loads - independent of geometry 2. C/
I
and r/J are local geometries 3. Stress concentrations are local
•
Deflection is a function of geometry everywhere 1. Deflection and slope are integrated buildup of bending/stiffness
Attachments
§11.6 Keys Keyways are usually sledrunner (K
t
~ 1.4), end milled (K
t
1.8), or W oodruff Key Failure Modes (P. 460):
§
Shear on key = Torque / (key wi dth length shaft radius) <S
sy
§
Compression = Torque / (0.5 key height length radius) <0.9 S
y
= S
cy
Pins are limited by the shear area of the pin §10.5&6 Press/Shrink Fits (P. 420) Torque capability = Pressure
interface
x Area
interface
x
µ
interface
x radius
interface
K
t
is up to 2 at entrance and exit. Set Screws (Cup Point; Source: SPS Unbrako) Tables show holding "power" (force) in LBs vs screw diameter. Torque capability = holding "power" x radius of contact Splines
•
Usually 4 or more splines, so loading is lower than keyways and more uniform
•
There are 2 pieces (not 3) so is less relative motion
•
Accurate for controlled fit 1. Side fit 2. Major diameter fit
•
Hardened surfaces
•
Indexable
•
Lengths = 0.75xDia to 1.25xDia
•
$$$
Screw Holding
Size Force
8 385 LB
10 540 LB ¼ 1000 LB 5/16 1500 LB
adDownload to read ad-free
C:\Bill\BEI\ME312\2005\ShaftsHam.doc
Shaft Analysis Procedure
§11.2
Draw t he Free Body Diagram 2. Calculate the reaction loads 3. Draw the Shear, V, and Moment, M, diagrams I N EACH PLANE that loads are in. 4. Draw the Total Moment diagram, vectorially summing the moments in each load plane: 5. Draw the Torque (Axial Moment) di agram 6. Apply stress concentrations
•
See Figures 6.5 &6.6 on Pages 231-232 7. Establish the location of the critical cross section where torque and moment are the largest.
Shaft Design
Static Loading If axial loads are small, use this simple equation [Eq. 11.17] to compute the shaft diameter by MSS (Max Shear Stress) given the bending moment, M, the torque, T, the factor of safety, n
s
, and the material yield strength, S
y
. Can also use this equation to estimate a shaft size early in the design process. 2. Fatigue Loading – General Case, MSS criterion [Eq. 11.35] where M
m
& M
a
are mean and alternating moments, T
m
& T
a
are mean and alternating torques, and K
f
& K
fs
are bending and shear fatigue stress concentration factors, respectively. 3. Fatigue Loading - Steady Torsion and Reversed Bending
•
Use the ANSI/ASME method, shown here: (Ref: ANSI/ASME B106.1M-1985 Standard)
•
See §7.7 for t he endurance-limit modifying factors, but DON'T include the Stress Concentration factor in S
e
it is booked separately in this equation.
•
See Fig. 7.9 on P. 283 for notch sensitivity!
2 2
y x tot
M M M
+=
3 1 2 2
32
+=
T M S n d
y s
π
3 1 2 2
32
+++=
a fs e y m a f e y m y s
T K S S T M K S S M S n d
π
3 1 2 2
4 3 32
+=
y m e a f s
S T S M K n d
π
Read this document in other languages
English
Română
Share this document
Share on Facebook, opens a new window
Share on LinkedIn, opens a new window
Share with Email, opens mail client
Copy link
Millions of documents at your fingertips, ad-free Subscribe with a free trial
You might also like
Thermodynamics: An Engineering Approach 9th Edition in SI Units 100% (1) Thermodynamics: An Engineering Approach 9th Edition in SI Units 61 pages
PHY 131 hw14 No ratings yet PHY 131 hw14 6 pages
Ansi Agma 2101 d04 100% (2) Ansi Agma 2101 d04 66 pages
DIN 6892 TRANSLATION English Parallel Keys No ratings yet DIN 6892 TRANSLATION English Parallel Keys 47 pages
AISE No. 7 Specifications For Ladle Hooks 100% (1) AISE No. 7 Specifications For Ladle Hooks 7 pages
Din 00740-1 1986 (En) No ratings yet Din 00740-1 1986 (En) 15 pages
ANSI AGMA 6113-B16 Factor de Servicio 0% (1) ANSI AGMA 6113-B16 Factor de Servicio 84 pages
Intro To DIN 743 No ratings yet Intro To DIN 743 8 pages
Ansi Asme B106-1 1985 No ratings yet Ansi Asme B106-1 1985 32 pages
Ansi Agma6001 d97 No ratings yet Ansi Agma6001 d97 48 pages
Press Fit Pressure Calculator 100% (1) Press Fit Pressure Calculator 4 pages
Examples - Shafts: B. Free Body Diagram of Forces A. Chain Drive Assembly No ratings yet Examples - Shafts: B. Free Body Diagram of Forces A. Chain Drive Assembly 4 pages
10.1 - Calculation of Strength For Spur and Helical Gears - TechnicalData - KGSTOCKGEARS PDF No ratings yet 10.1 - Calculation of Strength For Spur and Helical Gears - TechnicalData - KGSTOCKGEARS PDF 14 pages
Interference Fit According To DIN 7190 PDF No ratings yet Interference Fit According To DIN 7190 PDF 23 pages
Din 15020-1 1974 PDF 100% (3) Din 15020-1 1974 PDF 12 pages
JIS B0401-1986 System of Limits and Fits No ratings yet JIS B0401-1986 System of Limits and Fits 42 pages
ANSI AGMA 6004 F88 1988 Gear Power Rating For Cylindrical Grinding Mills Kilns Coolers and Dryers No ratings yet ANSI AGMA 6004 F88 1988 Gear Power Rating For Cylindrical Grinding Mills Kilns Coolers and Dryers 27 pages
Acrylic As A Structural Material No ratings yet Acrylic As A Structural Material 8 pages
Shaft Design and Stress Analysis 100% (1) Shaft Design and Stress Analysis 65 pages
Ansi Asme B106.1M-1985 No ratings yet Ansi Asme B106.1M-1985 14 pages
Calculation of Load Capacity of Shafts and Axles: German Standard October 2000 No ratings yet Calculation of Load Capacity of Shafts and Axles: German Standard October 2000 6 pages
Standard For Industrial Enclosed Gear Drives (Metric Edition) 50% (2) Standard For Industrial Enclosed Gear Drives (Metric Edition) 67 pages
Calculation of Load Capacity of Shafts and Axles: German Standard October 2000 100% (2) Calculation of Load Capacity of Shafts and Axles: German Standard October 2000 16 pages
Design of Shafts 100% (1) Design of Shafts 19 pages
Shaft Design PDF 100% (1) Shaft Design PDF 27 pages
DIN ISO 12128-1998 Plain Bearings - Lubrication Holes, Grooves and Pockets No ratings yet DIN ISO 12128-1998 Plain Bearings - Lubrication Holes, Grooves and Pockets 10 pages
Bevel Gear No ratings yet Bevel Gear 28 pages
Agma 915-3-A99 100% (1) Agma 915-3-A99 16 pages
Din743 Intro English 100% (1) Din743 Intro English 9 pages
Agma 9003-A91 100% (1) Agma 9003-A91 28 pages
Ansi Agma 2001 D04 100% (4) Ansi Agma 2001 D04 66 pages
Din 3964 English 100% (1) Din 3964 English 4 pages
A675 No ratings yet A675 4 pages
AGMA 9002-A86 Inch Bore and ANSI B17.1 Square Keyway Tolerances 100% (1) AGMA 9002-A86 Inch Bore and ANSI B17.1 Square Keyway Tolerances 6 pages
SSW-Q1R Class 2 No ratings yet SSW-Q1R Class 2 2 pages
Aise Standard No.7 (Ladle Hooks) 100% (1) Aise Standard No.7 (Ladle Hooks) 7 pages
ISO 5049 - 2 Design of Mobile Continuos Bulk Handling Equipment 100% (2) ISO 5049 - 2 Design of Mobile Continuos Bulk Handling Equipment 48 pages
Spur Gear No ratings yet Spur Gear 3 pages
Design of Shafts and Keys 100% (1) Design of Shafts and Keys 106 pages
Axle Holders No ratings yet Axle Holders 1 page
DIN 743 Shaft Calculation Guide No ratings yet DIN 743 Shaft Calculation Guide 7 pages
Coupling Selection DIN740 Part2-1 No ratings yet Coupling Selection DIN740 Part2-1 4 pages
SEW Project Planning of Drives No ratings yet SEW Project Planning of Drives 160 pages
Heppenstall Brochure 2016 No ratings yet Heppenstall Brochure 2016 6 pages
Fit Between Impeller (With Key) and Pump Shaft No ratings yet Fit Between Impeller (With Key) and Pump Shaft 3 pages
SKF Industrial Gearbox 2 100% (2) SKF Industrial Gearbox 2 78 pages
Calculating What Kind of Power I Need For Rack and Pinion 4x8 Table No ratings yet Calculating What Kind of Power I Need For Rack and Pinion 4x8 Table 21 pages
Spline Program Ver 1.1 100% (1) Spline Program Ver 1.1 7 pages
B29 24 2002 No ratings yet B29 24 2002 18 pages
Standards Related To Bearings 50% (2) Standards Related To Bearings 6 pages
Pressure Vessel Design Guide 50% (2) Pressure Vessel Design Guide 42 pages
6 Shafts PDF No ratings yet 6 Shafts PDF 23 pages
Shaft Design and Components Guide No ratings yet Shaft Design and Components Guide 7 pages
ME311 - Figures - CH7 - Shafts No ratings yet ME311 - Figures - CH7 - Shafts 25 pages
Chapter 1 # Machine Element at 2016 No ratings yet Chapter 1 # Machine Element at 2016 24 pages
Shaft, Keys and Coupling No ratings yet Shaft, Keys and Coupling 29 pages
Shaft Desing 2 No ratings yet Shaft Desing 2 53 pages
Bearings Shafts No ratings yet Bearings Shafts 74 pages
Design Shaft Key No ratings yet Design Shaft Key 32 pages
Concordia University: MECH 390 - Mechanical Engineering Design Project No ratings yet Concordia University: MECH 390 - Mechanical Engineering Design Project 58 pages
Chapter 1 No ratings yet Chapter 1 65 pages
Unit 8 Air Conditioning No ratings yet Unit 8 Air Conditioning 26 pages
1 s2.0 S0029549311006182 Main No ratings yet 1 s2.0 S0029549311006182 Main 5 pages
A Review of The Extended Finite Element Method On Macrocrack and Microcrack Growth Simulations 100% (1) A Review of The Extended Finite Element Method On Macrocrack and Microcrack Growth Simulations 39 pages
Falling Ball Viscometer No ratings yet Falling Ball Viscometer 17 pages
QP1 PDF No ratings yet QP1 PDF 3 pages
Engineering Lab Guide No ratings yet Engineering Lab Guide 53 pages
M320 Excavator Hydraulic System - Attachment: Item Component No 100% (1) M320 Excavator Hydraulic System - Attachment: Item Component No 2 pages
Report Mom Torsion) No ratings yet Report Mom Torsion) 4 pages
AE 3210 Week 7 Fanno FLow Revised No ratings yet AE 3210 Week 7 Fanno FLow Revised 22 pages
Truncated Perfect Nozzles in Optimum Nozzle Design No ratings yet Truncated Perfect Nozzles in Optimum Nozzle Design 7 pages
Unit I No ratings yet Unit I 5 pages
Calibration and Use of Pressure and Temperature Measuring Instruments 67% (3) Calibration and Use of Pressure and Temperature Measuring Instruments 52 pages
7100 Formula Sheet No ratings yet 7100 Formula Sheet 1 page
THERMODYNAMICS 1 Module 6 No ratings yet THERMODYNAMICS 1 Module 6 39 pages
Greggdrilling Webinars HnLod wT8M4g No ratings yet Greggdrilling Webinars HnLod wT8M4g 22 pages
Chevron Gray Pump Manual 100% (3) Chevron Gray Pump Manual 595 pages
Hydraulic Valve Specs & Features No ratings yet Hydraulic Valve Specs & Features 6 pages
EC4-lateral Torsional Buckling No ratings yet EC4-lateral Torsional Buckling 6 pages
Tbi11p01 Mep CA 009 Calculation For Hvac Ra Ifr No ratings yet Tbi11p01 Mep CA 009 Calculation For Hvac Ra Ifr 10 pages
Earthquake Study Worksheet No ratings yet Earthquake Study Worksheet 1 page
0089 - Bourdon Effect: Description No ratings yet 0089 - Bourdon Effect: Description 3 pages
Practical 7 No ratings yet Practical 7 2 pages
Diesel Engine No ratings yet Diesel Engine 4 pages
CIVL 3740 Spring Term 2022 Mid Term Exam No ratings yet CIVL 3740 Spring Term 2022 Mid Term Exam 3 pages
Short Notes Thermodynamics No ratings yet Short Notes Thermodynamics 23 pages
Innovative Composite Concrete Filled Plastic Tubes in Compression No ratings yet Innovative Composite Concrete Filled Plastic Tubes in Compression 15 pages
ad
Footer menu
Back to top
About
About Scribd, Inc.
Everand: Ebooks & Audiobooks
Slideshare
Join our team!
Contact us
Support
Help / FAQ
Accessibility
Purchase help
AdChoices
Legal
Terms
Privacy
Copyright
Cookie Preferences
Do not sell or share my personal information
Social
Instagram Instagram
Facebook Facebook
Pinterest Pinterest
Get our free apps
About
About Scribd, Inc.
Everand: Ebooks & Audiobooks
Slideshare
Join our team!
Contact us
Legal
Terms
Privacy
Copyright
Cookie Preferences
Do not sell or share my personal information
Support
Help / FAQ
Accessibility
Purchase help
AdChoices
Social
Instagram Instagram
Facebook Facebook
Pinterest Pinterest
Get our free apps
Documents
Language:
English
Copyright © 2025 Scribd Inc.
We take content rights seriously. Learn more in our FAQs or report infringement here.
We take content rights seriously. Learn more in our FAQs or report infringement here.
Language:
English
Copyright © 2025 Scribd Inc.
576648e32a3d8b82ca71961b7a986505
scribd.scribd.scribd.scribd.scribd.scribd.scribd. |
189246 | https://study.com/academy/lesson/sieve-of-eratosthenes-lesson-for-kids.html | Sieve of Eratosthenes: Lesson for Kids | Study.com
Log In
Sign Up
Menu
Plans
Courses
By Subject
College Courses
High School Courses
Middle School Courses
Elementary School Courses
By Subject
Arts
Business
Computer Science
Education & Teaching
English (ELA)
Foreign Language
Health & Medicine
History
Humanities
Math
Psychology
Science
Social Science
Subjects
Art
Business
Computer Science
Education & Teaching
English
Health & Medicine
History
Humanities
Math
Psychology
Science
Social Science
Art
Architecture
Art History
Design
Performing Arts
Visual Arts
Business
Accounting
Business Administration
Business Communication
Business Ethics
Business Intelligence
Business Law
Economics
Finance
Healthcare Administration
Human Resources
Information Technology
International Business
Operations Management
Real Estate
Sales & Marketing
Computer Science
Computer Engineering
Computer Programming
Cybersecurity
Data Science
Software
Education & Teaching
Education Law & Policy
Pedagogy & Teaching Strategies
Special & Specialized Education
Student Support in Education
Teaching English Language Learners
English
Grammar
Literature
Public Speaking
Reading
Vocabulary
Writing & Composition
Health & Medicine
Counseling & Therapy
Health
Medicine
Nursing
Nutrition
History
US History
World History
Humanities
Communication
Ethics
Foreign Languages
Philosophy
Religious Studies
Math
Algebra
Basic Math
Calculus
Geometry
Statistics
Trigonometry
Psychology
Clinical & Abnormal Psychology
Cognitive Science
Developmental Psychology
Educational Psychology
Organizational Psychology
Social Psychology
Science
Anatomy & Physiology
Astronomy
Biology
Chemistry
Earth Science
Engineering
Environmental Science
Physics
Scientific Research
Social Science
Anthropology
Criminal Justice
Geography
Law
Linguistics
Political Science
Sociology
Teachers
Teacher Certification
Teaching Resources and Curriculum
Skills Practice
Lesson Plans
Teacher Professional Development
For schools & districts
Certifications
Teacher Certification Exams
Nursing Exams
Real Estate Exams
Military Exams
Finance Exams
Human Resources Exams
Counseling & Social Work Exams
Allied Health & Medicine Exams
All Test Prep
Teacher Certification Exams
Praxis Test Prep
FTCE Test Prep
TExES Test Prep
CSET & CBEST Test Prep
All Teacher Certification Test Prep
Nursing Exams
NCLEX Test Prep
TEAS Test Prep
HESI Test Prep
All Nursing Test Prep
Real Estate Exams
Real Estate Sales
Real Estate Brokers
Real Estate Appraisals
All Real Estate Test Prep
Military Exams
ASVAB Test Prep
AFOQT Test Prep
All Military Test Prep
Finance Exams
SIE Test Prep
Series 6 Test Prep
Series 65 Test Prep
Series 66 Test Prep
Series 7 Test Prep
CPP Test Prep
CMA Test Prep
All Finance Test Prep
Human Resources Exams
SHRM Test Prep
PHR Test Prep
aPHR Test Prep
PHRi Test Prep
SPHR Test Prep
All HR Test Prep
Counseling & Social Work Exams
NCE Test Prep
NCMHCE Test Prep
CPCE Test Prep
ASWB Test Prep
CRC Test Prep
All Counseling & Social Work Test Prep
Allied Health & Medicine Exams
ASCP Test Prep
CNA Test Prep
CNS Test Prep
All Medical Test Prep
College Degrees
College Credit Courses
Partner Schools
Success Stories
Earn credit
Sign Up
Copyright
Math Courses
Math for Kids
Sieve of Eratosthenes: Lesson for Kids
Contributors: Nick Rogers
Instructor
NR
Instructor:
Nick Rogers
Show more
Cite this lesson
Prime numbers are important to mathematicians, and there are many ways to find them. The sieve of Eratosthenes is one of the first that was ever developed, which you'll explore here.
Table of Contents
Eratosthenes and His Sieve
Factors
Prime Numbers
How the Sieve Works
Lesson Summary
Show
Create an account
LessonQuizCourse
QuizCourse3.6K views
Eratosthenes and His Sieve --------------------------
Eratosthenes was a Greek scientist who lived nearly 2,000 years ago, studied geography, and made some of the first maps of the entire known world. He is credited with inventing a method for finding prime numbers known as a sieve. Lets review some information about factors and prime numbers that can help us understand his sieve.
Factors -------
Numbers are made up of other numbers called factors. We can think of factors as the building blocks of a number. Consider the number 12, which can be written as 3 x 4, or 6 x 2, or 12 x 1. The factors of 12 are 1, 2, 3, 4, 6, and 12. Numbers always have themselves and '1' as factors since any number can be written as itself times 1.
One reason that it is useful to break a number into its factors is because it makes multiplication easier. If we multiply 12 x 36, we can use the fact that 12 = 3 x 4, 36 = 6 x 6, and 6 = 3 x 2 to see that:
12 x 36 = (3 x 4) x (6 x 6)
12 x 36 = 3 x 4 x 3 x 2 x 3 x 2
Do you see how we've just written a complicated multiplication problem as many small problems? This is one of the benefits of factoring.
Prime Numbers -------------
Prime numbers are numbers that can be divided cleanly by only themselves and 1. The first few prime numbers are 2, 3, 5, and 7. Can you find some more? Prime numbers are the essential building blocks of all numbers.
If we try to break apart the multiplication problem 7 x 13 as we did above, we will not be able to. This is because 7 and 13 are both prime numbers. There is no way to simplify them any further. In fact, any number can always be written as a product of primes. Why not try some out for yourself?
How the Sieve Works -------------------
Sieves or 'sifters' are typically used for separating solid materials from liquids. They usually have a metal mesh, and you may have seen one in a kitchen. The sieve of Eratosthenes works in much the same way - except instead of pouring liquid through it, you pour numbers. In order to apply the sieve, just follow these simple steps:
Begin with a large set of numbers that you would like to remove non-primes from, and put them in order.
Take the first non-prime that you find (such as 2), and remove all of the multiples of that number from the set ( 4, 6, 8...).
Repeat the second step with the next prime number.
Example
Sounds pretty simple right? Let's try an example. Suppose we want to find all of the primes in the numbers from 1 to 15. We'll begin with {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}.
Start by removing 1, since it is not considered a prime. Now remove all of the multiples of 2 because if something is a multiple of 2, it cannot be a prime. You'll have: {2, 3, 5, 7, 9, 11, 13, 15}.
Now remove all the multiples of 3, but leave 3 alone: {2, 3, 5, 7, 11, 13, 15}.
Now remove all of the multiples of 5. There is only one: 15. This leaves you with: {2, 3, 5, 11, 13}.
Since there are no multiples of 11 or 13, these are all of the primes that are less than 15. The sieve method can be used on much larger sets of numbers as well; it works perfectly no matter how many numbers you begin with.
Bonus Question: Does it matter if we put the numbers in order first? What would happen if we didn't?
Lesson Summary --------------
In this lesson, you learned about using the sieve of Eratosthenes, a method for finding prime numbers. We also discussed factors and how numbers can be broken down as the product of smaller numbers. Any number always has itself and '1' as factors.
Register to view this lesson
Are you a student or a teacher?
I am a student I am a teacher
Unlock your education
See for yourself why 30 million people use Study.com
Become a Study.com member and start learning now.
Become a member
Already a member? Log in
Go back
Resources created by teachers for teachers
Over 30,000 video lessons & teaching resources—all in one place.
Video lessons
Quizzes and worksheets
Classroom integration
Lesson plans
I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline.
Jennifer B.
Teacher
Try it now
Go back
Create an account to start this course today
Start today. Try it now
Math for Kids
23 chapters
325 lessons
Chapter 1
Numbers for Elementary School
Ordinal Numbers: Lesson for Kids 3:30 min
Binary Numbers: Lesson for Kids
Triangular Numbers: Lesson for Kids
Prime Numbers: Lesson for Kids
Whole Numbers: Lesson for Kids 2:57 min
Pi Lesson for Kids Facts & History
Composite Number Lesson for Kids
Fibonacci Numbers in Nature: Lesson for Kids
Fibonacci Sequence Lesson for Kids: Numbers & Facts
Number Line: Lesson for Kids
Place Value Games for Kids
Place Value: Lesson for Kids 2:48 min
Roman Numerals: Lesson for Kids 5:59 min
Rounding Numbers: Lesson for Kids
Digit in Math | Meaning & Examples 3:30 min
9/4 as a Mixed Number: How-to & Steps 2:49 min
Place Value and Patterns with Rounding
Number Line for Addition & Subtraction | Overview & Examples 2:57 min
How to Use a Number Line
Counting on the Number Line
Numbers 1-50: Lesson for Kids 3:17 min
Numbers 1-20: Lesson for Kids
Numbers 1-100: Lesson for Kids 2:53 min
Numbers 1-30: Lesson for Kids
How to Count: Lesson for Kids
Chapter 2
Geometry for Elementary School
Surface Area Lesson for Kids 3:58 min
Types of Angles: Lesson for Kids 3:17 min
Types of Triangles: Lesson for Kids
Geometric Shapes: Lesson for Kids
Isosceles Triangle Lesson for Kids
How to Use a Protractor: Lesson for Kids
3-D Shapes: Lesson for Kids 2:59 min
Pentagon Shapes: Lesson for Kids 2:16 min
Polygon Shapes: Lesson for Kids
Diamond Shapes: Lesson for Kids 2:42 min
Measuring Angles: Lesson for Kids
Circle Definition: Lesson for Kids 3:10 min
Circumference Lesson for Kids: Definition & Formula
Cubes: Lesson for Kids
Properties of a Circle: Lesson for Kids 2:21 min
Rectangle Definition: Lesson for Kids
Triangle Lesson for Kids: Definition & Facts 3:06 min
2-D Shapes: Lesson for Kids
How to Find the Area of a Triangle: Lesson for Kids 3:01 min
How to Find the Volume of a Cylinder: Lesson for Kids
Pentagon Shape Facts: Lesson for Kids 2:39 min
Vertices: Lesson for Kids 3:22 min
What is a Regular Polygon? - Lesson for Kids 3:01 min
Pythagorean Theorem Proof: Lesson for Kids
Pythagorean Theorem Lesson for Kids: Definition & Examples 4:06 min
Perimeter of a Pentagon | Overview, Formula & Examples 2:43 min
Pentagon Polygon | Definition, Types & Examples 3:31 min
Geometry Definition: Lesson for Kids
Names of Polygons 3:18 min
What is a Venn Diagram? - Lesson for Kids 3:32 min
Triangle Sum Theorem | Definition, Proof & Diagrams
Facts About Right Angles 3:09 min
Octagon Lesson for Kids: Definition & Facts 2:38 min
Hexagon Lesson for Kids: Definition & Facts 2:29 min
Perimeter Lesson for Kids: Definition & Examples 3:42 min
4-Sided Polygons | Overview & Shapes 3:55 min
Dodecahedron | Definition, Faces & Examples 2:46 min
3-4-5 Triangle | Definition, Rules & Examples 4:11 min
Triangle Sum Theorem | Definition & Examples
Dodecahedron | Definition, Properties & Examples
How to Calculate the Area of a Hexagon 3:38 min
How to Divide an Angle into Two Equal Angles
What is a Congruent Angle? - Lesson for Kids
Chapter 3
Fractions for Elementary School
Comparing Fractions: Lesson for Kids
Equivalent Fractions: Lesson for Kids 3:21 min
How to Add Fractions: Lesson for Kids
Fractions Games for Kids
Fractions to Decimals: Lesson for Kids 4:13 min
Fractions: Lesson for Kids 2:39 min
Numerator & Denominator Lesson for Kids 3:16 min
Decimals: Lesson for Kids 3:42 min
Multiplying Fractions by Whole Numbers: Lesson for Kids 3:04 min
How to Simplify Fractions: Lesson for Kids
Improper Fractions: Lesson for Kids
Benchmark Fractions | Definition, Uses & Examples 3:48 min
Adding Mixed Numbers | Steps & Examples
Multiplying Compound Fractions
Simplifying Compound Fractions
Dividing Improper Fractions | Keep-Change-Flip Method & Examples 4:21 min
Adding Compound Fractions
Subtracting Compound Fractions
How to Subtract Fractions with Variables | Examples 5:55 min
Multiplying Fractions with Common Denominators | Steps & Examples 5:05 min
Subtracting Fractions with Like Denominators
Reducing Fractions: Rules & Practice
Definition of Simplest Form: Lesson for Kids
Adding Fractions with Unlike Denominators | Overview & Examples 4:51 min
Adding & Subtracting Fractions with Unlike Denominators 3:50 min
Subtracting Fractions | Rules & Examples
Rules for Multiplying Fractions | Steps & Examples 4:03 min
Rules for Dividing Fractions
Plotting Fractions on a Number Line | Methods & Examples
Locating Fractions on a Number Line
How to Add & Subtract Two Fractions with Like Denominators
How to Find Equivalent Fractions on a Number Line 5:22 min
How to Subtract Mixed Fractions with Unlike Denominators
Chapter 4
Math Basics for Elementary School
Subtraction: Lesson for Kids
Expanded Form: Lesson for Kids 3:26 min
Logic Problems: Lesson for Kids
How to Do Long Multiplication: Lesson for Kids
What is Multiplication? - Lesson for Kids
Addition: Lesson for Kids 3:30 min
Multiplication Properties Lesson for Kids
Math Card Games for Kids
Telling Time: Activities & Games for Kids
Telling Time: Lesson for Kids 3:32 min
Counting Money: Lesson for Kids
Telling Time to the Minute: Lesson for Kids
Divisibility Rules for 10: Lesson for Kids
Divisibility Rules for 2: Lesson for Kids 2:51 min
Divisibility Rules for 5: Lesson for Kids
Division Lesson for Kids: Definition & Method 3:04 min
Long Division Steps: Lesson for Kids 3:44 min
Estimation: Lesson for Kids 3:44 min
Lattice Method of Multiplication | Definition, Steps & Examples 3:50 min
Multiplication Methods & Types 3:52 min
Ways to Divide & Types of Division | Steps & Examples 3:47 min
Theorem | Meaning, Types & Examples 3:11 min
Subtraction with Regrouping | Overview & Examples 3:01 min
Dividing a Whole Number by a Decimal | Steps & Examples 4:46 min
Regrouping in Addition | Definition & Examples 3:09 min
Finding the Missing Addend 3:06 min
Metric vs. Imperial System | Units & Measurement 3:46 min
History of the Metric System: Lesson for Kids
How Money is Made: Lesson for Kids 3:54 min
Chinese Multiplication Method | Overview & Examples 3:46 min
Column Addition Method | Concept & Examples
4 Digit Subtraction with Regrouping 4:19 min
How to Multiply Using Expanded Form 3:52 min
Writing Numbers in Expanded Form | Overview & Examples 3:58 min
Famous Mathematicians: Lesson for Kids
Writing Decimals in Expanded Form | Steps & Examples 3:30 min
Length Lesson for Kids: Definition & Measurement 3:12 min
4 Digit Addition with Regrouping
Expanded Notation Method for Division 4:37 min
Expanded Notation Method for Multiplication 3:23 min
Multiplication & Division Fact Families | Overview & Uses 3:01 min
Fact Families: Addition & Subtraction | Definition & Examples 3:32 min
4 Digit by 1 Digit Multiplication
Metric Conversions: Lesson for Kids 3:22 min
How to Divide by Double Digit Numbers 5:58 min
Two-Step Math Word Problems 4:03 min
How to Borrow in Math
Math Word Problems: Lesson for Kids
How to Learn Times Tables
Box Method Multiplication | Definition, Steps & Examples 2:29 min
4 Digit by 2 Digit Multiplication 4:06 min
Equivalent in Math | Definition, Numbers & Fractions 4:13 min
1st Grade Math Vocabulary: Lesson for Kids
Chapter 5
Statistics for Elementary School
Algorithm Lesson for Kids
Arrays: Lesson for Kids
Probability Lesson for Kids: Examples & Definition 3:52 min
What Is Range in Math? - Lesson for Kids 3:11 min
Line Graphs: Lesson for Kids 3:39 min
Different Types of Graphs: Lesson for Kids 3:37 min
Statistics: Lesson for Kids
Five Number Summary | Definition, Conditions & Calculation
Chapter 6
Number Properties for Elementary School
Associative Property Lesson for Kids 3:55 min
Commutative Property: Lesson for Kids 3:45 min
What is a Factor? - Lesson for Kids 3:38 min
What is a Prime Factor? - Lesson for Kids 3:07 min
What are Factors & Multiples? 3:29 min
Chapter 7
Algebra for Elementary School
History of Algebra: Lesson for Kids
Math Formulas: Lesson for Kids
Distributive Property: Lesson for Kids
Algebra Lesson for Kids: Facts & Rules 4:18 min
Foil Method in Math | Definition & Examples 3:26 min
Chapter 8
Math Patterns for Elementary School
Fractals Lesson for Kids
Fractal Art | Meaning, Types & Uses
Semi-Regular Tessellation | Definition, Types & Examples
Tessellation Shapes, Patterns & Examples 3:39 min
Patterns in Math | Overview, Rule & Types 3:36 min
Pascal's Triangle Lesson for Kids: Definition & History
Chapter 9
History of Math for Elementary School
Al Khwarizmi Lesson for Kids: Biography & Facts
Abacus Lesson for Kids: History & Uses
Pythagoras Lesson for Kids: Biography & Facts
Chapter 10
Math Terms for Elementary School
Definition of Like Denominators
Attribute in Math | Definition, Shape & Examples
What Is Breadth in Math? 3:01 min
Chord in Math | Definition, Theorems & Calculation
Cluster in Math | Overview & Examples 2:36 min
Difference in Math | Overview, Subtraction & Examples 2:17 min
What is Dimension in Math? | Concept and Examples 2:09 min
Edges in Math | Definition, Identification & Examples 2:20 min
Elements of a Set | Definition & Examples 3:06 min
Equation Lesson for Kids: Definition & Examples 3:49 min
Exponent Definition: Lesson for Kids
Face in Math | Overview & Shapes
Function in Math | Definition, Types & Examples 2:15 min
Infinity in Math | Definition, Symbol & Signs 2:52 min
Integers Lesson for Kids
Interval in Math | Definition & Examples 3:40 min
Median Definition: Lesson for Kids
What Does Mode Mean in Math? - Lesson for Kids
Net in Math | Definition & Examples
Outcome in Math | Definition & Events
Point in Math | Definition, Uses & Examples 2:14 min
Proper Factor in Mathematics | Overview, Facts & Examples
Proportion Lesson for Kids: Definition & Examples
Quantity in Math | Definition, Uses & Examples 2:50 min
Quotient Lesson for Kids: Definition & Rule
Revolution in Math | Definition, Types & Angles
5th Grade Math Vocabulary: Lesson for Kids
4th Grade Math Vocabulary: Lesson for Kids
3rd Grade Math Vocabulary: Lesson for Kids
2nd Grade Math Vocabulary: Lesson for Kids
Chapter 11
Working with Numbers for Elementary School
Standard Algorithm for Addition 3:32 min
Standard Algorithm in Math | Meaning & Examples 2:53 min
Standard Algorithm for Division 3:25 min
Absolute Value: Lesson for Kids
Ascending Order Definition & Examples 3:07 min
Decomposing in Math | Definition, Process & Examples 3:24 min
Descending Order | Definition, Numbers & Examples 3:35 min
The Fundamental Theorem of Arithmetic 4:03 min
Golden Mean Definition, Uses & Examples
Power of a Number | Overview & Examples 3:38 min
Logical Thinking & Reasoning Questions: Lesson for Kids 2:51 min
Vertical Addition & Subtraction 3:39 min
Scale Factor Lesson for Kids: Definition & Examples
Square Roots: Lesson for Kids
Chapter 12
Types of Numbers for Elementary School
What is an Abundant Number?
What is a Base Number? 2:32 min
Cardinal Numbers | Overview, Definition & Examples 3:06 min
Consecutive Numbers | Definition, Example & Formulas 3:22 min
Finding the Sum of Consecutive Numbers 4:10 min
What is a Cubed Number?
Deficient Numbers | Definition, Properties & Examples
What Are Figurate Numbers? - Definition & Examples
Opposite Numbers | Definition & Facts 2:53 min
Rectangular Numbers | Definition, Properties & Types 3:05 min
Like & Unlike Terms
Is Zero an Integer? 2:17 min
What is 0? | Definition & Types 2:37 min
Viewing now Sieve of Eratosthenes: Lesson for Kids
Up next What Are Twin Prime Numbers? 3:22 min Watch next lesson
Chapter 13
Measurements for Elementary School
Acre | Definition, Area & Measurement 3:10 min
Capacity Lesson for Kids: Definition & Facts 2:25 min
Centigrade Definition, Conversion & Facts
Centimeter Definition, Symbol & Conversion 2:06 min
Cubic Centimeter | Definition & Conversion 3:20 min
Cubic Meter | Definition, Formula & Conversion 4:53 min
Hectare Definition, History & Conversion 4:56 min
Hectometer | Overview, Conversion & Examples 3:03 min
Grams & Kilograms: Lesson for Kids 2:43 min
Kilometer | Definition, Measurement & Examples 3:23 min
Meter in Math | Definition, Conversion & Examples 3:43 min
Milliliter Definition, Abbreviation & Conversion 2:58 min
Millimeter | Meaning, Conversion & Measurement 3:12 min
Ounce | oz Meaning & Examples 2:43 min
Pint Definition, Measurement & Conversion 2:13 min
Pound | Definition & Measurement 3:21 min
Quart | Definition, Measurement & Conversion 2:43 min
Volume & Capacity: Lesson for Kids 4:19 min
International Date Line Lesson for Kids: Definition & Facts
Chapter 14
Working with Data for Elementary School
Axis Definition: Lesson for Kids
Box & Whisker Plot: Lesson for Kids
What is a Carroll Diagram? - Definition & Examples
What is a Class Interval? - Definition & Example 3:45 min
How to Find a Class Interval 3:00 min
What is a Column Graph? - Definition & Example 2:42 min
Frequency Definition: Lesson for Kids 2:27 min
Frequency Distribution in Statistics | Definition & Examples 3:57 min
Frequency Distribution in Statistics | Table & Examples 3:05 min
Frequency Histogram | Parts & Calculation 3:12 min
Frequency Histogram | Definition, Purpose & Examples 3:57 min
Histogram Lesson for Kids
Ogive | Definition, Graph & Examples 3:33 min
Ratios Lesson for Kids: Definition & Examples
Sets in Math | Symbols, Definition & Examples 3:02 min
Chapter 15
Representing Numbers for Elementary School
Bar Notation Overview & Examples | What Does a Line Over a Number Mean? 2:54 min
Base Ten Blocks | Definition, Names & Examples 3:57 min
Base Ten System, Chart & Examples 3:21 min
Fractional Notation | Conversion & Examples 3:57 min
Hexadecimal | Definition, System & Examples
Creating Graphs: Lesson for Kids
What is a Sector Graph?
Stem-and-Leaf Plots with Decimals | Definition, Steps & Examples 4:00 min
Stem and Leaf Plot | Definition, Steps & Examples
3-Digit Stem-and-Leaf Plots 3:40 min
Stem-and-Leaf Display | Plot, Graph & Diagram 3:38 min
Chapter 16
Types of Data for Elementary School
Categorical Data Lesson for Kids: Definition & Examples
Census Lesson for Kids
Discrete Data in Math | Definition, Representation & Examples 3:27 min
Population Lesson for Kids: Definition & Facts
Qualitative Data | Definition & Examples 3:22 min
Triple Venn Diagram Overview, Uses & Examples 2:56 min
Chapter 17
Math Strategies for Elementary School
Jump Strategy Definition, Uses & Steps
Solving a 3x3 Magic Square | Overview, Formula & Examples 3:29 min
Magic Squares in Math | History & Examples 3:05 min
Trigonometry: Lesson for Kids
Chapter 18
Fraction Operations for Elementary School
Multiply Fractions with Different Denominators | Steps & Examples 3:38 min
Dividing Fractions with Unlike Denominators | Overview & Examples 4:56 min
Comparing Fractions with Unlike Denominators 3:35 min
How to Add Fractions & Whole Numbers 3:33 min
Order of Operations with Fractions | PEMDAS, Examples & Practice 3:39 min
Solving One-Step Equations with Fractions | Methods & Examples
Chapter 19
Shapes for Elementary School
Vertex Angle of an Isosceles Triangle | Definition & Examples 3:50 min
Vertex Angles of a Kite: Lesson for Kids
Geometry in Nature | Shapes, Types & Examples 3:13 min
Perimeter & Area of a Square | Definition, Formula & Relationship 3:10 min
Geometric Shapes in The Real World | Types & Examples 2:36 min
Octagon Shape | Area & Angles 3:29 min
Right Angle Shapes | Overview, Types & Examples 2:53 min
Surface Area of Composite Figures 4:27 min
Rotational & Radial Symmetry: Lesson for Kids
Cross Section Overview & Examples 2:07 min
Chapter 20
Negative Numbers for Elementary School
Zero Pair in Math | Definition & Examples 2:46 min
History of Negative Numbers | Invention & Uses 3:07 min
Subtraction of Positive & Negative Numbers 4:34 min
Chapter 21
Decimals for Elementary School
Long Division with Decimals | Steps & Examples 3:43 min
Dividing Decimals by Whole Numbers
Comparing & Ordering Fractions, Decimals & Percents 4:16 min
Dividing Decimals by Decimals
Chapter 22
Lines & Angles for Elementary School
Line Segment: Lesson for Kids 2:54 min
Angles Formed by Intersecting Lines | Overview, Types & Examples 3:42 min
Coincident Lines Definition, Conditions & Examples 2:46 min
What is an Oblique Line? 2:35 min
Chapter 23
Multiplication for Elementary School
Three Digit Multiplication
Three Digit Multiplication Word Problems 5:19 min
Multiplication with Lines | Steps & Examples 5:51 min
Related Study Materials
Sieve of Eratosthenes: Lesson for Kids
CoursesTopics
##### MEGA Elementary Education Study Guide and Test Prep
##### Math Review for Teachers: Study Guide & Help
##### Common Core Math - Number & Quantity: High School Standards
##### Common Core Math - Algebra: High School Standards
##### Common Core Math - Statistics & Probability: High School Standards
##### Common Core Math - Geometry: High School Standards
##### Common Core Math - Functions: High School Standards
##### Common Core Math Grade 8 - Expressions & Equations: Standards
##### Common Core Math Grade 8 - Functions: Standards
##### MEGA Elementary Education Study Guide and Test Prep
##### Math Review for Teachers: Study Guide & Help
##### Common Core Math - Number & Quantity: High School Standards
##### Common Core Math - Algebra: High School Standards
##### Common Core Math - Statistics & Probability: High School Standards
##### Common Core Math - Geometry: High School Standards
##### Common Core Math - Functions: High School Standards
##### Common Core Math Grade 8 - Expressions & Equations: Standards
##### Common Core Math Grade 8 - Functions: Standards
Browse by Courses
CLEP College Mathematics Study Guide and Exam Prep
CAHSEE Math Exam: Test Prep & Study Guide
TExES Mathematics 7-12 (235) Study Guide and Test Prep
Common Core Math Grade 8 - Functions: Standards
Algebra I: High School
Study.com SAT Study Guide and Test Prep
Geometry: High School
Common Core Math - Geometry: High School Standards
Common Core Math - Functions: High School Standards
Math 104: Calculus
Math 102: College Mathematics
Math 103: Precalculus
GED Math: Quantitative, Arithmetic & Algebraic Problem Solving
CLEP Calculus Study Guide and Exam Prep
Contemporary Math
Browse by Lessons
Sieve of Eratosthenes | History, Application & Examples
Proper Factor in Mathematics | Overview, Facts & Examples
Relatively Prime Numbers | Definition, List & Examples
52 as a Product of Prime Factors: Steps & How-to
Prime Numbers Lesson Plan
Factor Pairs | Definition, Facts & Examples
Finding the Prime Factorization with Exponents
Composite Number | Definition, Types & Examples
Prime Factorization Lesson Plan
Prime & Composite Numbers | Differences & Identification
What is a Prime Factor? - Lesson for Kids
Finding Common Factors of 2 Given Numbers: Lesson for Kids
Factors and Multiples Lesson Plan
Finding Factors of Whole Numbers
Finding the Prime Factorization of a Number | Meaning & Examples
Browse by Courses
CLEP College Mathematics Study Guide and Exam Prep
CAHSEE Math Exam: Test Prep & Study Guide
TExES Mathematics 7-12 (235) Study Guide and Test Prep
Common Core Math Grade 8 - Functions: Standards
Algebra I: High School
Study.com SAT Study Guide and Test Prep
Geometry: High School
Common Core Math - Geometry: High School Standards
Common Core Math - Functions: High School Standards
Math 104: Calculus
Math 102: College Mathematics
Math 103: Precalculus
GED Math: Quantitative, Arithmetic & Algebraic Problem Solving
CLEP Calculus Study Guide and Exam Prep
Contemporary Math
Browse by Lessons
Sieve of Eratosthenes | History, Application & Examples
Proper Factor in Mathematics | Overview, Facts & Examples
Relatively Prime Numbers | Definition, List & Examples
52 as a Product of Prime Factors: Steps & How-to
Prime Numbers Lesson Plan
Factor Pairs | Definition, Facts & Examples
Finding the Prime Factorization with Exponents
Composite Number | Definition, Types & Examples
Prime Factorization Lesson Plan
Prime & Composite Numbers | Differences & Identification
What is a Prime Factor? - Lesson for Kids
Finding Common Factors of 2 Given Numbers: Lesson for Kids
Factors and Multiples Lesson Plan
Finding Factors of Whole Numbers
Finding the Prime Factorization of a Number | Meaning & Examples
Create an account to start this course today
Used by over 30 million students worldwide
Create an account
Explore our library of over 88,000 lessons
Search
Browse
Browse by subject
College Courses
Business
English
Foreign Language
History
Humanities
Math
Science
Social Science
See All College Courses
High School Courses
AP
Common Core
GED
High School
See All High School Courses
Other Courses
College & Career Guidance Courses
College Placement Exams
Entrance Exams
General Test Prep
K-8 Courses
Skills Courses
Teacher Certification Exams
See All Other Courses
Study.com is an online platform offering affordable courses and study materials for K-12, college, and professional development. It enables flexible, self-paced learning.
Plans
Study Help
Test Preparation
College Credit
Teacher Resources
Working Scholars®
Online Tutoring
About us
Blog
Careers
Teach for Us
Press Center
Ambassador
Scholarships
Support
FAQ
Site Feedback
Terms of Use
Privacy Policy
DMCA Notice
ADA Compliance
Honor Code for Students
Mobile Apps
Contact us by phone at (877) 266-4919, or by mail at 100 View Street #202, Mountain View, CA 94041.
© Copyright 2025 Study.com. All other trademarks and copyrights are the property of their respective owners. All rights reserved.
× |
189247 | https://www.youtube.com/watch?v=-rGvhSBqlxg | Derivative Examples Problems | Calculus | MathSux
MathSux
2500 subscribers
8 likes
Description
719 views
Posted: 5 Jan 2022
In this video, we will go over derivative examples problems by going over the following:
-power rule
-product rule
-quotient rule
-chain rule
We use the derivative to find the rate of change of a function with respect to a variable. You can find out more about what a derivative is and its proper notation here at
🧮
✏️ Practice Questions:
🧮
Social:
🐦 Twitter:
✫Facebook:
TikTok:
📽Patreon:
🧮
Intro 0:00
What is a derivative? 0:27
Derivative Cheat Sheet 0:44
Power Rule 1:00
Product Rule 2:51
Quotient Rule 6:59
Chain Rule 12:07
Practice Questions 16:14
🧮
If you have any questions, please don’t hesitate to comment below. And if this video has helped you, please hit the like button and subscribe for more math videos! Thank you and happy calculating! :)
🧮
Hi and welcome to my Youtube channel, Mathsux. I’m Laura a Math Teacher/Tutor with a Masters in Mathematics Education. I will be covering topics in Algebra, Geometry, Algebra 2, Trigonometry, and Statistics every Wednesday with complimentary practice questions for every video. I hope this channel makes math suck just a little bit less and make life (or math class) just a little bit easier. And if these videos help you, please like and subscribe. Happy calculating! :)
🧮
Music:
Afternoon Tea (extended) by Mona Wonderlick Creative Commons — Attribution 3.0 Unported — CC BY 3.0 Free Download / Stream: Music promoted by Audio Library
mathsux #Calculus
1 comments
Transcript:
Intro hi everyone and welcome to math sucks this video is going to help you pass calculus in this video we're going to go over derivative example problems by going over the power rule the product rule the quotient rule the chain rule then you can try the practice problems at the end of this video i also have a derivatives rules cheat sheet that you can download on my blog mathsocks.org so be sure to check that out at the end of this video i'll put all the links in the description below so what is the derivative we use the derivative to find What is a derivative? the rate of change of a function with respect to our variable so you can find out more about what that means and its proper notation here on this website mathisfun.com i'll put that in the link in the description below as well before we begin here's just a quick Derivative Cheat Sheet overview of all the derivative rules we're gonna go over and we're gonna have three examples of each so don't worry if these don't make sense to you just now and yeah you could totally download this on the cheat sheet below so now let's look at some examples first let's start with the power rule Power Rule this is the easiest rule this happens when we have x raised to a power so a variable raised to a power and then we bring down that power multiply it by the variable and then subtract that original exponent so we have three examples here let's look at the first one we have f of x is equal to x to the fourth so when we apply the power rule and find the derivative of this function we're going to get f prime of x is equal to now we're just going to bring down that exponent 4 times x the variable and then we're just going to subtract 1 from that exponent so this becomes 4 minus 1 which is just 3 and that's our answer so we're going to just do the same pattern with the second example we have g of x is equal to 3x squared plus 1. so this time we're going to write g prime of x and then we're going to do the same thing we're going to bring down 2 but this time we have a 3 in front of our variable so we're just going to multiply 2 times 3 which gives us 6 leaving us with x and we're going to subtract this exponent 2 minus 1 it'll just give us 1 so we're just going to leave that the same whenever we have whole numbers we're just going to they're just we're going to become 0. so this will be our final answer for number two and the last one for the power rule here we have h of x equal to five x to the fourth plus two x minus one so we're gonna just do the same thing and we're gonna get h prime of x is equal to 20 x cubed plus two and remember when we have a whole number it just zeros out so this is our answer Product Rule up next we have the product rule so that's when we have two numbers or variables being multiplied together and then what we do is we find the the derivative of one times the original and then we're going to add it to the derivative times the other times the original of the other one so so let's look at how this works so first for number one we have x to the fourth y to the fifth so let's just say this is equal to f of x and let's say this is equal to g of y so we have f of x is equal to x to the fourth and then f prime or the derivative of f is equal to 4x cubed and then when we find g of y we're just saying that's equal to the other part of this which is y to the fifth and then the derivative of g to the y is going to be five y to the fourth so these are all the pieces to the puzzle that we're going to need to fill this in so so now um you can do this in your head but if you're just starting out you're going to want to follow this formula here for the product rule so we're going to look at f so we're going to leave and now we're going to fill in our formula so we have f which is x to the fourth g prime which is 5 y to the fourth and then plus f prime for x cubed times g which is y to the fifth so you can leave that or you can realize maybe you want to move this 5 outside so i'm just going to do that to make it nice and pretty 5x to the fourth y to the fourth plus four x cubed y to the fifth and that's our answer so okay let's try another one so here is our f of x here's our g of y and let's find them on the side here we have f of x equals x cubed f prime of x is equal to three x squared and then we have g of y which is y cubed so this is a very similar problem g y g prime of y is equal to three y squared and now let's just so we have other pieces let's look again at our product rule so we have f which is x cubed x cubed times g of g prime of y which is three y squared plus f prime which is three x squared times g which is y cubed so again i'm just going to move out that 3 to make this nice and pretty 3 x cubed y squared plus 3x squared y cubed and that's our answer for the second example so last one so this one looks a little different so let's see how this works out so we have f of x is equal to x plus one f prime of x is equal to one so we just only have that one variable here so we only get one g of x is equal to x squared plus three g prime of x is equal to 2x so we're just using that power rule that we learned with that first derivative rule so now let's follow along with our formula so we get f which is x plus one times g prime which is two x plus f prime which is just one times g which is x squared plus 3. so notice we can distribute it and do a little algebra here to clean this up so i'm just going to do that 2x squared plus 2x and then plus x squared plus 3. notice we can combine 2x squared and x squared and this will give us 3x squared plus 2x plus 3. and that's our answer onto the quotient rule so here we have Quotient Rule you know we get a ratio a fraction uh f to g and then we want to to get the derivative of f over g we're gonna take the derivative of f times g minus the derivative of g times f all over g squared so again you know these rules can look intimidating but once you get f and f prime and g and g prime you're just filling in it so let's say that our three is our f and x plus one is our g so we get f of x equals three f prime of x is equal to zero this time because we only have that whole number three g of x is equal to x plus one g prime of x is equal to just one and now we're just going to fill this in so we have f prime zero times g which is x plus one even though it doesn't matter because we're gonna have a zero and this is going to become 0 minus g prime which is 1 times f of x which is 3 all over g squared so g is x plus 1 and we're just going to square that so this is going to give us 0 minus 3 which is just negative 3 all over x plus 1 squared so remember that we just remembering notation that we found d dx of this function here which is negative 3 over x plus 1 squared so let's try another one so we have f this is our g so we have f of x is equal to x squared f prime of x is equal to two x g of x is equal to x plus two g prime of x is just the derivative of that which is one and now we have everything we need and let's plug it plug our stuff into this so we have f prime is two x times g which is x plus two minus g prime which is just one times f which is x squared all over g squared which is x plus 2 squared so i could stop here or you could realize we could distribute a bit around so let's do that we got 2x squared plus 4x minus x squared over x plus 2 squared so again we can combine these two 2x squared minus x squared will just give us x squared plus 4x all over x plus 2 squared so again we can simplify this a little bit more so let's just see um we can take out an x from the top here so when we do that we'll get x on the outside and then x plus 4 on the inside all over x plus 2 squared and we can't reduce this anymore so this is our final answer here okay looking good okay so next up we have the quotient rule again this time we're doing a trig function with cos of x over x so this is the first time we're doing a trig function today so here's just a little reminder about um derivatives of trig functions so they're they're nothing scary they're just like something you get used to with practice that you kind of end up memorizing [Music] so we're saying that our cosine is f of x and the derivative of cosine of x is negative sine of x and now our g of x is equal to just x and g prime of x is just equal to one so now let's plug in our stuff into our quotient rule so we get f prime which is negative sine of x times g which is just x minus g prime which is just one times f which is cosine of x all over g squared or x squared so i'm just gonna so this is our answer but just to rework it so it looks nicer and a bit easier to understand i'm just gonna take out i'm just gonna put this x to the end here so we get minus x sine of x minus cosine of x all over x squared you could even rework this again if you realize we can move the cosine of x to the other side and add x sine of x because we're basically kind of negating everything and moving things around a bit over x squared so if you don't know what i just did you could just leave your answer like this that's totally okay Chain Rule last but not least we have our chain rule so there's so many different rules to remember with derivatives but as long as you recognize them they shouldn't be so hard so here's our last one the chain rule so that's when we kind of have these nested functions so we have f of g of x so that's how we recognize the chain rule and what we do here is we're going to find the derivative of the whole thing on the outside and then whatever is on the inside so that's like nesting finding the derivative on the outside times the derivative of the inside function so so let's look at our first example we have cosine of x squared another trig function here so we have f of x is equal to cosine of x and then our derivative of that is going to be negative sine of x which we just did in the last example and then our g of x is equal to x squared and g for that right just like this inside here so this is our g and then the whole outside is the f and then our g prime of x is going to be equal to 2x so now we're just going to find the f prime which is negative sine of x but instead of this x we're going to put g of x so inside we have x squared and then we're going to find g prime of x which we found here which is just 2x so remember we're finding the derivative ddx just putting that notation out there and then i'm just going to reorder this and put 2x where it belongs so it doesn't confuse you it's going to go outside that sine of x so negative 2 x sine of x squared just because we're multiplying it by sine of x we want to make sure we're aware of it so that's our answer so let's try another one this one looks a little different again f of x is equal to x minus two so we're saying that in here is f and the derivative of that is just 1 and then g of x is going to be the whole thing so this is going to be x minus 2 squared and the derivative of that is just going to be 2 times x minus 2. so we're just using that power rule for this g of x here and then we're going to do the same thing so we have f prime of g of x so f prime of g of x so that's just going to be 1. and then the next part we want to find g prime of x which we have right here which is just 2 times x minus 2. so our answer here is just 2 times x minus 2. okay we have another similar question for our last question of the chain rule and our last question for this video where f of x is going to be we're going to take use that inside again this is f so we get 4 x squared minus 2 and the derivative of that is equal to 8 x and then for our outside we have the entire thing which is going to be 4x squared minus 2 raised to the fourth power and the derivative of that is going to be 4 times 4x squared minus 2 raised to the third power so now let's again look at our rule we want f prime of g of x so we have f prime right here which is just going to be 8x and then g prime of x is this whole guy right here so that's four times four x squared minus two raised to the third power and notice we can just multiply these two whole numbers together and we do that we'll get 32 x times 4x squared minus 2 raised to the third power and that's our answer Practice Questions so if you're looking for more check out the practice questions right here the answers are up on the blog massox.org and remember i also have a derivatives cheat sheet that goes over each type of derivative that you can download for free the link is in the description below and if this video helped you please give it a thumbs up and subscribe thanks so much for stopping by and happy calculating need more practice check out mathdocs.org for more questions link below also don't forget to subscribe happy calculating |
189248 | https://study.com/learn/lesson/range-function-steps-examples.html | Range of a Function | Overview, Examples & Graph | Study.com
Log In
Sign Up
Menu
Plans
Courses
By Subject
College Courses
High School Courses
Middle School Courses
Elementary School Courses
By Subject
Arts
Business
Computer Science
Education & Teaching
English (ELA)
Foreign Language
Health & Medicine
History
Humanities
Math
Psychology
Science
Social Science
Subjects
Art
Business
Computer Science
Education & Teaching
English
Health & Medicine
History
Humanities
Math
Psychology
Science
Social Science
Art
Architecture
Art History
Design
Performing Arts
Visual Arts
Business
Accounting
Business Administration
Business Communication
Business Ethics
Business Intelligence
Business Law
Economics
Finance
Healthcare Administration
Human Resources
Information Technology
International Business
Operations Management
Real Estate
Sales & Marketing
Computer Science
Computer Engineering
Computer Programming
Cybersecurity
Data Science
Software
Education & Teaching
Education Law & Policy
Pedagogy & Teaching Strategies
Special & Specialized Education
Student Support in Education
Teaching English Language Learners
English
Grammar
Literature
Public Speaking
Reading
Vocabulary
Writing & Composition
Health & Medicine
Counseling & Therapy
Health
Medicine
Nursing
Nutrition
History
US History
World History
Humanities
Communication
Ethics
Foreign Languages
Philosophy
Religious Studies
Math
Algebra
Basic Math
Calculus
Geometry
Statistics
Trigonometry
Psychology
Clinical & Abnormal Psychology
Cognitive Science
Developmental Psychology
Educational Psychology
Organizational Psychology
Social Psychology
Science
Anatomy & Physiology
Astronomy
Biology
Chemistry
Earth Science
Engineering
Environmental Science
Physics
Scientific Research
Social Science
Anthropology
Criminal Justice
Geography
Law
Linguistics
Political Science
Sociology
Teachers
Teacher Certification
Teaching Resources and Curriculum
Skills Practice
Lesson Plans
Teacher Professional Development
For schools & districts
Certifications
Teacher Certification Exams
Nursing Exams
Real Estate Exams
Military Exams
Finance Exams
Human Resources Exams
Counseling & Social Work Exams
Allied Health & Medicine Exams
All Test Prep
Teacher Certification Exams
Praxis Test Prep
FTCE Test Prep
TExES Test Prep
CSET & CBEST Test Prep
All Teacher Certification Test Prep
Nursing Exams
NCLEX Test Prep
TEAS Test Prep
HESI Test Prep
All Nursing Test Prep
Real Estate Exams
Real Estate Sales
Real Estate Brokers
Real Estate Appraisals
All Real Estate Test Prep
Military Exams
ASVAB Test Prep
AFOQT Test Prep
All Military Test Prep
Finance Exams
SIE Test Prep
Series 6 Test Prep
Series 65 Test Prep
Series 66 Test Prep
Series 7 Test Prep
CPP Test Prep
CMA Test Prep
All Finance Test Prep
Human Resources Exams
SHRM Test Prep
PHR Test Prep
aPHR Test Prep
PHRi Test Prep
SPHR Test Prep
All HR Test Prep
Counseling & Social Work Exams
NCE Test Prep
NCMHCE Test Prep
CPCE Test Prep
ASWB Test Prep
CRC Test Prep
All Counseling & Social Work Test Prep
Allied Health & Medicine Exams
ASCP Test Prep
CNA Test Prep
CNS Test Prep
All Medical Test Prep
College Degrees
College Credit Courses
Partner Schools
Success Stories
Earn credit
Sign Up
Math Courses / Algebra I: High School Course
Range of a Function | Overview, Examples & Graph
Lesson
Additional Info
Antonette Dela Cruz, Mervin Clyde Cananes
Author Antonette Dela Cruz
Antonette Dela Cruz is a veteran teacher of Mathematics with 25 years of teaching experience. She has a bachelor’s degree in Chemical Engineering (cum laude) and a graduate degree in Business Administration (magna cum laude) from the University of the Philippines. She’s currently teaching Analysis of Functions and Trigonometry Honors at Volusia County Schools in Florida.
View bio
Instructor Mervin Clyde Cananes
I help improve Study.com page rankings by conducting keyword mapping, keyword prioritizing, and adding interlinks to Study.com pages.
View bio
Learn about functions and explore how to find the range of a function. Study examples of the range of the function and see the domain and range of a graph. Updated: 11/21/2023
Table of Contents
What Is a Function?
What Is the Range of a Function?
How to Find the Range of a Function
Lesson Summary
Show
Frequently Asked Questions
What is range in math?
Range in math is all of the function's outputs, dependent variables, or y-values. When a function is represented as a graph, we can find its range by finding the y-values where there is a graph.
How do you find the range of a function?
The range of a function is the y-values of the equation or graph. To find the range of the function graphically, inspect the graph from the bottom to the top. If the graph is continuous, the range is all real numbers. If there are holes or breaks in between, then the range does not include those y-values. Algebraically, the range of the graph is found by substituting the x-values into the equation and solving for y.
Create an account
Table of Contents
What Is a Function?
What Is the Range of a Function?
How to Find the Range of a Function
Lesson Summary
Show
What Is a Function? -------------------
In mathematics, members of a set can come together. The members of these sets are called elements. When the elements of each group are paired up together, this correspondence between the two groups is called a relation. Thus, a relation is a set of ordered pairs of elements between two sets.
A special kind of relation is called a function, where each of the elements in the first group, called the domain, corresponds to exactly one element in the second group, called the range. Thus, a function is a pairing between two sets where the elements in the domain pair to exactly one element in the range.
Lesson
Quiz
Course
76K views
What Is the Range of a Function? --------------------------------
Figure 1 shows a mapping or correspondence of two groups.
Fig. 1: Mapping or pairing up of two sets showing a function behavior.
The first group, the domain, is the set of x variables in the function. In other sciences, this set is also called the input. The second group, called therange, has the function's y-values. This can also be called dependent variables or the output. The arrows represent the pairing of elements from the first group to the second group. Note that there has to be only one arrow coming from each element of the first group for the correspondence to be labeled as a function.
Using figure 1 as a reference, the range is {55, 78, 80, 86, 89, 92}.
Presented in another way, the function may also be shown as a list of data pairings:
| Function: {(1,2),(2,5),(3,7),(4,0),(5,−3),(7,0)} |
| Not a Function: {(1,2),(1,5),(3,7),(4,0),(5,−3),(7,0)} |
In the point distribution, the range of the function is the set of y-values of each point in the list.
Therefore, for the function, the range is {-3, 0, 2, 5, 7}.
Range can also be represented by non-numbers, such as names, colors, shapes, or letters. Anything paired up in a relation, the second group's elements paired with the first group are all members of the set of y-values, hence the range. Using figure 2 as a reference, the range is {blue, yellow, white}.
Fig. 2: Correspondence between two groups (Days and Colors).
How to Find the Range of a Function -----------------------------------
The range of a function is the set of all y values or outputs. When function equations are written, the notation f(x) is commonly used. This means ''function with respect to the x variable.'' This f(x) is equal to y, and is the range of the function.
For the equation y = x, x can take the value of any real number. Since y is equal to x, the values of y will also be any real numbers.
To find the range of a function, here are the five basic steps:
Label the function as y = f(x).
Express x as a function of y.
Find all values of y where f(y) is defined.
Eliminate the values of y that will make the function undefined.
Write the range.
Example 1:
Find the range of f(x)= x+3.
Solution:
Step 1: y = x + 3.
Step 2: To isolate x, -3 is added to both sides of the equation: x = y - 3.
Step 3: For this example, y can be any value and f(y) will still be defined.
Step 4: Nothing to eliminate for this function.
Step 5: Range = all real numbers.
Example 2:
Find the range of f(x)=x 2+2.
Solution:
Step 1: y=x 2+2.
Step 2: To isolate x, -2 is added to both sides and then square root on both sides: x=y−2.
Step 3: The square root can only be greater than or equal to zero, which means y≥2.
Step 4: Since y values need to be greater than 2, all the y-values less than 2 will be eliminated.
Step 5: The range for this function is all real numbers that are greater than or equal to 2.
Domain and Range of a Graph
The equation of a function may also be presented as a graph. Figure 3 shows a graph of a function.
Fig. 3 The range of a function is the set of y-values on a graph.
In this graph, the domain is the set of all x-values, and the range is the set of all y-values. To get the domain from a graph, inspect the graph starting from the leftmost point to the rightmost point. Since the graph has no limit on both ends, the domain continues endlessly and this is symbolized by ∞, which means the x-values go on forever in that direction.
The range from the graph is the set of all y-values on the graph. The range is evaluated by inspecting the graph from the bottom part of the graph to the top part and reading the y-values. In Fig. 3, the y-values at the bottom and top are endless. Thus, the range for this particular function is all real numbers. This is once again symbolized by ∞.
Lesson Summary --------------
A function is a correspondence between two sets where there is exactly one element from the second set pairing up with the elements from the first set. A function assigns exactly one output to each of its inputs. This correspondence may be presented using set mapping, point distribution, equations, or graphs. The range of a function is the set of all paired elements from the second set. These elements are also called outputs, dependent variables, or y-values. Thus, the range of a function is the set of all its outputs.
Finding the range of a function from a mapping or point distribution is simply listing down the elements corresponding to the y component, the output, or the dependent variable. If some days of the week, for example, are paired up with colors, such as Mondays are for reds, Wednesdays are for yellows, and Fridays are for blues, the range for this pairing is the set of colors: {reds, yellows, blues}. Finding the range of a function from an equation involves figuring out the y-values that correspond to each x-value. When a function is represented as a graph, we can find its range by finding the y-values where there is a graph. Finding the range of a function from a graph is done by inspecting the y-values of the graph from the bottom to the top. Sometimes graphs start at a definite point, and at other times graphs are continuous and go to infinity. The function y = x takes an input and gives the same number as its output. Since we can plug any number into this function for x, we can get any number out. Thus, the range consists of all real numbers.
Additional Info
Functions, Inputs, and Outputs
Before getting to the range of a function let's have a quick review of functions and their inputs and outputs. To say that 'a is a function of b', is the same as saying that 'a is determined by b'. For instance, consider a drink menu at a fast food restaurant. Assume a small drink is $1.50, a medium drink is $2.50, and a large drink is $3.50. Notice that the price of the drink is determined by the size of the drink. Therefore, the price of a drink is a function of the size.
A function can be thought of as a rule that assigns a set of inputs to a set of outputs. In our example, the inputs would be the drink size, and the outputs would be the price. Therefore, our inputs are {small, medium, large}, and our outputs are {$1.50, $2.50, $3.50}.
Function Rule
In a function, it is imperative that each input is assigned to only one output. The reason for this is because if an input were assigned to two different outputs, then given the input, we could not determine the output. Given an input, we must be able to determine the output. Thus, each input of a function has only one output.
A function can be represented in different ways. We can use words to represent a function, as we did in our drink menu example. We can also use an equation, a graph, or a table to represent a function.
Different Representations of a Function
Definition of the Range of a Function
As we said, functions have inputs and outputs. The inputs are what we put in the function, and the outputs are what come out. We call the set of outputs of a function the range of a function. For example, consider the function defined by the rule that we take an input and raise it to the third power. This can be represented in equation form as y = x^3, and when this function is given input values of {-2, -1, 0, 1, 2}, we can find the corresponding outputs by plugging those inputs in for 'x' in the equation.
For instance, if we input -2, we have y = (-2)^3 = -8, so when the input is -2, the output is -8, and -8 is in our range. When we find each of the corresponding outputs to our inputs, we have our range. In our example, the range is {-8, -1, 0, 1, 8}, because these are the outputs corresponding to -2, -1, 0, 1, and 2 respectively.
Finding Range of a Function
To find the range of a function, we simply need to find the functions outputs, based on its inputs. As we mentioned, functions can be represented in different ways. Let's look at some different examples of finding the range of a function for different representations of a function.
1.First, let's consider our initial drink menu example. We described this function using words. The function's rule assigns a small drink to $1.50, a medium drink to $2.50, and a large drink to $3.50. The inputs are the drink size, and the outputs are the drink price. The range of this function is the set of all the outputs. Therefore, the range for this function is the set of all outputs, or {$1.50, $2.50, $3.50}.
2.Consider the following function represented in the table.
| Input | Output |
--- |
| -3 | -6 |
| -2 | -4 |
| -1 | -2 |
| 0 | 0 |
| 1 | 2 |
| 2 | 4 |
| 3 | 6 |
By definition the range is the set of all the outputs of a function, so to find the range, we simply list the outputs {-6, -4, -2, 0, 2, 4, 6}.
3.Observe this graph:
Example
The function represented by this graph has its x-values as its inputs, and its y-values as its outputs. That is, given an x-value, we can determine the corresponding y-value by looking at the graph. Therefore, the range of this function consists of all the y-values where the function is defined (this is where the graph has a y-value, or the y-values where there is a picture). We can see that the graph takes on y-values from 1 to 4. Therefore, the range of this function consists of all the numbers from 1 to 4, including 1 and 4.
4.Lastly, let's look at the function represented by the equation y = x^2. The inputs of this function consist of all real numbers. This function takes an input and squares it to give the output. Notice that when we square a number, we always end up with a positive number. Therefore, our range will consist of all positive real numbers. It will also contain the number 0, because if we input 0, we get 0 as an output. This can also be observed by looking at the graph of y = x^2.
Graph of y = x^2
Notice the graph has a picture for all y values greater than or equal to zero. Therefore the range consists of all real numbers greater than or equal to zero.
Lesson Summary
The range of a function is the set of all outputs of that function. To find the range of a function, we simply find the outputs of the function. It is useful to know the range of a function, because this allows us to know what values will come out of the function and what to expect. This lets us predict how certain phenomena represented by functions will behave. As we can see, knowing what the range of a function is and how to find that range is extremely useful.
Register to view this lesson
Are you a student or a teacher?
I am a student I am a teacher
Unlock Your Education
See for yourself why 30 million people use Study.com
Become a Study.com member and start learning now.
Become a Member
Already a member? Log In
Back
Resources created by teachers for teachers
Over 30,000 video lessons& teaching resources‐all in one place.
Video lessons
Quizzes & Worksheets
Classroom Integration
Lesson Plans
I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline.
Jennifer B.
Teacher
Try it now
Back
Start today. Try it now
Algebra I: High School
19 chapters | 163 lessons | 1 flashcard set
Ch 1. High School Algebra: Solving Math Word Problems
Solving Word Problems: Steps & Examples 5:54
Solving Word Problems with Multiple Steps 5:30
Restating Word Problems Using Words or Images 4:32
Personalizing a Word Problem to Increase Understanding 4:50
Ch 2. High School Algebra: Percent...
Ch 3. High School Algebra: Calculations,...
Ch 4. High School Algebra: Real...
Ch 5. High School Algebra: Exponents and...
Ch 6. High School Algebra: Properties of...
Ch 7. High School Algebra: Radical...
Ch 8. High School Algebra: Algebraic...
Ch 9. High School Algebra: Algebraic...
Ch 10. High School Algebra: Properties of...
Ch 11. High School Algebra: Working With...
Ch 12. High School Algebra: Linear...
Ch 13. High School Algebra: Factoring
Ch 14. High School Algebra: Quadratic...
Ch 15. High School Algebra: Graphing and...
Ch 16. High School Algebra: Properties of...
Ch 17. High School Algebra: Rational...
Ch 18. High School Algebra: Matrices and...
Ch 19. High School Algebra: Data, Statistics,...
Range of a Function | Overview, Examples & Graph Related Study Materials
Related Topics
Browse by Courses
Study.com ACT Study Guide and Test Prep
Study.com SAT Study Guide and Test Prep
Math Review for Teachers: Study Guide & Help
GRE Test Study Guide and Test Prep
Common Core Math - Number & Quantity: High School Standards
Geometry: High School
Common Core Math - Functions: High School Standards
Prentice Hall Geometry: Online Textbook Help
Amsco Geometry: Online Textbook Help
McDougal Littell Geometry: Online Textbook Help
Common Core Math Grade 8 - Expressions & Equations: Standards
Common Core Math Grade 8 - Functions: Standards
ELM: CSU Math Study Guide
McDougal Littell Pre-Algebra: Online Textbook Help
Math 103: Precalculus
Browse by Lessons
Domain & Range of Composite Functions | Steps & Examples
Discrete & Continuous Domains: Definition & Examples
Domain & Range Activities & Games
Finding the Domain & Range of Functions with Inequalities
Domain of a Function | Definition & Examples
Interval Notation | Definition, Rules & Examples
Domain & Range | Differences & Examples
How to Find the Domain of a Graph | Overview & Examples
Function Operations Activities
Input & Output Function | Definition, Rules & Examples
Equivalence Relation | Definition & Examples
Properties of Functions Activities for High School Math
Combining Functions Using Arithmetic Operations
Relations & Functions Lesson Plan
Mapping Diagram | Overview, Function & Graphs
Create an account to start this course today
Used by over 30 million students worldwide
Create an account
Like this lessonShare
Explore our library of over 88,000 lessons
Search
Browse
Browse by subject
College Courses
Business
English
Foreign Language
History
Humanities
Math
Science
Social Science
See All College Courses
High School Courses
AP
Common Core
GED
High School
See All High School Courses
Other Courses
College & Career Guidance Courses
College Placement Exams
Entrance Exams
General Test Prep
K-8 Courses
Skills Courses
Teacher Certification Exams
See All Other Courses
Upgrade to enroll×
Upgrade to Premium to enroll in Algebra I: High School
Enrolling in a course lets you earn progress by passing quizzes and exams.
Track course progress
Take quizzes and exams
Earn certificates of completion
You will also be able to:
Create a Goal
Create custom courses
Get your questions answered
Upgrade to Premium to add all these features to your account!
Upgrade Now
Upgrade to Premium to add all these features to your account!
Download the app
Plans
Student Solutions
Teacher Solutions
Study.com for Schools
Working Scholars® Solutions
Online tutoring
About Us
Blog
Careers
Teach For Us
Press Center
Ambassador
Scholarships
Support
FAQ
Site Feedback
Download the app
Working Scholars® Bringing Tuition-Free College to the Community
© Copyright 2025 Study.com. All other trademarks and copyrights are the property of their respective owners. All rights reserved.
Terms of UsePrivacy PolicyDMCA NoticeADA ComplianceHonor Code For Students
× |
189249 | https://www.yumpu.com/en/document/view/7231477/nelson-textbook-of-pediatrics-richard-e-behrman-md | 404 Not Found
EN
EnglishDeutschFrançaisEspañolPortuguêsItalianoRomânNederlandsLatinaDanskSvenskaNorskMagyarBahasa IndonesiaTürkçeSuomiLatvianLithuaniančeskýрусскийбългарскиالعربيةUnknown
Login to YUMPU NewsLogin to YUMPU Publishing
TRY ADFREE
Discover
productsNewsPublishing
Magazines
Create ePaper
Login to YUMPU NewsLogin to YUMPU Publishing
Oops, 404!
404 Not Found
The requested resource could not be found but may be available again in the future.
Need help?
Recommended for You
Tropicana Sep-Oct 2025 #162 The Heritage Issue
TROPICANA.LIFESTYLE
×
Embed
Loading...
Close
products
FREE
adFREE
WEBKiosk
APPKiosk
PROKiosk
Resources
Blog
API
Help & Support
Status
tuxbrain.com
ooomacros.org
nubuntu.org
Direct Marketing
Company
Contact us
Careers
Terms of service
Privacy policy
Cookie policy
Cookie settings
Accessibility Statement
Imprint
Terms of service
Privacy policy
Cookie policy
Cookie settings
Accessibility Statement
Imprint
Change language
Made with love in Switzerland
© 2025 Yumpu.com, all rights reserved
Choose your language
Main languages
Further languages
English
Deutsch
Français
Italiano
Español
العربية
български
český
Dansk
Nederlands
Suomi
Magyar
Bahasa Indonesia
Latina
Latvian
Lithuanian
Norsk
Português
Român
русский
Svenska
Türkçe
Unknown
×
Done
Performing this action will revert the following features to their default settings:
Revert Cancel
Hooray! Your file is uploaded and ready to be published.
Saved successfully!
Ooh no, something went wrong! |
189250 | https://emedicine.medscape.com/article/187857-treatment | Upper Gastrointestinal Bleeding (UGIB) Treatment & Management: Approach Considerations, Proton-Pump Inhibitors, Therapeutic Endoscopy
[x]
X
No Results
No Results
For You
News & Perspective
Tools & Reference
CME/CE
Video
Events
Specialties
Topics
Edition English
Medscape Editions
English
Deutsch
Español
Français
Português
UK
Invitations
About You
Scribe
Professional Information
Newsletters & Alerts
Your Watch List
Formulary Plan Manager
Log Out
Register
Log In
For You
News & Perspective
Tools & Reference
CME/CE
More
Video
Events
Specialties
Topics
EN
Medscape Editions
English
Deutsch
Español
Français
Português
UK
X
Univadis from Medscape
Welcome to Medscape
About YouScribeProfessional InformationNewsletters & AlertsYour Watch ListFormulary Plan ManagerLog Out
RegisterLog In
X
No Results
No Results
close
Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log outCancel
processing....
Tools & Reference>Gastroenterology
Upper Gastrointestinal Bleeding (UGIB) Treatment & Management
Updated: Sep 01, 2021
Author: Bennie Ray Upchurch, III, MD, FACP, AGAF, FACG, FASGE; Chief Editor: BS Anand, MDmore...;)
71
Share
Print
Feedback
Close
Facebook
Twitter
LinkedIn
WhatsApp
Email
Sections
Upper Gastrointestinal Bleeding (UGIB)
Sections Upper Gastrointestinal Bleeding (UGIB)
Overview
Practice Essentials
Background
Etiology
Epidemiology
Prognosis
Show All
Presentation
History
Physical Examination
Show All
DDx
Workup
Approach Considerations
Hemoglobin Value and Type and Crossmatch Blood
BMP, BUN, and Coagulation
Risk Scoring
Endoscopy
Chest Radiography
Computed Tomography Scanning
Nuclear Medicine Scanning
Angiography
Nasogastric Lavage
Histologic Findings
Show All
Treatment
Approach Considerations
Proton-Pump Inhibitors
Therapeutic Endoscopy
Bleeding Peptic Ulcer Treatment
Bleeding Gastric Ulcer Treatment
Stress Ulcer Treatment
Mallory-Weiss Syndrome Treatment
Dieulafoy Lesion Treatment
Angiodysplasia Treatment
Aortoenteric Fistula
Treatment Complications
Posttreatment Monitoring and Care
Deterrence and Prevention of UGIB
Show All
Guidelines
Perforated and Bleeding Peptic Ulcer Clinical Practice Guidelines (2020)
ASGE Guidelines for Sedation and Anesthesia in Gastrointestinal Endoscopy (2018)
Additional Resources
Show All
Medication
Medication Summary
Proton Pump Inhibitors
Prokinetic Agents
H pylori Agents
Non-steroidal Anti-inflammatory Drug
Histamine H2 Antagonists
Iron Products
Antibiotics
Show All
Media Gallery;)
Tables)
References;)
Treatment
Approach Considerations
The goal of medical therapy in upper gastrointestinal (GI) bleeding (UGIB) is to correct shock and coagulation abnormalities and to stabilize the patient so that further evaluation and treatment can proceed. In addition to intravenous (IV) fluids, patients may need transfusion of packed red blood cells. High doses of proton pump inhibitors (PPIs) may reduce the need for endoscopic therapy (see Treatment with proton pump inhibitors).
Various methodologies have been proposed to quantitate the rebleeding risk (eg, Rockall score, Baylor score), with several instruments gaining more widespread acceptance. The Blatchford score (range 0-23) in particular can reliably triage patients with presumed UGIB, to hospital discharge for outpatient management with a score of 0, representing a less than 1% chance of requiring intervention, rather than inpatient admission and performance of endoscopy.
The 2008 Scottish Intercollegiate Guidelines Network (SIGN) guideline on the management of acute upper and lower GI bleeding recommends that an initial (pre-endoscopic) Rockall score be calculated for all patients presenting with an acute UGIB. In patients with an initial Rockall score >0, endoscopy is recommended for a full assessment of bleeding risk.
Resuscitation of a hemodynamically unstable patient begins with assessing and addressing the "ABCs" (ie, airway, breathing, circulation) of initial management. (Baradarian et al demonstrated that early, aggressive resuscitation can reduce mortality in acute UGIB. )
Patients presenting with severe blood loss and hemorrhagic shock present with mental status changes and confusion. In such circumstances, patients cannot protect their airway, especially when hematemesis is present. In these cases, patients are at an increased risk for aspiration, which is a potentially avoidable complication that can significantly affect morbidity and mortality. This situation must be recognized early, and patients should be electively intubated in a controlled setting. A multidisciplinary approach, with pulmonologists or intensivists in an intensive care unit (ICU) setting is desirable. Some centers have a dedicated GI bleeding team that utilizes treatment protocols beginning from the emergency department, progressing to the acute-care bed, assuring a controlled setting for endoscopy and delivering optimized in-hospital and postdischarge care.
Intravenous access must be obtained. Bilateral, 16-gauge (minimum), upper extremity, peripheral intravenous lines are adequate for volume resuscitative efforts. Poiseuille’s law states that the rate of flow through a tube is proportional to the fourth power of the radius of the cannula and is inversely related to its length. Thus, short, large-bore, peripheral intravenous lines are adequate for rapid fluid infusion. The 2008 SIGN guideline indicates either colloid or crystalloid solutions may be used to attain volume restoration prior to administering blood products.
There are remarkably few data on optimal fluid resuscitation pathways or algorithms, and very often the approach is based on the patient’s clinical presentation. The choice of fluids, infusion rates, and various in points are largely individual or institutionally driven.
A study published by Kaplan et al indicated that skin temperature on physical examination in combination with serum bicarbonate levels correlated well with the level of systemic perfusion. Foley catheter placement is helpful to allow a continuous evaluation of the urinary output as a guide to renal perfusion.
Patients with severe coexisting medical illnesses, such as cardiovascular and pulmonary diseases, may require pulmonary artery catheter insertion to closely monitor hemodynamic cardiac performance profiles during the early resuscitative phase.
After the ABCs have been addressed, assess the patient's response to resuscitation, based on evidence of end organ perfusion and oxygen delivery.
Consultation with a surgeon should be considered for all patients with GI hemorrhage. Depending on the patient’s comorbidities, subspecialty consultation is often needed and mirrors the trend in healthcare toward a multidisciplinary team approach.
Once the maneuvers to resuscitate are underway, it is often very helpful to insert a nasogastric tube and perform an aspirate and lavage procedure. This should be the first procedure performed to determine whether the GI bleeding is emanating from above or below the ligament of Treitz. If the stomach contains bile but no blood, UGIB is less likely. If the aspirate reveals clear gastric fluid, a duodenal site of bleeding may still be possible.
In a retrospective review of 1190 patients, Luk et al found that positive nasogastric-tube aspirate findings were 93% predictive of an upper GI source of bleeding.
According to a study performed by the American Society for Gastrointestinal Endoscopy (ASGE), however, a nasogastric-tube aspirate finding can be negative even in the setting of a large duodenal bleeding ulcer. The study compared nasogastric-tube aspirate findings with endoscopic findings of the bleeding source. The investigation revealed that 15.9% of patients with a clear nasogastric-tube aspirate, 29.9% of patients with coffee-ground aspirate, and 48.2% of patients with red blood aspirate had an active upper GI source of bleeding at the time of endoscopy.
A study correlated mortality with the color of the fluid from the nasogastric-tube aspirate and the color of the stool. As shown in the following table, the color of the nasogastric-tube aspirate can be a prognostic indicator.
Table 4.Effect of the Color of the Nasogastric Aspirate and of the Stool on UGIB Mortality Rate(Open Table in a new window))
Nasogastric Aspirate ColorStool ColorMortality, %
Clear Brown or red 6
Coffee-ground Brown or black 8.2
Red 19.1
Red blood Black 12.3
Brown 19.4
Red 28.7
UGIB = upper gastrointestinal bleeding.
See Pediatric Gastrointestinal Bleeding for more information on this topic.
Surgery
Primary surgical intervention should be considered in patients with a perforated viscus (eg, from perforated duodenal ulcer, perforated gastric ulcer, or Boerhaave syndrome). In patients who are poor operative candidates, conservative treatment with nasogastric suction and broad-spectrum antibiotics can be instituted. Endoscopic clipping or sewing techniques have also been used in such patients.
Emergency surgery in UBIG typically entails oversewing the bleeding vessel in the stomach or duodenum (usually preoperatively identified by endoscopy), vagotomy with pyloroplasty, or partial gastrectomy. Angiographic obliteration of the bleeding vessel is often considered a favorable modality in patients who are poor surgical candidates. (See Angiography.)
Treatment-related contraindications and precautions
Contraindications to upper endoscopy include an uncooperative or obtunded patient, severe cardiac decompensation, acute myocardial infarction (unless active, life-threatening hemorrhage is present), and perforated viscus (eg, esophagus, stomach, intestine). Expert subspecialty consultation may be beneficial to optimize the patient and establish a best “window of opportunity” should endoscopy proceed.
Contraindications to emergency surgery include impaired cardiopulmonary status and bleeding diathesis.
Esophagogastroduodenoscopy (EGD) may be more difficult or impossible if the patient has had previous oropharyngeal surgery or radiation therapy to the oropharynx. Altered upper GI tract anatomy from previous surgery (eg, Roux-en-Y gastric bypass) may pose unique challenges to endoscopic management of bleeding.
The presence of aZenker diverticulumcan make intubation of the esophagus more difficult.
Patients with Down syndrome are more sensitive to conscious sedation and, when possible, should be monitored by an anesthesiologist and/or intubated prophylactically prior to the procedure. Monitored anesthesia care has been increasingly used in more challenging and sometimes prolonged cases such as an EGD for active UGIB.
Hypotension may be exacerbated by sedation; therefore, patients who are clinically unstable should be carefully sedated. Continuous monitoring in the ICU is warranted and monitored anesthesia care by an anesthesia provider may improve the safety of endoscopy, particularly if there is decompensation or compromise of the airway.
Patients with massive bleeding should be considered for intubation to reduce the increased risk of aspiration. Such patients should be treated in an intensive care setting. As suggested, subspecialty consultation with a pulmonologist or an intensivist may be prudent, and sedation might be best managed with the subspecialists in attendance. Anesthesia assistance should be considered even with intubated patients depending on patients' comorbidities and overall stability.
Ideally, the patient should be stabilized prior to endoscopy and abnormalities in coagulation should be corrected. When this is not possible, the judgment of an experienced endoscopist is vital. The merits of the multidisciplinary team approach in critically ill patients cannot be overemphasized.
Next: Proton-Pump Inhibitors
Proton-Pump Inhibitors
The relative efficacy of proton-pump inhibitors (PPIs) may be due to their superior ability to maintain a gastric pH at a level above 6.0, thereby protecting an ulcer clot from fibrinolysis. Current guidelines recommend a regimen of an intravenous (IV) PPI 80-mg bolus, followed by a continuous infusion of 8 mg/hour for 72 hours. [79, 80, 81, 82, 83]
Lau et al demonstrated that high-dose IV omeprazole can accelerate the resolution of the stigmata of recent hemorrhage and reduce the need for endoscopic therapy. Barkun et al showed this therapy to be cost-effective. Laine et al demonstrated that high-dose IV lansoprazole, as well as orally administered high-dose lansoprazole, can maintain the intragastric pH above 6.
A meta-analysis of 24 randomized controlled trials that evaluated PPIs for bleeding ulcers (with or without endoscopic therapy) found a significant reduction in the risk of rebleeding, the need for repeat endoscopic hemostasis, and surgery. An improvement in mortality was also seen in Asian trials and in patients with active bleeding or nonbleeding visible vessels.
A systematic review of six randomized trials comprising 2223 patients to assess the use of a PPI before endoscopic evaluation found that pre-endoscopy PPI therapy did not significantly reduce mortality, rebleeding, or the requirement for surgery. However, there was a significantly lower proportion of peptic ulcers with high-risk stigmata at endoscopy and significantly lower rates of endoscopic treatment.
The 2010 international consensus guidelines on upper gastrointestinal (GI) bleeding (UGIB) recommended the use of IV PPIs in all patients with high-risk lesions post endoscopic therapy; PPI therapy might downgrade the lesion if given pre-endoscopy.
Standard daily-dose oral PPIs may be used in patients who do not have active bleeding or other high-risk stigmata for recurrent bleeding (eg, a visible vessel, adherent clots); in such patients, the risk of recurrent bleeding is low. The goal of treatment in these patients (following resuscitation) should be directed at healing the ulcers and at eliminating precipitating factors (eg, H pylori, nonsteroidal anti-inflammatory drugs [NSAIDs]).
When possible, it is important to take biopsy samples to test for H pylori at the initial endoscopy procedure. Because starting high-dose IV PPI therapy is the mainstay of initial management in UGIB, it is difficult to obtain the initial endoscopic biopsies without the presence of ongoing proton-pump inhibition. It is not well understood whether short-duration proton-pump inhibition alters the sensitivity of biopsies for H pylori. Biopsy specimens should be histologically evaluated when the rapid urease test is negative.
A combined analysis of five studies that evaluated oral dosing with PPI (with or without endoscopic therapy) found a significant reduction in the risk of rebleeding and surgery.
NOTE:Patients with liver cirrhosis may have an increased mortality if treated with PPIs. [91, 92]
PPIs in patients taking antiplatelet or anticoagulant therapy
Standard dose PPI therapy is advised for gastroprotection in all patients on antiplatelet therapy who are at increased risk of GI bleeding (age >65 years or concomitant use of corticosteroids or anticoagulants or history of peptic ulcer). International normalized ratio (INR) monitoring is required when starting or stopping PPI therapy in users of vitamin K antagonists.
No demonstrated interaction exists between PPIs and the novel oral anticoagulants. Based on a documented efficacy, antiplatelet therapy (aspirin < 300 mg/daily, ticlopidine 100 mg/daily, clopidogrel 75 mg/daily) is widely used for both primary and secondary prevention of cardiovascular and cerebrovascular ischemic events. [93, 94] However, antiplatelet drugs may cause adverse GI events (gastroduodenal ulcerations/erosions, overt bleeding, occult bleeding, and rare perforation), with a definite probability of death, particularly in the elderly. Therefore, gastroprotection is advised in those patients at increased GI risk during antiplatelet therapy.
Increased risk factors include age 65 years and older, concurrent use of steroid/anticoagulant therapy, or a history of peptic ulcer disease. The presence of relevant comorbidities (heart failure, renal impairment, stroke, diabetes, ongoing malignancy) and tobacco use are additional risk factors for both GI events and related mortality. Standard PPI dosing is the most effective gastroprotective therapy.
PPI therapy is not effective in preventing bleeding lesions induced by antiplatelet drugs in either the small intestine or the colon.
Anticoagulants, either vitamin K antagonists or novel oral anticoagulants (NOACs) (including dabigatran, rivaroxaban, and apixaban), do not cause gastroduodenal mucosa injury in and of themselves. These agents may, however, facilitate bleeding of preexisting peptic ulcers. Although gastroprotection with a PPI is currently routinely recommended, unless a concomitant antiplatelet or nonsteroidal anti-inflammatory drug (NSAID) therapy is prescribed, data exist that show PPI cotherapy is associated with reduced risk of warfarin-related UGIB.
In patients under acid suppression because of gastroprotection for any acid-related disease, intensified INR monitoring is recommended because PPIs may potentiate vitamin K antagonist-induced anticoagulation, most likely due to facilitated gastric absorption of warfarin.
Previous
Next: Proton-Pump Inhibitors
Therapeutic Endoscopy
For several decades, endoscopy has been the primary method of evaluating and managing upper gastrointestinal (GI) bleeding (UGIB).
Several observational studies, randomized clinical trials, and meta-analyses have demonstrated and supported the idea that early endoscopic hemostatic therapy significantly reduces the rates of recurrent bleeding, the need for emergent surgery, and mortality in patients with acute nonvariceal UGIB. This effect has been more evident in higher-risk patients. [57, 65, 98, 99, 100, 101]
Challenges, such as increasing patient age and comorbidity, the extremely effective antithrombotic or anticoagulant agents, and significant side-effect concerns of the most effective therapeutic agents for ulcer disease, have fortunately been balanced with various endoscopic technological developments and dissemination of evidence-based treatment pathways. Yet the mortality from peptic ulcer bleeding has changed very minimally over this time. [38, 102]
Three significant technological advances have been developed: (1) endoscopic application of Doppler probes to evaluate arteries in the ulcer base, (2) endoscopic application of hemostatic powders, and (3) over-the-scope clips, with enhanced capability over the standard endoclips.
Aside from ulcer hemorrhaging, other causes of GI bleeding, including mucosal tears in the esophagus or upper stomach due to vomiting (Mallory-Weiss tears), venous blebs, and vascular ectasias, can also be treated with endoscopic coagulation.
The bleeding from gastric cancers and ulcers in leiomyomas does not usually respond to endoscopic therapy; surgical or radiologic intervention is needed.
Much debate has focused on the significance of the nonbleeding visible vessel (ie, color, size, diagnostic characteristics, risk of rebleeding) in ulcer hemorrhage. These matters became clarified after the characteristics and significance of the visible vessel in the ulcer crater were defined and the evidence for endoscopic therapy was established, demonstrating that patients requiring therapy to control bleeding or rebleeding could be diagnosed and treated at the time of the upper endoscopy.
The use of Doppler probes to evaluate the arterial flow in the base of ulcers to assess for the rebleeding risk and adequacy of hemostasis may prove to be more accurate than the visual assessment of bleeding stigmata.
Patients should be considered for upper endoscopy if blood loss from the upper GI tract is suspected.
The patient should undergo upper endoscopy prior to any operative intervention in order to diagnose and localize the bleeding site. Most patients (85%-90%) respond to endoscopic therapy.
During the endoscopy, the patient is monitored according to the analgesia and sedation guidelines formulated by the American Society of Anesthesiology. The characteristics of the bleeding lesion are noted, and appropriate therapy is applied when necessary for high-risk lesions or active bleeding.
Urgent endoscopy
Urgent endoscopy is indicated when patients present with hematemesis, melena, or postural changes in blood pressure. Studies have demonstrated a lower rate of rebleeding and shorter length of stay when endoscopy was performed within 24 hours of admission. [77, 103] However, observational studies have not shown a benefit in clinical outcomes when endoscopy was performed within 2 to 12 hours of presentation. [58, 59]
Early endoscopy
Cooper et al studied the effectiveness of performing an early endoscopy within the first 24 hours of an acute UGIB episode and found it to be associated with reductions in the length of hospital stay, rate of recurrent bleeding, and the need for emergent surgical intervention.
According to the 2010 international consensus on nonvariceal UGIB, early endoscopy (within 24 hours of presentation) is appropriate for most patients with UGIB. In a retrospective review involving more than 30,000 cases, Yavorski et al showed that the mortality rates were more than twice as high in patients who did not undergo an early endoscopic procedure than for those who did undergo the procedure early on (11.1% vs 5.2%, respectively). More recent data also suggest endoscopy within 24 hours may have more favorable outcomes. [57, 98]
Studies on the effectiveness of endoscopy on weekends or off-hours have inconsistently shown poorer outcomes. [105, 106]
Endoscopic techniques
There are several currently widely accepted hemostatic treatment options. These include injection of epinephrine and tissue adhesives such as cyanoacrylate, ablative therapy with contact modalities such as thermal coagulation with heater probe and bipolar hemostatic forceps, noncontact modalities such as hemostatic power sprays and argon plasma coagulation, and mechanical hemostasis with band ligation, endoscopic hemoclips, and over-the-scope clips.
The following endoscopic techniques have been developed for achieving hemostasis :
Injection of epinephrine or sclerosants
Bipolar electrocoagulation
Band ligation
Heater probe coagulation
Constant probe pressure tamponade
Bipolar/soft coagulation hemostatic forceps
Argon plasma coagulator (APC)
Laser photocoagulation
Rubber band ligation
Application of hemostatic materials, including biologic glue and tissue adhesives
Application of hemoclips/endoclips or over-the-scope clips
Application of hemostatic powder/spray
Doppler ultrasonographic assessment, pre- and postendotherapy
Treatment using a combination of endoscopic therapies has become more common. For example, injection therapy can be applied first to better clarify the bleeding site with at least partial hemostasis, especially in the actively bleeding patient, followed by the application of heater probe or bipolar (gold) probe coagulation with coaptation for definitive hemostatic management. Injection therapy can also be performed prior to endoscopic placement of hemoclips.
According to the 2008 Scottish Intercollegiate Guidelines Network (SIGN) guideline, combinations of endoscopy with an injection of at least 13 mL of 1:10,000 adrenaline, coupled with either a thermal or mechanical treatment, are more effective than single modalities.
The 2010 international consensus guidelines on UGIB recommended the use of endoscopic clips or thermal therapy for high-risk lesions.
Heater probe coagulation
The heater probe consists of a resistor electrode enveloped by a titanium capsule and covered by Teflon (to reduce sticking to the mucosa by the probe). The probe temperature rises to 250°C (482°F).
Bipolar electrocoagulation
The bipolar probe consists of alternating bands of electrodes producing an electrical field that heats the mucosa and the vessel. The electrodes are coated with gold to reduce adhesiveness. The probes are stiff in order to allow adequate pressure to be applied to the vessel to appose the walls and thus produce coaptive coagulation when the electrical-field energy is transmitted. Careful technique is required to heat-seal the perforated vessel.
Injection therapy
Injection therapy involves the use of several different solutions injected into and around the bleeding lesion. Solutions available for injection include epinephrine, sclerosants, and various clot-producing materials, such as fibrin and cyanoacrylate glues.
The epinephrine used for injection is diluted (1:10,000) and injected as 0.5- to 1-mL aliquots. Debate continues over whether the hemostatic effect of epinephrine is due to induced vessel vasoconstriction and subsequent platelet aggregation or to the tamponade effect produced by injecting the volume of drug into the tissue surrounding the bleeding lesion.
Epinephrine injection is often used to reduce the volume of bleeding so that the lesion can be better localized and then treated with a coaptive technique (ie, heater probe, gold probe).
Combining epinephrine injections with human thrombin (600-1000 IU) reduces the risk of bleeding.
Although the epinephrine administered in injection therapy is absorbed into the systemic circulation, this does not appear to have any adverse effects on the hemodynamic status.
Injecting a volume of sterile isotonic sodium chloride solution and providing a tamponade effect also leads to hemostasis, although not as effectively as does epinephrine.
The sclerosant solutions such as ethanol, polidocanol, and sodium tetradecyl sulfate are not frequently administered relative to the use of other available techniques for hemostasis in nonvariceal GI bleeding. Band ligation of esophageal varices is currently used more commonly than sclerosants.
The sclerosants create hemostasis by inducing thrombosis, tissue necrosis, and inflammation at the site of injection. When large volumes are injected, the area of tissue necrosis can produce an increased risk of local complications, such as perforation. Combining the various agents into a single injection has not been shown to be more beneficial than a single-agent therapy alone.
The use of fibrin glue in injection therapy has been shown to be successful, with results similar to those of epinephrine injections.
Cyanoacrylate is effective in achieving hemostasis, with success rates similar to that of hemoclips. [111, 112]
Laser therapy
Laser phototherapy is a noncontact thermal method that uses an Nd:YAG (neodymium-doped yttrium aluminium garnet) laser to create hemostasis by generating heat and direct vessel coagulation. It is not as effective as coaptive coagulation, because it lacks the use of compression to create a tamponade effect. An additional deterrent to its use is expense. Laser therapy has largely been replaced with other endoscopic hemostatic methods.
Hemostatic clips and endoclips
Hemostatic clips are widely available and used in the United States.
With careful placement of the clip, closing the defect in the vessel is possible. Often, depending on the lesion and progress to affect hemostasis, multiple clips are applied. Typically, they become detached and pass from the GI tract within 2 weeks. Hemostatic clips are considered magnetic resonance imaging (MRI)-conditional because they are metallic, and they can serve as radiopaque markers to direct the interventional radiologist during angiography to the relevant area if endoscopy fails to achieve adequate hemostasis. These clips vary in their size and strength. Numerous manufacturers have produced hemostatic clips, with the most significant advancements being the ability to rotate for accurate placement and the ability to reopen and reapply when necessary.
There are substantial data documenting the efficacy of hemoclips, which is similar to that of thermal coagulation methods. [24, 113]
One report, concerning 113 patients with major stigmata of ulcer hemorrhage, found no difference between the use of hemoclips and photocoagulation with regard to hemostasis, 30-day mortality, and the need for emergency surgery. Patients randomized to the endoclip group had significantly lower rebleeding rates (2% vs 21%). However, only 60% of active bleeders were successfully treated with the heater probe, a rate much lower than in previous reports.
A study of 80 patients found a higher rate of control of initial bleeding with the heater probe compared with the Olympus endoclip (100% vs 85%). Rebleeding rates were not significantly different.
No significant differences in procedure duration, initial hemostasis, or rebleeding rates were found in a study of 47 patients comparing combination therapy with epinephrine injection plus monopolar electrocoagulation versus hemoclips.
There are some clinical settings in which endoclips may be preferred over other hemostatic methods. These include the treatment of ulcers in patients who are coagulopathic or who require ongoing anticoagulation; in such patients, electrocoagulation will increase the size, depth, and healing time of treated lesions. Endoclips may also be preferable in the retreatment of lesions that rebleed after initial thermal hemostasis. Finally, some endoscopists, including this author, may choose endoclips as their method of choice for hemostasis to avoid the potential for tissue destruction altogether, thus allowing a potentially safer setting to use any method of choice if repeat endoscopy for hemostasis is required.
Ulcers on the lesser curvature, the posterior duodenum, or the cardia increase the difficulty of clip deployment and clip failure rates.
Larger endoclips, such as the over-the-scope clips, have advantages over smaller hemoclips for the hemostasis of chronic ulcers, fibrotic lesions, and the closure of larger lesions. However, the use of the over-the-scope clips can be cumbersome in upper GI bleeding and, therefore, these clips have been more commonly used in refractory bleeding or as a salvage maneuver, but there are data showing efficacy of these clips as a primary modality. [117, 118]
Argon plasma coagulation
APC is a technique in which a stream of electrons flows along a stream of argon gas. The coagulation is similar to monopolar cautery, with the current flow going from a point of high current density (the point of contact of the gas with the mucosa) to an area of low current density (the conductive pad on the patient's body). The current flows through the body in an erratic path to the pad.
This monopolar cautery technique is similar to the laser technique in that energy is delivered to the vessel for coagulation with apposition of the vessel walls. This technique was found not to be effective for visible vessels larger than 1 mm, owing to the limited depth of the burn. Small, superficial vessels such as arteriovenous malformations, telangiectasias, and particularly gastric antral vascular ectasia (GAVE) respond well to treatment by APC.
Hemostatic powder/spray
Hemostatic powders are a novel technique. Thus, data are available, but no randomized controlled trials evaluating this technique have been conducted yet. These agents have primarily been used as a second-line option when other endoscopic hemostasis techniques have failed. In a literature review of several reported cases, a hemostatic powder spray (Hemospray) was successfully used for hemostasis in 88.5% of 234 cases of UGIB.
The hemosprays have the advantage of excellent initial hemostasis, but they can also obscure the endoscopic views of the underlying lesion. Intuitively, as a purely topical agent, hemosprays would not have the durability of mechanical clips or thermal coactive techniques.
In a retrospective study (2013-2017) that evaluated the effectiveness of a hemospray (Hemospray) for managing diffuse or refractory UGIB in 52 patients treated for peptic ulcer bleeding (n = 18), postinterventional bleeding (n = 13), or other UGIB (n = 21), there was 100% efficacy without adverse effects related to therapy, with immediate hemostasis in 51 patients. Twenty-two patients (43.1%) had recurrent bleeding within 3 days, with a 56.9% overall clinical success, and 25 patients had recurrent bleeding within 7 days (49%), with a 51% overall clinical success. In total, eight patients died (15.4%), two of which were related to bleeding (3.8%). Thus, although these findings indicate a high technical success of the hemospray for treating diffuse or refractory UGIB, the investigators acknowledge there was a high rebleeding risk and further investigation is needed.
Doppler ultrasonographic probes
There is increasing use of Doppler ultrasonographic probe-guided lesion assessment to detect significant arterial flow in the vessel at the ulcer base, thanks to the development of relatively low cost, easy-to-use Doppler units and disposable endoscopic probes. [112, 122] Two studies have shown Doppler probe assessment is more accurate than classic endoscopic scoring of stigmata in the base of ulcers, at predicting rebleeding risk. [123, 124]
Endoscopic treatment decisions
The choice of treatment modality is influenced by the size of the vessel. Animal studies have demonstrated that the heater probe and bipolar probe are effective for vessels as large as 2 mm in diameter.
Other techniques (eg, clips, band ligation) or a combination of techniques are needed for larger vessels or vessels that are not approachable by the heater probe or bipolar probe. (Surgical intervention should be considered when dealing with vessels larger than 2 mm in diameter, discounting an enlargement due to the development of a pseudoaneurysm.) The over-the-scope clips are able to grasp larger areas and apply more mechanical force to larger vessels, making them an option when standard endoclips may not suffice.
It is important to remember when planning endoscopic therapy for a large vessel, that the 7-French heater probe or gold probe catheter can traverse the accessory channel of a standard adult upper endoscope, but a therapeutic endoscope with a larger accessory channel is necessary to pass a 10-French thermal probe.
Ulcers with an overlying clot
In the patient who has an ulcer with an overlying clot, attempting to remove the clot by target washing is critical to allow treatment of the underlying stigmata if appropriate. Endoscopic removal of the clot by washing or cold snare has been demonstrated to be effective in reducing the recurrence of bleeding.
The findings under the clot (eg, bleeding vessel, visible vessel, clean base, examples of which are seen in the images below) help to determine the therapy needed and to improve efficacy by allowing treatment to be applied directly to the vessel. (See also the table below.)
Upper gastrointestinal bleeding (UGIB). Ulcer with active bleeding.
View Media Gallery)
Upper gastrointestinal bleeding (UGIB). Ulcer with a clean base.
View Media Gallery)
Upper gastrointestinal bleeding (UGIB). Diagram of an ulcer with a clean base.
View Media Gallery)
Upper gastrointestinal bleeding (UGIB). Ulcer with an overlying clot.
View Media Gallery)
Upper gastrointestinal bleeding (UGIB). Ulcer with a visible vessel.
View Media Gallery)
Upper gastrointestinal bleeding (UGIB). Diagram of an ulcer with a visible vessel.
View Media Gallery)
Table 5. Ulcer Characteristics and Correlations(Open Table in a new window))
Ulcer CharacteristicsPrevalence Rate, %Rebleeding Rate, %Surgery Rate, %Mortality, %
Clean base 42 5 0.5 2
Flat spot 20 10 6 3
Adherent clot 17 22 10 7
Visible vessel 17 43 34 11
Active bleeding 18 55 35 11
If the clot cannot be removed by washing, then using a cold snare can be effective in dislodging and removing the clot.
Vigorous washing of the clot formed after therapy is useful in determining the adequacy of coagulation. A combination of injection with heater probe or bipolar coaptive coagulation is often used and has been shown to be more effective in patients with active bleeding.
The patient is monitored under the protocol for conscious sedation, also called analgesia and sedation (ie, per the American Society of Anesthesiologists [ASA] and the American Society for Gastrointestinal Endoscopy [ASGE] guidelines). In selected cases, monitored anesthesia care has become the preferred option for sedation and endoscopy.
Active bleeding and rebleeding
Attempting to control active bleeding using the recommended techniques with the appropriate equipment or instituting appropriate therapy for a high-risk lesion is important. The large-channel therapeutic endoscope should be used so that the 10-French thermal probe can be employed for adequate coaptation.
Endoscopists should use the technique with which they have the most familiarity. The endoscopy should not be started unless the endoscopist is equipped for any potential lesions (eg, ulcer, varix, angioectasia, mucosal tear, tumor). The patient should be monitored for recurrent bleeding and treated a second time if appropriate. A surgical consultation should be considered for all patients with GI hemorrhage. Subspecialty consultation for multidisciplinary management should be considered, particularly in severe active bleeding.
Rebleeding occurs in 10%-30% of endoscopically treated patients.
Second-look endoscopy
A second attempt at endoscopic control is warranted if the initial endoscopy fails to control the bleeding. Some authorities have concerns about the perils of a second esophagogastroduodenoscopy (EGD), which may result in delayed surgery, perforation, and increased morbidity and mortality. However, this approach has been validated in a large, randomized, controlled trial that showed decreased morbidity and mortality.
Owing to the relatively high rebleeding rate associated with ulcers, some clinicians advocate scheduled second-look endoscopy, with the intent of identifying and proactively managing persistent or recurrent bleeding. This would be a strategy directed at individuals who are very likely to benefit from a second invasive procedure; however, no current guidelines recommend this strategy. A systematic review and meta-analysis of randomized trials assessing outcomes of second-look endoscopy reported a small but significant reduction in rebleeding in patients undergoing the procedure (P< 0.01) but no significant benefit in reducing surgery or death.
In a prospective multicenter study that evaluated the efficacy of scheduled second-look endoscopy (24-36 hours after initial hemostasis) in 319 patients with endoscopically confirmed bleeding peptic ulcer treated unsuccessfully with hemoclip application, thermal coagulation, and/or epinephrine injection, investigators found noninferiority of single endoscopy relative to second-look endoscopy for the rate of rebleeding (P = 0.132). Independent risk factors for rebleeding included endoscopists’ estimation of poorer success of the initial hemostasis, a patient history of NSAID use, and higher transfusion requirement (4 units of red blood cells). Thus, the researchers concluded that repeat endoscopy may only be beneficial in patients with less-than-satisfactory initial hemostasis at endoscopy, NSAIDs history, or higher transfusion requirement.
Specific characteristics at endoscopy can predict rebleeding. Rebleeding occurs in 55% of patients who have active bleeding (pulsatile, oozing), in 43% who have a nonbleeding visible vessel, in 22% who have an ulcer with an adherent clot, and in 0-5% who have an ulcer with a clean base.
At endoscopy, the prevalence rate for a clean base is 42%, for a flat spot is 20%, for an adherent clot is 17%, for a visible vessel is 17%, and for active bleeding is 18%. See the images below.
Upper gastrointestinal bleeding (UGIB). Diagram of an ulcer with a clean base.
View Media Gallery)
Upper gastrointestinal bleeding (UGIB). Ulcer with a visible vessel.
View Media Gallery)
Freeman et al have described a pale, visible vessel that appears to have a very high risk for rebleeding. This must be differentiated from the presence of a clean ulcer base.
Good visualization is important. The uncleared fundal pool may obscure an ulcer, mucosal tear, gastric varices, portal gastropathy, or tumor (eg, leiomyoma, adenocarcinoma, lymphoma). Endoscopic therapy is recommended for ulcers at increased risk for rebleeding.
Using a combination of techniques is prudent when re-treating the ulcer site because the first therapy may have produced necrosis and weakening of the intestinal wall. Ulcers on the anterior surface of the stomach and duodenum are at an increased risk for perforation. Using injection as the first step increases the thickness of the submucosal layer, thus providing an extra margin of safety.
Even operative techniques can have a significant rebleeding rate with significant mortality, as noted in the study of Poxon et al. In this investigation, the rebleeding rate was 10% (80% mortality for rebleeders) in patients who underwent a conservative surgical technique in which the ulcer base was undersewn. This more conservative approach was compared with the standard surgical technique (ie, vagotomy and pyloroplasty or partial gastrectomy). The comparison of the conservative approach with a standard gastrectomy resulted in similar mortality rates, ie, 26% versus 19%, respectively, with no rebleeding after partial gastrectomy.
Postendoscopic monitoring
Postoperatively, the patient is monitored for recovery from conscious sedation or receives monitored anesthesia care after endoscopy and fromgeneral anesthesiaafter abdominal surgery. Observe and follow the patient's mental status, vital signs, chest, cardiac, and abdominal findings to ascertain that the patient's clinical status has stabilized and that no complications (eg, aspiration, perforation, recurrent bleeding, myocardial infarction due to hypotension) have occurred. Monitor the hemoglobin level.
Previous
Next: Proton-Pump Inhibitors
Bleeding Peptic Ulcer Treatment
Upper gastrointestinal (GI) endoscopy is the most effective diagnostic tool for peptic ulcer disease and has become the method of choice for controlling active ulcer hemorrhage. Failure of endoscopy to maintain hemostasis is one of the indications to initiate surgical intervention, especially in high-risk patients.
In a randomized, prospective trial that included 92 patients with recurrent peptic ulcer bleeding after initial endoscopic therapy for hemostasis, patients who underwent a second endoscopic attempt to control bleeding were found to have decreased transfusion requirements, 30-day mortality rates, and duration of intensive care unit (ICU) stay in comparison with the surgical group.
With the exception of a patient in shock who has a life-threatening recurrent hemorrhage, this study supports attempting another trial of endoscopy to control a bleeding ulcer.
Regardless of the endoscopic therapy, however, 10%-12% of patients with acute ulcerous hemorrhage require an operation as the definitive procedure to control the bleeding ulcer. In most circumstances, the operation is performed emergently, and the associated mortality rate is as high as 15%-25%.
Medical therapy used in conjunction with endoscopy involves proton-pump inhibitor (PPI) administration. PPIs decrease the rebleeding rates in patients with bleeding ulcers associated with an overlying clot or visible, nonbleeding vessel in the base of the ulcer. [130, 131] Consider transcatheter angiographic embolization in patients who are poor surgical candidates. Because of the extensive collateral circulation of the upper GI tract, ischemic complications are rare.
Surgical treatment
If two attempts at endoscopic control of the bleeding vessel are unsuccessful, avoid further attempts (ie, because of increased complication risks and mortality) and pursue surgical intervention. The indications for surgery in patients with bleeding peptic ulcers are as follows:
Severe, life-threatening hemorrhage not responsive to resuscitative efforts
Failure of medical therapy and endoscopic hemostasis with persistent recurrent bleeding
A coexisting reason for surgery, such as perforation, obstruction, or malignancy
Prolonged bleeding, with loss of 50% or more of the patient's blood volume
A second hospitalization for peptic ulcer hemorrhage
The operative treatment options for a bleeding duodenal ulcer historically include vagotomy, gastric resection, and drainage procedures. Each specific operative option is associated with its own incidence of ulcer recurrence, postgastrectomy syndrome, and mortality. When making an intraoperative judgment on how to best manage the bleeding ulcer, it is extremely important for the surgeon to be aware of these differences.
The three most common operations performed for a bleeding duodenal ulcer are as follows :
Truncal vagotomy and pyloroplasty with suture ligation of the bleeding ulcer
Truncal vagotomy and antrectomy with resection or suture ligation of the bleeding ulcer
Proximal (highly selective) gastric vagotomy with duodenostomy and suture ligation of the bleeding ulcer
The purpose of the vagotomy is to divide the nerves to the acid-producing body and fundus of the stomach. This inhibits acid production that occurs during the cephalic phase of gastric secretion, thereby decreasing the risk for recurrent ulceration.
In addition to having the same effects as a highly selective vagotomy in the proximal stomach, a truncal vagotomy also has marked effects on distal gastric motor function. It weakens distal gastric peristalsis, thus requiring the creation of a pyloroplasty to decrease the resistance to outflow from the stomach. Proximal vagotomy abolishes gastric receptive relaxation and impairs storage in the proximal stomach. As a result, a more rapid gastric emptying of liquids occurs. A drainage procedure is not required, because the innervation of the antrum and pylorus is still intact. Consequently, the gastric emptying of solid food is not altered. The antropyloric mechanism still functions normally and continues to prevent duodenogastric reflux.
Truncal vagotomy and suture ligation of a bleeding ulcer is a frequently used operation for treating upper GI bleeding in elderly patients with life-threatening hemorrhage and shock. The procedure can be performed rapidly, minimizing the time spent in the operating room under general anesthesia.
Much of what is now known about the operations performed for bleeding duodenal ulcers came from the era before the etiologic role for H pylori and nonsteroid anti-inflammatory drugs (NSAIDs) in the development of peptic ulcers was understood. Reducing gastric acidity has been proven to be beneficial, with lower rebleeding rates when using high-dose omeprazole. Although PPIs seem to have an advantage, they have no effect on mortality.
The diagnosis of H pylori infection is important in the management of patients with a complicated bleeding peptic ulcer. If a patient with a bleeding ulcer requires surgery, then knowledge of the patient's H pylori status becomes pertinent, because it may help guide the decision to choose a particular surgical procedure (eg, simply oversewing the ulcer as opposed to performing an antiulcer operation).
Many studies support the decision to manage the bleeding ulcer in conjunction with eradication of H pylori.
The 2008 Scottish Intercollegiate Guidelines Network (SIGN) guideline recommends testing for H pylori in patients with peptic ulcer bleeding. Eradication therapy should be prescribed for those who test positive for an active infection. In those who take NSAIDs, maintenance antisecretory therapy should consist of daily PPI for prevention of recurrent ulceration after successful healing of the ulcer and H pylori eradication, if the NSAIDs cannot be discontinued. [45, 132, 133]
Previous
Next: Proton-Pump Inhibitors
Bleeding Gastric Ulcer Treatment
The surgical management of bleeding gastric ulcers is slightly different from that of duodenal ulcers, but the concepts are identical. The three most common complications of a gastric ulcer that mandate emergent surgical intervention are hemorrhage, perforation, and obstruction. The goals of surgery are to correct the underlying emergent problem, prevent recurrent bleeding or ulceration, and exclude malignancy.
A bleeding gastric ulcer is most commonly managed by a distal gastrectomy that includes the ulcer, with a gastroduodenostomy or a gastrojejunostomy reconstruction.
The common operations for the management of a bleeding gastric ulcer include (1) truncal vagotomy and pyloroplasty with a wedge resection of the ulcer, (2) antrectomy with wedge excision of the proximal ulcer, (3) distal gastrectomy to include the ulcer, with or without truncal vagotomy, and (4) wedge resection of the ulcer only.
Types of gastric ulcers
The choice of operation for a bleeding gastric ulcer depends on the location of the ulcer and the hemodynamic stability of the patient to withstand an operation. Five types of gastric ulcers occur, based on their location and acid-secretory status.
Type 1 gastric ulcers are located on the lesser curvature of the stomach, at or near the incisura angularis. These ulcers are not associated with a hypersecretory acid state.
Type 2 ulcers represent a combination of 2 ulcers that are associated with a hypersecretory acid state. The ulcer locations occur in the body of the stomach in the region of the incisura. The second ulcer occurs in the duodenum.
Type 3 ulcers are prepyloric ulcers. They are associated with high acid output and are usually within 3 cm of the pylorus.
Type 4 ulcers are located high on the lesser curvature of the stomach and (as with type 1 ulcers) are not associated with high acid output.
Type 5 ulcers are related to the ingestion of NSAIDs or aspirin. These ulcers can occur anywhere in the stomach.
Surgical management according to ulcer type
A vagotomy is added to manage type 2 or type 3 gastric ulcers.
Patients who are hemodynamically stable but have intermittent bleeding requiring blood transfusions should undergo a truncal vagotomy and distal gastric resection to include the ulcer for types 1, 2, and 3 ulcers.
In patients who present with life-threatening hemorrhage and a type 1, 2, or 3 ulcer, biopsy and oversew or excision of the ulcer in combination with a truncal vagotomy and a drainage procedure should be considered.
Patients with type 4 ulcers usually present with hemorrhage. The left gastric artery should be ligated, and a biopsy should be performed on the ulcer. Then, the ulcer should be oversewn through a high gastrotomy.
Rebleeding rates for the procedures that keep the ulcer in situ range from 20% to 40%.
Gastric bleeding in the immediate postoperative period from recurrent peptic ulcer disease is initially best managed by endoscopic or angiographic means. If reoperation is required, gastric resection is usually indicated, because a repeat vagotomy is not reliable and a more definitive solution is warranted.
Previous
Next: Proton-Pump Inhibitors
Stress Ulcer Treatment
Knowledge of the predisposing conditions for stress ulceration allows the clinician to identify patients at risk for developing stress ulceration and gastrointestinal (GI) bleeding—respiratory failure with mechanical ventilation and coagulopathy being very prominent risk factors. Treatment in this group of high-risk patients should focus on prevention. This is best accomplished by treating the underlying causes of ulceration.
Aggressive support of hemodynamic parameters ensures adequate mucosal blood flow. In addition, several strategies have evolved to treat gastric luminal acidity.
Stress-related bleeding usually occurs 7-10 days after the initial insult but may manifest sooner. Initially, endoscopy is the most important diagnostic tool. The acute superficial erosions are multiple, begin in the fundus, and progress toward the antrum. Ninety percent of patients stop bleeding with conservative medical therapy that includes gastric acid–controlling medications to maintain the gastric luminal pH above 5.0.
PPIs are the drugs of choice for acid suppression in stress ulcer prophylaxis (SUP). The risk of bleeding in an intensive care unit (ICU) is reduced by some 60% in patients receiving SUP compared to those treated with placebo or no prophylaxis. Current evidence does not substantiate routine prophylaxis. Therefore, withhold SUP in the majority of hospitalized patients, unless they have multiple risk factors and are likely to benefit from preventative strategies. Both cost and potential side effects from unnecessary proton-pump inhibitor (PPI) use can be reduced from following these guidelines. [19, 91]
Endoscopic hemostasis is attempted using traditional techniques, including electrocoagulation, argon plasma coagulation (APC), or injection therapy. Selective angiographic catheterization of the left gastric artery may be attempted with selective infusion of vasopressin (48-72 h) or embolization using Gelfoam, coils, or autologous clot to embolize the left gastric artery. Regardless of the angiographic technique used, it is often unsuccessful because of the rich and extensive submucosal plexus and collateral circulation within the stomach.
Surgical treatment
Surgical intervention becomes necessary if nonoperative therapy fails and blood loss continues. The goals of operative treatment are to control bleeding and to reduce recurrent bleeding and mortality. These patients are at extremely high risk, and the most expeditious procedure is the best option.
Simply oversewing an actively bleeding erosion is sometimes effective enough to control the bleeding. In the setting of life-threatening hemorrhage not amenable to endoscopic control, gastric resection with or without vagotomy with reconstruction may be necessary.
The type of gastric resection depends on the location of the gastric erosions, ie, whether they are proximal or distal. The options are antrectomy and subtotal, near total, or total gastrectomy. Operative mortality rates range from 4% to 17%. The choice of the initial operation must be made with an understanding of the patient's condition, the amount and location of gastric disease, and an accurate assessment of one's technical ability to rapidly and safely perform a gastric resection. The trend has been to perform less surgery in general and to minimalize the type of surgical procedure performed.
Managing the underlying insult causing the gastric stress ulcerations is also important. This involves supportive measures to maintain acceptable hemodynamic parameters, to provide adequate nutritional support in the critically ill patient, and to treat sepsis (if present).
Previous
Next: Proton-Pump Inhibitors
Mallory-Weiss Syndrome Treatment
Distinguishing Mallory-Weiss syndrome from Boerhaave syndrome is critical. Although both entities share a common pathogenesis, their management is completely different.
Boerhaave syndrome represents a full-thickness transmural laceration with perforation of the esophagus. A Gastrografin swallow helps to confirm the presence of the perforation in most cases, and prompt surgical intervention is necessary to prevent mediastinitis and sepsis.
However, surgical intervention in Mallory-Weiss syndrome is required to achieve hemostasis in only 10% of cases. The bleeding from a Mallory-Weiss tear spontaneously ceases in over 80% of patients by the time endoscopy is performed. [23, 22]
For patients in whom bleeding is visualized at endoscopy, the endoscopic treatment options are electrocoagulation, heater-probe application, hemoclips, epinephrine injection, or sclerotherapy.
In a series published by Bataller et al, hemostasis was achieved in 100% of patients with Mallory-Weiss tears by using endoscopic sclerotherapy with epinephrine (1:10,000) and 1% polidocanol. Other nonoperative therapies are reserved for cases in which endoscopic attempts at creating hemostasis have failed.
Other available options are angiographic intra-arterial infusion of vasopressin and transcatheter embolization of branches of the left gastric artery using Gelfoam. Avoid the balloon tamponade technique using the Sengstaken-Blakemore tube in this particular circumstance, because this apparatus may extend the mucosal laceration into a transmural laceration with perforation.
Surgical intervention is indicated in patients with continued bleeding after failed attempts at nonoperative therapies.
Bleeding from the gastroesophageal junction is visualized through an anterior gastrotomy. Once the tear is localized, the bleeding is controlled by oversewing the lesion.
The overall mortality rates for patients who require emergent surgery is 15%-25%, in contrast to a mortality of 3% or less for patients whose bleeding stops by the time of the initial endoscopy.
SeeMallory-Weiss Tear for more information on this topic.
Previous
Next: Proton-Pump Inhibitors
Dieulafoy Lesion Treatment
The initial endoscopic management of a Dieulafoy lesion can be highly successful. In a report by Norton et al describing their experience with 90 Dieulafoy lesions, endoscopic management achieved primary hemostasis in 96% of cases. The 30-day mortality was 13%, which is a reflection of the severe comorbid conditions associated with patients who have bleeding from a Dieulafoy lesion.
Contact thermal ablation with a heater probe is a very effective technique, with or without the combined use of epinephrine to slow or stop the bleeding prior to applying the heater probe. Argon plasma coagulation (APC) and endoclips have also been used successfully for hemostasis. No studies have been performed that compare surgical and endoscopic therapy for Dieulafoy lesions.
Although surgical intervention may be required after failed endoscopic therapy, endoscopy is still an important adjunct for management, because a nonbleeding Dieulafoy lesion may be undetectable through a gastrotomy.
Because of this potential problem, a combined endoscopic and surgical approach has been adopted. The vascular malformation can be marked with India ink through the endoscope.
Rebleeding after endoscopic therapy occurs in 11%-15% of cases, with most cases of rebleeding controlled at repeat endoscopy. (Repeat endoscopy in patients who have rebleeding has been validated in controlled studies of endoscopy and surgery.)
Previous
Next: Proton-Pump Inhibitors
Angiodysplasia Treatment
Bleeding from angiodysplasias can range from occult blood loss to life-threatening hemorrhage. Because the lesions are small and superficial, endoscopic therapy is highly successful. Endoscopic treatments and devices used for hemostasis include argon plasma coagulation (APC), contact heat probes, electrocoagulation, and injection therapy.
The contact probe coagulators have been the most common form of endoscopic treatment because of their proven success and ability to target a bleeding lesion tangentially. Similarly, a noncontact option, APC, is very effective with options of a straight firing, side firing, and circumferential firing probes that result in an increased ease of use with targeting these flat and sometimes broad areas of involvement. APC would be the treatment of choice when treating gastric antral vascular ectasia (GAVE), as it allows the endoscopist to apply prompt and effective “painting” of the angiodysplastic lesions in the distal stomach.
Recurrent bleeding can occur from the mucosal injury caused by the coagulation. To overcome the possibility of a delayed hemorrhage, endoscopic band ligation has been applied for hemostasis in nonvariceal gastrointestinal bleeding, including angiodysplasias.
When endoscopic techniques fail, surgical resection becomes necessary. When pangastric involvement is the source of bleeding, a total gastrectomy may be required; however, this is extremely rare. Available nonsurgical options include angiography with catheter-directed vasopressin. Combined hormonal therapy with estrogen and progesterone for patients in whom the diagnosis is unknown and vascular lesions are suggested has not been demonstrated to be effective.
Previous
Next: Proton-Pump Inhibitors
Aortoenteric Fistula
Patients with an aortoenteric fistula most often present with a self-limiting sentinel hemorrhage that is then followed by an exsanguinating, massive gastrointestinal (GI) bleed. For the warning lesser sentinel bleed in a patient with a history of an abdominal aortic aneurysm repair or a known aortic aneurysm, the possibility of a graft-enteric fistula should be considered.
An upper endoscopy is the procedure of choice to help diagnose the fistula. It should be performed to the ligament of Treitz. Upper endoscopy findings also help to exclude other sources of upper GI bleeding (UGIB).
Once the diagnosis of aortoenteric fistula is confirmed or seriously considered, emergency surgical intervention is required. In most instances, the aortic graft is removed after debridement and closure of the duodenum, followed by an extra-anatomic vascular bypass in order to bypass the ligated aorta and revascularize the lower extremities.
Perioperative mortality is 22%-75%, and major complications are common. Published opinions state that graft excision is not necessary as long as no gross contamination and purulence are present at the time of laparotomy. Under these circumstances, antibiotics are administered long-term.
Another option in the surgical literature is the use of endovascular stents to repair the fistula. [142, 143] Endovascular stent management is technically feasible and may be used as a bridge to more definitive treatment after hemodynamic stabilization in high-risk surgical patients. Stent grafting controls hemorrhage immediately; however, because the device is placed in an infected field, adjunctive measures, such as long-term antibiotic use, percutaneous drainage, and bowel diversion, may be required.
Although endovascular stents have been shown to be effective in treating aortoenteric fistulas, case reports have described aortoenteric fistulas in patients with abdominal aortic aneurysm treated initially with stent grafts as well.
Previous
Next: Proton-Pump Inhibitors
Treatment Complications
Complications of endoscopic therapy include aspiration pneumonia and perforation (1% for the first endoscopic therapy, 3% for the second). Bleeding can be caused by drilling into the vessel with contact thermal probes, by perforating the vessel with an injection, or by removing the clot with failure to coagulate the vessel.
Endoscopy is safe and effective in patients who present with upper gastrointestinal (GI) bleeding (UGIB). Careful consideration of the patient’s underlying comorbidities must be used to optimize the performance of endoscopy. As has been stated, mortality in UGIB is most often from the patient’s underlying comorbidities. Therefore, strategies to manage the bleeding episode and in preventing rebleeding need to include the management of the comorbidities. The value of a multidisciplinary team approach in managing these patients is strongly recommended.
The role of therapeutic angiography following failed endoscopic management should be considered as a favorable option to emergency surgery because there are similar success rates with hemostasis but potentially lower morbidity and mortality. (See Angiography)
Complications from emergency abdominal surgery include ileus, sepsis, poor wound healing, and myocardial infarction.
Salvage surgery is associated with a high mortality rate, reflecting the comorbidities of patients who rebleed or continue to bleed.
Previous
Next: Proton-Pump Inhibitors
Posttreatment Monitoring and Care
The 2010 international consensus guidelines on upper gastrointestinal (GI) bleeding (UGIB) state that selected low-risk patients may be discharged immediately following endoscopy, but high-risk patients should remain hospitalized for at least 72 hours.
According to the 2008 Scottish Intercollegiate Guidelines Network (SIGN) guideline, patients with a post-endoscopic Rockall score of less than 3 have a low risk of rebleeding or death and are candidates for early discharge and outpatient follow up.
More recent international validation of the Glasgow-Blatchford bleed score (GBS) has confirmed that a score of 0 or 1 is associated with a very low risk of intervention and that hospital admission and emergency endoscopy are not required.
The goal is to maintain the intragastric pH above 6 to maintain the clot. This is most easily achieved by intravenous proton pump inhibitor (PPI) therapy. After the acute phase, 72 hours, the coagulated vessel should be stable and the patient can be switched to oral therapy. If the patient rebleeds or has ongoing bleeding, repeat endoscopic therapy is suggested. If this is not successful, interventional radiology is performed to clot the bleeding vessel. If this fails, surgery would be considered.
The greatest risk for perforation is usually within the first 48 hours after endoscopic therapy. In the subsequent 48-72 hours after endoscopic therapy, the patient should receive acid-suppressive therapy to maintain a high gastric pH (above 6). A high gastric pH can be achieved by a continuous infusion of high-dose intravenous PPI therapy.
Tachyphylaxis may develop within 24 hours if H2-receptor antagonists are administered.
Patients who do not require endoscopic therapy and do not have other comorbidities should be considered for discharge.
Patients who did not require endoscopic treatment should receive routine, oral dosing of a PPI, ie, daily dosing prior to breakfast. Whether high-dose intravenous proton-pump inhibitor (PPI) therapy is advantageous in this setting remains controversial.
Oral PPI therapy can be used with any of the oral PPI preparations.
Patients should be tested for H pylori either by histology of gastric biopsy specimens taken on initial upper endoscopy or by other tests of active infection. Serologic testing should be avoided as it cannot be used to diagnose an active infection. If H pylori testing is positive,H pylori therapy should be instituted after the patient has been discharged and is in stable condition. Moreover,H pylori eradication should be confirmed 4-6 weeks later in patients with UGIB. This can be done by checking the stool for the H pylori antigen, or an H pylori breath test. The accuracy of eradication testing is much more reliable 2-4 weeks after the therapy has been completed and the patient has had no further antibiotics or antisecretory therapy. [146, 147]
Data on acid suppression via oral PPI therapy in order to produce a reduction in rebleeding are limited. High-dose intravenous PPI therapy appears to reduce rebleeding, but PPIs are not currently approved by the US Food and Drug Administration (FDA) for such treatment. The patient may be fed after recovery from local and intravenous anesthesia.
Some patients may require further endoscopic therapy. If repeat endoscopic therapy is needed, the stomach will usually empty liquids without residue within 2-3 hours. The 2011 American Society of Anesthesiologists (ASA) guidelines recommend a minimum of 2 hours without oral intake before performing endoscopy.
The 2008 SIGN guideline recommends repeat endoscopy and endotherapy within 24 hours when initial endoscopic treatment is deemed suboptimal or in patients in whom rebleeding will likely be life threatening.
If the patient remains stable, the patient can then be started on therapy for ulcer healing.
The patient should continue oral therapy for ulcer disease noted on endoscopy or for ulcers caused by cautery techniques during endoscopic therapy, only for a duration long enough to heal the ulceration.
Long-term acid suppression to prevent ulcer recurrence or its complications may not be required in patients with low-risk endoscopic findings and should be individualized in those with the need for continued nonsteroidal anti-inflammatory drug (NSAID) use, aspirin or other antiplatelet therapies, or anticoagulation. [149, 65]
Aspirin and NSAID therapies should be avoided in view of their adverse effect on platelet aggregation and ulcer healing. However, according to the 2010 international consensus guidelines, resumption of aspirin therapy in patients who require anticlotting prophylaxis should not be delayed as cardiovascular risks outweigh the risk of rebleeding.
The 2008 SIGN guidelines state that patients with healed bleeding ulcers who are negative for H pylori require concomitant PPI therapy at the usual daily dose if NSAIDs, aspirin, or cyclooxygenase (COX)-2 inhibitors are indicated.
If patients must remain on NSAIDs or low-dose aspirin, secondary prophylaxis against NSAID-induced ulcers should be given. According to the 2010 international consensus guidelines on UGIB, postdischarge use of aspirin or NSAIDs requires cotherapy with a PPI.
The patient's hemoglobin value should be monitored to assess the efficacy of iron therapy as an outpatient; further improvement should be noted. Oral or parenteral iron supplementation to treat posthemorrhagic anemia are both effective options. Erythropoietin analogues have been shown to be effective in increasing the rate of hemoglobin production after ulcer hemorrhage, but they may not have a favorable cost-benefit ratio.
Repeat endoscopy should be done at follow-up in patients with gastric ulcers to document ulcer healing and to exclude cancer.
Previous
Next: Proton-Pump Inhibitors
Deterrence and Prevention of UGIB
H pylori eradication therapy should be given if H pylori is present in the setting of any history of ulcer disease. Eradication of H pylori has been demonstrated to reduce the risk of recurrent ulcers and, therefore, recurrent ulcer hemorrhages.
Avoid nonsteroidal anti-inflammatory drugs (NSAIDs). If this is not possible, use the lowest dose and duration.
Proton-pump inhibitors (PPIs) or misoprostol cotherapy should be used along with NSAIDs.
The use of cyclooxygenase (COX)-2 inhibitors has been shown to reduce the risk of ulcer hemorrhage, although only when not combined with aspirin therapy. Concerns have been raised about an increase in myocardial infarction and stroke in patients taking selective COX-2 inhibitors. More recent data suggest this may be a risk with all NSAIDs. [151, 152]
As demonstrated in the study by al-Assi et al, the combination of H pylori infection and NSAID use may increase the risk of ulcer hemorrhage.
Previous
Guidelines
;)
References
al-Assi MT, Genta RM, Karttunen TJ, Graham DY. Ulcer site and complications: relation to Helicobacter pylori infection and NSAID use. Endoscopy. 1996 Feb. 28(2):229-33. [QxMD MEDLINE Link].
Frattaroli FM, Casciani E, Spoletini D, et al. Prospective study comparing multi-detector row CT and endoscopy in acute gastrointestinal bleeding. World J Surg. 2009 Oct. 33(10):2209-17. [QxMD MEDLINE Link].
Lam KL, Wong JC, Lau JY. Pharmacological treatment in upper gastrointestinal bleeding. Curr Treat Options Gastroenterol. 2015 Dec. 13(4):369-76. [QxMD MEDLINE Link].
Curdia-Goncalves T, Rosa B, Cotter J. New insights on an old medical emergency: non-portal hypertension related upper gastrointestinal bleeding. Rev Esp Enferm Dig. 2016 Oct. 108(10):648-56. [QxMD MEDLINE Link].
Lirio RA. Management of upper gastrointestinal bleeding in children: variceal and nonvariceal. Gastrointest Endosc Clin N Am. 2016 Jan. 26(1):63-73. [QxMD MEDLINE Link].
Fallah MA, Prakash C, Edmundowicz S. Acute gastrointestinal bleeding. Med Clin North Am. 2000 Sep. 84(5):1183-208. [QxMD MEDLINE Link].
Pongprasobchai S, Nimitvilai S, Chasawat J, Manatsathit S. Upper gastrointestinal bleeding etiology score for predicting variceal and non-variceal bleeding. World J Gastroenterol. 2009 Mar 7. 15(9):1099-104. [QxMD MEDLINE Link]. [Full Text].
Straube S, Tramer MR, Moore RA, Derry S, McQuay HJ. Mortality with upper gastrointestinal bleeding and perforation: effects of time and NSAID use. BMC Gastroenterol. 2009 Jun 5. 9:41. [QxMD MEDLINE Link]. [Full Text].
Yavorski RT, Wong RK, Maydonovitch C, Battin LS, Furnia A, Amundson DE. Analysis of 3,294 cases of upper gastrointestinal bleeding in military medical facilities. Am J Gastroenterol. 1995 Apr. 90(4):568-73. [QxMD MEDLINE Link].
Stabile BE, Stamos MJ. Surgical management of gastrointestinal bleeding. Gastroenterol Clin North Am. 2000 Mar. 29(1):189-222. [QxMD MEDLINE Link].
Cheung FK, Lau JY. Management of massive peptic ulcer bleeding. Gastroenterol Clin North Am. 2009 Jun. 38(2):231-43. [QxMD MEDLINE Link].
Tiriveedhi K, Simon J, Cerulli MA. Does gastric lavage reduce the detection of Helicobacter pylori in the biopsy specimens?. Gastrointest Endosc. 2007 Apr. 65(5):AB312.
Boonpongmanee S, Fleischer DE, Pezzullo JC, et al. The frequency of peptic ulcer as a cause of upper-GI bleeding is exaggerated. Gastrointest Endosc. 2004 Jun. 59(7):788-94. [QxMD MEDLINE Link].
Elmunzer BJ, Young SD, Inadomi JM, Schoenfeld P, Laine L. Systematic review of the predictors of recurrent hemorrhage after endoscopic hemostatic therapy for bleeding peptic ulcers. Am J Gastroenterol. 2008 Oct. 103(10):2625-32; quiz 2633. [QxMD MEDLINE Link].
Sung JJ, Tsoi KK, Ma TK, Yung MY, Lau JY, Chiu PW. Causes of mortality in patients with peptic ulcer bleeding: a prospective cohort study of 10,428 cases. Am J Gastroenterol. 2010 Jan. 105(1):84-9. [QxMD MEDLINE Link].
Corson JD, Williamson RCN, eds. Surgery. London, UK: Mosby-Year Book; 2001.
Stanley AJ, Laine L. Management of acute upper gastrointestinal bleeding. BMJ. 2019 Mar 25. 364:l536. [QxMD MEDLINE Link]. [Full Text].
Stollman N, Metz DC. Pathophysiology and prophylaxis of stress ulcer in intensive care unit patients. J Crit Care. 2005 Mar. 20(1):35-45. [QxMD MEDLINE Link].
Barletta JF, Bruno JJ, Buckley MS, Cook DJ. Stress ulcer prophylaxis. Crit Care Med. 2016 Jul. 44(7):1395-405. [QxMD MEDLINE Link].
Bardou M, Quenot JP, Barkun A. Stress-related mucosal disease in the critically ill patient. Nat Rev Gastroenterol Hepatol. 2015 Feb. 12(2):98-107. [QxMD MEDLINE Link].
Quenot JP, Thiery N, Barbar S. When should stress ulcer prophylaxis be used in the ICU?. Curr Opin Crit Care. 2009 Apr. 15(2):139-43. [QxMD MEDLINE Link].
Michel L, Serrano A, Malt RA. Mallory-Weiss syndrome. Evolution of diagnostic and therapeutic patterns over two decades. Ann Surg. 1980 Dec. 192(6):716-21. [QxMD MEDLINE Link]. [Full Text].
Cameron JL, ed. Current Surgical Therapy. 5th ed. St. Louis, Mo: Mosby-Year Book; 1995.
Jensen DM, Machicado GA, Hirabayashi K. Randomized controlled study of 3 different types of hemoclips for hemostasis of bleeding canine acute gastric ulcers. Gastrointest Endosc. 2006 Nov. 64(5):768-73. [QxMD MEDLINE Link].
Larson G, Schmidt T, Gott J, Bond S, O'Connor CA, Richardson JD. Upper gastrointestinal bleeding: predictors of outcome. Surgery. 1986 Oct. 100(4):765-73. [QxMD MEDLINE Link].
Reilly HF 3rd, al-Kawas FH. Dieulafoy's lesion. Diagnosis and management. Dig Dis Sci. 1991 Dec. 36(12):1702-7. [QxMD MEDLINE Link].
Pilotto A, Maggi S, Noale M, Franceschi M, Parisi G, Crepaldi G. Development and validation of a new questionnaire for the evaluation of upper gastrointestinal symptoms in the elderly population: a multicenter study. J Gerontol A Biol Sci Med Sci. 2010 Feb. 65(2):174-8. [QxMD MEDLINE Link].
Jairath V, Desborough MJ. Modern-day management of upper gastrointestinal haemorrhage. Transfus Med. 2015 Dec. 25(6):351-7. [QxMD MEDLINE Link].
Lanas A, Perez-Aisa MA, Feu F, et al. A nationwide study of mortality associated with hospital admission due to severe gastrointestinal events and those associated with nonsteroidal antiinflammatory drug use. Am J Gastroenterol. 2005 Aug. 100(8):1685-93. [QxMD MEDLINE Link].
D'Souza D, et al. Upper gastrointestinal bleeding. Radiopaedia. Available at Accessed: March 24, 2019.
Patel SD, Desai R, Patel U, et al. Thirty-day readmissions after upper and lower gastrointestinal hemorrhage: a national perspective in the United States. J Clin Gastroenterol. 2019 Sep. 53(8):582-90. [QxMD MEDLINE Link].
Peter DJ, Dougherty JM. Evaluation of the patient with gastrointestinal bleeding: an evidence based approach. Emerg Med Clin North Am. 1999 Feb. 17(1):239-61, x. [QxMD MEDLINE Link].
[Guideline] Adler DG, Leighton JA, Davila RE, et al, for ASGE. ASGE guideline: The role of endoscopy in acute non-variceal upper-GI hemorrhage. Gastrointest Endosc. 2004 Oct. 60(4):497-504. [QxMD MEDLINE Link].
Laine L, Shah A. Randomized trial of urgent vs. elective colonoscopy in patients hospitalized with lower GI bleeding. Am J Gastroenterol. 2010 Dec. 105(12):2636-41; quiz 2642. [QxMD MEDLINE Link].
Huang ES, Strate LL, Ho WW, Lee SS, Chan AT. Long-term use of aspirin and the risk of gastrointestinal bleeding. Am J Med. 2011 May. 124(5):426-33. [QxMD MEDLINE Link]. [Full Text].
American College of Surgeons Committee on Trauma. Advanced Trauma Life Support Course Manual. Chicago, Ill: American College of Surgeons; 1997.
Kim BS, Li BT, Engel A, et al. Diagnosis of gastrointestinal bleeding: a practical guide for clinicians. World J Gastrointest Pathophysiol. 2014 Nov 15. 5(4):467-78. [QxMD MEDLINE Link]. [Full Text].
Peery AF, Crockett SD, Barritt AS, et al. Burden of gastrointestinal, liver, and pancreatic diseases in the United States. Gastroenterology. 2015 Dec. 149(7):1731-1741.e3. [QxMD MEDLINE Link]. [Full Text].
Schenker MP, Majdalany BS, Funaki BS, et al. ACR appropriateness criteria on upper gastrointestinal bleeding. J Am Coll Radiol. 2010 Nov. 7(11):845-53. [QxMD MEDLINE Link].
Norton JA, Foster DS, Ito T, Jensen RT. Gastrinomas: medical or surgical treatment. Endocrinol Metab Clin North Am. 2018 Sep. 47(3):577-601. [QxMD MEDLINE Link].
Bornman PC, Theodorou NA, Shuttleworth RD, Essel HP, Marks IN. Importance of hypovolaemic shock and endoscopic signs in predicting recurrent haemorrhage from peptic ulceration: a prospective evaluation. Br Med J (Clin Res Ed). 1985 Jul 27. 291(6490):245-7. [QxMD MEDLINE Link]. [Full Text].
Silverstein FE, Gilbert DA, Tedesco FJ, Buenger NK, Persing J. The national ASGE survey on upper gastrointestinal bleeding. II. Clinical prognostic factors. Gastrointest Endosc. 1981 May. 27(2):80-93. [QxMD MEDLINE Link].
Villanueva C, Colomo A, Bosch A, et al. Transfusion strategies for acute upper gastrointestinal bleeding. N Engl J Med. 2013 Jan 3. 368(1):11-21. [QxMD MEDLINE Link]. [Full Text].
Waknine Y. GI bleeds: withholding transfusions boosts survival. Medscape Medical News. January 4, 2013. Available at Accessed: January 15, 2013.
[Guideline] Scottish Intercollegiate Guidelines Network (SIGN). Management of acute upper and lower gastrointestinal bleeding. A national clinical guideline. Edinburgh (Scotland): Scottish Intercollegiate Guidelines Network (SIGN); 2008 Sep. (SIGN publication; no. 105): [Full Text].
Peloquin JM, Seraj SM, King LY, Campbell EJ, Ananthakrishnan AN, Richter JM. Diagnostic and therapeutic yield of endoscopy in patients with elevated INR and gastrointestinal bleeding. Am J Med. 2016 Jun. 129(6):628-34. [QxMD MEDLINE Link].
Stanley AJ, Laine L, Dalton HR, et al, for the International Gastrointestinal Bleeding Consortium. Comparison of risk scoring systems for patients presenting with upper gastrointestinal bleeding: international multicentre prospective study. BMJ. 2017 Jan 4. 356:i6432. [QxMD MEDLINE Link]. [Full Text].
Abougergi MS, Charpentier JP, Bethea E, et al. A prospective, multicenter study of the AIMS65 Score compared with the Glasgow-Blatchford Score in predicting upper gastrointestinal hemorrhage outcomes. J Clin Gastroenterol. 2016 Jul. 50(6):464-9. [QxMD MEDLINE Link].
Saltzman JR, Tabak YP, Hyett BH, Sun X, Travis AC, Johannes RS. A simple risk score accurately predicts in-hospital mortality, length of stay, and cost in acute upper GI bleeding. Gastrointest Endosc. 2011 Dec. 74(6):1215-24. [QxMD MEDLINE Link].
Alexandrino G, Carvalho R, Reis J. Comparison of the AIMS65 score with other risk stratification scores in upper variceal and nonvariceal gastrointestinal bleeding. Gut Liver. 2018 Jan 15. 12(1):111-3. [QxMD MEDLINE Link]. [Full Text].
Gu L, Xu F, Yuan J. Comparison of AIMS65, Glasgow-Blatchford and Rockall scoring approaches in predicting the risk of in-hospital death among emergency hospitalized patients with upper gastrointestinal bleeding: a retrospective observational study in Nanjing, China. BMC Gastroenterol. 2018 Jun 28. 18(1):98. [QxMD MEDLINE Link]. [Full Text].
Robertson M, Majumdar A, Boyapati R, et al. Risk stratification in acute upper GI bleeding: comparison of the AIMS65 score with the Glasgow-Blatchford and Rockall scoring systems. Gastrointest Endosc. 2016 Jun. 83(6):1151-60. [QxMD MEDLINE Link].
Contreras-Omana R, Alfaro-Reynoso JA, Cruz-Chavez CE, et al. The Progetto Nazionale Emorragia Digestiva (PNED) system vs. the Rockall score as mortality predictors in patients with nonvariceal upper gastrointestinal bleeding: A multicenter prospective study. Rev Gastroenterol Mex. 2017 Apr - Jun. 82(2):123-8. [QxMD MEDLINE Link].
Shahrami A, Ahmadi S, Safari S. Full and modified Glasgow-Blatchford Bleeding Score in predicting the outcome of patients with acute upper gastrointestinal bleeding; a diagnostic accuracy study. Emerg (Tehran). 2018. 6(1):e31. [QxMD MEDLINE Link]. [Full Text].
Aoki T, Nagata N, Yamada A, et al. Next endoscopic approach for acute lower gastrointestinal bleeding without an identified source on colonoscopy: upper or capsule endoscopy?. Endosc Int Open. 2019 Mar. 7(3):E337-E346. [QxMD MEDLINE Link]. [Full Text].
Micic D, Gaetano JN, Nigam N, et al. Risk factors for small bowel bleeding in an overt gastrointestinal bleeding presentation after negative upper and lower endoscopy. PLoS One. 2019. 14(2):e0212509. [QxMD MEDLINE Link]. [Full Text].
Cai JX, Saltzman JR. Initial assessment, risk stratification, and early management of acute nonvariceal upper gastrointestinal hemorrhage. Gastrointest Endosc Clin N Am. 2018 Jul. 28(3):261-75. [QxMD MEDLINE Link].
Spiegel BM, Vakil NB, Ofman JJ. Endoscopy for acute nonvariceal upper gastrointestinal tract hemorrhage: is sooner better? A systematic review. Arch Intern Med. 2001 Jun 11. 161(11):1393-404. [QxMD MEDLINE Link]. [Full Text].
Tsoi KK, Ma TK, Sung JJ. Endoscopy for upper gastrointestinal bleeding: how urgent is it?. Nat Rev Gastroenterol Hepatol. 2009 Aug. 6(8):463-9. [QxMD MEDLINE Link].
Chandran S, Testro A, Urquhart P, et al. Risk stratification of upper GI bleeding with an esophageal capsule. Gastrointest Endosc. 2013 Jun. 77(6):891-8. [QxMD MEDLINE Link].
[Guideline] Schenker MP, Majdalany BS, Funaki BS, et al; and Expert Panel on Vascular Imaging and Interventional Radiology. ACR Appropriateness Criteria® upper gastrointestinal bleeding. [online publication]. Reston (VA): American College of Radiology (ACR); 2010. [Full Text].
Wortman JR, Landman W, Fulwadhva UP, Viscomi SG, Sodickson AD. CT angiography for acute gastrointestinal bleeding: what the radiologist needs to know. Br J Radiol. 2017 Jul. 90(1075):20170076. [QxMD MEDLINE Link]. [Full Text].
Lee HH, Park JM, Chun HJ, Oh JS, Ahn HJ, Choi MG. Transcatheter arterial embolization for endoscopically unmanageable non-variceal upper gastrointestinal bleeding. Scand J Gastroenterol. 2015 Jul. 50(7):809-15. [QxMD MEDLINE Link].
Kyaw M, Tse Y, Ang D, Ang TL, Lau J. Embolization versus surgery for peptic ulcer bleeding after failed endoscopic hemostasis: a meta-analysis. Endosc Int Open. 2014 Mar. 2(1):E6-E14. [QxMD MEDLINE Link]. [Full Text].
Laine L, Jensen DM. Management of patients with ulcer bleeding. Am J Gastroenterol. 2012 Mar. 107(3):345-60; quiz 361. [QxMD MEDLINE Link].
Houist G, Guillaumot MA, Soyer P, Coriat R. [May CT scan replace endoscopy to diagnose upper GI bleeding?] [French]. Presse Med. 2018 Sep. 47(9):831-3. [QxMD MEDLINE Link].
Millward SF. ACR appropriateness criteria on treatment of acute nonvariceal gastrointestinal tract bleeding. J Am Coll Radiol. 2008 Apr. 5(4):550-4. [QxMD MEDLINE Link].
Shukla PA, Zybulewski A, Kolber MK, Berkowitz E, Silberzweig J, Hayim M. No catheter angiography is needed in patients with an obscure acute gastrointestinal bleed and negative CTA. Clin Imaging. 2017 May - Jun. 43:106-19. [QxMD MEDLINE Link].
Iwasaki H, Shimura T, Yamada T, et al. Novel nasogastric tube-related criteria for urgent endoscopy in nonvariceal upper gastrointestinal bleeding. Dig Dis Sci. 2013 Sep. 58(9):2564-71. [QxMD MEDLINE Link].
Aljebreen AM, Fallone CA, Barkun AN. Nasogastric aspirate predicts high-risk endoscopic lesions in patients with acute upper-GI bleeding. Gastrointest Endosc. 2004 Feb. 59(2):172-8. [QxMD MEDLINE Link].
Tu TC, Lee CL, Wu CH, et al. Comparison of invasive and noninvasive tests for detecting Helicobacter pylori infection in bleeding peptic ulcers. Gastrointest Endosc. 1999 Mar. 49(3 Pt 1):302-6. [QxMD MEDLINE Link].
Liao CC, Lee CL, Lai YC, et al. Accuracy of three diagnostic tests used alone and in combination for detecting Helicobacter pylori infection in patients with bleeding gastric ulcers. Chin Med J (Engl). 2003 Dec. 116(12):1821-6. [QxMD MEDLINE Link].
Choi YJ, Kim N, Lim J, et al. Accuracy of diagnostic tests for Helicobacter pylori in patients with peptic ulcer bleeding. Helicobacter. 2012 Apr. 17(2):77-85. [QxMD MEDLINE Link].
Monteiro S, Goncalves TC, Magalhaes J, Cotter J. Upper gastrointestinal bleeding risk scores: Who, when and why?. World J Gastrointest Pathophysiol. 2016 Feb 15. 7(1):86-96. [QxMD MEDLINE Link].
Baradarian R, Ramdhaney S, Chapalamadugu R, et al. Early intensive resuscitation of patients with upper gastrointestinal bleeding decreases mortality. Am J Gastroenterol. 2004 Apr. 99(4):619-22. [QxMD MEDLINE Link].
Kaplan LJ, McPartland K, Santora TA, Trooskin SZ. Start with a subjective assessment of skin temperature to identify hypoperfusion in intensive care unit patients. J Trauma. 2001 Apr. 50(4):620-7; discussion 627-8. [QxMD MEDLINE Link].
Sarin N, Monga N, Adams PC. Time to endoscopy and outcomes in upper gastrointestinal bleeding. Can J Gastroenterol. 2009 Jul. 23(7):489-93. [QxMD MEDLINE Link]. [Full Text].
Green FW Jr, Kaplan MM, Curtis LE, Levine PH. Effect of acid and pepsin on blood coagulation and platelet aggregation. A possible contributor prolonged gastroduodenal mucosal hemorrhage. Gastroenterology. 1978 Jan. 74(1):38-43. [QxMD MEDLINE Link].
[Guideline] Fujishiro M, Iguchi M, Kakushima N, et al. Guidelines for endoscopic management of non-variceal upper gastrointestinal bleeding. Dig Endosc. 2016 May. 28(4):363-78. [QxMD MEDLINE Link]. [Full Text].
Yachimski PS, Farrell EA, Hunt DP, Reid AE. Proton pump inhibitors for prophylaxis of nosocomial upper gastrointestinal tract bleeding: effect of standardized guidelines on prescribing practice. Arch Intern Med. 2010 May 10. 170(9):779-83. [QxMD MEDLINE Link]. [Full Text].
Laine L. Clinical practice. Upper gastrointestinal bleeding due to a peptic ulcer. N Engl J Med. 2016 Jun 16. 374(24):2367-76. [QxMD MEDLINE Link].
Worden JC, Hanna KS. Optimizing proton pump inhibitor therapy for treatment of nonvariceal upper gastrointestinal bleeding. Am J Health Syst Pharm. 2017 Feb 1. 74(3):109-16. [QxMD MEDLINE Link].
[Guideline] Garcia-Iglesias P, Botargues JM, Feu Caballe F, et al. Management of non variceal upper gastrointestinal bleeding: position statement of the Catalan Society of Gastroenterology. Gastroenterol Hepatol. 2017 May. 40(5):363-74. [QxMD MEDLINE Link]. [Full Text].
Lau JY, Leung WK, Wu JC, et al. Omeprazole before endoscopy in patients with gastrointestinal bleeding. N Engl J Med. 2007 Apr 19. 356(16):1631-40. [QxMD MEDLINE Link].
Barkun AN, Herba K, Adam V, Kennedy W, Fallone CA, Bardou M. High-dose intravenous proton pump inhibition following endoscopic therapy in the acute management of patients with bleeding peptic ulcers in the USA and Canada: a cost-effectiveness analysis. Aliment Pharmacol Ther. 2004 Mar 1. 19(5):591-600. [QxMD MEDLINE Link].
Laine L, Shah A, Bemanian S. Intragastric pH with oral vs intravenous bolus plus infusion proton-pump inhibitor therapy in patients with bleeding ulcers. Gastroenterology. 2008 Jun. 134(7):1836-41. [QxMD MEDLINE Link].
Leontiadis GI, Sharma VK, Howden CW. Proton pump inhibitor therapy for peptic ulcer bleeding: Cochrane collaboration meta-analysis of randomized controlled trials. Mayo Clin Proc. 2007 Mar. 82(3):286-96. [QxMD MEDLINE Link].
Sreedharan A, Martin J, Leontiadis GI, et al. Proton pump inhibitor treatment initiated prior to endoscopic diagnosis in upper gastrointestinal bleeding. Cochrane Database Syst Rev. 2010 Jul 7. CD005415. [QxMD MEDLINE Link].
Lau JY, Sung JJ, Lam YH, et al. Endoscopic retreatment compared with surgery in patients with recurrent bleeding after initial endoscopic control of bleeding ulcers. N Engl J Med. 1999 Mar 11. 340(10):751-6. [QxMD MEDLINE Link]. [Full Text].
Leontiadis GI, Sharma VK, Howden CW. Systematic review and meta-analysis of proton pump inhibitor therapy in peptic ulcer bleeding. BMJ. 2005 Mar 12. 330(7491):568. [QxMD MEDLINE Link]. [Full Text].
Buendgens L, Koch A, Tacke F. Prevention of stress-related ulcer bleeding at the intensive care unit: Risks and benefits of stress ulcer prophylaxis. World J Crit Care Med. 2016 Feb 4. 5(1):57-64. [QxMD MEDLINE Link]. [Full Text].
Nardelli S, Gioia S, Ridola L, Farcomeni A, Merli M, Riggio O. Proton pump inhibitors are associated with minimal and overt hepatic encephalopathy and increased mortality in patients with cirrhosis. Hepatology. 2019 Aug. 70(2):640-9. [QxMD MEDLINE Link].
Weisman SM, Graham DY. Evaluation of the benefits and risks of low-dose aspirin in the secondary prevention of cardiovascular and cerebrovascular events. Arch Intern Med. 2002 Oct 28. 162(19):2197-202. [QxMD MEDLINE Link]. [Full Text].
[Guideline] US Preventive Services Task Force. Aspirin for the prevention of cardiovascular disease: U.S. Preventive Services Task Force recommendation statement. Ann Intern Med. 2009 Mar 17. 150(6):396-404. [QxMD MEDLINE Link]. [Full Text].
Ray WA, Chung CP, Murray KT, et al. Association of proton pump inhibitors with reduced risk of warfarin-related serious upper gastrointestinal bleeding. Gastroenterology. 2016 Dec. 151(6):1105-1112.e10. [QxMD MEDLINE Link]. [Full Text].
Agewall S, Cattaneo, Collet JP, et al, for the ESC Working Group on Cardiovascular Pharmacology and Drug Therapy and ESC Working Group on Thrombosis. Expert position paper on the use of proton pump inhibitors in patients with cardiovascular disease and antithrombotic therapy. Eur Heart J. 2013 Jun. 34(23):1708-13, 1713a-1713b. [QxMD MEDLINE Link]. [Full Text].
[Guideline] Hwang JH, Fisher DA, Ben-Menachem T, et al, for the Standards of Practice Committee of the American Society for Gastrointestinal Endoscopy. The role of endoscopy in the management of acute non-variceal upper GI bleeding. Gastrointest Endosc. 2012 Jun. 75(6):1132-8. [QxMD MEDLINE Link]. [Full Text].
Jeong N, Kim KS, Jung YS, Kim T, Shin SM. Delayed endoscopy is associated with increased mortality in upper gastrointestinal hemorrhage. Am J Emerg Med. 2019 Feb. 37(2):277-80. [QxMD MEDLINE Link].
Lim LG, Ho KY, Chan YH, et al. Urgent endoscopy is associated with lower mortality in high-risk but not low-risk nonvariceal upper gastrointestinal bleeding. Endoscopy. 2011 Apr. 43(4):300-6. [QxMD MEDLINE Link].
Tai CM, Huang SP, Wang HP, et al. High-risk ED patients with nonvariceal upper gastrointestinal hemorrhage undergoing emergency or urgent endoscopy: a retrospective analysis. Am J Emerg Med. 2007 Mar. 25(3):273-8. [QxMD MEDLINE Link].
Schacher GM, Lesbros-Pantoflickova D, Ortner MA, Wasserfallen JB, Blum AL, Dorta G. Is early endoscopy in the emergency room beneficial in patients with bleeding peptic ulcer? A "fortuitously controlled" study. Endoscopy. 2005 Apr. 37(4):324-8. [QxMD MEDLINE Link].
[Guideline] Gralnek IM, Dumonceau JM, Kuipers EJ, et al. Diagnosis and management of nonvariceal upper gastrointestinal hemorrhage: European Society of Gastrointestinal Endoscopy (ESGE) Guideline. Endoscopy. 2015 Oct. 47(10):a1-46. [QxMD MEDLINE Link]. [Full Text].
Cooper GS, Chak A, Way LE, Hammar PJ, Harper DL, Rosenthal GE. Early endoscopy in upper gastrointestinal hemorrhage: associations with recurrent bleeding, surgery, and length of hospital stay. Gastrointest Endosc. 1999 Feb. 49(2):145-52. [QxMD MEDLINE Link].
[Guideline] Barkun AN, Bardou M, Kuipers EJ, et al. International consensus recommendations on the management of patients with nonvariceal upper gastrointestinal bleeding. Ann Intern Med. 2010 Jan 19. 152(2):101-13. [QxMD MEDLINE Link]. [Full Text].
Gupta A, Agarwal R, Ananthakrishnan AN. "Weekend effect" in patients with upper gastrointestinal hemorrhage: A systematic review and meta-analysis. Am J Gastroenterol. 2018 Jan. 113(1):13-21. [QxMD MEDLINE Link].
Liu L, Hao D, Liu W, Wang L, Wang X. Does weekend hospital admission affect upper gastrointestinal hemorrhage outcomes?: A systematic review and network meta-analysis. J Clin Gastroenterol. 2018 Aug 16. [QxMD MEDLINE Link].
Matsui S, Kamisako T, Kudo M, Inoue R. Endoscopic band ligation for control of nonvariceal upper GI hemorrhage: comparison with bipolar electrocoagulation. Gastrointest Endosc. 2002 Feb. 55(2):214-8. [QxMD MEDLINE Link].
Giday SA, Kim Y, Krishnamurty DM, et al. Long-term randomized controlled trial of a novel nanopowder hemostatic agent (TC-325) for control of severe arterial upper gastrointestinal bleeding in a porcine model. Endoscopy. 2011 Apr. 43(4):296-9. [QxMD MEDLINE Link].
Laine L, McQuaid KR. Endoscopic therapy for bleeding ulcers: an evidence-based approach based on meta-analyses of randomized controlled trials. Clin Gastroenterol Hepatol. 2009 Jan. 7(1):33-47; quiz 1-2. [QxMD MEDLINE Link].
Vargas HE, Gerber D, Abu-Elmagd K. Management of portal hypertension-related bleeding. Surg Clin North Am. 1999 Feb. 79(1):1-22. [QxMD MEDLINE Link].
Jiang Y, Hu J, Li P, Jiang W, Liang W, Wei H. A retrospective analysis of cyanoacrylate injection versus hemoclip placement for bleeding Dieulafoy's lesion in duodenum. Gastroenterol Res Pract. 2018. 2018:3208690. [QxMD MEDLINE Link]. [Full Text].
Martinez-Alcala A, Monkemuller K. Emerging endoscopic treatments for nonvariceal upper gastrointestinal hemorrhage. Gastrointest Endosc Clin N Am. 2018 Jul. 28(3):307-20. [QxMD MEDLINE Link].
Jensen DM, Machicado GA. Hemoclipping of chronic canine ulcers: a randomized, prospective study of initial deployment success, clip retention rates, and ulcer healing. Gastrointest Endosc. 2009 Nov. 70(5):969-75. [QxMD MEDLINE Link]. [Full Text].
Cipolletta L, Bianco MA, Marmo R, et al. Endoclips versus heater probe in preventing early recurrent bleeding from peptic ulcer: a prospective and randomized trial. Gastrointest Endosc. 2001 Feb. 53(2):147-51. [QxMD MEDLINE Link].
Lin HJ, Hsieh YH, Tseng GY, Perng CL, Chang FY, Lee SD. A prospective, randomized trial of endoscopic hemoclip versus heater probe thermocoagulation for peptic ulcer bleeding. Am J Gastroenterol. 2002 Sep. 97(9):2250-4. [QxMD MEDLINE Link].
Saltzman JR, Strate LL, Di Sena V, et al. Prospective trial of endoscopic clips versus combination therapy in upper GI bleeding (PROTECCT--UGI bleeding). Am J Gastroenterol. 2005 Jul. 100(7):1503-8. [QxMD MEDLINE Link].
Chan SM, Chiu PW, Teoh AY, Lau JY. Use of the Over-the-Scope Clip for treatment of refractory upper gastrointestinal bleeding: a case series. Endoscopy. 2014 May. 46(5):428-31. [QxMD MEDLINE Link].
Manno M, Mangiafico S, Caruso A, et al. First-line endoscopic treatment with OTSC in patients with high-risk non-variceal upper gastrointestinal bleeding: preliminary experience in 40 cases. Surg Endosc. 2016 May. 30(5):2026-9. [QxMD MEDLINE Link].
Haddara S, Jacques J, Lecleire S, et al. A novel hemostatic powder for upper gastrointestinal bleeding: a multicenter study (the "GRAPHE" registry). Endoscopy. 2016 Dec. 48(12):1084-95. [QxMD MEDLINE Link].
Changela K, Papafragkakis H, Ofori E, et al. Hemostatic powder spray: a new method for managing gastrointestinal bleeding. Therap Adv Gastroenterol. 2015 May. 8(3):125-35. [QxMD MEDLINE Link]. [Full Text].
Cahyadi O, Bauder M, Meier B, Caca K, Schmidt A. Effectiveness of TC-325 (Hemospray) for treatment of diffuse or refractory upper gastrointestinal bleeding - a single center experience. Endosc Int Open. 2017 Nov. 5(11):E1159-E1164. [QxMD MEDLINE Link]. [Full Text].
Ghassemi KA, Jensen DM. The cutting edge: Doppler probe in guiding endoscopic hemostasis. Gastrointest Endosc Clin N Am. 2018 Jul. 28(3):321-30. [QxMD MEDLINE Link].
Jensen DM, Ohning GV, Kovacs TO, et al. Doppler endoscopic probe as a guide to risk stratification and definitive hemostasis of peptic ulcer bleeding. Gastrointest Endosc. 2016 Jan. 83(1):129-36. [QxMD MEDLINE Link]. [Full Text].
Jensen DM, Kovacs TOG, Ohning GV, et al. Doppler endoscopic probe monitoring of blood flow improves risk stratification and outcomes of patients with severe nonvariceal upper gastrointestinal hemorrhage. Gastroenterology. 2017 May. 152(6):1310-1318.e1. [QxMD MEDLINE Link]. [Full Text].
Bini EJ, Cohen J. Endoscopic treatment compared with medical therapy for the prevention of recurrent ulcer hemorrhage in patients with adherent clots. Gastrointest Endosc. 2003 Nov. 58(5):707-14. [QxMD MEDLINE Link].
Marmo R, Rotondano G, Bianco MA, Piscopo R, Prisco A, Cipolletta L. Outcome of endoscopic treatment for peptic ulcer bleeding: Is a second look necessary? A meta-analysis. Gastrointest Endosc. 2003 Jan. 57(1):62-7. [QxMD MEDLINE Link].
Park SJ, Park H, Lee YC, et al. Effect of scheduled second-look endoscopy on peptic ulcer bleeding: a prospective randomized multicenter trial. Gastrointest Endosc. 2018 Feb. 87(2):457-65. [QxMD MEDLINE Link].
Freeman ML, Cass OW, Peine CJ, Onstad GR. The non-bleeding visible vessel versus the sentinel clot: natural history and risk of rebleeding. Gastrointest Endosc. 1993 May-Jun. 39(3):359-66. [QxMD MEDLINE Link].
Poxon VA, Keighley MR, Dykes PW, Heppinstall K, Jaderberg M. Comparison of minimal and conventional surgery in patients with bleeding peptic ulcer: a multicentre trial. Br J Surg. 1991 Nov. 78(11):1344-5. [QxMD MEDLINE Link].
Leontiadis GI, Howden CW. The role of proton pump inhibitors in the management of upper gastrointestinal bleeding. Gastroenterol Clin North Am. 2009 Jun. 38(2):199-213. [QxMD MEDLINE Link].
Targownik LE, Bolton JM, Metge CJ, Leung S, Sareen J. Selective serotonin reuptake inhibitors are associated with a modest increase in the risk of upper gastrointestinal bleeding. Am J Gastroenterol. 2009 Jun. 104(6):1475-82. [QxMD MEDLINE Link].
Ali Khan M, Howden CW. The role of proton pump inhibitors in the management of upper gastrointestinal disorders. Gastroenterol Hepatol (N Y). 2018 Mar. 14(3):169-75. [QxMD MEDLINE Link]. [Full Text].
Scarpignato C, Gatta L, Zullo A, Blandizzi C, for the SIF-AIGO-FIMMG Group, on behalf of the Italian Society of Pharmacology, et al. Effective and safe proton pump inhibitor therapy in acid-related diseases - A position paper addressing benefits and potential harms of acid suppression. BMC Med. 2016 Nov 9. 14(1):179. [QxMD MEDLINE Link]. [Full Text].
Liu B, Liu S, Yin A, Siddiqi J. Risks and benefits of stress ulcer prophylaxis in adult neurocritical care patients: a systematic review and meta-analysis of randomized controlled trials. Crit Care. 2015 Nov 17. 19:409. [QxMD MEDLINE Link]. [Full Text].
So JB, Yam A, Cheah WK, Kum CK, Goh PM. Risk factors related to operative mortality and morbidity in patients undergoing emergency gastrectomy. Br J Surg. 2000 Dec. 87(12):1702-7. [QxMD MEDLINE Link].
Sadic J, Borgstrom A, Manjer J, Toth E, Lindell G. Bleeding peptic ulcer - time trends in incidence, treatment and mortality in Sweden. Aliment Pharmacol Ther. 2009 Aug 15. 30(4):392-8. [QxMD MEDLINE Link]. [Full Text].
Norton ID, Petersen BT, Sorbi D, Balm RK, Alexander GL, Gostout CJ. Management and long-term prognosis of Dieulafoy lesion. Gastrointest Endosc. 1999 Dec. 50(6):762-7. [QxMD MEDLINE Link].
Jo Y, Matsumoto T, Aoyagi K, Yano Y, Kawasaki A, Fujishima M. Endoscopic band ligation device for bleeding gastric angiodysplasia. Gastrointest Endosc. 1999 Oct. 50(4):599. [QxMD MEDLINE Link].
Mascarenhas A, Mendo R, Barreiro P, Chagas C. An uncommon cause of gastrointestinal bleeding: the calm before the storm. Rev Esp Enferm Dig. 2021 Aug 23. [QxMD MEDLINE Link].
Socrate AM, Rosati L, Locati P. Surgical treatment of aorto-enteric fistulas. Minerva Cardioangiol. 2001 Feb. 49(1):37-45. [QxMD MEDLINE Link].
Young RM, Cherry KJ Jr, Davis PM, et al. The results of in situ prosthetic replacement for infected aortic grafts. Am J Surg. 1999 Aug. 178(2):136-40. [QxMD MEDLINE Link].
Deshpande A, Lovelock M, Mossop P, Denton M, Vidovich J, Gurry J. Endovascular repair of an aortoenteric fistula in a high-risk patient. J Endovasc Surg. 1999 Nov. 6(4):379-84. [QxMD MEDLINE Link].
Lonn L, Dias N, Veith Schroeder T, Resch T. Is EVAR the treatment of choice for aortoenteric fistula?. J Cardiovasc Surg (Torino). 2010 Jun. 51(3):319-27. [QxMD MEDLINE Link].
Burks JA Jr, Faries PL, Gravereaux EC, Hollier LH, Marin ML. Endovascular repair of bleeding aortoenteric fistulas: a 5-year experience. J Vasc Surg. 2001 Dec. 34(6):1055-9. [QxMD MEDLINE Link].
Abou-Zamzam AM Jr, Bianchi C, Mazraany W, et al. Aortoenteric fistula development following endovascular abdominal aortic aneurysm repair: a case report. Ann Vasc Surg. 2003 Mar. 17(2):119-22. [QxMD MEDLINE Link].
[Guideline] Chey WD, Leontiadis GI, Howden CW, Moss SF. ACG clinical guideline: treatment of Helicobacter pylori Infection. Am J Gastroenterol. 2017 Feb. 112(2):212-39. [QxMD MEDLINE Link].
[Guideline] Malfertheiner P, Megraud F, O'Morain CA, et al, for the European Helicobacter and Microbiota Study Group and Consensus panel. Management of Helicobacter pylori infection-the Maastricht V/Florence Consensus Report. Gut. 2017 Jan. 66(1):6-30. [QxMD MEDLINE Link]. [Full Text].
[Guideline] Apfelbaum JL, Caplan RA, Connis RT, et al, for the American Society of Anesthesiologists Committee. Practice guidelines for preoperative fasting and the use of pharmacologic agents to reduce the risk of pulmonary aspiration: application to healthy patients undergoing elective procedures: an updated report by the American Society of Anesthesiologists Committee on Standards and Practice Parameters. Anesthesiology. 2011 Mar. 114(3):495-511. [QxMD MEDLINE Link]. [Full Text].
Penston JG, Wormsley KG. Review article: maintenance treatment with H2-receptor antagonists for peptic ulcer disease. Aliment Pharmacol Ther. 1992 Feb. 6(1):3-29. [QxMD MEDLINE Link].
Podolsky I, Storms PR, Richardson CT, Peterson WL, Fordtran JS. Gastric adenocarcinoma masquerading endoscopically as benign gastric ulcer. A five-year experience. Dig Dis Sci. 1988 Sep. 33(9):1057-63. [QxMD MEDLINE Link].
Pirlamarla P, Bond RM. FDA labeling of NSAIDs: Review of nonsteroidal anti-inflammatory drugs in cardiovascular disease. Trends Cardiovasc Med. 2016 Nov. 26(8):675-80. [QxMD MEDLINE Link].
Nissen SE, Yeomans ND, Solomon DH, et al, for the PRECISION Trial Investigators. Cardiovascular safety of celecoxib, naproxen, or ibuprofen for arthritis. N Engl J Med. 2016 Dec 29. 375(26):2519-29. [QxMD MEDLINE Link]. [Full Text].
[Guideline] Tarasconi A, Coccolini F, Biffl WL, et al. Perforated and bleeding peptic ulcer: WSES guidelines. World J Emerg Surg. 2020. 15:3. [QxMD MEDLINE Link]. [Full Text].
[Guideline] Early DS, Lightdale JR, Vargo JJ 2nd, et al, for the ASGE Standards of Practice Committee. Guidelines for sedation and anesthesia in GI endoscopy. Gastrointest Endosc. 2018 Feb. 87(2):327-337. [QxMD MEDLINE Link].
Lau JY, Leung WK, Wu JC, et al. Omeprazole before endoscopy in patients with gastrointestinal bleeding. N Engl J Med. 2007 Apr 19. 356(16):1631-40. [QxMD MEDLINE Link]. [Full Text].
Leontiadis GI, Sharma VK, Howden CW. Proton pump inhibitor therapy for peptic ulcer bleeding: Cochrane collaboration meta-analysis of randomized controlled trials. Mayo Clin Proc. 2007 Mar. 82(3):286-96. [QxMD MEDLINE Link].
Levine JE, Leontiadis GI, Sharma VK, Howden CW. Meta-analysis: the efficacy of intravenous H2-receptor antagonists in bleeding peptic ulcer. Aliment Pharmacol Ther. 2002 Jun. 16(6):1137-42. [QxMD MEDLINE Link].
Barkun AN, Bardou M, Martel M, Gralnek IM, Sung JJ. Prokinetics in acute upper GI bleeding: a meta-analysis. Gastrointest Endosc. 2010 Dec. 72(6):1138-45. [QxMD MEDLINE Link].
Carbonell N, Pauwels A, Serfaty L, Boelle PY, Becquemont L, Poupon R. Erythromycin infusion prior to endoscopy for acute upper gastrointestinal bleeding: a randomized, controlled, double-blind trial. Am J Gastroenterol. 2006 Jun. 101(6):1211-5. [QxMD MEDLINE Link].
Coffin B, Pocard M, Panis Y, et al, for the Groupe des endoscopistes de garde á l'AP-HP. Erythromycin improves the quality of EGD in patients with acute upper GI bleeding: a randomized controlled study. Gastrointest Endosc. 2002 Aug. 56(2):174-9. [QxMD MEDLINE Link].
Frossard JL, Spahr L, Queneau PE, et al. Erythromycin intravenous bolus infusion in acute upper gastrointestinal bleeding: a randomized, controlled, double-blind trial. Gastroenterology. 2002 Jul. 123(1):17-23. [QxMD MEDLINE Link].
Theivanayagam S, Lim RG, Cobell WJ, et al. Administration of erythromycin before endoscopy in upper gastrointestinal bleeding: a meta-analysis of randomized controlled trials. Saudi J Gastroenterol. 2013 Sep-Oct. 19(5):205-10. [QxMD MEDLINE Link]. [Full Text].
Szary NM, Gupta R, Choudhary A, et al. Erythromycin prior to endoscopy in acute upper gastrointestinal bleeding: a meta-analysis. Scand J Gastroenterol. 2011 Jul. 46(7-8):920-4. [QxMD MEDLINE Link].
Pateron D, Vicaut E, Debuc E, et al, for the HDUPE Collaborative Study Group. Erythromycin infusion or gastric lavage for upper gastrointestinal bleeding: a multicenter randomized controlled trial. Ann Emerg Med. 2011 Jun. 57(6):582-9. [QxMD MEDLINE Link].
Daram SR, Garretson R. Erythromycin is preferable to metoclopramide as a prokinetic in acute upper GI bleeding. Gastrointest Endosc. 2011 Jul. 74(1):234; author reply 234-5. [QxMD MEDLINE Link].
Ehrenpreis ED, Deepak P, Sifuentes H, Devi R, Du H, Leikin JB. The metoclopramide black box warning for tardive dyskinesia: effect on clinical practice, adverse event reporting, and prescription drug lawsuits. Am J Gastroenterol. 2013 Jun. 108(6):866-72. [QxMD MEDLINE Link].
[Guideline] Acosta RD, Abraham NS, Chandrasekhara V, et al, for the ASGE Standards of Practice Committee. The management of antithrombotic agents for patients undergoing GI endoscopy. Gastrointest Endosc. 2016 Jan. 83(1):3-16. [QxMD MEDLINE Link]. [Full Text].
Sung JJ, Lau JY, Ching JY, et al. Continuation of low-dose aspirin therapy in peptic ulcer bleeding: a randomized trial. Ann Intern Med. 2010 Jan 5. 152(1):1-9. [QxMD MEDLINE Link]. [Full Text].
Bager P, Dahlerup JF. Randomised clinical trial: oral vs. intravenous iron after upper gastrointestinal haemorrhage--a placebo-controlled study. Aliment Pharmacol Ther. 2014 Jan. 39(2):176-87. [QxMD MEDLINE Link]. [Full Text].
Tseng PH, Liou JM, Lee YC, et al. Emergency endoscopy for upper gastrointestinal bleeding in patients with coronary artery disease. Am J Emerg Med. 2009 Sep. 27(7):802-9. [QxMD MEDLINE Link].
Lai KC, Lam SK, Chu KM, et al. Lansoprazole for the prevention of recurrences of ulcer complications from long-term low-dose aspirin use. N Engl J Med. 2002 Jun 27. 346(26):2033-8. [QxMD MEDLINE Link].
Raskin JB, White RH, Jackson JE, et al. Misoprostol dosage in the prevention of nonsteroidal anti-inflammatory drug-induced gastric and duodenal ulcers: a comparison of three regimens. Ann Intern Med. 1995 Sep 1. 123(5):344-50. [QxMD MEDLINE Link]. [Full Text].
Bai Y, Guo JF, Li ZS. Meta-analysis: erythromycin before endoscopy for acute upper gastrointestinal bleeding. Aliment Pharmacol Ther. 2011 Jul. 34(2):166-71. [QxMD MEDLINE Link].
Brooks M. Poorer peptic ulcer outcomes without H. pylori etiology. Reuters Health Information. July 9, 2013. [Full Text].
Brooks M. SSRIs linked to upper GI bleeds. Medscape Medical News. September 24, 2013. Available at Accessed: September 30, 2012.
Chason RD, Reisch JS, Rockey DC. More favorable outcomes with peptic ulcer bleeding due to Helicobacter pylori. Am J Med. 2013 Sep. 126(9):811-18.e1. [QxMD MEDLINE Link].
Wang YP, Chen YT, Tsai CF, et al. Short-term use of serotonin reuptake inhibitors and risk of upper gastrointestinal bleeding. Am J Psychiatry. 2014 Jan. 171(1):54-61. [QxMD MEDLINE Link].
Bonnet N, Paul J, Helleputte T, et al. Novel insights into the assessment of risk of upper gastrointestinal bleeding in decompensated cirrhotic children. Pediatr Transplant. 2019 Jun. 23(4):e13390. [QxMD MEDLINE Link].
Toews I, George AT, Peter JV, et al. Interventions for preventing upper gastrointestinal bleeding in people admitted to intensive care units. Cochrane Database Syst Rev. 2018 Jun 4. 6:CD008687. [QxMD MEDLINE Link]. [Full Text].
Mavrogenis G, Del Natale M. Hemostatic clips and magnetic resonance imaging. Are there any compatibility issues?. Endoscopy. 2013 Nov. 45(11):933. [QxMD MEDLINE Link].
Honegger C, Valli PV, Wiegand N, Bauerfeind P, Gubler C. Establishment of Over-The-Scope-Clips (OTSC) in daily endoscopic routine. United European Gastroenterol J. 2017 Mar. 5(2):247-54. [QxMD MEDLINE Link]. [Full Text].
Kim JW, Jang JY, Lee CK, Shim JJ, Chang YW. Comparison of hemostatic forceps with soft coagulation versus argon plasma coagulation for bleeding peptic ulcer--a randomized trial. Endoscopy. 2015 Aug. 47(8):680-7. [QxMD MEDLINE Link].
Jensen DM, Markovic D, Jensen ME, Gornbein J. 784 Red cell transfusions do not increase 30 day mortality rates for unselected patients with severe UGI hemorrhage. Gastroenterology. 2015 Apr. 148(4) Suppl 1:S-154.
Holst LB, Petersen MW, Haase N, Perner A, Wetterslev J. Restrictive versus liberal transfusion strategy for red blood cell transfusion: systematic review of randomised trials with meta-analysis and trial sequential analysis. BMJ. 2015 Mar 24. 350:h1354. [QxMD MEDLINE Link]. [Full Text].
Media Gallery
Upper gastrointestinal bleeding (UGIB). Ulcer with active bleeding.
Upper gastrointestinal bleeding (UGIB). Ulcer with a clean base.
Upper gastrointestinal bleeding (UGIB). Diagram of an ulcer with a clean base.
Upper gastrointestinal bleeding (UGIB). Ulcer with an overlying clot.
Upper gastrointestinal bleeding (UGIB). Ulcer with a visible vessel.
Upper gastrointestinal bleeding (UGIB). Diagram of an ulcer with a visible vessel.
of 6
Tables
Table 1. Probable Source of GI Bleeding Within the Gut;)
Table 2.Estimated Fluid and Blood Losses in Shock;)
Table 3. Effect of Number of Packed Erythrocyte Transfusions on Need for Surgery and Mortality from UGIB;)
Table 4.Effect of the Color of the Nasogastric Aspirate and of the Stool on UGIB Mortality Rate;)
Table 5. Ulcer Characteristics and Correlations;)
Table 1. Probable Source of GI Bleeding Within the Gut
Clinical IndicatorProbability of Upper GI SourceProbability of Lower GI Source
Hematemesis Almost certain Rare
Melena Probable Possible
Hematochezia Possible Probable
Blood-streaked stool Rare Almost certain
Occult blood in stool Possible Possible
GI = gastrointestinal.
Table 2.Estimated Fluid and Blood Losses in Shock
Class 1Class 2Class 3Class 4
Blood Loss, mLUp to 750 750-1500 1500-2000>2000
Blood Loss, % blood volumeUp to 15 15-30 30-40>40
Pulse Rate, bpm< 100>100>120>140
Blood PressureNormal Normal Decreased Decreased
Respiratory RateNormal or Increased Decreased Decreased Decreased
Urine Output, mL/h>35 30-40 20-30 14-20
CNS/Mental StatusSlightly
anxious Mildly
anxious Anxious,
confused Confused,
lethargic
Fluid Replacement, 3-for-1 ruleCrystalloid Crystalloid Crystalloid and blood Crystalloid and blood
bpm = beats per minute; CNS = central nervous system
Table 3. Effect of Number of Packed Erythrocyte Transfusions on Need for Surgery and Mortality from UGIB
Number of Units Transfused Need for Surgery, %Mortality, %
0 4 4
1-3 6 14
4-5 17 28
5 57 43
Table 4.Effect of the Color of the Nasogastric Aspirate and of the Stool on UGIB Mortality Rate
Nasogastric Aspirate ColorStool ColorMortality, %
Clear Brown or red 6
Coffee-ground Brown or black 8.2
Red 19.1
Red blood Black 12.3
Brown 19.4
Red 28.7
UGIB = upper gastrointestinal bleeding.
Table 5. Ulcer Characteristics and Correlations
Ulcer CharacteristicsPrevalence Rate, %Rebleeding Rate, %Surgery Rate, %Mortality, %
Clean base 42 5 0.5 2
Flat spot 20 10 6 3
Adherent clot 17 22 10 7
Visible vessel 17 43 34 11
Active bleeding 18 55 35 11
Back to List
;)
Contributor Information and Disclosures
Author
Bennie Ray Upchurch, III, MD, FACP, AGAF, FACG, FASGE Staff Physician, Department of Gastroenterology, Licking Memorial Health Systems
Bennie Ray Upchurch, III, MD, FACP, AGAF, FACG, FASGE is a member of the following medical societies: American College of Gastroenterology, American College of Physicians, American Gastroenterological Association, American Society for Gastrointestinal Endoscopy
Disclosure: Nothing to disclose.
Chief Editor
BS Anand, MD Professor, Department of Internal Medicine, Division of Gastroenterology, Baylor College of Medicine
BS Anand, MD is a member of the following medical societies: American Association for the Study of Liver Diseases, American College of Gastroenterology, American Gastroenterological Association, American Society for Gastrointestinal Endoscopy
Disclosure: Nothing to disclose.
Additional Contributors
Maurice A Cerulli, MD, FACP, FACG, FASGE, AGAF Associate Professor of Clinical Medicine, Albert Einstein College of Medicine of Yeshiva University; Associate Professor of Clinical Medicine, Hofstra Medical School
Maurice A Cerulli, MD, FACP, FACG, FASGE, AGAF is a member of the following medical societies: American Association for the Study of Liver Diseases, American College of Gastroenterology, American College of Physicians, New York Society for Gastrointestinal Endoscopy, American Gastroenterological Association, American Medical Association, American Society for Gastrointestinal Endoscopy
Disclosure: Nothing to disclose.
Shahzad Iqbal, MD Advanced Endoscopy Fellow, Department of Gastroenterology, Columbia University Medical Center
Shahzad Iqbal, MD is a member of the following medical societies: American Association for the Study of Liver Diseases, American College of Gastroenterology, American College of Physicians, American Gastroenterological Association, American Society for Gastrointestinal Endoscopy
Disclosure: Nothing to disclose.
Acknowledgements
James de Caestecker, DO Instructor, Department of Surgery, MCP Hahnemann University
James de Caestecker, DO is a member of the following medical societies: American College of Surgeons
Disclosure: Nothing to disclose.
Michael A Grosso, MD Consulting Staff, Department of Cardiothoracic Surgery, St Francis Hospital
Michael A Grosso, MD is a member of the following medical societies: American College of Surgeons, Society of Thoracic Surgeons, and Society of University Surgeons
Disclosure: Nothing to disclose.
Douglas M Heuman, MD, FACP, FACG, AGAF Chief of Hepatology, Hunter Holmes McGuire Department of Veterans Affairs Medical Center; Professor, Department of Internal Medicine, Division of Gastroenterology, Virginia Commonwealth University School of Medicine
Douglas M Heuman, MD, FACP, FACG, AGAF is a member of the following medical societies: American Association for the Study of Liver Diseases, American College of Physicians, and American Gastroenterological Association
Disclosure: Novartis Grant/research funds Other; Bayer Grant/research funds Other; Otsuka Grant/research funds None; Bristol Myers Squibb Grant/research funds Other; Scynexis None None; Salix Grant/research funds Other; MannKind Other
Alex Jacocks, MD Program Director, Professor, Department of Surgery, University of Oklahoma School of Medicine
Disclosure: Nothing to disclose.
Jason Straus, MD Staff Physician, Department of Surgery, Wright State University School of Medicine
Jason Straus, MD is a member of the following medical societies: American College of Surgeons, American Medical Association, and Society of American Gastrointestinal and Endoscopic Surgeons
Disclosure: Nothing to disclose.
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Medscape Salary Employment
Close;)
What would you like to print?
What would you like to print?
Print this section
Print the entire contents of
Print the entire contents of article
Top Picks For You
encoded search term (Upper Gastrointestinal Bleeding (UGIB)) and Upper Gastrointestinal Bleeding (UGIB)
What to Read Next on Medscape
Related Conditions and Diseases
Upper Gastrointestinal Bleeding Imaging
Lower Gastrointestinal Bleeding
Gastric Ulcer Imaging
Small Intestinal Diverticulosis
Angiodysplasia of the Colon
Hemorrhagic Shock
News & Perspective
Smoking Tied to Higher Readmission for Upper GI Bleeding
Skip Antibiotic Prophylaxis for Upper GI Bleeding in Cirrhosis?
Blood Detection Capsule Helpful in Suspected Upper GI Bleeding
United European Gastroenterology (UEG) Week 2024
Real-Time, Blood-Sensing Capsule Accurately Diagnoses UGIB in 7 Minutes and May Help Patient Triage
Management of Gastrointestinal Bleeding and Resumption of Oral Anticoagulant Therapy in Patients With Atrial Fibrillation
Tools
Drug Interaction Checker
Pill Identifier
Calculators
Formulary
Sections Upper Gastrointestinal Bleeding (UGIB)
Overview
Practice Essentials
Background
Etiology
Epidemiology
Prognosis
Show All
Presentation
History
Physical Examination
Show All
DDx
Workup
Approach Considerations
Hemoglobin Value and Type and Crossmatch Blood
BMP, BUN, and Coagulation
Risk Scoring
Endoscopy
Chest Radiography
Computed Tomography Scanning
Nuclear Medicine Scanning
Angiography
Nasogastric Lavage
Histologic Findings
Show All
Treatment
Approach Considerations
Proton-Pump Inhibitors
Therapeutic Endoscopy
Bleeding Peptic Ulcer Treatment
Bleeding Gastric Ulcer Treatment
Stress Ulcer Treatment
Mallory-Weiss Syndrome Treatment
Dieulafoy Lesion Treatment
Angiodysplasia Treatment
Aortoenteric Fistula
Treatment Complications
Posttreatment Monitoring and Care
Deterrence and Prevention of UGIB
Show All
Guidelines
Perforated and Bleeding Peptic Ulcer Clinical Practice Guidelines (2020)
ASGE Guidelines for Sedation and Anesthesia in Gastrointestinal Endoscopy (2018)
Additional Resources
Show All
Medication
Medication Summary
Proton Pump Inhibitors
Prokinetic Agents
H pylori Agents
Non-steroidal Anti-inflammatory Drug
Histamine H2 Antagonists
Iron Products
Antibiotics
Show All
Media Gallery;)
Tables)
References;)
Recommended
2003 /viewarticle/1002454 Gearing Up to Manage Gastroesophageal Reflux Disease 0.75 CME / ABIM MOC Credits You are being redirected to Medscape Education Yes, take me there0.75 CME / ABIM MOC Gearing Up to Manage Gastroesophageal Reflux Disease
2001 /viewarticle/997407 Real-Time, Blood-Sensing Capsule Accurately Diagnoses UGIB in 7 Minutes and May Help Patient Triage
Medscape
Log in or register for free to unlock more Medscape content
Unlimited access to our entire network of sites and services
Log in or Register
Log in or register for free to unlock more Medscape content
Unlimited access to industry-leading news, research, resources, and more
Email
Continue
or
Log in with Google
Log in with Facebook
Log in with Apple
Policies
Medscape
About
For Advertisers
Privacy PolicyEditorial PolicyAdvertising PolicyYour Privacy ChoicesTerms of UseCookies
News & PerspectivesTools & ReferenceCME/CEVideoEventsSpecialtiesTopicsAccount InformationScribeNewsletters & Alerts
About MedscapeMedscape StaffMarket ResearchHelp CenterContact Us
Advertise with UsAdvertising Policy
Get the Medscape App
Download on the App Store
Get it on Google Play
All material on this website is protected by copyright, Copyright © 1994-2025 by WebMD LLC. This website also contains material copyrighted by 3rd parties.
Close |
189251 | https://es.wikipedia.org/wiki/Secci%C3%B3n_eficaz | Sección eficaz - Wikipedia, la enciclopedia libre
Ir al contenido
[x] Menú principal
Menú principal
mover a la barra lateral ocultar
Navegación
Portada
Portal de la comunidad
Actualidad
Cambios recientes
Páginas nuevas
Página aleatoria
Ayuda
Notificar un error
Páginas especiales
Buscar
Buscar
[x] Apariencia
Apariencia
mover a la barra lateral ocultar
Texto
Pequeño Estándar Grande
Esta página siempre usa texto pequeño
Anchura
Estándar Ancho
Esta página siempre es ancha
Color (beta)
Automático Claro Oscuro
Esta página está siempre en modo claro
Donaciones
Crear una cuenta
Acceder
[x] Herramientas personales
Donaciones
Crear una cuenta
Acceder
Páginas para editores desconectados más información
Contribuciones
Discusión
[x] Cambiar a la tabla de contenidos
Contenidos
mover a la barra lateral ocultar
Inicio
1 Introducción
2 Dependencia de la energía de σ(E)
3 Sección eficaz macroscópica
4 Véase también
Sección eficaz
[x] 36 idiomas
العربية
Беларуская
Català
Čeština
Deutsch
English
فارسی
Suomi
Français
Gaeilge
עברית
हिन्दी
Magyar
Հայերեն
Bahasa Indonesia
Italiano
日本語
한국어
Nederlands
Norsk nynorsk
Norsk bokmål
Polski
پنجابی
Português
Română
Русский
Slovenčina
Svenska
ไทย
Türkçe
Татарча / tatarça
Українська
اردو
Oʻzbekcha / ўзбекча
Tiếng Việt
中文
Editar enlaces
Artículo
Discusión
[x] español
Leer
Editar
Ver historial
[x] Herramientas
Herramientas
mover a la barra lateral ocultar
Acciones
Leer
Editar
Ver historial
General
Lo que enlaza aquí
Cambios en enlazadas
Subir archivo
Enlace permanente
Información de la página
Citar esta página
Obtener URL acortado
Descargar código QR
Editar enlaces interlingüísticos
Imprimir/exportar
Crear un libro
Descargar como PDF
Versión para imprimir
En otros proyectos
Elemento de Wikidata
De Wikipedia, la enciclopedia libre
La sección eficaz es una medida de la interacción entre proyectiles o partículas lanzadas contra un centro dispersor. Es una magnitud escalar que se mide en unidades de superficie. En ciertos casos representa la sección transversal efectiva vista por los proyectiles al aproximarse al blanco. Usualmente se representa con la letra griega sigma minúscula inicial o intermedia (σ). Se suele medir en barns o «barnios»:
1 b=10−28 m 2=10−24 c m 2{\displaystyle 1\ \mathrm {b} =10^{-28}\ \mathrm {m} ^{2}=10^{-24}\ \mathrm {cm} ^{2}}
Introducción
[editar]
Estadísticamente, a los núcleos de los átomos de una placa se les puede considerar círculos diminutos de radio r distribuidos a lo largo de un plano de superficie A. En el diagrama siguiente se representa un grupo de partículas a que inciden a velocidad V sobre un grupo de partículas X que actúan como blanco de las primeras. Así la probabilidad de impactar contra una de esas partículas distribuidas en la lámina es (nπr 2)/A, donde n representa el número de partículas X distribuidas en la superficie A.
El radio nuclear típico es de unos 10−12 cm. Por ello las secciones eficaces entre núcleos son del orden de 10−24 cm², valor que devino en unidad propia: el barn, o barnio. Dependiendo de qué reacciones se trate, las dimensiones de las secciones eficaces pueden variar enormemente: desde 0.001 barn hasta 1 000 barns.
El resultado de las partículas X al recibir el impacto de los corpúsculos a es un núcleo excitado que tras la fusión se desintegra y propicia una serie de posibilidades distintas o canales de salida, según la probabilidad de ocurrencia de cada uno.
a+X→C∗→Y+b{\displaystyle a+X\rightarrow C^{}\rightarrow Y+b}
La sección eficaz de las reacciones entre dichas partículas se calcula como sigue:
σ a x b=num. de reacciones por blanco X y por segundo Flujo de proyectiles=r e a c c i o n e s/c m 3/s p a r t.X/c m 3 p a r t.a c m 3⋅V(c m/s)=π λ 2 g Γ a Γ b Γ 2 f(E){\displaystyle {\sigma {ax}^{b}}={{\hbox{num. de reacciones por blanco X y por segundo}} \over {\hbox{Flujo de proyectiles}}}={\frac {\frac {reacciones/cm^{3}/s}{part.X/cm^{3}}}{{\frac {part.a}{cm^{3}}}\cdot V(cm/s)}}=\pi \lambda ^{2}g{\frac {\Gamma {a}\Gamma {b}}{\Gamma ^{2}}}f(E)}
Γ a{\displaystyle \Gamma {a}} representa la anchura del nivel de energía de la partícula a y Γ{\displaystyle \Gamma } la anchura de desintegración total. λ{\displaystyle \lambda } es la longitud de onda de De Broglie y f(E) es el factor de forma. Su valor dependerá de si ocurre –o no– resonancia nuclear. En caso de que no suceda este fenómeno su valor será constante. Así pues:
λ=ℏ p=ℏ(2 m E)1/2→π λ 2=0,657 A⋅E(M e V)b a r n{\displaystyle \lambda ={\frac {\hbar }{p}}={\frac {\hbar }{(2mE)^{1/2}}}\rightarrow \pi \lambda ^{2}={\frac {0,657}{A\cdot E(MeV)}}barn}A=(A a A x)/(A a+A x){\displaystyle A=(A_{a}A_{x})/(A_{a}+A_{x})}
En caso de que la energía de fusión entre las partículas a y X coincida con la de alguno de los niveles de energía acontece la resonancia nuclear. Entonces el factor de forma se torna dependiente de la energía y vale:
f(E)=Γ 2(E−E r e s)2+(Γ/2)2{\displaystyle f(E)={\frac {\Gamma ^{2}}{(E-E_{res})^{2}+(\Gamma /2)^{2}}}}
E res simboliza la energía de resonancia. Como se infiere fácilmente, a poco que E se aleje de E res el término dejará de contribuir. Por ello se le puede considerar pico de Dirac.
Dependencia de la energía de σ(E)
[editar]
La sección eficaz es un parámetro altamente dependiente de la energía. Por ello resulta complicado especular sus valores a bajas energías, más allá de donde se obtienen datos experimentales. A altas energías no es difícil recabar datos, ya que la probabilidad de ocurrencia de las reacciones es alta, pero a bajas energías la probabilidad es tan baja que con las muestras de partículas con las que se trabaja nunca ocurre algo.
Según la fórmula que se ha aportado de la sección eficaz, la dependencia de la energía sería λ 2∝1/E{\displaystyle \lambda ^{2}\propto 1/E}. Este es el recorrido libre medio. Γ a/Γ∝exp(−b/E 1/2){\displaystyle \Gamma {a}/\Gamma \propto \exp(-b/E^{1/2})} es el factor de penetración de la barrera coulombiana (más información en: Pico de Gamow). La razón de Γ b/Γ{\displaystyle \Gamma {b}/\Gamma \,!} es prácticamente constante. f(E){\displaystyle f(E)\,!} depende sólo en un margen estrecho, en las cercanías de la resonancia nuclear. Normalmente es constante.
Para resolver este problema, a partir de la sección eficaz se ha creado el factor astrofísico (S(E)), mucho menos dependiente de E, lo cual le permite ser más extrapolable. Se usa, sobre todo, en astrofísica, porque cambia poco a lo largo de la «vida» de una estrella.
S(E)=σ(E)exp(b E 1/2){\displaystyle S(E)=\sigma (E)\ \exp \left({\frac {b}{E^{1/2}}}\right)}
Como queda manifiesto, lo que se ha hecho es quitarle la dependencia respecto del factor de penetración.
Sección eficaz macroscópica
[editar]
Al producto (segundo miembro de la ecuación) Σ=N σ(E){\displaystyle \Sigma =N\sigma (E)} se le denomina sección eficaz macroscópica. N es la densidad de partículas blanco que pueden interaccionar. Las unidades resultantes para esta sección son de «longitud inversa».
Véase también
[editar]
Pico de Gamow
| Control de autoridades | Proyectos Wikimedia Datos:Q17128025 Identificadores BNF:11951137r(data) GND:4190024-8 LCCN:sh85034281 NLI:987007533512405171 Ontologías Número IEV:113-06-38 |
Datos:Q17128025
Obtenido de «
Categoría:
Física nuclear y de partículas
Categorías ocultas:
Wikipedia:Artículos con identificadores BNF
Wikipedia:Artículos con identificadores GND
Wikipedia:Artículos con identificadores LCCN
Esta página se editó por última vez el 27 mar 2025 a las 21:49.
El texto está disponible bajo la Licencia Creative Commons Atribución-CompartirIgual 4.0; pueden aplicarse cláusulas adicionales. Al usar este sitio aceptas nuestros términos de uso y nuestra política de privacidad.
Wikipedia® es una marca registrada de la Fundación Wikimedia, una organización sin ánimo de lucro.
Política de privacidad
Acerca de Wikipedia
Limitación de responsabilidad
Código de conducta
Desarrolladores
Estadísticas
Declaración de cookies
Versión para móviles
Edita configuración de previsualizaciones
Buscar
Buscar
[x] Cambiar a la tabla de contenidos
Sección eficaz
36 idiomasAñadir tema |
189252 | https://www.whitman.edu/documents/Academics/Mathematics/2016/Jacobson.pdf | Chaos: From Seeing to Believing Tate Jacobson Whitman Mathematics Department Spring 2016 Contents Introduction: Motivating the Study of Chaos 2 A Small Adjustment with Big Implications . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Our Goals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 1 Dynamical Systems 5 1.1 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 1.2 Types of Points in a Dynamical System . . . . . . . . . . . . . . . . . . . . . . . . .
8 1.3 Limits of Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 1.4 Hyperbolicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 1.5 Dynamical Systems: A Wide Field of Study . . . . . . . . . . . . . . . . . . . . . . .
12 2 Chaos 13 2.1 Our Definition of Chaos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 2.2 A Straightforward Example of a Chaotic Dynamical System . . . . . . . . . . . . . .
14 3 Topological Conjugacy: A Powerful Tool 16 3.1 The Properties Shared by Topologically Conjugate Maps . . . . . . . . . . . . . . . .
16 3.1.1 Fixed and Periodic Points in Topologically Conjugate Maps . . . . . . . . . .
16 3.1.2 Asymptotic Orbits in Topologically Conjugate Maps . . . . . . . . . . . . . .
18 3.1.3 How a Topological Conjugacy Maps Sets . . . . . . . . . . . . . . . . . . . . .
18 3.2 Topological Conjugacy and Chaos . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 4 Proving a Dynamical System is Chaotic Using Topological Conjugacy 21 4.1 The Shift Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 4.1.1 The Shift Map’s Domain: Sequence Space . . . . . . . . . . . . . . . . . . . .
21 4.1.2 The Shift Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 4.1.3 The Shift Map is Chaotic . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 4.2 Logistic Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24 4.3 Building our Topological Conjugacy: The Itinerary . . . . . . . . . . . . . . . . . . .
26 4.4 S : Λ →Σ2 is a Topological Conjugacy . . . . . . . . . . . . . . . . . . . . . . . . . .
26 4.5 Chaotic Logistic Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29 4.6 F4 : A Special Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29 Conclusion: Believing in Order 31 Appendix: Necessary Background for Chaos 32 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 Topology in a Metric Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36 Alphabetical Index 38 1 Introduction: Motivating the Study of Chaos In popular culture, the term “chaos” is most often a generic stand-in for “disorder.” It calls to mind countless headlines and over-the-top action movies—at the time of writing, searching for “chaos” on Google produces articles about recent Nascar wrecks, drama among the Kardashians, and “Captain American: Civil War.” These popular renderings of chaos as mere disorder fail to capture the central premise of the mathematical study of chaos: that chaos is not disorder, but rather order in apparent disorder.
Before cracking into any rigorous definitions, let us take a na¨ ıve look at an example of this sort of disorderly order.
A Small Adjustment with Big Implications Imagine that we are scientists tracking the size of a single population of rabbits over time. Further-more, suppose that, unbeknownst to us, the size of that population perfectly follows the logistic growth model described by the function f(x) = 2x(1 −x), where x denotes the population (in millions) of rabbits at the beginning of a given year and f(x) denotes the population by the end of that year.
Since the end of each year is the beginning of the next we see that we are dealing with a recursive process: for any integer n, if we let xn denote the population at the end of the nth year, then we see that xn = f(xn−1).
Suppose, then, that x0 = 0.01, where x0 denotes the initial population. We find that x1 = f(x0) = 2(0.01)(0.99) = 0.0198 and that x2 = f(x1) = 2(0.0198)(.9802) = 0.03881592.
Repeating this process, we produce the following table and graph which relate n and xn for the first few values of n: n xn 0 0.01000000 1 0.01980000 2 0.03881592 3 0.07461848 4 0.13810113 5 0.23805842 6 0.36277322 7 0.46233762 8 0.49716309 9 0.49998390 10 0.49999999 11 0.50000000 12 0.50000000 Figure 1: A sequence plot of {xn} under f.
Looking at this raw data we see that the population grows quickly and seems to level offat 0.5 (it is actually just approaching 0.5, but our calculators round offthe tiny differences past a certain point). Intuitively, this is how we would expect a population to grow: exploding initially, then reaching a stable value as the scarcity of resources forces members of the population to compete with each other. In our above formulation we might redundantly call this “orderly” order. Since this behavior fits our intuition we take the data at face-value.
2 Now suppose that the size of the population instead follows the function g(x) = 4x(1 −x), and suppose again that x0 = .01. Applying g a few times, we see: n xn 0 0.01000000 1 0.03960000 2 0.15212736 3 0.51593851 4 0.99898386 5 0.00406043 6 0.01617577 7 0.06365646 8 0.23841726 9 0.72629788 10 0.79515708 11 0.65152919 12 0.90815562 Figure 2: A sequence plot of {xn} under g.
Seeing this data, we will likely assume that some catastrophe occurred during the fifth year or that the population’s growth is entirely random.
When we slightly change our input value, we find that g has another bizarre characteristic. Let x0 = .01 as above and let y0 = .0099. The following image shows the sequence plots of {xn} and {yn} up to n = 25, with the former plotted in red and the latter in blue: Figure 3: Overlaid sequence plots of {xn} and {yn} under g, where x0 = .01 and y0 = .0099.
While the plots of {xn} and {yn} cleave to each other at first, past n = 14 their behavior seems to completely diverge.
As readers, we know that a simple rule is guiding the growth of this population, yet our data seems to defy any predictable patterns. How do we account for this behavior?
Our experience tracking the behavior of {xn} and {yn} under g echoes meteorologist Edward Lorenz’s experience in analyzing his primitive weather simulations in the early 1960s. Though he based his model on a handful of equations, he found that even the slightest changes in his initial values led to remarkably different outcomes. He initially assumed that something was wrong with 3 his computer or with his code, yet found that everything was in working order. As in our own experiment, the models themselves produced the seemingly random behavior. It was with this dis-covery of such disorderly order that Lorenz first sparked interest in what eventually became Chaos Theory. In observing the behavior of our seemingly bizarre logistic map, we too are beginning our journey into chaos in much the same way. (For a more complete and dramatic account of Lorenz’s story and the origins of Chaos theory, see Gleick .) Our Goals In this paper we will characterize the differences between non-chaotic and chaotic dynamics. The first section will introduce dynamical systems in general, providing vocabulary and highlighting why f in our na¨ ıve example is so predictable. The second section will then provide a general definition of chaos and give an example of a chaotic dynamical system. The third and fourth sections will then explore a powerful method for proving that dynamical systems are chaotic: topological conjugacy.
We will then close by using this method to prove that many logistic maps, like g, are chaotic.
As I am aiming this paper at a fairly general audience, I have included a lengthy appendix introduc-ing metric spaces and topology, as a basic understanding of both is essential to some of the deeper results that I cover. If at any point you encounter an unfamiliar term or question an assumption in a proof, check the appendix for details.
4 1 Dynamical Systems Before we discuss any particulars, we need to establish a broad definition of the sorts of systems we will be talking about.
Definition 1.0.1. A dynamical system consists of a set of possible states along with a rule that determines the present state as a function of past states.
It is important to note at this point that the dynamical systems which we will be exploring have a few restrictions. First, they are deterministic rather than stochastic. Simply put, this means that our rule will always return the same output for a given input, meaning that it is in no way random.
Second, they are discrete, as opposed to continuous, dynamical systems.
In our first example the set of states was the set of values the population of rabbits could take on and the rule which determined the present state as a function of past states was f. We used the output from one application of our rule as the input value for the next application of our rule. We call each re-application of our rule an iteration. As is clear from our example, with each iteration we are composing our rule with itself. To streamline our notation we write the second iterate of f, namely f(f(x)), as f 2(x), the third iterate of f, namely f(f(f(x))), as f 3(x), and the nth iterate of f as f n(x).
1.1 Orbits Definition 1.1.1. Given a map f and a point x in the domain of f, we call the set of points {x, f(x), f 2(x), . . .} the forward orbit of x under f and denote it O+(x). As x is the starting point of the orbit, we call it the initial value or seed of the orbit.
Returning to our examples, we see that our first table provides the forward orbit of .01 under f and that our second table provides the forward orbit of .01 under g.
While all of this vocabulary is helpful, a visual representation of orbits helps solidify the concept.
We call these diagrams cobweb plots and construct them as follows: Let x0 be the seed of our orbit. In our plot we graph both our function f(x) and the line g(x) = x.
With these guidelines, we first trace a line, in red, from (x0, 0) to (x0, f(x0)), then from (x0, f(x0)) to (f(x0), f(x0)) (this is where plotting g(x) = x is useful). From there we can trace a line to (f(x0), f 2(x0)), then to (f 2(x0), f 2(x0)), and so on. With these plots, we can find f n(x) for any n and, perhaps more importantly, see how the orbit of x got to f n(x).
Figure 4 provides an easy-to-follow, albeit somewhat bland, example of a cobweb plot.
With a basic understanding of cobweb plots, we can start to visual the behavior of f(x) = 2x(1 −x) and g(x) = 4x(1 −x) for larger values of n.
Figure 5 and Figure 6 show us that the orbit of x0 = .01 under f continues to approach .5.
Figures 7, 8, 9, and 10 reveal that the orbit of x0 = .01 under g seems to travel all over the interval [0, 1].
5 Figure 4: The cobweb plot of x0 = .75 under the map f(x) = x2 up to 5 iterations.
Figure 5: The cobweb plot of x0 = .01 under the map f(x) = 2x(1 −x) up to 12 iterations.
Figure 6: The cobweb plot of x0 = .01 under the map f(x) = 2x(1 −x) up to 100 iterations.
We see that nothing unexpected happens with the orbit of .01.
6 Figure 7: The cobweb plot of x0 = .01 under the map f(x) = 4x(1 −x) up to 5 iterations.
We see the sudden drop after the fifth iteration as we did in our table.
Figure 8: The cobweb plot of x0 = .01 under the map f(x) = 4x(1 −x) up to 25 iterations.
Clearly the orbit of .01 is covering a fair amount of the interval [0, 1].
Figure 9: The cobweb plot of x0 = .01 under the map f(x) = 4x(1 −x) up to 100 iterations.
We see that the orbit of .01 continues to hit new points.
Figure 10: The cobweb plot of x0 = .01 under the map f(x) = 4x(1−x) up to 1000 iterations.
The orbits of .01 is still covering new ground.
7 1.2 Types of Points in a Dynamical System Now that we are familiar with what dynamical systems are, we might wonder whether we can determine how the orbits of certain points will behave. The most predictable points in a dynamical system are undoubtedly fixed points, followed closely by periodic points.
Definition 1.2.1. Given a point p in the domain of f, if f(p) = p we call p a fixed point of the map f. We denote the set of fixed points of f by Fix(f).
Example 1.2.2. We can rather easily find the fixed points of f(x) = 2x(1 −x). We simply need to solve the equation f(p) = p for p: 2p(1 −p) = p (1) 2p −2p2 = p (2) p −2p2 = 0 (3) p(1 −2p) = 0 .
(4) Clearly Fix(f) = {0, 1 2}.
Example 1.2.3. We can find the fixed points of f(x) = 4x(1 −x) in a similar manner. We see: 4p(1 −p) = p (5) 4p −4p2 = p (6) 3p −4p2 = 0 (7) p(3 −4p) = 0 , (8) so that Fix(f) = {0, 3 4}.
Pulling together a few concepts from calculus and this definition, we have the following theorem which provides conditions on f and its domain which guarantee that f has a fixed point.
Theorem 1.2.4. If f : [a, b] →[a, b] is continuous, then f has at least one fixed point in [a, b].
Proof. Let g(x) = f(x) −x. As g is the difference of continuous functions, it is likewise continuous.
We note that a ≤f(a) and that f(b) ≤b as [a, b] is the range of f. If either f(a) = a or f(b) = b then we are already done. Suppose then that a < f(a) and f(b) < b. Then g(a) > 0 and g(b) < 0, so the Intermediate Value Theorem provides that there exists a point c between a and b such that g(c) = 0. Therefore f(c) = c and we are done.
We can show that if f has one additional property, then f has a unique fixed point. Note that in the following discussion, f ′ denotes the derivative of f.
Theorem 1.2.5. Let I be a closed interval. If f : I →I and |f ′(x)| < 1 for all x in I, then there exists a unique fixed point for f in I.
Proof. As f is differentiable on I it is continuous on I. Therefore our previous theorem guarantees that f has at least one fixed point on I. Suppose, then, that both x and y are fixed points and x ̸= y. By the Mean Value Theorem, there exists a point c between x and y such that f ′(c) = f(y) −f(x) y −x = y −x y −x = 1 a clear contradiction of our assumption that |f ′(x)| < 1 for all x in I. Thus x = y.
8 While fixed points have a number of unique qualities, they ultimately fall into the much larger category of periodic points.
Definition 1.2.6. We say that a point x0 is periodic of period n if f n(x0) = x0 for some n > 0.
We call the orbit of x0 a periodic orbit or cycle in this case.
We denote the set of periodic points of period n under f by Pern(f) and set of all periodic points under f by Per(f).
With this definition, we can see that Fix(f) = Per1(x).
It is important to note that the set of periodic points of period n might contain points with differ-ent periods. For example, if x1 has period 6 and x2 has period 4, then both are members of Per12(f).
Example 1.2.7. Let f(x) = x2 −1.
We see that x = −1 is a periodic point of period 2 as f 2(−1) = f(f(−1)) = f(0) = −1.
Moreover, it is interesting to note that if x is periodic of period n under f, then for every i ∈{1, . . . , n −1} f i(x) is likewise periodic of period n. We see that this is the case as f n(f i(x)) = f n+i(x) = f i+n(x) = f i(x).
Therefore, as x = −1 is periodic of period 2 under f(x) = x2 −1 and f(−1) = 0, we know that 0 is periodic of period 2 under f as well.
As you can imagine, the cobweb plot of the forward orbit of a fixed point is not too terribly exciting.
The cobweb plot of the forward orbit a periodic point, however, can prove a little more informative.
Figures 11 and 12 depict cycles of period 2 and 3 respectively.
Figure 11: The forward orbit of x = 0 under f(x) = x2 −1. As we can see, 0 is periodic of period 2 under this map.
Figure 12: The forward orbit of x = 0 under f(x) = −3 2x2 + 5 2x + 1.
As we can see, 0 is periodic of period 3 under this map.
Note that a general dynamical system, might not have any non-trivial periodic points at all.
Consider f : R →R defined by f(x) = 2x. Clearly only x0 = 0 is a periodic point under f.
9 A point that is not fixed or periodic itself may have points in its forward orbit that are either fixed or periodic.
Definition 1.2.8. We say that a point x is eventually fixed if x is not fixed but there exists an m > 0 such that f i+1(x) = f i(x) for all i ≥m (that is, f i(x) is fixed for i ≥m).
As we noted in the introduction, the orbit of x0 = .01 under f(x) = 2x(1 −x) seemed to level out at .5, but was actually just approaching 0.5. As such, .01 is not eventually fixed under f.
Definition 1.2.9. We say that a point x is eventually periodic of period n if x is not periodic but there exists an m > 0 such that f n+i(x) = f i(x) for all i ≥m (that is, f i(x) is periodic for i ≥m).
Example 1.2.10. Let f(x) = x2 −1 again. We see that x = 1 is an eventually periodic point of period 2 as f 2(1) = f(f(1)) = f(0) = −1 and f 4(1) = f(f(f 2(1))) = f(f(−1)) = f(0) = −1.
1.3 Limits of Orbits While not all points in a dynamical system have orbits which eventually cycle through a finite set of values, many do have orbits which approach certain points (again, .01 under f(x) = 2x(1 −x)).
Just as we study the limit of a function f(x) as x approaches infinity, we may also study the limit of the orbit of a particular point in our dynamical system as n approaches infinity.
Definition 1.3.1. Let p be periodic of period n. A point x is forward asymptotic to p if lim i→∞f in(x) = p .
The stable set of p, denoted by W s(p), consists of all points forward asymptotic to p.
Note that in this paper we will primarily concern ourselves with points which are forward asymptotic to fixed points.
Definition 1.3.2. Suppose that f is invertible. We say that x is backward asymptotic to p if lim i→−∞f in(x) = p .
We call the set of backward asymptotic points to p the unstable set of p and denote it W u(p).
Example 1.3.3. In Figure 6 we see that even though x0 = .01 is relatively close close 0, the forward orbit of x0 climbs towards 0.5, suggesting that the stable set of f is the interval (0, 1].
1.4 Hyperbolicity While we might understand how points can be forward asymptotic or backward asymptotic to a periodic point p, we still need a way to tell whether p has points converging to it. Whether p has this property has to do with whether p is hyperbolic.
10 Definition 1.4.1. Let p be a periodic point of period n. We say p is hyperbolic if |(f n)′(p)| ̸= 1.
We call (f n)′(p) the multiplier of p.
Example 1.4.2. Consider the function f(x) = x2. Clearly f has fixed points at x = 0 and x = 1.
As f ′(0) = 0 and f ′(1) = 2 clearly both fixed points are hyperbolic.
Example 1.4.3. As a counterexample, consider the function f(x) = x. Clearly every point x is a fixed point. However, none these fixed points is hyperbolic as f ′(x) = 1 for all x.
With the concept of hyperbolicity established, we can develop a way of telling whether a pe-riodic point p has points converging to it. For the following problems we are assuming that f is differentiable and that its derivative is continuous.
Theorem 1.4.4. Let p be a hyperbolic fixed point. If |f ′(p)| < 1, then there is an open interval U about p such that if x ∈U, then lim n→∞f n(x) = p .
Proof. We know that there exists a real number A such that |f ′(p)| < A < 1. We will show by induction that as p is a fixed point and |f ′(p)| < A, |(f n)′(p)| < An for all n. For our base case we see that the statement is obviously true for n = 1. Suppose it is true for n −1. We see that |(f n)′(p)| = |(f(f n−1))′(p)| = |f ′(f n−1(p))||(f n−1)′(p)| (9) = |f ′(p)||(f n−1)′(p)| < AAn−1 = An (10) as f n−1(p) = p since p is a fixed point under f. We know that the derivative of f n is continuous. Let ϵ = An−|(f n)′(p)|. As (f n)′ is continuous at p there exists a δ > 0 such that |(f n)′(x)−(f n)′(p)| < ϵ if x ∈(p −δ, p + δ). We then see that |(f n)′(x)| = |(f n)′(x) −(f n)′(p) + (f n)′(p)| (11) ≤ |(f n)′(x) −(f n)′(p)| + |(f n)′(p)| (12) < An −|(f n)′(p)| + |(f n)′(p)| = An (13) if x ∈(p −δ, p + δ).
Let x ∈(p −δ, p + δ). By the Mean Value Theorem |f n(x) −f n(p)| |x −p| < An so that |f n(x) −p| = |f n(x) −f n(p)| < An|x −p| < Anδ .
As δ is fixed by ϵ and |A| < 1, clearly limn→∞Anδ = 0. Thus lim n→∞f n(x) = p .
A direct consequence of this is that the interval (p −δ, p + δ) is a subset of the stable set of p.
Definition 1.4.5. If p is a hyperbolic periodic point of period n with |(f n)′(p)| < 1 we say that p is an attracting periodic point or a sink.
11 Example 1.4.6. Returning to f(x) = 2x(1−x), we note that f ′(x) = 2−4x. As such f ′(1/2) = 0.
Therefore 1 2 is a sink, which helps explain the behavior we saw in Figure 6.
Definition 1.4.7. If p is a fixed point with |f ′(p)| > 1 we call p a repelling fixed point or source.
Theorem 1.4.8. Let f be a function which is infinitely differentiable where all of its derivatives are continuous and let p be a hyperbolic fixed point with |f ′(p)| > 1. Then there is an open interval U around p such that, if x ∈U and x ̸= p, then there exists an integer k such that f k(x) / ∈U.
The proof of this theorem is similar enough to our last proof that we will not bother with it here.
Example 1.4.9. Returning to f(x) = 2x(1 −x), we note that f ′(x) = 2 −4x. As such f ′(0) = 2.
Therefore 0 is a source, which again helps explain the behavior we saw in Figure 6.
As our analysis has suggested so far, the map f(x) = 2x(1−x) seems incredibly predictable. We can generalize this behavior to large class of logistic maps. Recall that logistic maps are functions of the form Fµ(x) = µx(1 −x).
Theorem 1.4.10. Let 1 < µ < 3.
1. Fµ has a sink at pµ = µ−1 µ and a source at 0.
2. If 0 < x < 1, then lim n→∞F n µ (x) = pµ.
Unfortunately, this theorem is quite difficult to prove. Given that non-chaotic systems are not our primary focus, we will leave it without proof here.
1.5 Dynamical Systems: A Wide Field of Study Having now established a firm understanding of dynamical systems we can see what a wide variety of forms they can take. As such, we cannot make too many general statements about the behavior of dynamical systems. We did find, however, that any logistic map Fµ with 1 < µ < 3 behaves quite predictably. As we delve into chaos and, ultimately, chaotic logistic maps, we will see how diverse this seemingly mundane family of functions really is.
12 2 Chaos Note that the following discussion of chaos presumes a general knowledge of metric spaces and topology.
If you are unfamiliar with these concepts or would like a brief refresher, consult the Appendix.
Chaotic dynamics are defined by a few key properties which we must explore in isolation before pulling together a complete definition of chaos. Note that I have adapted the following definitions from Devaney , generalizing his results in (R, d1) to any metric space (X, d).
2.1 Our Definition of Chaos Definition 2.1.1. Let (X, d) be a metric space and let J ⊆X.
f : J →J has sensitive dependence on initial conditions if there exists a δ > 0 such that, for any x ∈J and any open set N containing x, there exists a y ∈N and an n ≥0 such that d(f n(x), f n(y)) > δ.
Simply put, f has sensitive dependence if even the slightest change in initial conditions eventually leads to a substantially different outcome. This is the very behavior which we saw with g(x) = 4x(1 −x) and which first alarmed Lorenz (see Introduction). Moreover, this property is frequently referred to as The Butterfly Effect. Among the properties of a chaotic dynamical system, this is perhaps the most well-known. Some popular depictions of chaos even equate it solely with the Butterfly Effect, ignoring the other two key properties of a chaotic dynamical system entirely.
Example 2.1.2. Consider the map f : R →R defined by f(x) = 2x.
Let s ∈R and let N be an open set containing s. Since N is open, we know that there exists a real number ϵ > 0 such that (s −ϵ, s + ϵ) = Bd1(s, ϵ) ⊆N. Let y = s + ϵ 2. Clearly y ∈(s −ϵ, s + ϵ).
We know that there exists a positive integer M such that 2Mϵ > 1. We see that for any positive integer n and any t ∈R, f n(t) = 2nt. Therefore |f M+1(s) −f M+1(y)| = |2M+1s −2M+1y| = 2M+1|s −y| = 2Mϵ > 1 .
Therefore f has sensitive dependence on initial conditions.
Definition 2.1.3. Let (X, d) be a metric space and let J ⊆X.
f : J →J is said to be topologically transitive if for any pair of open sets U, V ⊆J there exists a k >0 such that f k(U) ∩V ̸= ∅.
This is to say that if you take two open sets U and V in J and look far enough in the orbits of all of the elements of U under f you will eventually find some element in V . In even more general terms: wherever you look in J you will be able to find a point which has an orbit which travels all over J. We saw this behavior in the orbit of .01 under g(x) = 4x(1 −x) in Figures 8, 9, and 10.
Example 2.1.4. As a counterexample note that the map f : R →R defined by f(x) = 2x is certainly not topologically transitive.
Let V = (0, 1) and U = (1, 2).
Clearly if x ∈U, then f k(x) > 1 for any k. As such, for every k we see that f k(U) ∩V = ∅.
Our third property requires no new definition. It is simply that periodic points are dense under f. If you are unfamiliar with the concept of density, see the Appendix. Essentially, this means that wherever you look in J, you will find a periodic point under f.
13 Example 2.1.5. We will once again use the map f : R →R defined by f(x) = 2x as a coun-terexample. Clearly f does not have any periodic points other than 0 as f k+1(x) > f k(x) for every k.
With all of the pieces in place we can finally formalize our definition of chaos: Definition 2.1.6. Let (X, d) be a metric space and let V ⊆X.
We say that f : V →V is chaotic on V if 1. f has sensitive dependence on initial conditions, 2. f is topologically transitive, 3. periodic points under f are dense in V .
2.2 A Straightforward Example of a Chaotic Dynamical System With this definitions of chaos we can show that the following system is chaotic: Consider the function f(θ) = 2θ defined on the unit circle, denoted S1, under the metric |x −y|. We note that θ = θ + 2kπ for any integer k.
To see that f has sensitive dependence on initial conditions, let δ = 1 and consider two points θ1, θ2 ∈S1 where θ1 < θ2. We recognize that f n(θ) = 2nθ.
Therefore |f n(θ2) −f n(θ1)| = |2nθ2 −2nθ1| = 2n(θ2 −θ1) .
By the Archimedean property there exists a real number r such that 1 < r(θ2−θ1). As the sequence {2n} is unbounded above, there exists a positive integer N such that 2N > r. Thus |f N(θ2) −f N(θ1)| = 2N(θ2 −θ1) > r(θ2 −θ1) > 1 .
We will now show that f is topologically transitive on S1. We note that any open subset of S1 must contain an open interval, say (θa, θb). We see that the codomain of (θa, θb) under f n(θ) will be (2nθa, 2nθb). It is not hard to see, using the same argument we used in our discussion of the sensitive dependence of f, that there exists an integer M such that 2Mθa −2Mθb > 2π. Thus eventually the orbits of elements from (θa, θb) will cover S1 and, by extension, any subset of S1. To further elucidate this concept, I have included the following image. (θa, θb) in the argument is the blue arc below, (f(θa), f(θb)) is the green arc, and (f 2(θa), f 2(θb)) is the yellow arc.
14 (0, 0) (1, 0) x (0, 1) y The density of periodic points under f is perhaps the most difficult to prove.
We note that f n(θ) = 2nθ so θ is a periodic point of period n if and only if 2nθ = θ + 2kπ for some integer n, that is, if and only if θ = 2kπ 2n −1 .
From here we only need to see that numbers of the form k 2n−1 where k and n are integers are dense in the interval [0, 1].
Consider any open subset of [0, 1]. We know that it will contain some open interval (a, b). By the Archimedean property there exists a positive integer m such that 1 < m(b −a). Moreover, since {2n −1} is unbounded above, there exists a positive integer p such that 2p −1 > m. Therefore 1 < m(b −a) < (2p −1)(b −a) so that 1 2p −1 < b −a and a < 1 2p −1 + a < b .
As 1 2p −1 + a = 1 + a(2p −1) 2p −1 and p and 1 + a(2p −1) are both integers, we have found a number of the form k 2n−1, where k and n are integers, in a generic open subset of [0, 1]. Thus numbers of this form are dense in [0, 1] and, by extension, numbers of the form 2kπ 2n−1 are dense in S1.
Therefore f is chaotic.
15 3 Topological Conjugacy: A Powerful Tool All told, directly using the definition of chaos to prove that the doubling map on the unit circle is chaotic is not too difficult: The orbits of two distinct points diverge predictably, the images of subsets of S1 clearly move around S1, and periodic points are relatively easy to find.
Unfortunately, there are many chaotic maps whose chaotic properties are not so immediately ap-parent. In order to prove that such maps are chaotic, we show that they are topologically conjugate to other maps which we know are chaotic.
First, however, we must unpack what it means for two maps to be topologically conjugate. To this end, we define a homeomorphism.
Definition 3.0.1. Let f : I →J.
We say that f is a homeomorphism if f is one-to-one, onto, and continuous, and f −1 is also continuous.
Example 3.0.2. The function f(x) = √x, where x ∈R+ is a homeomorphism.
With that definition in place, we can define topological conjugacy: Definition 3.0.3. Let f : A →A and g : B →B be two maps.
We say that f and g are topologically conjugate if there exists a homeomorphism h : A →B such that h ◦f = g ◦h. We call h a topological conjugacy.
As a visual reference, we have the following diagram: A A B B f g h h 3.1 The Properties Shared by Topologically Conjugate Maps It is not immediately apparent that two topologically conjugate maps, such as f : A →A and g : B →B, have the same dynamics. Thus before we can use topological conjugacy to prove that a dynamical system is chaotic, we need to explore the implications of topological conjugacy at some length.
3.1.1 Fixed and Periodic Points in Topologically Conjugate Maps We will starting by exploring the many relationships between the dynamics of x ∈A under f and the dynamics of h(x) ∈B under g.
16 Theorem 3.1.1. Suppose that f : A →A and g : B →B are topologically conjugate under the homeomorphism h : A →B. Then a point p in A is a fixed point under f if and only if h(p) is a fixed point under g.
Proof. We will start with the forward direction.
Let p ∈A be a fixed point under f. Then f(p) = p. Therefore h(p) = (h ◦f)(p) = (g ◦h)(p) , making h(p) a fixed point under g by definition.
Moving in the opposite direction, let p ∈A and suppose that h(p) is a fixed point under g. Then (g ◦h)(p) = h(p). Therefore p = (h−1 ◦h)(p) = (h−1 ◦g ◦h)(p) = (h−1 ◦h ◦f)(p) = f(p) , making p a fixed point under f by definition.
We can use the same logic to find a similar correspondence between periodic points under f and periodic points under g.
Theorem 3.1.2. Suppose that f : A →A and g : B →B are topologically conjugate under the homeomorphism h : A →B. Then a point x in A is a periodic point of period n under f if and only if h(x) is a periodic point of period n under g.
Proof. We will start with the forward direction.
Let x ∈A be a periodic point of period n under f. Then f n(x) = x. We see that h(x) = (h ◦f n)(x) = (g ◦h ◦f n−1)(x) = (g2 ◦h ◦f n−2)(x) .
Continuing the process of substituting g ◦h for h ◦f, we find that h(x) = (gn ◦h)(x) .
Therefore h(x) is a periodic point of period n under g by definition.
Moving in the opposite direction, let x ∈A and suppose that h(x) is a periodic point of period n under g. Then (gn ◦h)(x) = h(x). Therefore x = (h−1 ◦h)(x) = (h−1 ◦gn ◦h)(x) = (h−1 ◦gn−1 ◦h ◦f)(x) = (h−1 ◦gn−2 ◦h ◦f 2)(x) .
Continuing the process of substituting h ◦f for g ◦h, we find that x = (h−1 ◦h ◦f n)(x) = f n(x) .
Therefore x is a periodic point of period n under f by definition.
Therefore h gives a one-to-one correspondence between Pern(f) and Pern(g).
Recall that a point x is eventually fixed if x is not fixed but there exists a positive integer m > 0 such that f i+1(x) = f i(x) for all i ≥m. Also recall that a point x is eventually periodic of period n 17 if x is not periodic but there exists a positive integer m > 0 such that f n+i(x) = f i(x) for all i ≥m.
It does not take too much imagination to adapt the argument used in the proof of the previous theorem to show that a point x ∈A is eventually fixed under f if and only if h(x) is eventually fixed under g, and that x is eventually periodic under f if and only if h(x) is eventually periodic under g.
Thus there are direct correspondences between the most predictable types of points in two topo-logically conjugate maps.
3.1.2 Asymptotic Orbits in Topologically Conjugate Maps As in our general discussion of dynamical systems, we might be curious about the end-behavior of points under f and g.
Let p be periodic of period n. Recall that a point x is forward asymptotic to p if limi→∞f in(x) = p.
Theorem 3.1.3. Let (A, d1) and (B, d2) be metric spaces and let H : A →B be a homeomorphism.
Suppose that p is a period point of period n under f. Then a point x ∈A is forward asymptotic to p if and only if h(x) is forward asymptotic to h(p).
Proof. We will prove the forward direction here and leave the other direction without proof as it requires essentially the same argument. Let ϵ > 0. We note that as h is continuous, there exists a δ > 0 such that d2(h(s), h(t)) < ϵ if d1(s, t) < δ. As x is forward asymptotic to p, we know that there exists a positive integer M such that d1(f in(x), p) < δ if i ≥M.
Therefore d2((h ◦f in)(x), h(p)) < ϵ if i ≥M. Moreover, we note by the argument employed in the proof of Theorem 2 that (h◦f in)(x) = (gin ◦h)(x), so that d2((gin ◦h)(x), h(p)) < ϵ if i ≥M. Therefore h(x) is forward asymptotic to h(p) under g by definition.
Provided f and g are invertible, we can apply this same argument to show an equivalence between backward asymptotic orbits as well.
3.1.3 How a Topological Conjugacy Maps Sets While we have discussed how a topological conjugacy acts on individual points, we have yet to discuss how it acts on subsets of a generic metric space (X, d). We need to explore these concepts before discussing topological conjugacy and chaos because both topological transitivity and density are properties which describe how a map acts on subsets of its domain.
Theorem 3.1.4. Let (X, d1) and (Y, d2) be metric spaces and let h : X →Y be a continuous function. If M ⊆Y is an open set, then {x ∈X : f(x) ∈M}, the preimage of M under f, is also an open set.
18 Proof. Let x0 ∈{x ∈X : f(x) ∈M}. We know that f(x0) ∈M. Since M is open we know that f(x0) is an interior point of M. As such there exists a real number r > 0 such that Bd2(f(x0), r) ⊆ M.
Let 0 < ϵ ≤r. Because f(x) is continuous at x0 we know that there exists a δ > 0 such that f(x) ∈Bd2(f(x0), ϵ) if x ∈Bd1(x0, δ).
Let x ∈Bd1(x0, δ). Then f(x) ∈Bd2(f(x0), ϵ) ⊆M, meaning that x ∈{x ∈X : f(x) ∈M}.
Thus Bd1(x0, δ) ⊆{x ∈X : f(x) ∈M}, making x0 is an interior point of {x ∈X : f(x) ∈M}.
Therefore {x ∈X : f(x) ∈M} is an open set.
With this theorem about continuous maps established, we can discuss how homeomorphisms map sets. Recall that both a homeomorphism and its inverse are continuous.
Theorem 3.1.5. Let (X, d1) and (Y, d2) be metric spaces, let E ⊆X, and let h : X →Y be a homeomorphism. Then E is an open subset of (X, d1) if and only if h(E) is an open subset of (Y, d2).
Proof. Let E be an open subset of X. Since h−1 is a continuous map we know from our last theorem that {y ∈Y : h−1(y) ∈E} = h(E) is open in Y .
Let E be a subset of X with the property that h(E) is an open subset of y. Then {x ∈X : h(x) ∈ W} = (h−1 ◦h)(E) = E is open in X.
We know that if E is open in X, then X \ E is closed (see Appendix). Since a homeomorphism maps open sets to open sets, we can conclude that a homeomorphism also maps closed sets to closed sets.
3.2 Topological Conjugacy and Chaos Having discussed how a topological conjugacy acts on sets we can show that if two maps f and g are topologically conjugate, then f is chaotic if and only if g is chaotic.
Theorem 3.2.1. Let (X, d1) and (Y, d2) be metric spaces and let f : X →X and g : Y →Y be two maps with a topological conjugacy h : X →Y between them. Then f is topologically transitive if and only if g is topologically transitive.
Proof. Suppose that f is topologically transitive and let U and V be two open sets in Y . We know that h−1(U) and h−1(V ) are both open sets in X. Since f is topologically transitive, we know that there exists a positive integer k such that (f k ◦h−1)(U) ∩h−1(V ) ̸= ∅.
We see that (h ◦f k ◦h−1)(U) = (gk ◦h ◦h−1)(U) = gk(U). Therefore gk(U) ∩V ̸= ∅.
Hence g is topologically transitive.
Suppose that g is topologically transitive and let U and V be two open sets in X. We know that h(U) and h(v) are both open sets in Y . Since g is topologically transitive, we know that there exists a positive integer j such that (gj ◦h)(U) ∩h(V ) ̸= ∅.
19 We see that (h−1 ◦gj ◦h)(U) = (f j ◦h−1 ◦h)(U) = f j(U). Therefore f j(U) ∩V ̸= ∅.
Hence f is topologically transitive.
Theorem 3.2.2. Let (X, d1) and (Y, d2) be metric spaces and let f : X →X and g : Y →Y be two maps with a topological conjugacy h : X →Y between them. Then periodic points are dense in X under f if and only if periodic points are dense in Y under g.
Proof. Suppose that periodic points are dense in X under f. Let U be an open set in Y . We then know that h−1(U) is an open set in X. Since periodic points are dense in X under f we know that there exists a periodic point p ∈h−1(U) under f. We know that since p is a periodic point under f, h(p) is a periodic point under g. Since h(p) is an element of U, we can see that every open set in Y contains a periodic point under g.
Suppose that periodic points are dense in Y under g. Let U be an open set in X. We then know that h(U) is an open set in Y . Since periodic points are dense in Y under g we know that there exists a periodic point p ∈h(U) under g. We know that since p is a periodic point under f, h−1(p) is a periodic point under f. Since h−1(p) is an element of U, we can see that every open set in X contains a periodic point under f.
Theorem 3.2.3. Let (X, d1) and (Y, d2) be metric spaces and let f : X →X and g : Y →Y be two maps with a topological conjugacy h : X →Y between them. Then f has sensitive dependence on initial conditions if and only if g has sensitive dependence on initial conditions.
Proof. Suppose that f has sensitive dependence on initial conditions and let ϵ > 0 be the real number guaranteed to exist by that property. Let x ∈Y and let N be a neighborhood of x. We know that there exists some point y ∈h−1(N) ⊆X and some positive integer n such that d1((f n ◦h−1)(x), f n(y)) > ϵ .
Since h−1 is continuous we know that there exists some δ > 0 such that d1(h−1(s), h−1(t)) < ϵ if d2(s, t) ≤δ. Note that since ϵ is fixed, δ is as well.
Suppose that d2(gn(x), (gn ◦h)(y)) ≤δ. Then d1((h−1 ◦gn)(x), (h−1 ◦gn ◦h)(y)) = d1((f n ◦h−1)(x), f n(y)) < ϵ , a contradiction. Therefore d2(gn(x), (gn ◦h)(y)) > δ. Thus we have shown that given any point x ∈Y and any neighborhood N around it, we can find a point w ∈N (the h(y) we found using the sensitive dependence of f) and a positive integer n such that d2(gn(x), gn(w)) > δ. As such, g has sensitive dependence on initial conditions.
Given that the proof of the converse is essentially the same (the only difference being that it relies on the fact that h is continuous instead of the fact that h−1 is continuous) we will omit it here.
With all of these pieces in place we have our powerful tool for proving that sets are chaotic: Theorem 3.2.4. Suppose that f and g are topologically conjugate. Then f is chaotic if and only if g is chaotic.
20 4 Proving a Dynamical System is Chaotic Using Topological Conjugacy As we noted at the beginning of our discussion of topological conjugacy, given two topologically conjugate chaotic maps, f : A →A and g : N →B, it can often be far easier to prove that f is chaotic than that g is chaotic. As such, we choose to prove directly that f is chaotic in order to show that g is chaotic. In this section we will be doing exactly that, with the shift map, σ : Σ2 →Σ2, playing the role of f in our scheme and any logistic map, Fµ = µx(1−x) where µ > 2+ √ 5, playing the role of g (we will cover F4, from the introduction, later as it is a special case).
4.1 The Shift Map 4.1.1 The Shift Map’s Domain: Sequence Space Before we can discuss the shift map, we need to discuss the set on which it acts, the sequence space on the symbols 0 and 1.
Definition 4.1.1. The sequence space on the symbols 0 and 1 is the set Σ2 = {s = (s0s1s2...)|sj = 0 or 1} .
Elements of Σ2 are infinite strings of 0’s and 1’s, such as (001101 . . .) or (101001 . . .) .
As our previous discussions have indicated, we need Σ2 to be a metric space in order to talk about chaos and topological conjugacy. To this end, we develop the following definition of the distance between elements in the sequence space.
Definition 4.1.2. We define the distance between two sequences s = (s0s1s2 . . .) and t = (t0t1t2 . . .) to be d(s, t) = ∞ X i=0 |si −ti| 2i .
Since |si−ti| 2i ≤ 1 2i for every positive integer i and since P∞ i=0 1 2i , a geometric series, converges, we know that P∞ i=0 |si−ti| 2i converges. As such d(s, t) ∈R.
Example 4.1.3. We see that if r = (000000 . . .), s = (010101 . . .), and t = (111111 . . .), then d(r, t) = ∞ X i=0 1 2i = 2 and d(r, s) = d(s, t) = ∞ X i=0 1 22i = 4 3 .
It is not too difficult to verify that d(s, t) is a metric: 21 Theorem 4.1.4. d(s, t) is a metric.
Proof. We will work through the definition of a metric systematically to show that d(s, t) is a metric.
1. It is clear that |a −b| ≥0 for any points a, b ∈{0, 1}. As such d(s, t) ≥0 for any s, t ∈Σ2.
2. We note that s = t if and only if si = ti for each i, that si = ti if and only if |si −ti| = 0, and that |si −ti| = 0 for each i if and only if d(s, t) = P∞ i=0 |si−ti| 2i = 0.
3. As |si −ti| = |ti −si| for each i, it is clear that d(s, t) = d(t, s).
4. Let r, s, t ∈Σ2. We know that |ri −si| ≤|ri −ti| + |ti −si| for each i. As such d(r, s) ≤ d(r, t) + d(t, s).
As may be clear from our discussions of chaos and topological conjugacy, we will often want to find an upper bound for d(s, t). The following theorem helps us to find such an upper bound fairly easily.
Theorem 4.1.5. Let s, t ∈Σ2. If si = ti for all i ∈{0, 1, . . . , n}, then d(s, t) ≤ 1 2n .
Conversely, if d(s, t) < 1 2n , then si = ti for all i ∈{0, 1, . . . , n}.
Proof. Suppose si = ti for all i ∈{0, 1, . . . , n}. We see d(s, t) = n X i=0 |si −ti| 2i + ∞ X i=n+1 |si −ti| 2i ≤ ∞ X i=n+1 |si −ti| 2i = 1 2n+1 ∞ X i=0 |si −ti| 2i = 1 2n .
We will prove the converse by contrapositive. Suppose that sj ̸= tj for some j ≤n. Then d(s, t) ≥1 2j ≥1 2n .
Thus if d(s, t) < 1 2n , then si = ti for all i ∈{0, 1, . . . , n}.
4.1.2 The Shift Map With all of this background information established, we can finally talk about the shift map.
Definition 4.1.6. The shift map σ : Σ2 →Σ2 is defined by σ(s0s1s2 . . .) = (s1s2s3 . . .) .
Intuitively, the shift map just “shifts” every entry in the sequence one entry to the left and cuts offthe first entry.
22 4.1.3 The Shift Map is Chaotic By design, it is relatively easy to see that the shift map is chaotic even though we hardly know anything about it.
Theorem 4.1.7. The map σ : Σ2 →Σ2 is chaotic.
Proof. We will show that the set of periodic points under σ, Per(σ), is dense in Σ2 by showing that any point s ∈Σ2 is a limit point of Per(σ), and therefore Σ2 ⊆Per(σ).
Let s = (s0s1s2 . . .) ∈Σ2 and let ϵ > 0. We know that there exists a positive integer N such that 1 2N < ϵ.
Let τn = (s0s1 . . . sns0s1 . . . sns0 . . .) for each n. Clearly each τn is periodic of period n.
From our previous theorem, since for a given n τn and s agree in their first n + 1 entries, we know that d(s, τn) ≤1 2n .
Thus we see that if m ≥N, then d(s, τm) ≤1 2m ≤1 2N < ϵ .
As such {τn} converges to s.
Turning our focus towards sensitive dependence, we consider a sequence s in Σ2 and an open set N containing it. We know that there exists some ϵ > 0 such that Bd(s, ϵ) ⊆N. Moreover, we know that there exists some integer M such that 1 2M < ϵ. Let r ∈Bd(s, 1 2M ) with rm ̸= sm for some m > M + 1 (our previous theorem provides that so long as ri = si for i ∈{0, . . . , M + 1}, we are guaranteed that d(s, r) < 1 2M+1 < 1 2M ).
We note that σm(s) = (smsm+1 . . .) and σm(r) = (rmrm+1 . . .).
As sm ̸= rm, we know that d(σm(s), σm(r)) ≥1 2 .
Finally, we will show that σ is topologically transitive by finding an element s′ of Σ2 such that given two open sets U and V in Σ2, σn(s′) ∈U and σm(s′) ∈V where m ≥n.
Consider the sequence s′ = (0 1|00 01 10 11|000 001 . . . | . . .) .
We see that s′ consists of all possible strings of 0’s and 1’s of length 1, followed by all strings of length 2, and so on. Let U and V be open sets containing points u and v, respectively. Since U and V are open, there exist real numbers ϵu > 0 and ϵv > 0 such that Bd(u, ϵu) ⊆U and Bd(v, ϵv) ⊆V .
Moreover, we know that there exist positive integers Nu and Nv such that 1 2Nu < ϵu and 1 2Nv < ϵv.
Given how we constructed s′, we know that there exists an integer Mu such that σMu(s′) agrees with u in its first Nu+1 entries. As such d(σMu(s′), u) ≤ 1 2Nu < ϵu, 23 so that σMu(s′) ∈Bd(u, ϵu) ⊆U.
Likewise, we know that there exists integer Mv > Mu such that σMv(s′) agrees with v in its first Nv+1 entries. As such d(σMv(s′), v) ≤ 1 2Nv < ϵv, so that σMv(s′) ∈Bd(v, ϵv) ⊆V .
Therefore σMv(s′) ∈σMv−Mu(U) ∩V .
As in our discussion of the doubling map on the unit circle, we see that chaotic properties come to the shift map fairly easily. It is for this very reason that we use it to show that other maps are chaotic.
4.2 Logistic Maps Recall that logistic maps are functions of the form Fµ(x) = µx(1−x). In the following few sections, we will show that if µ > 2 + √ 5, then Fµ defined on a subset of [0, 1] and σ are topologically conjugate, and, by extension, Fµ is chaotic.
In order to see this we have to discuss the subset of [0, 1] in question.
We will be exploring the set of elements whose orbits under Fµ do not leave [0, 1]. To start, we have the following theorem: Theorem 4.2.1. Suppose µ > 1.
If x < 0, then limn→∞F n µ (x) = −∞, and if x > 1, then limn→∞F n µ (x) = −∞.
Proof. We see that if x < 0, then µ(1 −x) > 1, so that Fµ(x) = µx(1 −x) < x. Therefore the sequence {F n µ (x)} is strictly decreasing. Clearly the sequence does not converge to a point y, for if it did then the sequence {F n+1 µ (x)} would converge to Fµ(y) < y. Therefore {F n µ (x)} diverges to −∞.
If x > 1, then Fµ(x) < 0. The argument that we just used thus applies to Fµ(x) and {F n+1 µ (x)} diverges to −∞.
We find some interesting behavior when µ > 4. We will denote I = [0, 1] for convenience. We find that F ′ µ(x) = µ −2µx, so that Fµ attains its maximum at 1 2. We note that Fµ( 1 2) = µ 1 2(1 −1 2) = µ 4 is that maximum. As such, the maximum value of Fµ on I is greater than 1 when µ > 4.
Since Fµ is continuous at 1 2, we know that there exists some open interval A0 centered at 1 2, such that if x ∈A0, then Fµ(x) > 1. This means that for any point x ∈A0, the orbit of x leaves I after a single iteration of Fµ and from there tends inexorably towards −∞.
Let A1 = {x ∈I : Fµ(x) ∈A0}. We see that if x ∈A1, then F 2 µ(x) > 1, so that the orbit of x tends towards −∞from there. The following diagram depicts A0 and A1 for F5: 24 (1, 0) A1 A0 A1 (0, 1) With this basic idea in place, we can define An = {x ∈I : F n µ (x) ∈A0} for any positive integer n. We recognize that An consists of all points in I for which F i µ(x) ∈I if i ≤n, but F n+1 µ (x) / ∈I. Clearly S∞ n=0 An consists of every point in I which eventually escapes I and tends towards −∞.
We will thus focus on the set I \ S∞ n=0 An which we will denote Λ.
The set Λ has one key quality which helps us show that Fµ is topologically conjugate to the shift map.
Definition 4.2.2. A set is totally disconnected if it contains no intervals.
Theorem 4.2.3. If µ > 2 + √ 5, then Λ is totally disconnected.
Proof. We need |F ′ µ(x)| > 1 for all x ∈I \ A0. We will not bother to prove it here, but Fµ has this property if µ > 2 + √ 5. Nevertheless, as this is the case, we know that there exists some real number r such that |F ′ µ(x)| > r > 1 for all x ∈Λ as Λ ⊂I \ A0. As we showed in our discussion of hyperbolicity, this implies that |(F n µ )′(x)| > rn. Now suppose that Λ contains some interval [x, y].
Then |(F n µ )′(t)| > rn for all t ∈[x, y]. We know that there exists a positive integer N such that rN|x −y| > 1. By the Mean Value Theorem, we know that |F N(x) −F N(y)| ≥rN|x −y| > 1, which implies that at least one of F N(x) or F N(y) is not contained in I, contradicting the fact that x and y are in Λ. Thus Λ does not contain any intervals.
To further understand the concept of a totally disconnected set, we look at perhaps the most familiar totally disconnected set: the Cantor Middle Third set. The Cantor set is constructed as fol-lows: Start with the interval [0, 1]; remove the open interval (1/3, 2/3), that being the middle third; remove the middle thirds of two intervals that remain; repeat. The following diagram illustrates how we construct the set: 25 Note that we construct Λ in much the same way that we construct the Cantor set: by removing open intervals from the middles of larger intervals.
4.3 Building our Topological Conjugacy: The Itinerary With strong understandings of σ and Fµ we can now connect the two maps with a topological conjugacy: the itinerary.
Definition 4.3.1. The itinerary of a point x ∈Λ is a sequence S(x) = s0s1s2 . . . where sj = 0 if F j µ(x) ∈I0 and sj = 1 if Fµj(x) ∈I1.
(1, 0) I1 A0 I0 (0, 1) We recall that points in Λ never escape I. As Λ ⊆I0 ∪I1 this means that we can follow the orbit of any point x ∈Λ as it moves between I0 and I1. As the name suggests, the itinerary could thus be thought of as a way to record where the orbit of x ∈Λ goes using a sequence S(x) ∈Σ2.
4.4 S : Λ →Σ2 is a Topological Conjugacy Theorem 4.4.1. S : Λ →Σ2 is a homeomorphism.
Proof. As the task of showing that a function is a homeomorphism entails many smaller arguments, we will break this section up into more manageable pieces.
1. We will start by using a contradiction to prove that S is one-to-one.
Let x, y ∈Λ, where x ̸= y and S(x) = S(y). If this is the case, then for each integer n, F n µ (x) and F n µ (y) are both in either I0 or I1. This means that both are on the same side of 1 2. As such Fµ is monotonic on the interval between F n µ (x) and F n µ (y).
Therefore, if z is between x and y, then Fµ(z) is between Fµ(x) and Fµ(y), F 2 µ(z) is between F 2µ(x) and F 2µ(y), and so on. Thus the orbit of z remains in I for all n, making z an element of Λ. Since z is a generic element of the interval between x and y, this implies that 26 Λ contains the interval between x and y. This contradicts that Λ is totally disconnected.
2. We will now show that S is onto.
Let s = (s0s1s2 . . .). We want to find an point x ∈Λ such that S(x) = s.
As one might expect, given a closed interval J ⊆I we might want to talk about the preimage of J and more generally about the set F −n µ (J) = {x ∈I : F n µ (x) ∈J}, for each integer n. We note that as J is a closed interval, its preimage F −1 µ (J) must have components in I0 and I1 which are themselves closed intervals as Fµ is symmetrical. For ease of notation, we define I(s0s1...sn) = {x ∈I : x ∈Is0, Fµ(x) ∈Is1, . . . , F n µ (x) ∈Isn} = Is0 ∩F −1 µ (Is1) ∩. . . ∩F −n µ (Isn) .
The following graph depicts the four possibilities for I(s0s1): (1, 0) I10 I11 I01 I00 (0, 1) We will use proof by induction to show that the I(s0s1s2...sn) form a nested sequence of nonempty closed intervals. We note that I(s0s1s2...sn) = Is0 ∩F −1 µ (I(s1s2...sn)) .
By our induction hypothesis, I(s1s2...sn) is nonempty and closed.
As we showed above, F −1 µ (I(s1s2...sn)) consists of two closed intervals, one in each of I0 and I1.
Thus Is0 ∩ F −1 µ (I(s1s2...sn)) is a single closed interval, as Is0 is really one of I0 or I1. These intervals are nested because I(s0s1s2...sn) = I(s0s1s2...sn−1) ∩F −n µ (I(s1s2...sn)) ⊂I(s0s1s2...sn−1) .
27 Thus we find that T n≥0 I(s0s1s2...sn) is nonempty.
From this we can conclude that if x ∈T n≥0 I(s0s1s2...sn) then x ∈Is0, Fµ(x) ∈Is1, and so on for any integer n. As such, S(x) = (s0s1 . . .). Since s is a generic point in Σ2, this implies that S is onto.
It is important to note here that since S is one to one, T n≥0 I(s0s1s2...sn) = {x}.
3. To show that S is continuous let x ∈Λ, let S(x) = (s0s1s2 . . .), and let ϵ > 0. We know that there exists a positive integer N such that 1 2N < ϵ. Consider all possible sequences (t0t1 . . . tN) and the associated intervals I(t0t1...tN). We note that all of these intervals are disjoint and Λ is a subset of their union. We find combinatorially that there are 2N+1 such intervals and note that I(s0s1s2...sN) is among them. Thus we may find a δ > 0 such that if |x −y| < δ and y ∈Λ, then y ∈I(s0s1s2...sN). As such, S(x) and S(y) agree in their first N +1 terms, so that, by our earlier theorem, d(S(x), S(y)) ≤1 2N < ϵ .
4. Finally, we want to show that S−1 is continuous.
Let ϵ > 0. Consider a sequence s = (s0s1 . . .). Since the I(s0s1...sn) form a nested sequence of intervals that converges to {S−1(s)}, we know that there exists some integer N such that I(s0s1...sN) ⊂(S−1(s) −ϵ, S−1(s) + ϵ).
Let r be a sequence where d(s, r) < 1 2N . We know that r agrees with s in its first N +1 terms.
As such S−1(r) ∈I(s0s1...sN). Therefore |S−1(s) −S−1(r)| < ϵ.
Theorem 4.4.2. S ◦Fµ = σ ◦S.
Proof. Recall that F −n µ (J) = {x ∈I : F n µ (x) ∈J}, and that Is0s1...sn = Is0 ∩F −1 µ (Is1) ∩. . . ∩F −n µ (Isn) .
Since S is one-to-one we know that T n≥0 Is0s1...sn consists of a single point in Λ which we will call x.
Since Is0s1...sn = Is0 ∩F −1 µ (Is1) ∩. . . ∩F −n µ (Isn) and Fµ(Is0) = I, we see that Fµ(Is0s1...sn) = Is1 ∩F −1 µ (Is2) ∩. . . ∩F −n+1 µ (Isn) = Is1s2...sn .
Thus S ◦Fµ(x) = S ◦Fµ ∞ \ n=0 Is0s1...sn !
= S ∞ \ n=1 Is1...sn !
= s1s2s3 . . .
= σ ◦S(x) .
28 4.5 Chaotic Logistic Maps Thus our big result falls out: Theorem 4.5.1. A logisitic map Fµ with µ > 2 + √ 5 is chaotic.
Proof. The shift map σ is chaotic. Any logistic map Fµ with µ > 2 + √ 5 is topologically conjugate with σ. As σ is chaotic, so too is Fµ.
4.6 F4 : A Special Case Since we started by looking at the chaotic behavior of F4(x) = 4x(1 −x)(simply called g at the time), it behooves us to prove that F4 is actually chaotic. Unfortunately, the itinerary does not serve as a topological conjugacy between F4 and the shift map, as points do not escape I under F4 (it is for this very reason that I used F4 for the pictures in the first place). Instead, we must prove this one from the ground up, albeit with a few mappings very similar to conjugacies. Given the similarities of the following proof to several previous proofs, we will not dwell on the details for too long.
Theorem 4.6.1. F4 is chaotic.
Proof. Recall that the function g(θ) = 2θ defined on the unit circle, denoted S1, is chaotic. Define h1 : S1 →[−1, 1] by h1(θ) = cos(θ) and define q : [−1, 1] →[−1, 1] by q(x) = 2x2 −1. We see that (h1 ◦g)(θ) = cos(2θ) = 2 cos(θ)2 −1 = (q ◦h1)(θ) .
Define h2 : [−1, 1] →[0, 1] by h2(t) = 1 2(1 −t), we see that (F4 ◦h2)(t) = 4 1 2(1 −t) 1 − 1 2(1 −t) = 4 1 2(1 −t) 1 2(1 + t) = (1 −t)(1 + t) = 1 −t2 = 1 2(2 −2t2) = 1 2(1 −(2t2 −1)) = (h2 ◦q)(t) .
This gives us: 29 S1 S1 [−1, 1] [−1, 1] [0, 1] [0, 1] g q F4 h1 h1 h2 h2 We will start by showing that F4 is topologically transitive.
We note that as h1 and h2 are both surjective and continuous, (h2 ◦h1) is also surjective and continuous. As such, if U and V are two open intervals in [0, 1], we can find open arcs U ′ and V ′ in S1 such that (h2 ◦h1)(U ′) = U and (h2 ◦h1)(V ′) = V . As g is topologically transitive there exists a positive integer N such that gN(U ′) ∩V ′ ̸= ∅. Therefore we see that F N 4 (U) = (F N 4 ◦h2 ◦h1)(U ′) = (h2 ◦qN ◦h1)(U ′) = (h2 ◦h1 ◦gN)(U ′) .
Thus F N 4 (U) ∩V = (h2 ◦h1)(gN(U ′) ∩V ′), so that F N 4 (U) ∩V ̸= ∅.
To see that f has sensitive dependence on initial conditions, let x ∈[0, 1] and let U be an open set containing x. We know that the preimage of U under (h2 ◦h1), let’s call it U ′, is an open set.
Recall from our proof that g is chaotic that since U ′ is an open set there exists a positive integer M such that gM(U ′) covers S1. Therefore F M 4 (U) = (h2 ◦h1 ◦gM)(U ′) covers [0, 1]. Thus we know that there certainly exists some point y ∈U such that |F M 4 (x) −F M 4 (y)| ≥1 2.
To show that periodic points are dense under F4, we will first show that (h2 ◦h1) maps periodic points to periodic points. Let p ∈be a periodic point of period n under g. Then gn(x) = x. We see that (h2 ◦h1)(p) = (h2 ◦h1 ◦gn)(p) = (F n 4 ◦h2 ◦h1)(p) .
Therefore (h2 ◦h1)(p) is a periodic point of period n under F4 by definition.
Therefore if we pick an open interval U in [0, 1], we know that its preimage U ′, also an open set, will contain some periodic point p. Thus U contains the periodic point (h2 ◦h1)(p).
As such, periodic points are dense under F4.
Note that (h2 ◦h1) : S1 →[0, 1] is not a topological conjugacy, as h1 is a two-to-one function at most points. Nevertheless, the method we used to show that F4 is chaotic relies on the same basic principles as our method of using a topological conjugacy.
30 Conclusion: Seeing Order Thus we have come full circle, showing why our two original functions F2 and F4 have such dras-tically different behavior. Like Edward Lorenz, we started by looking at some alarming data and tried to unravel why it was occurring. While we returned to the behavior of logistic maps a number of times, along the way we also developed a vocabulary for describing discrete dynamical systems in general, rigorously defined chaos, and constructed a powerful tool for proving that dynamical systems are chaotic. In the end, we saw that chaotic systems have a very particular kind of order underlying their seemingly disorderly behavior: their very nature makes any two sets of distinct initial conditions lead to drastically different outcomes, and causes the orbits of points travel all over their domains.
As an application-driven field, chaos theory invites us to seek new patterns to unravel. Weather forecasting, for instance, has improved dramatically thanks to developments in chaos theory. Chaos might be hiding where we never expect to find it; perhaps we just need to run a few simulations to start our search.
Further Research It is my hope that a motivated calculus student could read this paper, albeit with frequent visits to the Appendix. Perhaps it will spark their interest for their own eventual senior project. I feel that using the itinerary as a topological conjugacy for a wider variety of systems could be a particularly ripe topic. Additionally, given how much work I put into generalizing this paper’s results to metric spaces (unlike any of the sources I was drawing on), I would hope that future students could find other chaotic maps defined on strange metric spaces.
Acknowledgments I would like to thank Alec Foote for carefully reading and editing this paper every single week this semester, Professor Doug Hundley for guiding me through a topic that was entirely foreign to me before I started this project, and Professor Barry Balof for teaching the Senior Project course this semester.
31 Appendix: Necessary Background for Chaos This appendix is intended as a primer on some basic concepts about metric spaces and topology which we require to discuss chaos. As it is intended for readers entirely unfamiliar with these ideas, this appendix features many examples explained at length. Many of the following definitions have been adapted from Gordon .
Metric Spaces Before we can discuss whether a map over a space is chaotic we need to have a means of measuring the distance between two elements in that space. To that end we have the concept of a metric space: Definition 4.6.2. (Gordon ) A metric space (X, d) consists of a set X and a function d : X × X →R, called a metric, that satisfies the following four properties.
1. d(x, y) ≥0 for all x, y ∈X.
2. d(x, y) = 0 if and only if x = y.
3. d(x, y) = d(y, x) for all x, y ∈X.
4. d(x, y) ≤d(x, z) + d(y, z) for all x, y, z ∈X.
Example 4.6.3. Perhaps the most familiar metric space is (R, d1) where R denotes the set of real numbers and d1(x, y) = |x −y|, where x and y are real numbers. Let us verify that d1(x, y) is a metric by checking each property: 1. We know that |t| ≥0 for any real number t, so clearly d1(x, y) = |x −y| ≥0 for all x, y ∈R.
2. We recall that |0| = 0, so that d1(x, x) = |x −x| = 0. Likewise, we recall that |t| > 0 when t ̸= 0, so that if x ̸= y then d1(x, y) = |x −y| > 0.
3. We know that |x −y| = |y −x| for all x, y ∈R.
4. We know that |a + b| ≤|a| + |b| for all a, b ∈R. As such, |x −y| ≤|x −z| + |z −y| for all x, y, z ∈R.
Topology in a Metric Space While thus far we have focused on how our maps act on individual points in a dynamical system, we also need a firm understanding of how our maps act on sets of points. Before we can do this, however, we need to develop a vocabulary for discussing different types of sets.
Definition 4.6.4. (Gordon ) Let (X, d) be a metric space.
Let v ∈X and let r > 0. The open ball centered at v with radius r is defined by Bd(v, r) = {x ∈X : d(x, v) < r} .
32 Example 4.6.5. Consider the familiar metric space (R, d1) where d1(x, y) = |x−y| for all x, y ∈R.
We see that Bd1(0, 1) = (−1, 1).
Example 4.6.6. Consider the metric space (R2, d2) where d2(x, y) = p (x1 −y1)2 + (x2 −y2)2 for all points x = (x1, x2) and y = (y1, y2) in R2. We can see that Bd2((0, 0), 1) is the interior of the cir-cle of radius 1 centered at the origin. In the following image we can see that d2((0, 0), (x1, x2)) < 1, so that (x1, x2) ∈Bd1((0, 0), 1).
(0, 0) (1, 0) x (0, 1) y (x1, x2) Definition 4.6.7. (Gordon ) Let (X, d) be a metric space, let E ⊆X, and let x ∈X.
1. The point x is an interior point of E if there exists an r > 0 such that Bd(x, r) ⊆E.
2. The point x is a limit point of E if for each r > 0, the set E ∩Bd(x, r) contains a point of E other than x.
An equivalent definition is that x is a limit point of E if there exists a sequence in E \ {x} that converges to x.
3. The set E is open in (X, d) if each point of E is an interior point of E.
4. The set E is closed in (X, d) if E contains all of its limit points.
Example 4.6.8. Consider the metric space (R, d1). Let E = (0, 1). We see that .5 is an interior point of E as Bd1(.5, .25) = (.5 −.25, .5 + .25) ⊆(0, 1).
Example 4.6.9. Consider the metric space (R, d1). Let G = { 1 n : n ∈Z+}. We will show that 0 is a limit point of G.
Let ϵ > 0. We note that there exists a positive integer N such that 1 N < ϵ. Therefore 1 N ∈Bd1(0, ϵ).
Thus for every ϵ > 0, Bd1(0, ϵ) contains a point in E other than 0.
33 Example 4.6.10. We will show that (a, b), where a, b ∈R, is an open set in (R, d1).
Let x ∈(a, b). Since a < x < b, we know that b −x > 0 and x −a > 0. Let r = min{b −x, x −a}.
We see that Bd1(x, r) = (x −r, x + r) ⊆(a, b).
Example 4.6.11. As a counterexample, we will show that (0, 1] is not an open as 1 is not an interior point of (0, 1] Let ϵ > 0. Clearly 1 + ϵ 2 ∈Bd1(x, ϵ). We note, however, that 1 + ϵ 2 > 1, so that 1 + ϵ 2 / ∈(0, 1].
Example 4.6.12. We will show that [a, b], where a, b ∈R, is a closed set in (R, d1).
Let x / ∈[a, b]. We want to show that x is not a limit point of [a, b].
Since x / ∈[a, b], either x < a or b < x. Suppose that x < a. Let ϵ = a−x and consider the open ball Bd1(x, ϵ) = (x −ϵ, x + ϵ). Let y ∈Bd1(x, ϵ). We know that y < x + ϵ = x + (a −x) = a. Therefore y / ∈[a, b]. As such, the open ball Bd1(x, ϵ) does not contain any point in [a, b], meaning that y is clearly not a limit point of [a, b].
We can see rather easily that a similar argument applies if b < x, so we will not bother with that case here.
Thus [a, b] contains all of its limit points, and is therefore closed.
Example 4.6.13. As a counterexample, we will show that the set G = { 1 n : n ∈Z+} is not closed.
As we showed above, 0 is a limit point of G. We see, however, that 0 / ∈G.
In order to test your comfort with these terms, attempt to prove the following helpful theorem.
Theorem 4.6.14. Any open ball is an open set.
As we will be discussing mappings of open and closed sets, it is helpful to know the relationship between the two in a generic metric space.
Theorem 4.6.15. Let (X, d) be a metric space and let E ⊆X.
1. The set E is open in (X, d) if and only if X \ E is closed in (X, d).
2. The set E is closed in (X, d) if and only if X \ E is open in (X, d).
Proof. We will start by showing that if E is open in (X, d), then X \ E is closed in (X, d).
Let x ∈E. As x is necessarily an interior point of E we know that there exists some r > 0 such that Bd(x, r) ⊆E. This implies that Bd(x, r) ∩X \ E = ∅. Therefore x is clearly not a limit point of X \ E. Thus X \ E must contain all of its limit points and be closed.
Moving to showing that if X \ E is closed, then E is open, suppose that x is a point in E that is not an interior point of E. This means that for every r > 0, Bd(x, r) ∩X \ E ̸= ∅. Since x / ∈X \ E, this means that for every r > 0, Bd(x, r) contains a point in X \ E other than x. Thus x is a limit point of X \ E. As X \ E is closed, however, this means that x ∈X \ E, a contradiction. Thus every point in E is an interior point of E, making E open.
The proof of the second statement follows immediately from the first statement and the fact that X \ (X \ E) = E.
Density Definition 4.6.16. (Gordon ) Let (X, d) be a metric space and let E ⊆X.
34 1. The derived set of E, denoted E′, is the set of all limit points of E.
2. The closure of E, denoted E, is the set E ∪E′.
Example 4.6.17. Consider the metric space (R, d1).
If E = (a, b), then E′ = [a, b] and E = [a, b].
Example 4.6.18. Consider the metric space (R, d1).
If G = { 1 n : n ∈Z+}, then G′ = {0} and G = { 1 n : n ∈Z+} ∪{0}.
Definition 4.6.19. (Trench ) A set U is dense in a set S if U ⊆S ⊆U.
While this definition is certainly intuitive, there are easier ways of showing that a set U is dense in a set S. To this end we have the following theorem: Theorem 4.6.20. Let (X, d) be a metric space and let S ⊆X.
If each open ball that intersects S contains a point in U, then U is dense in S.
Proof. Suppose that every open ball that intersects S contains a point in U. We want to show that every point in S is either a point in U or a limit point of U.
Let s ∈S. If s is a point in U, then we’re finished, so suppose that s is not a point in U. Consider a sequence {ϵn}, where each ϵn > 0, that converges to 0. As every open which intersects S contains a point in U, for each integer n there exists an xn ∈U such that xn ∈Bd(s, ϵn) ∩S. Consider, then, the sequence {xn}. Let ϵ > 0. Since {ϵn} converges to 0 there exists a positive integer N such that ϵn < ϵ if n ≥N. Therefore d(xn, s) < ϵn < ϵ if n ≥N. As such, s is the limit point of a sequence of points in U. Therefore every point in S is either a point in U or a limit point of U, that is U ⊆S ⊆U.
As an example, we have the following theorem and proof adapted from Trench .
Theorem 4.6.21 (Trench ). The rational numbers are dense in the reals.
Proof. Let a and b be two real numbers. We want to show that there exists a rational number p/q ∈(a, b).
By the Archimedean Property, we know that there exists a positive integer q such that q(b−a) > 1.
Let p be the smallest integer such that p > qa. Then p −1 ≤qa, so that qa < p ≤qa + 1.
As q(b −a) > 1, clearly qa < p < qa + q(b −a) = qb.
Thus a < p/q < b .
35 Continuity Finally, it is helpful to develop a vocabulary for describing how functions map points which are close together to other points which are close together. To this end, we formulate a generic concept of continuity in a metric space.
Definition 4.6.22. Let (X, d1) and (Y, d2) be metric spaces.
1. A function f : X →Y is continuous at x0 ∈X if for each ϵ > 0 there exists a δ > 0 such that f(x) ∈Bd2(f(x0), ϵ) for all x ∈Bd1(x0, δ).
2. A function f : X →Y is continuous on X if it is continuous at each point of X.
We can see fairly easily how this concept carries over to (R, d1). We may, however, want a more interesting example. As such, we will show that the shift map σ : Σ2 →Σ2 (see Section 4.1.2) is continuous.
Theorem 4.6.23. The shift map σ : Σ2 →Σ2 is continuous.
Proof. Let ϵ > 0 and s = (s0s1s2 . . .). We know that there exists a positive integer N such that 1 2N < ϵ. Let δ = 1 2N+1 . If t = (t0t1t2 . . .) is a sequence in Σ2. such that d(s, t) < δ, then si = ti for i ≤N + 1. Therefore, the ith entries of σ(s) and σ(t) agree for i ≤N, so that d(σ(s), σ(t)) ≤1 2N < ϵ.
36 References Kathleen T. Alligood, Tim D. Sauer, and James A. Yorke, Chaos: An Introduction to Dynamical Systems, Springer-Verlag, New York, 1997.
Robert L. Devaney, A First Course in Chaotic Dynamical Systems: Theory and Experiment, Westview Press, Boulder, Co., 1992.
Robert L. Devaney, An Introduction to Chaotic Dynamical Systems, Addison-Wesley Publishing Company, Reading, Ma., 1989.
James Gleick, Chaos: Making a New Science, Penguin Books, New York, 1987.
Russell A. Gordon, Real Analysis: A First Course, Addison-Wesley Publishing Company, 2002.
William F. Trench, Introduction to Real Analysis, The American Institute of Mathematics’ Open Textbook Initiative, August 2015.
37 Alphabetical Index attracting periodic point, 11 backward asymptotic, 10 closed, 33 closure, 35 cobweb plot, 5 continuous, 36 dense, 35 derived set, 35 dynamical system, 5 fixed point, 8 forward asymptotic, 10 forward orbit, 5 homeomorphism, 16 hyperbolic, 11 initial value, 5 interior point, 33 iteration, 5 limit point, 33 metric space, 32 open, 33 open ball, 32 periodic, 9 repelling fixed point, 12 seed, 5 sensitive dependence, 13 shift map, 22 sink, 11 source, 12 stable set, 10 topologically conjugate, 16 topologically transitive, 13 totally disconnected, 25 unstable set, 10 38 |
189253 | https://www.eng-tips.com/threads/thermal-expansion-of-carbon-steel-at-temperature.351594/ | Published Time: Thu, 03 Jul 2025 10:15:58 GMT
thermal expansion of carbon steel at temperature | Eng-Tips
[x] Search titles and first posts only
[x] Search titles only
By:
SearchAdvanced search…
Home
ForumsNew postsSearch forumsForumFAQsLinksMVPs
What's newFeatured contentNew postsNew profile postsLatest activity
MembersCurrent visitorsNew profile postsSearch profile posts
Log inRegister
DefaultAlternate
What's newSearch
Search
[x] Search titles and first posts only
[x] Search titles only
By:
Search Advanced search…
New posts
Search forums
Forum
FAQs
Links
MVPs
Menu
Log in
Register
Navigation
Install the app
Install
How to install the app on iOS
Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
More options
Style variation
DefaultAlternate
Contact us
Close Menu
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
RegisterLog in
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Home
Forums
People
Materials Engineers
Material engineering general discussion
thermal expansion of carbon steel at temperature
Thread startersydneyjongleur
Start dateSep 11, 2013
Status Not open for further replies.
Sep 11, 2013
#1
sydneyjongleur
Materials
Jul 22, 2011 39GB
Hi,
Does anyone know the thermal expansion rate at 1000°C on Carbon steel. A rough figure will do.
Sort by dateSort by votes
Sep 11, 2013
#2
TVP
Materials
Apr 12, 2002 4,592US
I don't have any data up to 1000 C, because it almost certainly does not exist. Carbon steels cannot be used at that temperature for any sustained amount of time. The average value from 20 to 700 C is ~ 15 μm/m·K.
Upvote0Downvote
Sep 11, 2013
#3
cloa
Petroleum
Jul 18, 2008 1,071JP
I think there would such data for severe upset conditions like 9-11 type fires where you'd want to estimate the limited life of a skyscraper in severe fire for escape planning.
Otherwise it would need to be in non-oxidising environment and accept maybe 30% room temperature
strength for rather short periods as it will creep badly.
extrapolate it from this data.
Use translation assistance for Engineers forum
Note the rules include No Student posting
Upvote0Downvote
Sep 11, 2013
#4
btrueblood
Mechanical
May 26, 2004 10,033US
CRC Handbook, 40th ed., 1958 lists for steel, 1.2% carbon, from Le Chatlier in 1899:
Temp. (C) Coeff. (x10^-6)
0-100 10.5
100-200 11.5
200-300 13.
300-400 15.
400-500 14.
500-600 16.
600-700 16.
above 900 29.
As TVP noted, and visible in the data, is the phase transition to austenite occurring above 900 C, which tends to honk up the numbers. Note this is also a fairly high carbon steel.
Upvote0Downvote
Sep 11, 2013
Thread starter
#5
sydneyjongleur
Materials
Jul 22, 2011 39GB
Thanks guys,
The reason I was asking is because it was suggested to use this as a fixture and clamps used in a heat teatment process to remove distortion. It was suggested that this material would be used as a base fixture to locate a nickel alloy ring onto, and also use the steel material as clamps to hold the nickel alloy ring down, and heat treat to 1000°C in order to remove distortion. I was looking at the thermal expansion rates to prove that this wasn't possible. In any case I think the carbon steel would be so soft at that temp that the nickel alloy would be welded into it.
Upvote0Downvote
Sep 11, 2013
#6
EdStainless
Materials
May 20, 2004 16,357Waukesha WI USA
We made furnace muffles out of low C steel (1005 or so) for use at 2200F (1200C).
It is difficult to find a metal with any usable strength at that temp.
According to ASM, High temperature property data: ferrous alloys
all in 10exp-6/C
0-600C 14.6
0-700C 14.9
0-800C 16.6
0-1000 13.7
These are averages for the entire range.
Yes there is a peak in the curve.
= = = = = = = = = = = = = = = = = = = =
Plymouth Tube
Upvote0Downvote
Sep 12, 2013
#7
MagBen
Materials
Jun 7, 2012 728US
Ed, useful info on low C steel as furnace muffle at 1200C. The muffles we are using are made of a high temperature alloy, but the wedling area is the weak point. After 100-200 times of serve at 1200C, we have to change the muffle due to the leakage at welding. I am thinking welding quality may be able to be improved if using low C steel.
For the thermal expansion coeffient data, what could be the down peak at 0-1000C? somthing to do with the phase trasformation to anstenite? we know austenite is fcc, ferrite is bcc, the former has large lattice parameters. also, austenite has a higher coefficient about 24ppm/C. I expected a larger coefficient at 0-1000C, compared with 0-800C!
Carbon content is another factor to consider: carbon obviously increases lattice parameters, but the coefficient normally decreases with increasing C content, i.e. high carbon steel has a lower thermal expansion coefficient.
Upvote0Downvote
Sep 12, 2013
#8
arunmrao
Materials
Oct 1, 2000 4,758IN
sydneyjongleur,
I am reminded of a gear manufacturing company, several years ago, who had a serious problem . They were facing distortion problem in their machined gears and there was not enough stock to correct them. The problem was they had mild steel fabrications as fixture for oil quench and temper cycle. The fixtures were severely scaled, twisted and cracked. It is not a wise option to use a mild steel fixture for heat treatment applications.
Upvote0Downvote
Sep 12, 2013
#9
Compositepro
Chemical
Oct 22, 2003 7,841US
Placing a bead or corrugation (think bellows) in the sheet metal at appropriate spots will do more to solve your problem than changing alloys. Usually the problem is not the temperature change but the differences in temperature between one spot and another. Hot areas cannot expand freely while attached to cold areas. There has to be flexibility in the structure to accommodate this or it will crack.
Upvote0Downvote
Sep 12, 2013
Thread starter
#10
sydneyjongleur
Materials
Jul 22, 2011 39GB
thanks guys, I will take on board what you are saying. I agree mild steel is a bad idea.
Upvote0Downvote
Sep 13, 2013
#11
MagBen
Materials
Jun 7, 2012 728US
distortion is not a big deal for my muffle applications. leakage is my main concern after long cycles (no quench).we can not get muffles made by a whole piece,and have to weld for a sealed muffle for vacuum heat treatment. leakage always started with the welding area. low carbon steel may be a good choice for a better welding!
Upvote0Downvote
Sep 13, 2013
#12
EdStainless
Materials
May 20, 2004 16,357Waukesha WI USA
At high temp (>95% of MP) plain Fe has more strength than Ni alloys.
Weird but that is how it works.
We were running pure Hydrogen on the inside.
The heating elements were wrapped on the outside (with a thin layer of ceramic paper underneath).
= = = = = = = = = = = = = = = = = = = =
Plymouth Tube
Upvote0Downvote
Sep 18, 2013
#13
racookpe1978
Nuclear
Feb 1, 2007 5,994US
Could a ceramic or poured non-metallic solid (concrete ?) hold up the material being treated without itself being distorted by the heat or losing strength so the support effect fails to hold the load without contaminating the heat-treated materiel?
Upvote0Downvote
Sep 19, 2013
#14
NickClark
Materials
Mar 6, 2013 25GB
Vacuum brazing companies use graphite plates and jigs to support parts.its dimensionally stable and won't react.
Upvote0Downvote
Status Not open for further replies.
Similar threads
N
Question
Overly of ERCuAl-A2 over low alloy steel
Nanda Kishore Kandregula
Apr 29, 2025
Material engineering general discussion
Replies 6 Views 766
May 1, 2025
weldstan
K
Question
How is thermal analysis used in your metal post-processing workflow?
Kunal@2412
May 6, 2025
Material engineering general discussion
Replies 0 Views 257
May 6, 2025
Kunal@2412
K
K
Question
White liquor and steel 316
Krillo
Apr 22, 2025
Material engineering general discussion
Replies 1 Views 2K
Apr 24, 2025
lucky-guesser
Question
New Patented High Speed Steel Alloys
Maui
Apr 25, 2025
Material engineering general discussion
Replies 16 Views 3K
May 5, 2025
Maui
Question
Stainless Steel 300 series
FrancoisY
Feb 10, 2025
Material engineering general discussion
Replies 13 Views 10K
Tuesday at 9:04 PM
EdStainless
Share:
FacebookXBlueskyLinkedInRedditPinterestTumblrWhatsAppEmailShareLink
Log in
Your name or email address
Password
Forgot your password?
- [x] Stay logged in
Log in
Don't have an account? Register now
Part and Inventory Search
Sponsor
Home
Forums
People
Materials Engineers
Material engineering general discussion
DefaultAlternate
Change width
Contact us
Terms and rules
Privacy policy
Help
Home
RSS
Copyright © 2025 WTWH Media, LLC. All Rights Reserved.
Back
Top |
189254 | https://en.wikipedia.org/wiki/Maxwell%27s_equations | Jump to content
Search
Contents
1 History of the equations
2 Conceptual descriptions
2.1 Gauss's law
2.2 Gauss's law for magnetism
2.3 Faraday's law
2.4 Ampère–Maxwell law
3 Formulation in terms of electric and magnetic fields (microscopic or in vacuum version)
3.1 Key to the notation
3.1.1 Differential equations
3.1.2 Integral equations
3.2 Formulation in the SI
3.3 Formulation in the Gaussian system
4 Relationship between differential and integral formulations
4.1 Flux and divergence
4.2 Circulation and curl
5 Charge conservation
6 Vacuum equations, electromagnetic waves and speed of light
7 Macroscopic formulation
7.1 Bound charge and current
7.2 Auxiliary fields, polarization and magnetization
7.3 Constitutive relations
8 Alternative formulations
9 Solutions
10 Overdetermination of Maxwell's equations
11 Maxwell's equations as the classical limit of QED
12 Variations
12.1 Magnetic monopoles
13 See also
14 Explanatory notes
15 References
16 Further reading
16.1 Historical publications
17 External links
17.1 Modern treatments
17.2 Other
Maxwell's equations
Afrikaans
Alemannisch
العربية
Asturianu
Azərbaycanca
تۆرکجه
বাংলা
Беларуская
Беларуская (тарашкевіца)
Български
Bosanski
Català
Чӑвашла
Čeština
Dansk
Deutsch
Eesti
Ελληνικά
Español
Esperanto
Euskara
فارسی
Français
Galego
한국어
Հայերեն
हिन्दी
Hrvatski
Bahasa Indonesia
Interlingua
Íslenska
Italiano
עברית
ಕನ್ನಡ
ქართული
Қазақша
Kreyòl ayisyen
Latina
Latviešu
Lietuvių
Limburgs
Magyar
Македонски
मराठी
Bahasa Melayu
မြန်မာဘာသာ
Nederlands
नेपाली
日本語
Norsk bokmål
Norsk nynorsk
Oʻzbekcha / ўзбекча
ਪੰਜਾਬੀ
Polski
Português
Romnă
Русский
Shqip
Simple English
Slovenčina
Slovenščina
Српски / srpski
Srpskohrvatski / српскохрватски
Suomi
Svenska
Tagalog
தமிழ்
Татарча / tatarça
తెలుగు
ไทย
Türkçe
Українська
اردو
Tiếng Việt
吴语
ייִדיש
粵語
中文
Edit links
Article
Talk
Read
Edit
View history
Tools
Actions
Read
Edit
View history
General
What links here
Related changes
Upload file
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Print/export
Download as PDF
Printable version
In other projects
Wikimedia Commons
Wikibooks
Wikiquote
Wikiversity
Wikidata item
Appearance
From Wikipedia, the free encyclopedia
Equations describing classical electromagnetism
For thermodynamic relations, see Maxwell relations.
| Electromagnetism |
| Electricity Magnetism Optics History Computational Textbooks Phenomena |
| Electrostatics Charge density Conductor Coulomb law Electret Electric charge Electric dipole Electric field Electric flux Electric potential Electrostatic discharge Electrostatic induction Gauss's law Insulator Permittivity Polarization Potential energy Static electricity Triboelectricity |
| Magnetostatics Ampère's law Biot–Savart law Gauss's law for magnetism Magnetic dipole Magnetic field Magnetic flux Magnetic scalar potential Magnetic vector potential Magnetization Permeability Right-hand rule |
| Electrodynamics Bremsstrahlung Cyclotron radiation Displacement current Eddy current Electromagnetic field Electromagnetic induction Electromagnetic pulse Electromagnetic radiation Faraday's law Jefimenko equations Larmor formula Lenz's law Liénard–Wiechert potential London equations Lorentz force Maxwell's equations Maxwell tensor Poynting vector Synchrotron radiation |
| Electrical network Alternating current Capacitance Current density Direct current Electric current Electric power Electrolysis Electromotive force Impedance Inductance Joule heating Kirchhoff's laws Network analysis Ohm's law Parallel circuit Resistance Resonant cavities Series circuit Voltage Watt Waveguides |
| Magnetic circuit AC motor DC motor Electric machine Electric motor Gyrator–capacitor Induction motor Linear motor Magnetomotive force Permeance Reluctance (complex) Reluctance (real) Rotor Stator Transformer |
| Covariant formulation Electromagnetic tensor Electromagnetism and special relativity Four-current Four-potential Mathematical descriptions Maxwell equations in curved spacetime Relativistic electromagnetism Stress–energy tensor |
| Scientists Ampère Biot Coulomb Davy Einstein Faraday Fizeau Gauss Heaviside Helmholtz Henry Hertz Hopkinson Jefimenko Joule Kelvin Kirchhoff Larmor Lenz Liénard Lorentz Maxwell Neumann Ohm Ørsted Poisson Poynting Ritchie Savart Singer Steinmetz Tesla Thomson Volta Weber Wiechert |
| v t e |
Maxwell's equations, or Maxwell–Heaviside equations, are a set of coupled partial differential equations that, together with the Lorentz force law, form the foundation of classical electromagnetism, classical optics, electric and magnetic circuits. The equations provide a mathematical model for electric, optical, and radio technologies, such as power generation, electric motors, wireless communication, lenses, radar, etc. They describe how electric and magnetic fields are generated by charges, currents, and changes of the fields.[note 1] The equations are named after the physicist and mathematician James Clerk Maxwell, who, in 1861 and 1862, published an early form of the equations that included the Lorentz force law. Maxwell first used the equations to propose that light is an electromagnetic phenomenon. The modern form of the equations in their most common formulation is credited to Oliver Heaviside.
Maxwell's equations may be combined to demonstrate how fluctuations in electromagnetic fields (waves) propagate at a constant speed in vacuum, c (299792458 m/s). Known as electromagnetic radiation, these waves occur at various wavelengths to produce a spectrum of radiation from radio waves to gamma rays.
In partial differential equation form and a coherent system of units, Maxwell's microscopic equations can be written as (top to bottom: Gauss's law, Gauss's law for magnetism, Faraday's law, Ampère-Maxwell law) With the electric field, the magnetic field, the electric charge density and the current density. is the vacuum permittivity and the vacuum permeability.
The equations have two major variants:
The microscopic equations have universal applicability but are unwieldy for common calculations. They relate the electric and magnetic fields to total charge and total current, including the complicated charges and currents in materials at the atomic scale.
The macroscopic equations define two new auxiliary fields that describe the large-scale behaviour of matter without having to consider atomic-scale charges and quantum phenomena like spins. However, their use requires experimentally determined parameters for a phenomenological description of the electromagnetic response of materials.
The term "Maxwell's equations" is often also used for equivalent alternative formulations. Versions of Maxwell's equations based on the electric and magnetic scalar potentials are preferred for explicitly solving the equations as a boundary value problem, analytical mechanics, or for use in quantum mechanics. The covariant formulation (on spacetime rather than space and time separately) makes the compatibility of Maxwell's equations with special relativity manifest. Maxwell's equations in curved spacetime, commonly used in high-energy and gravitational physics, are compatible with general relativity.[note 2] In fact, Albert Einstein developed special and general relativity to accommodate the invariant speed of light, a consequence of Maxwell's equations, with the principle that only relative movement has physical consequences.
The publication of the equations marked the unification of a theory for previously separately described phenomena: magnetism, electricity, light, and associated radiation. Since the mid-20th century, it has been understood that Maxwell's equations do not give an exact description of electromagnetic phenomena, but are instead a classical limit of the more precise theory of quantum electrodynamics.
History of the equations
[edit]
Main article: History of Maxwell's equations
Conceptual descriptions
[edit]
Gauss's law
[edit]
Main article: Gauss's law
Gauss's law describes the relationship between an electric field and electric charges: an electric field points away from positive charges and towards negative charges, and the net outflow of the electric field through a closed surface is proportional to the enclosed charge, including bound charge due to polarization of material. The coefficient of the proportion is the permittivity of free space.
Gauss's law for magnetism
[edit]
Main article: Gauss's law for magnetism
Gauss's law for magnetism states that electric charges have no magnetic analogues, called magnetic monopoles; no north or south magnetic poles exist in isolation. Instead, the magnetic field of a material is attributed to a dipole, and the net outflow of the magnetic field through a closed surface is zero. Magnetic dipoles may be represented as loops of current or inseparable pairs of equal and opposite "magnetic charges". Precisely, the total magnetic flux through a Gaussian surface is zero, and the magnetic field is a solenoidal vector field.[note 3]
Faraday's law
[edit]
Main article: Faraday's law of induction
The Maxwell–Faraday version of Faraday's law of induction describes how a time-varying magnetic field corresponds to the negative curl of an electric field. In integral form, it states that the work per unit charge required to move a charge around a closed loop equals the rate of change of the magnetic flux through the enclosed surface.
The electromagnetic induction is the operating principle behind many electric generators: for example, a rotating bar magnet creates a changing magnetic field and generates an electric field in a nearby wire.
Ampère–Maxwell law
[edit]
Main article: Ampère's circuital law
The original law of Ampère states that magnetic fields relate to electric current. Maxwell's addition states that magnetic fields also relate to changing electric fields, which Maxwell called displacement current. The integral form states that electric and displacement currents are associated with a proportional magnetic field along any enclosing curve.
Maxwell's modification of Ampère's circuital law is important because the laws of Ampère and Gauss must otherwise be adjusted for static fields.[clarification needed] As a consequence, it predicts that a rotating magnetic field occurs with a changing electric field. A further consequence is the existence of self-sustaining electromagnetic waves which travel through empty space.
The speed calculated for electromagnetic waves, which could be predicted from experiments on charges and currents,[note 4] matches the speed of light; indeed, light is one form of electromagnetic radiation (as are X-rays, radio waves, and others). Maxwell understood the connection between electromagnetic waves and light in 1861, thereby unifying the theories of electromagnetism and optics.
Formulation in terms of electric and magnetic fields (microscopic or in vacuum version)
[edit]
In the electric and magnetic field formulation there are four equations that determine the fields for given charge and current distribution. A separate law of nature, the Lorentz force law, describes how the electric and magnetic fields act on charged particles and currents. By convention, a version of this law in the original equations by Maxwell is no longer included. The vector calculus formalism below, the work of Oliver Heaviside, has become standard. It is rotationally invariant, and therefore mathematically more transparent than Maxwell's original 20 equations in x, y and z components. The relativistic formulations are more symmetric and Lorentz invariant. For the same equations expressed using tensor calculus or differential forms (see § Alternative formulations).
The differential and integral formulations are mathematically equivalent; both are useful. The integral formulation relates fields within a region of space to fields on the boundary and can often be used to simplify and directly calculate fields from symmetric distributions of charges and currents. On the other hand, the differential equations are purely local and are a more natural starting point for calculating the fields in more complicated (less symmetric) situations, for example using finite element analysis.
Key to the notation
[edit]
Symbols in bold represent vector quantities, and symbols in italics represent scalar quantities, unless otherwise indicated. The equations introduce the electric field, E, a vector field, and the magnetic field, B, a pseudovector field, each generally having a time and location dependence. The sources are
the total electric charge density (total charge per unit volume), ρ, and
the total electric current density (total current per unit area), J.
The universal constants appearing in the equations (the first two ones explicitly only in the SI formulation) are:
the permittivity of free space, ε0, and
the permeability of free space, μ0, and
the speed of light,
Differential equations
[edit]
In the differential equations,
the nabla symbol, ∇, denotes the three-dimensional gradient operator, del,
the ∇⋅ symbol (pronounced "del dot") denotes the divergence operator,
the ∇× symbol (pronounced "del cross") denotes the curl operator.
Integral equations
[edit]
In the integral equations,
Ω is any volume with closed boundary surface ∂Ω, and
Σ is any surface with closed boundary curve ∂Σ,
The equations are a little easier to interpret with time-independent surfaces and volumes. Time-independent surfaces and volumes are "fixed" and do not change over a given time interval. For example, since the surface is time-independent, we can bring the differentiation under the integral sign in Faraday's law: Maxwell's equations can be formulated with possibly time-dependent surfaces and volumes by using the differential version and using Gauss' and Stokes' theorems as appropriate.
is a surface integral over the boundary surface ∂Ω, with the loop indicating the surface is closed
is a volume integral over the volume Ω,
is a line integral around the boundary curve ∂Σ, with the loop indicating the curve is closed.
is a surface integral over the surface Σ,
The total electric charge Q enclosed in Ω is the volume integral over Ω of the charge density ρ (see the "macroscopic formulation" section below): where dV is the volume element.
The net magnetic flux ΦB is the surface integral of the magnetic field B passing through a fixed surface, Σ:
The net electric flux ΦE is the surface integral of the electric field E passing through Σ:
The net electric current I is the surface integral of the electric current density J passing through Σ: where dS denotes the differential vector element of surface area S, normal to surface Σ. (Vector area is sometimes denoted by A rather than S, but this conflicts with the notation for magnetic vector potential).
Formulation in the SI
[edit]
| Name | Integral equations | Differential equations |
---
| Gauss's law | | |
| Gauss's law for magnetism | | |
| Maxwell–Faraday equation (Faraday's law of induction) | | |
| Ampère–Maxwell law | | |
Formulation in the Gaussian system
[edit]
Main article: Gaussian units
The definitions of charge, electric field, and magnetic field can be altered to simplify theoretical calculation, by absorbing dimensioned factors of ε0 and μ0 into the units (and thus redefining these). With a corresponding change in the values of the quantities for the Lorentz force law this yields the same physics, i.e. trajectories of charged particles, or work done by an electric motor. These definitions are often preferred in theoretical and high energy physics where it is natural to take the electric and magnetic field with the same units, to simplify the appearance of the electromagnetic tensor: the Lorentz covariant object unifying electric and magnetic field would then contain components with uniform unit and dimension.: vii Such modified definitions are conventionally used with the Gaussian (CGS) units. Using these definitions, colloquially "in Gaussian units", the Maxwell equations become:
| Name | Integral equations | Differential equations |
---
| Gauss's law | | |
| Gauss's law for magnetism | | |
| Maxwell–Faraday equation (Faraday's law of induction) | | |
| Ampère–Maxwell law | | |
The equations simplify slightly when a system of quantities is chosen in the speed of light, c, is used for nondimensionalization, so that, for example, seconds and lightseconds are interchangeable, and c = 1.
Further changes are possible by absorbing factors of 4π. This process, called rationalization, affects whether Coulomb's law or Gauss's law includes such a factor (see Heaviside–Lorentz units, used mainly in particle physics).
Relationship between differential and integral formulations
[edit]
The equivalence of the differential and integral formulations are a consequence of the Gauss divergence theorem and the Kelvin–Stokes theorem.
Flux and divergence
[edit]
According to the (purely mathematical) Gauss divergence theorem, the electric flux through the boundary surface ∂Ω can be rewritten as
The integral version of Gauss's equation can thus be rewritten as Since Ω is arbitrary (e.g. an arbitrary small ball with arbitrary center), this is satisfied if and only if the integrand is zero everywhere. This is the differential equations formulation of Gauss equation up to a trivial rearrangement.
Similarly rewriting the magnetic flux in Gauss's law for magnetism in integral form gives
which is satisfied for all Ω if and only if everywhere.
Circulation and curl
[edit]
By the Kelvin–Stokes theorem we can rewrite the line integrals of the fields around the closed boundary curve ∂Σ to an integral of the "circulation of the fields" (i.e. their curls) over a surface it bounds, i.e. Hence the Ampère–Maxwell law, the modified version of Ampère's circuital law, in integral form can be rewritten as Since Σ can be chosen arbitrarily, e.g. as an arbitrary small, arbitrary oriented, and arbitrary centered disk, we conclude that the integrand is zero if and only if the Ampère–Maxwell law in differential equations form is satisfied. The equivalence of Faraday's law in differential and integral form follows likewise.
The line integrals and curls are analogous to quantities in classical fluid dynamics: the circulation of a fluid is the line integral of the fluid's flow velocity field around a closed loop, and the vorticity of the fluid is the curl of the velocity field.
Charge conservation
[edit]
The invariance of charge can be derived as a corollary of Maxwell's equations. The left-hand side of the Ampère–Maxwell law has zero divergence by the div–curl identity. Expanding the divergence of the right-hand side, interchanging derivatives, and applying Gauss's law gives: i.e., By the Gauss divergence theorem, this means the rate of change of charge in a fixed volume equals the net current flowing through the boundary:
In particular, in an isolated system the total charge is conserved.
Vacuum equations, electromagnetic waves and speed of light
[edit]
Further information: Electromagnetic wave equation, Inhomogeneous electromagnetic wave equation, Sinusoidal plane-wave solutions of the electromagnetic wave equation, and Helmholtz equation
In a region with no charges (ρ = 0) and no currents (J = 0), such as in vacuum, Maxwell's equations reduce to:
Taking the curl (∇×) of the curl equations, and using the curl of the curl identity we obtain
The quantity has the dimension (T/L)2. Defining , the equations above have the form of the standard wave equations
Already during Maxwell's lifetime, it was found that the known values for and give , then already known to be the speed of light in free space. This led him to propose that light and radio waves were propagating electromagnetic waves, since amply confirmed. In the old SI system of units, the values of and are defined constants, (which means that by definition ) that define the ampere and the metre. In the new SI system, only c keeps its defined value, and the electron charge gets a defined value.
In materials with relative permittivity, εr, and relative permeability, μr, the phase velocity of light becomes which is usually[note 5] less than c.
In addition, E and B are perpendicular to each other and to the direction of wave propagation, and are in phase with each other. A sinusoidal plane wave is one special solution of these equations. Maxwell's equations explain how these waves can physically propagate through space. The changing magnetic field creates a changing electric field through Faraday's law. In turn, that electric field creates a changing magnetic field through Maxwell's modification of Ampère's circuital law. This perpetual cycle allows these waves, now known as electromagnetic radiation, to move through space at velocity c.
Macroscopic formulation
[edit]
The above equations are the microscopic version of Maxwell's equations, expressing the electric and the magnetic fields in terms of the (possibly atomic-level) charges and currents present. This is sometimes called the "general" form, but the macroscopic version below is equally general, the difference being one of bookkeeping.
The microscopic version is sometimes called "Maxwell's equations in vacuum": this refers to the fact that the material medium is not built into the structure of the equations, but appears only in the charge and current terms. The microscopic version was introduced by Lorentz, who tried to use it to derive the macroscopic properties of bulk matter from its microscopic constituents.: 5
"Maxwell's macroscopic equations", also known as Maxwell's equations in matter, are more similar to those that Maxwell introduced himself.
| Name | Integral equations (SI) | Differential equations (SI) | Differential equations (Gaussian system) |
--- --- |
| Gauss's law | | | |
| Ampère–Maxwell law | | | |
| Gauss's law for magnetism | | | |
| Maxwell–Faraday equation (Faraday's law of induction) | | | |
In the macroscopic equations, the influence of bound charge Qb and bound current Ib is incorporated into the displacement field D and the magnetizing field H, while the equations depend only on the free charges Qf and free currents If. This reflects a splitting of the total electric charge Q and current I (and their densities ρ and J) into free and bound parts:
The cost of this splitting is that the additional fields D and H need to be determined through phenomenological constituent equations relating these fields to the electric field E and the magnetic field B, together with the bound charge and current.
See below for a detailed description of the differences between the microscopic equations, dealing with total charge and current including material contributions, useful in air/vacuum;[note 6] and the macroscopic equations, dealing with free charge and current, practical to use within materials.
Bound charge and current
[edit]
Main articles: Current density, Bound charge, and Bound current
When an electric field is applied to a dielectric material its molecules respond by forming microscopic electric dipoles – their atomic nuclei move a tiny distance in the direction of the field, while their electrons move a tiny distance in the opposite direction. This produces a macroscopic bound charge in the material even though all of the charges involved are bound to individual molecules. For example, if every molecule responds the same, similar to that shown in the figure, these tiny movements of charge combine to produce a layer of positive bound charge on one side of the material and a layer of negative charge on the other side. The bound charge is most conveniently described in terms of the polarization P of the material, its dipole moment per unit volume. If P is uniform, a macroscopic separation of charge is produced only at the surfaces where P enters and leaves the material. For non-uniform P, a charge is also produced in the bulk.
Somewhat similarly, in all materials the constituent atoms exhibit magnetic moments that are intrinsically linked to the angular momentum of the components of the atoms, most notably their electrons. The connection to angular momentum suggests the picture of an assembly of microscopic current loops. Outside the material, an assembly of such microscopic current loops is not different from a macroscopic current circulating around the material's surface, despite the fact that no individual charge is traveling a large distance. These bound currents can be described using the magnetization M.
The very complicated and granular bound charges and bound currents, therefore, can be represented on the macroscopic scale in terms of P and M, which average these charges and currents on a sufficiently large scale so as not to see the granularity of individual atoms, but also sufficiently small that they vary with location in the material. As such, Maxwell's macroscopic equations ignore many details on a fine scale that can be unimportant to understanding matters on a gross scale by calculating fields that are averaged over some suitable volume.
Auxiliary fields, polarization and magnetization
[edit]
The definitions of the auxiliary fields are: where P is the polarization field and M is the magnetization field, which are defined in terms of microscopic bound charges and bound currents respectively. The macroscopic bound charge density ρb and bound current density Jb in terms of polarization P and magnetization M are then defined as
If we define the total, bound, and free charge and current density by and use the defining relations above to eliminate D, and H, the "macroscopic" Maxwell's equations reproduce the "microscopic" equations.
Constitutive relations
[edit]
Main article: Constitutive equation § Electromagnetism
In order to apply 'Maxwell's macroscopic equations', it is necessary to specify the relations between displacement field D and the electric field E, as well as the magnetizing field H and the magnetic field B. Equivalently, we have to specify the dependence of the polarization P (hence the bound charge) and the magnetization M (hence the bound current) on the applied electric and magnetic field. The equations specifying this response are called constitutive relations. For real-world materials, the constitutive relations are rarely simple, except approximately, and usually determined by experiment. See the main article on constitutive relations for a fuller description.: 44–45
For materials without polarization and magnetization, the constitutive relations are (by definition): 2 where ε0 is the permittivity of free space and μ0 the permeability of free space. Since there is no bound charge, the total and the free charge and current are equal.
An alternative viewpoint on the microscopic equations is that they are the macroscopic equations together with the statement that vacuum behaves like a perfect linear "material" without additional polarization and magnetization. More generally, for linear materials the constitutive relations are: 44–45 where ε is the permittivity and μ the permeability of the material. For the displacement field D the linear approximation is usually excellent because for all but the most extreme electric fields or temperatures obtainable in the laboratory (high power pulsed lasers) the interatomic electric fields of materials of the order of 1011 V/m are much higher than the external field. For the magnetizing field , however, the linear approximation can break down in common materials like iron leading to phenomena like hysteresis. Even the linear case can have various complications, however.
For homogeneous materials, ε and μ are constant throughout the material, while for inhomogeneous materials they depend on location within the material (and perhaps time).: 463
For isotropic materials, ε and μ are scalars, while for anisotropic materials (e.g. due to crystal structure) they are tensors.: 421: 463
Materials are generally dispersive, so ε and μ depend on the frequency of any incident EM waves.: 625: 397
Even more generally, in the case of non-linear materials (see for example nonlinear optics), D and P are not necessarily proportional to E, similarly H or M is not necessarily proportional to B. In general D and H depend on both E and B, on location and time, and possibly other physical quantities.
In applications one also has to describe how the free currents and charge density behave in terms of E and B possibly coupled to other physical quantities like pressure, and the mass, number density, and velocity of charge-carrying particles. E.g., the original equations given by Maxwell (see History of Maxwell's equations) included Ohm's law in the form
Alternative formulations
[edit]
For the equations in special relativity, see Classical electromagnetism and special relativity and Covariant formulation of classical electromagnetism.
For the equations in general relativity, see Maxwell's equations in curved spacetime.
For an overview, see Mathematical descriptions of the electromagnetic field.
For the equations in quantum field theory, see Quantum electrodynamics.
Following are some of the several other mathematical formalisms of Maxwell's equations, with the columns separating the two homogeneous Maxwell equations from the two inhomogeneous ones. Each formulation has versions directly in terms of the electric and magnetic fields, and indirectly in terms of the electrical potential φ and the vector potential A. Potentials were introduced as a convenient way to solve the homogeneous equations, but it was thought that all observable physics was contained in the electric and magnetic fields (or relativistically, the Faraday tensor). The potentials play a central role in quantum mechanics, however, and act quantum mechanically with observable consequences even when the electric and magnetic fields vanish (Aharonov–Bohm effect).
Each table describes one formalism. See the main article for details of each formulation.
The direct spacetime formulations make manifest that the Maxwell equations are relativistically invariant, where space and time are treated on equal footing. Because of this symmetry, the electric and magnetic fields are treated on equal footing and are recognized as components of the Faraday tensor. This reduces the four Maxwell equations to two, which simplifies the equations, although we can no longer use the familiar vector formulation. Maxwell equations in formulation that do not treat space and time manifestly on the same footing have Lorentz invariance as a hidden symmetry. This was a major source of inspiration for the development of relativity theory. Indeed, even the formulation that treats space and time separately is not a non-relativistic approximation and describes the same physics by simply renaming variables. For this reason the relativistic invariant equations are usually called the Maxwell equations as well.
Each table below describes one formalism.
Tensor calculus
| Formulation | Homogeneous equations | Inhomogeneous equations |
| Fields Minkowski space | | |
| Potentials (any gauge) Minkowski space | | |
| Potentials (Lorenz gauge) Minkowski space | | |
| Fields any spacetime | | |
| Potentials (any gauge) any spacetime (with §topological restrictions) | | |
| Potentials (Lorenz gauge) any spacetime (with topological restrictions) | | |
Differential forms
| Formulation | Homogeneous equations | Inhomogeneous equations |
| Fields any spacetime | | |
| Potentials (any gauge) any spacetime (with topological restrictions) | | |
| Potentials (Lorenz gauge) any spacetime (with topological restrictions) | | |
In the tensor calculus formulation, the electromagnetic tensor Fαβ is an antisymmetric covariant order 2 tensor; the four-potential, Aα, is a covariant vector; the current, Jα, is a vector; the square brackets, [ ], denote antisymmetrization of indices; ∂α is the partial derivative with respect to the coordinate, xα. In Minkowski space coordinates are chosen with respect to an inertial frame; (xα) = (ct, x, y, z), so that the metric tensor used to raise and lower indices is ηαβ = diag(1, −1, −1, −1). The d'Alembert operator on Minkowski space is ◻ = ∂α∂α as in the vector formulation. In general spacetimes, the coordinate system xα is arbitrary, the covariant derivative ∇α, the Ricci tensor, Rαβ and raising and lowering of indices are defined by the Lorentzian metric, gαβ and the d'Alembert operator is defined as ◻ = ∇α∇α. The topological restriction is that the second real cohomology group of the space vanishes (see the differential form formulation for an explanation). This is violated for Minkowski space with a line removed, which can model a (flat) spacetime with a point-like monopole on the complement of the line.
In the differential form formulation on arbitrary space times, F = 1/2Fαβdxα ∧ dxβ is the electromagnetic tensor considered as a 2-form, A = Aαdxα is the potential 1-form, is the current 3-form, d is the exterior derivative, and is the Hodge star on forms defined (up to its orientation, i.e. its sign) by the Lorentzian metric of spacetime. In the special case of 2-forms such as F, the Hodge star depends on the metric tensor only for its local scale. This means that, as formulated, the differential form field equations are conformally invariant, but the Lorenz gauge condition breaks conformal invariance. The operator is the d'Alembert–Laplace–Beltrami operator on 1-forms on an arbitrary Lorentzian spacetime. The topological condition is again that the second real cohomology group is 'trivial' (meaning that its form follows from a definition). By the isomorphism with the second de Rham cohomology this condition means that every closed 2-form is exact.
Other formalisms include the geometric algebra formulation and a matrix representation of Maxwell's equations. Historically, a quaternionic formulation was used.
Solutions
[edit]
Maxwell's equations are partial differential equations that relate the electric and magnetic fields to each other and to the electric charges and currents. Often, the charges and currents are themselves dependent on the electric and magnetic fields via the Lorentz force equation and the constitutive relations. These all form a set of coupled partial differential equations which are often very difficult to solve: the solutions encompass all the diverse phenomena of classical electromagnetism. Some general remarks follow.
As for any differential equation, boundary conditions and initial conditions are necessary for a unique solution. For example, even with no charges and no currents anywhere in spacetime, there are the obvious solutions for which E and B are zero or constant, but there are also non-trivial solutions corresponding to electromagnetic waves. In some cases, Maxwell's equations are solved over the whole of space, and boundary conditions are given as asymptotic limits at infinity. In other cases, Maxwell's equations are solved in a finite region of space, with appropriate conditions on the boundary of that region, for example an artificial absorbing boundary representing the rest of the universe, or periodic boundary conditions, or walls that isolate a small region from the outside world (as with a waveguide or cavity resonator).
Jefimenko's equations (or the closely related Liénard–Wiechert potentials) are the explicit solution to Maxwell's equations for the electric and magnetic fields created by any given distribution of charges and currents. It assumes specific initial conditions to obtain the so-called "retarded solution", where the only fields present are the ones created by the charges. However, Jefimenko's equations are unhelpful in situations when the charges and currents are themselves affected by the fields they create.
Numerical methods for differential equations can be used to compute approximate solutions of Maxwell's equations when exact solutions are impossible. These include the finite element method and finite-difference time-domain method. For more details, see Computational electromagnetics.
Overdetermination of Maxwell's equations
[edit]
Maxwell's equations seem overdetermined, in that they involve six unknowns (the three components of E and B) but eight equations (one for each of the two Gauss's laws, three vector components each for Faraday's and Ampère's circuital laws). (The currents and charges are not unknowns, being freely specifiable subject to charge conservation.) This is related to a certain limited kind of redundancy in Maxwell's equations: It can be proven that any system satisfying Faraday's law and Ampère's circuital law automatically also satisfies the two Gauss's laws, as long as the system's initial condition does, and assuming conservation of charge and the nonexistence of magnetic monopoles. This explanation was first introduced by Julius Adams Stratton in 1941.
Although it is possible to simply ignore the two Gauss's laws in a numerical algorithm (apart from the initial conditions), the imperfect precision of the calculations can lead to ever-increasing violations of those laws. By introducing dummy variables characterizing these violations, the four equations become not overdetermined after all. The resulting formulation can lead to more accurate algorithms that take all four laws into account.
Both identities , which reduce eight equations to six independent ones, are the true reason of overdetermination.
Equivalently, the overdetermination can be viewed as implying conservation of electric and magnetic charge, as they are required in the derivation described above but implied by the two Gauss's laws.
For linear algebraic equations, one can make 'nice' rules to rewrite the equations and unknowns. The equations can be linearly dependent. But in differential equations, and especially partial differential equations (PDEs), one needs appropriate boundary conditions, which depend in not so obvious ways on the equations. Even more, if one rewrites them in terms of vector and scalar potential, then the equations are underdetermined because of gauge fixing.
Maxwell's equations as the classical limit of QED
[edit]
Maxwell's equations and the Lorentz force law (along with the rest of classical electromagnetism) are extraordinarily successful at explaining and predicting a variety of phenomena. However, they do not account for quantum effects, and so their domain of applicability is limited. Maxwell's equations are thought of as the classical limit of quantum electrodynamics (QED).
Some observed electromagnetic phenomena cannot be explained with Maxwell's equations if the source of the electromagnetic fields are the classical distributions of charge and current. These include photon–photon scattering and many other phenomena related to photons or virtual photons, "nonclassical light" and quantum entanglement of electromagnetic fields (see Quantum optics). E.g. quantum cryptography cannot be described by Maxwell theory, not even approximately. The approximate nature of Maxwell's equations becomes more and more apparent when going into the extremely strong field regime (see Euler–Heisenberg Lagrangian) or to extremely small distances.
Finally, Maxwell's equations cannot explain any phenomenon involving individual photons interacting with quantum matter, such as the photoelectric effect, Planck's law, the Duane–Hunt law, and single-photon light detectors. However, many such phenomena may be explained using a halfway theory of quantum matter coupled to a classical electromagnetic field, either as external field or with the expected value of the charge current and density on the right hand side of Maxwell's equations. This is known as semiclassical theory or self-field QED and was initially discovered by de Broglie and Schrödinger and later fully developed by E.T. Jaynes and A.O. Barut.
Variations
[edit]
Popular variations on the Maxwell equations as a classical theory of electromagnetic fields are relatively scarce because the standard equations have stood the test of time remarkably well.
Magnetic monopoles
[edit]
Main article: Magnetic monopole
Maxwell's equations posit that there is electric charge, but no magnetic charge (also called magnetic monopoles), in the universe. Indeed, magnetic charge has never been observed, despite extensive searches,[note 7] and may not exist. If they did exist, both Gauss's law for magnetism and Faraday's law would need to be modified, and the resulting four equations would be fully symmetric under the interchange of electric and magnetic fields.: 273–275
See also
[edit]
Electronics portal
Physics portal
Algebra of physical space
Fresnel equations
Gravitoelectromagnetism
Interface conditions for electromagnetic fields
Moving magnet and conductor problem
Riemann–Silberstein vector
Spacetime algebra
Wheeler–Feynman absorber theory
Explanatory notes
[edit]
^ Electric and magnetic fields, according to the theory of relativity, are the components of a single electromagnetic field.
^ In general relativity, however, they must enter, through its stress–energy tensor, into Einstein field equations that include the spacetime curvature.
^ The absence of sinks/sources of the field does not imply that the field lines must be closed or escape to infinity. They can also wrap around indefinitely, without self-intersections. Moreover, around points where the field is zero (that cannot be intersected by field lines, because their direction would not be defined), there can be the simultaneous begin of some lines and end of other lines. This happens, for instance, in the middle between two identical cylindrical magnets, whose north poles face each other. In the middle between those magnets, the field is zero and the axial field lines coming from the magnets end. At the same time, an infinite number of divergent lines emanate radially from this point. The simultaneous presence of lines which end and begin around the point preserves the divergence-free character of the field. For a detailed discussion of non-closed field lines, see L. Zilberti "The Misconception of Closed Magnetic Flux Lines", IEEE Magnetics Letters, vol. 8, art. 1306005, 2017.
^ The quantity we would now call (ε0μ0)−1/2, with units of velocity, was directly measured before Maxwell's equations, in an 1855 experiment by Wilhelm Eduard Weber and Rudolf Kohlrausch. They charged a leyden jar (a kind of capacitor), and measured the electrostatic force associated with the potential; then, they discharged it while measuring the magnetic force from the current in the discharge wire. Their result was 3.107×108 m/s, remarkably close to the speed of light. See Joseph F. Keithley, The story of electrical and magnetic measurements: from 500 B.C. to the 1940s, p. 115.
^ There are cases (anomalous dispersion) where the phase velocity can exceed c, but the "signal velocity" will still be ≤ c
^ In some books—e.g., in U. Krey and A. Owen's Basic Theoretical Physics (Springer 2007)—the term effective charge is used instead of total charge, while free charge is simply called charge.
^ See magnetic monopole for a discussion of monopole searches. Recently, scientists have discovered that some types of condensed matter, including spin ice and topological insulators, display emergent behavior resembling magnetic monopoles. (See sciencemag.org and nature.com.) Although these were described in the popular press as the long-awaited discovery of magnetic monopoles, they are only superficially related. A "true" magnetic monopole is something where ∇ ⋅ B ≠ 0, whereas in these condensed-matter systems, ∇ ⋅ B = 0 while only ∇ ⋅ H ≠ 0.
References
[edit]
^ Hampshire, Damian P. (29 October 2018). "A derivation of Maxwell's equations using the Heaviside notation". Philosophical Transactions of the Royal Society A: Mathematical, Physical and Engineering Sciences. 376 (2134). arXiv:1510.04309. Bibcode:2018RSPTA.37670447H. doi:10.1098/rsta.2017.0447. ISSN 1364-503X. PMC 6232579. PMID 30373937.
^ "2022 CODATA Value: speed of light in vacuum". The NIST Reference on Constants, Units, and Uncertainty. NIST. May 2024. Retrieved 2024-05-18.
^ a b c Jackson, John. "Maxwell's equations". Science Video Glossary. Berkeley Lab. Archived from the original on 2019-01-29. Retrieved 2016-06-04.
^ J. D. Jackson, Classical Electrodynamics, section 6.3
^ Principles of physics: a calculus-based text, by R. A. Serway, J. W. Jewett, page 809.
^ Bruce J. Hunt (1991) The Maxwellians, chapter 5 and appendix, Cornell University Press
^ "Maxwell's Equations". Engineering and Technology History Wiki. 29 October 2019. Retrieved 2021-12-04.
^ Šolín, Pavel (2006). Partial differential equations and the finite element method. John Wiley and Sons. p. 273. ISBN 978-0-471-72070-6.
^ a b c J. D. Jackson (1975-10-17). Classical Electrodynamics (3rd ed.). Wiley. ISBN 978-0-471-43132-9.
^ Littlejohn, Robert (Fall 2007). "Gaussian, SI and Other Systems of Units in Electromagnetic Theory" (PDF). Physics 221A, University of California, Berkeley lecture notes. Retrieved 2008-05-06.
^ David J Griffiths (1999). Introduction to electrodynamics (Third ed.). Prentice Hall. pp. 559–562. ISBN 978-0-13-805326-0.
^ Kimball Milton; J. Schwinger (18 June 2006). Electromagnetic Radiation: Variational Methods, Waveguides and Accelerators. Springer Science & Business Media. ISBN 978-3-540-29306-4.
^ See David J. Griffiths (1999). "4.2.2". Introduction to Electrodynamics (third ed.). Prentice Hall. ISBN 9780138053260. for a good description of how P relates to the bound charge.
^ See David J. Griffiths (1999). "6.2.2". Introduction to Electrodynamics (third ed.). Prentice Hall. ISBN 9780138053260. for a good description of how M relates to the bound current.
^ a b c d Andrew Zangwill (2013). Modern Electrodynamics. Cambridge University Press. ISBN 978-0-521-89697-9.
^ a b c Kittel, Charles (2005), Introduction to Solid State Physics (8th ed.), USA: John Wiley & Sons, Inc., ISBN 978-0-471-41526-8
^ Jack, P. M. (2003). "Physical Space as a Quaternion Structure I: Maxwell Equations. A Brief Note". arXiv:math-ph/0307038.
^ A. Waser (2000). "On the Notation of Maxwell's Field Equations" (PDF). AW-Verlag.
^ a b Peter Monk (2003). Finite Element Methods for Maxwell's Equations. Oxford UK: Oxford University Press. p. 1 ff. ISBN 978-0-19-850888-5.
^ Thomas B. A. Senior & John Leonidas Volakis (1995-03-01). Approximate Boundary Conditions in Electromagnetics. London UK: Institution of Electrical Engineers. p. 261 ff. ISBN 978-0-85296-849-9.
^ a b T Hagstrom (Björn Engquist & Gregory A. Kriegsmann, Eds.) (1997). Computational Wave Propagation. Berlin: Springer. p. 1 ff. ISBN 978-0-387-94874-4.
^ Henning F. Harmuth & Malek G. M. Hussain (1994). Propagation of Electromagnetic Signals. Singapore: World Scientific. p. 17. ISBN 978-981-02-1689-4.
^ David M Cook (2002). The Theory of the Electromagnetic Field. Mineola NY: Courier Dover Publications. p. 335 ff. ISBN 978-0-486-42567-2.
^ Jean-Michel Lourtioz (2005-05-23). Photonic Crystals: Towards Nanoscale Photonic Devices. Berlin: Springer. p. 84. ISBN 978-3-540-24431-8.
^ S. G. Johnson, Notes on Perfectly Matched Layers, online MIT course notes (Aug. 2007).
^ S. F. Mahmoud (1991). Electromagnetic Waveguides: Theory and Applications. London UK: Institution of Electrical Engineers. Chapter 2. ISBN 978-0-86341-232-5.
^ John Leonidas Volakis, Arindam Chatterjee & Leo C. Kempel (1998). Finite element method for electromagnetics : antennas, microwave circuits, and scattering applications. New York: Wiley IEEE. p. 79 ff. ISBN 978-0-7803-3425-0.
^ Bernard Friedman (1990). Principles and Techniques of Applied Mathematics. Mineola NY: Dover Publications. ISBN 978-0-486-66444-6.
^ Taflove A & Hagness S C (2005). Computational Electrodynamics: The Finite-difference Time-domain Method. Boston MA: Artech House. Chapters 6 & 7. ISBN 978-1-58053-832-9.
^ H Freistühler & G Warnecke (2001). Hyperbolic Problems: Theory, Numerics, Applications. Springer. p. 605. ISBN 9783764367107.
^ J Rosen (1980). "Redundancy and superfluity for electromagnetic fields and potentials". American Journal of Physics. 48 (12): 1071. Bibcode:1980AmJPh..48.1071R. doi:10.1119/1.12289.
^ J. A. Stratton (1941). Electromagnetic Theory. McGraw-Hill Book Company. pp. 1–6. ISBN 9780470131534. {{cite book}}: ISBN / Date incompatibility (help)
^ B Jiang & J Wu & L. A. Povinelli (1996). "The Origin of Spurious Solutions in Computational Electromagnetics". Journal of Computational Physics. 125 (1): 104. Bibcode:1996JCoPh.125..104J. doi:10.1006/jcph.1996.0082. hdl:2060/19950021305.
^ Weinberg, Steven (1972). Gravitation and Cosmology. John Wiley. pp. 161–162. ISBN 978-0-471-92567-5.
^ Courant, R. & Hilbert, D. (1962), Methods of Mathematical Physics: Partial Differential Equations, vol. II, New York: Wiley-Interscience, pp. 15–18, ISBN 9783527617241 {{citation}}: ISBN / Date incompatibility (help)
Further reading
[edit]
See also: List of textbooks in electromagnetism
Imaeda, K. (1995), "Biquaternionic Formulation of Maxwell's Equations and their Solutions", in Ablamowicz, Rafał; Lounesto, Pertti (eds.), Clifford Algebras and Spinor Structures, Springer, pp. 265–280, doi:10.1007/978-94-015-8422-7_16, ISBN 978-90-481-4525-6
Historical publications
[edit]
On Faraday's Lines of Force – 1855/56. Maxwell's first paper (Part 1 & 2) – Compiled by Blaze Labs Research (PDF).
On Physical Lines of Force – 1861. Maxwell's 1861 paper describing magnetic lines of force – Predecessor to 1873 Treatise.
James Clerk Maxwell, "A Dynamical Theory of the Electromagnetic Field", Philosophical Transactions of the Royal Society of London 155, 459–512 (1865). (This article accompanied a December 8, 1864 presentation by Maxwell to the Royal Society.)
A Dynamical Theory Of The Electromagnetic Field – 1865. Maxwell's 1865 paper describing his 20 equations, link from Google Books.
J. Clerk Maxwell (1873), "A Treatise on Electricity and Magnetism":
Maxwell, J. C., "A Treatise on Electricity And Magnetism" – Volume 1 – 1873 – Posner Memorial Collection – Carnegie Mellon University.
Maxwell, J. C., "A Treatise on Electricity And Magnetism" – Volume 2 – 1873 – Posner Memorial Collection – Carnegie Mellon University.
Developments before the theory of relativity
Larmor Joseph (1897). "On a dynamical theory of the electric and luminiferous medium. Part 3, Relations with material media". Phil. Trans. R. Soc. 190: 205–300.
Lorentz Hendrik (1899). "Simplified theory of electrical and optical phenomena in moving systems". Proc. Acad. Science Amsterdam. I: 427–443.
Lorentz Hendrik (1904). "Electromagnetic phenomena in a system moving with any velocity less than that of light". Proc. Acad. Science Amsterdam. IV: 669–678.
Henri Poincaré (1900) "La théorie de Lorentz et le Principe de Réaction" (in French), Archives Néerlandaises, V, 253–278.
Henri Poincaré (1902) "La Science et l'Hypothèse" (in French).
Henri Poincaré (1905) "Sur la dynamique de l'électron" (in French), Comptes Rendus de l'Académie des Sciences, 140, 1504–1508.
Catt, Walton and Davidson. "The History of Displacement Current" Archived 2008-05-06 at the Wayback Machine. Wireless World, March 1979.
External links
[edit]
Wikimedia Commons has media related to Maxwell's equations.
Wikiquote has quotations related to Maxwell's equations.
Wikiversity discusses basic Maxwell integrals for students.
"Maxwell equations", Encyclopedia of Mathematics, EMS Press, 2001
maxwells-equations.com — An intuitive tutorial of Maxwell's equations.
The Feynman Lectures on Physics Vol. II Ch. 18: The Maxwell Equations
Wikiversity Page on Maxwell's Equations
Modern treatments
[edit]
Electromagnetism (ch. 11), B. Crowell, Fullerton College
Lecture series: Relativity and electromagnetism, R. Fitzpatrick, University of Texas at Austin
Electromagnetic waves from Maxwell's equations on Project PHYSNET.
MIT Video Lecture Series (36 × 50 minute lectures) (in .mp4 format) – Electricity and Magnetism Taught by Professor Walter Lewin.
Other
[edit]
Silagadze, Z. K. (2002). "Feynman's derivation of Maxwell equations and extra dimensions". Annales de la Fondation Louis de Broglie. 27: 241–256. arXiv:hep-ph/0106235.
Nature Milestones: Photons – Milestone 2 (1861) Maxwell's equations
| v t e Major branches of physics |
| Divisions | Pure Applied + Engineering |
| Approaches | Experimental Theoretical + Computational |
| Classical | Classical mechanics + Newtonian + Analytical + Celestial + Continuum Acoustics Classical electromagnetism Classical optics + Ray + Wave Thermodynamics + Statistical + Non-equilibrium |
| Modern | Relativistic mechanics + Special + General Nuclear physics Particle physics Quantum mechanics Atomic, molecular, and optical physics + Atomic + Molecular + Modern optics Condensed matter physics + Solid-state physics + Crystallography |
| Interdisciplinary | Astrophysics Atmospheric physics Biophysics Chemical physics Geophysics Materials science Mathematical physics Medical physics Ocean physics Quantum information science |
| Related | History of physics Nobel Prize in Physics Philosophy of physics Physics education + research Timeline of physics discoveries |
| v t e |
| Specialrelativity | | | | --- | | Background | Principle of relativity (Galilean relativity Galilean transformation) Special relativity Doubly special relativity | | Fundamentalconcepts | Frame of reference Speed of light Hyperbolic orthogonality Rapidity Maxwell's equations Proper length Proper time Proper acceleration Relativistic mass | | Formulation | Lorentz transformation Textbooks | | Phenomena | Time dilation Mass–energy equivalence (E=mc2) Length contraction Relativity of simultaneity Relativistic Doppler effect Thomas precession Ladder paradox Twin paradox Terrell rotation | | Spacetime | Light cone World line Minkowski diagram Biquaternions Minkowski space | |
| Generalrelativity | | | | --- | | Background | Introduction Mathematical formulation | | Fundamentalconcepts | Equivalence principle Riemannian geometry Penrose diagram Geodesics Mach's principle | | Formulation | ADM formalism BSSN formalism Einstein field equations Linearized gravity Post-Newtonian formalism Raychaudhuri equation Hamilton–Jacobi–Einstein equation Ernst equation | | Phenomena | Black hole Event horizon Singularity Two-body problem Gravitational waves: astronomy detectors (LIGO and collaboration Virgo LISA Pathfinder GEO) Hulse–Taylor binary Other tests: precession of Mercury lensing (together with Einstein cross and Einstein rings) redshift Shapiro delay frame-dragging / geodetic effect (Lense–Thirring precession) pulsar timing arrays | | Advancedtheories | Brans–Dicke theory Kaluza–Klein Quantum gravity | | Solutions | Cosmological: Friedmann–Lemaître–Robertson–Walker (Friedmann equations) Lemaître–Tolman Kasner BKL singularity Gödel Milne Spherical: Schwarzschild (interior Tolman–Oppenheimer–Volkoff equation) Reissner–Nordström Axisymmetric: Kerr (Kerr–Newman) Weyl−Lewis−Papapetrou Taub–NUT van Stockum dust discs Others: pp-wave Ozsváth–Schücking Alcubierre Ellis In computational physics: Numerical relativity | |
| Scientists | Poincaré Lorentz Einstein Hilbert Schwarzschild de Sitter Weyl Eddington Friedmann Lemaître Milne Robertson Chandrasekhar Zwicky Wheeler Choquet-Bruhat Kerr Zel'dovich Novikov Ehlers Geroch Penrose Hawking Taylor Hulse Bondi Misner Yau Thorne Weiss others |
| |
| Authority control databases |
| International | |
| National | United States France BnF data Czech Republic Israel |
| Other | Yale LUX |
Retrieved from "
Categories:
Maxwell's equations
Electromagnetism
Eponymous equations of physics
Functions of space and time
James Clerk Maxwell
Partial differential equations
Scientific laws
Hidden categories:
CS1 errors: ISBN date
Articles with short description
Short description is different from Wikidata
Wikipedia articles needing clarification from May 2022
Articles with French-language sources (fr)
Webarchive template wayback links
Commons category link from Wikidata
Maxwell's equations
Add topic |
189255 | https://math.stackexchange.com/questions/2497977/a-basic-question-about-the-fixed-points-of-fx-x2 | real analysis - A basic question about the fixed points of $f(x) = x^2$ - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
A basic question about the fixed points of f(x)=x 2 f(x)=x 2
Ask Question
Asked 7 years, 11 months ago
Modified7 years, 11 months ago
Viewed 354 times
This question shows research effort; it is useful and clear
1
Save this question.
Show activity on this post.
I am given the following exercise:
The function f(x)=x 2 f(x)=x 2 has two obvious fixed points: p 0=0 p 0=0 and p 1=1 p 1=1. Show that there is a 0<δ<1 0<δ<1 such that |f(x)−p 0|<|x−p 0||f(x)−p 0|<|x−p 0|, whenever |x−p o|<δ,x≠p 0|x−p o|<δ,x≠p 0. Conclude that f n(x)→p 0 f n(x)→p 0 whenever |x−p 0|<δ,x≠p 0|x−p 0|<δ,x≠p 0.
In contrast, find a δ>0 δ>0 such that if |x−p 1|<δ,x≠p 1|x−p 1|<δ,x≠p 1, then |f(x)−p 1|>|x−p 1||f(x)−p 1|>|x−p 1|. That means that p 1 p 1 is a repelling fixed point for f f; orbits that start out near 1 are pushed away from 1. In fact, given any x≠1 x≠1, we have f n(x)↛1 f n(x)↛1.
Question:
I'm a bit confused by this exercise because it seems to me that 'finding' the δ′s δ′s is in fact quite trivial. In the first case every 0<δ<1 0<δ<1 will do and in the second case any δ>0 δ>0 will do, right? Is this exercise just about introducing different kind of fixed points, or am I missing something?
real-analysis
convergence-divergence
fixed-point-theorems
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Oct 31, 2017 at 10:44
titusAdamtitusAdam
asked Oct 31, 2017 at 10:37
titusAdamtitusAdam
2,987 2 2 gold badges 18 18 silver badges 47 47 bronze badges
1
1 In the second case you'll want 1≥δ>0 1≥δ>0 since if x=0 x=0 then |f(x)−p 1|=|x−p 1||f(x)−p 1|=|x−p 1|, since 0 0 is a fixed point.snulty –snulty 2017-10-31 10:42:40 +00:00 Commented Oct 31, 2017 at 10:42
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
1
Save this answer.
Show activity on this post.
You're missing nothing (except for the constraint that @snulty notes). The author is showing that two fixed points of a particularly simple function have certain properties, and is hoping to generalize these later (I expect).
It's worth noting that if you add some perturbation like −10 x 4−10 x 4 to your function --- a perturbation that's "small" near p 0=x p 0=x, then the same claim (about p 0 p 0) holds, but finding the exact delta becomes considerably harder.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Oct 31, 2017 at 10:43
John HughesJohn Hughes
102k 4 4 gold badges 88 88 silver badges 160 160 bronze badges
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
real-analysis
convergence-divergence
fixed-point-theorems
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
5How to prove that the following iteration process converges?
0Fixed points and periodic orbits of F(x)=x 2−1.1 F(x)=x 2−1.1
1Show that 1 1 is a repelling fixed point for f(x)=x 2 f(x)=x 2.
2Stability of a fixed point of a discrete dynamical system
7Brouwer's fixed point theorem and continuous functional dependence on the fixed point.
0Convergence of fixed point iteration
7Can at least one fixed point of a correspondence be approximated by fixed points of nearby continuous functions?
Hot Network Questions
What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left?
Xubuntu 24.04 - Libreoffice
Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term?
How different is Roman Latin?
Why multiply energies when calculating the formation energy of butadiene's π-electron system?
Overfilled my oil
Is encrypting the login keyring necessary if you have full disk encryption?
What were "milk bars" in 1920s Japan?
Bypassing C64's PETSCII to screen code mapping
How to start explorer with C: drive selected and shown in folder list?
Why does LaTeX convert inline Python code (range(N-2)) into -NoValue-?
Numbers Interpreted in Smallest Valid Base
Where is the first repetition in the cumulative hierarchy up to elementary equivalence?
Origin of Australian slang exclamation "struth" meaning greatly surprised
Exchange a file in a zip file quickly
How to convert this extremely large group in GAP into a permutation group.
ConTeXt: Unnecessary space in \setupheadertext
What is the meaning and import of this highlighted phrase in Selichos?
Is existence always locational?
Is direct sum of finite spectra cancellative?
Who is the target audience of Netanyahu's speech at the United Nations?
Drawing the structure of a matrix
Making sense of perturbation theory in many-body physics
The rule of necessitation seems utterly unreasonable
more hot questions
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
189256 | https://askfilo.com/user-question-answers-smart-solutions/x-3-y-3-x-y-x-y-3132333036313330 | x ^ { 3 } + y ^ { 3 } + x y ( x + y ) | Filo
World's only instant tutoring platform
Instant TutoringPrivate Courses
Tutors
Explore TutorsBecome Tutor
Login
StudentTutor
CBSE
Smart Solutions
x ^ { 3 } + y ^ { 3 } + x y ( x + y )
Question
Question asked by Filo student
x 3+y 3+x y(x+y)
Views: 5,623 students
Updated on: Sep 3, 2024
Not the question you're searching for?
Ask your question
Ask your question
Or
Upload the image of your question
Get Solution
Text solutionVerified
Concepts:
Algebra, Polynomial factorization
Explanation:
The given expression is x 3+y 3+x y(x+y). To simplify this expression, we can use the identity for the sum of cubes and factorization techniques.
Step by Step Solution:
Step 1
Notice that x 3+y 3 can be factored using the sum of cubes formula: x 3+y 3=(x+y)(x 2−x y+y 2).
Step 2
Rewrite the given expression using the sum of cubes formula: x 3+y 3+x y(x+y)=(x+y)(x 2−x y+y 2)+x y(x+y).
Step 3
Factor out (x+y) from the expression: (x+y)(x 2−x y+y 2+x y)=(x+y)(x 2+y 2). Hence, the simplified form of the given expression is (x+y)(x 2+y 2).
Final Answer:
(x+y)(x 2+y 2)
Ask your next question
Or
Upload the image of your question
Get Solution
Get instant study help from an expert tutor 24/7 Download Filo
Found 8 tutors discussing this question
Charlotte Discussed
x 3+y 3+x y(x+y)
12 mins ago
Discuss this question LIVE
12 mins ago
One destination to cover all your homework and assignment needs
Learn Practice Revision Succeed
Instant 1:1 help, 24x7
60, 000+ Expert tutors
Textbook solutions
Big idea maths, McGraw-Hill Education etc
Essay review
Get expert feedback on your essay
Schedule classes
High dosage tutoring from Dedicated 3 experts
Download AppExplore now
Trusted by 4 million+ students
Students who ask this question also asked
Question 1
Views: 5,881
Write an application to the Principal of your school requesting remission (waiver) of fees as you belong to the economically weaker section of society.
Topic: Smart Solutions
View solution
Question 2
Views: 5,281
Half past 7 and quarter to 8 ?
Topic: Smart Solutions
View solution
Question 3
Views: 5,148
[3 3 1 of 2 1+2+2×{2−(2−5 1)}
Topic: Smart Solutions
View solution
Question 4
Views: 5,982
Represent 3 and 5 on the number line using geometrical construction. (Use compass and scale.) Write the decimal expansions of: (a) 8 1 (b) 7 22 (c) 3 1 And state whether the decimals are terminating or non-terminating recurring.
Topic: Smart Solutions
View solution
View more
Video Player is loading.
Play Video
Play Skip Backward
Mute
Current Time 0:00
/
Duration-:-
Loaded: 0%
Stream Type LIVE
Seek to live, currently behind live LIVE
Remaining Time-0:00
1x
Playback Rate
2.5x
2x
1.5x
1x, selected
0.75x
Chapters
Chapters
Descriptions
descriptions off, selected
Captions
captions settings, opens captions settings dialog
captions off, selected
Audio Track
Picture-in-Picture Fullscreen
This is a modal window.
Beginning of dialog window. Escape will cancel and close the window.
Text Color Opacity Text Background Color Opacity Caption Area Background Color Opacity
Font Size Text Edge Style Font Family
Reset restore all settings to the default values Done
Close Modal Dialog
End of dialog window.
Stuck on the question or explanation?
Connect with our 216 tutors online and get step by step solution of this question.
Talk to a tutor now
400 students are taking LIVE classes
Question Text x 3+y 3+x y(x+y)
Updated On Sep 3, 2024
Topic All topics
Subject Smart Solutions
Class Class 8
Answer Type Text solution:1
Are you ready to take control of your learning?
Download Filo and start learning with your favorite tutors right away!
Questions from top courses
Algebra 1
Algebra 2
Geometry
Pre Calculus
Statistics
Physics
Chemistry
Advanced Math
AP Physics 2
Biology
Smart Solutions
College / University
Explore Tutors by Cities
Tutors in New York City
Tutors in Chicago
Tutors in San Diego
Tutors in Los Angeles
Tutors in Houston
Tutors in Dallas
Tutors in San Francisco
Tutors in Philadelphia
Tutors in San Antonio
Tutors in Oklahoma City
Tutors in Phoenix
Tutors in Austin
Tutors in San Jose
Tutors in Boston
Tutors in Seattle
Tutors in Washington, D.C.
World's only instant tutoring platform
Connect to a tutor in 60 seconds, 24X7
27001
Filo is
ISO 27001:2022 Certified
Become a Tutor
Instant Tutoring
Scheduled Private Courses
Explore Private Tutors
Filo Instant Ask Button
Instant tutoring API
High Dosage Tutoring
About Us
Careers
Contact Us
Blog
Knowledge
Privacy Policy
Terms and Conditions
© Copyright Filo EdTech INC. 2025
This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply. |
189257 | https://www.cmu.edu/biolphys/smsl/teaching/IntermedOptics/IntOptics_data/lab%20manuals/Lab3%20-%20Polarization.pdf | 33-353 – Intermediate Optics Lab 3 – Fall 2010 POLARIZATION (1 Lab Period) Objectives: To confirm Malus’s Law: I(θ) = I(0) cos2 θ To acquire skill at analyzing polarization states of light References: Hecht, Chapter 8, especially Sections 8.1, 8.2, 8.7 and 8.8 Overview: These experiments will provide some experience with the phenomenon of polarization. The first experiment, on Malus’s Law, is clearly specified in terms of the measurements you should make and how you should plot and analyze the data. The second experiment, on polarization states, has a specific objective, but gives you considerable flexibility in choosing your approach. Equipment: Optical rail and riders, or optical table and post-holders, polarized laser, light bulb, pair of linear-polarizing filters (mounted in calibrated rotation holders), λ/4 retardation plate, laser power meter, various unknown polarization filters. I. Malus’s Law Preliminary setup: Place one of the mounted linear-polarizing filters, which we will refer to as the polarizer, in front of the laser, and set the polarization angle to 0° (if you have the Edmund polarizing filters) or 90° (if you have the Leybold polarizing filters). The easy-pass direction of the polarizer is now vertical. Turn on the laser power meter, but DO NOT yet place it in the path of the laser beam. Check the reading on the meter, and adjust the zero, if necessary. Set up the laser power meter to receive the transmitted laser beam. Rotate the laser to maximize the reading on the power meter (plane of polarization of the laser is parallel to the polarizer). Insert the second linear-polarizing filter, which we will call the analyzer, between the polarizer and the power meter. Measurements and analysis: • Keep the orientation of the polarizer fixed and vary the angle, θ, of the analyzer, with respect to the polarizer, over the range from -90° to +90°. • Measure the intensity of the transmitted light over this range of angles, and estimate the uncertainty in the measured intensity for each angle. • Plot the measured intensity, I(θ). Include error bars on the measured intensity. • For each value of θ, calculate the predicted intensity according to Malus’s Law: I(θ) = I(0) cos2 θ. • Plot the predicted intensities on the same graph as your measurements. How well does the predicted function fit your data? 33-353 – Intermediate Optics Lab 3 – Fall 2010 II. Polarization States • In this experiment you will be presented with three “unknown” filters, which will affect the polarization state of the laser in different ways. Note that some filters have two different effects, depending on which side of the filter faces the light source. • Using the materials and equipment provided, make appropriate observations and measurements to enable you to determine the polarization state produced by each of the filters. Be sure to make observations and measurements for both sides of each filter. • For each side of each filter, write a convincing argument, based on your measurements, to justify your conclusion about the resulting polarization state. If the filter exhibits a difference in effect, depending on which side faces the light source, how do you account for the observed difference? |
189258 | https://askfilo.com/user-question-answers-smart-solutions/if-a-b-c-12-and-a2-b2-c2-66-find-value-of-ab-bc-ca-3135353832323835 | If a+b+c = 12 and a² +b² + c² =66 , find value of ab + bc + ca
Not the question you're searching for?
Ask your question
Ask your question
Or
Upload the image of your question
Text solutionVerified
Concepts:
Algebra, Equations, Symmetric sums
Explanation:
To find the value of ab + bc + ca, we can use the identities involving the sums of the variables. We know that: 1. (a + b + c)² = a² + b² + c² + 2(ab + ac + bc). We can rearrange this to find ab + ac + bc. 2. We substitute the known values into the equation.
Step by Step Solution:
Step 1
Start with the equation (a + b + c)² = a² + b² + c² + 2(ab + ac + bc).
Step 2
Substitute the known values: (12)² = 66 + 2(ab + ac + bc).
Step 3
Calculate (12)² which is 144. So, we have 144 = 66 + 2(ab + ac + bc).
Step 4
Rearranging gives us 2(ab + ac + bc) = 144 - 66 = 78, thus ab + ac + bc = 39.
Final Answer:
39
Ask your next question
Or
Upload the image of your question
Get instant study help from an expert tutor 24/7
Download Filo
Found 3 tutors discussing this question
Lucas Discussed
If a+b+c = 12 and a² +b² + c² =66 , find value of ab + bc + ca
14 mins ago
Discuss this question LIVE
14 mins ago
One destination to cover all your homework and assignment needs
Learn Practice Revision Succeed
Instant 1:1 help, 24x7
60, 000+ Expert tutors
Textbook solutions
Big idea maths, McGraw-Hill Education etc
Essay review
Get expert feedback on your essay
Schedule classes
High dosage tutoring from Dedicated 3 experts
Trusted by 4 million+ students
Students who ask this question also asked
Question 1
Views: 5,211
Who were considered desirable according to the Nazis?
Topic:
Smart Solutions
View solution
Question 2
Views: 5,477
The altitude of a right-angled triangle is 5 cm less than its base. If its area is 150 cm², find the measures of the base and the altitude.
Topic:
Smart Solutions
View solution
Question 3
Views: 5,876
Reproduction in human being . Explain?
Topic:
Smart Solutions
View solution
Question 4
Views: 5,691
7.50 m sq
from #fa
feremua; (ii) za
(1)
farm #affara
(iii) Tr #20 sua fat farm
-(i) 125 m (ii) 125 m (iii) -150 m/s
Topic:
Smart Solutions
View solution
View more
Video Player is loading.
Current Time
/
Duration
Loaded: 0%
Stream Type LIVE
Remaining Time
1x
2.5x
2x
1.5x
1x
, selected
0.75x
Chapters
descriptions off
, selected
captions settings
, opens captions settings dialog
captions off
, selected
This is a modal window.
Beginning of dialog window. Escape will cancel and close the window.
End of dialog window.
Stuck on the question or explanation?
Connect with our 418 Smart Solutions tutors online and get step by step solution of this question.
363 students are taking LIVE classes
| | |
--- |
| Question Text | If a+b+c = 12 and a² +b² + c² =66 , find value of ab + bc + ca |
| Updated On | Dec 14, 2024 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 9 |
| Answer Type | Text solution:1 |
Are you ready to take control of your learning?
Download Filo and start learning with your favorite tutors right away! |
189259 | https://www.cfm.brown.edu/people/dobrush/cs52/Mathematica/Part1/affine.html | es
Recall that 𝔽 denotes one of the following fields of numbers: ℤ, integers, ℚ, rational numbers, ℝ, real numbers; hence we exclude from our consideration ℂ, complex numbers in this section. It is caused by applications of affine transformations in geometry and computer graphics that utilize only real numbers. We denote by 𝔽m×n or 𝔽m,n the vector space of m-by-n matrices with entries from field 𝔽.
According to Wikipedia, the term linear function can refer to two distinct concepts, based on the context:
In Calculus, a linear function is a polynomial function of degree zero or one; in other words, a function of the form f(x) = m x + b for some constants m and b ∈ ℝ.
In Linear Algebra, a linear function is a linear mapping, or linear transformation: f(λx + y) = λf(x) + f(y). for any scalar λ and any two vectors x and y.
A matrix A of size m x n defines a linear map upon multiplication from left:
Fn×1⟶Fm×1,Fn×1∋x⟶Ax∈Fm×1,Fn×1⟶Fm×1,Fn×1∋x⟶Ax∈Fm×1,
also denoted by A : 𝔽m×1 ⇾ 𝔽n×1. The same matrix defines a linear transformation between row vectors upon multiplication from right,
F1×n⟵F1×m,F1×n∋vA⟵v∈F1×m.F1×n⟵F1×m,F1×n∋vA⟵v∈F1×m.
Such a map has the basic property A 0 = 0 for column vectors and 0 A = 0 for row vectors.
$Post :=
If[MatrixQ[#1],
MatrixForm[#1], #1] & ( outputs matricies in MatrixForm)
Remove[ "Global`"] // Quiet ( remove all variables )
Affine Transformations
An affine transformation or affinity (in 1748, Leonhard Euler introduced the term affine, which stems from the Latin, affinis, "connected with") is a geometric transformation that preserves the parallelism of lines and the ratio of distances between points.
Affine transformation is closely related to projective transformation---this technique is widely used in computer graphics, image processing, machine learning, and neural networks to perform geometric transformations in a
simple way using transformation matrices.
Although there are several open computer vision libraries for affine transformations such as openGL and
openCV, we prefer to use Mathematica and its build-in commands: AffineTransform and TransformationMatrix.
The following definition does not satisfy mathematical scrutiny because it does not say explicitly what are domain and codomain of this transformation---formal definition will be given later. However, it gives us an idea where affine transformations come from.
(Naïve Representation)
Any map f : 𝔽n×1 ↦ 𝔽m×1
of the form
Fn×1∋x⟶f(x)=Ax+bFn×1∋x⟶f(x)=Ax+b(1)
for some column vector b ∈ 𝔽m×1, is called an affine map. Similarly, a mapping between row vectors
F1×m∋v⟶f(v)=vA+wF1×m∋v⟶f(v)=vA+w(2)
for some row vector w ∈ 𝔽1×n, is also called an affine transformation.
Both formulae (1)(1) and (2)(2) are just short cuts of the general transformation of the form
{y1=a1,2x1+a1,2x2+⋯+a1,nxn+b1,y2=a2,2x1+a2,2x2+⋯+a2,nxn+b2, ⋮⋮⋮ym=am,2x1+am,2x2+⋯+am,nxn+bm.⎧⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪⎩y1=a1,2x1+a1,2x2+⋯+a1,nxn+b1,y2=a2,2x1+a2,2x2+⋯+a2,nxn+b2, ⋮⋮⋮ym=am,2x1+am,2x2+⋯+am,nxn+bm.
Affine map is a geometric transformation that preserves lines and parallelism, but not necessarily Euclidean distances and angles.
Since f(0) = b, such a map can be
be linear only when b = 0 in Eq.(1)(1) or w = 0 in Eq.(2)(2).
Basically, there are four affine transformations or their compositions:
Translate moves a set of points a fixed distance in each coordinate.
Scale scales a set of points up or down in each coordinate.
Rotate rotates a set of points about the origin,
Shear offsets a set of points a distance proportional to their x and y coordinates.
Note that only shear and non-uniform scale change the shape determined by a set of points. A subclass of affine transformations that locally preserves angles, but not necessarily lengths is called the set of conformal maps.
Example 1:
Let us consider an affine transformation
[xy]↦[xy]+[−1−1][xy]↦[xy]+[−1−1]
It is a “shear” followed by a translation. The
effect of this shear on the square (𝑎, b, c, d) is shown in the following figure. The
image of this square is the parallelogram.
First, we rewrite the transformation in coordinates explicitly:
x↦x+y−1,y↦2y+1.x↦x+y−1,y↦2y+1.
Then vertical line x = −1 is mapped into the line
x↦y−2,y↦2y+1.x↦y−2,y↦2y+1.
The images of vertices become
(−1,−1)↦(−3,−1),(−1,1)↦(−1,3),(1,1)↦(1,3),(1,−1)↦(−1,−1).(−1,−1)↦(−3,−1),(−1,1)↦(−1,3),(1,1)↦(1,3),(1,−1)↦(−1,−1).
Graphics[{Opacity[.35], Blue, Rectangle[{-1, -1}, {1, 1}], Red,
Rectangle[{-3, -1}, {1, 3}]}, Axes -> True]
| | | |
---
| | | |
Clear[A, x, y, xy]
A = {{1, 1}, {0, 2}};
b = {1, -1};
xy = {x, y}
A . xy
{x + y, 2 y}
Solve[{(A . xy) == 1, (A . xy) == -1}, {x, y}]
(x→32 y→−12)(x→32 y→−12)
t[{x, y}]
{1 + x + y, -1 + 2 y}
TransformationMatrix@AffineTransform[{A, b}] // MatrixForm
(11102−1001)⎛⎜⎝11102−1001⎞⎟⎠
■
End of Example 1
Theorem 1:
If an affine transformation has an inverse, then it is also an affine
transformation.
Let q = A x + b be an affine transformation written for column vectors. It has an inverse only when A is nonsingular matrix, so det(A) ≠ 0. Then A x = q − b. Application of inverse matrix (which exists for nonsingular matrices) to the latter, we obtain x = A−1q − A−1b. `Hence, x = B q + v, where B = A−1 and v = − A−1b.
Example 2:
Matrix A=A= maps all points to the x-axis, so
it is a projection on this axis. The area of any closed
region will become zero. We have det(A) = 0, which verifies that any closed
region’s area will be scaled by zero.
In general, for any given closed region, the area under an affine transformation A x + b is scaled by det(A). This result is valid for any linear mapping y = A x.
■
End of Example 2
Corollary 1:
A composition of affine transformations is an affine transformation.
Let f(x) = A x + a and g(x) = B x + b be affine transformations. Then (g∘f)(x) = g(f)(x)) = B(A x + a) + b = (B A)x + (B a + b)
Example 3:
■
End of Example 3
The basic properties of affine transformations are summarize in the following statement.
Theorem 2:
Let f(x) = A x + b be an affine transformation. Then f
maps a line to a line,
maps a line segment to a line segment,
preserves the property of parallelism among lines and line segments
maps an n-gon to an n-gon,
maps a parallelogram to a parallelogram,
preserves the ratio of lengths of two parallel segments, and
preserves the ratio of areas of two figures.
Let L be aline and let L: p + tm, t ∈ ℝ, be an equation of L in vector form. Then for every t ∈ ℝ,
f(p+tm)=A(p+tm)+b=p1+tm1,f(p+tm)=A(p+tm)+b=p1+tm1,
where p₁ = A p + b and m₁ = A m. Hence, f(L) = L₁, where L₁ : p₁ + tm₁, t ∈ ℝ, is again a line.
The proof is the same as that for (1), with t restricted to [0, 1].
Suppose that L: p + tm and L₁ : p₁ + tm₁, t ∈ ℝ, are parallel lines. Then m₁ = km for some k ∈ ℝ. Therefore,
f(p+tm)=A(p+tm)+b=(Ap+b)+t(Am)=q+tn,f(p1+tm1)=f(p1+tkm)=A(p1+tkm)+b=(Ap1+b)+t(Akm)=p2+tm2.f(p+tm)=A(p+tm)+b=(Ap+b)+t(Am)=q+tn,f(p1+tm1)=f(p1+tkm)=A(p1+tkm)+b=(Ap1+b)+t(Akm)=p2+tm2.
That is, L and L₁ are mapped to lines that are parallel.
It is clear that for two line segments or a line and a line segment the proof is absolutely
analogous.
4. We prove this by strong induction on n. For the base case, when n = 3, consider a triangle T.
Then T and its interior can be represented in vector form as T :
u + sv + tw, where s, t ∈ [0, 1],
s + t ≤ 1, and the vectors v and w are not collinear. Then
f(T)=F(u+sv+tw)=A(u+sv+tw)+b=(Au+b)+s(Av)+t(Aw)=u1+sv1+tw1,f(T)=F(u+sv+tw)=A(u+sv+tw)+b=(Au+b)+s(Av)+t(Aw)=u1+sv1+tw1,
where s, t ∈ [0, 1], s + t ≤ 1. By part3, v₁ = A v and w₁ = A w are not parallel. Thus, T is mapped
to a triangle T₁, which completes the proof of the base case.
Now suppose that f maps each n-gon to an n-gon for all n, 3 ≤ n ≤ k, and let P be a
polygon with k + 1 sides. We know that every polygon with
at least 4 sides has a diagonal contained completely in its interior. Let
¯AB¯¯¯¯¯¯¯¯AB be such a diagonal
in P. This diagonal divides P into two polygons, P₁ and P₂ containing t and k + 1 − t sides,
respectively, for some t, 3 ≤ t ≤ k. By the inductive hypothesis, f(P₁) and f(P₂) will be
t-sided and (k + 3 − t)-sided polygons, respectively. Since each of these polygons will have
the segment from f(A) to f(B) as a diagonal, the union of P₁ and P₂ will form a polygon with k + 1 sides, which concludes the proof.
5. The proof that a parallelogram is mapped to a parallelogram is analogous to the proof
that triangles get mapped to triangles in part (4), by simply dropping the condition that s + t ≤ 1.
6. Consider parallel line segments, S₁ and S₂, given in vector form as Si : pi + t ui, t ∈ [0, 1].
Because they are parallel, u₂ = k u₁ for some k ∈ ℝ. As |ui| is the length of Si , the ratio of
lengths of S₂ and S₁ is |k|. From parts (1) and (2), Si is mapped into a segment of length |A ui|. Since A u₂ = A(ku₁) = k(A&thinsppu), |A u₂| = |k| |A u₁|, which shows that the ratio of lengths
of f(S₂) and f(S₁) is also |k|.
7. We postpone discussion of the proof of this property until the end of this section
Example 4: ■
End of Example 4
Affine Spaces
Recall that the Cartesian product of two sets A and B, denoted
A ×B, is the set of all ordered pairs (a, b), where a is in A and b is in B. Since our main object of interest is ℝ, the set of real numbers, its direct product ℝ² = ℝ × ℝ inherits a linear structure from field ℝ. This space provides the main historical example of the Cartesian plane in analytic geometry.
The set of all such pairs (i.e., the Cartesian product ℝ × ℝ, denoted by ℝ²) is assigned to the set of all points in the plane as well as to the set of all free vectors. All these three sets (the set of points on the plane, the set of 2-tuples, and the set of free vectors in ℝ²) are in one-to-one and onto correspondence between each other. Therefore, they traditionally are denoted by ℝ², and content specifies which of these sets is in use.
One can similarly define the Cartesian product of n sets, also known as an n-fold Cartesian product, which can be represented by an n-dimensional array, where each element is an n-tuple.
In Euclidean space, points and vectors are usually identified with n-tuples.
Theorem 3:
A subset of 𝔽n is an affine set if and only if it is the solution set to
a system of linear equations over 𝔽.
If S ⊆ 𝔽n is an affine set, it has the form S = b + W, where W is a subspace of 𝔽n and b is a vector in 𝔽n. Given such a set, let W₁ be a direct complement to W in 𝔽n. Note that we can assume b is in W₁.
Let A be the standard matrix representation of the projection onto W₁ along W. For w₁ in W₁ and w in W, we have
A(w1+w)=w1.A(w1+w)=w1.
In particular, NullSpace(A) = W and A b = b. Lemma 2 (the general solution of A b = b is the sum of a particular solution and the general solution of the homogeneous equation) in section thus ensures that S is the
solution set to A b = b. This completes the proof in one direction.
The same Lemma 2 gives us the proof in the other direction.
Example 5:
Let us consider the affine transformation S : ℝ³ ↦ ℝ4 given by S(x) = A x + b, where
A=[−23537−15277−9216],b=[3−6−1215].A=⎡⎢
⎢
⎢⎣−23537−15277−9216⎤⎥
⎥
⎥⎦,b=⎡⎢
⎢
⎢⎣3−6−1215⎤⎥
⎥
⎥⎦.
First, we find solution A u = −b, so S(u) = 0, withe aid of Mathematica.
A = {{-2, 3, 5}, { 3, 7, -1}, {5, 27, 7}, {-9, 2, 16}};
b = {{3}, {-6}, {-12}, {15}};
RowReduce[Join[A, b, 2]] // MatrixForm
(10−3823−3923011323−32300000000)⎛⎜
⎜
⎜
⎜
⎜⎝10−3823−3923011323−32300000000⎞⎟
⎟
⎟
⎟
⎟⎠
From solution given above, we can reconstruct the solution as a line given parametrically
L={(x,y,z) : x=3823t−3923,y=−1323t−323,z=t,∀t∈R}.L={(x,y,z) : x=3823t−3923,y=−1323t−323,z=t,∀t∈R}.
We can decompose L into a sum of two components: the first is the line L₀, which passes through origin; the second is a translation by a particular vector vp. To find the particular vector vp, notice that all we have to do is set t = 0 in the parametric definition of L given above, which yields
vp=[−3923, −323, 0]vp=[−3923, −323, 0] Once we know vp, the line L₀ is simply the remaining portion of the solution
L0={(x,y,z) : x=3823t,y=−1323t,z=t,∀t∈R}.L0={(x,y,z) : x=3823t,y=−1323t,z=t,∀t∈R}.
Clearly, L₀ is a line through the origin, and is thus a subspace of ℝ³. The line L can be realized as a translate of the line L₀ by the particular solution xp. Now let us plot these two lines along with the particular solution
xp.
lineL = ParametricPlot3D[{38/23t - 39/23, 13/23t - 3/23, t}, {t, -3,
3}, PlotStyle -> {Thickness[0.007], Blue}];
lineL0 =
ParametricPlot3D[{38/23t, 13/23t, t}, {t, -3, 3},
PlotStyle -> {Thickness[0.007], Red}];
pts = Graphics3D[{Black, Sphere[{0, 0, 0}, 0.13], Black,
Sphere[{-39/23, -3/23, 0}, 0.13]}];
arrow = Graphics3D[{Arrowheads[0.05], Thickness[0.007], Purple,
Arrow[{{0, 0, 0}, {-39/23, -3/23, 0}}]}];
txt = Graphics3D[{Text[Style["O", Black, 20], {0, 0, 0.6}],
Text[Style["P", Black, 20], {-39/23, -3/23, 0.6}]}];
Show[lineL, lineL0, txt, arrow, pts]
■
End of Example 5
In particular, an Euclidean plane contains points P(x, y) and vectors v(x, y) simultaneously because they both have the same coordinates.
In computer graphics, the main problem is to render or display a three-dimensional objects (or models) by projecting or mapping them into two-dimensional images.
Then the two-dimensional data must be converted into a form that the computer
can display (rasterization) and then be displayed. This requires a viewpoint or direction of projection and a viewing or projection plane. Fortunately, a monitor is just a two-dimensional array of finite number of pixels, short for picture elements.
This practical situation with rastering data in computer graphics shows that we
need to distinct points from vectors. It is important because points and vectors have some mutually exclusive properties. A point has location but no extent while a vector in ℝn has both direction and magnitude (norm) but
its location is independent.
In order to define an affine plane (which is a two-dimensional geometric object), we need to separate points from vectors (that are also called lines) according to their gender. Namely, we identify points with extra integer "1," but we mark vectors with "0." Hence, we write points as P(x, y, 1) and vectors as v(x, y, 0). However, you can move vectors to the point plane and attach them to points. This allows us to move a point P into another position Q along vector v. In coordinates, it can be written as
P(x,y,1)+v(a,b,0)=Q(x+a,y+b,1)⟹v==Q−P=¯PQ.P(x,y,1)+v(a,b,0)=Q(x+a,y+b,1)⟹v==Q−P=¯¯¯¯¯¯¯¯PQ.
From this prospective, we are not allowed to add points, but we can add points and vectors, as well as vectors and vectors. For any two points P and Q from the inhabited set A, there exists a unique vector v ∈ V such that Q = P + v; so we can identify this vector as PQ or Q − P. In general, an affine space consists of an inhabited set of points A together with a vector space V and subtraction operation of two points, producing a unique vector.
In order to visualize an affine plane, we
consider the vector space ℝ³. Inside ℝ³, we can choose two planes, as in the picture below. We'll call the yellow one the vector space V ≌ ℝ² and the blue one as the point plane A. The plane V passes through the origin since it is a vector space, but the blue plane A does not. However, the inhabited set A looks almost exactly the same as V, having the exact same, flat geometry, and in fact A and V are simply translates of one another. This plane A is a classical example of an affine space. You will learn in Part 3 that A is a coset of V.
| | | |
---
| | | |
The left picture shows an attempt to introduce vector structure in the inhabited set A. Let T : A ↦ V be a translation of the point set to the vector space. You may try to define addition of two points as
P(+)Q=T−1(T(P)+T(Q)).P(+)Q=T−1(T(P)+T(Q)).
However, the resulting vector (P(+)Q) does not belong to the inhabited set A. It is impossible to introduce a vector structure into an affine space---it is not a vector space.
Now we are ready to make general definition of an affince space,
according to Wikipedia.
An affine space is a geometric structure that generalizes some of the properties of Euclidean spaces in such a way that these are independent of the concepts of distance and measure of angles, keeping only the properties related to parallelism and ratio of lengths for parallel line segments. Affine space is the setting for affine geometry.
In context of linear algebra, an affine space is a set of points A equipped with a set of transformations (that is bijective mappings); the translations, which forms a vector space (over a given field, commonly the set of real numbers), and such that for any given ordered pair of points there is a unique translation sending the first point to the second one; the composition of two translations is their sum in the vector space of the translations.
An affine space with vector space V is a nonempty set A of
points and a vector valued map d : A × A ↦ V called a difference function, such
that for all P, Q, R ∈ A
d(P, Q) + d(Q, R) = d(P, R),
the restricted map d₁ = d{P}×A : {P} × A ↦ V defined as mapping (P, Q) ↦ d(P, Q) is a bijection.
The first condition (i) is just the usual “parallelogram property” of the addition of vectors. From the second condition, it follows that for every pair of points P and Q from A, there exits a unique vector v ∈ V such that P + v = Q, which is naturally denoted by ¯PQor→PQ.¯¯¯¯¯¯¯¯PQor→PQ. From properties of a vector space, we derive
P+0=P,(P+v)+u=P+(v+u)∀v,u∈V.P+0=P,(P+v)+u=P+(v+u)∀v,u∈V.
Example 6:
Any finite dimensional vector space V has an affine space structure specified by choosing the inhabited set A = V and letting d be subtraction in the vector space V. We will refer
to the affine structure (V, V, d) on a vector space V as the canonical affine structure on V. In particular, the vector space ℝn can be viewed as the affine space (ℝn, ℝn, d), denoted by 𝔸n. The affine space 𝔸n is called the real affine space of dimension n.
■
End of Example 6
Lemma 1:
In an affine space (A, V, d) with difference function d we have
d(P, P) = 0 for all points P ∈ A,
d(P, Q) = −d(Q, P) for all points P, Q ∈ A.
In an affine space (A, V, d), for any three points P, Q, R ∈ A, and any real number λ addition of points and scalar multiplication is defined via
(P,Q)+(P,R)=d−1(d(P,Q)+d(P,R)),λ(P,Q)=d−1(λd((P,Q))).(P,Q)+(P,R)=d−1(d(P,Q)+d(P,R)),λ(P,Q)=d−1(λd((P,Q))).
This vector space is the
tangent space to A at point P, denoted TP(A). For v ∈ TP(A) ≌ V, we denote P + d−1(v) as P + v.
Example 7:
Let us consider the subset A of 𝔸³ consisting of all
points (x, y, z) satisfying the equation
x2+y2−z=0.x2+y2−z=0.
The set of points A is a paraboloid of revolution, with axis Oz. The surface A can be made into an official affine space by defining the action of addition of points and vectors (which is equivalent to the difference operation d) s : A × ℝ² → A
of ℝ² on A defined such that for every point (x, y, x² + y²) on A and any vector v = (v, u) ∈ ℝ²,
(x,y,x2+y2)+[vu]=(x+v,y+u,(x+v)2+(y+u)2).(x,y,x2+y2)+[vu]=(x+v,y+u,(x+v)2+(y+u)2).
■
End of Example 7
As the notion of parallel lines is one of the main properties that is independent of any metric, affine geometry is often considered as the study of parallel lines.
Affine Transformations
Let P₁ and P₂ be two arbitrary points in
inhabited set A of an affine space (A, V, d). An affinite combination of points P₁ and P₂ is
P=αP1+(1−α)P2∀α∈[0,1].P=αP1+(1−α)P2∀α∈[0,1].
In general, a linear combination of m points P₁, P₂, … , Pm is
∑iαiPi=P0+m∑i=1αi(Pi−P0),m∑i=1αi=1.∑iαiPi=P0+m∑i=1αi(Pi−P0),m∑i=1αi=1.
Example 8:
We start with two points ina two-dimensional affince space.
ar = Graphics[{Blue, Thick, Arrow[{{0, 0}, {1, 0.5}}]}];
ar2 = Graphics[{Black, Dashed, Thick, Arrow[{{1, 0.5}, {2, 1}}]}];
txt = Graphics[{Text[Style[Subscript[P, 1], Black, 20], {0, -0.15}],
Text[Style[Subscript[P, 2], Black, 20], {2.1, 0.9}]}];
dot = Graphics[{{Purple, Disk[{1, 0.5}, 0.02]}, {Purple,
Disk[{0, 0}, 0.02]}, {Purple, Disk[{2, 1}, 0.02]}}];
txt2 = Graphics[{Text[Style["+ t (", Black, 20], {1.3, 0.45}],
Text[Style["-", Black, 20], {1.6, 0.45}],
Text[Style[")", Black, 20], {1.81, 0.45}]}];
txt3 = Graphics[{Text[
Style[Subscript[P, 1], Black, 20], {1.1, 0.45}],
Text[Style[Subscript[P, 2], Black, 20], {1.46, 0.45}],
Text[Style[Subscript[P, 1], Black, 20], {1.74, 0.45}]}];
Show[ar, ar2, txt, dot, txt2, txt3]
Now we consider three points; the following figure shows a combination of these points:
P=α1P1+α2P2+α3P3,\alpja1+α2+α3=1.P=α1P1+α2P2+α3P3,\alpja1+α2+α3=1.
ar = Graphics[{Blue, Thick, Arrow[{{0, 0}, {1, 0.5}}]}];
ar2 = Graphics[{Black, Dashed, Thick, Arrow[{{1, 0.5}, {2, 1}}]}];
ar3 = Graphics[{Blue, Thick, Arrow[{{1, 0.5}, {1.7, 0}}]}];
line = Graphics[{Black, Dashed, Thick,
Line[{{2, 1}, {3, -0.7}, {0, 0}}]}];
dot = Graphics[{{Purple, Disk[{1, 0.5}, 0.02]}, {Purple,
Disk[{0, 0}, 0.02]}, {Purple, Disk[{2, 1}, 0.02]}}];
dot2 = Graphics[{{Purple, Disk[{3, -0.7}, 0.02]}, {Red,
Disk[{1.7, 0}, 0.03]}}];
■
End of Example 8
Formally, an affine transformation is a mapping from one affine space to
another (which may be, and in fact usually is, the same space) that preserves affine combinations.
The properties of affine transformations on points and vectors are summarized in the following
theorem.
Theorem 4:
Let P and Q be points and u and v be vectors in an affine space 𝔸 = (A,V, d). Let F : 𝔸 → 𝔹
be an affine transformation from 𝔸 to another affine space 𝔹. Then for all scalars α and β
F(αP + βQ) = α F(P) + β F(Q),
F(v) = F(P − Q) = F(P) − F(Q) for v = P − Q,
F(P + αv) = F(P) + αF(v),
F(u + v) = F(u) + F(v),
F(αv) = αF(v).
The first two properties are the definition of an affine transformation of a point and a vector.
Showing part (d) is straight forward if P and Q are points in 𝔸 such that u = P − Q and
v = Q − R and the head-to-tail axiom is applied several times.
F(u−v)=F((P−Q)+(Q−R))=F(P−R)=F(P)−F(R)=F(P)−F(Q)+F(Q)−F(R)=F(P−Q)+F(Q−R)=F(u)+F(v).
Example 9:
■
End of Example 9
Recall that a linear transformation T : ℝ² ⇾ ℝ² is uniquely determined by taking a line segment (or its endpoints) in the domain and map it into another line segment (or its endpoints) in the codomain. This is no longer the case for an affine map f : 𝔸² ↦ 𝔸². It turns out that an affine transformation f : 𝔸² ↦ 𝔸² is uniquely determined by taking a triangle (or three points) in the domain and mapping it into another triangle (or three points) in the codomain. To see how this works, let the triangle in the domain be defined as the interior of the three points
T1={(x1,y1), (x2,y2), (x3,y3)}.
Similarly, s[[pse these points are mapped into
T2={(z1,w1), (z2,w2), (z3,w3)}.
Then
f(T1)=T2⟺f(xj,yj)=(zj,wj),j=1,2,3.
Given that f is determined by formula f(x) = A x + b, with A and b given respectively by
A=[abcd],b=[αβ],
we get the following system of equations:
[abcd]⋅[xjyj]=[αβ],j=1,2,3.
This gives the new single matrix equation
[x1y1001000x1y101x2y2001000x2y201x3y3001000x3y301][abcdαβ]=[z1w1z2w2z3w3].
This matrix/vector equation can be solved for our six unknowns
{𝑎, b, c, d, α, β}, which determine the affine map uniquely. Therefore, a two-dimensional affine space has six degrees of freedom.
Theorem 5:
Given two ordered sets of
three non-collinear points each, there exists a unique affine transformation f mapping one set
onto the other.
We first show that the special (ordered) triple of vectors,
{0=,i=,j=,}
can be mapped by an appropriate affine transformation to an arbitrary (ordered) triple of vectors
{p=[p1p2],q=[q1q2],r=[r1r2]},
which corresponds to three non-collinear points. Let
A=[q1−p1r1−p1q2−p2r2−p2]andb=[p1p2].
One can immediately verify that
A0+b=p,Ai+b=q,Aj+b=r.
Note that the columns of A correspond to the vectors q − p and r − p. Since the points (p₁, p₂),
(q₁, q₂), and (r₁, r₂) are non-collinear, the vectors q − p and r − p are non-parallel vectors. Hence, the determinant of A is nonzero. Thus, A is invertible, and
f(x) = A x + b is an affine transformation by definition.
Let (p, q, r) and (p₁, q₁, r₁) be two ordered triples of position vectors representing two arbitrary
triples of non-collinear points. Using the result we have just proven, there exist affine transformations
f and g mapping the special triple {0, i, j} to {p, q, r} and to {p₁, q₁, r₁}, respectively. Then g ∘ f−1 is an affine transformation that maps {p, q, r} into {p₁, q₁, r₁}. The uniqueness of this transformation is left to you.
Example 10:
Let us consider two sets of points on the plane ℝ²:
T1={(−5,−3), (2,10), (3,−5)},T2={(−4,1), (−3,11), (1,9)}.
We want to find an affine transformation that maps points from T₁ into T₂. So we use Mathematica.
X1 ={-5, 2, 3};
Y1 = {-3, 10, -5};
X2 = {-4, -3, 1}; Y2 = {1, 11, 9};
R = {{X2}, {Y2}, {X2}, {Y2}, {X2},
{Y2}} ;
d = {{X1, Y1, 0, 0, 1, 0}, {0, 0, X1, Y1, 0,
1}, {X1, Y1, 0, 0, 1, 0}, {0, 0, X1, Y1, 0,
1}, {X1, Y1, 0, 0, 1, 0}, {0, 0, X1, Y1, 0, 1}};
% // MatrixForm
(−5−3001000−5−301210001000210013−50010003−501)
AB = Inverse[d] . R
{{67/118}, {-(27/118)}, {62/59}, {12/59}, {-(109/59)}, {405/59}}
A = {{AB, AB}, { AB, AB}};
% // MatrixForm
(67118−2711862591259)
B = {{AB}, {AB}};
% // MatrixForm
(−1095940959)
A . {X1, Y1};
% // MatrixForm
(−1095940559)
A . {X1, Y1};
% // MatrixForm
(−685924459)
A . {X1, Y1};
% // MatrixForm
(1685912659)
The last three Mathematica commands are simply verifications that the vectors (xk, yk) determine the corners of triangle T₁ were sent to their corresponding counterparts (zk, wk) of T₂.
=================== check
■
End of Example 10
Homogeneous Coordinates
An n-dimensional affine space (A, V, d) is specified by the vector space V and the inhabited set of points A. The n-dimensional vector space V is completely described by providing an ordered basis for it.
From the definition of an affine space, it is known that for every pair of points in A there exists a
vector in V that “connects” them. Once a particular point O is selected from A, every other
point in A can be obtained by adding a vector from V to O. Therefore, supplying an ordered basis for V and a single point in A is sufficient to specify the affine space A.
A frame for the n-dimensional affine space 𝔸 = (A, V, d) consists of the set of basis vectors e₁, e₂, … , en for V and a point O from A. The point O locates the origin of
the frame within A. We use the notation ϕ = (e₁, e₂, … , en, O) to denote a frame. Every vector
u in V can be expressed as
u=c1e1+c2e2+⋯+cnen,
and every point P in A can be written as
P=k1e1+k2e2+⋯+knen+O.
Specifying a frame for an affine space is equivalent to providing a coordinate system for it;
once a frame has been determined any point or vector in the affine space can be described by
a set of scalar values. To do this in matrix notation, however, the following definition must be
made. This is often specified as a third axiom to definition of the affine space:
0⋅P=Oand1⋅P=P∈A.
We start demonstration of affine transformations with plane case. So we choose a frame ϕ = (e₁, e₂, O) for an affine space 𝔸 = (A, ℝ², d). Any vector u in ℝ² can be written in either column form or row form:
u=[e1e2O][α1α20]=(α1:α2:0)(e1e2O).
Hence, column vector [α1α20] and the corresponding row vector (α1:α2:0) are coordinate vectors written in column form and in row form, respectively. Traditionally, coordinate vectors are written in row form where components are separated by ":" in projective geometry. Similarly, point P in the inhabited set A can be expressed as
P=[e1e2O][β1β21]=(β1:β2:1)(e1e2O).
Similar expression are valid for three-dimensional affine spaces, and in general, they are extended for arbitrary n-dimensional case.
Since there is no standard notation for affine coordinates, some authors prefer column notation while others use row form. Therefore, we place both notations together and let the reader deside which one is preferable.
Example 11:
Given the frame
ϕ=([−2−3], , (62)),
determine the point Q that has the coordinates (-3, 2, 1).
Solution:
We use the coordinates to form a linear combination of the vectors in the frame
that we then add to the frame’s origin. Because we are adding a vector to a point the result will indeed be a point.
Q=−3[−2−3]+2+(62)=.
-3{2, -3} + 2 {1, 4} + {6, 2}
{2, 19}
■
End of Example 11
Often it is desirable to find the coordinates of a point relative to one frame given the coordinates of that point relative to another frame. This operation, called a change of frames, is
analogous to the change of basis operation in vector spaces. Let β = (v₁, v₂, v₃, O) and
ϕ = (e₁, e₂, e₃, Q) be two frames for the 3-dimensional affine space 𝔸. To find coordinate vector of arbitrary point in frame ϕ, denoted by ⟦P⟧ϕ given ⟦P⟧β = [α₁, α₂, α₃, 1], we
must first write the basis vectors and point in β in terms of the basis vectors and point in ϕ:
v1=a1e1+b1e2+c1e3,v2=a2e1+b2e2+c2e3,v3=a3e1+b3e2+c3e3,O=a4e1+b4e2+c4e3+Q.
Then
[P]]ϕ=ϕ+ϕ=α1ϕ+α2ϕ+α3ϕ+ϕ=ϕϕϕϕ][α1α2α31]=[a1a2a3a4b1b2b3b4c1c2c3c40001][α1α2α31]
is the change of frame matrix.
Example 12:
Let β and ϕ be two frames for the same affine space such that
β=(, , (5, 2)),ϕ=(, [−3−2], (3, −2)).
If ⟦Q⟧β = (-3, 1, 1), then find ⟦Q⟧ϕ.
Solution: The basis vectors in β can be written as
=0⋅−[−3−2]+0⋅(3, −2),=−2⋅−5⋅[−3−2]+0⋅(3, −2),(5, 2)=−85⋅−225⋅[−3−2]+1⋅(3, −2),
Inverse[{{7, -3}, {3, -2}}] . {3, 2}
{0, -1}
Inverse[{{7, -3}, {3, -2}}] . {1, 4}
{-2, -5}
Inverse[{{7, -3}, {3, -2}}] . {2, 4}
{-(8/5), -(22/5)}
so the change of frame matrix M is
M=[0−1−85−2−3−225001].
Knowing M we can compute ⟦Q⟧ϕ:
ϕ=[0−1−85−2−3−225001][−311]=[−135−751].
{{0,-1,-8/5}, {-2,-3,-22/5}, {0,0,1}} . {-3, 1, 1}
{-(13/5), -(7/5), 1}
■
End of Example 12
Compared to Euclidean geometry, projective geometry has a different setting and has extra points for a given dimension. This allows translation to be
described as a linear transformation, thereby allowing all the transformations we would like to
affect to be represented by matrix multiplication.
Recall that a linear translation is not a linear transformation in vector spaces. The way out of this dilemma is
to turn the n-dimensional problem into a (n+1)-dimensional problem, but in homogeneous coordinates, introduced by the German mathematician August Ferdinand Möbius (1790--1868) in his 1827 work Der barycentrische Calcul.
The real projective plane ℙ² can be given in terms of equivalence classes. For non-zero elements of ℝ³, define (x₁, y₁, z₁) ~ (x₂, y₂, z₂) to mean there is a non-zero λ so that (x₁, y₁, z₁) = (λx₂, λy₂, λz₂). Then ~ is an equivalence relation and the projective plane can be defined as the equivalence classes of ℝ³ ∖ {0}. If (x, y, z) is one of the elements of the equivalence class p, then these are taken to be homogeneous coordinates of p. The homogeneous coordinates or projective coordinates of the point are denoted with columns, either (x:y:z) or [x:y:z].
Homogenous coordinates for a 𝑛-dimensional space consist of tuples with 𝑛+1 coordinates, where the extra coordinate is kept at a special value.
When z ≠ 0, the point [x:y:z] represents the point (x/z, y/z) in the Euclidean plane ℝ².
Homogeneous coordinates of the form (x, y, 0) do not correspond to a point in the Cartesian plane.
Instead, they correspond to the unique point at infinity in the direction (x, y). Hence, the projective plane ℙ² can be seen as the plane ℝ² plus all the points at infinity, each
of which along a different direction. The plane ℙ² also makes sense of the notion that two parallel lines intersect at infinity,
The projective transformation does not preserve parallelism, length, and angle. But it still preserves collinearity and incidence.
Projective transformation can be represented as transformation of an arbitrary quadrangle (i.e. system of four points) into another one.
Example 13:
■
End of Example 13
Affine Matrices
Suppose that 𝔸 and 𝔹 are n-dimensional and m-dimensional affine spaces, respectively. Let
α = (a₁, a₂, … , an, Oα) and β = (b₁, b₂, … , bm, Oβ) be frames for 𝔸 and 𝔹. Suppose further that there esists
an affine transformation F such that F : 𝔸 → 𝔹 so that if P is a point in inhabited set A, then
Q = F(P) is a point in inhabited set B. Finally, let ⟦P⟧α = [α₁, α₂, … , αn, 1]. Then
Q=F(P)=F(α1a1+α2a2+⋯+αnan+Oα)=α1F(a1)+α2F(a2)+⋯+αnF(an)+F(Oα)
where the last step is possible because of properties (c), (d) and (e) of theorem 5. Hence,
β=[α1F(a1)+α2F(a2)+⋯+αnF(an)+F(Oα)]β=[[F(a1)]β⋯[F(an)]β[Oα]β][α1α2⋮αn1]=[a1,1a1,2⋯a1,na1,n+1a2,1a2,2⋯a2,na2,n+1⋮⋮⋱⋮⋮am,1am,2⋯am,nam,n+100⋯01][α1α2⋮αn1]
since
[F(a1)]β=[a1,1a2,1⋮am,10].[F(a2)]β=[a1,2a2,2⋮am,20]⋯,[F(Oα)]β=[a1,n+1a2,n+1⋮an,m+11].
The matrix
M=[a1,1a1,2⋯a1,na1,n+1a2,1a2,2⋯a2,na2,n+1⋮⋮⋱⋮⋮am,1am,2⋯am,nam,n+100⋯01]
is the standard matrix of the affine transformation. In the common cases of two- and three-
dimensional affine spaces M has the form
M=[a1a2a3b1b2b3001],M=[a1a2a3a4b1b2b3b4c1c2c3c40001]
In case of row vectors, these matrices become
M=(a1a20b1b20c1c21),M=(a1a2a30b1b2b30c1c2c30d1d2d31)
Example 14:
Let 𝔸 = (A, ℝ², d) be an affine space with frame α = [e₁, e₂, O], where O = (0, 0). Let T : 𝔸 → 𝔸 be defined as T(P) = P + t, where t = [Δx, Δy]. Find the 3 × 3 matrix T that mplements this transformation.
Solution: Since only frame α is used, we have
[T(e1)]α==and[T(e2)]α==
while
[T(O)]α=[O+t]α=[O]α+[t]α=+[ΔxΔy0]=[ΔxΔy1].
Thus the matrix T is given by
T=[10Δx01Δy001].
■
End of Example 14
Upon embedding n dimensional case into (n+1)-dimensional, we can define an affine transformation as regular linear transformation via matrix/vector multiplication.
Since matrix form is so handy for building up complicated transforms from simpler ones,
it would be very useful to be able to represent all of the affine transforms by matrices.
We also extend our augmented m-by-(n+1) matrix [A ∣ b] from Eq.(1) into (m+1) × n+1)) matrix
[A∣b]⟹Ab=[Ab0⋯01]orAb=(A0b1)or
define maps 𝔽m×(n+1) → 𝔽(m+1)×(n+1) .
On the subset V ⊂ 𝔽(n+1)×1 consisting of vectors with last
component 1, we recover the affine maps
Ab0⋯01=[Ax+b1].
Since V does not contain the zero vector, it is not a vector subspace. But if V₀ denotes the subspace consisting of vectors having last component 0, then
V={t+v| v∈V0}=t+V0,
where t denotes any vector having last component 1. We view it as a translate of
a vector subspace. Any subset of a vector space which is obtained by translation
from a vector subspace is called affine subspace. For example, the set of solutions
of a linear system is an affine subspace: It is a translate of the subspace of
solutions of the associated homogeneous system.
Composition of affine maps is expressed by the following formula:
AbBc=[Ab0⋯01]⋅[Bc0⋯01]=[ABAc+b0⋯01].
Example 14:
■
End of Example 14
2D Affine Transformations
Matrix Representation of
2D Affine Transformations:
Translation:
T=[10Δx01Δy001].
Scale:
S=[sx000sy0001].
Reflection in the plane is given a line and maps points
by flipping the plane about this line.
Fx=[−100010001].
Rotation in positive (counterclockwise) direction by angle θ (in radians):
R[θ]=[cosθ−sinθ0sinθcosθ0001].
Shear:
H=[1a0b10001].
Example 6:
■
End of Example 15
Example 16:
■
End of Example 16
Example 17:
■
End of Example 17
Example 18:
■
End of Example 18
Rotation: In its most general form, rotation is defined to take place about some fixed point.
We will consider the simplest case where the fixed point is the origin of the coordinate frame.
Example 19:
■
End of Example 19
3D Affine Transformations
Now, we can extend all of previously disucced ideas to 3D in the following way. First, we convert all 3D points to homogeneous coordinates of point P(x, y, z), written in either row form or column form:
(x:y:z)∈R1×4or[xyz1]∈R4×1.
The following matrices constitute the basic affine transforms in 3D, expressed in homogeneous form:
Translate: [100Δx010Δy001Δz0001],Scale: [sx0000sy0000sz00001],
Shear: [1hxyhxz0hyx1hyz0hzxhzy100001].
Example 20:
■
End of Example 20
Reflection in 3-space is given a plane, and flips
points in space about this plane. In this case, reflection is just a special case of scaling,
but where the scale factor is negative. A common simple version of this is when the plane
about which the reflection is performed is one of the coordinate planes (corresponding
to x = 0, y = 0, or z = 0).
For example, to reflect points about the xz-coordinate plane (that is, the plane y = 0),
we can scale the y-coordinate by −1. Using the scaling matrix above, we
have the following transformation matrix:
Fy=[10000−10000100001].
The cases for the other two coordinate frames are similar.
Example 21:
■
End of Example 21
Rotation: n its most general form, rotation is defined to take place about some fixed vector in space &Ropf'³.
We will consider the simplest case where the fixed vector is one of the coordinate axes. There are three basic rotations: about the x, y and z-axes. In each case, the rotation
is counterclockwise through an angle θ (given in radians). The rotation is assumed to
be in accordance with a right-hand rule: if your right thumb is aligned with the axes of
rotation, then positive rotation is indicated by the direction in which the fingers of this
hand are pointing. To produce a clockwise rotation, simply negate the angle involved.
Consider a rotation about the z-axis. The z-unit vector and origin are unchanged.
The x-unit vector is mapped to (cos θ, sin θ, 0, 0), and the y-unit vector is mapped to (− sin θ, cos θ, 0, 0). Thus the rotation matrix is:
Rotation about z axis: [cosθz−sinθz00sinθzcosθz0000100001].
Example 22:
■
End of Example 22
RotationMatrix[[Theta], {0, 0, 1}] // MatrixForm
(cosθz−sinθz0sinθzcosθz0001)
Observe that both points and vectors are altered by rotation. For the other two axes we have:
Rotation about x axis: [10000cosθx−sinθx00sinθxcosθx00001],
Example 23:
■
End of Example 23
RotationMatrix[[Theta], {1, 0, 0}] // MatrixForm
(1000cosθx−sinθx0sinθxcosθx)
Rotation about y axis: [cosθy0sinθy00100−sinθy0cosθy00001],
A rotation by angle θ about an arbitrary axis
can be decomposed into the concatenation
of rotations about the x, y, and z axes.
Since a quaternion q basically store the axis vector w and angle of rotation θ, it is not surprisong that we can write the components of a rotation matrix based on quaternion data:
q=(cosθ,sinθw)=(a,(x,y,z)),‖w‖=1,
Then the rotation matrix by angle 2θ is given by
Rq=[1−2y2−2z22xy−2az2xz+2ay02xy+2az1−2x2−2z22yz−2ax02xz−2ay2az+2ax1−2x2−2y200001]
Example 24:
Let us consider a line that goes through point P(7, 11, -5), which is parallel to the vector w = (3, 1, 8). We want to rotate the point Q(6, -9, 15) about this line through angels θ that are multiples of 5° until we are back at the point Q. Each rotation through a fixed angle θ is one application of an affiner map. The plot of these points is essentially the circle of rotation for Q. The formula for the rotated point Qnew is given by
■
End of Example 24
Suppose that an affine space A has two frames
β=([2−2], , (2,−4))
and
ϕ=(, , (−2,5))
Find the change of frame matrix M and use it
to compute ⟦Q⟧β = (5, -3, 1), then find ⟦Q⟧ϕ.
Suppose M is the change of frame matrix that
transforms coordinates relative to frame β to
coordinates relative to frame ϕ. Prove that
M−1 exists.
Determine the matrix representation of the
affine transformation S : 𝔸 → 𝔸 if 𝔸 = (A, ℝ²) and S(P) = Q where
Q = (x+2y, y) if P = (x, y).
What type of transformation is this?
Anton, Howard (2005), Elementary Linear Algebra (Applications Version) (9th ed.), Wiley International
Dunn, F. and Parberry, I. (2002). 3D math primer for graphics and game development. Plano, Tex.: Wordware Pub.
Foley, James D.; van Dam, Andries; Feiner, Steven K.; Hughes, John F. (1991), Computer Graphics: Principles and Practice (2nd ed.), Reading: Addison-Wesley, ISBN 0-201-12110-7
Matrices and Linear Transformations
Rogers, D.F., Adams, J. A., Mathematical Elements for Computer Graphics, McGraw-Hill Science/Engineering/Math, 1989.
Szeliski, R., Computer Vision: Algorithms and Applications, 2nd edition, Springer,
Watt, A., 3D Computer Graphics, Addison-Wesley; 3rd edition, 1999. |
189260 | https://justmaths.co.uk/wp-content/uploads/2015/12/Statistics-H-Cumulative-Frequency-v2-SOLUTIONS-1.pdf | www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 Cumulative Frequency (H) A collection of 9-1 Maths GCSE Sample and Specimen questions from AQA, OCR, Pearson-Edexcel and WJEC Eduqas. 1. The times taken by customer service operators to answer 120 telephone calls are illustrated in the cumulative frequency diagram shown below. (a) Calculate an estimate for the percentage of telephone calls that were answered within 50 seconds. (b) The customer service team was given a target to answer 80% of the telephone calls within 70 seconds. Did the team meet their target? Give a reason for your answer and state any assumption you have made when calculating your answer. You must show all your working. Name: Total Marks: www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 2. The table shows the marks gained by 150 students taking an examination. (a) (i) Construct a cumulative frequency table. (ii) Draw the cumulative frequency graph on the grid below. www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 (b) Students are to be awarded Gold, Silver, Bronze or Fail. The students’ teacher wishes to award the top 10% of students Gold, the next 60% Silver and the next 20% Bronze. Use your graph to estimate the lowest mark that Silver will be awarded for. (b) ............................. (c) Explain why the teacher’s method will not necessarily award Gold to exactly 10% of the students. 3. The cumulative frequency table shows the marks some students got in a test. (a) On the grid, plot a cumulative frequency graph for this information. www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 (b) Find the median mark. ....................................................... Students either pass the test or fail the test. The pass mark is set so that 3 times as many students fail the test as pass the test. (c) Find an estimate for the lowest possible pass mark. ....................................................... 4. Here are the examination marks for 60 pupils. www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 Molly drew this cumulative frequency graph to show the data. Make two criticisms of Molly’s graph. Criticism 1 Criticism 2 5. Gavin measures the heights of 80 plants he has grown. This table summarises his results. a) (i) Complete the cumulative frequency table below. www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 (ii) Draw the cumulative frequency graph. b) Ted asks if Gavin has 10 plants over 120 cm in height. Explain why Gavin cannot be certain that he has 10 plants over this height. www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 c) Gavin sells these 80 plants using the price list below. Each plant costs him 60p to grow. Estimate the total profit Gavin will receive when he sells all these plants. c) £ ......................................... 6. The cumulative frequency graphs show information about the times taken by 100 male runners and by 100 female runners to finish the London marathon. www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 A male runner is chosen at random. a) Find an estimate for the probability that this runner took less than 4 hours to finish the London marathon. ....................................................... b) Use medians and interquartile ranges to compare the distribution of the times taken by the male runners with the distribution of the times taken by the female runners. 7. The table shows the running times of some films. a) Draw a cumulative frequency graph on the grid to represent the data. www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 b) Estimate the number of these films with a running time of less than 2 1 2 hours. 8. What percentage of a distribution is covered by the inter-quartile range? Circle your answer. 25% 37.5% 50% 75% www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 9. The cumulative frequency diagram shows the times taken by runners to complete a half-marathon. On the grid opposite, draw a histogram to represent the data. Use this table to help you. www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 www.justmaths.co.uk Cumulative Frequency (H) - Version 2 January 2016 CREDITS AND NOTES Question Awarding Body 1 WJEC Eduqas 2 OCR 3 Pearson Edexcel 4 AQA 5 OCR 6 Pearson Edexcel 7 AQA 8 AQA 9 AQA Notes: These questions have been retyped from the original sample/specimen assessment materials and whilst every effort has been made to ensure there are no errors, any that do appear are mine and not the exam board s (similarly any errors I have corrected from the originals are also my corrections and not theirs!). Please also note that the layout in terms of fonts, answer lines and space given to each question does not reflect the actual papers to save space. These questions have been collated by me as the basis for a GCSE working party set up by the GLOW maths hub - if you want to get involved please get in touch. The objective is to provide support to fellow teachers and to give you a flavour of how different topics “could” be examined. They should not be used to form a decision as to which board to use. There is no guarantee that a topic will or won’t appear in the “live” papers from a specific exam board or that examination of a topic will be as shown in these questions. Links: AQA OCR Pearson Edexcel WJEC Eduqas Contents: This version contains questions from: AQA – Sample Assessment Material, Practice set 1 and Practice set 2 OCR – Sample Assessment Material and Practice set 1 Pearson Edexcel – Sample Assessment Material, Specimen set 1 and Specimen set 2 WJEC Eduqas – Sample Assessment Material |
189261 | https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Active_Calculus_(Boelkins_et_al.)/06%3A_Using_Definite_Integrals/6.05%3A_Improper_Integrals | 6.5: Improper Integrals - Mathematics LibreTexts
Skip to main content
Table of Contents menu
search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode
school Campus Bookshelves
menu_book Bookshelves
perm_media Learning Objects
login Login
how_to_reg Request Instructor Account
hub Instructor Commons
Search
Search this book
Submit Search
x
Text Color
Reset
Bright Blues Gray Inverted
Text Size
Reset
+-
Margin Size
Reset
+-
Font Type
Enable Dyslexic Font - [x]
Downloads expand_more
Download Page (PDF)
Download Full Book (PDF)
Resources expand_more
Periodic Table
Physics Constants
Scientific Calculator
Reference expand_more
Reference & Cite
Tools expand_more
Help expand_more
Get Help
Feedback
Readability
x
selected template will load here
Error
This action is not available.
chrome_reader_mode Enter Reader Mode
6: Using Definite Integrals
Book: Active Calculus (Boelkins et al.)
{ }
{ "6.01:Using_Definite_Integrals_to_Find_Area_and_Length" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.02:_Using_Definite_Integrals_to_Find_Volume" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.03:_Density_Mass_and_Center_of_Mass" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.04:_Physics_Applications-Work_Force_and_Pressure" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.05:_Improper_Integrals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.E:_Using_Definite_Integrals(Exercises)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
{ "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Understanding_the_Derivative" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Computing_Derivatives" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Using_Derivatives" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_The_Definite_Integral" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Finding_Antiderivatives_and_Evaluating_Integrals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Using_Definite_Integrals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Differential_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Sequences_and_Series" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Multivariable_and_Vector_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Derivatives_of_Multivariable_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Multiple_Integrals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Appendices" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
Thu, 28 Sep 2023 23:55:34 GMT
6.5: Improper Integrals
107835
107835
Paul Seeburger
{ }
Anonymous
Anonymous
2
false
false
[ "article:topic", "improper integral", "divergence", "convergence", "license:ccbysa", "showtoc:no", "authorname:activecalc", "autonumheader:yes2", "licenseversion:40", "source@ ]
[ "article:topic", "improper integral", "divergence", "convergence", "license:ccbysa", "showtoc:no", "authorname:activecalc", "autonumheader:yes2", "licenseversion:40", "source@ ]
Search site Search Search Go back to previous article
Sign in
Username Password Sign in
Sign in
Sign in
Forgot password
Expand/collapse global hierarchy
1. Home
2. Bookshelves
3. Calculus
4. Book: Active Calculus (Boelkins et al.)
5. 6: Using Definite Integrals
6. 6.5: Improper Integrals
Expand/collapse global location
6.5: Improper Integrals
Last updated Sep 28, 2023
Save as PDF
6.4: Physics Applications - Work, Force, and Pressure
6.E: Using Definite Integrals (Exercises)
Page ID 107835
Matthew Boelkins, David Austin & Steven Schlicker
Grand Valley State University via ScholarWorks @Grand Valley State University
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Motivating Questions
2. Preview Activity 6.5.1
3. Improper Integrals Involving Unbounded Intervals
1. Activity 6.5.2
Convergence and Divergence
Note
Activity 6.5.3
Improper Integrals Involving Unbounded Integrands
Activity 6.5.4
Summary
Motivating Questions
What are improper integrals and why are they important?
What does it mean to say that an improper integral converges or diverges?
What are some typical improper integrals that we can classify as convergent or divergent?
Another important application of the definite integral measures the likelihood of certain events. For instance, consider a company that manufactures incandescent light bulbs. Based on a large volume of test results, they have determined that the fraction of light bulbs that fail between times t=a and t=b of use (where t is measured in months) is given by
∫a b 0.3e−0.3t d t.
For example, the fraction of light bulbs that fail during their third month of use is given by
∫2 3 0.3e−0.3t d t=−e−0.3t|2 3=−e−0.9+e−0.6≈0.1422.
Thus about 14.22% of all lightbulbs fail between t=2 and t=3. Clearly we could adjust the limits of integration to measure the fraction of light bulbs that fail during any time period of interest.
Preview Activity 6.5.1
A company with a large customer base has a call center that receives thousands of calls a day. After studying the data that represents how long callers wait for assistance, they find that the function p(t)=0.25e−0.25t models the time customers wait in the following way: the fraction of customers who wait between t=a and t=b minutes is given by
∫a b p(t)d t.
Use this information to answer the following questions.
Determine the fraction of callers who wait between 5 and 10 minutes.
Determine the fraction of callers who wait between 10 and 20 minutes.
Next, let's study the fraction who wait up to a certain number of minutes:
What is the fraction of callers who wait between 0 and 5 minutes?
What is the fraction of callers who wait between 0 and 10 minutes?
Between 0 and 15 minutes? Between 0 and 20?
Let F(b) represent the fraction of callers who wait between 0 and b minutes. Find a formula for F(b) that involves a definite integral, and then use the First FTC to find a formula for F(b) that does not involve a definite integral.
What is the value of the limit lim b→∞F(b)? What is its meaning in the context of the problem?
Improper Integrals Involving Unbounded Intervals
In view of the above examples, we see that we may want to integrate over an interval whose upper limit grows without bound. For example, to find the fraction of light bulbs that fail eventually, we wish to find
lim b→∞∫0 b 0.3e−0.3t d t,
for which we will also use the notation
(6.5.1)∫0∞0.3e−0.3t d t.
Such an integral can be interpreted as the area of an unbounded region, as pictured at right in Figure 6.5.1.
Figure 6.5.1. At left, the area bounded by p(t)=0.3e−0.3t on the finite interval [0,b]; at right, the result of letting b→∞. By “⋯” in the righthand figure, we mean that the region extends to the right without bound.
We call an integral for which the interval of integration is unbounded improper. For instance, the integrals
∫1∞1 x 2 d x,∫−∞0 1 1+x 2 d x,and∫−∞∞e−x 2 d x
are all improper because they have limits of integration that involve ∞. To evaluate an improper integral we replace it with a limit of proper integrals. That is,
∫0∞f(x)d x=lim b→∞∫0 b f(x)d x.
We first attempt to evaluate ∫0 b f(x)d x using the First FTC, and then evaluate the limit. Is it even possible for the area of an unbounded region to be finite? The following activity explores this issue and others in more detail.
Activity 6.5.2
In this activity we explore the improper integrals ∫1∞1 x d x and ∫1∞1 x 3/2 d x.
First we investigate ∫1∞1 x d x.
Use the First FTC to determine the exact values of ∫1 10 1 x d x,∫1 1000 1 x d x, and ∫1 100000 1 x d x. Then, use your computational device to compute a decimal approximation of each result.
Use the First FTC to evaluate the definite integral ∫1 b 1 x d x (which results in an expression that depends on b).
Now, use your work from (ii.) to evaluate the limit given by lim b→∞∫1 b 1 x d x.
Next, we investigate ∫1∞1 x 3/2 d x.
Use the First FTC to determine the exact values of ∫1 10 1 x 3/2 d x,∫1 1000 1 x 3/2 d x, and ∫1 100000 1 x 3/2 d x. Then, use your calculator to compute a decimal approximation of each result.
Use the First FTC to evaluate the definite integral ∫1 b 1 x 3/2 d x (which results in an expression that depends on b).
Now, use your work from (ii.) to evaluate the limit given by lim b→∞∫1 b 1 x 3/2 d x.
Plot the functions y=1 x and y=1 x 3/2 on the same coordinate axes for the values x=0…10. How would you compare their behavior as x increases without bound? What is similar? What is different?
How would you characterize the value of ∫1∞1 x d x? of ∫1∞1 x 3/2 d x? What does this tell us about the respective areas bounded by these two curves for x≥1?
Convergence and Divergence
Activity 6.5.2 suggests that lim b→∞∫1 b f(x)d x is either finite or infinite (or it doesn't exist). With these possibilities in mind, we introduce the following terminology.
Note
If f(x) is nonnegative for x≥a, then we say that the improper integral ∫a∞f(x)d xconverges provided that
lim b→∞∫a b f(x)d x
exists and is finite. Otherwise, we say that ∫a∞f(x)d xdiverges.
We will restrict our interest to improper integrals for which the integrand is nonnegative. Also, we require that lim x→∞f(x)=0, for if f does not approach 0 as x→∞, then it is impossible for ∫a∞f(x)d x to converge.
Activity 6.5.3
Determine whether each of the following improper integrals converges or diverges. For each integral that converges, find its exact value.
∫1∞1 x 2 d x
∫0∞e−x/4 d x
∫2∞9(x+5)2/3 d x
∫4∞3(x+2)5/4 d x
∫0∞xe−x/4 d x
∫1∞1 x p d x, where p is a positive real number
Improper Integrals Involving Unbounded Integrands
An integral is also called improper if the integrand is unbounded on the interval of integration. For example, consider
∫0 1 1 x d x.
Because f(x)=1 x has a vertical asymptote at x=0,f is not continuous on [0,1], and the integral represents the area of the unbounded region shown at right in Figure 6.5.2.
Figure 6.5.2. At left, the area bounded by f(x)=1 x on the finite interval [a,1]; at right, the result of letting a→0+, where we see that the shaded region will extend vertically without bound.
We address the problem of the integrand being unbounded by replacing the improper integral with a limit of proper integrals. For example, to evaluate ∫0 1 1 x d x, we replace 0 with a and let a approach 0 from the right. Thus,
∫0 1 1 x d x=lim a→0+∫a 1 1 x d x.
We evaluate the proper integral ∫a 1 1 x d x, and then take the limit. We will say that the improper integral converges if this limit exists, and diverges otherwise. In this example, we observe that
∫0 1 1 x d x=lim a→0+∫a 1 1 x d x=lim a→0+2x|a 1=lim a→0+21−2a=2,
so the improper integral ∫0 1 1 x d x converges (to the value 2).
We have to be particularly careful with unbounded integrands, for they may arise in ways that may not initially be obvious. Consider, for instance, the integral
∫1 3 1(x−2)2 d x.
At first glance we might think that we can simply apply the Fundamental Theorem of Calculus by antidifferentiating 1(x−2)2 to get −1 x−2 and then evaluating from 1 to 3. Were we to do so, we would be erroneously applying the FTC because f(x)=1(x−2)2 fails to be continuous throughout the interval, as seen in Figure 6.5.3.
Figure 6.5.3. The function f(x)=1(x−2)2 on an interval including x=2.
Such an incorrect application of the FTC leads to an impossible result (−2), which would itself suggest that something we did must be wrong. Instead, we must address the vertical asymptote at x=2 by writing
∫1 3 1(x−2)2 d x=lim a→2−∫1 a 1(x−2)2 d x+lim b→2+∫b 3 1(x−2)2 d x.
We then evaluate two separate limits of proper integrals. For instance, doing so for the integral with a approaching 2 from the left, we find
∫1 2 1(x−2)2 d x=lim a→2−∫1 a 1(x−2)2 d x=lim a→2−−1(x−2)|1 a=lim a→2−−1(a−2)+1 1−2=∞,
since 1 a−2→−∞ as a approaches 2 from the left. Thus, the improper integral ∫1 2 1(x−2)2 d x diverges; similar work shows that ∫2 3 1(x−2)2 d x also diverges. From either of these two results, we can conclude that that the original integral, ∫1 3 1(x−2)2 d x diverges, too.
Activity 6.5.4
For each of the following definite integrals, decide whether the integral is improper or not. If the integral is proper, evaluate it using the First FTC. If the integral is improper, determine whether or not the integral converges or diverges; if the integral converges, find its exact value.
∫0 1 1 x 1/3 d x
∫0 2 e−x d x
∫1 4 1 4−x d x
∫−2 2 1 x 2 d x
∫0 π/2 tan(x)d x
∫0 1 1 1−x 2 d x
Summary
An integral ∫a b f(x)d x can be improper if at least one of a or b is ±∞, making the interval unbounded, or if f has a vertical asymptote at x=c for some value of c that satisfies a≤c≤b. One reason that improper integrals are important is that certain probabilities can be represented by integrals that involve infinite limits.
When we encounter an improper integral, we work to understand it by replacing the improper integral with a limit of proper integrals. For instance, we write ∫a∞f(x)d x=lim b→∞∫a b f(x)d x,
and then work to determine whether the limit exists and is finite. For any improper integral, if the resulting limit of proper integrals exists and is finite, we say the improper integral converges. Otherwise, the improper integral diverges.
An important class of improper integrals is given by ∫1∞1 x p d x
where p is a positive real number. We can show that this improper integral converges whenever p>1, and diverges whenever 0<p≤1. A related class of improper integrals is ∫0 1 1 x p d x, which converges for 0<p<1, and diverges for p≥1.
This page titled 6.5: Improper Integrals is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Matthew Boelkins, David Austin & Steven Schlicker (ScholarWorks @Grand Valley State University) via source content that was edited to the style and standards of the LibreTexts platform.
Back to top
6.4: Physics Applications - Work, Force, and Pressure
6.E: Using Definite Integrals (Exercises)
Was this article helpful?
Yes
No
Recommended articles
3.7: Improper IntegralsIn this section, we define integrals over an infinite interval as well as integrals of functions containing a discontinuity on the interval. Integrals...
3.9: Chapter 3 Review ExercisesThis page reviews integral calculus concepts through exercises on integration methods, numerical techniques for approximating integrals, and improper ...
1.7E: Exercises for Section 1.8This page provides exercises on the Gauss-Seidel method for solving systems of equations, detailing convergence criteria based on the coefficient matr...
4.3: More on the Fourier SeriesWe have computed the Fourier series for a 2??-periodic function, but what about functions of different periods.
5.6: Stochastic MatricesThis page explores stochastic matrices and their applications in difference equations and Markov chains, particularly in Google's PageRank algorithm. ...
Article typeSection or PageAuthorActive CalculusAutonumber Section Headingstitle with colon delimitersLicenseCC BY-SALicense Version4.0Show Page TOCno
Tags
convergence
divergence
improper integral
source@
© Copyright 2025 Mathematics LibreTexts
Powered by CXone Expert ®
?
The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org.
Support Center
How can we help?
Contact Support Search the Insight Knowledge Base Check System Status×
contents readability resources tools
☰
6.4: Physics Applications - Work, Force, and Pressure
6.E: Using Definite Integrals (Exercises) |
189262 | https://www.uptodate.com/contents/evaluation-and-management-of-secondary-amenorrhea | Your Privacy
To give you the best possible experience we use cookies and similar technologies. We use data collected through these technologies for various purposes, including to enhance website functionality, remember your preferences, and show the most relevant content. You can select your preferences by clicking the link. For more information, please review our Privacy & Cookie Notice
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device. Because we respect your right to privacy, you can choose not to allow certain types of cookies on our website. Click on the different category headings to find out more and manage your cookie preferences. However, blocking some types of cookies may impact your experience on the site and the services we are able to offer.
Privacy & Cookie Notice
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function. They are usually set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, this may have an effect on the proper functioning of (parts of) the site.
Performance Cookies
These cookies support analytic services that measure and improve the performance of our site. They help us know which pages are the most and least popular and see how visitors move around the site. |
189263 | https://byjus.com/maths/quadratics/ | Published Time: 2017-09-08T16:09:54+05:30
Quadraticscan be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations. The general form of the quadratic equation is:
ax² + bx + c = 0
where x is an unknown variable and a, b, c are numerical coefficients. For example, x 2 + 2x +1 is a quadratic or quadratic equation. Here, a ≠ 0 because if it equals zero then the equation will not remain quadratic anymore and it will become a linear equation, such as:
bx+c=0
Thus, this equation cannot be called a quadratic equation.
The terms a, b and c are also called quadratic coefficients.
The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations. Theroots of any polynomial are the solutions for the given equation.
What is Quadratic Equation?
The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of:
ax² + bx + c = 0
where x is the unknown variable and a, b and c are the constant terms.
Standard Form of Quadratic Equation
Since the quadratic includes only one unknown term or variable, thus it is called univariate. The power of variable x is always non-negative integers. Hence the equation is a polynomial equation with the highest power as 2.
The solution for this equation is the values of x, which are also called zeros. Zeros of the polynomial are the solution for which the equation is satisfied. In the case of quadratics, there are two roots or zeros of the equation. And if we put the values of roots or x on the left-hand side of the equation, it will equal to zero. Therefore, they are called zeros.
Quadratics Formula
The formula for a quadratic equation is used to find the roots of the equation. Since quadratics have a degree equal to two, therefore there will be two solutions for the equation. Suppose ax² + bx + c = 0 is the quadratic equation, then the formula to find the roots of this equation will be:
x = [-b±√(b 2-4ac)]/2a
The sign of plus/minus indicates there will be two solutions for x. Learn in detail the quadratic formula here.
Examples of Quadratics
Beneath are the illustrations of quadratic equations of the form (ax² + bx + c = 0)
x² –x – 9 = 0
5x² – 2x – 6 = 0
3x² + 4x + 8 = 0
-x² +6x + 12 = 0
Examples of a quadratic equation with the absence of a ‘ C ‘- a constant term.
-x² – 9x = 0
x² + 2x = 0
-6x² – 3x = 0
-5x² + x = 0
-12x² + 13x = 0
11x² – 27x = 0
Following are the examples of a quadratic equation in factored form
(x – 6)(x + 1) = 0 [ result obtained after solving is x² – 5x – 6 = 0]
–3(x – 4)(2x + 3) = 0 [result obtained after solving is -6x² + 15x + 36 = 0]
(x − 5)(x + 3) = 0 [result obtained after solving is x² − 2x − 15 = 0]
(x – 5)(x + 2) = 0 [ result obtained after solving is x² – 3x – 10 = 0]
(x – 4)(x + 2) = 0 [result obtained after solving is x² – 2x – 8 = 0]
(2x+3)(3x – 2) = 0 [result obtained after solving is 6x² + 5x – 6]
Below are the examples of a quadratic equation with an absence of linear co – efficient ‘ bx’
2x² – 64 = 0
x² – 16 = 0
9x² + 49 = 0
-2x² – 4 = 0
4x² + 81 = 0
-x² – 9 = 0
How to Solve Quadratic Equations?
There are basically four methods of solving quadratic equations. They are:
Factoring
Completing the square
Using Quadratic Formula
Taking the square root
Factoring of Quadratics
Begin with a equation of the form ax² + bx + c = 0
Ensure that it is set to adequate zero.
Factor the left-hand side of the equation by assuming zero on the right-hand side of the equation.
Assign each factor equal to zero.
Now solve the equation in order to determine the values of x.
Suppose if the main coefficient is not equal to one then deliberately, you have to follow a methodology in the arrangement of the factors.
Example:
2x²-x-6=0
(2x+3)(x-2)=0
2x+3=0
x=-3/2
x=2
Learn more about the factorization of quadratic equations here.
Completing the Square Method
Let us learn this method with example.
Example: Solve 2x 2 – x – 1 = 0.
First, move the constant term to the other side of the equation.
2x 2 – x = 1
Dividing both sides by 2.
x 2 – x/2 = ½
Add the square of half of the coefficient of x, (b/2a)2, on both the sides, i.e., 1/16
x 2 – x/2 + 1/16 = ½ + 1/16
Now we can factor the right side,
(x-¼)2 = 9/16 = (¾)2
Taking root on both sides;
X – ¼ = ±3/4
Add ¼ on both sides
X = ¼ ± ¾
Therefore,
X = ¼ + ¾ = 4/4 = 1
X = ¼ – ¾ = -2/4 = -½
To learn more about completing the square method, click here.
Using Quadratic Formula
For the given Quadratic equation of the form, ax² + bx + c = 0
Therefore the roots of the given equation can be found by:
x=−b±b 2−4 a c 2 a
where ± (one plus and one minus) represent two distinct roots of the given equation.
Taking the Square Root
We can use this method for the equations such as:
x 2 + a 2 = 0
Example: Solve x 2 – 50 = 0.
x 2 – 50 = 0
x 2 = 50
Taking the roots both sides
√x 2 = ±√50
x = ±√(2 x 5 x 5)
x = ±5√2
Thus, we got the required solution.
Related Articles
Zeros Of polynomial
Quadratic Formula
Quadratic Equation For Class 10
Quadratic Equations Class 11
Quadratic Equation Solver
Quadratic Equation Worksheet
Video Lesson on Quadratic Equations
Range of Quadratic Equations
19,348
Solved Problems on Quadratic Equations
Lets Work Out:
Example 1:
3 x 2–5 x+2=0
Solution:
3 x 2–5 x+2=0
Solving the quadratic equation using the above method:
x=−b±b 2−4 a c 2 a
x=−(−5)±(−5)2−4×3×2 2×3
x=5±1 6
x=6 6 o r 4 6
or,
x=1 o r 2 3
Example 2: Solve x 2 – 6 x = 16.
Solution: x 2 – 6 x = 16.
x 2 – 6 x – 16 = 0
By factorisation method,
( x – 8)( x + 2) = 0
Therefore,
x = 8 and x = -2
Example 3: Solve x 2 – 16 = 0. And check if the solution is correct.
Solution: x 2 – 16 = 0.
x 2 – 4 2 = 0 [By algebraic identities]
(x-4) (x+4) = 0
x = 4 and x = -4
Check:
Putting the values of x in the LHS of the given quadratic equation
If x = 4
X 2 – 16 = (4)2 – 16 = 16 – 16 = 0
If x = -4,
X 2 – 16 = (-4)2 – 16 = 16 – 16 = 0
Example 4: Solve for y: y 2 = –2y + 2.
Solution: Given,
y 2 = –2y + 2
Rewriting the given equation;
y 2 + 2 y – 2 = 0
Using quadratic formula,
y=−b±b 2−4 a c 2 a
y=−(2)±(2)2−4(1)(−2)2(1)
y=−2±4+8 2
y=−2±12 2
Therefore,
y = -1 + √3 or y = -1 – √3
Applications of Quadratic Equations
Many real-life word problems can be solved using quadratic equations. While solving word problems, some common quadratic equation applications include speed problems and Geometry area problems.
Solving the problems related to finding the area of quadrilateral such as rectangle, parallelogram and so on
Solving Word Problems involving Distance, speed, and time, etc.,
Example: Find the width of a rectangle of area 336 cm2 if its length is equal to the 4 more than twice its width.
Solution:
Let x cm be the width of the rectangle.
Length = (2x + 4) cm
We know that
Area of rectangle = Length x Width
x(2x + 4) = 336
2x 2 + 4x – 336 = 0
x 2 + 2x – 168 = 0
x 2 + 14x – 12x – 168 = 0
x(x + 14) – 12(x + 14) = 0
(x + 14)(x – 12) = 0
x = -14, x = 12
Measurement cannot be negative.
Therefore, Width of the rectangle = x = 12 cm
Practice Questions
Solve x 2 + 2 x + 1 = 0.
Solve 5x 2 + 6x + 1 = 0
Solve 2x 2 + 3 x + 2 = 0.
Solve x 2 − 4x + 6.25 = 0
Frequently Asked Questions on Quadratics
Q1
What is a quadratic equation?
The polynomial equation whose highest degree is two is called a quadratic equation. The equation is given by ax² + bx + c = 0, where a ≠ 0.
Q2
What are the methods to solve a quadratic equation?
There are majorly four methods of solving quadratic equations. They are:
Factorisation
Using Square roots
Completing the square
Using quadratic formula
Q3
Is x 2 – 1 a quadratic equation?
Since the degree of the polynomial is 2, therefore, given equation is a quadratic equation.
Q4
What is the solution of x 2 + 4 = 0?
The solution of quadratic equation x 2 – 4 is x = 2 or x = -2.
Q5
Write the quadratic equation in the form of sum and product of roots.
If α and β are the roots of a quadratic equation, then;
Sum of the roots = α+β
Product of the roots = αβ
Therefore, the required equation is:
x 2 – (α+β)x + (αβ) = 0
Comments
Leave a Comment Cancel reply
Your Mobile number and Email id will not be published.Required fields are marked
Send OTP
Did not receive OTP?
Request OTP on Voice Call
Website
Post My Comment
Anjana BrijeshSeptember 14, 2020 at 2:38 pm Thanks a lot ,This was very useful for me Reply
lavanyaDecember 28, 2020 at 10:13 am x=√9
Squaring both the sides,
x^2 = 9
x^2 – 9 = 0
It is a quadratic equation. Reply
Register with BYJU'S & Download Free PDFs
Send OTP
Download Now
Register with BYJU'S & Watch Live Videos
Send OTP
Watch Now
×
To continue watching the video please share the details.
Send OTP
Did not receive OTP?
Request OTP on Voice Call
Play Now |
189264 | https://www.ni.com/docs/en-US/bundle/labwindows-cvi/page/advancedanalysisconcepts/lvac_polynomial_composition.html?srsltid=AfmBOopZ5y8Lrr3-y1rgOuCTBh-oPAwaY-SoHU8Ysz3oKHfp5vC859Ty | Modular Data Acquisition
Distributed Measurement and Control
High-Performance Test
Automated Test System Development Software
Perspectives showcases how NI sees what’s next in the world of test and technology.
You can request repair, RMA, schedule calibration, or get technical support. A valid service agreement may be required.
Provides support for NI data acquisition and signal conditioning devices.
Provides support for Ethernet, GPIB, serial, USB, and other types of instruments.
Provides support for NI GPIB controllers and NI embedded controllers with GPIB ports.
LabWindows/CVI
Table of Contents
Polynomial Composition
Polynomial composition involves replacing the variable x in a polynomial with another polynomial.
For example, the following equations define two polynomials.
Replacing x in P(x) with the polynomial Q(x) results in the following equation.
where P(Q(x)) denotes the composite polynomial.
Previous
Next
Previous
Next
Log in to get a better experience |
189265 | https://ditki.com/course/usmle-comlex-step-2/glossary/pathophysiologic-disorder/pericarditis-for-usmle-step-2-comlex-usa-level-2 | USMLE / COMLEX - Step 2 Glossary: Pericarditis for USMLE Step 2 & COMLEX-USA Level 2 | ditki medical & biological sciences
Back
medical & biological sciences
Pericarditis for USMLE Step 2 & COMLEX-USA Level 2
Start your One-Week Free Trial
Already subscribed? Log in »
Overview of Pericarditis
Definition: Inflammation of the pericardium, the fibrous sac surrounding the heart.
Etiology
Causes:
Infectious: Viral (e.g., coxsackievirus, echovirus) is most common.
Non-infectious: Autoimmune diseases (e.g., lupus, rheumatoid arthritis), post-myocardial infarction (Dressler's syndrome), uremia, and radiation therapy.
Idiopathic: Often presumed viral in origin.
Clinical Presentation
Chest Pain: Sharp and pleuritic, typically worsens with inspiration and improves with sitting up and leaning forward.
Pericardial Friction Rub: Scratchy, high-pitched sound best heard at the left sternal border during end expiration with the patient leaning forward.
Other Symptoms: Dyspnea, fever, and general malaise.
Diagnosis
ECG Findings:
Diffuse ST elevation and PR depression are typical in the acute phase.
Echocardiography:
Useful for detecting pericardial effusion and evaluating for cardiac tamponade.
Laboratory Tests:
Elevated inflammatory markers (CRP, ESR), but specific etiology often requires further investigation based on clinical context.
Management
Medical Treatment:
NSAIDs: First-line treatment for pain and inflammation.
Colchicine: Reduces inflammation and prevents recurrence.
Corticosteroids: Reserved for severe cases or those refractory to first-line treatments, or specific causes like autoimmune disorders.
Monitoring and Follow-Up:
Monitor for signs of cardiac tamponade and constrictive pericarditis, especially in severe cases.
Essential Points for USMLE Step 2
Pain Management: NSAIDs combined with colchicine is effective for reducing symptoms and preventing recurrence.
Identify Serious Complications: Be vigilant for cardiac tamponade, which presents with hypotension, jugular venous distension, and muffled heart sounds (Beck's triad).
Differentiation from MI: Unlike myocardial infarction, pericarditis causes diffuse ST elevation without reciprocal ST depression, and troponin may be mildly elevated or normal unless myopericarditis is present. |
189266 | https://study.com/skill/learn/how-to-add-and-subtract-fractions-with-different-denominators-explanation.html | Add and Subtract Fractions with Different Denominators | Algebra | Study.com
Log In
Sign Up
Menu
Plans
Courses
By Subject
College Courses
High School Courses
Middle School Courses
Elementary School Courses
By Subject
Arts
Business
Computer Science
Education & Teaching
English (ELA)
Foreign Language
Health & Medicine
History
Humanities
Math
Psychology
Science
Social Science
Subjects
Art
Business
Computer Science
Education & Teaching
English
Health & Medicine
History
Humanities
Math
Psychology
Science
Social Science
Art
Architecture
Art History
Design
Performing Arts
Visual Arts
Business
Accounting
Business Administration
Business Communication
Business Ethics
Business Intelligence
Business Law
Economics
Finance
Healthcare Administration
Human Resources
Information Technology
International Business
Operations Management
Real Estate
Sales & Marketing
Computer Science
Computer Engineering
Computer Programming
Cybersecurity
Data Science
Software
Education & Teaching
Education Law & Policy
Pedagogy & Teaching Strategies
Special & Specialized Education
Student Support in Education
Teaching English Language Learners
English
Grammar
Literature
Public Speaking
Reading
Vocabulary
Writing & Composition
Health & Medicine
Counseling & Therapy
Health
Medicine
Nursing
Nutrition
History
US History
World History
Humanities
Communication
Ethics
Foreign Languages
Philosophy
Religious Studies
Math
Algebra
Basic Math
Calculus
Geometry
Statistics
Trigonometry
Psychology
Clinical & Abnormal Psychology
Cognitive Science
Developmental Psychology
Educational Psychology
Organizational Psychology
Social Psychology
Science
Anatomy & Physiology
Astronomy
Biology
Chemistry
Earth Science
Engineering
Environmental Science
Physics
Scientific Research
Social Science
Anthropology
Criminal Justice
Geography
Law
Linguistics
Political Science
Sociology
Teachers
Teacher Certification
Teaching Resources and Curriculum
Skills Practice
Lesson Plans
Teacher Professional Development
For schools & districts
Certifications
Teacher Certification Exams
Nursing Exams
Real Estate Exams
Military Exams
Finance Exams
Human Resources Exams
Counseling & Social Work Exams
Allied Health & Medicine Exams
All Test Prep
Teacher Certification Exams
Praxis Test Prep
FTCE Test Prep
TExES Test Prep
CSET & CBEST Test Prep
All Teacher Certification Test Prep
Nursing Exams
NCLEX Test Prep
TEAS Test Prep
HESI Test Prep
All Nursing Test Prep
Real Estate Exams
Real Estate Sales
Real Estate Brokers
Real Estate Appraisals
All Real Estate Test Prep
Military Exams
ASVAB Test Prep
AFOQT Test Prep
All Military Test Prep
Finance Exams
SIE Test Prep
Series 6 Test Prep
Series 65 Test Prep
Series 66 Test Prep
Series 7 Test Prep
CPP Test Prep
CMA Test Prep
All Finance Test Prep
Human Resources Exams
SHRM Test Prep
PHR Test Prep
aPHR Test Prep
PHRi Test Prep
SPHR Test Prep
All HR Test Prep
Counseling & Social Work Exams
NCE Test Prep
NCMHCE Test Prep
CPCE Test Prep
ASWB Test Prep
CRC Test Prep
All Counseling & Social Work Test Prep
Allied Health & Medicine Exams
ASCP Test Prep
CNA Test Prep
CNS Test Prep
All Medical Test Prep
College Degrees
College Credit Courses
Partner Schools
Success Stories
Earn credit
Sign Up
Copyright
Add and Subtract Fractions with Different Denominators
Pre-Algebra Skills Practice
Click for sound
5:07
You must c C reate an account to continue watching
Register to access this and thousands of other videos
Are you a student or a teacher?
I am a student I am a teacher
Try Study.com, risk-free
As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed.
Get unlimited access to over 88,000 lessons.
Try it risk-free
It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time.
Already registered? Log in here for access
Back
What teachers are saying about Study.com
Try it risk-free for 30 days
Already registered? Log in here for access
00:04 How to add and…
03:27 How to add and…
Jump to a specific example
Speed Normal
0.5x
Normal
1.25x
1.5x
1.75x
2x
Speed
Nidhi Agarwal, AMY MAYERS
Instructors Nidhi Agarwal
Nidhi holds a Bachelor's degree in Secondary Education with a teaching major Biological Sciences and Master's degree in education from Lucknow University and has taught middle and high school math. I have over 15 years of experience working with students of different ages including two years of teaching experience internationally.
View bio
AMY MAYERS
Amy has taught middle school math and algebra for over seven years. She has a Bachelor's degree in Mathematical Sciences from the University of Houston and a Master's degree in Curriculum and Instruction from The University of St. Thomas. She is a Texas certified teacher for grades 4-12 in Mathematics.
View bio
Example SolutionsPractice Questions
Adding and Subtracting Fractions with Different Denominators
Step 1: Identify the denominator of the fractions that are being added or subtracted.
Step 2: Find the least common multiple (LCM) of the denominators. To find the LCM of the denominators, find the multiples of each denominator. Then, look for the lowest multiple that is common to both denominators. That will be the least common multiple.
Step 3: Rewrite each fraction to its equivalent fraction with a denominator equal to the least common multiple(LCM).
Step 4: Add or subtract the equivalent fractions obtained in step 3 to get the final answer. Add or subtract the numerator as required in the problem. The denominator of the answer will be the same as the denominator of the fractions that are being added or subtracted.
Adding and Subtracting Fractions with Different Denominators Vocabulary
Fractions: Fractions represent a part of the whole. They can be written as p q, where q≠0. The number p is the numerator in the fraction, and the number q is the denominator in the fraction.
Equivalent fraction: Equivalent fraction to a given fraction is a fraction with a different numerator and a different denominator but it has the same value as the given fraction.
We will take a look at two examples to get a clear understanding of the concept of adding or subtracting fractions with different denominators. The first example will show the addition of fractions with different denominators while the second example will show the subtraction of fractions with different denominators.
Adding and Subtracting Fractions with Different Denominators: Adding the Fractions Example
Find the sum: 3 5+6 7
Step 1: Identify the denominator of the fractions that are being added or subtracted.
The denominators of the fractions that are being added are 5 and 7.
Step 2: Find the least common multiple (LCM) of the denominators.
List the multiples of 5 and 7 and look for the lowest multiple that is common in the lists.
Multiples of 5 are: 5, 10, 15, 20, 15, 30, 35, 40
Multiples of 7 are: 7, 14, 21, 28, 35
The lowest multiple that is common in the list is 35.
Therefore, the LCM is 35.
Step 3: Rewrite each fraction to its equivalent fraction with a denominator equal to the least common multiple(LCM).
Rewrite 3 5 to its equivalent fraction with a denominator of 35.
Think of a number which when multiplied by 5 gives 35.
5×7=35
We multiply the numerator and the denominator by 7 to get an equivalent fraction.
3×7 5×7=21 35
Rewrite 6 7 to its equivalent fraction with a denominator of 35
Think of a number which when multiplied by 7 gives 35.
7×5=35
We multiply the numerator and the denominator by 5 to get an equivalent fraction.
6×5 7×5=30 35
Step 4: Add or subtract the equivalent fractions obtained in step 3 to get the final answer.
21 35+30 35=51 35
Adding and Subtracting Fractions with Different Denominators: Subtracting the Fractions Example
Find the difference: 2 5−3 8
Step 1: Identify the denominator of the fractions that are being added or subtracted.
The denominators of the fractions that are being subtracted are 5 and 8.
Step 2: Find the least common multiple (LCM) of the denominators.
List the multiples of 5 and 8 and look for the lowest multiple that is common in the lists.
Multiples of 5 are: 5, 10, 15, 20, 15, 30, 35, 40
Multiples of 8 are: 8, 16, 24, 32, 40
The lowest multiple that is common in the list is 40
Therefore, the LCM is 40
Step 3: Rewrite each fraction to its equivalent fraction with a denominator equal to the least common multiple (LCM).
Rewrite 2 5 to its equivalent fraction with a denominator of 40.
Think of a number that when multiplied by 5 gives 40.
5×8=40
We multiply the numerator and the denominator by 8 to get an equivalent fraction.
2×8 5×8=16 40
Rewrite 3 8 to its equivalent fraction with a denominator of 40.
We multiply the numerator and the denominator by 5 to get an equivalent fraction.
3×5 8×5=15 40
Step 4: Add or subtract the equivalent fractions obtained in step 3 to get the final answer.
16 40−15 40=1 40
Get access to thousands of practice questions and explanations!
Create an account
Table of Contents
Adding and Subtracting Fractions with Different Denominators
Adding and Subtracting Fractions with Different Denominators Vocabulary
Adding the Fractions Example
Subtracting the Fractions Example
Test your current knowledge
Practice
Adding and Subtracting Fractions with Different Denominators
Recently updated on Study.com
Videos
Courses
Lessons
Articles
Quizzes
Concepts
Teacher Resources
The First & Second Balkan Wars | Background & Consequences
Ethnic Groups in Indonesia | Demographics & People
Gods of the Winter Solstice
Holes by Louis Sachar | Themes, Quotes & Analysis
Aurangzeb | Empire, Achievements & Failures
Libya Ethnic Groups | Demographics, Population & Cultures
Cold War Lesson for Kids: Facts & Timeline
The Chronicles of Narnia Series by C.S. Lewis | Overview...
Isotherms Definition, Maps & Types
What is Antidiuretic Hormone? - Definition & Function
Karl Marx's Economic Determinism | Overview & History
General Studies Literature: Help & Review
GACE 713 Study Guide - English (6-12) Exam Prep
GACE 711 Study Guide - Mathematics (6-12) Exam Prep
GACE 706 Study Guide - Middle Grades Language Arts (4-8)...
GACE 704 Study Guide - Middle Grades Math (4-8) Exam Prep
GACE 735 Study Guide - Special Education General...
Physics 112L: Physics II with Lab
Science 101L: Intro to Life Sciences with Lab
UExcel Science of Nutrition: Study Guide & Test Prep
MTLE Elementary Education Study Guide and Test Prep
World Population
Mental Health Study Guide
MTTC Early Childhood Education (106) Study Guide and Test...
Immunology: Help & Review
TExES Core Subjects EC-6 (291) Prep
Praxis 5165 Study Guide - Mathematics Exam Prep
FSA - Grade 8 ELA: Test Prep & Practice
History of Islam: Help & Review
Rock in Performing Arts | History, Characteristics & Facts
Reproductive Endocrinology Meaning & Specialization
Drug Rehabilitation Types, Effects & Importance
Cardiology Definition, Tests & Specializations
Neurology | Definition, Importance & Disorders
Anaerobic Bacteria Definition, Types & Examples
Seasonal Festivals | List & Examples
Clinical Assessment Definition, Importance & Types
John Dunstable | Biography, Music & Significance
Why Are Yawns Contagious?
Leopold Mozart | Biography, Music & Timeline
Disorganized Schizophrenia | Symptoms, Treatment & Examples
Embodied Cognition | Definition, History & Examples
The Beginnings of Opera: Influences and Components
Spanish Conquistadors | Purpose, Facts & History
Bipolar Disorder & Cognitive Impairment
Cooperative Learning: Roles & Activities
William Henry Fox Talbot Life & Photography
How to Use & Cite AI Tools in College Saver Course...
Understanding Generative AI as a Student: Uses, Benefits...
WEST Prep Product Comparison
What Are the Features of My Institutional Student Account...
How to Pass the SLLA Exam
How to Pass the TExES Exams
CHPN Certification Requirements
Special Education Private Schools in California
How Long is the GMAT?
Benefits of Study.com vs. Traditional College
Adult Community Education
Quiz & Worksheet - Periodic Functions
Quiz & Worksheet - Legacy of the Neo-Babylonian Empire
Quiz & Worksheet - Barbados Ethnic Groups
Quiz & Worksheet - The Mule in Their Eyes Were Watching God
Quiz & Worksheet - Aslan in The Lion, the Witch & the...
Quiz & Worksheet - Who Dies in Bridge to Terabithia?
Quiz & Worksheet - Past, Present & Future Business Trends
Math
Social Sciences
Science
Business
Humanities
Education
Art and Design
History
Tech and Engineering
Health and Medicine
Plans
Study help
Test prep
College credit
Teacher resources
Working Scholars®
School group plans
Online tutoring
About us
Blog
Careers
Teach for us
Press Center
Ambassador
Scholarships
Support
Contact support
FAQ
Site feedback
Resources and Guides
Download the app
Study.com on Facebook Study.com on YouTube Study.com on Instagram Study.com on Twitter Study.com on LinkedIn
© Copyright 2025 Study.com. All other trademarks and copyrights are the property of their respective owners. All rights reserved.
Contact us by phone at (877)266-4919, or by mail at 100 View Street#202, Mountain View, CA 94041.
About Us
Terms of Use
Privacy Policy
DMCA Notice
ADA Compliance
Honor Code For Students
Support
× |
189267 | https://emedicine.medscape.com/article/2121028-overview | close
Please confirm that you would like to log out of Medscape.
If you log out, you will be required to enter your username and password the next time you visit.
Log out
Cancel
processing....
Tools & Reference>Endocrinology
Exocrine Pancreatic Insufficiency
Updated: Mar 01, 2024
Author: Samer Al-Kaade, MD; Chief Editor: Romesh Khardori, MD, PhD, FACP more...;)
23
Share
Print
Feedback
Close
Facebook
Twitter
LinkedIn
WhatsApp
Email
Sections
Exocrine Pancreatic Insufficiency
Sections Exocrine Pancreatic Insufficiency
Overview
Practice Essentials
Anatomy
Pathophysiology
Etiology
Epidemiology and Prognosis
Show All
Presentation
History
Physical Examination
Show All
DDx
Workup
Laboratory Studies
Pancreatic Function Tests
Abdominal Imaging
Show All
Treatment
Approach Considerations
Pancreatic Enzyme Replacement Therapy
Show All
Guidelines
Medication
Medication Summary
Pancreatic Enzyme Products
Show All
Media Gallery;)
Tables)
References;)
Overview
Practice Essentials
Exocrine pancreatic insufficiency (EPI) is a condition characterized by deficiency of exocrine pancreatic enzymes, resulting in the inability to digest food properly, or maldigestion. The etiology of EPI includes pancreatic and nonpancreatic causes (see Etiology). [1, 2]
The exocrine pancreas produces three main types of enzymes: amylase, protease, and lipase, which are responsible for the digestion of carbohydrates, protein, and fat. Under normal physiologic conditions, lipase breaks undigested triglycerides into fatty acids and monoglycerides, which are then solubilized by bile salts (see Pathophysiology). Because the exocrine pancreas retains a large reserve capacity for enzyme secretion, fat digestion is not clearly impaired until lipase output decreases to below 10% of the normal level.
The diagnosis of EPI is largely clinical. It may go undetected because the signs and symptoms are similar to those of other gastrointestinal (GI) diseases or because signs and symptoms are not always evident, due to dietary restrictions (see Presentation and Differential Diagnosis).
Signs of exocrine pancreatic insufficiency
Major symptoms of EPI include steatorrhea (oily stool) and weight loss. Although steatorrhea is the most common symptomatic complaint, sometimes the stool can be watery, reflecting the osmotic load received by the intestine.
Workup in exocrine pancreatic insufficiency
Blood tests
These can include the following:
Complete blood count (CBC)
Antigliadin and antiendomysial antibodies - To rule out celiac disease
Stool tests
Determination of fecal elastase (a protease produced by the pancreas) can be used to support the diagnosis of EPI.
Malabsorption tests
These can include the following:
Fat absorption tests
D-xylose test
Carbohydrate absorption test
Bile salt absorption test
Schilling test
13C-D-xylose breath test
Pancreatic function tests
These can include the following:
Direct testing - Secretin test, cholecystokinin (CCK) test, secretin-CCK test
Indirect testing - Qualitative fecal fat analysis, fecal elastase level analysis
Abdominal imaging
Abdominal imaging can help in identifying features of chronic pancreatitis, which is the most common cause of EPI.
Management of exocrine pancreatic insufficiency
Management strategies for EPI include the following:
Lifestyle modifications - Eg, avoidance of fatty foods, limitation of alcohol intake, cessation of smoking, and consumption of a well-balanced diet
Vitamin supplementation - Primarily the fat-soluble vitamins A, D, E, and K
Pancreatic enzyme replacement therapy (PERT)
PERT is the basis of treatment for EPI; the endpoints of treatment are normalization of gut absorption and correction of nutritional deficiencies. The typical indications for initiating PERT are progressive weight loss and steatorrhea.
Next: Anatomy
Anatomy
The pancreas, named for the Greek words pan (all) and kreas (flesh), is a soft, lobulated, retroperitoneal organ that is 12-15 cm long and roughly J-shaped (like a hockey stick). It lies transversely, though a bit obliquely, on the posterior abdominal wall behind the stomach, across the lumbar (L1-2) spine (see the image below). The pancreas is prismoid in shape and appears triangular in cut section, with superior, inferior, and anterior borders as well as anterosuperior, anteroinferior, and posterior surfaces.
Pancreas anatomy.
View Media Gallery)
The head of the pancreas lies in the duodenal C loop in front of the inferior vena cava (IVC) and the left renal vein (see the images below). The uncinate process is an extension of the lower (inferior) half of the head toward the left; it is of varying size and is wedged between the superior mesenteric vessels in front (the vein on the right and the artery on the left) and the aorta behind. The pancreatic head constitutes about 50% of the pancreatic parenchymal mass.
The duodenum and pancreas.
View Media Gallery)
The pancreas and duodenum, posterior view.
View Media Gallery)
The body and tail of the pancreas run obliquely upward to the left in front of the aorta and the left kidney. The pancreatic neck is the arbitrary junction between the head and body of the pancreas. The narrow tip of the tail of the pancreas reaches the splenic hilum in the splenorenal (lienorenal) ligament. The body and tail make up the remaining 50% of the pancreatic parenchymal mass.
The transverse mesocolon (with the middle colic vessels in it) is attached to the anterior surface of the lower (inferior) surface of the body and tail; thus, most of the gland is located in the supracolic compartment. The body and tail of the pancreas lie in the lesser sac (omental bursa) behind the stomach.
Previous
Next: Anatomy
Pathophysiology
The GI tract is responsible for digesting and absorbing food. Lipids provide the richest source of energy for the body, with 9 calories in every gram of fat; in comparison, carbohydrate and protein contains 4 calories per gram. Whereas protein and carbohydrate begin to undergo digestion in the stomach, triglycerides remain mostly unchanged until they reach the small intestine. Intragastric breakdown accounts for approximately 10% of total lipid digestion.
The pancreatic enzymes responsible for lipid digestion are inactivated when the pH drops below 5; thus, before digestion can continue in the duodenum, the acidic contents of the stomach must be neutralized. Fortunately, the pancreas also secretes bicarbonate, which increases the pH of the duodenal contents.
The exocrine pancreas produces 3 main types of enzymes: amylase, protease, and lipase. Under normal physiologic conditions, lipase breaks the undigested triglycerides into fatty acids and monoglycerides. Bile salts then solubilize these breakdown products to form micelles, which are vehicles for absorbing lipid breakdown products. Normal fat digestion also depends on postprandial synchrony between delivery of nutrients to the duodenum and discharge of pancreatic enzymes.
Pancreatic secretion is governed by neural and hormonal mechanisms. The hormones responsible for regulation are secretin and cholecystokinin (CCK). Secretin is secreted in response to acid in the duodenum, causing duct cells to release water and bicarbonate; CCK is secreted in response to protein and fat in the small intestine, stimulating acinar cells to release the pancreatic enzymes (see the image below).
Factors controlling release of pancreatic secretions. Image courtesy of Wikimedia Commons.
View Media Gallery)
EPI is characterized by a deficiency of these exocrine pancreatic enzymes, which results in inability to digest food properly (ie, maldigestion). Because pancreatic lipase accounts for up to 90% of fat digestion, maldigestion of fat is more profound in EPI than maldigestion of proteins and carbohydrates is. Because the exocrine pancreas retains a large reserve capacity for enzyme secretion, fat digestion is not clearly impaired until lipase output decreases to below 10% of the normal level.
Fat malabsorption precedes malabsorption of other macronutrients. Bile salt precipitation and subsequent adsorption to undigested food reduces the bile salt pool, and this reduction further impairs fat digestion. Undigested fat, rather than being absorbed, is excreted in the feces, leading to steatorrhea.
Another factor that contributes to pancreatic steatorrhea is the presence of neurohormonal disturbances, which result in gall bladder hypomotility and accelerated gastric and intestinal transit. Malabsorption of fat-soluble vitamins A, D, E, and K may accompany EPI.
Previous
Next: Anatomy
Etiology
The etiology of EPI can be classified into pancreatic and nonpancreatic causes. [12, 13]
Pancreatic causes
These include the following:
Chronic pancreatitis (the most common cause of EPI) - This condition has a number of possible causes, but the end result is a metabolic insult to the pancreatic exocrine cells that leads to necrosis, fibrosis and loss of function (see the image below).
Residual islets in dense fibrous stroma secondary to loss of exocrine pancreatic tissue in chronic pancreatitis (hematoxylin-eosin stain, medium magnification). Image courtesy of Dr. Rose Anton.
View Media Gallery)
Acute pancreatitis - A literature review by Huang et al found that during first admission for acute pancreatitis (ie, the time between the commencement of oral refeeding and discharge) in 370 patients, exocrine pancreatic insufficiency (EPI) had a pooled prevalence of 62%. At follow-up (1 month or more following discharge for a first attack of acute pancreatitis) in 1795 patients, the pooled prevalence of EPI was 35%. The investigators also found at follow-up that compared with mild acute pancreatitis, the pooled prevalence for EPI was twice as great for severe acute pancreatitis (21% vs 42%, respectively).
Cystic fibrosis - In this condition, reduced chloride transport in the pancreas leads to reduced water content of secretions, precipitation of proteins, and plugging of ductules and acini, preventing the pancreatic enzymes from reaching the gut; autodigestion of the pancreas occasionally leads to pancreatitis.
Obstructions of the pancreatic duct (eg, from pancreatic cancer or ampullary tumors) - These hinder pancreatic exocrine secretions from reaching the gut.
Shwachman-Diamond syndrome (SDS) - This is a rare autosomal recessive disorder characterized by EPI, bone marrow dysfunction, leukemia predisposition, and skeletal abnormalities. [15, 16]
Diabetes - A study by Yatchenko et al suggested that EPI in type 2 diabetes mellitus derives from the effects of high insulin levels on pancreatic acinar cells. The investigators found evidence that high insulin concentrations impact naïve acinar cells via the activating transcription factor 6 (ATF6) and inositol-requiring enzyme 1 (IRE1) pathways, leading to activation of the endoplasmic reticulum–stress unfolded protein response (UPR).
Cystic fibrosis is caused by defects in the cystic fibrosis gene, which codes for a protein transmembrane conductance regulator (CFTR) that functions as a chloride channel (see the image below) and is regulated by cyclic adenosine monophosphate (cAMP). Mutations in the CFTR gene result in abnormalities of cAMP-regulated chloride transport across epithelial cells on mucosal surfaces.
Defective protein transmembrane conductance regulator (CFTR) in cystic fibrosis.
View Media Gallery)
Nonpancreatic causes
Celiac disease
Celiac disease (secondary to decreased pancreatic stimulation) leads to EPI in about one third of patients and may be an unrecognized cause of treatment failure.
Crohn disease
Crohn disease is associated with pancreatic autoantibodies that lead to impaired pancreatic exocrine function.
Autoimmune pancreatitis
Autoimmune pancreatitis is often caused by immunoglobulin G4 (IgG4)-related disease and can progress to EPI.
Zollinger-Ellison syndrome
Zollinger-Ellison syndrome can produce EPI through acid inactivation of pancreatic enzymes; it is corrected by controlling the acid secretion
GI and pancreatic surgical procedures
Any such procedures that lead to loss of postprandial synchrony, decreased pancreatic stimulation, and loss of pancreatic parenchyma can cause EPI. A study by Huddy et al found that EPI contributes to postoperative morbidity in patients who undergo esophagectomy. Using multivariate analysis, a study by Dhar et al indicated that in patients with chronic pancreatitis who undergo duodenum-sparing head resection or pancreaticoduodenectomy to relieve abdominal pain, EPI and narcotic requirement are the only predictors of the need for revision surgery.
A study by Okano et al suggested that following pancreatectomy, a remnant pancreatic volume of under 24 mL is the only independent predictor of postoperative EPI. The study included 227 patients.
However, a study by Hallac et al indicated that in patients who undergo distal pancreatectomy, the chance of developing de-novo exocrine pancreatic insufficiency (EPI) is greater in those with an underlying obstructive pancreatic pathology and in individuals who present with a history of acute pancreatitis. The investigators found that new-onset EPI arose in 38 of 324 patients (11.7%), while EPI existed preoperatively in 22 (6.8%) of patients, suggesting that patients set to undergo pancreatectomy may not uncommonly have preexisting EPI.
Previous
Next: Anatomy
Epidemiology and Prognosis
Because EPI has multiple possible causes and is not usually recorded as a medical statistic, its prevalence and demographics cannot be established with certainty at present. In a German-based study, one of the most common causes of EPI had an age-adjusted prevalence of 8 per 100,000 for males and 2 per 100,000 for women; these numbers are probably relatively close to the prevalence of EPI in most developed countries. No other reliable data are currently available.
The natural history and progression of EPI depend on the underlying etiology. For example, patients with autoimmune pancreatitis or cystic fibrosis may progress to almost complete insufficiency, whereas those with alcohol-induced EPI may recover from or at least halt the progression of pancreatic insufficiency if they abstain from alcohol. Even with complete loss of exocrine function, however, protease and lipase supplements are effective in restoring normal digestion of dietary nutrients.
A prospective, longitudinal cohort study by de la Iglesia et al indicated that in patients with chronic pancreatitis, EPI is an independent risk factor for cardiovascular events, with the incidence rate ratio for such events in patients with EPI compared with those without being 3.67. In addition, the odds ratio for cardiovascular events in individuals with a combination of EPI and diabetes mellitus was higher than for EPI patients without diabetes.
A study by Vujasinovic et al indicated that in patients with chronic pancreatitis, EPI is an independent risk factor for nephrolithiasis. The investigators found that the adjusted hazard ratio (aHR) for kidney stones in chronic pancreatitis patients with EPI is 4.95. Male sex and an increase in body mass index (BMI) were also determined to be risk factors for stone formation, the aHRs being 4.51 and 1.16, respectively.
Previous
Clinical Presentation
References
Struyvenberg MR, Martin CR, Freedman SD. Practical guide to exocrine pancreatic insufficiency - breaking the myths. BMC Med. 2017 Feb 10. 15 (1):29. [QxMD MEDLINE Link]. [Full Text].
Perbtani Y, Forsmark CE. Update on the diagnosis and management of exocrine pancreatic insufficiency. F1000Res. 2019. 8:[QxMD MEDLINE Link]. [Full Text].
Keller J, Layer P. Human pancreatic exocrine response to nutrients in health and disease. Gut. 2005 Jul. 54 Suppl 6:vi1-28. [QxMD MEDLINE Link]. [Full Text].
DiMagno EP, Go VL, Summerskill WH. Relations between pancreatic enzyme ouputs and malabsorption in severe pancreatic insufficiency. N Engl J Med. 1973 Apr 19. 288(16):813-5. [QxMD MEDLINE Link].
Fieker A, Philpott J, Armand M. Enzyme replacement therapy for pancreatic insufficiency: present and future. Clin Exp Gastroenterol. 2011. 4:55-73. [QxMD MEDLINE Link]. [Full Text].
Leeds JS, Oppong K, Sanders DS. The role of fecal elastase-1 in detecting exocrine pancreatic disease. Nat Rev Gastroenterol Hepatol. 2011 May 31. 8(7):405-15. [QxMD MEDLINE Link].
Shandro BM, Nagarajah R, Poullis A. Challenges in the management of pancreatic exocrine insufficiency. World J Gastrointest Pharmacol Ther. 2018 Oct 25. 9 (5):39-46. [QxMD MEDLINE Link]. [Full Text].
Ferrone M, Raimondo M, Scolapio JS. Pancreatic enzyme pharmacotherapy. Pharmacotherapy. 2007 Jun. 27(6):910-20. [QxMD MEDLINE Link].
Krishnamurty DM, Rabiee A, Jagannath SB, Andersen DK. Delayed release pancrelipase for treatment of pancreatic exocrine insufficiency associated with chronic pancreatitis. Ther Clin Risk Manag. 2009 Jun. 5(3):507-20. [QxMD MEDLINE Link]. [Full Text].
Bruno MJ, Rauws EA, Hoek FJ, Tytgat GN. Comparative effects of adjuvant cimetidine and omeprazole during pancreatic enzyme replacement therapy. Dig Dis Sci. 1994 May. 39(5):988-92. [QxMD MEDLINE Link].
Mizushima T, Ochi K, Ichimura M, Kiura K, Harada H, Koide N. Pancreatic enzyme supplement improves dysmotility in chronic pancreatitis patients. J Gastroenterol Hepatol. 2004 Sep. 19(9):1005-9. [QxMD MEDLINE Link].
Dominguez-Munoz JE. Pancreatic enzyme therapy for pancreatic exocrine insufficiency. Curr Gastroenterol Rep. 2007 Apr. 9(2):116-22. [QxMD MEDLINE Link].
Report of the expert committee on the diagnosis and classification of diabetes mellitus. Diabetes Care. 2003 Jan. 26 Suppl 1:S5-20. [QxMD MEDLINE Link].
Huang W, de la Iglesia-Garcia D, Baston-Rey I, et al. Exocrine Pancreatic Insufficiency Following Acute Pancreatitis: Systematic Review and Meta-Analysis. Dig Dis Sci. 2019 Jul. 64 (7):1985-2005. [QxMD MEDLINE Link]. [Full Text].
Sanklecha M, Balani K. Chronic pancreatic insufficiency-think of Shwachmann Diamond Syndrome. Indian Pediatr. 2012 May. 49(5):417-8. [QxMD MEDLINE Link].
Nelson A, Myers K, Adam MP, et al. Shwachman-Diamond Syndrome. 2008 Jul 17 [updated 2018 Oct 18]. [QxMD MEDLINE Link]. [Full Text].
Yatchenko Y, Horwitz A, Birk R. Endocrine and exocrine pancreas pathologies crosstalk: Insulin regulates the unfolded protein response in pancreatic exocrine acinar cells. Exp Cell Res. 2019 Jan 6. [QxMD MEDLINE Link].
Sah RP, Chari ST. Autoimmune pancreatitis: an update on classification, diagnosis, natural history and management. Curr Gastroenterol Rep. 2012 Apr. 14(2):95-105. [QxMD MEDLINE Link].
Ali S, T N, Gagloo M, Dhar S. Revisiting the problem of pancreatic exocrine insufficiency in surgical patients. Internet J Surg. 2012. 28(2):[Full Text].
Huddy JR, Macharg FM, Lawn AM, Preston SR. Exocrine pancreatic insufficiency following esophagectomy. Dis Esophagus. 2012 Nov 30. [QxMD MEDLINE Link].
Dhar VK, Levinsky NC, Xia BT, et al. The natural history of chronic pancreatitis after operative intervention: The need for revisional operation. Surgery. 2016 Jul 20. [QxMD MEDLINE Link].
Okano K, Murakami Y, Nakagawa N, et al. Remnant pancreatic parenchymal volume predicts postoperative pancreatic exocrine insufficiency after pancreatectomy. Surgery. 2016 Mar. 159 (3):885-92. [QxMD MEDLINE Link].
Hallac A, Aleassa EM, Rogers M, Falk GA, Morris-Stiff G. Exocrine pancreatic insufficiency in distal pancreatectomy: incidence and risk factors. HPB (Oxford). 2019 Jul 18. [QxMD MEDLINE Link].
Lowenfels AB, Sullivan T, Fiorianti J, Maisonneuve P. The epidemiology and impact of pancreatic diseases in the United States. Curr Gastroenterol Rep. 2005 May. 7(2):90-5. [QxMD MEDLINE Link].
de la Iglesia D, Vallejo-Senra N, Lopez-Lopez A, et al. Pancreatic exocrine insufficiency and cardiovascular risk in patients with chronic pancreatitis: a prospective, longitudinal cohort study. J Gastroenterol Hepatol. 2018 Aug 29. [QxMD MEDLINE Link].
Vujasinovic M, Dugic A, Thiel T, Kjellman A, Yang C, Lohr JM. Pancreatic exocrine insufficiency is a risk factor for kidney stones in patients with chronic pancreatitis. Pancreatology. 2023 Apr. 23 (3):294-8. [QxMD MEDLINE Link].
Lindkvist B, Dominguez-Munoz JE, Luaces-Regueira M, Castineiras-Alvarino M, Nieto-Garcia L, Iglesias-Garcia J. Serum nutritional markers for prediction of pancreatic exocrine insufficiency in chronic pancreatitis. Pancreatology. 2012 Jul-Aug. 12(4):305-10. [QxMD MEDLINE Link].
Ghoshal UC, Kumar S, Chourasia D, Misra A. Lactose hydrogen breath test versus lactose tolerance test in the tropics: does positive lactose tolerance test reflect more severe lactose malabsorption?. Trop Gastroenterol. 2009 Apr-Jun. 30(2):86-90. [QxMD MEDLINE Link].
Casterton PL, Verbeke KA, Brouns F, Dammann KW. Evaluation of sucromalt digestion in healthy children using breath hydrogen as a biomarker of carbohydrate malabsorption. Food Funct. 2012 Apr. 3(4):410-3. [QxMD MEDLINE Link].
Hope HB, Tveito K, Aase S, Messelt E, Utzon P, Skar V. Small intestinal malabsorption in chronic alcoholism determined by 13C-D-xylose breath test and microscopic examination of the duodenal mucosa. Scand J Gastroenterol. 2010. 45(1):39-45. [QxMD MEDLINE Link].
Chowdhury RS, Forsmark CE. Review article: Pancreatic function testing. Aliment Pharmacol Ther. 2003 Mar 15. 17(6):733-50. [QxMD MEDLINE Link].
Hammer HF. Pancreatic exocrine insufficiency: diagnostic evaluation and replacement therapy with pancreatic enzymes. Dig Dis. 2010. 28(2):339-43. [QxMD MEDLINE Link].
Gonzalez-Sanchez V, Amrani R, Gonzalez V, Trigo C, Pico A, de-Madaria E. Diagnosis of exocrine pancreatic insufficiency in chronic pancreatitis: 13C-Mixed Triglyceride Breath Test versus Fecal Elastase. Pancreatology. 2017 Mar 6. [QxMD MEDLINE Link].
Saad M, Vitale DS, Lin TK, et al. Image or scope: magnetic resonance imaging and endoscopic testing for exocrine and endocrine pancreatic insufficiency in children. Pancreatology. 2023 Apr 16. [QxMD MEDLINE Link].
Sankararaman S, Schindler T, Sferra TJ. Management of Exocrine Pancreatic Insufficiency in Children. Nutr Clin Pract. 2019 Oct. 34 Suppl 1:S27-S42. [QxMD MEDLINE Link]. [Full Text].
Fieker A, Philpott J, Armand M. Enzyme replacement therapy for pancreatic insufficiency: present and future. Clin Exp Gastroenterol. 2011. 4:55-73. [QxMD MEDLINE Link]. [Full Text].
Dominguez-Munoz JE. Pancreatic enzyme replacement therapy for pancreatic exocrine insufficiency: when is it indicated, what is the goal and how to do it?. Adv Med Sci. 2011. 56(1):1-5. [QxMD MEDLINE Link].
de la Iglesia-Garcia D, Huang W, Szatmary P, et al. Efficacy of pancreatic enzyme replacement therapy in chronic pancreatitis: systematic review and meta-analysis. Gut. 2016 Dec 9. [QxMD MEDLINE Link]. [Full Text].
Erchinger F, Ovre AKN, Aarseth MM, et al. Fecal fat and energy loss in pancreas exocrine insufficiency: the role of pancreas enzyme replacement therapy. Scand J Gastroenterol. 2018 Sep 7. 1-7. [QxMD MEDLINE Link].
Lewis DM, Rieke JG, Almusaylim K, Kanchibhatla A, Blanchette JE, Lewis C. Exocrine Pancreatic Insufficiency Dosing Guidelines for Pancreatic Enzyme Replacement Therapy Vary Widely Across Disease Types. Dig Dis Sci. 2024 Feb. 69 (2):615-33. [QxMD MEDLINE Link].
Ultresa. Drugs.com. Available at Updated August 23, 2023; Accessed: February 22, 2024.
Whitcomb DC, Lehman GA, Vasileva G, et al. Pancrelipase delayed-release capsules (CREON) for exocrine pancreatic insufficiency due to chronic pancreatitis or pancreatic surgery: A double-blind randomized trial. Am J Gastroenterol. 2010 Oct. 105(10):2276-86. [QxMD MEDLINE Link].
Gubergrits N, Malecka-Panas E, Lehman GA, Vasileva G, Shen Y, Sander-Struckmeier S, et al. A 6-month, open-label clinical trial of pancrelipase delayed-release capsules (Creon) in patients with exocrine pancreatic insufficiency due to chronic pancreatitis or pancreatic surgery. Aliment Pharmacol Ther. 2011 May. 33(10):1152-61. [QxMD MEDLINE Link].
Toskes PP, Secci A, Thieroff-Ekerdt R. Efficacy of a novel pancreatic enzyme product, EUR-1008 (Zenpep), in patients with exocrine pancreatic insufficiency due to chronic pancreatitis. Pancreas. 2011 Apr. 40(3):376-82. [QxMD MEDLINE Link].
Trapnell BC, Strausbaugh SD, Woo MS, et al. Efficacy and safety of PANCREAZE® for treatment of exocrine pancreatic insufficiency due to cystic fibrosis. J Cyst Fibros. 2011 Sep. 10(5):350-6. [QxMD MEDLINE Link].
Jablonska B. Is endoscopic therapy the treatment of choice in all patients with chronic pancreatitis?. World J Gastroenterol. 2013 Jan 7. 19(1):12-6. [QxMD MEDLINE Link]. [Full Text].
Borowitz D, Stevens C, Brettman LR, Campion M, Wilschanski M, Thompson H. Liprotamase long-term safety and support of nutritional status in pancreatic-insufficient cystic fibrosis. J Pediatr Gastroenterol Nutr. 2012 Feb. 54(2):248-57. [QxMD MEDLINE Link].
Phillips ME, Hopper AD, Leeds JS, et al. Consensus for the management of pancreatic exocrine insufficiency: UK practical guidelines. BMJ Open Gastroenterol. 2021 Jun. 8 (1):[QxMD MEDLINE Link].
[Guideline] Whitcomb DC, Buchner AM, Forsmark CE. AGA Clinical Practice Update on the Epidemiology, Evaluation, and Management of Exocrine Pancreatic Insufficiency: Expert Review. Gastroenterology. 2023 Nov. 165 (5):1292-301. [QxMD MEDLINE Link]. [Full Text].
Karnik NP, Jan A. Pancrelipase. 2020 Jan. [QxMD MEDLINE Link]. [Full Text].
Venkatesh P, Kasi A. Pancrelipase Therapy. 2020 Jan. [QxMD MEDLINE Link]. [Full Text].
Media Gallery
Complete replacement of the pancreas with cystic disease and intrahepatic biliary dilation caused by extrinsic compression of the common bile duct. Note also the renal cysts and masses. This patient had exocrine pancreatic insufficiency. Image courtesy of Wikimedia Commons.
Residual islets in dense fibrous stroma secondary to loss of exocrine pancreatic tissue in chronic pancreatitis (hematoxylin-eosin stain, medium magnification). Image courtesy of Dr. Rose Anton.
Pancreas anatomy.
The duodenum and pancreas.
The pancreas and duodenum, posterior view.
Factors controlling release of pancreatic secretions. Image courtesy of Wikimedia Commons.
Defective protein transmembrane conductance regulator (CFTR) in cystic fibrosis.
of 7
Tables
Table 1. Lipase Content of Currently Available Pancreatic Enzyme Products;)
Table 1. Lipase Content of Currently Available Pancreatic Enzyme Products
| | |
--- |
| Pancreatic Enzyme Product | Lipase Content, units |
| Creon 1203 | 3000 |
| Creon 1206 | 6000 |
| Creon 1212 | 12,000 |
| Creon 1224 | 24,000 |
| Zenpep EURAND 3 | 3000 |
| Zenpep EURAND 5 | 5000 |
| Zenpep EURAND 10 | 10,000 |
| Zenpep EURAND 15 | 15,000 |
| Zenpep EURAND 20 | 20,000 |
| Zenpep EURAND 25 | 25,000 |
| Pancreaze MT 4 | 4200 |
| Pancreaze MT 10 | 10,500 |
| Pancreaze MT 16 | 16,800 |
| Pancrease MT 24 | 21,000 |
| Viokace 9111 | 10,440 |
| Viokace 9116 | |
| Pertzye 8 | |
| Pertzye 16 | |
Back to List
encoded search term (Exocrine Pancreatic Insufficiency) and Exocrine Pancreatic Insufficiency
What to Read Next on Medscape |
189268 | https://en.wikipedia.org/wiki/John_Playfair | John Playfair - Wikipedia
Jump to content
[x] Main menu
Main menu
move to sidebar hide
Navigation
Main page
Contents
Current events
Random article
About Wikipedia
Contact us
Contribute
Help
Learn to edit
Community portal
Recent changes
Upload file
Special pages
Search
Search
[x] Appearance
Donate
Create account
Log in
[x] Personal tools
Donate
Create account
Log in
Pages for logged out editors learn more
Contributions
Talk
Contents
move to sidebar hide
(Top)
1 Life
2 Family
3 Honours
4 Works
5 Critical bibliography
6 References
7 External links
[x] Toggle the table of contents
John Playfair
[x] 27 languages
العربية
تۆرکجه
Български
Català
Deutsch
Español
فارسی
Français
Galego
Italiano
ქართული
مصرى
Nederlands
日本語
Norsk bokmål
Norsk nynorsk
Piemontèis
Polski
Português
Română
Русский
Simple English
Suomi
Svenska
Türkçe
Українська
中文
Edit links
Article
Talk
[x] English
Read
Edit
View history
[x] Tools
Tools
move to sidebar hide
Actions
Read
Edit
View history
General
What links here
Related changes
Upload file
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Print/export
Download as PDF
Printable version
In other projects
Wikimedia Commons
Wikiquote
Wikisource
Wikidata item
Appearance
move to sidebar hide
From Wikipedia, the free encyclopedia
Scottish minister and scientist (1748–1819)
| John Playfair FRS, FRSE |
| |
| Born | (1748-03-10)10 March 1748[citation needed] Benvie, Forfarshire, Scotland |
| Died | 20 July 1819(1819-07-20) (aged 71) Burntisland, Fife, Scotland |
| Resting place | Old Calton Burial Ground |
| Alma mater | University of St Andrews University of Edinburgh |
| Known for | Playfair's axiom Playfair's law Playfair (lunar crater) Playfair (Martian crater) playfairite |
| Scientific career |
| Fields | Mathematics, natural philosophy, geology |
| Institutions | University of Edinburgh |
| |
John PlayfairFRSE, FRS (10[citation needed] March 1748 – 20 July 1819) was a Church of Scotland minister, remembered as a scientist and mathematician, and a professor of natural philosophy at the University of Edinburgh. He is best known for his book Illustrations of the Huttonian Theory of the Earth (1802), which summarised the work of James Hutton. It was through this book that Hutton's principle of uniformitarianism, later taken up by Charles Lyell, first reached a wide audience. Playfair's textbook Elements of Geometry made a brief expression of Euclid's parallel postulate known now as Playfair's axiom.
In 1783 he was a co-founder of the Royal Society of Edinburgh. He served as General Secretary to the society 1798–1819.
Life
[edit]
Born at Benvie, slightly west of Dundee to Margaret Young (1719/20 – 1805) and Reverend James Playfair (died 1772), the kirk minister of Liff and Benvie.
Playfair was educated at home until the age of 14, when he entered the University of St Andrews to study divinity. He also did further studies at Edinburgh University. In 1766, when only 18, he was a candidate for the chair of mathematics in Marischal College (now part of the University of Aberdeen), and, although he was unsuccessful, his claims were admitted to be high.
Six years later (1772) he applied for the chair of natural philosophy (physics) at St Andrews University, but again without success. In 1773 he was licensed to preach by the Church of Scotland and was offered the united parishes of Liff and his home parish of Benvie (made vacant by the death of his father). However, Playfair chose to continue his studies in mathematics and physics, and in 1782 he resigned his charge to become the tutor of Adam Ferguson. By this arrangement Playfair regularly visited Edinburgh and went on to cultivate the literary and scientific society for which the city was at that time specially distinguished. In particular, he attended the natural history course of John Walker. Through Nevil Maskelyne, whose acquaintance he had first made in the course of the celebrated Schiehallion experiments in 1774, he also gained access to the scientific circles of London. In 1785 when Dugald Stewart succeeded Ferguson in the University of Edinburgh Chair of Moral Philosophy, Playfair succeeded the former to become the chair of mathematics.
Sir John Playfair by Sir Francis Chantrey
In 1795 Playfair published an alternative, more stringent formulation of Euclid's parallel postulate, which is now called Playfair's axiom. Although the axiom bears Playfair's name, he did not create it, but credited others, in particular William Ludlam with its prior use.
In 1802 Playfair published his celebrated volume entitled Illustrations of the Huttonian Theory of the Earth. The influence exerted by James Hutton on the development of geology is thought to be largely due to its publication. In 1805 Playfair exchanged the Chair of Mathematics for that of natural philosophy in succession to John Robison, whom also he succeeded as general secretary to the Royal Society of Edinburgh. He took a prominent part, on the liberal side, in the ecclesiastical controversy that arose in connection with Sir John Leslie's appointment to the post he had vacated, and published a satirical letter (1806).
He moved from 6 Buccleuch Place to a new house at 2 Albany Street (then called Albany Row) in 1807.
Playfair was an opponent of Gottfried Leibniz's vis viva principle, an early version of the conservation of energy. In 1808, he launched an attack on John Smeaton and William Hyde Wollaston's work championing the theory. In 1808 he also published a review of Laplace's Traité de Mécanique Celeste.
He died at 2 Albany Street on 20 July 1819. He is buried nearby in Old Calton Burial Ground (a secular burial ground).
Family
[edit]
Memorial to John Playfair, Old Calton Burial Ground, Edinburgh
Playfair's brothers were architect James Playfair, solicitor Robert Playfair and engineer William Playfair. His nephew, William Henry Playfair (1790–1857) was an eminent architect in Scotland.
In later life he admired and proposed to the wealthy widow Jane Apreece. She turned him down and married Sir Humphry Davy.
He died of strangury on 20 July 1819, and, although an eminent man, was buried in an unmarked grave in Old Calton Burial Ground, on Waterloo Place in Edinburgh. His, and his brother, James's graves were marked by a plaque unveiled in 2011 following a local campaign. The monument to his memory by William Henry Playfair, on Calton Hill, is visible from the spot.
Honours
[edit]
Monument to John Playfair on Calton Hill, Edinburgh
Fellow of the Royal Society of Edinburgh
Fellow of the Royal Society of London, 1807
Craters on Mars and the Moon were named in his honour.
The mineral playfairite was named in his honour.
Explication de Playfair sur la Théorie de la Terre, 1815
Works
[edit]
Illustrations of the Huttonian theory of the Earth (in French). London: Pierre François Jean Baptiste Leblanc. 1815.
Critical bibliography
[edit]
A four-volume collected edition of Playfair's works, with a memoir by James G. Playfair, appeared at Edinburgh in 1822.
His writings include a number of essays contributed to the Edinburgh Review from 1804 onwards, various papers in the Philosophical Transactions of the Royal Society (including his earliest publication, "On the Arithmetic of Impossible Quantities", 1779, and an "Account of the Lithological Survey of Schehallion", 1811) and in the Transactions of the Royal Society of Edinburgh ("On the Causes which Affect the Accuracy of Barometrical Measurements" and others), the articles "Aepinus" and "Physical Astronomy", and a "Dissertation on the Progress of Mathematical and Physical Science since the Revival of Learning in Europe" in the Encyclopædia Britannica (Supplement to fourth, fifth and sixth editions). He also took an interest in Indian astronomy and compared them with traditional and ancient astronomy from Egypt and Greece. He also examined Indian concepts in trigonometry.
His Elements of Geometry first appeared in 1795 and has passed through many editions; his Outlines of Natural Philosophy (2 vols., 1812–1816) consist of the propositions and formulae which were the basis of his class lectures. Playfair's contributions to pure mathematics were not considerable, his papers "On the Arithmetic of Impossible Quantities" and "On the Causes which Affect the Accuracy of Barometrical Measurements", and his Elements of Geometry, all already referred to, being the most important. His lives of Matthew Stewart, Hutton, and Robison, many of his reviews, and above all his "Dissertation" are of the utmost value.
References
[edit]
^Playfair, John (1802). Illustration of the Huttonian Theory. Edinburgh: Cadell & Davies – via Internet Archive.
^Biographical Index of Former Fellows of the Royal Society of Edinburgh 1783–2002(PDF). The Royal Society of Edinburgh. July 2006. ISBN0-902-198-84-X. Archived from the original(PDF) on 4 March 2016. Retrieved 8 January 2018.
^C D Waterston; A Macmillan Shearer (July 2006). Former Fellows of The Royal Society of Edinburgh, 1783–2002: Part 2 (K–Z)(PDF). Royal Society of Edinburgh. ISBN090219884X. Retrieved 13 February 2022.
^Morrell, Jack (2004). Playfair, John (1748–1819), mathematician and geologist. Oxford: Oxford University Press.
^ abcdeChisholm 1911.
^J. Playfair and Euclid, Elements of geometry; containing the first six books of Euclid, with two books on the geometry of solids. To which are added, elements of plane and spherical trigonometry, J.B. Lippincott & Co, 1860, p. 291. Available online from Google Books. See also Cajori's A History of Mathematics.
^Edinburgh Post Office Directories 1805 to 1810
^grant's Old and New Edinburgh vol III
^Edinburgh Review, 12, 1808, 120–130
^Smith, Sydney. "Review of Traité de Mécanique Céleste par P. S. Laplace". Edinburgh Review. 11 (22): 249–284.
^Grant's Old and New Edinburgh vol III
^Sophie Forgan, 'Davy, Jane, Lady Davy (1780–1855)', Oxford Dictionary of National Biography, Oxford University Press, 2004; online edn, May 2008 accessed 17 Dec 2014
^"Edinburgh World Heritage - News Article". Archived from the original on 29 August 2013. Retrieved 2 June 2013.
^Biographical Dictionary of Eminent Scotsmen (1856), reproduced in Significant Scots
^Information about Playfairite on Mindat database, Retrieved on 19 April 2012.
^J.L. Jambor (1967) New lead sulfantimonides from Madoc, Ontario; Part 2, Mineral descriptions, Canadian Mineralogist, vol. 9, 194–6
^See the Collected Works of John Playfair on the Internet Archive (www.archive.org)
^Playfair, John (1790). "XIII. Remarks on the Astronomy of the Brahmins". Earth and Environmental Science Transactions of the Royal Society of Edinburgh. 2 (2): 135–192. doi:10.1017/s0263593300027322. ISSN2053-5945. S2CID251576277.
^Burgess, James (1893). "Art. XVIII.—Notes on Hindu Astronomy and the History of our Knowledge of it". Journal of the Royal Asiatic Society. 25 (4): 717–761. doi:10.1017/s0035869x00022553. ISSN1474-0591. S2CID163252270.
External links
[edit]
Wikiquote has quotations related to John Playfair.
Wikisource has the text of the 1885–1900 Dictionary of National Biography's article about Playfair, John.
Dictionary of Scientific Biography
O'Connor, John J.; Robertson, Edmund F., "John Playfair", MacTutor History of Mathematics Archive, University of St Andrews
Significant Scots: John Playfair
National Portrait Gallery
John Playfair (1836) Elements of Geometry from Google books, see page 22 for parallel axiom.
The works of John Playfair. Edinburgh: A. Constable & Co. 1822. (from HathiTrust)
From the Linda Hall Library
Playfair's (1802) Illustrations of the Huttonian Theory of the Earth
Playfair's (1815) Explication de Playfair sur la théorie de la terre par Hutton
This article incorporates text from a publication now in the public domain:Chisholm, Hugh, ed. (1911). "Playfair, John". Encyclopædia Britannica. Vol.21 (11th ed.). Cambridge University Press. p.831.
| v t e Age of Enlightenment |
| | Topics | | Atheism Capitalism Civil liberties Classicism Counter-Enlightenment Critical thinking Deism Democracy Empiricism Encyclopédistes Enlightened absolutism Enlightenment philosophy Haskalah Humanism Human rights Individualism Liberalism Liberté, égalité, fraternité Methodological skepticism Midlands Modernity Natural philosophy Objectivity Progressivism Rationality Rationalism Reason Reductionism Sapere aude Science Scientific method Spanish America Universality Utopianism | |
| | Thinkers | | | Austria | Sonnenfels Zinzendorf Swieten | | England | Addison Ashley-Cooper Bacon Bentham Collins Gibbon Godwin Harrington Hooke Johnson Locke Milton Newton Pope Price Priestley Reynolds Sidney Tindal Wollstonecraft | | France | d'Alembert d'Argenson Bayle Beaumarchais Chamfort Châtelet Condillac Condorcet Descartes Diderot Fontenelle Gouges Helvétius d'Holbach Jaucourt La Mettrie Lavoisier Leclerc Mably Maréchal Meslier Montesquieu Morelly Pascal Quesnay Raynal Sade Turgot Voltaire | | Geneva | Abauzit Bonnet Burlamaqui Prévost Rousseau Saussure | | Germany | Goethe Herder Humboldt Kant Leibniz Lessing Lichtenberg Mendelssohn Pufendorf Schiller Thomasius Weishaupt Wieland Wolff | | Greece | Farmakidis Feraios Kairis Korais | | Ireland | Berkeley Boyle Burke Swift Toland | | Italy | Beccaria Galiani Galvani Genovesi Pagano Verri Vico Volta | | Netherlands | Bekker de la Court Grotius Huygens Koerbagh Leeuwenhoek Mandeville Meyer Nieuwentyt Spinoza Swammerdam | | Poland | Kołłątaj Konarski Krasicki Niemcewicz Poniatowski Śniadecki Staszic Wybicki | | Portugal | Carvalho e Melo | | Romania | Budai-Deleanu Cantemir Golescu Maior Micu-Klein Pann Șincai Văcărescu | | Russia | Catherine II Fonvizin Kantemir Kheraskov Lomonosov Novikov Radishchev Vorontsova-Dashkova | | Scotland | Beattie Black Blair Boswell Burnett Burns Cullen Ferguson Hume Hutcheson Hutton Mill Playfair Reid Smith Stewart | | Serbia | Obradović Mrazović | | Spain | Cadalso Charles III Feijóo y Montenegro Moratín Jovellanos Villarroel | | United States | Franklin Jefferson Madison Mason Paine | | |
| Romanticism → Category |
| Authority control databases |
| International | ISNI VIAF GND FAST WorldCat |
| National | United States France BnF data Italy Australia Portugal Netherlands Norway Greece Korea Poland Vatican Israel Belgium |
| Academics | CiNii Mathematics Genealogy Project zbMATH MathSciNet |
| People | Trove Deutsche Biographie DDB |
| Other | IdRef SNAC Yale LUX |
Retrieved from "
Categories:
1748 births
1819 deaths
Clergy from Dundee
Scientists from Dundee
Alumni of the University of St Andrews
Academics of the University of Edinburgh
Founder fellows of the Royal Society of Edinburgh
Fellows of the Royal Society
People of the Scottish Enlightenment
Scottish physicists
Scottish geologists
Scottish scholars and academics
Burials at Old Calton Burial Ground
18th-century ministers of the Church of Scotland
18th-century Scottish Presbyterian ministers
18th-century Scottish mathematicians
19th-century British mathematicians
19th-century ministers of the Church of Scotland
19th-century Scottish Presbyterian ministers
Fellows of the Royal Society of Edinburgh
Hidden categories:
Articles with short description
Short description is different from Wikidata
Use British English from August 2014
All Wikipedia articles written in British English
Use dmy dates from August 2014
All articles with unsourced statements
Articles with unsourced statements from January 2022
Articles with hCards
CS1 French-language sources (fr)
Wikipedia articles incorporating a citation from the 1911 Encyclopaedia Britannica with Wikisource reference
Wikipedia articles incorporating text from the 1911 Encyclopædia Britannica
Articles containing French-language text
Articles containing Hebrew-language text
Articles containing Latin-language text
This page was last edited on 31 May 2025, at 01:29(UTC).
Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization.
Privacy policy
About Wikipedia
Disclaimers
Contact Wikipedia
Code of Conduct
Developers
Statistics
Cookie statement
Mobile view
Search
Search
[x] Toggle the table of contents
John Playfair
27 languagesAdd topic |
189269 | https://webspace.maths.qmul.ac.uk/p.j.cameron/class_gps/ch3.pdf | 3 Polarities and forms 3.1 Sesquilinear forms We saw in Chapter 1 that the projective space PG n 1 F is isomorphic to its dual if and only if the field F is isomorphic to its opposite. More precisely, we have the following. Let σ be an anti-automorphism of F, and V an F-vector space of rank n. A sesquilinear form B on V is a function B : V V F which satisfies the following conditions: (a) B c1x1 c2x2 y c1B x1 y c2B x2 y , that is, B is a linear function of its first argument; (b) B x c1y1 c2y2 B x y1 cσ 1 B x y2 cσ 2, that is, B is a semilinear function of its second argument, with field anti-automorphism σ.
(The word ‘sesquilinear’ means ‘one-and-a-half’.) If σ is the identity (so that F is commutative), we say that B is a bilinear form.
The left radical of B is the subspace x V : y V B x 0 , and the right radical is the subspace y V : x V B x y 0 .
Exercise 3.1 (a) Prove that the left and right radicals are subspaces.
(b) Show that the left and right radicals have the same rank (if V has finite rank).
(c) Construct a bilinear form on a vector space of infinite rank such that the left radical is zero and the right radical is no-zero.
The sesquilinear form B is called non-degenerate if its left and right radicals are zero. (By the preceding exercise, it suffices to assume that one of the radicals is zero.) A non-degenerate sesquilinear form induces a duality of PG n 1 F (an iso-morphism from PG n 1 F to PG n 1 F ) as follows: for any y V, the map x B x y is a linear map from V to F, that is, an element of the dual space V (which is a left vector space of rank n over F ); if we call this element βy, then the map y βy is a σ-semilinear bijection from V to V , and so induces the required duality.
Theorem 3.1 For n 3, any duality of PG n 1 F is induced in this way by a non-degenerate sesquilinear form on V Fn.
27 Proof By the Fundamental Theorem of Projective Geometry, a duality is induced by a σ-semilinear bijection φ from V to V , for some anti-automorphism σ. Set B x y x yφ We can short-circuit the passage to the dual space, and write the duality as U U x V : B x y 0 for all y U Obviously, a duality applied twice is a collineation. The most important types of dualities are those whose square is the identity. A polarity of PG n F is a duality which satisfies U U for all flats U of PG n F .
It will turn out that polarities give rise to a class of geometries (the polar spaces) with properties similar to those of projective spaces, and define groups analogous to the projective groups. If a duality is not a polarity, then any collineation which respects it must commute with its square, which is a collineation; so the group we obtain will lie inside the centraliser of some element of the collineation group. So the “largest” subgroups obtained will be those preserving polarities.
A sesquilinear form B is reflexive if B x y 0 implies B y x 0.
Proposition 3.2 A duality is a polarity if and only if the sesquilinear form defining it is reflexive.
Proof B is reflexive if and only if x ! y " $# y ! x " . Hence, if B is reflexive, then U % U for all subspaces U. But by non-degeneracy, dimU & dimV dimU dimU; and so U U & for all U. Conversely, given a polarity , if y ! x " , then x ' x " & % y " (since inclusions are reversed).
We now turn to the classification of reflexive forms. For convenience, from now on F will always be assumed to be commutative. (Note that, if the anti-automorphism σ is an automorphism, and in particular if σ is the identity, then F is automatically commutative.) The form B is said to be σ-Hermitian if B y x B x y σ for all x y V. If B is a non-zero σ-Hermitian form, then (a) for any x, B x x lies in the fixed field of σ; (b) σ2 1. For every scalar c is a value of B, say B x y c; then cσ2 B x y σ2 B y x σ B x y c 28 If σ is the identity, such a form (which is bilinear) is called symmetric.
A bilinear form b is called alternating if B x x ( 0 for all x V. This implies that B x y ) B y x for all x y V. For 0 B x y x y B x x B x y B y x B y y ( B x y B y x Hence, if the characteristic is 2, then any alternating form is symmetric (but not conversely); but, in characteristic different from 2, only the zero form is both symmetric and alternating.
Clearly, an alternating or Hermitian form is reflexive. Conversely, we have the following: Theorem 3.3 A non-degenerate reflexive σ-sesquilinear form is either alternat-ing, or a scalar multiple of a σ-Hermitian form. In the latter case, if σ is the identity, then the scalar can be taken to be 1.
Proof I will give the proof just for a bilinear form. Thus, it must be proved that a non-degenerate reflexive bilinear form is either symmetric or alternating.
We have B u v B u w B u w B u v 0 by commutativity; that is, using bilinearity, B u B u v w B u w v ( 0 By reflexivity, B B u v w B u w v u & 0 whence bilinearity again gives B u v B w u ( B u w B v u (1) Call a vector u good if B u v B v u + 0 for some v. By Equation (1), if u is good, then B u w , B w u for all w. Also, if u is good and B u v - 0, then v is good. But, given any two non-zero vectors u1 u2, there exists v with B ui v . 0 for i 1 2. (For there exist v1 v2 with B ui vi + 0 for i 1 2, by non-degeneracy; and at least one of v1 v2 v1 v2 has the required property.) So, if some vector is good, then every non-zero vector is good, and B is symmetric.
But, putting u w in Equation (1) gives B u u B u v B v u / 0 for all u v. So, if u is not good, then B u u 0; and, if no vector is good, then B is alternating.
29 Exercise 3.2 (a) Show that the left and right radicals of a reflexive form are equal.
(b) Assuming Theorem 3.3, prove that the assumption of non-degeneracy in the theorem can be removed.
Exercise 3.3 Let σ be a (non-identity) automorphism of F of order 2. Let E be the subfield Fix σ .
(a) Prove that F is of degree 2 over E, i.e., a rank 2 E-vector space.
[See any textbook on Galois theory. Alternately, argue as follows: Take λ F 0 E. Then λ is quadratic over E, so E λ has degree 2 over E. Now E λ contains an element ω such that ωσ 1 ω (if the characteristic is not 2) or ωσ ω 1 (if the characteristic is 2). Now, given two such elements, their quotient or difference respectively is fixed by σ, so lies in E.] (b) Prove that λ F : λλσ 1 2) ε 3 εσ : ε F [The left-hand set clearly contains the right. For the reverse inclusion, separate into cases according as the characteristic is 2 or not.
If the characteristic is not 2, then we can take F E ω , where ω2 α E and ωσ 4 ω. If λ 1, then take ε 1; otherwise, if λ a bω, take ε bα a 1 ω.
If the characteristic is 2, show that we can take F E ω , where ω2 ω α 0, α E, and ωσ ω 1. Again, if λ 1, set ε 1; else, if λ a bω, take ε a 1 bω.] Exercise 3.4 Use the result of the preceding exercise to complete the proof of Theorem 3.3 in general.
[If B u u 0 for all u, the form B is alternating and bilinear. If not, suppose that B u u . 0 and let B u u σ λB u u . Choosing ε as in Exercise 3.3 and re-normalising B, show that we may assume that λ 1, and (with this choice) that B is Hermitian.] 3.2 Hermitian and quadratic forms We now change ground slightly from the last section. On the one hand, we restrict things by excluding some bilinear forms from the discussion; on the other, we 30 introduce quadratic forms. The loss and gain exactly balance if the characteristic is not 2; but, in characteristic 2, we make a net gain.
Let σ be an automorphism of the commutative field F, of order dividing 2. Let Fix σ 5 λ F : λσ λ be the fixed field of σ, and Tr σ 1 λ λσ : λ F the trace of σ. Since σ2 is the identity, it is clear that Fix σ 76 Tr σ . Moreover, if σ is the identity, then Fix σ F, and Tr σ 98 0 if F has characteristic 2, F otherwise.
Let B be a σ-Hermitian form. We observed in the last section that B x x : Fix σ for all x V. We call the form B trace-valued if B x x 2 Tr σ for all x V.
Exercise 3.5 Let σ be an automorphism of a commutative field F such that σ2 is the identity.
(a) Prove that Fix σ is a subfield of F.
(b) Prove that Tr σ is closed under addition, and under multiplication by ele-ments of Fix σ .
Proposition 3.4 Tr σ 7 Fix σ unless the characteristic of F is 2 and σ is the identity.
Proof E Fix σ is a field, and K Tr σ is an E-vector space contained in E (Exercise 3.5). So, if K E, then K 0, and σ is the map x x. But, since σ is a field automorphism, this implies that the characteristic is 2 and σ is the identity.
Thus, in characteristic 2, symmetric bilinear forms which are not alternating are not trace-valued; but this is the only obstruction. We introduce quadratic forms to repair this damage. But, of course, quadratic forms can be defined in any char-acteristic. However, we note at this point that Theorem 3.3 depends in a crucial way on the commutativity of F; this leaves open the possibility of additional types of polar spaces defined by so-called pseudoquadratic forms. We will not pursue this here: see Tits’s classification of spherical buildings.
Let V be a vector space over F. A quadratic form on V is a function q : V F satisfying 31 (a) q λx λ2 f x for all λ F, x V; (b) q x y q x q y B x y , where B is bilinear.
Now, if the characteristic of F is not 2, then B is a symmetric bilinear form.
Each of q and B determines the other, by B x y ; q x y q x q y q x ; 1 2B x x the latter equation coming from the substitution x y in (b). So nothing new is obtained.
On the other hand, if the characteristic of F is 2, then B is an alternating bi-linear form, and q cannot be recovered from B. Indeed, many different quadratic forms correspond to the same bilinear form. (Note that the quadratic form does give extra structure to the vector space; we’ll see that this structure is geometri-cally similar to that provided by an alternating or Hermitian form.) We say that the bilinear form B is obtained by polarisation of q.
Now let B be a symmetric bilinear form over a field of characteristic 2, which is not alternating. Set f x B x x . Then we have f λx < λ2 f x f x y < f x f y since B x y B y x & 0. Thus f is “almost” a semilinear form; the map λ λ2 is a homomorphism of the field F with kernel 0, but it may fail to be an automor-phism. But in any case, the kernel of f is a subspace of V, and the restriction of B to this subspace is an alternating bilinear form. So again, in the spirit of the vague comment motivating the study of polarities in the last section, the structure provided by the form B is not “primitive”. For this reason, we do not consider symmetric bilinear forms in characteristic 2 at all. However, as indicated above, we will consider quadratic forms in characteristic 2.
Now, in characteristic different from 2, we can take either quadratic forms or symmetric bilinear forms, since the structural content is the same. For consistency, we will take quadratic forms in this case too. This leaves us with three “types” of forms to study: alternating bilinear forms; σ-Hermitian forms where σ is not the identity; and quadratic forms.
We have to define the analogue of non-degeneracy for quadratic forms. Of course, we could require that the bilinear form obtained by polarisation is non-32 degenerate; but this is too restrictive. We say that a quadratic form q is non-degenerate if q x 0 & y V B x y 0 # x 0 where B is the associated bilinear form; that is, if the form q is non-zero on every non-zero vector of the radical.
If the characteristic is not 2, then non-degeneracy of the quadratic form and of the bilinear form are equivalent conditions.
Now suppose that the characteristic is 2, and let W be the radical of B. Then B is identically zero on W; so the restriction of q to W satisfies q x y q x q y q λx ; λ2q x As above, f is very nearly semilinear.
The field F is called perfect if every element is a square. If F is perfect, then the map x x2 is onto, and hence an automorphism of F; so q is indeed semilinear, and its kernel is a hyperplane of W. We conclude: Theorem 3.5 Let q be a non-singular quadratic form, which polarises to B, over a field F.
(a) If the characteristic of F is not 2, then B is non-degenerate.
(b) If F is a perfect field of characteristic 2, then the radical of B has rank at most 1.
Exercise 3.6 Let B be an alternating bilinear form on a vector space V over a field F of characteristic 2. Let vi : i I be a basis for V, and ci : i I any function from I to F. Show that there is a unique quadratic form q with the properties that q vi ci for every i I, and q polarises to B.
Exercise 3.7 (a) Construct an imperfect field of characteristic 2.
(b) Construct a non-singular quadratic form with the property that the radical of the associated bilinear form has rank greater than 1.
Exercise 3.8 Show that finite fields of characteristic 2 are perfect.
Exercise 3.9 Let B be a σ-Hermitian form on a vector space V over F, where σ is not the identity. Set f x ( B x x . Let E Fix σ , and let V = be V regarded as an E-vector space by restricting scalars. Prove that f is a quadratic form onV = , which polarises to the bilinear form Tr B defined by Tr B x y > B x y B x y σ.
Show further that Tr B is non-degenerate if and only if B is.
33 3.3 Classification of forms As explained in the last section, we now consider a vector space V of finite rank equipped with a form of one of the following types: a non-degenerate alternating bilinear form B; a non-degenerate trace-valued σ-Hermitian form B, where σ is not the identity; or a non-singular quadratic form q. In the third case, we let B be the bilinear form obtained by polarising q; then B is alternating or symmetric according as the characteristic is or is not 2, but B may be degenerate. We also let f denote the function q. In the other two cases, we define a function f : V F by f x ? B x x — this is identically zero if b is alternating. See Exercise 3.10 for the Hermitian case.
Such a pair V B or V q will be called a formed space.
Exercise 3.10 Let B be a σ-Hermitian form on a vector spaceV over F, where σ is not the identity. Set f x ( B x x . Let E Fix σ , and let V = be V regarded as an E-vector space by restricting scalars. Prove that f is a quadratic form onV = , which polarises to the bilinear form Tr B defined by Tr B x y > B x y B x y σ.
Show further that Tr b is non-degenerate if and only if B is.
We say that V is anisotropic if f x @ 0 for all x 0. Also, V is a hyperbolic plane if it is spanned by vectors v and w with f v A f w 7 0 and B v w A 1.
(The vectors v and w are linearly independent, so V has rank 2.) Theorem 3.6 A non-degenerate formed space is the direct sum of a number r of hyperbolic lines and an anisotropic space U. The number r and the isomorphism type of U are invariants of V.
Proof If V is anisotropic, then there is nothing to prove, since V cannot contain a hyperbolic plane. So suppose that V contains a vector v 0 with f v 0.
We claim that there is a vector w with B v w B 0. In the alternating and Hermitian cases, this follows immediately from the non-degeneracy of the form.
In the quadratic case, if no such vector exists, then v is in the radical of B; but v is a singular vector, contradicting the non-degeneracy of f.
Multiplying w by a non-zero constant, we may assume that B v w 1.
Now, for any value of λ, we have B v w λv 1. We wish to choose λ so that f w λv 0; then v and w will span a hyperbolic line. Now we distinguish cases.
(a) If B is alternating, then any value of λ works.
34 (b) If B is Hermitian, we have f w λv C f w λB v w λσB w v λλσ f v f w λ λσ ; and, since B is trace-valued, there exists λ with Tr λ f w .
(c) Finally, if f q is quadratic, we have f w λv < f w λB w v λ2 f v f w λ so we choose λ f w .
Now let W1 be the hyperbolic line v w λv " , and let V1 W 1 , where orthog-onality is defined with respect to the form B. It is easily checked that V V1 D W1, and the restriction of the form to V1 is still non-degenerate. Now the existence of the decomposition follows by induction.
The uniqueness of the decomposition will be proved later, as a consequence of Witt’s Lemma (Theorem 3.15).
The number r of hyperbolic lines is called the polar rank of V, and (the iso-morphism type of) U is called the germ of V.
To complete the classification of forms over a given field, it is necessary to determine all the anisotropic spaces. In general, this is not possible; for exam-ple, the study of positive definite quadratic forms over the rational numbers leads quickly into deep number-theoretic waters. I will consider the cases of the real and complex numbers and finite fields.
First, though, the alternating case is trivial: Proposition 3.7 The only anisotropic space carrying an alternating bilinear form is the zero space.
In combination with Theorem 3.6, this shows that a space carrying a non-degenerate alternating bilinear form is a direct sum of hyperbolic planes.
Over the real numbers, Sylvester’s theorem asserts that any quadratic form in n variables is equivalent to the form x2 1 // x2 r x2 r E 1 !//F x2 r E s for some r s with r s G n. If the form is non-singular, then r s n. If both r and s are non-zero, there is a non-zero singular vector (with 1 in positions 1 and r 1, 0 elsewhere). So we have: 35 Proposition 3.8 If V is a real vector space of rank n, then an anisotropic form on V is either positive definite or negative definite; there is a unique form of each type up to invertible linear transformation, one the negative of the other.
The reals have no non-identity automorphisms, so Hermitian forms do not arise.
Over the complex numbers, the following facts are easily shown: (a) There is a unique non-singular quadratic form (up to equivalence) in n vari-ables for any n. A space carrying such a form is anisotropic if and only if n G 1.
(b) If σ denotes complex conjugation, the situation for σ-Hermitian forms is the same as for quadratic forms over the reals: anisotropic forms are positive or negative definite, and there is a unique form of each type, one the negative of the other.
For finite fields, the position is as follows.
Theorem 3.9 (a) An anisotropic quadratic form in n variables over GF q ex-ists if and only if n G 2. There is a unique form for each n except when n 1 and q is odd, in which case there are two forms, one a non-square multiple of the other.
(b) Let q r2 and let σ be the field automorphism α αr. Then there is an anisotropic σ-Hermitian form in n variables if and only if n G 1. The form is unique in each case.
Proof (a) Consider first the case where the characteristic is not 2. The multiplica-tive group of GF q is cyclic of even order q 1; so the squares form a subgroup of index 2, and if η is a fixed non-square, then every non-square has the form ηα2 for some α. It follows easily that any quadratic form in one variable is equivalent to either x2 or ηx2.
Next, consider non-singular forms in two variables. By completing the square, such a form is equivalent to one of x2 y2, x2 ηy2, ηx2 ηy2.
Suppose first that q H 1 mod 4 . Then 1 is a square, say 1 β2. (In the multiplicative group, 1 has order 2, so lies in the subgroup of even order 1 2 q 1 consisting of squares.) Thus x2 y2 x βy x βy , and the first and third forms are not anisotropic. Moreover, any form in 3 or more variables, when 36 converted to diagonal form, contains one of these two, and so is not anisotropic either.
Now consider the other case, q H4 1 mod 4 . Then 1 is a non-square (since the group of squares has odd order), so the second form is x y x y , and is not anisotropic. Moreover, the set of squares is not closed under addition (else it would be a subgroup of the additive group, but 1 2 q 1 doesn’t divide q); so there exist two squares whose sum is a non-square. Multiplying by a suitable square, there exist β γ with β2 γ2 5 1. Then x2 y2 βx γy 2 γx βy 2 and the first and third forms are equivalent. Moreover, a form in three variables is certainly not anisotropic unless it is equivalent to x2 y2 z2, and this form vanishes at the vector β γ 1 ; hence there is no anisotropic form in three or more variables.
The characteristic 2 case is an exercise (see below).
(b) Now consider Hermitian forms. If σ is an automorphism of GF q of order 2, then q is a square, say q r2, and ασ αr. We need the fact that every element of Fix σ GF r has the form αασ (see Exercise 3.3).
In one variable, we have f x A µxxσ for some non-zero µ Fix σ ; writing µ αασ and replacing x by αx, we can assume that µ 1.
In two variables, we can similarly take the form to be xxσ yyσ. Now 1 Fix σ , so 1 λλσ; then the form vanishes at 1 λ . It follows that there is no anisotropic form in any larger number of variables either.
Exercise 3.11 Prove that there is, up to equivalence, a unique non-degenerate al-ternating bilinear form on a vector space of countably infinite dimension (a direct sum of countably many isotropic planes).
Exercise 3.12 Let F be a finite field of characteristic 2.
(a) Prove that every element of F has a unique square root.
(b) By considering the bilinear form obtained by polarisation, prove that a non-singular form in 2 or 3 variables over F is equivalent to αx2 xy βy2 or αx2 xy βy2 γz2 respectively. Prove that forms of the first shape (with α β 0) are all equivalent, while those of the second shape cannot be anisotropic.
37 3.4 Polar spaces Polar spaces describe the geometry of vector spaces carrying a reflexive sesquilin-ear form or a quadratic form in much the same way as projective spaces describe the geometry of vector spaces. We now embark on the study of these geometries; the three preceding sections contain the prerequisite algebra.
First, some terminology. The polar spaces associated with the three types of forms (alternating bilinear, Hermitian, and quadratic) are referred to by the same names as the groups associated with them: symplectic, unitary, and orthogonal respectively. Of what do these spaces consist?
Let V be a vector space carrying a form of one of our three types. Recall that as well as a sesquilinear form b in two variables, we have a form f in one variable — either f is defined by f x 7 B x x , or b is obtained by polarising f — and we make use of both forms. A subspace of V on which B vanishes identically is called a B-flat subspace, and one on which f vanishes identically is called a f-flat subspace. (Note: these terms are not standard; in the literature, such spaces are called totally isotropic (t.i.) and totally singular (t.s.) respectively.) The unqualified term flat subspace will mean a B-flat subspace in the symplectic or unitary case, and a q-flat subspace in the orthogonal case.
The polar space associated with a vector space carrying a form is the geometry whose flats are the flat subspaces (in the above sense). Note that, if the form is anisotropic, then the only member of the polar space is the zero subspace. The polar rank of a classical polar space is the largest vector space rank of any flat subspace; it is zero if and only if the form is anisotropic. Where there is no confusion, polar rank will be called simply rank. (We will soon see that there is no conflict with our earlier definition of rank as the number of hyperbolic planes in the decomposition of the space.) We use the terms point, line, plane, etc., just as for projective spaces.
Polar spaces bear the same relation to formed spaces as projective spaces do to vector spaces.
We now proceed to derive some properties of polar spaces. Let Γ be a classical polar space of polar rank r.
(P1) Any flat, together with the flats it contains, is a projective space of dimen-sion at most r 1.
(P2) The intersection of any family of flats is a flat.
(P3) If U is a flat of dimension r 1 and p a point not in U, then the union of the 38 planes joining p to points of U is a flat W of dimension r 1; and U I W is a hyperplane in both U and W.
(P4) There exist two disjoint flats of dimension r 1.
(P1) is clear since a subspace of a flat subspace is itself flat. (P2) is also clear.
To prove (P3), let p 4 y " . The function x B x y on the vector space U is linear; let K be its kernel, a hyperplane in U. Then the line (of the projective space) joining p to a point q U is flat if and only if q K; and the union of all such flat lines is a flat space W 5 K y " , such that W I U K, as required.
Finally, to prove (P4), we use the hyperbolic-anisotropic decomposition again.
If L1 ///J Lr are the hyperbolic planes, and xi yi are the distinguished spanning vectors in Li, then the required flats are x1 /K/K xr " and y1 /K/J yr " .
The significance of the geometric properties (P1)–(P4) lies in the major result of Veldkamp and Tits which determines all the geometries of rank at least 3 which satisfy them. All these geometries are polar spaces (as we have defined them) or slight generalisations, together with a couple of exceptions of rank 3. In particular, the following theorem holds: Theorem 3.10 A finite geometry satisfying (P1)–(P4) with r 3 is a polar space.
Exercise 3.13 Let P PG 3 F for some (not necessarily commutative) division ring F. Construct a new geometry Γ as follows: (a) the ‘points’ of Γ are the lines of P; (b) the ‘lines’ of Γ are the plane pencils in P (consisting of all lines lying in a plane Π and containing a point p of Π); (c) the ‘planes’ of Γ are of two types: the pencils (consisting of all the lines through a point) and the dual planes (consisting of all the lines in a plane).
Prove that Γ satisfies (P1)–(P4) with r 3.
Prove that, if F is not isomorphic to its opposite, then Γ contains non-isomorphic planes.
(We will see later that, if F is commutative, then Γ is an orthogonal polar space.) Exercise 3.14 Prove the Buekenhout–Shult property of the geometry of points and lines in a polar space: if p is a point not lying on a line L, then p is collinear with one or all points of L.
39 You should prove this both from the analytic description of polar spaces, and using (P1)–(P4).
In a polar space Γ, given any set S of points, we let S denote the set of points which are perpendicular to (that is, collinear with) every point of S. Polar spaces have good inductive properties. Let G be a classical polar space. There are two natural ways of producing a “smaller” polar space from G: (a) Take a point x of G, and consider the quotient space x 3 x, the space whose points, lines, ... are the lines, planes, ... of G containing x.
(b) Take two non-perpendicular points x and y, and consider x y .
In each case, the space constructed is a classical polar space, having the same germ as G but with polar rank one less than that of G. (Note that, in (b), the span of x and y in the vector space is a hyperbolic plane.) Exercise 3.15 Prove the above assertions.
There are more general versions. For example, if S is a flat of dimension d 1, then S 3 S is a polar space of rank r d with the same germ as G. We will see below how this inductive process can be used to obtain information about polar spaces.
We investigate just one type in more detail, the so-called hyperbolic quadric, the orthogonal space which is a direct sum of hyperbolic planes (that is, having germ 0). The quadratic form defining this space can be taken to be x1x2 x3x4 /K x2r L 1x2r.
Proposition 3.11 The maximal flats of a hyperbolic quadric fall into two classes, with the properties that the intersection of two maximal flats has even codimension in each if and only if they belong to the same class.
Proof First, note that the result holds when r 1, since then the quadratic form is x1x2 and there are just two singular points, 1 0 /" and 0 1 /" . By the inductive principle, it follows that any flat of dimension r 2 is contained in exactly two maximal flats.
We take the r 1 -flats and r 2 -flats as the vertices and edges of a graph Γ, that is, we join two r 1 -flats if their intersection is an r 2 -flat. The theorem will follow if we show that Γ is connected and bipartite, and that the distance between two vertices of Γ is the codimension of their intersection. Clearly the 40 codimension of the intersection increases by at most one with every step in the graph, so it is at most equal to the distance. We prove equality by induction.
Let U be a r 1 -flat and K a r 2 -flat. We claim that the two r 1 -spaces W1 W2 containing K have different distances from U. Factoring out the flat subspace U I K and using induction, we may assume that U I K / 0. Then U I K is a point p, which lies in one but not the other of W1 W2; say p W1. By induction, the distance from U to W1 is r 1; so the distance from U to W2 is at most r, hence equal to r by the remark in the preceding paragraph.
This establishes the claim about the distance. The fact that Γ is bipartite also follows, since in any non-bipartite graph there exists an edge both of whose ver-tices have the same distance from some third vertex, and the argument given shows that this doesn’t happen in Γ.
In particular, the rank 2 hyperbolic quadric consists of two families of lines forming a grid, as shown in Figure 1. This is the so-called “ruled quadric”, famil-iar from models such as wastepaper baskets.
M M M M M M M M M M M M N N N N N N N N N N N N N N N N N N N N N N N N M M M M M M M M M M M M O O O O O O O O O O O O P P P P P P P P P P P P P P P P P P P P P P P P O O O O O O O O O O O O Q Q Q Q Q Q Q Q Q Q Q Q R R R R R R R R R R R R R R R R R R R R R R R R Q Q Q Q Q Q Q Q Q Q Q Q S S S S S S S S S S S T T T T T T T T T T T T T T T T T T T T T T S S S S S S S S S S S U U U U U U U U U U U V V V V V V V V V V V U U U U U U U U U U U V V V V V V V V V V V W W W W W W W W W W W W X X X X X X X X X X X X W W W W W W W W W W W W X X X X X X X X X X X X Figure 1: A ruled quadric Exercise 3.16 Show that Proposition 3.11 can be proved using only properties (P1)–(P4) of polar spaces together with the fact that an r 1 -flat lies in exactly two maximal flats.
3.5 Finite polar spaces The classification of finite classical polar spaces was achieved by Theorem 3.6.
We subdivide these spaces into six families according to their germ, viz., one 41 symplectic, two unitary, and three orthogonal. (Forms which differ only by a scalar factor obviously define the same polar space.) The following table gives some information about them. In the table, r denotes the polar space rank, and δ the vector space rank of the germ; the rank n of the space is given by n 2r δ.
The significance of the parameter ε will emerge shortly. This number, depending only on the germ, carries numerical information about all spaces in the family.
Note that, in the unitary case, the order of the finite field must be a square.
Type δ ε Symplectic 0 0 Unitary 0 1 2 Unitary 1 1 2 Orthogonal 0 1 Orthogonal 1 0 Orthogonal 2 1 Table 1: Finite polar spaces Theorem 3.12 The number of points in a finite polar space of rank 1 is q1 E ε 1, where ε is given in Table 1.
Proof Let V be a vector space carrying a form of rank 1 over GF q . Then V is the orthogonal direct sum of a hyperbolic line L and an anisotropic germ U of dimension k (say). Let nk be the number of points.
Suppose that k Y 0. If p is a point of the polar space, then p lies on the hyper-plane p ; any other hyperplane containing p is non-degenerate with polar rank 1 and having germ of dimension k 1. Consider a parallel class of hyperplanes in the affine space whose hyperplane at infinity is p . Each such hyperplane con-tains nk L 1 1 points, and the hyperplane at infinity contains just one, viz., p. So we have nk 1 q nk L 1 1 from which it follows that nk 1 n0 1 qk. So it is enough to prove the result for the case k 0, that is, for a hyperbolic line.
In the symplectic case, each of the q 1 projective points on a line is isotropic.
Consider the unitary case. We can take the form to be B / x1 y1 x2 y2 K x1y2 y1x2 42 where x xσ xr, r2 q. So the isotropic points satisfy xy yx 0, that is, Tr xy , 0. How many pairs x y satisfy this? If y 0, then x is arbitrary. If y 0, then a fixed multiple of x is in the kernel of the trace map, a set of size q1 Z 2 (since Tr is GF q1 Z 2 -linear). So there are q q 1 q1 Z 2 1 q 1 q1 Z 2 1 vectors, i.e., q1 Z 2 1 projective points.
Finally, consider the orthogonal case. The quadratic form is equivalent to xy, and has two singular points, 1 0 /" and 1 0 /" .
Theorem 3.13 In a finite polar space of rank r, there are qr 1 qr E ε 1 /3 q 1 points, of which q2r L 1 E ε are not perpendicular to a given point.
Proof We let F r be the number of points, and G r the number not perpen-dicular to a given point. (We do not assume that G r is constant; this constancy follows from the induction that proves the theorem.) We use the two inductive principles described at the end of the last section.
Claim 1: G r q2G r 1 .
Take a point x, and count pairs y z , where y x , z x , and z y . Choos-ing z first, there are G r choices; then x z " is a hyperbolic line, and y is a point in x z " , so there are F r 1 choices for y. On the other hand, choosing y first, the lines through y are the points of the rank r 1 polar space x 3 x, and so there are F r 1 of them, with q points different from x on each, giving qF r 1 choices for y; then x y " and y z " are non-perpendicular lines in y , i.e., points of y 3 y, so there are G r 1 choices for y z " , and so qG r 1 choices for y. thus G r [ F r 1 qF r 1 [ qG r 1 from which the result follows.
Since G 1 ? q1 E ε, it follows immediately that G r q2r L 1 E ε, as required.
Claim 2: F r 1 qF r 1 G r .
For this, simply observe (as above) that points perpendicular to x lie on lines of x 3 x.
Now it is just a matter of calculation that the function qr 1 qr E ε 1 /3 q 1 satisfies the recurrence of Claim 2 and correctly reduces to q1 E ε 1 when r 1.
43 Theorem 3.14 The number of maximal flats in a finite polar space of rank r is r ∏ i \ 1 1 qi E ε ] Proof Let H r be this number. Count pairs x U , where U is a maximal flat and x U. We find that F r [ H r 1 & H r [ qr 1 /3 q 1 so H r 1 qr E ε H r 1 Now the result is immediate.
It should now be clear that any reasonable counting question about finite polar spaces can be answered in terms of q r ε. We will do this for the associated classical groups at the end of the next section.
3.6 Witt’s Lemma Let V be a formed space, with sesquilinear form B and (if appropriate) quadratic form q. An isometry of V is a linear map g : V V which satisfies B xg yg A B x y for all x y V, and (if appropriate) q xg q x for all x V. (Note that, in the case of a quadratic form, the second condition implies the first.) The set of all isometries of V forms a group, the isometry group of V. This group is our object of study for the next few sections.
More generally, if V and W are formed spaces of the same type, an isometry from V to W is a linear map from V to W satisfying the conditions listed above.
Exercise 3.17 Let V be a (not necessarily non-degenerate) formed space of sym-plectic or Hermitian type, with radical V . Prove that the natural map from V to V 3 V is an isometry.
The purpose of this subsection is to prove Witt’s Lemma, a transitivity assertion about the isometry group of a formed space.
Theorem 3.15 Suppose that U1 and U2 are subspaces of the formed space V, and h : U1 U2 is an isometry. Then there is an isometry g of V which extends h if and only if U1 I V h U2 I V .
In particular, if V 0, then any isometry between subspaces of V extends to an isometry of V.
44 Proof Assume that h : U1 U2 is an isometry. Clearly, if h is the restriction of an isometry g of V, then V g V , and so U1 I V h U1 I V g U1g I V g U2 I V > We have to prove the converse.
First we show that we may assume that U1 and U2 contain V . Suppose not.
Choose a subspace W of V which is a complement to both U1 I V and U2 I V (see Exercise 3.18), and extend h to U1 D W by the identity map on W. This is easily checked to be an isometry to U2 D W.
The proof is by induction on rk U1 3 V . If U1 V U2, then choose any complement W for V in V and extend h by the identity on W. So the base step of the induction is proved. Assume that the conclusion of Witt’s Lemma holds for V = , U = 1, U = 2, h = whenever rk U = 1 3 V =^ 7_ rk U1 3 V .
Let H be a hyperplane of U1 containing V . Then the restriction f = of f to H has an extension to an isometry g = of V. Now it is enough to show that h g = L 1 extends to an isometry; in other words, we may assume that h is the identity on H.
Moreover, the conclusion is clear if h is the identity on U1; so suppose not. Then ker h 1 H, and so the image of h 1 is a rank 1 subspace P of U1.
Since h is an isometry, for all x y U1 we have B xh yh y C B xh yh B xh y B x y B xh y B x xh y So, if y H, then any vector xh x of P is orthogonal to y; that is, H G P .
Now suppose that P G U 1 . Then U1 I P U2 I P H. If W is a comple-ment to H in P , then we can extend h by the identity on W to obtain the required isometry. So we may assume further that U1 U2 G P . In particular, P G P .
Next we show that we may assume that U1 U2 P . Suppose first that U1 U2. If Ui ` H ui " for i 1 2, let W0 be a complement for U1 U2 in P , and W a W0 u1 u2 " ; then h can be extended by the identity on W to an isometry on P . If U1 U2, take any complement W to U1 in P . In either case, the extension is an isometry of P which acts as the identity on a hyperplane H = of P containing H. So we may replace U1 U2 H by P P H = .
Let P 9 x " and let x uh u for some u U1. We have B x x A 0. In the orthogonal case, we have q x q uh u q uh q u B uh u & 2q u B u u ( 0 45 (We have B uh u ( B u u because B uh u u 0.) So P is flat, and there is a hyperbolic plane u v " , with v 3 P . Our job is to extend h to the vector v.
To achieve this, we show first that there is a vector v = such that uh v =^" u v " . This holds because u v " is a hyperplane in uh " not containing V .
Next, we observe that uh v =^" is a hyperbolic plane, so we can choose a vector v =b= such that B uh v =b=c? 1 and (if relevant) Q v =b=^ 0.
Finally, we observe that by extending h to map v to v =d= we obtain the required isometry of V.
Exercise 3.18 Let U1 and U2 be subspaces of a vector space V having the same rank. Show that there is a subspace W of V which is a complement for both U1 and U2.
Corollary 3.16 (a) The ranks of maximal flat subspaces of a formed space are all equal.
(b) The Witt rank and isometry type of the germ of a non-degenerate formed space are invariants.
Proof (a) Let U1 and U2 be maximal flat subspaces. Then both U1 and U2 con-tains V . If rk U1 e_ rk U2 , there is an isometry h from U1 into U2. If g is the extension of h to V, then the image of U2 under g L 1 is a flat subspace properly containing U1, contradicting maximality.
(b) The result is clear if V is anisotropic. Otherwise, let U1 and U2 be hyper-bolic planes. Then U1 and U2 are isometric and are disjoint from V . An isometry of V carrying U1 to U2 takes U 1 to U 2 . Then the result follows by induction.
Theorem 3.17 Let Vr be a non-degenerate formed space with polar rank r and germ W over GF q . Let Gr be the isometry group of Vr. Then f Gr f g r ∏ i \ 1 qi 1 qi E ε 1 q2i L 1 E ε h f G0 f qr i r E ε j g r ∏ i \ 1 qi 1 qi E ε 1 h f G0 f 46 where f G0 f is given by the following table: Type δ ε f G0 f Symplectic 0 0 1 Unitary 0 1 2 1 Unitary 1 1 2 q1 Z 2 1 Orthogonal 0 1 1 Orthogonal 1 0 k 2 (q odd) 1 (q even) Orthogonal 2 1 2 q 1 Proof By Theorem 3.13, the number of choices of a vector x spanning a flat subspace is qr 1 qr E ε 1 . Then the number of choices of a vector y spanning a flat subspace and having inner product 1 with x is q2r L 1 E ε. Then x and y span a hyperbolic plane. Now Witt’s Lemma shows that Gr acts transitively on such pairs, and the stabiliser of such a pair is Gr L 1, by the inductive principle.
In the cases where δ 0, G0 is the trivial group on a vector space of rank 0.
In the unitary case with δ 1, G0 preserves the Hermitian form xxq1 l 2, so consists of multiplication by q1 Z 2 1 st roots of unity. In the orthogonal case with δ 1, G0 preserves the quadratic form x2, and so consists of multiplication by m 1 only. Finally, consider the orthogonal case with δ 2. Here we can represent the quadratic form as the norm from GF q2 to GF q , that is, N x xq E 1. The GF q -linear maps which preserve this form a dihedral group of order 2 q 1 : the cyclic group is generated by the q 1 st roots of unity in GF q2 , which is inverted by the non-trivial field automorphism over GF q (since, if xq E 1 1, then xq x L 1).
47 |
189270 | https://www.mathplanet.com/education/pre-algebra/probability-and-statistics/the-mean-the-median-and-the-mode | The mean, the median and the mode - Mathplanet
Mathplanet
A free service from Mattecentrum
Other tools
Mathplanet
Matteboken
Matteboken (Arabic)
Webmatematik (Danish)
MenuPre-Algebra/Probability and statistics/
The mean, the median and the mode
Do excercisesShow all 3 exercises
Mean median mode I
Mean median mode II
Mean median mode III
Mean, median and mode are numbers that represent a whole set of data or information. Mean, median and mode are together called the measures of central tendency.
The mean is often called the average. To find the mean you take a set of data and calculate the sum of the data, after that you divide the sum by the number of pieces in the set.
Example
Find the mean of the following data set:
56, 35, 45, 67, 12, 24, 48, 55, 58, 30
56+35+45+67+12+24+48+55+58+30 10=56+35+45+67+12+24+48+55+58+30 10=
=430 10=43=430 10=43
T h e m e a n=43 T h e m e a n=43
The median is the number in an ordered set of data that is in the middle.
If we have a set of data with an odd number of data points then the median is the data point in the middle.
1,2,3,4,5,6,7 1,2,3,4,5,6,7
If we have a set of data with an even number of data points, then the median is the mean of the two data points in the middle
1,2,3,4,5,6,7,8 1,2,3,4,5,6,7,8
4+5 2=9 2=4.5 4+5 2=9 2=4.5
The mode is the most common number in the set of data.
Example
2,5,6,2,2,2,5,6,2 2,5,6,2,2,2,5,6,2
T h e m o d e=2 T h e m o d e=2
Video lesson
Find mean, median and mode
You need to accept marketing cookies to play the video.
Accept marketing cookies
Do excercisesShow all 3 exercises
Mean median mode I
Mean median mode II
Mean median mode III
More classes on this subject
Pre-Algebra Probability and statistics: Calculating the outcomePre-Algebra Probability and statistics: Combinations and permutationsPre-Algebra Probability and statistics: Finding the oddsPre-Algebra Probability and statistics: Probability of events
Next Chapter:
PROBABILITY AND STATISTICS –Stem-and-Leaf Plots and Box-and-Whiskers Plot
Search
Math Playground
All courses
Math PlaygroundOverview
Programming
All courses
ProgrammingOverview
Pre-Algebra
All courses
Pre-AlgebraOverview
Introducing Algebra
Pre-Algebra
Introducing Algebra
Overview
Operations in the correct order
Evaluate expressions
Identify properties
Equations with variables
Coordinate system and ordered pairs
Inequalities
Explore and understand integers
Pre-Algebra
Explore and understand integers
Overview
Absolute value
Adding and subtracting integers
Multiplying and dividing with integers
Inequalities and one-step equations
Pre-Algebra
Inequalities and one-step equations
Overview
Different ways to solve equations
Calculating the area and the perimeter
Solving inequalities
Understanding inequalities and equations
Discover fractions and factors
Pre-Algebra
Discover fractions and factors
Overview
Monomials and adding or subtracting polynomials
Powers and exponents
Multiplying polynomials and binomials
Factorization and prime numbers
Finding the greatest common factor
Finding the least common multiple
More about the four rules of arithmetic
Pre-Algebra
More about the four rules of arithmetic
Overview
Integers and rational numbers
Learn how to estimate calculations
Calculating with decimals and fractions
Geometric sequences of numbers
Scientific notation
More about equation and inequalities
Pre-Algebra
More about equation and inequalities
Overview
Fundamentals in solving Equations in one or more steps
Calculating the circumference of a circle
Graphing and functions
Pre-Algebra
Graphing and functions
Overview
Linear equations in the coordinate plane
The slope of a linear function
Graphing linear inequalities
Solve systems of equations by graphing
Ratios and percent
Pre-Algebra
Ratios and percent
Overview
Rates and ratios
Proportions and percent
Solving problems with percentages
Probability and statistics
Pre-Algebra
Probability and statistics
Overview
The mean, the median and the mode
Stem-and-Leaf Plots and Box-and-Whiskers Plot
Calculating the outcome
Combinations and permutations
Finding the odds
Probability of events
About Mathplanet
Cookie settings)
Introducing geometry
Pre-Algebra
Introducing geometry
Overview
Geometry – fundamental statements
Circle graphs
Angles and parallel lines
Triangles
Quadrilaterals, polygons and transformations
Area and Volume
Pre-Algebra
Area and Volume
Overview
Measure areas
Pyramids, prisms, cylinders and cones
Right triangles and algebra
Pre-Algebra
Right triangles and algebra
Overview
Square roots and real numbers
The Pythagorean Theorem
Trigonometry
Algebra 1
All courses
Algebra 1Overview
Algebra 2
All courses
Algebra 2Overview
Geometry
All courses
GeometryOverview
SAT
All courses
SATOverview
ACT
All courses
ACTOverview
Mathplanet is licensed byCreative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens.
Cookie settings
Math Planet and our partners save information on your device to offer a customized experience, enable visitor statistics, and support personalized marketing. You decide which types of cookies you accept in addition to the necessary ones needed for the website to work.
Accept all
Only necessary - [x] Visitor statistics - [x] Marketing
Read more about our cookies
Necessary
These cookies are needed for the website to function technically and functionally and must therefore be accepted to visit the site.
CookieTractor:1
Description of _cc_cookieConsent set by CookieTractor| Cookie name: | _cc_cookieConsent |
| Purpose: | Used to save your given cookie consents between site visits. |
| Policy: | Link to policy (cookietractor.com) Open policy for CookieTractor in new window |
| Removed after: | 11 months, 30 days |
| Domain: | www.mathplanet.com |
Google reCAPTCHA:1
Description of _GRECAPTCHA set by Google reCAPTCHA| Cookie name: | _GRECAPTCHA |
| Purpose: | To provide spam protection in forms. |
| Policy: | Link to policy (business.safety.google) Open policy for Google reCAPTCHA in new window |
| Removed after: | 5 months, 27 days |
| Domain: | www.google.com |
Trinket:1
Description of session set by Trinket| Cookie name: | session |
| Purpose: | Used by Trinket, an embedded programming environment, to handle sessions for the visitor. |
| Policy: | Link to policy (trinket.io) Open policy for Trinket in new window |
| Removed after: | 1 year |
| Domain: | trinket.io |
YouTube:1
Description of VISITOR_PRIVACY_METADATA set by YouTube| Cookie name: | VISITOR_PRIVACY_METADATA |
| Purpose: | Used to track and enrich the users privacy settings on the Youtube platform |
| Policy: | Link to policy (policies.google.com) Open policy for YouTube in new window |
| Removed after: | 5 months, 27 days |
| Domain: | .youtube.com |
Visitor statistics
These cookies are used to collect statistics about visitors, which pages they visit, and what they do on the website.
Google Analytics:4
Description of _ga_. set by Google Analytics| Cookie name: | _ga_. |
| Purpose: | Used to store and track visits on websites. |
| Policy: | Link to policy (business.safety.google) Open policy for Google Analytics in new window |
| Removed after: | 1 year, 11 months |
| Domain: | .mathplanet.com |
Description of _ga set by Google Analytics| Cookie name: | _ga |
| Purpose: | This cookie is set by Google Analytics to distinguish unique users and track visitor interactions with the website. It enables data collection on user behaviour, such as page views and session duration. This information helps the website owner understand user engagement and optimise the site. |
| Policy: | Link to policy (business.safety.google) Open policy for Google Analytics in new window |
| Removed after: | 1 year, 11 months |
| Domain: | .mathplanet.com |
Description of ga_session_duration set by Google Analytics| Cookie name: | ga_session_duration |
| Purpose: | Average duration for Google Analytics sessions attributed to clicks on search or social objects in Search Ads 360. |
| Policy: | Link to policy (business.safety.google) Open policy for Google Analytics in new window |
| Removed after: | 2 hours, 31 minutes |
| Domain: | www.mathplanet.com |
Description of ar_debug set by Google Analytics| Cookie name: | ar_debug |
| Purpose: | Used for debugging purposes of implementation. |
| Policy: | Link to policy (business.safety.google) Open policy for Google Analytics in new window |
| Removed after: | 2 months, 29 days |
| Domain: | .region1.google-analytics.com |
YouTube:2
Description of YSC set by YouTube| Cookie name: | YSC |
| Purpose: | This cookie is used to store and track user interaction. |
| Policy: | Link to policy (policies.google.com) Open policy for YouTube in new window |
| Removed after: | When browser is closed |
| Domain: | .youtube.com |
Description of __Secure-YEC set by YouTube| Cookie name: | __Secure-YEC |
| Purpose: | Stores the user's video player preferences using embedded YouTube video. |
| Policy: | Link to policy (policies.google.com) Open policy for YouTube in new window |
| Removed after: | When browser is closed |
| Domain: | .youtube.com |
Marketing
These cookies are used to track visitors in different digital channels to be able to serve personalized ads and marketing.
YouTube:2
Description of VISITOR_INFO1_LIVE set by YouTube| Cookie name: | VISITOR_INFO1_LIVE |
| Purpose: | Used to remember user preferences and can also be used to personalization and advertising. |
| Policy: | Link to policy (policies.google.com) Open policy for YouTube in new window |
| Removed after: | 5 months, 27 days |
| Domain: | .youtube.com |
Description of __Secure-ROLLOUT_TOKEN set by YouTube| Cookie name: | __Secure-ROLLOUT_TOKEN |
| Purpose: | Registers a unique ID to keep statistics of what videos from YouTube the user has seen. |
| Policy: | Link to policy (policies.google.com) Open policy for YouTube in new window |
| Removed after: | 5 months, 27 days |
| Domain: | .youtube.com |
Your current consent
Information about your current cookie consent and when you set it.
Data about your interaction with CookieTractor| Consent: | |
| Date: | |
| Key: | |
Powered by CookieTractorRead more about CookieTractor (opens in new window) |
189271 | https://physics.uwo.ca/~mhoude2/courses/PDF%20files/physics1501/Ch6-Fluid_Mechanics.pdf | 123 - Chapter 6. Fluid Mechanics Notes: • Most of the material in this chapter is taken from Young and Freedman, Chap. 12. 6.1 Fluid Statics Fluids, i.e., substances that can flow, are the subjects of this chapter. But before we can delve into this topic, we must first define a few fundamental quantities. 6.1.1 Mass Density and Specific Gravity We have already encountered the mass density (often abbreviated to density) in previous chapters. Namely, the mass density is simply the ratio of the mass m of an object to its volume V ρ = m V (6.1) with units of kg/m3 . Evidently, the density of objects can vary greatly depending of the materials composing them. For example, the density of water is 1,000 kg/m3 at 4C, that of iron is 7,800 kg/m3 , while a neutron star has a mean density of approximately 1018kg/m3 ! The specific gravity of a substance is defined as the ratio of the mass density to that of water at 4οC (i.e., 1,000 kg/m3). It would probably be more precise to use the term relative density instead of specific gravity, but such is not the custom… 6.1.2 Pressure and Buoyance A fluid is composed at the microscopic level by molecules and/or atoms that are constantly wiggling around. When the fluid is contained in a vessel these particles will collide with the walls of the container, a process that will then change their individual momenta. The change of momentum that a particle experiences will impart an impulse over the time interval during which the collision takes place, as a result the walls of the vessel will “feel” a force. The pressure p at a given point on a wall is defined as the force component perpendicular to the wall at that point per unit area. That is, if dF ⊥ is this elemental perpendicular force applied to an infinitesimal area dA on a wall, then the pressure on that area is p ≡dF ⊥ dA . (6.2) When the pressure is the same at all points of a macroscopic, plane surface of area A , then the perpendicular force F ⊥ must also be the same everywhere on that surface and - 124 - p = F ⊥ A . (6.3) The pascal (Pa) is the unit of pressure with 1 Pa = 1 N/m2. (6.4) Related to the pascal is the bar, which equals 105Pa , and, accordingly, the millibar, which equals 100 Pa. The atmospheric pressure pa , i.e., the average atmospheric pressure at sea level, is 1 atmosphere (atm) with 1 atm = 101,325 Pa = 1,103.25 millibar. (6.5) It is important to note that motion of the particles that cause the pressure is random in orientation and pressure is therefore isotropic. That is, pressure at one point is the same in all directions. Also, since the pressure at a point is directly proportional to the force effected at that point, it should be clear that weight can be a source of pressure. For example, the pressure in the earth’s atmosphere decreases as one goes to higher altitude as the weight of the, or the amount of, fluid above is reduced. Similarly, an increase in pressure is felt by a diver who descends to greater depths in a body of water. We can quantify this effect by studying how pressure varies within a fluid contained in a vessel. Accordingly, referring to Figure 1, we consider a fluid of uniform density ρ under the effect of gravity g and consider a fluid element of thickness dy and area A . We assume that the bottom of the vessel is located at y = 0 and the position of the fluid element is at y (> 0 ; y thus increases upwards). If the pressure at the bottom of the element is p , then the pressure immediately on top of it will be p + dp . If we further assume that the fluid is in equilibrium, then this fluid element must be static and the different forces, say, at the bottom of the element must cancel each other out. That is, pA − p + dp ( )A + dw ⎡ ⎣ ⎤ ⎦= 0, (6.6) where dw is the weight of the fluid element dw = ρAdy ( )g. (6.7) The quantity between parentheses in equation (6.7) is simply the mass of the fluid element. Equation (6.6) then becomes dp A = −ρg Ady, (6.8) or - 125 - dp dy = −ρg. (6.9) Equation (6.9) is often called the equation of hydrostatic equilibrium. This result shows that pressure decreases as one moves upward in the vessel, as expected. We can integrate this equation to find the difference in pressure between two points y1 and y2 ( y2 > y1) with p2 −p1 = dp 1 2 ∫ = −ρg dy 1 2 ∫ = −ρg y2 −y1 ( ), (6.10) which we rewrite as (with Δy = y2 −y1 > 0 ) p1 = p2 + ρgΔy. (6.11) If we set y2 at the top surface of the fluid (i.e., near the opening of the vessel), then p2 ≡p0 , where ‘0’ means ‘zero depth’, equals the pressure at the exterior of the fluid. For example, if the vessel is located at sea level, then p0 = 1 atm, (6.12) and p1 = p0 + ρgΔy. (6.13) Figure 1 – The pressure as function of height in a fluid. - 126 - It is then convenient to think of Δy > 0 as the depth in the fluid where the pressure p1 is encountered. Equation (6.13) also implies that increasing p0 by some amount will increase the pressure at any point within the fluid by the same amount. This is the so-called Pascal’s Law Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. We can use equation (6.11) to explain the behavior of objects submerged (sometimes not completely) in a fluid, such as water. Let us consider Figure 2 where an objet of mass m , horizontal area A , and height h is immersed in a fluid of density ρ ; the whole apparatus is subject to gravity. We denote by p1 and p2 the pressures at the bottom and top surfaces of the object, respectively, likewise the force components perpendicular to those surfaces are F 1 and F 2 . But we know from equation (6.11) that p1A −p2A = F 1 −F 2 = ρghA, (6.14) or, while defining the volume of the object with V = hA , we have F 1 −F 2 = ρVg. (6.15) Since the F 1 −F 2 is net buoyancy force acting on the body and ρV is the mass of fluid displaced by the presence of the body, we are then led to Archimedes’ Principle The net upward, buoyancy force acting on a partially or completely immersed body equals the weight of fluid displaced by the body. It is important to note that the buoyancy force is independent of the weight of the object. Also, although we derived this result for an object of rectangular volume, it should be clear that it applies to any possible shape since only the net perpendicular forces on the areas spanned by the top and bottom surfaces of the object come into play. 1 h 2 1 p = F /A p = F /A 2 1 Figure 2 - An object immersed in a fluid. - 127 - 6.1.3 Exercises 1. Hydraulic lift. Consider the representation of a hydraulic lift shown in Figure 3. Use Pascal’s Law to explain the lift’s functioning. Solution. According to equation (6.11), two points in fluid located at the same height are subjected to the same pressure. If we consider points 1 and 2 in Figure 3 both located at the surface of the fluid, then we can write p1 = p2 F 1 A1 = F 2 A2 (6.16) and therefore F 2 = A2 A1 F 1. (6.17) It follows that we can multiply at point 2 the effect of the force F 1 applied at point 1 over an area A1 (with a piston, for example) if we use a surface A2 > A1 at point 2. The resulting force is multiplied by the ratio of the areas, as shown in equation (6.17). 2. Archimedes’ Principle and Buoyancy. We drop a rectangular piece of cork of mass m , volume Ah , and density ρm = 240 kg/m3 in a vessel containing water (of density ρ = 1000 kg/m3 ). Determine the depth at which the object’s bottom surface settles once equilibrium is reached. Figure 3 – A hydraulic lift. - 128 - Assume that an atmospheric pressure p0 = 1 atm is present at the surface of the water and that ρair = 1.2 kg/m3. Solution. If denser than water, an object would sink to the bottom of the vessel. We can understand this by considering Newton’s Second Law for the forces acting on the object p1A −p2A −mg = ma. (6.18) But by replacing p1A −p2A with the right-hand side of equation (6.14) and setting m = ρmAh we get ρ −ρm ( )gAh = ρmAha, (6.19) or a = g ρ ρm −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟< 0 (6.20) and the object sinks to the bottom. If on the other hand the object is less dense than water, as is the case here, then the object will float, as shown in Figure 4. We denote by h1 > 0 the depth at which the object’s bottom surface is located relative to the surface of the water and by h2 > 0 the portion above the water (i.e., h = h1 + h2). Using Newton’s Second Law we write p1A − p0 + dp0 dy h2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟A −mg = 0 (6.21) since we assume equilibrium (i.e., a = 0 ), and where p1 is the pressure at the object’s bottom surface, m = ρmAh (as before), and h2 ⋅dp0 dy is the change in atmospheric pressure from the water surface to the top of the object. 1 ρ p = 1 atm 0 ρ h2 m h1 Figure 4 – An object partially immersed in water. - 129 - From Archimedes’ buoyancy relation we have p0A − p0 + dp0 dy h2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟A = ρairAh2g, (6.22) or, as expected for hydrostatic equilibrium, dp0 dy = −ρairg. (6.23) Also from the equation of hydrostatic equilibrium (i.e., equation (6.13)), we can write p1 = p0 + ρgh1. (6.24) Inserting equations (6.23) and (6.24) into equation (6.21), while setting h2 = h −h1, we find that p0 + ρgh1 ( )−p0 −ρairg h −h1 ( ) ⎡ ⎣ ⎤ ⎦−ρmgh = 0, (6.25) and h1 = ρm −ρair ρ −ρair ⎛ ⎝ ⎜ ⎞ ⎠ ⎟h ρm ρ h 0.24h, (6.26) since the density of air is completely negligible compared to those of water and cork. It should be clear from this discussion that an object that is completely immersed in a fluid, and at equilibrium (as in Figure 2), must have the same density of the fluid. 3. (Prob. 12.33 in Young and Freedman.) A rock is suspended by a light string. When the rock is in the air, the tension in the string is 39.2 N. When the rock is totally immersed in water, the tension in the string is 28.4 N. When the rock is totally immersed in an unknown liquid, the tension is 18.6 N. What is the density of the unknown liquid? Solution. According to Archimedes’ Principle the buoyancy force acting on the rock equals the weight of the displaced volume of liquid. That is, F b = ρVg, (6.27) - 130 - where V is the volume of the rock. According to Newton’s Second Law T + F b −mg = 0, (6.28) with T the tension in the string and m the mass of the rock. Applying equation (6.28) to the cases of air and water and then equating them, we have Tair + ρairVg = Twater + ρwaterVg (6.29) and V = Tair −Twater ρwater −ρair ( )g = 39.2 −28.4 ( )N 1000 −1.2 ( )kg/m3 ⋅9.81 m/s2 = 1.10 ×10−3m3. (6.30) One way to proceed is to insert this result in equation (6.28) to find the mass of the rock (let us choose the case where it is immersed in water) m = Twater g + ρwaterV = 28.4 N 9.81 m/s2 +1000 kg/m3 ⋅1.10 ×10−3m3 = 4.00 kg. (6.31) Equation (6.28) for the unknown liquid then yields ρu = mg −Tu Vg = 4.00 kg⋅9.81 m/s2 −18.6 N 1.10 ×10−3m−3 ⋅9.81 m/s2 = 1907 kg/m3. (6.32) Alternatively, we could have written Tair + ρairVg = Tu + ρuVg, (6.33) and, with a similar outcome, ρu = ρair + Tair −Tu Vg . (6.34) - 131 - 6.2 Fluid Flows The kinematics and motions of fluids can be a very complicated affair. In what follows, we will limit our studies to ideal fluids, i.e., those that are incompressible and non-viscous (that have no internal friction). We will also concentrate on steady flows, when motions have taken place for a long enough period that no transient behavior remains. Finally, we will not consider turbulence (random and chaotic flow patterns) and limit ourselves to laminar flows where layers of fluid can smoothly flow next to one another without the creation of turbulence. 6.2.1 The Continuity Equation As a fluid is in steady motion and its flow progresses with time, it must be that the amount of mass flowing per unit time across to areas perpendicular to the flow is conserved. That is, if we consider the flow configuration shown in Figure 5, then we can write the following dm dt = ρA1 dx1 dt = ρA1v1 (6.35) for the amount of mass dm passing through area A1 in the time interval dt ( ρ is the density of the fluid). If the flow is steady, then the amount of mass flowing through some other area A2 during the same interval dt . More specifically, we can write ρA1v1 = ρA2v2, (6.36) or simply A1v1 = A2v2. (6.37) Figure 5 – Continuity and mass conservation in a flow. - 132 - Equation (6.37) is the so-called continuity equation for steady flows (of incompressible fluids). Generally, we can state that the volume flow rate is conserved dV dt = Av = constant. (6.38) 6.2.2 Bernoulli’s Equation We now seek to apply the mass continuity equation while taking into account any changes of pressure that can accompany the flow of fluids. Such changes in pressure are to be expected whenever the cross-sectional area A changes along a flow. This is because as the area varies, the velocity must also change according to the continuity equation (6.37); if the flow speed changes, then there must be forces acting on the flow to cause this acceleration. Finally, pressure variations must also occur since pressure is defined as the force per cross-sectional area. To derive the equation that relates these quantities, we will use the work-energy theorem defined in Chapter 2, which we write her for convenience Wother = ΔK + ΔUgrav, (6.39) where Wother is the work done by all forces other than gravity, ΔK is the change in kinetic energy, and ΔUgrav is the change in gravitational potential energy. In our case we will substitute Wother →Wpressure ≡Wp . Let us consider the tube of changing cross-section shown in Figure 6. We first concentrate on the section on the left of width v1dt and cross-section A1 , through which the flow speed is v1 and the pressure p1 . We can ask what amount of work dW1 was done by the pressure on a fluid that has traveled from the entrance to the exit of that section? The important fact to remember is that pressure is isotropic, meaning that the force p1A1 at the entrance has the same magnitude as the force at the exit but of opposite direction. It therefore follows that dWp,1 = Fnet,1 ⋅dr 1 = p1A1v1dt ( ) entrance + −p1A1v1dt ( ) exit = 0. (6.40) The pressure does no work when cross-section of the flow is constant. The same result dWp,2 = 0 would be found for the section, of width v2dt and cross-section A2 , through which the flow speed is v2 and the pressure p2 , on the right of the tube. The same cannot be said for the middle section of the tube, where the cross-section changes from A1 to A2 . In this case we find - 133 - dWp,12 = Fnet,12 ⋅dr 12 = p1A1v1dt ( ) entrance + −p2A2v2dt ( ) exit , (6.41) where dt is an infinitesimal time interval such that v1dt and v2dt are much smaller than the width of the section. We now use equation (6.37) for mass continuity and transform equation (6.41) to dWp,12 = p1 −p2 ( )dV, (6.42) with dV the volume element spanned in the interval dt (see equation (6.38)). This equation can be integrated over between any two points along a tube, and generalized to (for an incompressible and non-viscous fluid) Wp = p1 −p2 ( )dV. (6.43) We can now write down the corresponding changes in kinetic and gravitational potential energies (if there is a change in vertical position y between point 1 and 2) ΔK = 1 2 ρ v2 2 −v1 2 ( )dV ΔUgrav = ρg y2 −y1 ( )dV. (6.44) Combining equations (6.39), (6.43), and (6.44) we get the so-called Bernoulli’s Equation p1 + 1 2 ρv1 2 + ρgy1 = p2 + 1 2 ρv2 2 + ρgy2. (6.45) 2 1 p 1 p 1 A 1 v dt 1 v dt 2 p 2 p 2 p 2 A p Figure 6 – A tube of changing cross-section, through which an incompressible and non-viscous fluid is flowing. - 134 - Since equation (6.45) applies for any two points along the flow, we can write p + 1 2 ρv2 + ρgy = constant. (6.46) It is interesting to note that, although it is defined as an applied force per unit area, pressure can also be thought of as energy per unit volume. Finally, we note that when the fluid is not moving v1 = v2 = 0 in equation (6.45) we recover p1 = p2 + ρg y2 −y1 ( ), (6.47) which is the same as equation (6.11), previously derived for cases of hydrostatic equilibrium. 6.3 Exercises 4. (Prob. 12.40 in Young and Freedman.) Artery blockage. A medical technician is trying to determine what percentage of a patient’s artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 1.20 ×104Pa , while in the region of blockage it is 1.15 ×104Pa . Furthermore, she knows that the blood flowing through the normal artery just before the point of blockage is traveling at 30.0 cm/s, and that the specific gravity of this patient’s blood is 1.06. What percentage of cross-sectional area of the patient’s artery is blocked by the plaque? Solution. We use Bernoulli’s equation (i.e., equation (6.45)) while assuming that y1 = y2 , where ‘1’ and ‘2’ correspond to regions before and at the blockage, respectively. We first determine the speed of the blood in the blockage with v2 = 2 p1 −p2 ( ) ρ + v1 2 = 2 1.20 ×104 −1.15 ×104 ( )Pa 1,060 kg/m3 + 0.300 m/s ( ) 2 = 1 m/s. (6.48) We then use the continuity equation to find A2 A1 = v1 v2 = 0.30. (6.49) The artery is therefore 70% blocked by the plaque. - 135 - 5. (Prob. 12.55 in Young and Freedman.) A dam has the shape of a rectangular solid. The side facing the lake has an area A and a height h . The surface of the fresh water lake behind the dam is at the top of the dam. (a) Show that the net horizontal force exerted by the water on the dam is ρgAh 2 − that is, the average gauge pressure across the face of the dam times the area. (b) Show that the torque exerted by the water about an axis along the bottom of the dam is ρgAh2 6 . Solution. (a) The pressure at a given depth h −y ( y = 0 is at the bottom of the dam) is given by p = p0 + ρg h −y ( ) (6.50) with p0 the atmospheric pressure at the top of the lake (and dam). The force exerted by the water on a horizontal strip of the dam of width dy at that depth is dF ⊥= p A h dy = A h p0 + ρg h −y ( ) ⎡ ⎣ ⎤ ⎦dy, (6.51) with A h the width of the dam since it is rectangular. The total force exerted by the water on the dam will then be F ⊥= dF ⊥ 0 h ∫ = A h p0 + ρgh ( ) dy 0 h ∫ −ρg ydy 0 h ∫ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = A h p0 + ρgh ( )h −1 2 ρgh2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = p0A + 1 2 ρgAh. (6.52) However, the dam “feels” a force equal to p0A on its side opposing the lake from the atmosphere. The total force on the dam is therefore F total = 1 2 ρgAh. (6.53) (b) The torque about the axis at y = 0 acting on a horizontal strip of the wall at depth h −y is - 136 - dτ = dF ⊥−p0 A h dy ⎛ ⎝ ⎜ ⎞ ⎠ ⎟⋅y = A h ρgy h −y ( )dy. (6.54) The total torque on the dam is thus τ = dτ 0 h ∫ = A h ρg h ydy 0 h ∫ − y2 dy 0 h ∫ ( ) = A h ρg h3 2 −h3 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1 6 ρgAh2. (6.55) 6. (Prob. 12.98 in Young and Freedman.) A siphon, as shown in Figure 7, is a convenient device for removing liquids from containers. To establish the flow, the tube must be initially filled with fluid. Let the fluid have a density ρ , and let the atmospheric pressure be patm . Assume that the cross-sectional area of the tube is the same at all points along it. (a) If the lower end of the siphon is at a distance h below the surface of the liquid in the container, what is the speed of the fluid as it flows out of the lower end of the siphon? (Assume that the container has a very large diameter, and ignore any effect of viscosity.) (b) A curious feature of a siphon is that the fluid initially flows “uphill.” What is the greatest height H that the high point of the tube can have if flow is still to occur? Solution. (a) The pressure at the top of the liquid in the container is p0 = patm , the same as it is at the lower end of the tube. Applying Bernoulli’s equation with points “1” and “2” at the top of the liquid in the container and the lower end of the tube, respectively, we have p0 + 1 2 ρv1 2 + ρgh = p0 + 1 2 ρv2 2, (6.56) or v2 2 = v1 2 + 2gh. (6.57) But with a very large container we can assume that v1 ≈0 and v2 = 2gh. (6.58) - 137 - (b) The pressure at the high point of the tube can be related to that at its low end with p = p0 −ρg H + h ( ). (6.59) The liquid will not flow if the absolute pressure is negative anywhere in the tube, since a zero absolute pressure would imply that a perfect vacuum be present at that point. We then write p0 −ρg H + h ( ) > 0, (6.60) or H < p0 ρg −h. (6.61) We note that equation (6.61) also implies a limitation on H + h, which for water and normal atmospheric pressure yields H + h < patm ρg <10.3 m. (6.62) Figure 7 – A siphon, for removing liquids from a container. |
189272 | https://germanna.edu/sites/default/files/2022-03/Naming%20Compounds.pdf | Naming Compounds Provided by the Academic Center for Excellence 1 Updated October 2013 Naming Compounds Naming compounds is an important part of chemistry. Most compounds fall in to one of three categories- ionic compounds, molecular compounds, or acids. You can navigate to specific sections of this handout by clicking the links below. Naming Ionic Compounds: pg. 1 Naming Binary Molecular Compounds: pg. 3 Naming Acids: pg. 4 Practice Problems: pg. 6 Part One: Naming Ionic Compounds Identifying Ionic Compounds Ionic compounds consist of combinations of positively charged ions called cations (usually metals), and negatively charged ions called anions (usually non-metals). In general, you can identify an ionic compound because it contains a metal (these are usually found in the left and center areas of the periodic table) and a non-metal (these are generally found in the right hand area of the periodic table). Also, a compound will have no charge. For example, NaCl and Fe2O3 are ionic compounds; they each contain a metal (Na and Fe) and a non-metal (Cl and O), and they do not have charges. MnO4- is NOT an ionic compound; it does contain a metal (Mn) and a non-metal (O), but it has a charge. Thus, it is a polyatomic ion, not a compound. A compound will NEVER have a charge! Naming Ionic Compounds There are three steps involved in naming ionic compounds- naming the cation, naming the anion, and naming the entire compound. 1. Name the cation. i. Cations formed from metal atoms have the same name as the metal. Examples: Na+= sodium ion; Al3+= aluminum ion ii. If a metal can form ions of different charges (i.e., is one of the central transition metals), specify the charge with Roman numerals in parentheses. Examples: Fe+= iron (I) ion; Fe2+= iron (II) ion; Fe3+= iron (III) ion Provided by the Academic Center for Excellence 2 Naming Compounds iii. Cations formed from nonmetal ions have names ending in –ium. These are not common; the main ones are NH4+ (ammonium ion) and H3O+ (hydronium ion) 2. Name the anion. i. Monoatomic anions (those formed from a single atom) have names formed by replacing the end of the element name with – ide. Examples: F- = fluoride ion; O2-= oxide ion. A few simple polyatomic anions (those formed from multiples atoms) also have names ending in –ide. Examples: CN- = cyanide ion; OH- = hydroxide ion; O22-= peroxide ion. ii. Most polyatomic ions contain oxygen, and have names ending in -ate or -ite. They are known as oxyanions. The ending –ate is used for the most common oxyanion form. The ending –ite is used for an oxyanion that has the same charge, but one less oxygen atom. Examples: SO42- = sulfate; SO32- = sulfite (same charge, but one less oxygen) iii. The suffixes per- and hypo- are added to the names of oxyanions to show the addition or subtraction of additional oxygen atoms. Per- indicates the addition of one oxygen to the -ate form. Hypo- indicates the subtraction of one oxygen from the –ite form. Thus – ate is the most common form, per--ate has one extra oxygen, -ite has one less oxygen, and hypo--ite has two less oxygen. Example: ClO4- = perchlorate (one more oxygen than regular form) ClO3- = chlorate (regular form) ClO2- = chlorite (one less oxygen than regular form) ClO- = hypochlorite (two less oxygen than regular form) iv. Anions formed by adding H+ to an oxyanion have the word “hydrogen” in front of their names (or “dihydrogen,” if two hydrogens are present.) Examples: CO32- = carbonate ion; HCO3- = hydrogen carbonate ion (notice that the addition of hydrogen lessens the negative charge by one). PO43- = phosphate ion; H2PO4- = dihydrogen phosphate. 3. Name the compound. i. To name the compound, simply put the names of the ions together. The name of an ionic compound is always the cation name followed by the anion name. Examples: CaCl2= calcium chloride; Al(NO3)3= aluminum nitrate ii. If you are dealing with a transition metal, don’t forget to specify its charge. Provided by the Academic Center for Excellence 3 Naming Compounds iii. If you are dealing with an oxyanion, be sure you have the right name for the form you are using. Example: Cu(ClO4)2 = copper (II) perchlorate iv. If you are having trouble determining the charge on an ion, look at the subscript on the opposite ion. In the above example, we know that the charge on the copper ion is +2 because the subscript on the opposite ion, the perchlorate, is 2, and copper is a metal, so it always has a positive charge. The charge on the perchlorate is -1 because the subscript on the copper is 1 (subscripts of 1 are not written in formulas- thus, because the copper has no written subscript, we know that it is 1), and perchlorate is an anion, so it always has a negative charge. v. You can use this same method to determine the correct subscript when you are writing a chemical formula based on a name. Example: write the formula for magnesium bromide. This is a compound containing magnesium and bromine ions- Mg2+, and Br - . To determine what subscripts, if any, to use, look at the opposite charges. The subscript on bromine will be 2, because the charge on the magnesium is 2. The subscript on magnesium will be one, because the charge on bromine is -1. Thus the formula is MgBr2. (Remember, subscripts of 1 are not written). Likewise, given the name Iron (III) oxide, we can determine that the iron will have a subscript of 2, because the charge on oxygen ion is -2; the oxygen will have a subscript of 3, because we have been told we are dealing with iron with a charge of 3. So the formula is Fe2O3. The only time this rule is not true is when the charges on the ions are equal- for example, when oxygen, with a charge of -2, bonds with magnesium, which has a charge of +2. In this case, the charge on one oxygen ion is equal to the charge on one magnesium ion, so it will only take one oxygen ion and one magnesium ion to form a compound that has no charge. Thus, this compound has the formula MgO, not Mg2O2. The same thing happens when calcium and oxygen combine. Calcium has a charge of +2, and oxygen has a charge of -2. Because their charges are equal, it only takes one of each to form a compound with no charge, so the formula is CaO, not Ca2O2. Provided by the Academic Center for Excellence 4 Naming Compounds Part Two: Naming Binary Molecular Compounds Identifying Binary Molecular Compounds Molecular compounds consist of combinations of non-metals. Binary molecular compounds are composed of only two elements. They are easy to identify, as they consist merely of two non-metal elements. Examples: H2O (water), NF3, and N2O4. Naming Binary Molecular Compounds There are four steps to name binary molecular compounds: 1. The name of the element farthest to the left in the periodic table is written first. i. There are occasional exceptions to this rule. The main exception is oxygen. Oxygen, except when combined with fluorine, is always written last. 2. If both elements are in the same group in the table, the lower one is written first 3. The name of the second element is given an -ide ending. 4. Greek prefixes are used to indicate the number of atoms of each element. i. The prefixes are as follows: Mono-= one Di-= two Tri-= three Tetra-= four Penta-= five Hexa-= six Hepta-= seven Octa-= eight Nona-= nine Deca-= ten ii. The prefix mono- is never used with the first element. If only one atom of the first element is present, do not use a prefix. Examples of binary molecular compounds and their names: Cl2O= dichlorine monoxide NF3= nitrogen trifluoride N2O4= dinitrogen tetroxide P4S10= tetraphosphorus decasulfide. Provided by the Academic Center for Excellence 5 Naming Compounds Part Three: Naming Acids Identifying Acids Acids are hydrogen containing compounds. Acids are easy to recognize- they are composed of hydrogen and an anion (the hydrogen always comes first), and they have no charge. Examples: HCl and H2S04 are acids; they are made up of hydrogen and anions, and they do not have charges. HCO3- is NOT an acid; it is made up of hydrogen and an anion, but it has a charge, and so it is a polyatomic ion. Naming Acids There are two steps involved in naming acids. 1. Acids based on anions whose names end in –ide When an ion ending in -ide becomes an acid, its name changes- its suffix changes from –ide to –ic, and it gains a prefix, hydro-. Thus, Cl-, the chloride ion, becomes HCl, hydrochloric acid. S2-, the sulfide ion, becomes H2S hydrosulfuric acid (we add two hydrogen ions because the sulfide ion has a charge of 2-. We must add enough hydrogen ions, which have a charge of 1+, to cancel out the charge on the sulfide. One hydrogen ion would give us HS-, which is not an acid as it still has a charge). 2. Acids based on anions whose names in –ate or –ite When an ion ending in –ate becomes an acid, its suffix changes to –ic, but it does not gain a prefix. If it already contains the prefix per- (as in perchlorate), it will retain that prefix, and will be per_ic acid. When an ion ending in ite becomes an acid, its suffix changes to -ous. If it contains the prefix hypo- (as in hypochlorite), it retains that prefix, and will be hypoous acid. Thus, ClO3-, the chlorate becomes HClO3, chloric acid. Perchlorate (ClO4-) becomes HClO4, perchloric acid. Chlorite, ClO2-, becomes HClO2, chlorous acid, while hypochlorite, ClO-, becomes HClO, hypochlorous acid. The naming of acids can be summarized in the following chart: Provided by the Academic Center for Excellence 6 Naming Compounds -ide anion (chloride, Cl-) -ate anion (chlorate, ClO3-) perate anion (perchlorate, ClO4-) -ite anion (chlorite, ClO2-) hydro__ic acid (hydrochloric acid, HCl) _ic acid (chloric acid, HClO3) per_ic acid (perchloric acid, HClO4) _ous acid (chlorous acid, HClO2) hypo__ite anion (hypochlorite, ClO-) hypo__ous acid (hypochlorous acid, HClO) Provided by the Academic Center for Excellence 7 Naming Compounds Practice Problems- Ionic Compounds Name the following ionic compounds: 1. NH4Br 2. Cr2O3 3. Co(NO3)2 4. K2SO4 5. Ba(OH)2 6. FeCl3 7. AlF3 8. Fe(OH)2 9. Cu(NO3)2 10. Ba(ClO4)2 11. Li3PO4 12. Hg2S 13. Cr2(CO3)3 14. K2CrO4 15. (NH4)2SO4 16. Ca(C2H3O)2 Now go the other way- give the formulas for the following names: 17. Potassium sulfide 25. Cesium fluoride 18. Calcium carbonate 26. Magnesium iodide 19. Nickel (II) perchlorate 27. Iron (III) carbonate 20. Magnesium sulfate 28. Sodium hypobromite 21. Silver (I) sulfide 29. Cobalt (II) nitrate 22. Lead (II) nitrate 30. Chromium (II) acetate 23. Copper (I) oxide 31. Copper (II) perchlorate 24. Aluminum hydroxide 32. Calcium hydrogen carbonate Practice Problems- Molecular Compounds Name these binary molecular compounds: 1. SO2 2. PCl5 3. N2O3 4. SF6 5. IF5 6. XeO3 7. N2O5 8. BF3 9. CCl4 10. P4O6 11. SiO2 12. O2F2 13. XeF6 14. AsCl3 15. P2O5 16. AsBr3 Provide formulas for the following binary molecular compounds: 17. Silicon tetrabromide 18. Disulfur dichloride 19. Dinitrogen tetroxide 20. Tetraphosphorus hexasulfide 21. Sulfur hexafluoride 22. Phosphorus tribromide 23. Carbon tetraiodide 24. Dihydrogen monoxide 25. Phosphorus triiodide 26. Iodine monobromide 27. Diboron trioxide 28. Nitrogen trichloride 29. Carbon monoxide Provided by the Academic Center for Excellence 8 Naming Compounds 30. Silicon tetrachloride 31. Dinitrogen pentoxide 32. Nitrogen dioxide Practice Problems- Acids Name the following acids: 1. HCN 2. HNO3 3. H2SO4 4. H2SO3 5. HF 6. HBr 7. HI 8. H3PO4 9. HC2H3O2 10. HNO2 11. HBrO3 12. HBrO4 13. H2Se 14. H3PO3 15. HCl 16. H2CO3 NOTE: Problems 11-14 all use ions that are not common. The ion in problem 11, BrO3-, is bromate. The ion in problem 12, BrO4- is perbromate. The ion in problem 13, Se2-, is selenide, the ion formed by element 34, selenium. The ion in problem 14, PO33-, is phosphate. Provide formulas for the following acids: 17. Hypochlorous acid 18. Hydroiodic acid 19. Sulfurous acid 20. Hydrobromic acid 21. Hydrosulfuric acid 22. Nitrous acid 23. Perbromic acid 24. Acetic acid 25. Hydroselenic acid 26. Bromous acid 27. Hydrofluoric acid 28. Phosphoric acid 29. Nitric acid 30. Hydrocyanic acid 31. Sulfuric acid 32. Carbonic acid Provided by the Academic Center for Excellence 9 Naming Compounds Answer Key- Ionic Compounds Names from formulas: 1. Ammonium bromide 9. Copper (II) nitrate 2. Chromium (III) oxide 10. Barium perchlorate 3. Cobalt (II) nitrate 11. Lithium phosphate 4. Potassium sulfate 12. Mercury (I) sulfide 5. Barium hydroxide 13. Chromium (III) carbonate 6. Iron (III) chloride 14. Potassium chromate 7. Aluminum fluoride 15. Ammonium sulfate 8. Iron (II) hydroxide 16. Calcium acetate Formulas from names: 17. K2S 25. CsF 18. CaCO3 26. MgI2 19. Ni(ClO4)2 27. Fe2(CO3)3 20. MgSO4 28. NaBrO 21. Ag2S 29. Co(NO3)2 22. Pb(NO3)2 30. Cr(C2H3O2)2 23. Cu2O 31. Cu(ClO4)2 24. Al(OH)3 32. Ca(HCO3)2 Answer Key- Molecular Compounds Names from formulas: 1. Sulfur dioxide 2. Phosphorus pentachloride 3. Dinitrogen trioxide 4. Sulfur hexafluoride 5. Iodine pentafluoride 6. Xenon trioxide 7. Dinitrogen pentoxide 8. Boron trifluoride 9. Carbon tetrachloride 10. Tetraphosphorus hexoxide 11. Silicon dioxide 12. Dioxide difluoride 13. Xenon hexafluoride 14. Arsenic trichloride 15. Diphosphorus pentoxide 16. Arsenic tribromide Formulas from names: 17. SiBr4 18. S2Cl2 19. N2O4 20. P4S6 21. SF6 22. PBr3 23. CI4 24. H2O 25. PI3 26. IBr 27. B2O3 28. NCl3 29. CO 30. SiCl4 31. N2O5 32. NO2 Provided by the Academic Center for Excellence 10 Naming Compounds Answer Key- Acids Names from formulas: 1. Hydrocyanic acid 2. Nitric acid 3. Sulfuric acid 4. Sulfurous acid 5. Hydrofluoric acid 6. Hydrobromic acid 7. Hydroiodic acid 8. Phosphoric acid 9. Acetic acid 10. Nitrous acid 11. Bromic acid 12. Perbromic acid 13. Hydroselenic acid 14. Phosphorous acid 15. Hydrochloric acid 16. Carbonic acid Formulas from names: 17. HClO 18. HI 19. H2SO3 20. HBr 21. H2S 22. HNO2 23. HBrO4 24. HC2H3O2 25. H2Se 26. HBrO2 27. HF 28. H3PO4 29. HNO3 30. HCN 31. H2SO4 32. H2CO3 |
189273 | https://byjus.com/jee/nature-of-roots-depending-upon-coefficients-and-discriminant/ | We know that a quadratic equation is a second degree polynomial equation of the form ax2 + bx + c = 0, where a, b, c are real numbers, x is the unknown variable and a ≠ 0. For the equation ax2 + bx + c = 0, the discriminant is given by D = b2 – 4ac. It is also denoted by ∆. A quadratic equation has 2 roots. It will be real or imaginary. In this article we discuss the nature of roots depending upon coefficients and discriminant.
If α and β are the values of x which satisfy the quadratic equation, α and β are called the roots of the quadratic equation. Roots are given by the equation (-b±√(b2-4ac))/2a. The nature of the roots depends on the discriminant.
Nature of Roots depending upon Discriminant
According to the value of discriminant, we shall discuss the following cases about the nature of roots.
Case 1: D = 0
If the discriminant is equal to zero (b2 – 4ac = 0), a, b, c are real numbers, a≠0, then the roots of the quadratic equation ax2 + bx + c = 0, are real and equal. In this case, the roots are x = -b/2a. The graph of the equation touches the X axis at a single point.
Case 2: D > 0
If the discriminant is greater than zero (b2 – 4ac > 0), a, b, c are real numbers, a≠0, then the roots of the quadratic equation ax2 + bx + c = 0, are real and unequal. The graph of the equation touches the X-axis at two different points.
Case 3: D < 0
If the discriminant is less than zero (b2 – 4ac < 0), a, b, c are real numbers, a≠0, then the roots of the quadratic equation ax2 + bx + c = 0, are imaginary and unequal. The roots exist in conjugate pairs. The graph of the equation does not touch the X-axis.
Case 4: D > 0 and perfect square
If D > 0 and a perfect square, then the roots of the quadratic equation are real, unequal and rational.
Case 5: D > 0 and not a perfect square
If D > 0 and not a perfect square, then the roots of the quadratic equation are real, unequal and irrational.
We can summarize all the above cases in the table below.
| | |
--- |
| Discriminant | Nature of roots |
| D = 0 | Real and equal roots. |
| D > 0 | Real and unequal roots. |
| D < 0 | Unequal and imaginary |
| D > 0 and perfect square | Real, unequal and rational |
| D > 0 and not a perfect square | Real, unequal and irrational |
Nature of Roots depending upon coefficients
Depending upon the nature of the coefficients of the quadratic equation, we can summarize the following.
Bridge Course – Nature of Roots of Quadratic Equations
Also Read
Quadratic inequalities
Solved Examples
Example 1:
The roots of the quadratic equation 3x2-10x+3 = 0 are
a) real and equal
b) imaginary
c) real, unequal and rational
d) none of these
Solution:
Given equation 3x2-10x+3 = 0
Here discriminant, D = b2-4ac
=> (-10)2 – 4×3×3
= 100 – 36
= 64
D is positive and a perfect square.
So the roots of the quadratic equation are real, unequal and rational.
Hence option c is the answer.
Example 2:
Find the value of p if the equation 3x2-18x+p = 0 has real and equal roots.
a) 27
b) 18
c) 9
d) none of these
Solution:
Given 3x2-18x+p = 0 has real and equal roots.
=> b2-4ac = 0
=>(-18)2-4×3×p = 0
=> 324 – 12p = 0
=> p = 324/12
= 27
Hence option a is the answer.
Example 3:
The quadratic equation with real coefficients when one of its root is (3+2i) is
Solution:
Given one root is 3+2i.
Complex roots occur in conjugate pairs.
So other root = 3-2i
Sum of roots = 6
Product of roots = (3+2i)(3-2i) = 13
Required equation is x2-(Sum)x+Product = 0
=> x2-6x+13 = 0
Example 4:
Show that the equation 3x2+4x+6 = 0 has no real roots.
Solution:
Given equation 3x2+4x+6 = 0
Here a = 3, b = 4, c = 6
Discriminant D = b2-4ac
=> 42-4×3×6
= 16-72
= -56
Since D<0, the roots are imaginary.
Hence the equation has no real roots.
Video Lesson – Nature of Roots
Frequently Asked Questions
Give the equation for the discriminant of the quadratic equation.
The discriminant of a quadratic equation is given by D = b2 – 4ac.
What is the nature of the roots of a quadratic equation, if discriminant equals zero?
If discriminant, D = 0, then the roots are real and equal.
What is the nature of the roots of a quadratic equation, if the discriminant is greater than zero?
If discriminant, D>0, then the roots are real and unequal.
What is the nature of the roots of a quadratic equation, if the discriminant is less than zero?
If discriminant, D<0, then the roots are imaginary and unequal.
Comments
Leave a Comment Cancel reply
Your Mobile number and Email id will not be published. Required fields are marked
Request OTP on
Voice Call
Website
Post My Comment
Register with BYJU'S & Download Free PDFs
Register with BYJU'S & Watch Live Videos |
189274 | https://www.bohrium.com/paper-details/inventory-and-pricing-model-with-price-dependent-demand-time-varying-holding-cost-and-quantity-discounts/814555722195927041-3899 | Research Assistant - Bohrium | AI for Science with Global Scientists
New Chat
Academic Search
Explore
Subscription
Library
Scholars
Knowledge Base
Practice
Tools
Notebooks
Courses
Competitions
Uni-Lab
Computation
Jobs
Files
Nodes
Datasets
Images
Projects
Database
History
view all history
Upgrade
English
Log in
Back
Inventory and pricing model with price-dependent demand, time-varying holding cost, and quantity discounts
Page navigation
Details
AI Summary
Comprehensive info
Abstract
References
Cited by
Related articles
Journal information
Details
2016-04-01
704
Citations 84
Inventory and pricing model with price-dependent demand, time-varying holding cost, and quantity discounts
Save Cite Share
Computers and Industrial Engineering
H
Hesham K. Alfares
A
Ahmed M. Ghaithan
LinkOfficial website
IF 6.5 DOI 10.1016/j.cie.2016.02.009
OA No Research category-
Summary
Research Direction: Inventory model ｜ Price - dependent demand ｜ Variable holding cost
Introduction
The paper starts by stating that most inventory control models assume a fixed demand rate, but in reality, demand can be affected by variables like price and availability. Also, unit holding cost may vary over time. The objective is to develop an inventory model where demand rate depends on selling price, unit holding cost on storage time, and purchase cost on order size. The demand rate is assumed to be a linearly decreasing function of the unit selling price, the unit holding cost as a linearly increasing function of storage time, and the unit purchase cost as a decreasing step function of order quantity. A mathematical model will be formulated and an optimal solution methodology developed to maximize total profit. The food industry is given as an example of where these assumptions are applicable.
Literature Review
The relevant recent research is classified into four types of EOQ inventory models: those with variable demand, variable holding costs, both variable demand and variable holding costs, and those with quantity discounts. For variable demand rate models, demand can vary with price, stock level or both. There are various models developed by different researchers considering different relationships between demand, price, deterioration rate, etc. For variable holding cost models, different assumptions like non - linear dependence on storage time are considered. In models with both variable demand and holding costs, demand rate is often stock - dependent. For inventory models with quantity discounts, all - units and incremental quantity discounts are considered. However, no previous inventory model has simultaneously considered price - dependent demand, variable holding cost, and quantity discounts.
Problem Description and Formulation
The aim of the paper is to maximize the profit of the proposed inventory model with specific features. These include the demand rate being a decreasing linear function of the selling price, the holding cost being an increasing linear function of the storage time, and the purchase cost being a decreasing step function of the order size according to all - units quantity discounts. Assumptions are made such as the unit holding cost being proportional to the unit purchase cost, no shortages are allowed, a single item is considered which does not deteriorate in storage. The rate of decrease in inventory level is related to the demand rate. The objective function to maximize total profit per time unit is formulated considering sales revenue and cost components like ordering, purchasing and holding costs.
Solution Algorithm
The optimum solution procedure steps are described. Solutions obtained from certain equations need to be rounded. The process starts with initializing values. Then, for each purchase cost range, values are substituted into equations to solve for price and order size. If the order size is in the right range, the solution is feasible and profit is calculated. If not, alternative steps are followed. If the solution is not the best so far, the process is repeated for the next purchase cost range until the final optimal solution is found which includes optimal values of order size, price, cycle time, total cost and total profit.
Conclusions and Recommendation
The paper presented an inventory model with variable demand rate, variable holding cost, and variable purchase cost based on realistic assumptions. A mathematical model was formulated and an efficient solution algorithm developed to find the optimal order size and unit selling price. A numerical example was solved and sensitivity analysis done. The results show that the decision variables and profit function are mostly affected by demand parameters. Companies should first focus on increasing revenues through marketing, then reduce costs especially purchasing and ordering costs. Also, if these costs are reduced, the selling price can be decreased to boost demand and increase revenues. Possible extensions of the model include considering different forms of demand variability, shortages, deteriorating items, and the time value of money.
Comprehensive information
Keywords
All-units quantity discounts
Inventory model
Price-dependent demand
Variable holding cost
Abstract
In typical EOQ-based inventory models, the demand rate and the holding cost are assumed to have constant values and the unit purchase cost is assumed constant regardless of the order size. In actual applications, however, the demand rate for a specific item can be affected by many variables such as seasonality, selling price, and availability. Moreover, the unit holding cost tends to be higher for extended storage periods. Additionally, the unit purchase cost is generally lower for larger order sizes due to quantity discounts. The objective of this paper is to simultaneously consider the variability of the demand rate, the unit holding cost, and the unit purchase cost. An inventory model is presented with a selling price-dependent demand rate, a storage time-dependent holding cost, and an order size-dependent purchase cost based on all-units quantity discount. A mathematical model is constructed, and a solution methodology is developed for determining the optimal solution. A numerical example is solved, and sensitivity analysis is conducted to study the effect of various parameters on the optimal solution.
References
MAXIMUM-PROFIT INVENTORY MODEL WITH STOCK-DEPENDENT DEMAND, TIME-DEPENDENT HOLDING COST, AND ALL-UNITS QUANTITY DISCOUNTS
H
Hesham Kamal Alfares
2015-11-23Mathematical Modelling and Analysis(IF 1.4)
An Inventory Model with Price and Quality Dependent Demand Where Some Items Produced Are Defective
Tapan Kumar Datta
2013-01-01Advances in Operations Research(IF 0.8)
EOQ Models with Varying Holding Cost
N
Naser Ghasemi
Behrouz Afshar-Nadjafi
2013-05-30Journal of Industrial Mathematics
Expand More
Cited by
An application of Arctic puffin optimization algorithm of a production model for selling price and green level dependent demand with interval uncertainty
H
Hachen Ali
M
Md. Al-Amin Khan
+3
2025-07-28scientific reports(IF 3.9)
Application of interval Laplace transformation in a production system with price and quality dependent demand via water cycle algorithm
H
Hachen Ali
F
Fleming Akhtar
+3
2025-06-21Evolutionary Intelligence(IF 2.6)
Price-dependent fuzzy demand and shortages scenario with memory effect on inventory system incorporating optimal replenishment strategy
S
Shilpi Pal
P
P K Santra
+1
2025-05-30Engineering Research Express(IF 1.6)
Expand More
Related articles
Refresh
An Inventory Model for Perishable Items with Price-, Stock-, and Time-Dependent Demand Rate considering Shelf-Life and Nonlinear Holding Costs
Adrián Macías López
Leopoldo Eduardo Cárdenas-Barrón
+2
2021-04-09Mathematical Problems in Engineering
Selling price, time dependent demand and variable holding cost inventory model with two storage facilities
D
Deo Datta Aarya
Y
Yogendra Kumar Rajoria
+5
2022-01-01Materials Today: Proceedings
An Inventory Model with Time-Varying Demand and Comprehensive Cost
V
VIVEK VIJAY -
2024-04-30International Journal For Multidisciplinary Research
Inventory System with Time and Price Dependent Demand Ratio for Decreasing Items
D
Deep Kamal Sharma
O
Ompal Singh
2023-10-02International Journal of Membrane Science and Technology
An inventory model with price- and stock-dependent demand and time- and stock quantity-dependent holding cost under profitability maximization
V
Valentín Pando
L
Luis A. San-José
+2
2024-01-05Computers & Operations Research(IF 4.3)
Journal information
Journal name Computers & Industrial Engineering
Journal name abbreviation COMPUT IND ENG
Official website
IF 6.5 (JCR-Q1，中科院-工程技术1区)
Country/region ENGLAND
Annual articles 812
Percentage of Chinese authors 53.25%
Self-Citation Rate 10.8%
Layout fee US$2,950
Review period 平均11.1月
Open access 否
Warning Not in the warning list
Database Scopus收录/SCIE
Publication frequency Bimonthly
Publisher Elsevier Ltd
Review journal Yes
ISSN 0360-8352
Journal description Industrial engineering is one of the earliest fields to utilize computers in research, education, and practice. Over the years, computers and electronic communication have become an integral part of industrial engineering. Computers & Industrial Engineering (CAIE) is aimed at an audience of researchers, educators and practitioners of industrial engineering and associated fields. It publishes original contributions on the development of new computerized methodologies for solving industrial engineering problems, as well as the applications of those methodologies to problems of interest in the broad industrial engineering and associated communities. The journal encourages submissions that expand the frontiers of the fundamental theories and concepts underlying industrial engineering techniques.CAIE also serves as a venue for articles evaluating the state-of-the-art of computer applications in various industrial engineering and related topics, and research in the utilization of computers in industrial engineering education. Papers reporting on applications of industrial engineering techniques to real life problems are welcome, as long as they satisfy the criteria of originality in the choice of the problem and the tools utilized to solve it, generality of the approach for applicability to other problems, and significance of the results produced. A major aim of the journal is to foster international exchange of ideas and experiences among scholars and practitioners with shared interests all over the world.For more information, please visit the Journal's Editorial Office page.Benefits to authors We also provide many author benefits, such as free PDFs, a liberal copyright policy, special discounts on Elsevier publications and much more. Please click here for more information on our author services. Please see our Guide for Authors for information on article submission. If you require any further information or help, please visit our Support Center
IF
Annual articles
Self-Citation Rate
Cite Score
Hi, Scientist！
Inventory and pricing model with price-dependent demand, time-varying holding cost, and quantity discounts
LitTalk Upload files or ask any question about the current paper
Summarize
Extract Findings
Research Methods
Academic Concepts
To pick up a draggable item, press the space bar. While dragging, use the arrow keys to move the item. Press space again to drop the item in its new position, or press escape to cancel.
Would you recommend Bohrium to friends or peers?
0
1
2
3
4
5
6
7
8
9
10
Not at All Likely Extremely Likely
Submit |
189275 | https://m.youtube.com/watch?v=hqFE3X-YAZg | Lecture 15 - Binomial Theorem | Binomial Coefficient | Combinatorics | Discrete Mathematics | Deepak
GO Classes for GATE CS
116000 subscribers
147 likes
Description
12061 views
Posted: 11 Aug 2023
Annotated Notes - Lecture 14B-17 - Binomial Theorem, Permutation with repetition:
➤➤ Combinatorics Complete Course Playlist:
➤ Discrete Mathematics Complete Course(FREE for ALL):
Group Theory Complete Course(Abstract Algebra Course):
Graph Theory Complete Course:
Crack GATE Computer Science Exam with the Best Course.
➤ Join "GO Classes #GateCSE Complete Course":
Join GO Classes #Gate2024 Complete Course here:
➤ Download GO Classes Android APP:
GoClasses Website :
GOClasses ALL Links :
➤ Join GATEOverflow + GoClasses Combined TEST SERIES for Best Quality questions for GATE CSE Preparation, Here:
➤Join GATE Overflow & GO Classes #Telegram Groups for GATE CSE Doubt Discussions:
1.
2.
3.
➤ Watch Complete Discrete Mathematics and Complete Engineering Mathematics Courses on GO Classes( FREE for ALL learners) :
Complete #Discrete_Mathematics Course(FREE) Link :
Complete #Engineering_Mathematics Course(FREE) Link :
Know ALL about GO Classes GATE 2024 Course:
Download GATEOverflow GATE Previous Years Questions(GATE CSE PYQs) Books here:
Feel free to Contact Us for any query.
➤ GO Classes Contact :
(+91)63025 36274
(+91)9468930964
GO Classes Mail ID :
contact@goclasses.in
contact.goclasses@gmail.com
gatecse #gate2024 #goclasses #computerscience #computer_science
7 comments
Transcript:
now the next topic is binomial theorem this is your favorite from your school life this is the most easy thing what is binomial theorem first what is binomial what is binomial binomial is basically summation of two terms this is called binomial means a plus b this thing is called binomial okay A plus b where a comma B are real numbers they can be any real number so this thing is called a A plus b this is called binomial binomial means summation of two numbers summation of two numbers that is your binomial now you already know and what is binomial theorem what do we mean by binomial theorem binomial theorem says that if you have two values a plus b then what is their power n where N is a natural number n must be a natural number so for example what is their power one what will be the power one that will be a plus b what is power two this is your favorite formula from school life a square plus 2 a B plus b square what is power three that is AQ plus 3 A square B plus 3 a b square plus b Cube only these we remember these we don't have to remember only these we remember we need to remember these now okay so let's move on how we can how we can get a plus b power 10 for example how we can get a plus b power 10 power 10 how will we get this so this is the idea so let's understand each of them one by one if you do a plus b square a plus b square if you do what this means this means a plus b multiplied by a plus b correct this means a plus b multiplied by a plus b now what you can do see total two terms you have here in this product your first and you have second so let me call it one and let me call it two okay so you have two things one two now how to multiply from this one you can take a yes or no from this one you can take a from these two you can take a or or or what I can do from this one I can take a from this 2 I can take B or what I can do from this one I can take B from this 2 you can take a or what I can do from this one you take B and from this 2 you take B so this you will get okay yes or no so finally can you notice this and this they are same a B plus b a these are same so finally I can say a power 2 plus 2 a B plus b square this is your formula but a few interesting things you should understand you should understand few interesting things what finally how many terms we have finally how many terms we have we have total three terms finally we have total three terms can you notice in all the terms number of a plus number of B that is 2 right number of a plus number of B that is two like here number of a is two here number of a plus number of B that is two here number of a plus number of B that is true so number of a plus number of B that is always two why this is happening this is happening because of this you have a plus b you have a plus b now let's think about in some different way some different way we will think you know one thing one thing you can guarantee that whatever term I will get whatever term I will get number of there will be some a there will be some B's yes or no there will be some A's there will be some is and there will be some B's how many bees will be there if number of a is k then how many B's will be there 2 minus k right yes or no because if you are getting so okay there are total two terms so if you are getting K A's then you will get 2 minus K B's now how many ways you can create this term this term in how many ways you can create this term in how many ways for example this term this term how many ways you can create to create this term you need one a okay just understand please understand this how many ways we can create this term this is my question how many ways you can how many ways we can create this term we can create this term so to create this term if you want to create this term then from either from this or from this we need one a so from these two we need one a either from here you need a or from here you need a so from these two from one of them you need a from the remaining automatically B will come from the remaining automatically B will come so this term in how many ways you can create so determine how many ways you can create two ways this term you can create in two ways okay because you need one a so there are two things here either either from here you take a or from here you take a so there are two things from them from one of them you will take a from the remaining who will come from the remaining like from here if I take a then from here who will come B from here B will come okay so now tell me how many ways you can create this term this term in how many ways you create so you need two A's so from both of them you should take a so this term you can create on only one way and in how many ways we can create this term this B Square in how many ways we can create so from these two from this and this from these two you select two and from both of them you should take B from here B should come from here B should come so this term also I can create in only one way so this is your answer so finally answer will be what this term you can create in one way so this term you can create in one way from both of them you take a so I can say this term you can create in one way plus this term you can create in two ways plus this term you can create in one way let's see this one now let's see a plus b power a plus b power three what that will be a plus b a plus b a plus b okay you tell me how many ways I can create this term I want to create this term how many ways I can create this term to create this term you have total three things you have one two three so basically decide from where you are taking B see 1B you want how many B you want you want one B so either you can take B from here then from here who will come then from here who will come a from these two a will come so what you can do either from here you can take b or or you can take from here you can take b or from here you can take B so there are total three things from these three things you can decide who will give you B you can decide who will give you B so this term will come with this and how many ways you can create this term how many ways you can create this term this term how many ways you can create zero way you cannot create this term so this term you cannot create so this term will be created in zero ways this term will be created in zero ways how many ways you can create this term so from all of them all of them a should come from all of them a should come so from all of them a should come from all of them OS would come only one way so this term you can create in only one way how many ways you can create this term a b square this term again you can create in three ways like from okay you can decide who will give you a who will give you a you can decide like this will give you a or this will give you a or this will give you a so this you can decide so this is the idea now that's it this is the final answer this is your final step if you have a plus b power n then you have a plus b a plus b a plus b and so on a plus b so total how many a plus b you have total n one two three total n a plus b you have this term how many ways you can create a of a power r b power n minus r see if you are having a power R then automatically number of B will be from the remaining B will come if from here if from R places a is coming then from remaining places total n places you have if from n r plusing a is coming from remaining places B will come okay so in how many ways you can get this how many ways you can decide from these n you can decide who will give you a from these n you can decide you can decide R of them who will give you a so this is the idea and finally if I want to say that a plus b power n then how can I say I can say that I can say every term will have a power r b power n minus r every term will be like this and how many such terms we have every term will be like this if you have a power R then B power will be n minus r how many such terms you can create from n you can decide who will give you which R will give you a and this R can go from where to where this power can go from where to where this power number of a number of a can go number of a can go from 0 to n okay this is your answer do you know NCR is equal to n c n minus r you know that NCR is equal to n minus r so you can write like this also like this also you can write okay this is your binomial theorem this is the binomial theorem did you understand the proof of binomial theorem basically every term will be like this every term will be like this every term every term is like this is like this okay if you have R A then you will have n minus r b okay if your number of A's are if you are taking number of A's are then how many B's will be there if you are taking R is then automatically automatically n minus r B's will be there then automatically n minus B is okay now so every term will be like this and in how many ways I can create this term NCR these many ways I can create this term I can decide from these n there are total n from these n i can decide who will give me a which R of them will give me a so from this I and I can select all of them they will give me a NCR they will give me a they will give me is okay these are people they will give you is so finally I can say that this number of a this can go from 0 to n this is the binomial theorem so in the binomial theorem these are the important points so in a plus b power n in the expansion how many terms you will be there they can ask you how many terms will be there in this expansion a plus b power n how many terms will be there in the expansion n plus one terms will be there why there will be n plus one terms for example what is a plus b power 4. a plus b power 4 is like this you can have a power zero then how many B will be there if you have a power 0 then how many B will be there for example maybe if you have a power 4 then how many B will be there if you have a power 4 then number of B will be zero okay and this you can create in 4C this you can create in 4C maybe okay so maybe you can do like this anyway all our same okay so if you have a power 0 then B power will be four so you can decide who will give you a none of them will give you a you can decide who will give you a none of them will give you a if a power is one if number of a is one then how many B's will be there three you can decide who will give you one a you can decide who will give you one a here you can decide who will give you 0 a if a power 2 then B power automatically will be two you can decide who will give you two A's from these four from these four who will give you two A's plus if a power 3 then B power will be 1 you can decide who will give you three A's if a power 4 then B power zero you can decide who will give you four is that's it no other term is Possible only these are possible so you can notice this is nothing but summation R is going from 0 to 4 you have NCR you have four C 4cr and if you have a power r then you have B power 4 minus r so this is the expression is it clear to everyone so how many terms we have in this a in this a plus b power 4 in this we have one two three four five so total five terms we have so this will have total five terms okay so similarly this n plus once this uh this a plus b power n this a plus b power n this will have how many terms total because this formula is like this the formula is like this nck if you have K is then number of B's will be n minus K and this K is going from 0 to n so because this K is going from 0 to n so how many terms because this K is going from 0 to n so how many terms so number of terms will be because this K is going from 0 to n so number of terms will be n plus one so these many terms you will have okay let's move on so this is the binomial theorem and these are the statements so first of all there will be n plus one terms in every term you will have okay uh every term you will have some coefficient every turn will be like this some coefficient followed by power of a followed by power of B and number of summation of exponent summation of power of a comma B that will always be n and if a power is increasing then B power will decrease so this is your binomial theorem isn't it simple this is the binomial theorem X Plus y power n that will be just decide if you are taking for example if you are taking X power n then y power zero so just decide who will give you power 0. who will give you a y power 0 so NC 0 who will give you y power 1 so nc1 who will give you y power 2 nc2 if you who will give you y power n minus 1 n c n minus one who will give you y power n n c n so if x power is n minus J then y power will be J and this j y power this can go from 0 to n and you can decide who will give you y power J so from n j will give you from n you can select J those will give you y this is your X Plus y power 4 tell me the answer for this what is the coefficient of x power 12 y power 12 in the expansion of X Plus y power 25 so X power 12 y power 13 first you tell me is this possible X power 12 y power 13 is this possible yes this is possible right because this is possible because 12 plus 13 that is equal to 25 so this is definitely possible and next okay so how many terms so total 25 terms you have you can decide who will give you X you can decide who will give you X and this is same as you can also decide you can also decide who will give you why you can also decide who will give you y so the answer will be answer will be you can decide who will give you X and that is same as you can also decide who will give you y both are same so your answer will be 25 C 12. if I ask you now tell me what is the answer X power 12 y power 14 so what is the coefficient of this so the coefficient of this that will be 0. 0 will be the coefficient of this so here what is the answer answer will be 0. because this is never possible see this is never possible because if you notice then this 12 plus 14 this is not equal to 25. okay tell me the answer for this what is the coefficient of x power 12 y power 13 in the expansion of this I want first is it possible is this possible yes of course it is possible you can check it will be possible now what is the answer so you can think like this you can think a as 2X and you can think b as minus 3y this is how you can think so you have a plus b plus power 25 this is what you have this is what you have okay so in this basically what we want we want a power we want a power 12 we want B power 30. this is what we want so that will be 25 C 12. now what is your a now what is your a is basically 2x 2x power 12 what is your b b is minus 3 by power 13. 25 C12 the question is asking what is the coefficient of x power 12 y power 30 so X power 12 y power 13 the coefficient will be 2 power 12 multiplied by minus 3 power 13 multiplied by 25 C 12. this will be the answer okay because we are only interested in this this is what we want we only want X power 12 y power 13 so the answer will be so finally what is the final answer tell me the final answer final answer will be minus of minus of 2 power 12 3 power 13 multiplied by 25 C 12. this will be your answer okay is this clear because we want we want this this is what we want so the coefficient will be this next question next question so let's assume you have X power 3 plus 2 y power 2 power 10 okay then my question is what is the coefficient of what is the coefficient of I can ask you what is the coefficient of x power so basically this type of question they can ask you okay X power three so every term that you will create here every term that you will get that will be like this so This X power 3 power something will be there and 2 y Square power 10 minus r will be there so what I can do I can do maybe if I take so I can do 12 okay so tell me the answer for this tell me the answer for this what is the coefficient of what is the coefficient of this I am asking coefficient of this so tell me the answer I am asking coefficient of this coefficient of 2 x power 12 y power 12. what will be the answer so you can notice that first of all you want X power 12. you won't X power to all so what do you need if you want X power 12 then you need its power 4. its power 4 you want its power 4 you want then automatically what will be the if if you want Power 4 for this you want Power 4. then automatically this will have what power this will have watt power so 2 2 y Square this two y Square this power will be 10 minus 4. okay yes or no its power will be 10 minus 4. and how many and how many times you can create this that will be 10 C4 that will be 10c 4. because this is your r this is your R so I can say 10 C four times you can create this so from here what we got we've got X power 12 y power 12 what else we got we got 10 C 4 what else we got we got 2 power 2 power 6. so I can write like this now the question is asking what is the coefficient of this so the answer will be this is your answer answer will be 10 C4 to power five is it clear is this clear to everyone in the same question if I do in the same person let me change if I do y power 14 now what will be the answer in the same question if I do y power 14 now the coefficient will be what will be the coefficient here the answer will be 0. the coefficient will be 0 because see what you want you want X power 12 so this term you want four times because you want X power 12 this is what you want this is what you want so this term you want Total four times this you want then automatically automatically automatically this will come six times then automatically this will come six times so you can notice if you have X power to all then automatically y power will be 12 y power will not be 14 so here the answer will be 0. okay so for this answer will be 0. so this is the idea so you can get only y power 12. like this you can handle all these things what is 1 plus X power n like if one of them has fixed then what that will be N C 0 x power 0 and c 1 x power 1 NC 2 x power 2 and so on ncn X power n right because this is already 1 this is fixed this is one so I can say like how many X will be there so you can think like this so if x is 0 then you can decide who will give you 0x so NC 0 if x is 1 then nc1 if x is true nc2 if x is power n then so what is this finally what is this this is basically X part this will be 1 plus okay it will be like this you will have 1 plus and c 1 X plus n c 2 x square nc2 x square plus nc3 X Cube and so on ncn X power n so this will be your formula now if I put if I put x equal to 1 in this if I put x equal to 1 what will happen put x equal to 1 what will happen put x equal to 1 so put x equal to 1. if you put x equal to then 1 what will happen you will get 2 power n this will become 2 power n and this will become N C 0 plus n c 1 plus n c 2 n c n and this is your summation of binomial coefficients yes or no this is your summation of binomial coefficient we have already seen this this is another way the same thing you can do in many different ways so this is your NCR summation NCR where R is going from n to 0 to n so this will be 2 power n just put x equal to one if you put x equal to 1 this will become 2 power n and if you put x equal to 1 then this will become summation of binomial coefficients put x equal to 1. so remember in 1 plus X power n this you already know this is your NC 0 plus n c 1 X plus NC 2 x square and so on n c n x power n put x equal to minus 1. then you will get 0 0 equal to 0 equal to because you are putting x equal to minus 1 so this will be like this nc2 minus nc3 plus alternative plus minus will go on alternative plus minus alternative plus minus now what you can do you can take all the negative terms you can take on this side so this will be NC 0 plus nc2 plus nc4 all the even terms that is equal to all the old terms all the even terms will be equal to all the odd terms summation of even terms will be equal to summation of all terms if you put n equal to -1 okay is this clear so if you put a x equal to minus 1 if you put x equal to minus 1 this will become 0 0 is equal to Alternative plus minus plus minus will go on alternative so I can say that I can say the summation of even coefficients summation of even even binomial coefficient is equal to summation of odd binomial coefficient is this clear summation of even binomial coefficient is equal to summation of odd binomial coefficient |
189276 | https://www.learnalberta.ca/content/memg/division03/repeating%20decimal/index.html | Repeating Decimal
Definition
A repeating decimal is a decimal number that contains a digit or group of digits that repeat endlessly. Another name for repeating decimal is "recurring" decimal.
The number of digits that repeat is called the period of the repeating decimal.
There are different ways to represent a repeating decimal. The most common method is to put a segment over the first digit (or group of digits) that repeat.
Examples
More
All repeating decimals are rational numbers that can be written as reduced fractions with denominators containing at least one prime number factor other than two or five.
The number of digits that repeat (the period) is always less than the denominator of the reduced fraction.
Some examples are listed below: |
189277 | http://content.nroc.org/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U18_L3_T1_text_final.html | Introduction to Natural and Common Logarithms
Learning Objective(s)
· Use a calculator to find logarithms or powers of base e.
· Graph exponential and logarithmic functions of base e.
· Find logarithms to bases other than e or 10 by using the change of base formula.
In both exponential functions and logarithms, any number can be the base. However, there are two bases that are used so frequently that mathematicians have special names for their logarithms, and scientific and graphing calculators include keys specifically for them! These are the common and natural logarithms.
Common Logarithms and e
A common logarithm is any logarithm with base 10. Recall that our number system is base 10; there are ten digits from 0-9, and place value is determined by groups of ten. You can remember a “common logarithm,” then, as any logarithm whose base is our “common” base, 10.
Natural logarithms are different than common logarithms. While the base of a common logarithm is 10, the base of a natural logarithm is the special number e. Although this looks like a variable, it represents a fixed irrational number approximately equal to 2.718281828459. (Like pi, it continues without a repeating pattern in its digits.) e is sometimes called Euler′s number or Napier’s constant, and the letter e was chosen to honor the mathematician Leonhard Euler (pronounced oiler).
e is a complicated but interesting number. Let’s take a closer look at it through the lens of a formula you have seen before: compound interest.
The formula for compound interest is , where A is the amount of money after t years, P is the principal or initial investment, r is the annual interest rate (expressed as a decimal, not a percent), m is the number of compounding periods in a year, and t is the number of years.
Imagine what happens when the compounding happens frequently. If interest is compounded annually, then m = 1. If compounded monthly, then m = 12. Compounding daily would be represented by m = 365; hourly would be represented by m = 8,760. You can see that as the frequency of the compounding periods increases, the value of m increases quickly. Imagine the value of m if interest were compounded each minute or each second!
You can even go more frequently than each second, and eventually get compounding continuously. Look at the values in this table, which looks a lot like the expression multiplied by P in the above formula. As x gets greater, the expression more closely resembles continuous compounding.
| | |
--- |
| | |
| | |
| | |
| | |
| | |
| | |
| 100,000 | 2.71826… |
| 1,000,000 | 2.71828… |
Notice that although x is increasing a lot (multiplying by 10 each time!), the value of is not increasing wildly. In fact, it is getting closer and closer to 2.718281828459…or the value now called e.
The function f(x) = ex has many applications in economics, business, and biology. e is an important number for this reason.
Working with Bases of e and 10
Scientific and graphing calculators all have keys that help you work with e. Look on your calculator and find one labeled “e” or “exp.” (Some graphing calculators may require you to use a menu to find e. If you can’t see the key, consult your manual or ask your instructor.)
How to evaluate exponential expressions using e (such as e3) depends on your calculator. On some calculators you press the [ex] key first then enter the exponent and press enter. On others you enter the exponent first then press the [ex] key. It is important that you know how your calculator works. With your calculator, try finding e3. The result should be 20.0855369… (the number of digits displayed will also depend on your calculator).
| | | |
---
| Example | | |
| Problem | Find e1.5 using a calculator. Round your answer to the nearest hundredth. | |
| | | Enter the keystrokes needed for your calculator. If you are having trouble getting the correct answer, consult your manual or instructor. |
| | 4.4816890… | Calculator result. Then round the answer to the nearest hundredth. |
| Answer | 4.48 | To see this worked out on a calculator, see the Worked Examples for this topic. |
You can find powers of 10 (the common base) in the same way. Some calculators have a [10^] or [10x] key that you can use to find powers of 10. Another way to find powers of 10 is to use the [xy] or the [yx] key that will work with any base (although if you use this method, you will have to key in two numbers—the base, 10, and whatever exponent you are raising it to).
| | | |
---
| Example | | |
| Problem | Find 101.5, using a calculator. Round your answer to the nearest hundredth. | |
| | | Enter the keystrokes needed for your calculator. If you are having trouble getting the correct answer, consult your manual or instructor. |
| | 31.6227766… | Calculator result. Then round the answer to the nearest hundredth. |
| Answer | 31.62 | To see this worked out on a calculator, see the Worked Examples for this topic. |
Natural logarithms (using e as the base) and common logarithms (using 10 as the base) are also available on scientific and graphing calculators. When a logarithm is written without a base, you should assume the base is 10. For example:
log 100 = log10100 = 2
Natural logarithms also have their own symbol: ln.
ln 100 = loge100 = 4.60517…
The logarithm keys are often easier to find, but they may work differently from one calculator to the next. Most handheld scientific calculators require you to provide the input first, then press the [log] (common) or [ln] (natural) key. Other calculators work in reverse: press the [log] or [ln] key, and then provide the input and press [Enter] or [=].
On your calculator, find the common logarithm ([log] or [log10]) key and the natural logarithm ([ln]) key and verify that ln 100 = loge100 = 4.60517….
| | | |
---
| Example | | |
| Problem | Find ln 3, using a calculator. Round your answer to the nearest hundredth. | |
| | | Remember ln means “natural logarithm,” or loge. Enter the keystrokes needed for your calculator. If you are having trouble getting the correct answer, consult your manual or instructor. |
| | 1.098612… | Calculator result. Then round the answer to the nearest hundredth. |
| Answer | 1.10 | To see this worked out on a calculator, see the Worked Examples for this topic. |
| | | |
---
| Example | | |
| Problem | Find log 34, using a calculator. Round your answer to the nearest hundredth. | |
| | | Remember, when no base is specified, this is the common logarithm (base 10). Enter the keystrokes needed for your calculator. If you are having trouble getting the correct answer, consult your manual or instructor. |
| | 1.5314789… | Calculator result. Then round the answer to the nearest hundredth. |
| Answer | 1.53 | To see this worked out on a calculator, see the Worked Examples for this topic. |
| |
| Use a calculator to find ln 7. A) 0.845098… B) 1.945910… C) 1096.633… D) 10,000,000 Show/Hide Answer A) 0.845098… Incorrect. You found the value of log 7, that is, log107. The correct answer is 1.945910…. B) 1.945910… Correct. You correctly identified the keys on your calculator and found the natural log of 7. C) 1096.633… Incorrect. You found the value of e7. The correct answer is 1.945910…. D) 10,000,000 Incorrect. You found the value of 107. The correct answer is 1.945910…. |
Graphing Exponential and Logarithmic Functions of Base e
Graphing functions with the base e is no different than graphing other exponential and logarithmic functions: Create a table of values, plot the points, and connect them with a smooth curve. You will want to use a calculator when creating the table.
| | | |
---
| Example | | |
| Problem | Graph f(x) = ex. | |
| | | | | --- | | x | f(x) | | −2 | 0.1353… | | −1 | 0.3678… | | 0 | 1 | | 1 | 2.7182… | | 2 | 7.3890… | | Start with a table of values. Don’t forget to choose positive and negative values for x. Use a calculator to find the f(x) values. |
| | | | | (x, y) pairs | | (−2, 0.1353…) | | (−1, 0.3678…) | | (0, 1) | | (1, 2.7182…) | | (2, 7.3890…) | | If you think of f(x) as y, each row forms an ordered pair that you can plot on a coordinate grid. |
| | | Plot the points. |
| Answer | | Connect the points as best you can, using a smooth curve (not a series of straight lines). Use the shape of an exponential graph to help you: the graph gets close to the x-axis on the left, and gets steeper and steeper on the right. |
The same process works for logarithmic functions. Choose x values and use a calculator to find the y values.
| | | |
---
| Example | | |
| Problem | Graph f(x) = ln x. | |
| | | | | | --- | x | f(x) | | | 0.1 | −2.30… | | | 0.5 | −0.69… | | | 1 | 0 | | | e | 1 | | | 5 | 1.60… | | | 10 | 2.30… | | | Start with a table of values. If you choose x values, remember that x must be greater than 0. Choose values greater than and less than the base. The base and 1 are also good choices for x values. |
| | | | | (x, y) pairs | | (0.1, −2.30…) | | (0.5, −0.69…) | | (1, 0) | | (e, 1) | | (5, 1.60…) | | (10, 2.30…) | | If you think of f(x) as y, each row forms an ordered pair that you can plot on a coordinate grid. |
| | | Plot the points. |
| Answer | | Connect the points as best you can, using a smooth curve (not a series of straight lines). Use the shape of a logarithmic graph to help you: the graph gets close to the y-axis for x near 0. |
Sometimes the inputs to the logarithm, or the exponent on the base, will be more complicated than just a single variable. In those cases, be sure to use the correct input on the calculator.
Note: If your calculator uses the “input last” method for logarithms, either calculate the input separately and write it down, or use parentheses to be sure the correct input is used. For example, when calculating log(3x) when x = 4, the correct answer is 1.079… . If you don’t use the parentheses, the calculator will find log 3, and multiply that by 4 to get 1.908… .
| | | |
---
| Example | | |
| Problem | Graph f(x) = ln 4x. | |
| | | | | | --- | x | 4x | f(x) | | 0.1 | 0.4 | −0.91… | | 0.5 | 2 | 0.69… | | 1 | 4 | 1.38… | | 3 | 12 | 2.48… | | 10 | 40 | 3.68… | | Create a table of values. Although everything could be done using the calculator, let’s include a column for the input of the logarithm. This helps you avoid calculator errors. |
| | | Use the table pairs to plot points. You may want to choose additional values for the table to give a better idea for the entire visible graph. |
| Answer | | Connect the points as best you can, using a smooth curve. |
| |
| Which of the following is a graph for f(x) =e0.5x? A) B) C) D) Show/Hide Answer A) Incorrect. This is a linear graph. It’s actually f(x) = (e0.5)x. The correct answer is Graph C. B) Incorrect. This graph is decreasing, while f(x)= e0.5x is increasing. The correct answer is Graph C. C) Correct. This graph accurately shows f(x) =e0.5x. D) Incorrect. This is a graph of f(x) = ln(0.5x). The correct answer is Graph C. |
Finding Logarithms of Other Bases
Now you know how to find base 10 and base e logarithms of any number. What if you wanted to calculate log736? Converting to an exponential equation, you have 7x = 36. You know 71 is 7, and 72 is 49, so you can reason that x must be between 1 and 2, probably very close to 2. But how close? You don’t have a key for base 7, so you use a Change of Base formula to change the base of a log function to another base.
| |
| Change of Base formula |
Notice that a appears as the base in both logarithms on the right side of the formula. For example,, using a new base of 10. You could also say , or even . Of course, that last one isn’t any easier to calculate than the original expression—but using the [log] or [ln] keys on a calculator, you can use or to find log7 36.
| | | |
---
| Example | | |
| Problem | Find log7 36. | |
| | | Use the Change of Base formula. You can use common logarithms or natural logarithms. For this example, let’s use common logarithms. |
| | | Use the calculator to evaluate the quotient. |
| Answer | | |
If you had used natural logarithms, you would have gotten the same answer:
| | | |
---
| Example | | |
| Problem | Find log3 25.9. | |
| | | Use the Change of Base formula. This time, let’s use natural logarithms. |
| | | Evaluate the quotient. |
| Answer | | |
| |
| Find log5 200. A) 40 B) 0.303… C) 3.292… D) 2.301… Show/Hide Answer A) 40 Incorrect. You found the value of 200 ÷ 5. The correct answer is 3.292…. B) 0.303… Incorrect. When using the change of base formula, the log of the original base is the denominator: . The correct answer is 3.292…. C) 3.292… D) 2.301… Incorrect. You found the value of log 200. The correct answer is 3.292…. |
Common logarithms (base 10, written log x without a base) and natural logarithms (base e, written ln x) are used often. Scientific and graphing calculators have keys or menu items that allow you to easily find log x and ln x, as well as 10x and ex. Using these keys and the change of base formula, you can find logarithms in any base. |
189278 | https://www.khanacademy.org/math/precalculus-tx/xecf67014514b60db:rational-functions-equations-and-inequalities | Use of cookies
Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account.
For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session.
Functional Cookies
These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences.
Targeting Cookies
These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission.
We do not use cookies to serve third party ads on our Service.
Performance Cookies
These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates.
For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users.
We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. |
189279 | https://math.stackexchange.com/questions/3281275/remainder-is-less-than-divisor | modular arithmetic - Remainder is less than divisor - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Remainder is less than divisor
Ask Question
Asked 6 years, 3 months ago
Modified6 years, 3 months ago
Viewed 102 times
This question shows research effort; it is useful and clear
0
Save this question.
Show activity on this post.
I'm reading a book and it says the equation
a mod n=a−n⌊a n⌋a mod n=a−n⌊a n⌋
follows that
0≤a mod n<n.0≤a mod n<n.
I understand that the remainder is less than divisor, but I can't understand how the author got it from the first equation. Could someone, please, explain it to me?
modular-arithmetic
Share
Share a link to this question
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this question to receive notifications
edited Jul 2, 2019 at 21:58
Bill Dubuque
284k 42 42 gold badges 339 339 silver badges 1k 1k bronze badges
asked Jul 2, 2019 at 21:49
E. ShcherboE. Shcherbo
533 2 2 silver badges 11 11 bronze badges
1
2 Suggestion: look at definition of greatest integer function J. W. Tanner –J. W. Tanner 2019-07-02 22:00:43 +00:00 Commented Jul 2, 2019 at 22:00
Add a comment|
2 Answers 2
Sorted by: Reset to default
This answer is useful
3
Save this answer.
Show activity on this post.
As ⌊x⌋≤x<⌊x⌋+1⌊x⌋≤x<⌊x⌋+1, we have
0≤a n−⌊a n⌋<1 0≤a n−⌊a n⌋<1
and after multiplication with n n the claim.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
answered Jul 2, 2019 at 22:01
Hagen von EitzenHagen von Eitzen
384k 33 33 gold badges 379 379 silver badges 686 686 bronze badges
1
Oh, thank you very much for very easy explanation :)E. Shcherbo –E. Shcherbo 2019-07-02 22:14:38 +00:00 Commented Jul 2, 2019 at 22:14
Add a comment|
This answer is useful
1
Save this answer.
Show activity on this post.
Since we have that
a n−1<⌊a n⌋≤a n a n−1<⌊a n⌋≤a n
Then we can bound the RHS by
a−n(a n)≤a−n⌊a n⌋<a−n(a n−1)a−n(a n)≤a−n⌊a n⌋<a−n(a n−1)
0≤a−n⌊a n⌋<n 0≤a−n⌊a n⌋<n
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
answered Jul 2, 2019 at 22:02
Peter ForemanPeter Foreman
20.2k 2 2 gold badges 13 13 silver badges 43 43 bronze badges
1
Now I got it. Thanks much!E. Shcherbo –E. Shcherbo 2019-07-02 22:15:04 +00:00 Commented Jul 2, 2019 at 22:15
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
modular-arithmetic
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
12How to solve system of equations with mod?
3The Chinese remainder theorem
2Congruent iff Same Remainder (CISR) Confusion
0How does author reach step of s a+t m≡1(mod m)s a+t m≡1(mod m)?
6What is the remainder of 2 2015 36 2 2015 36?
1Showing that ⌊n−1 2⌋+⌊n+2 4⌋+⌊n+4 4⌋=n⌊n−1 2⌋+⌊n+2 4⌋+⌊n+4 4⌋=n
10Does Euclidean division still produce unique quotient and remainder when the dividend and divisor are real numbers?
2Name for integer "quotient" rounded up (ceiling) instead of down (floor), and its negative or complementary "remainder"
4Solving a system like ⌊x/a i⌋=b i(mod M)⌊x/a i⌋=b i(mod M)
Hot Network Questions
With with auto-generated local variables
Storing a session token in localstorage
The geologic realities of a massive well out at Sea
How to start explorer with C: drive selected and shown in folder list?
How to locate a leak in an irrigation system?
How to solve generalization of inequality problem using substitution?
Alternatives to Test-Driven Grading in an LLM world
An odd question
Discussing strategy reduces winning chances of everyone!
Gluteus medius inactivity while riding
What’s the usual way to apply for a Saudi business visa from the UAE?
how do I remove a item from the applications menu
Lingering odor presumably from bad chicken
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
Cannot build the font table of Miama via nfssfont.tex
Why include unadjusted estimates in a study when reporting adjusted estimates?
Vampires defend Earth from Aliens
Xubuntu 24.04 - Libreoffice
For every second-order formula, is there a first-order formula equivalent to it by reification?
Is encrypting the login keyring necessary if you have full disk encryption?
How can the problem of a warlock with two spell slots be solved?
Implications of using a stream cipher as KDF
Passengers on a flight vote on the destination, "It's democracy!"
The rule of necessitation seems utterly unreasonable
more hot questions
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
189280 | https://www.mathway.com/popular-problems/Trigonometry/303455 | Enter a problem...
Trigonometry Examples
Popular Problems
Trigonometry
Verify the Identity sin(2x)=2sin(x)cos(x)
Step 1
Start on the right side.
Step 2
Apply the sine double-angle identity.
Step 3
Because the two sides have been shown to be equivalent, the equation is an identity.
is an identity
| |
| |
Please ensure that your password is at least 8 characters and contains each of the following:
a number
a letter
a special character: @$#!%?& |
189281 | https://www.bbc.co.uk/bitesize/guides/z897dp3/revision/3 | Finding the original quantity after a % increase - Calculating percentage changes - 3rd level Maths Revision - BBC Bitesize
BBC Homepage
Skip to content
Accessibility Help
Sign in
Home
News
Sport
Earth
Reel
Worklife
Travel
Culture
Future
Music
TV
Weather
Sounds
More menu
More menu
Search Bitesize
Home
News
Sport
Earth
Reel
Worklife
Travel
Culture
Future
Music
TV
Weather
Sounds
Close menu
Bitesize
Menu
Home
Learn
Study support
Careers
Teachers
Parents
Trending
My Bitesize
More
England
Early years
KS1
KS2
KS3
GCSE
Functional Skills
Northern Ireland
Foundation Stage
KS1
KS2
KS3
GCSE
Scotland
Early Level
1st Level
2nd Level
3rd Level
4th Level
National 4
National 5
Higher
Core Skills
An Tràth Ìre
A' Chiad Ìre
An Dàrna Ìre
3mh ìre
4mh ìre
Nàiseanta 4
Nàiseanta 5
Àrd Ìre
Wales
Foundation Phase
KS2
KS3
GCSE
WBQ
Essential Skills
Cyfnod Sylfaen
CA2
CA3
CBC
TGAU
International
KS3
IGCSE
More from Bitesize
About us
All subjects
All levels
Primary games
Secondary games
3rd level
Calculating percentage changes Finding the original quantity after a % increase
When a quantity is increased or decreased by a percentage, you can calculate how much it has changed by and what the new amount is. You can also work backwards from a final number to find the original quantity (100%).
Part ofMathsFractions, decimals and percentages
Save to My Bitesize Save to My Bitesize Saving Saved Removing Remove from My Bitesize
Save to My Bitesize
close panel
In this guide
Revise
Video
Test
Pages
Percentage change
Using percentages for discount calculations
Finding the original quantity after a % increase
Finding the original quantity after a % decrease
Finding the original quantity after a percentage increase
Sometimes you have a final amount after a percentage increase or decrease. You can use this to find the original amount.
Example
A bottle of lemonade advertises that it’s now 50% bigger. If it is now 750 ml, what was its original volume?
Answer
Start with the original volume as 100%
The new volume is 50% more than this: 100% + 50% = 150%
150% = 750 ml. We can divide by 150 to find 1%.
1% = 5 ml. We can now multiply by 100 to find 100%, which was the original volume.
100% = 500 ml
Question
After a price increase of 10% a television costs £495. How much did it cost originally?
Show answer Hide answer
Start with the original volume as 100%.
The new price is an increase of 10%
£495 = 100% + 10% = 110%
1% = 495 ÷ 110 = £4.50
100% = £4.50 x 100 = £450
The original price of the television was £450.
Next page Finding the original quantity after a % decrease
Previous page Using percentages for discount calculations
More guides on this topic
Introducing fractions
Equivalent fractions
Mixed numbers and improper fractions
How to add and subtract fractions
Introduction to percentages
Calculating fractions and percentages
Ratios
Language:
Home
News
Sport
Earth
Reel
Worklife
Travel
Culture
Future
Music
TV
Weather
Sounds
Terms of Use
About the BBC
Privacy Policy
Cookies
Accessibility Help
Parental Guidance
Contact the BBC
BBC emails for you
Advertise with us
Do not share or sell my info
Copyright © 2025 BBC. The BBC is not responsible for the content of external sites. Read about our approach to external linking. |
189282 | https://www.eag.com/wp-content/uploads/2020/04/M-044220_w.pdf | By Dr. Daniel J. D. Sullivan, Patrick Davis, Greg Strossman, and Dr. Daniel Tseng INTRODUCTION Thin films are widely used in many high-tech devices and consumer products.1,2 The reasons for their use varies widely but is predominantly due to their physical, mechanical, electrical, thermal or decorative properties.3,4 The quality and consistency of these films is of great interest to manufacturers and users of these materials. One of the key film properties of interest is density.5,6 This paper addresses the processes best used to measure the density of such films. The elemental composition and film thickness can also be determined and are also discussed.
The use of X-ray Fluorescence Spectroscopy (XRF) to determine film thickness and composition is coupled with Rutherford Back Scattering Spectrometry (RBS) which determines the film’s areal density and elemental composition. The third technique of interest is Dual Beam Focused Ion Beam (DB FIB) which can directly measure film thickness.7,8,9 These techniques combine to provide the film’s elemental composition, thickness and density.
SAMPLE PREPARATION The sample we used in this study is a US 10 cent coin that was coated with a thin layer of gold. This film is used for demonstration purposes only and is not as uniform as state-of-the-art films used for demanding electronic or semiconductor applications, but it is a good example for the techniques and measurements that can be made on real films and film stacks. No sample preparation was required for any of the techniques. XRF measurements were made on both sides of the coin at several locations. The RBS measurement and the DB FIB cross section were made in one location only.
INFORMATION OBTAINED Optical microscopy allows inspection of devices for major defects such as large cracks, discolorations, and gross contamination. It is often used to document the samples prior to analysis. Bright field and dark field or a mix can be used to examine and document specific surface features of different types, e.g. pits, scratches or particles.
XRF uses X-rays to excite fluorescence in materials. The wavelength of the fluorescence shows the species present in the sample and the intensity of the fluorescence can be used to determine the elemental composition. Layer thicknesses can also be determined on flat samples. Detection limits are typically in the PPM (parts per million) range for most elements, but this varies with atomic number.
In Rutherford Back Scattering Spectrometry (RBS) MeV alpha particles (He nuclei) bombard the sample and the energy of the COPYRIGHT © 2020 EUROFINS SCIENTIFIC | Rev. 10.20.21 EAG.COM M-044220 APPLICATION NOTE Determining Density of Thin Films Table 1: Test Matrix Method Information acquired Comments Optical microscopy Color images of sample, dimensional measurements.
Limited to ~ 2500×. Only detects visible light.
WDXRF Elemental composition and film thickness Minimum sampling area is 0.5 mm diameter RBS Elemental composition up to 2 microns depth. Elemental areal density and calculated film thickness.
No standards required.
DB FIB Allows direct imaging of a cross section of the sample. Grey scale images with high depth of field with magnifications up to 100,000× on standard tools. Small area cut. Only shows that one plane in the sample. Limited depth of approximately 50 microns.
Figure 1: Optical images of the sample after DB FIB scattered ions are measured. The date obtained are modeled to determine the elemental composition and layer structure. The areal density of the film can also be determined. With this technique, if the film thickness is known, then the density can be determined and vice versa: if the density is known, then the film thickness can be calculated.
Dual Beam FIB can be used to cross section through the thin film layers to directly measure the film thickness(es) using subsequent in situ SEM inspection. The interface and the entire film can also be imaged to investigate the sharpness of the interface and to determine if any voids/defects are present. DB FIB cross-sections can be made at any accessible location on a sample. RESULTS OF EACH TEST Optical imaging in this case was used just to document the sample and inspect it prior to analysis.
XRF is used to detect the elements present in the film and also in the underlying substrate, with ppm detection limits. With this data and knowing the gold is in the overlying film, calculations can be made to determine the thickness of the film. Several locations were analyzed to determine the overall uniformity of the gold coating.
Table 2. Gold coating (front side) Element Coating (wt%) Substrate (wt%) Au 100 Cu 73.9 Ni 25.7 Co 0.019 Fe 0.059 Mn 0.30 Zn 0.042 Film Substrate Substrate Au Cu Ni nm St. Dev.
RSD (%) wt% wt% Spot 1 98 0.29 0.30 75.4 24.6 Spot 2 87 0.49 0.56 75.1 24.9 Spot 3 112 0.33 0.29 75.2 24.8 The coating thickness is nonuniform. Each analysis was repeated 3 times at each location (note the RSD determined from the repeat measurements). WDXRF has excellent precision and can measure sub-nm scale differences in thickness. Determining Density of Thin Films COPYRIGHT © 2020 EUROFINS SCIENTIFIC | Rev. 10.20.21 EAG.COM PAGE 2 OF 3 Figure 2: The XRF scan shows the elements present in the sample Table 3: Elemental composition of the surface layer and the substrate, the coating is pure Au.
Table 4: Quantified using calibration from Au thickness standards over a 1mm area 0 5000 10000 15000 20000 25000 30000 35000 40000 45000 50000 0 100 200 300 400 500 Yield Channel (backscattered energy) Expt Theor 2.275 MeV 4He++ Au film, 82nm on dime Figure 3: RBS data with the best fit modeled data. Using the best fit to the data, a depth profile of the film and the areal density are then calculated.
RBS Thickness [nm] Atomic Concentrations [at%] Assumed Density [at/cc] Ni Cu Au Layer 1 82 --100 5.90E22 BULK -26.5 73.5 -8.63E22 The average thickness for this film was measured to be 70 nm by Dual Beam FIB. The film is rough with a standard deviation of 8 nm, as measured directly using the SEM mountd on the DB FIB. The density of the gold film can be calculated using the areal density obtained from RBS and the layer thicknesses from either XRF or DB FIB. The XRF gives an average thickness over a larger area while the DB FIB shows the film thickness and variation in a much smaller specific location.
Using the atomic mass of gold, 3.2707x10-22 g/atom, this results in a range of 22.6-24.0 g/cm3 which compared to the common density of gold of 19.3 g/cm3 (see table below). Aerial density (atom/cm2) Thickness Atom Density (atm/cm3) Calculated Density g/cm3 4.85x1017 70-87 nm 6.91 X1022 22.6-24.0 SUMMARY: The use of thin films in products requires a knowledge of the film properties. This paper illustrates a procedure to determine the composition, thickness and density of a film. The Three techniques used on different locations on the sample provided a range of thicknesses. This sample was checked and the gold coating thickness ws found to be very variable. The analyses of the techniques were repeated and the individual locations provide very repeatable results however the sample variation across the analyzed regions showed the non-uniformity of the coating.
While there are alternative techniques to determine film thickness and composition this approach has the advantages of direct measurement of the thickness (via DB-FIB) at a particular location and elemental composition determination with excellent detection limits (via XRF) along with averaged thickness over a larger area. The RBS then adds the capability to measure the aerial density and ultimately provide a film density. This combination of techniques enables average thickness measurements and specific area measurements to be made to better understand the thickness variations in the film(s).
Additional properties of thin films can be investigated with additional techniques: roughness with AFM, Stress and texture by XRD, trace contaminations with SIMS, grain structure with EBSD and surface composition with XPS or Auger Electron Spectroscopy.
REFERENCES 1. A. A. Voevodin and J. S. Zabinski, Thin Solid Films 370 (2000) 223 2. Tomoaki Matsushima, ... Kiyotaka Wasa, in Handbook of Sputtering Technology (Second Edition), 2012.
3. Carretero, E.; Alonso, R. Semitransparent Decorative Coatings Based on Optical Interference of Metallic and Dielectric Thin Films for High Temperature Applications. Coatings 2018, 8, 183.
4. Gould R.D., Kasap S., Ray A.K. (2017) Thin Films. In: Kasap S., Capper P. (eds) Springer Handbook of Electronic and Photonic Materials. Springer Handbooks. Springer, Cham.
5. Gould R.D., Kasap S., Ray A.K. (2017) Thin Films. In: Kasap S., Capper P. (eds) Springer Handbook of Electronic and Photonic Materials. Springer Handbooks. Springer, Cham 6. LOVELL, S., ROLLINSON, E. Density of Thin Films of Vacuum Evaporated Metals. Nature 218, 1179–1180 (1968).
7. Thin-film characterization by x-ray fluorescence, T. C. Huang, Invited paper presented at the Federation of Analytical Chemistry and Spectroscopy Societies Annual Meeting, Chicago, IL, 1–6 October, 1989.
8. Abiodun E. Adeoye, Emmanuel Ajenifuja, Bidini A. Taleatu, and A. Y. Fasasi, “Rutherford Backscattering Spectrometry Analysis and Structural Properties of Thin Films Deposited by Chemical Spray Pyrolysis,” Journal of Materials, vol. 2015, Article ID 215210, 8 pages, 2015.
9. ] J. Orloff, M. Utlaut, L. Swanson, High Resolution Focused Ion Beams: FIB and its Applications, first ed., Kluwer Academic/Plenum Publishers, New York, 2003 Determining Density of Thin Films COPYRIGHT © 2020 EUROFINS SCIENTIFIC | Rev. 10.20.21 EAG.COM PAGE 3 OF 3 Table 5: RBS elemental composition, showing an aerial density of 4.85e17 atoms/cm2 FIgure 4: A DB FIB cross section of the gold layer (see red arrow) on the sample allows measurement of the layer thickness.
Top of dime with Pt protective coat Au layer Substrate |
189283 | https://www.youtube.com/watch?v=x_aax6LSRtE | Area circumradius formula proof | AIME | Math for fun and glory | Khan Academy
Khan Academy
9030000 subscribers
367 likes
Description
80557 views
Posted: 7 Jan 2011
Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now:
Proof of the formula relating the area of a triangle to its circumradius. Created by Sal Khan.
Watch the next lesson:
Missed the previous lesson?
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
For free. For everyone. Forever. #YouCanLearnAnything
Subscribe to Khan Academy’s Math For Fun and Glory channel:
Subscribe to Khan Academy:
30 comments
Transcript:
What i want to do in this video is to come up with a relationship between the area of a triangle and the triangle's circumscribed circle or circum-circle. So before we think about the circum-circle let's just think about the area of the triangle. So let's say that the triangle looks something like this. Actually I don't want to make it look isosceles. So let me make it a little bit so it doesn't look like any particular type of triangle and let's call this traingle "ABC". That's the vertices and then the length of the side opposite "A" is "a" "b" over here, and then "c" We know how to calculate the area of this triangle if we know its height. If we drop an altitude right here and if this altitude has length "h" we know that the area of [ABC] - and we write [ABC] with the brackets around it means the area of the traingle [ABC] - is equal to 1/2 times the base, which is "b" times the height. Fair enough. We have an expression for the area. Let's see if we can somehow relate some of these things with the area to the radius of the triangle's circumscribed circle. So the circumscribed circle is a circle that passes through all of the vertices of the triangle and every triangle has a circumscribed circle. So let me try to draw it. This is the hard part, right over here so it might look something like this That's fair enough. That's close enough to a circle I think you get the general idea That is the circum-circle for this triangle. or this triangle's circumscribed circle. Let me label it. This is the circum-circle for this triangle. Now let's think about the center of that circum-circle sometimes refer to as the circumcenter. So looks like it would be sitting I don't know, just eyeballing it right on this little "b" here. So that's the circum-circle of the circle Let's draw a diameter through that circumcircle and draw a diameter from vertex "B" through that circumcenter. So then we go there, and we just keep going over here Let's call this point over here "D". Now let's create a triangle with vertices A, B, and D. So we can just draw another line over here and we have triangle ABD Now we proved in the geometry play - and it's not actually a crazy prove at all - that any triangle that's inscribed in a circle where one of the sides of the triangle is a diameter of the circle then that is going to be a right triangle and the angle that is going to be 90 degrees is the angle opposite the diameter So this is the right angle right here. You can derive that, pretty straightforward. You have this arc here that is 180 degrees. because obviously this is a diameter. And it subtends this inscribed angle. We've also proved that an inscribed angle that is subtended by the arc will be half of the arc length This is an 180 degree arc so this is going to be a 90 degree angle. So either way this's going to be 90 degrees over there The other thing we see is that we have this arc right over here that I'm drawing in magenta the arc that goes from "A" to "B" That arc subtends two different angles in our drawing - it subtends this angle right over here, angle ACB it subtends that right over there - but it also subtends angle ADB that's why we construct it this way So it also subtends this So these two angles are going to be congruent. They'll both have half the degree measure of this arc over here because they're both inscribed angles subtended by the same exact arc. Something interesting is popping up. We have two triangles here we have triangle ABD and triangle BEC They have two angles that resemble They have right angle and this magenta angle and their third angle must be the same. We'll do it in yellow The third angle must be congruent to that angle. They have three angles that are the same. They must be similar triangles. or the ratio between the corresponding sides must be the same. So we can use that information now to relate the length of this side which is really the diameter, is two times the radius to the height of this smaller triangle. We know the relationship between the height of the smaller triangle and the area and we essentially are in the home stretch. So let's do that So these are two similar triangles We know that the ratio of C to this diameter right here What's the length of the diameter? The length of the diameter is 2 times the radius This is the radius. We know that the ratio, C to two times the radius is going to be the same exact thing as the ratio of "h" - and we want to make sure we're using the same side - to the hypoteneuse of that triangle to the ratio of "H" to "A". And the way we figured that out we look at corresponding sides. "C" and the hypoteneuse are both the sides adjacent to this angle right over here So you have "H" and "A". So "C" is to "2r "as "H" is to "a". Or, we could do a lot of things. 1, we could solve for h over here and substitute an expression that has the area Actually let's just do that So if we use this first expression for the area. We could multiply both sides by two. And divide both sides by B. That cancels with that. We get that H is equal to 3 times the area over B. We can rewrite this relationship as c/2r is equals to h which is 2 times the area of our triangle over B and then all of that is going to be over A. Or, we could rewrite that second part over here as two times the area over - we're dividing by "b" and then divided by "a", that's the same thing as dividing by ab So we can ignore this right here. So we have c/2r is equals to 2 times the area over ab And now we can cross-multiply ab times c is going to be equal to 2r times 2abc. So that's going to be 4r times the area of our triangle. I just cross multiply this times this is going to be equal to that times that. We know that cross multiplication is just multiplying both sides of the equation by 2r and multiplying both sides of the equation by ab. So we did that on the left hand side we also did that on the right hand side 2r and ab obviously that cancels with that, that cancels with that So we get ABC is equal to 2r times 2abc. Or 4r times the area of our triangle. And now we're in the home stretch. We divide both sides of this by 4 times the area and we're done. This cancels with that, that cancels with that and we have our relationship The radius, or we can call it the circumradius. The radius of this triangle's circumscribed circle is equal to the product of the side of the triangle divided by 4 times the area of the triangle. That's a pretty neat result. |
189284 | https://files.ele-math.com/articles/mia-25-53.pdf | Mathematical Inequalities & Applications Volume 25, Number 3 (2022), 839–850 doi:10.7153/mia-2022-25-53 INEQUALITIES FOR TRIGONOMETRIC SUMS IN TWO VARIABLES V. E. S. SZAB ´ O (Communicated by M. Praljak) Abstract. We prove sharp inequalities for the two-variable versions of the sum n ∑ k=1 sin(kx) k .
1. Introduction Inequalities for trigonometric sums and polynomials have many applications in approximation theory, special functions theory, see e.g. . The classical one is Sn(x) = n ∑ k=1 sin(kx) k > 0 (n ⩾1, 0 < x < π).
(1) The validity of (1) was conjectured by Fej´ er in 1910 and proved one year later by Jackson . Many refinements, generalization can be found in the literature .
In 1932, Koschmieder showed that the Fej´ er-Jackson inequality (1) can be applied to prove an extension for two variables: Fn(x,y) = n ∑ k=1 sin(kx)sin(ky) k2 > 0 (n ⩾1, 0 < x,y < π).
(2) Tur´ an published in 1938 an upper bound for Fn(x,y). Let 0 < y < π . Then Fn(x,y) < (π −y)x, if 0 ⩽x ⩽y, (π −x)y, if y ⩽x < π.
(3) In 2009 Alzer and Shi proved that Fn(x,y) < π2/8 (n ⩾1, x,y ∈(0,π)), (4) where π2/8 is the best possible constant. They also proved for n ⩾1, x,y ∈(0,π) −1 < Gn(x,y) = n ∑ k=1 cos(kx)cos(ky) k2 < π2 6 , (5) Mathematics subject classification (2020): 26D05, 42A05.
Keywords and phrases: Trigonometric sum, two variables, inequalities.
c ⃝ , Zagreb Paper MIA-25-53 839 840 V. E. S. SZAB ´ O − 4 √ 3(3 + √ 3) 4 √ 2 < Hn(x,y) = n ∑ k=1 cos(kx)sin(ky) k2 < 4 √ 3(3 + √ 3) 4 √ 2 , (6) where the constants are the best possible.
In 1932, Koschmieder showed that for n ⩾1, 0 < x−y < π , 0 < x+ y < π −π2 12 < Gn(x,y) = 2n ∑ k=1 cos(kx)cos(ky) k2 < π2 6 .
The aim of this paper is to improve these results and to give some other estimations for trigonometric sums.
We prove the following estimations in Theorem 1-3: Fn(x,y) ⩽1 2(x(π −x)y(π −y))1/2.
Here (x(π −x)y(π −y))1/2/2 ⩽π2/8, (see (4)), and for πy 4π −3y ⩽x ⩽y < π, or πx 4π −3x ⩽y ⩽x < π, it is sharper than (3).
The next one is Gn(x,y) < π2 6 −1 2x(π −x) 1/2π2 6 −1 2y(π −y) 1/2 , which is sharper than π2/6, (see (5)).
At last, we have Hn(x,y) < π2 6 −1 2x(π −x) 1/21 2y(π −y) 1/2 , that is sharper than (6) for certain numbers x,y ∈(0,π).
In Theorem 4 a refinement of (6) is given. In Theorem 5 a lower estimation is given for a generalization of Fn(x,y). In Theorem 6 we extend the validity of (3) for a generalization of Fn(x,y). In Theorem 7 a lower estimation of Gn(x,y) is derived.
2. Known and new Lemmas We define for positive integers n and real numbers x the function gn(x) = n ∑ k=1 sin(kx) k2 .
LEMMA 1. For n ⩾1 and x ∈(0,π) we have gn(x) > 0.
INEQUALITIES FOR TRIGONOMETRIC SUMS IN TWO VARIABLES 841 Proof. See Lemma 2.1.
□ In the next lemma we give an improvement of Lemma 2.3.
LEMMA 2. Let s,t be real numbers with 0 ⩽s ⩽2π , −π ⩽t ⩽π , and t ⩽s.
Then we have for n ⩾1 gn(s)+ gn(t) 2 ⩽ ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ gn(s) 2 , if −π ⩽t ⩽0, 0 ⩽s ⩽π, 0, if −π ⩽t ⩽0, π ⩽s ⩽2π, gn(s)+gn(t) 2 , if 0 ⩽t ⩽s ⩽π, gn(t) 2 , if 0 ⩽t ⩽π, π ⩽s ⩽2π.
Proof. We use the idea of Lemma 2.3, modifying it where it is needed.
Case 1. −π ⩽t ⩽0.
Let 0 ⩽s ⩽π . Lemma 1 gives gn(t) ⩽0. Thus gn(s)+ gn(t) 2 ⩽gn(s) 2 .
Let π ⩽s ⩽2π . Since gn(s) = −gn(2π −s) ⩽0 we have gn(s)+ gn(t) 2 ⩽0.
Case 2. 0 ⩽t ⩽π .
Let π ⩽s. Then π ⩽s ⩽2π and gn(s) ⩽0. We obtain gn(s)+ gn(t) 2 ⩽gn(t) 2 .
This completes the proof of lemma.
□ LEMMA 3. For n ⩾1 and 0 ⩽x ⩽π we have n ∑ k=1 sin(kx) k ⩽αn(π −x) with the best possible constant factor αn = 1, if n is odd, 0.663959..., if n is even.
Proof. See □ The next lemma is well-known.
842 V. E. S. SZAB ´ O LEMMA 4. For n ⩾1 we have n ∑ k=1 1 k = log n + γ + 1 2n − p−1 ∑ k=1 B2k 2kn2k −θ1B2p 2pn2p, where 0 < θ1 < 1, γ is the Euler’s constant, and B j ’s are the Bernoulli numbers, B2 = 1/6, B4 = −1/30.
LEMMA 5. For n ⩾1 and 0 ⩽x ⩽π we have gn(x) = n ∑ k=1 sin(kx) k2 ⩽βn(π −x), where βn = ⎧ ⎪ ⎨ ⎪ ⎩ 1, if n = 1, 5/6, if n = 3, (1 −α2)log 2 + α2 = 0.896884..., if n ̸= 1,3.
Proof. Using summation by parts we obtain gn(x) = n ∑ k=1 Sk(x) 1 k − 1 k + 1 + Sn(x) n + 1.
Case 1. n = 2m.
Applying Lemma 3 we can write g2m(x) ⩽ 2m ∑ k=1 αk(π −x) 1 k − 1 k + 1 + α2m(π −x) 2m+ 1 = (π −x) 2m ∑ k=1 αk 1 k − 1 k + 1 + α2 2m+ 1 .
Here 2m ∑ k=1 αk 1 k − 1 k + 1 + α2 2m+ 1 = α1 m ∑ j=1 1 2 j −1 −1 2 j + α2 m ∑ j=1 1 2 j − 1 2 j + 1 + α2 2m+ 1 = (α1 −α2) m ∑ j=1 1 2 j −1 + α2 + α2 −α1 2 m ∑ j=1 1 j .
Since m ∑ j=1 1 2 j + m ∑ j=1 1 2 j −1 = 2m ∑ k=1 1 k , INEQUALITIES FOR TRIGONOMETRIC SUMS IN TWO VARIABLES 843 that is, m ∑ j=1 1 2 j −1 = 2m ∑ k=1 1 k −1 2 m ∑ j=1 1 j , we obtain 2m ∑ k=1 αk 1 k − 1 k + 1 + α2 2m+ 1 = (α1 −α2) 2m ∑ k=1 1 k + (α2 −α1) m ∑ j=1 1 j + α2.
Using Lemma 4 with p = 1, we get for m ⩾1 (α1 −α2) 2m ∑ k=1 1 k + (α2 −α1) m ∑ j=1 1 j + α2 = (α1 −α2) log 2 + log(m)+ γ + 1 4m −θ1B2 23m2 −(α1 −α2) log(m)+ γ + 1 2m −θ2B2 2m2 + α2 = (α1 −α2) log 2 −1 4m + 4θ2 −θ1 48m2 + α2 ⩽(α1 −α2)log 2 + α2 = 0.896884....
Thus g2m(x) ⩽((α1 −α2)log 2 + α2)(π −x).
Case 2. n = 2m−1.
Applying Lemma 3 we can write g2m−1(x) ⩽ 2m−1 ∑ k=1 αk(π −x) 1 k − 1 k + 1 + α2m−1(π −x) 2m = (π −x) 2m−1 ∑ k=1 αk 1 k − 1 k + 1 + α1 2m .
Here 2m−1 ∑ k=1 αk 1 k − 1 k + 1 + α1 2m = α1 m ∑ j=1 1 2 j −1 −1 2 j + α1 2m + α2 m−1 ∑ j=1 1 2 j − 1 2 j + 1 = α2 + (α1 −α2) m ∑ j=1 1 2 j −1 + α2 −α1 2 m−1 ∑ j=1 1 j = α2 + (α1 −α2) m ∑ j=1 1 2 j −1 −α2 −α1 2m + α2 −α1 2 m ∑ j=1 1 j .
844 V. E. S. SZAB ´ O Since m ∑ j=1 1 2 j −1 = 2m ∑ k=1 1 k −1 2 m ∑ j=1 1 j , we obtain 2m−1 ∑ k=1 αk 1 k − 1 k + 1 + α1 2m = α2 + α1 −α2 2m + (α1 −α2) 2m ∑ k=1 1 k + (α2 −α1) m ∑ j=1 1 j .
Using Lemma 4 with p = 1, we get for m ⩾3 α1 −α2 2m + α2 + (α1 −α2) 2m ∑ k=1 1 k + (α2 −α1) m ∑ j=1 1 j = (α1 −α2) log 2 + α1 −α2 2m −1 4m + 4θ2 −θ1 48m2 + α2 ⩽(α1 −α2)log 2 + α2 = 0.896884....
For m = 1 we obtain α2 + α1 −α2 2m + (α1 −α2) 2m ∑ k=1 1 k + (α2 −α1) m ∑ j=1 1 j = α1 = 1.
This constant estimation is sharp, because the function f : [0,π) →R, f(x) := g1(x) π −x = sin(x) π −x is strictly increasing and lim x→π−0 f(x) = 1.
For m = 2 we obtain g3(x) = 3 ∑ k=1 sin(kx) k2 .
Define the function f : [0,π) →R, f(x) := g3(x) π −x.
We prove that the function f is strictly increasing.
Let us write f(x) = 1 36 sin(x) π −x (32 + 18cos(x)+ 16cos2(x)).
Then f ′(x) = 1 36(π −x)2 F(x), INEQUALITIES FOR TRIGONOMETRIC SUMS IN TWO VARIABLES 845 where F(x) = (π −x)(48cos3(x)+ 36cos2(x)−18) + sin(x)(32 + 18cos(x)+ 16cos2(x)).
We have to show that F(x) > 0 for x ∈[0,π). Calculation gives F′(x) = −72(π −x)sin(x)cos(x)(2cos(x)+ 1).
It yields that F is strictly increasing in [0,π/2] and [2π/3,π], strictly increasing in [π/2,2π/3]. Since F(0) = 66π , F(π/2) = −9π + 32, F(2π/3) = −5π + 27 √ 3/2 are positive numbers and F(π) = 0 it follows that F is positive in [0,π).
From the strictly monotonicity of f on [0,π) we get f(x) < lim x→π−0 f(x) = 5 6 = 0.833333..., that is, g3(x) ⩽5 6(π −x).
□ LEMMA 6. Let ak (k = 1,...,n) be real numbers such that a1 ⩾a2 ⩾... ⩾an ⩾ 0. Then we have for all integers n ⩾1 and x ∈(0,π) n ∑ k=1 ak sin(kx) k ⩾a1 2 x2 cot x 2 −π −x 2 .
Proof. See Corollary 1.
□ LEMMA 7. The Taylor series of tangent function is tan(z) = ∞ ∑ n=1 2(22n −1)ζ(2n)π−2nz2n−1, |z| < π/2, where ζ is the zeta function of Riemann.
Proof. See e.g. 1.20(4).
□ LEMMA 8. For n ⩾1 and 0 ⩽x ⩽π we have hn(x) = n ∑ k=1 cos(kx) k2 ⩾τn −αn 2 x(2π −x), where τn = n ∑ k=1 1 k2 .
846 V. E. S. SZAB ´ O Proof. Applying the inequality in Lemma 3 we obtain hn(x) = n ∑ k=1 cos(kx)−cos(k0) k2 + n ∑ k=1 1 k2 = τn − x 0 n ∑ k=1 sin(kt) k dt ⩾τn −αn x 0 (π −t)dt = τn −αn 2 x(2π −x).
□ 3. New Theorems In this section we prove our sharp estimations for the trigonometric sums in two variables, which we defined in the first section.
THEOREM 1. For all integers n ⩾1 and real numbers x,y ∈(0,π) we have Fn(x,y) = n ∑ k=1 sin(kx)sin(ky) k2 < 1 2(x(π −x)y(π −y))1/2.
Proof. By the Cauchy-Schwarz inequality we have n ∑ k=1 sin(kx)sin(ky) k2 ⩽ n ∑ k=1 sin2(kx) k2 1/2 n ∑ k=1 sin2(ky) k2 1/2 ⩽ ∞ ∑ k=1 sin2(kx) k2 1/2 ∞ ∑ k=1 sin2(ky) k2 1/2 = 1 2(x(π −x)y(π −y))1/2, because ∞ ∑ k=1 sin2(kx) k2 = 1 2 ∞ ∑ k=1 1 −cos(2kx) k2 = 1 2 π2 6 − ∞ ∑ k=1 1 k2 cos(2kx) .
□ THEOREM 2. For all integers n ⩾1 and real numbers x,y ∈(0,π) we have Gn(x,y) = n ∑ k=1 cos(kx)cos(ky) k2 < π2 6 −1 2x(π −x) 1/2π2 6 −1 2y(π −y) 1/2 .
INEQUALITIES FOR TRIGONOMETRIC SUMS IN TWO VARIABLES 847 Proof. By the Cauchy-Schwarz inequality we have n ∑ k=1 cos(kx)cos(ky) k2 ⩽ ∞ ∑ k=1 cos2(kx) k2 1/2 ∞ ∑ k=1 cos2(ky) k2 1/2 = ∞ ∑ k=1 1 −sin2(kx) k2 1/2 ∞ ∑ k=1 1 −sin2(ky) k2 1/2 = π2 6 −1 2x(π −x) 1/2π2 6 −1 2y(π −y) 1/2 .
□ THEOREM 3. For all integers n ⩾1 and real numbers x,y ∈(0,π) we have Hn(x,y) = n ∑ k=1 cos(kx)sin(ky) k2 < π2 6 −1 2x(π −x) 1/2 1 2y(π −y) 1/2 .
Proof. By the Cauchy-Schwarz inequality we have n ∑ k=1 cos(kx)sin(ky) k2 ⩽ ∞ ∑ k=1 cos2(kx) k2 1/2 ∞ ∑ k=1 sin2(kx) k2 1/2 = π2 6 −1 2x(π −x) 1/21 2y(π −y) 1/2 .
□ THEOREM 4. For all integers n ⩾1 and real numbers x,y ∈(0,π) we have Hn(x,y) = n ∑ k=1 cos(kx)sin(ky) k2 ⩽ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ βn 2 (π −x−y), {π/2 ⩽x, y ⩽π −x}, or {y ⩽x ⩽π/2}; 0, {π/2 ⩽x, π −x ⩽y ⩽x}; βn(π −y), {x ⩽π/2, x ⩽y ⩽π −x}; βn 2 (π −y−x), {x ⩽π/2, π −x ⩽y}, or {π/2 ⩽x ⩽y}; and Hn(x,y) ⩾ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ −βn 2 (π + y−x), {π/2 ⩽x, y ⩽π −x}, or {y ⩽x ⩽π/2}; −βny, {π/2 ⩽x, π −x ⩽y ⩽x}; 0, {x ⩽π/2, x ⩽y ⩽π −x}; −βn 2 (x+ y−π), {x ⩽π/2, π −x ⩽y}, or {π/2 ⩽x ⩽y}.
848 V. E. S. SZAB ´ O Proof. First consider the upper estimation. Let 0 < x,y < π , s = x + y, and t = y−x. Then we get Hn(x,y) = gn(s)+ gn(t) 2 .
Since 0 ⩽s ⩽2π , −π ⩽t ⩽π , and t ⩽s we can apply Lemma 2. Solving the inequal-ities for x and y we have the following cases.
Case 1. π/2 ⩽x ⩽π and 0 ⩽y ⩽π −x, or 0 ⩽y ⩽x ⩽π/2.
In this case Hn(x,y) ⩽1 2gn(x+ y) ⩽βn 2 (π −x−y).
Case 2. π/2 ⩽x ⩽π , and π −x ⩽y ⩽x.
In this case Hn(x,y) ⩽0.
Case 3. 0 ⩽x ⩽π/2 and x ⩽y ⩽π −x.
In this case Hn(x,y) ⩽βn(π −y).
Case 4. 0 ⩽x ⩽π/2 and π −x ⩽y ⩽π , or π/2 ⩽x ⩽y ⩽π .
In this case Hn(x,y) ⩽βn 2 (π −y+ x).
Now consider the lower estimation. Since Hn(x,y) = −Hn(π −x,π −y), from the upper estimation we immediately obtain the lower estimation.
□ THEOREM 5. Let ak (k = 1,...,n) be real numbers such that a1 ⩾a2 ⩾... ⩾ an ⩾0. Then we have for all integers n ⩾1 and x,y ∈(0,π), x ⩾y n ∑ k=1 ak sin(kx)sin(ky) k2 ⩾ a1p(x,y), x+ y ⩽π; a1p(π −x,π −y), x+ y > π, where p(x,y) = 1 180y 15x4(3π −x)−(45π2+ 50y2)x3 +15π(π2 + 6y2)x2 −15y2(3π2 + y2)x+ 5π3y2 + 9πy4 .
Proof. Since sin(k(π −x))sin(k(π −y)) = sin(kx)sin(ky) we may assume that x+ y ⩽π . Obviously 2sin(kx)sin(ky) k2 = cos(k(x−y)) k2 −cos(k(x+ y)) k2 = x+y x−y sin(kt) k dt.
INEQUALITIES FOR TRIGONOMETRIC SUMS IN TWO VARIABLES 849 It yields n ∑ k=1 ak sin(kx)sin(ky) k2 = 1 2 x+y x−y n ∑ k=1 ak sin(kt) k dt.
Applying Lemma 6 we obtain n ∑ k=1 ak sin(kx)sin(ky) k2 ⩾a1 4 x+y x−y t2 cot t 2 −π −t 2 dt.
Here 0 ⩽x+ y ⩽π , 0 ⩽x−y < π . Using Lemma 7 it implies x+y x−y t2 cot t 2 −π −t 2 dt = x+y x−y t2 tan π −t 2 −π −t 2 dt ⩾ x+y x−y t2 (π −t)3 24 dt = 1 180y 15x4(3π −x)−(45π2+ 50y2)x3 +15π(π2+6y2)x2−15y2(3π2+y2)x+5π3y2+9πy4 .
Now the proof is complete.
□ COROLLARY 1. If we assume that x ⩽π/2 in Theorem 5, then the sum is positive.
By numerical investigation we can extend the restriction beyond π/2.
THEOREM 6. Let ak (k = 1,...,n) be real numbers such that a1 ⩾a2 ⩾... ⩾ an ⩾0. Then we have for all integers n ⩾1 and x,y ∈[0,π] n ∑ k=1 ak sin(kx)sin(ky) k2 ⩽ a1y(π −x), y ⩽x; a1x(π −y), x ⩽y.
Proof. Since sin(k(π −x))sin(k(π −y)) = sin(kx)sin(ky) we may assume that x + y ⩽π . In addition, we may assume that x ⩾y. As we have seen in the proof of previous theorem it holds n ∑ k=1 ak sin(kx)sin(ky) k2 = 1 2 x+y x−y n ∑ k=1 ak sin(kt) k dt, where 0 ⩽x−y ⩽x+ y ⩽π . Here we have the identity (see also (3.13)) n ∑ k=1 ak sin(kt) k = n ∑ k=1 (ak −ak+1) k ∑ j=1 sin(jt) j , (an+1 = 0).
Using the estimation of Tur´ an k ∑ j=1 sin(jt) j ⩽π −t, 850 V. E. S. SZAB ´ O and ak −ak+1 ⩾0 (k = 1,...,n) it follows n ∑ k=1 ak sin(kx)sin(ky) k2 ⩽1 2 x+y x−y a1(π −t)dt = a1y(π −x).
The other cases, x ⩽y, x+ y > π can be handled similarly, see , p. 281.
□ THEOREM 7. For all integers n ⩾1 and real numbers x,y ∈[0,π], x ⩾y, x+y ⩽ π we have Gn(x,y) = n ∑ k=1 cos(kx)cos(ky) k2 ⩾τn + αn 2 (−2πx+ x2+ y2).
Proof. Since 2cos(kx)cos(ky) = cos(x−y)+ cos(x+ y), we can write Gn(x,y) = hn(x−y)+ hn(x+ y) 2 .
Applying Lemma 8 we get Gn(x,y) ⩾1 2 τn −αn 2 (x−y)(2π −x+ y)+ τn−αn 2 (x+ y)(2π −x−y = τn + αn 2 (−2πx+ x2 + y2).
□ R E F E R E N C E S H. ALZER, S. KOUMANDOS, Sharp inequalities for trigonometric sums, Math. Proc. Cambridge Philos. Soc. 134, 1 (2003), 139–52.
H. ALZER, S. KOUMANDOS, Inequalities of Fej´ er-Jackson Type, Monatsh. Math. 139, 2 (2003), 89– 103.
H. ALZER, X. SHI, Sharp bounds for trigonometric polynomials in two variables, Anal. and Appl. 7, 4 (2009), 341–350.
A. ERD´ ELYI, W. MAGNUS, F. OBERHETTINGER, F. G. TRICOMI,Higher Transcendental Functions, Vol. 1, McGraw-Hill, New York, 1953.
D. JACKSON, ¨ Uber eine trigonometrische Summe, Rend. Circ. Mat. Palermo 32, 1 (1911), 257–262.
L. KOSCHMIEDER,Vorzeicheneigenschaften der Abschnitte einiger physikalisch bedeutsamer Reihen, Monatsh. Math. Phys. 39, 1 (1932), 321–344.
G. V. MILOVANOVI ´ C, D. S. MITRINOVI ´ C, TH. M. RASSIAS,Topics in Polynomials: Extremal Prob-lems, Inequalities, Zeros, World Sci. Publ., Singapore, 1994.
P. TUR ´ AN, ¨ Uber die Partialsummen der Fourierreihe, J. London. Math. Soc. 13, 1 (1938), 278–282.
(Received April 14, 2022) V. E. S. Szab´ o Department of Analysis, Institute of Mathematics Budapest University of Technology and Economics M˝ uegyetem rkp. 3., Budapest, H-1111, Hungary e-mail: sszabo@math.bme.hu Mathematical Inequalities & Applications www.ele-math.com mia@ele-math.com |
189285 | https://anndermatol.org/DOIx.php?id=10.5021/ad.23.052 | pISSN 1013-9087 eISSN 2005-3894
Open Access, Peer-reviewed
Ann Dermatol. 2024 Apr;36(2):99-111. English.Published online Feb 05, 2024.
© 2024 The Korean Dermatological Association and The Korean Society for Investigative Dermatology
Original Article
Skin Barrier Function Assessment: Electrical Impedance Spectroscopy Is Less Influenced by Daily Routine Activities Than Transepidermal Water Loss
Lisa Huygen,1,2 Pauline Marie Thys,1 Andreas Wollenberg,2,3,4 Jan Gutermuth,1,2 and Inge Kortekaas Krohn1,2
Author information
Author notes
Copyright and License
1Vrije Universiteit Brussel (VUB), Skin Immunology & Immune Tolerance (SKIN) Research Group, Brussels, Belgium.
2Vrije Universiteit Brussel (VUB), Universitair Ziekenhuis Brussel (UZ Brussel), Department of Dermatology, Brussels, Belgium.
3Department of Dermatology and Allergy, Ludwig-Maximilian-University, Munich, Germany.
4Department of Dermatology and Allergy, University Hospital Augsburg, Augsburg, Germany.
Corresponding Author: Lisa Huygen. Vrije Universiteit Brussel (VUB), Skin Immunology & Immune Tolerance (SKIN) Research Group, Laarbeeklaan 103, 1090, Brussels, Belgium. Email: lisa.huygen@vub.be
Received July 03, 2023; Revised October 27, 2023; Accepted November 21, 2023.
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Go to:
Abstract
Background
Skin barrier function assessment is commonly done by measuring transepidermal water loss (TEWL). An important limitation of this method is the influence of intrinsic and extrinsic factors. Electrical impedance spectroscopy (EIS) is a lesser-established method for skin barrier function assessment. Some influential factors have been described, but no guidelines exist regarding the standardization of these measurements.
Objective
To evaluate the effect size of daily routine activities on TEWL and EIS, as well as their correlation with age and anatomical differences.
Methods
Healthy participants (n=31) were stratified into three age groups (18–29, 30–49, and ≥50 years). In a climate-controlled room, EIS and TEWL measurements were performed on the left and right volar forearm and abdomen.
Results
Body cream application decreased TEWL and EIS values after 15 and 90 minutes. Skin washing decreased TEWL for 15 minutes and EIS values for at least 90 minutes. TEWL was increased 5 minutes after moderate to intense exercise. Coffee intake increased TEWL on the abdomen after 60 minutes. TEWL and EIS values did not correlate with participants’ age and no anatomical differences were observed. No correlation was observed between TEWL and EIS.
Conclusion
Body cream application and skin washing should be avoided at least 90 minutes prior to measurements of TEWL and EIS. Exercise and coffee intake should also be avoided prior to TEWL measurements. EIS may be a promising tool for skin barrier function assessment as it is less affected by daily routine activities than TEWL.
Keywords
Activities of Daily Living; Electric Impedance; Hydration; Skin Barrier; Transepidermal water loss; Washing
Go to:
INTRODUCTION
An intact epidermal barrier protects against pathogen invasion, withstands chemical and physical assault, and controls physiological loss of water and solutes through the skin1, 2, 3, 4. The stratum corneum (SC) contributes significantly to this barrier function5. The SC consists of flattened, anucleate corneocytes, which are interconnected by desmosomes and surrounded by lamellar lipid sheets. These are composed of ceramides, cholesterol, and free fatty acids, which form a hydrophobic matrix3, 6, 7. Epidermal tight junctions, adherence junctions, and desmosomes also contribute to the skin barrier function3, 8.
Epidermal barrier disruption contributes to the pathogenesis of skin diseases like contact dermatitis, ichthyosis, and atopic dermatitis (AD)2, 7, 9, 10, 11. Skin barrier function assessment is important to evaluate skin health and the extent of skin diseases. Skin examination already provides clinically relevant information. However, measurements of biophysical skin properties are needed for the scientific assessment of this barrier function. The most frequently used method is the measurement of transepidermal water loss (TEWL). It is defined as the water amount diffusing across a fixed skin area per time unit12. TEWL reflects the inside-out perspective of the skin barrier and skin permeability formed by the SC thickness and compactness, SC lipid layers, and tight junction integrity3, 12, 13. Skin barrier disruption increases permeability and TEWL14.
Electrical impedance spectroscopy (EIS), a lesser-established method, measures the response of a skin region to a low-voltage electrical alternating current at various frequencies15, 16. EIS is a single value parameter, which is influenced by skin hydration, SC thickness, condition of water channels, and cell properties (size, shape, orientation, compactness, and structure of the cell membrane) at the same time17, 18, 19. A magnitude index (MIX) value can be calculated from EIS results, which is used for assessment of skin barrier function20, 21. This value is the ratio of the total skin impedance magnitude at two fixed frequencies of 20 kHz and 500 kHz (MIX = |Z20 kHz|/|Z500 kHz|)22.
Multiple extrinsic and intrinsic factors can influence skin barrier function measurements12, 18, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33. The influence of aging on the skin barrier remains controversial, whereas some studies observed decreased skin barrier function, which resulted in increased MIX values and decreased TEWL18, 33, 34, 35, 36. Differences in sweat gland activity, SC thickness, and skin hydration cause disparities in TEWL and EIS at different anatomical locations28, 33. Extended water exposure leads to disruption of intercellular lipid lamellae, and corneocyte swelling, which results in increased TEWL and skin impedance29, 30, 37, 38. Strenuous efforts trigger arteriolar vasodilatation, increase capillary blood flow, and stimulate sweat glands, leading to increased SC capacitance and TEWL32, 39 Finally, body cream/lotion application causes skin hydration, skin barrier improvement, and protection against water loss and pathogen invasion resulting in decreased TEWL and MIX values22, 38, 40. The effects of the above-mentioned influencing factors were mainly studied in well-controlled, and sometimes artificial settings.
The European Expert Group on Efficacy Measurement of Cosmetics and Other Topical Products (EEMCO) developed guidelines for TEWL assessment to standardize measurement results in clinical study contexts15, 23. An acclimatization period of 15–30 minutes in a room with a temperature of 20–22°C and relative humidity of 40–60% is advised. Eating, perspiration, water exposure, usage of skin products, and smoking before and during measurements should be avoided15, 23. Caffeine intake is also discouraged by current recommendations, but its effect on TEWL has not yet been studied. This extensive list of restrictions can prevent potential subjects from participating in a study.
Because many situations were initially examined in artificial settings, this study aimed to analyze the effect of relevant daily routine activities on TEWL measurements and EIS. Moreover, no guidelines exist regarding the standardization of EIS for skin barrier function assessment. We hypothesized that body cream application, skin washing, physical activity, and coffee intake are relevant daily activities that might affect these measurements. Additionally, the correlation between TEWL and MIX values with participants’ age and the presence of anatomical differences was examined.
Go to:
MATERIALS AND METHODS
Study participants
The study was approved by the local Committee of Medical Ethics of the Universitair Ziekenhuis Brussel (EC-2021-303). Ethical principles were performed according to the Declaration of Helsinki, Good Clinical Practice, and national Belgian law. All participants were informed about the study content. Eligible Dutch- or French-speaking subjects were enrolled if ≥18 years after giving written informed consent. 31 non-smoking healthy participants without known skin diseases were included from October 2021 until December 2021 at the Department of Dermatology, Vrije Universiteit Brussel/Universitair Ziekenhuis Brussel. Participants were stratified into three age groups, 18–29 (n=11), 30–49 (n=10), and ≥50 years old (n=10).
Study design
Participants had to refrain from applying skin products, washing their volar forearm and abdomen, making strenuous efforts, and drinking hot and/or caffeine/theine-containing beverages for at least 1 hour prior to their visit. Measurements took place after an acclimatization period of 30 minutes in a standardized relative humidity-controlled room (humidity of 50–55%, temperature of 22°C, and a roller blind to avoid direct sun exposure) thereby eliminating possible environmental influences.
Measurements were performed within two visits. Baseline TEWL and MIX values were obtained on the volar forearm and abdomen at each visit. At visit one, TEWL measurements and EIS were performed 15 and 90 minutes after the application of 1/4 fingertip unit (FTU) of water in oil emulsion Emollient Cream (Dexeryl®; Pierre Fabre Dermatologie, Paris, France) containing glycerin and petrolatum on the right volar forearm and abdomen (area: ±ere100 cm2). Next, the left forearm and abdomen were washed with tepid water (temperature of 36–38°C) and ½ FTU of shower cream containing glycerin and sweet almond oil (Pure Care Shower Cream, Creme Soft, Nivea®; Nivea, Hamburg, Germany) for one minute. Measurements took place after 15 and 90 minutes. On the second visit, participants had to walk for 5 minutes at low intensity in the corridor (climate provided by central ventilation system), after which measurements were performed at 5, 30, and 60 minutes. Next, TEWL measurements and EIS were performed 5, 30, and 60 minutes after walking up and down stairs for 5 minutes (moderate to high intensity). Finally, measurements were performed 5, 20, and 60 minutes after finishing a cup of coffee (Douwe Egberts Lungo 6 Dessert).
Transepidermal water loss
The Multi Skin Test Center MC 1000 (Courage+Khazaka electronic GmbH, Köln, Germany) and Complete Skin Investigation (CSI) software were used for assessing TEWL41. This device generates a TEWL index score (1-20) using an open-chamber probe with two sensors which is placed on the participant’s skin12, 15, 41, 42. An increasing score corresponds to an increasing TEWL, which reflects a skin barrier deterioration. Three reproducible measurements, each taking 15 seconds to complete, were used to calculate a final average score.
Electrical impedance spectroscopy
EIS was performed with the Nevisense device (SciBase AB, Stockholm, Sweden)43. Skin impedance is measured at 35 different frequencies, logarithmically distributed from 1 kHz to 2.5 MHz, at four different depths, and 10 different permutations. The applied voltage and resulting current are limited to 150 mV and 75 μA19. This device uses a spring-loaded probe with a disposable gold-plated electrode consisting of small micro-invasive pins19. Each measurement, taking approximately 10 seconds to complete, was preceded by soaking the test site with 0.9% saline solution (wound cleansing wipe, Salvequick®; Orkla Wound Care, Barcelona, Spain) for approximately 30 seconds to reduce the naturally high impedance of the stratum corneum and to obtain a good electrical connection between the probe and the skin19, 44. Apart from mild erythema, measurements did not cause harm or discomfort for participants19.
The Nevisense device generates a MIX value for skin barrier function assessment. This value is defined as the ratio of total skin impedance magnitude at two fixed frequencies of 20 kHz and 500 kHz (MIX=|Z20 kHz|/|Z500 kHz|)22. Impedance measurements at 20 kHz are confined to the extracellular space of the stratum corneum, while the impedance measured at 500 kHz reflects both intra- and extracellular properties and capacitive properties of cell membranes of deeper skin layers19, 33, 44. The final MIX value was calculated by the average of two reproducible measurements.
Statistical analysis
Statistical analyses were performed using SPSS Statistics version 28 (IBM Corp., Armonk, NY, USA) and GraphPad Prism 9 (GraphPad Software Inc., San Diego, CA, USA). The normality of distribution of data sets was determined using Kolmogorov-Smirnov test. A one-way Repeated-Measures analysis of variance or Friedman test was performed to evaluate the influence of daily routine activities on TEWL measurements and EIS. A simple linear regression (Pearson’s correlation coefficient) or Spearman correlation was performed to assess the correlation between age and TEWL and MIX values and between TEWL and MIX values. Lastly, TEWL and MIX values at different anatomical locations were compared by a Friedman test.
Go to:
RESULTS
In total, 31 healthy participants (13 men [41.9%], 18 women [58.1%]) with a median age of 36 years old (interquartile range: 24.0–56.0) were included in this study. Of those, 26 had a Fitzpatrick skin phototype of 2, four of them had type 3, and one had type 4 (Supplementary Table 1).
Body cream application and skin washing decreased TEWL and MIX values, while exercise and coffee intake increased TEWL
Body cream application decreased TEWL on the volar forearm (15 min: p<0.0001; 90 min: p=0.0077) and abdomen (15 min: p<0.0001; 90 min: p=0.0133) compared to baseline (Fig. 1A and B). The MIX value also decreased after body cream application on the forearm (15 min: p<0.0001; 90 min: p=0.0015) and abdomen (15 min: p<0.0001; 90 min: p<0.0001) (Fig. 2A and B). A lower TEWL was observed 15 minutes after washing the volar forearm (p<0.0001) and abdomen (p=0.0005) with a shower cream containing glycerin and sweet almond oil. This effect was not present after 90 minutes (Fig. 1C and D). Skin washing also decreased the MIX value on forearm (15 min: p<0.0001; 90 min: p<0.0001) and abdomen (15 min: p<0.0001; 90 min: p<0.0001) (Fig. 2C and D). Walking at low intensity did not influence TEWL measurements and EIS (Figs. 1E, 1F, 2E, and 2F). An increased TEWL was observed 5 minutes after moderate to intense exercise (walking up and down stairs) on the forearm (p=0.0007) and abdomen (p=0.0051). This increase was not significantly detectable after 30 and 60 minutes (Fig. 1G and H). EIS was not influenced by moderate to intense exercise (Fig. 2G and H). Finally, coffee intake increased TEWL after 60 minutes on the abdomen (p=0.0146) (Fig. 1I and J), but had no effect on EIS (Fig. 2I and J).
Fig. 1 The effect of daily routine activities on results of TEWL measurements on the volar forearm and abdomen. TEWL was measured in healthy subjects (n=31) at baseline and 15 and 90 minutes after body cream application on the volar forearm (A) and abdomen (B). Next, TEWL measurements were performed 15 and 90 minutes after skin washing on the volar forearm (C) and abdomen (D). Five, 30, and 60 minutes after walking the effect of low-intensity effort was measured on the volar forearm (E) and abdomen (F). The effect of moderate to intense effort was measured on the volar forearm (G) and abdomen (H). Five, 30, and 60 minutes after walking up and down stairs. Finally, TEWL was measured 5, 20, and 60 minutes after coffee intake on the volar forearm (I) and abdomen (J). Each individual point represents the mean value of the repeated individual measurements per participant. Data are given as median and interquartile range. A one-way repeated measure ANOVA or nonparametric Friedman test was performed.TEWL: transepidermal water loss, ANOVA: analysis of variance.
p≤0.05, p≤0.01, p≤0.001, p≤0.0001.
Click for larger image
Download as PowerPoint slide
Fig. 2 The effect of daily routine activities on results of EIS on the volar forearm and abdomen. MIX values were measured in healthy subjects (n=31) at baseline and 15 and 90 minutes after body cream application on the volar forearm (A) and abdomen (B). Next, EIS was performed 15 and 90 minutes after skin washing on the volar forearm (C) and abdomen (D). Five, 30, and 60 minutes after walking the effect of low-intensity effort was measured on the volar forearm (E) and abdomen (F). The effect of moderate to intense effort was measured on the volar forearm (G) and abdomen (H). Five, 30, and 60 minutes after walking up and down stairs. Finally, MIX values were measured 5, 20, and 60 minutes after coffee intake on the volar forearm (I) and abdomen (J). Each individual point represents the mean value of the repeated individual measurements per participant. Data are given as median and interquartile range. A one-way repeated measure ANOVA or nonparametric Friedman test was performed.EIS: electrical impedance spectroscopy, ANOVA: analysis of variance.
p≤0.01, p≤0.0001.
Click for larger image
Download as PowerPoint slide
Age does not correlate with TEWL and MIX values
No correlation was observed between participants’ age and TEWL at all anatomical locations. (Fig. 3). A non-significant trend toward negative correlation was observed between age and MIX values on all anatomical locations (Fig. 4).
Fig. 3 Correlation between participants’ age and TEWL. The relationship between age and TEWL was analyzed in healthy individuals (n=31) on the: (A) right volar forearm (B) left volar forearm, (C) right abdomen, and (D) left abdomen. Each individual point represents the mean value of the reproducible measurements per participant. A simple linear regression with the calculation of the Pearson’s correlation coefficient or nonparametric Spearman correlation was used. p-values of ≤0.05 are considered significant.TEWL: transepidermal water loss.
Click for larger image
Download as PowerPoint slide
Fig. 4 Correlation between participants’ age and MIX values. The relationship between age and MIX value was analyzed in healthy individuals (n=31) on the: (A) right volar forearm, (B) left volar forearm, (C) right abdomen, and (D) left abdomen. Each individual point represents the mean value of the reproducible measurements per participant. A simple linear regression with the calculation of the Pearson’s correlation coefficient or nonparametric Spearman correlation was used. p-values of ≤0.05 are considered significant.MIX: magnitude index.
Click for larger image
Download as PowerPoint slide
No differences in TEWL and MIX values were observed between volar forearm and abdomen
TEWL and MIX values did not significantly differ when comparing the volar forearm and abdomen on right and left bodyside (Fig. 5).
Fig. 5 Differences in TEWL and MIX values between the volar forearm and abdomen on the left and right side. (A) Anatomical differences of TEWL. (B) Anatomical differences of MIX values. Each individual point represents the mean value of the reproducible measurements per participant. Baseline data from healthy subjects (n=31) are given as median and interquartile range. A nonparametric Friedman test was performed.TEWL: transepidermal water loss, MIX: magnitude index.
Click for larger image
Download as PowerPoint slide
No correlation exists between TEWL and MIX values
No correlation was observed between TEWL and MIX values (Fig. 6).
Fig. 6 Correlation between TEWL and MIX value. The relationship between TEWL and MIX value was analyzed in healthy individuals (n=31) on the: (A) right volar forearm, (B) left volar forearm, (C) right abdomen, and (D) left abdomen. Each individual point represents the mean value of the reproducible measurements per participant. A simple linear regression with the calculation of the Pearson’s correlation coefficient or nonparametric Spearman correlation was used. p-values of ≤0.05 are considered significant.TEWL: transepidermal water loss, MIX: magnitude index.
Click for larger image
Download as PowerPoint slide
Go to:
DISCUSSION
TEWL measurements and EIS reflect different skin properties. Therefore, these measurement techniques can not readily be compared. TEWL mainly reflects skin permeability by measuring water evaporation12, 14. This permeability depends on the SC lipid layer and tight junctions of the viable epidermis3. A major drawback of TEWL measurements lies in its susceptibility to temperature, humidity, and other factors, which requires a dedicated sheltered space and therefore impedes its widescale use. EIS reflects the extra- and intracellular environment, cell membrane structure, and other skin cell properties like size, shape, orientation, and compactness17, 18, 19. To gain knowledge on the potential value of EIS for skin barrier function analysis, we compared TEWL measurements and EIS side by side for factors, such as daily routine activities, age, and anatomical location in homeostatic conditions.
Body cream application and skin washing with a shower cream containing glycerin and sweet almond oil reduced TEWL and MIX values in comparable patterns, while exercise and coffee intake increased TEWL measurements, but not MIX values.
TEWL and MIX values were decreased 15 and 90 minutes after body cream application. This could be explained by the hydrating and moisturizing effect of the body cream due to the presence of substances, such as glycerol, white soft paraffin, liquid paraffin, and dimethicone40, 45, 46, 47. Additionally, these substances protect against water loss through the skin, which contributes to a decreased TEWL40, 45, 47, 48. Skin hydration decreases skin impedance at all frequencies, most likely due to better current conductivity22. Body cream application will mainly lower extracellular resistive properties of superficial skin layers (measured at 20 kHz) by filling pre-existing air-filled voids like hair follicles, sweat glands, and furrows, while intracellular resistive properties and capacitive properties (measured at 500 kHz) are less influenced22. The greater influence of body cream application on the impedances measured at 20 kHz results in a decreased MIX value.
Skin washing with tepid water and shower cream for one minute decreased TEWL for 15 minutes and MIX values for 15 and at least 90 minutes on both anatomical locations. Our observation of decreased TEWL is in contrast to previously documented increased TEWL after water exposure, supposedly due to evaporation of remaining water and disruption of SC intercellular lipid layers29. These contradictory results may be explained by the limited duration of water exposure (one minute in our study versus 30 minutes for five consecutive days in previous studies) and the hydrating properties of glycerin and sweet almond oil present in the soap used in this study45. Decreased MIX values after skin washing may be due to the hydrating properties of glycerin and sweet almond oil present in the soap and the role of water in altering extracellular resistive properties of superficial skin layers by filling pre-existing voids like hair follicles, sweat glands, and furrows22.
Future studies should address the duration of the above-mentioned effects in order to optimize recommendations about body cream usage and skin washing prior to skin barrier function assessment.
Increased TEWL following exercise was in line with previous reports and might be triggered by vasodilatation, increased blood flow, increased body temperature, and stimulated sweat glands during and directly after exercise39. Perspiration will be detected by the TEWL probe together with the true TEWL, resulting in increased TEWL values32, 49. Since this influence was not observed after 30 and 60 minutes, the acclimatization period currently foreseen in the EEMCO guidelines is sufficient to eliminate influences of activity on TEWL measurements23. Physical activity did not influence EIS. This method might be less sensitive to perspiration, due to cleaning and moistening of the test area prior to measurements. The effect of physical activity on EIS has not been studied previously.
The effect of coffee intake on TEWL measurements and EIS was not yet investigated. In this study, coffee intake increased TEWL after 60 minutes on the abdomen. This effect may be explained by the high temperature of the drink leading to a slight increase in body temperature resulting in vasodilation and perspiration50, 51. Additionally, caffeine increases intra-arterial blood pressure and stimulates the sympathetic nervous system, resulting in increased sweat gland activity50, 51. On the volar forearm, TEWL did not increase after coffee intake. This may be explained by the fact that the evaporation of sweat on the forearm is not hindered by occlusion caused by clothes when compared to the abdomen. Further research is necessary to evaluate whether high temperature, caffeine content, or a combination of both factors are responsible for increased TEWL. EIS was not influenced by coffee intake, possibly due to its insensibility to evaporating sweat.
In this study, participants’ age did not correlate with TEWL. However, MIX values exhibited a trend toward a negative correlation with increasing age at all anatomical locations, but no significant correlation with the MIX value was found. This is in contrast with a previous study, where a significantly higher MIX value was observed in elderly participants (>60 years old) compared to younger ones (20–40 years old), probably due to increased corneocyte size, decreased SC lipid content, and decreased SC moisture content33. It is important to mention that participants’ age in our study only ranged from 22 to 65, while a potential decline in skin barrier function may only occur from the age of 60. In general, the effect of aging on skin barrier function remains controversial33-36.
Further, when comparing measurements on the volar forearm and abdomen, no differences in TEWL and MIX values were observed. These findings are in accordance with recent studies showing comparable results of skin barrier function measurements on the volar forearm and abdomen25, 28. Thus, both the volar forearm and abdomen may be useful anatomical locations to assess skin barrier function in clinical trial settings. From the aspect of practical handling, we found that EIS using the Nevisense device was easier to perform on the volar forearm, due to a lower amount of subcutaneous fat compared to the abdomen. This allowed for more accurate placement of the measuring probe.
No correlation was observed between TEWL and MIX value. Notably, a modest yet statistically significant positive correlation between TEWL and MIX value was observed solely on the right abdomen. In contrast, no significant or pronounced correlation manifested at the other anatomical locations. Given the small sample size, we hereby deduce that the observed correlation on the right abdomen, while statistically significant, may lack substantial clinical significance. These results are in contrast with a recent study conducted by Rinaldi et al.16, which demonstrates a negative correlation between TEWL and EIS at 1kHz measured after inducing skin barrier damage by the application of cysteine protease papain and trypsin and by tape stripping. The absence of correlation in our study may be attributed to the assessment of TEWL and MIX values on healthy, intact skin, unlike the previous study that focused on damaged skin. In our research, the MIX value was utilized as an indicator to evaluate the skin barrier function. Notably, compared to EIS at 1kHz, the MIX value also encompasses the barrier function of deeper skin layers. This study is the first to assess the correlation between TEWL and MIX value on healthy skin. Given the fact that TEWL measurements and EIS are two distinct biophysical methods for evaluating skin barrier functionality, we are convinced that a combination of these techniques, used side by side, can offer an optimal means of comprehending the functionality of the various elements contributing to skin barrier function. A recent study demonstrated the combined use of TEWL measurements and EIS for evaluating epithelial barrier impairment induced by laundry detergent and sodium dodecyl sulfate52.
An important limitation of this study was the limited age range (22 to 65) of participants. In future studies, a wider age range should be taken into account. Also, additional factors, including skin type, sex, smoking, and alcohol intake may have an important influence on skin barrier function measurements and should be further investigated. The present study only included healthy individuals, while future studies should also explore the added value of EIS in evaluating skin barrier dysfunction in diseased skin, e.g. ichthyosis, atopic dermatitis, and other forms of eczema. A recent study proved that EIS can detect skin barrier impairment in children with atopic dermatitis53.
This study confirmed the effect of body cream application, skin washing, moderate to intense physical activity, and coffee intake on TEWL. These factors should therefore be avoided at least 90 minutes prior to TEWL measurements. An acclimatization period of 30 minutes in the standardized room seemed to be sufficient to eliminate the effect of skin washing and physical activity. Body cream application and skin washing also influenced EIS and should therefore be avoided at least 90 minutes prior to these measurements. EIS may be a promising tool for skin barrier function assessment and is less affected by daily routine activities than TEWL.
Go to:
SUPPLEMENTARY MATERIAL
Supplementary Table 1
Demographics of the study population
Click here to view.(33K, xls)
Go to:
Notes
FUNDING SOURCE:None.
CONFLICTS OF INTEREST:The authors have nothing to disclose.
DATA SHARING STATEMENT:The data supporting the findings of this study are available upon request from the corresponding author, Lisa Huygen.
Go to:
ACKNOWLEDGMENT
We would like to thank SciBase AB (Stockholm, Sweden) for kindly providing the measuring electrodes of the Nevisense device.
Go to:
References
|
| |
Elias PM. Skin barrier function. Curr Allergy Asthma Rep 2008;8:299–305.
PubMed
CrossRef
Sano S. Psoriasis as a barrier disease. Zhonghua Pifuke Yixue Zazhi 2015;33:64–69.
Proksch E, Brandner JM, Jensen JM. The skin: an indispensable barrier. Exp Dermatol 2008;17:1063–1072.
PubMed
CrossRef
Lawton S. Skin 1: the structure and functions of the skin. Nurs Times 2019;115:30–33.
Darlenski R, Kazandjieva J, Tsankov N. Skin barrier function: morphological basis and regulatory mechanisms. J Clin Med 2011;4:36–45.
Elias PM, Menon GK. Structural and lipid biochemical correlates of the epidermal permeability barrier. Adv Lipid Res 1991;24:1–26.
PubMed
CrossRef
Yang G, Seok JK, Kang HC, Cho YY, Lee HS, Lee JY. Skin barrier abnormalities and immune dysfunction in atopic dermatitis. Int J Mol Sci 2020;21:2867.
PubMed
CrossRef
Nguyen AV, Soulika AM. The dynamics of the skin’s immune system. Int J Mol Sci 2019;20:1811.
PubMed
CrossRef
Taïeb A. Hypothesis: from epidermal barrier dysfunction to atopic disorders. Contact Dermat 1999;41:177–180.
CrossRef
Elias PM, Wood LC, Feingold KR. Epidermal pathogenesis of inflammatory dermatoses. Am J Contact Dermat 1999;10:119–126.
PubMed
CrossRef
Akiyama M. Corneocyte lipid envelope (CLE), the key structure for skin barrier function and ichthyosis pathogenesis. J Dermatol Sci 2017;88:3–9.
PubMed
CrossRef
Alexander H, Brown S, Danby S, Flohr C. Research techniques made simple: transepidermal water loss measurement as a research tool. J Invest Dermatol 2018;138:2295–2300.e1.
PubMed
CrossRef
Bäsler K, Bergmann S, Heisig M, Naegel A, Zorn-Kruppa M, Brandner JM. The role of tight junctions in skin barrier function and dermal absorption. J Control Release 2016;242:105–118.
CrossRef
Fluhr JW, Feingold KR, Elias PM. Transepidermal water loss reflects permeability barrier status: validation in human and rodent in vivo and ex vivo models. Exp Dermatol 2006;15:483–492.
PubMed
CrossRef
Berardesca E. European Group for Efficacy Measurements on Cosmetics and Other Topical Products (EEMCO). EEMCO guidance for the assessment of stratum corneum hydration: electrical methods. Skin Res Technol 1997;3:126–132.
PubMed
CrossRef
Rinaldi AO, Korsfeldt A, Ward S, Burla D, Dreher A, Gautschi M, et al. Electrical impedance spectroscopy for the characterization of skin barrier in atopic dermatitis. Allergy 2021;76:3066–3079.
PubMed
CrossRef
Lu F, Wang C, Zhao R, Du L, Fang Z, Guo X, et al. Review of stratum corneum impedance measurement in non-invasive penetration application. Biosensors (Basel) 2018;8:31.
PubMed
CrossRef
du Plessis J, Stefaniak A, Eloff F, John S, Agner T, Chou TC, et al. International guidelines for the in vivo assessment of skin properties in non-clinical settings: Part 2. transepidermal water loss and skin hydration. Skin Res Technol 2013;19:265–278.
PubMed
CrossRef
Nevisense: clinical reference Guide. Stockholm: SciBase; 2014.
Nicander I, Ollmar S. Clinically normal atopic skin vs. non-atopic skin as seen through electrical impedance. Skin Res Technol 2004;10:178–183.
PubMed
CrossRef
Nicander I, Ollmar S, Eek A, Lundh Rozell B, Emtestam L. Correlation of impedance response patterns to histological findings in irritant skin reactions induced by various surfactants. Br J Dermatol 1996;134:221–228.
PubMed
CrossRef
Morin M, Ruzgas T, Svedenhag P, Anderson CD, Ollmar S, Engblom J, et al. Skin hydration dynamics investigated by electrical impedance techniques in vivo and in vitro . Sci Rep 2020;10:17218.
PubMed
CrossRef
Rogiers V. EEMCO Group. EEMCO guidance for the assessment of transepidermal water loss in cosmetic sciences. Skin Pharmacol Appl Skin Physiol 2001;14:117–128.
PubMed
CrossRef
Mehta HH, Nikam VV, Jaiswal CR, Mehta HB. A cross-sectional study of variations in the biophysical parameters of skin among healthy volunteers. Indian J Dermatol Venereol Leprol 2018;84:521.
PubMed
CrossRef
Pinnagoda J, Tupker RA, Agner T, Serup J. Guidelines for transepidermal water loss (TEWL) measurement. A report from the Standardization Group of the European Society of Contact Dermatitis. Contact Dermat 1990;22:164–178.
CrossRef
Pinnagoda J, Tupker RA, Coenraads PJ, Nater JP. Transepidermal water loss with and without sweat gland inactivation. Contact Dermat 1989;21:16–22.
CrossRef
Taylor NA, Machado-Moreira CA. Regional variations in transepidermal water loss, eccrine sweat gland density, sweat secretion rates and electrolyte composition in resting and exercising humans. Extrem Physiol Med 2013;2:4.
PubMed
CrossRef
Darlenski R, Fluhr JW. Influence of skin type, race, sex, and anatomic location on epidermal barrier function. Clin Dermatol 2012;30:269–273.
PubMed
CrossRef
Firooz A, Aghazadeh N, Rajabi Estarabadi A, Hejazi P. The effects of water exposure on biophysical properties of normal skin. Skin Res Technol 2015;21:131–136.
PubMed
CrossRef
Fartasch M, Taeger D, Broding HC, Schöneweis S, Gellert B, Pohrt U, et al. Evidence of increased skin irritation after wet work: impact of water exposure and occlusion. Contact Dermat 2012;67:217–228.
CrossRef
Draelos ZD. The science behind skin care: moisturizers. J Cosmet Dermatol 2018;17:138–144.
PubMed
CrossRef
Mayrovitz HN. Effects of local forearm skin heating on skin properties. Clin Physiol Funct Imaging 2020;40:369–376.
PubMed
CrossRef
Nicander I, Nyren M, Emtestam L, Ollmar S. Baseline electrical impedance measurements at various skin sites - related to age and sex. Skin Res Technol 1997;3:252–258.
PubMed
CrossRef
Luebberding S, Krueger N, Kerscher M. Age-related changes in skin barrier function - quantitative evaluation of 150 female subjects. Int J Cosmet Sci 2013;35:183–190.
PubMed
CrossRef
Manuskiatti W, Schwindt DA, Maibach HI. Influence of age, anatomic site and race on skin roughness and scaliness. Dermatology 1998;196:401–407.
PubMed
CrossRef
Cotterill JA, Cunliffe WJ, Williamson B, Bulusu L. Age and sex variation in skin surface lipid composition and sebum excretion rate. Br J Dermatol 1972;87:333–340.
PubMed
CrossRef
Warner RR, Stone KJ, Boissy YL. Hydration disrupts human stratum corneum ultrastructure. J Invest Dermatol 2003;120:275–284.
PubMed
CrossRef
Hildebrandt D, Ziegler K, Wollina U. EIectrical impedance and transepidermal water loss of healthy human skin under different conditions. Skin Res Technol 1998;4:130–134.
PubMed
CrossRef
Patel PN, Zwibel H. In: Physiology, exercise. Treasure Island: StatPearls; 2022.
Danby SG, Andrew PV, Brown K, Chittock J, Kay LJ, Cork MJ. An investigation of the skin barrier restoring effects of a cream and lotion containing ceramides in a multi-vesicular emulsion in people with dry, eczema-prone, skin: the RESTORE study phase 1. Dermatol Ther (Heidelb) 2020;10:1031–1041.
PubMed
CrossRef
Multi Skin Test Center® MC 1000 & Software. Cologne: Courage+Khazaka electronic GmbH; 2023.
Berardesca E, Loden M, Serup J, Masson P, Rodrigues LM. The revised EEMCO guidance for the in vivo measurement of water in the skin. Skin Res Technol 2018;24:351–358.
PubMed
CrossRef
Malvehy J, Hauschild A, Curiel-Lewandrowski C, Mohr P, Hofmann-Wellenhof R, Motley R, et al. Clinical performance of the Nevisense system in cutaneous melanoma detection: an international, multicentre, prospective and blinded clinical trial on efficacy and safety. Br J Dermatol 2014;171:1099–1107.
PubMed
CrossRef
Nicander I, Ollmar S. Electrical impedance measurements at different skin sites related to seasonal variations. Skin Res Technol 2000;6:81–86.
PubMed
CrossRef
Fluhr JW, Darlenski R, Surber C. Glycerol and the skin: holistic approach to its origin and functions. Br J Dermatol 2008;159:23–34.
PubMed
CrossRef
O’lenick A Jr, O’lenick K. Silicone polymers in skin care. MRS Bull 2007;32:801–806.
CrossRef
Czarnowicki T, Malajian D, Khattri S, Correa da Rosa J, Dutt R, Finney R, et al. Petrolatum: barrier repair and antimicrobial responses underlying this “inert” moisturizer. J Allergy Clin Immunol 2016;137:1091–1102.e7.
PubMed
CrossRef
Kovács A, Péter-Héderi D, Perei K, Budai-Szűcs M, Léber A, Gácsi A, et al. Effects of formulation excipients on skin barrier function in creams used in pediatric care. Pharmaceutics 2020;12:729.
CrossRef
Paciência I, Rodolfo A, Leão L, Silva D, Cavaleiro Rufo J, Mendes F, et al. Effects of exercise on the skin epithelial barrier of young elite athletes-swimming comparatively to non-water sports training session. Int J Environ Res Public Health 2021;18:653.
CrossRef
Echeverri D, Montes FR, Cabrera M, Galán A, Prieto A. Caffeine’s vascular mechanisms of action. Int J Vasc Med 2010;2010:834060
PubMed
Kim TW, Shin YO, Lee JB, Min YK, Yang HM. Effect of caffeine on the metabolic responses of lipolysis and activated sweat gland density in human during physical activity. Food Sci Biotechnol 2010;19:1077–1081.
CrossRef
Rinaldi AO, Li M, Barletta E, D’Avino P, Yazici D, Pat Y, et al. Household laundry detergents disrupt barrier integrity and induce inflammation in mouse and human skin. Allergy 2024;79:128–141.
PubMed
CrossRef
Sasaki M, Sundberg M, Frei R, Ferstl R, Heye KN, Willems EP, et al. Electrical impedance spectroscopy detects skin barrier dysfunction in childhood atopic dermatitis. Allergy 2024;79:142–152.
PubMed
CrossRef
Cite
Article
PDF
Supplementary Materials
Cited by
Crossref 7
Google Scholar
PubMed 2
Scopus 8
Web of Science 3
Publication Types
Original Article
MeSH Terms
Abdomen
Activities of Daily Living
Coffee
Dielectric Spectroscopy
Electric Impedance
Forearm
Healthy Volunteers
Humans
Reference Standards
Skin
Water
Water
Since 2024/04/01
Page Views 342
PDF Downloads 90
Show all...
1 / 6
ORCID IDs
Lisa Huygen
Pauline Marie Thys
Andreas Wollenberg
Jan Gutermuth
Inge Kortekaas Krohn
Cited by Crossref
Is Cited by the Following Articles in
Citation successfully copied.
Copy and paste a formatted citation from below or use one of the hyperlinks at the bottom to download a file for import into a bibliography manager.
EndNote / ProCite / Reference Manager
RIS Format
Plain Text
EndNote / ProCite / Reference Manager
RIS Format
Plain Text
Permalink information copied to clipboard |
189286 | https://www.geeksforgeeks.org/maths/assumed-mean-method/ | Assumed Mean Method
Last Updated :
23 Jul, 2025
Suggest changes
Like Article
Assumed Mean Method is a statistical technique that is used to calculate the arithmetic mean of a group of data. It is particularly helpful when dealing with large numbers in grouped data. This method involves selecting a central value, known as the assumed mean, and then adjusting the calculations around this value to make the arithmetic more manageable. This technique is used in data analysis to estimate the central tendency of a dataset when the exact mean is not known.
For instance, if you have a data set with class intervals and their respective frequencies, the assumed mean method allows you to break down the problem into simpler steps, making it easier to find the mean without any hard calculations.
What is Mean?
Mean, often referred to as the average, is a measure of central tendency that provides a single value representing the center of a data set.
It is calculated by summing all the values in the data set and then dividing by the number of values. There are three different methods to find the mean of a group of data/%E2%88%91fi-,Methods%20for%20Calculating%20Mean%20for%20Grouped%20Data,-To%20calculate%20mean). These 3 methods are as follows:
Direct Method
Assumed Mean Method
Step-Deviation Method
In this article, we will discuss the Assumed Mean Method in detail.
Table of Content
What is Assumed Mean Method?
Steps to Find Mean using Assumed Mean Method
Difference between Direct, Assumed Mean and Step Deviation Method
Solved Examples on Assumed Mean Method
Practice Questions on Assumed Mean Method: Unsolved
What is Assumed Mean Method?
Assumed Mean Method, also known as the "Shortcut Method". Assumed Mean Method works by choosing an assumed mean (A) close to the actual mean, and then calculating deviations from this assumed mean to calculate the actual mean of the given dataset.
Assumed Mean Method Formula
Assumed Mean Method simplifies the calculation of the mean by using an assumed mean (A). The formula for the Assumed Mean Method is:
x̄ = a + ∑ƒidi /∑ƒi
a is assumed mean,
ƒi is frequency of ith class,
di = xi - a is derivation of ith class,
∑ƒi = n is total number of observations
xi is class mark and is equal to (upper class limit + lower class limit)/2
Steps to Find Mean using Assumed Mean Method
For process for calculating mean by using assumed mean method, are discussed below:
Step 1: For each class interval, we have to calculate the class mark by using the formula,
xi = (upper limit + lower limit)/2
Step 2: Choose an approximate and suitable value of mean, and denote it by "a" .
Step 3: Calculate the deviations using the following formula,
di = (xi - a) for each i
Step 4: Calculate the product by,
(ƒi × di ), for each i
Step 5: Find total frequency
n = ∑ƒi
Step 6: Finally, calculate the mean by using the following formula,
x̄= a + ∑ ƒidi/∑ƒi
Let's consider an example for better understanding.
Example: Find mean using assumed mean method for following data:
| Class Interval | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 |
--- --- ---
| Frequency (f) | 3 | 7 | 12 | 15 | 8 | 5 |
Solution:
Step 1: Calculate mid-point for each class-interval:
| | | |
---
| Class Interval | Midpoint (xi) | Frequency (f) |
| 10 - 20 | 15 | 3 |
| 20 - 30 | 25 | 7 |
| 30 - 40 | 35 | 12 |
| 40 - 50 | 45 | 15 |
| 50 - 60 | 55 | 8 |
| 60 - 70 | 65 | 5 |
Step 2: Select a midpoint close to the center of the data. Let's assume a = 45.
Step 3: For each class interval, calculate the deviation from the assumed mean using di = (xi - a) for each i.
Step 4: Multiply each deviation by the corresponding frequency.
| Class Interval | Midpoint (xi) | Frequency (fi) | Deviation (di) | fi × di |
--- ---
| 10 - 20 | 15 | 3 | 15 - 45 = -30 | 3 × (-30) = -90 |
| 20 - 30 | 25 | 7 | 25 - 45 = -20 | 7 × (20) = -140 |
| 30 - 40 | 35 | 12 | 35 - 45 = -10 | 12 × (-10) = -120 |
| 40 - 50 | 45 | 15 | 45 - 45 = 0 | 15 × 0 =0 |
| 50 - 60 | 55 | 8 | 55 - 45 = 10 | 8 × 10 =80 |
| 60 - 70 | 65 | 5 | 65 - 45 = 20 | 5 × 20 = 100 |
| | | ∑ƒi = 50 | | ∑ƒidi = -170 |
Step 5: Calculate the mean using formula: x̄= a + ∑ ƒidi/∑ƒi
x̄= 45 + (-170)/50
⇒ x̄= 45 - 3.4 = 41.6
Thus, mean of given dataset is 41.6
Difference between Direct, Assumed Mean and Step Deviation Method
Direct Method, Assumed Mean Method, and Step Deviation Method are three different techniques for calculating the mean (average) of a dataset. Each method has its own way of simplifying the calculations. Some of the common differences among these methods are:
| Direct Method | Assumed Mean Method | Step Deviation Method |
---
| Best method for simple problems and small data set. | Best method for large data sets with potentially large values. | Best method for large data sets with uniform class intervals. |
| This method is straightforward and easy to understand. | It simplifies large arithmetic calculations using the assumed mean. | Further simplifies calculations and arithmetic complexity by working with smaller numbers. |
| In this method we calculate midpoints(xi) of each class intervals and multiply them with corresponding frequency(ƒ), i.e., ƒixi . | In this method we calculate deviation(di) of each data point and multiply them with corresponding frequency(ƒ), i.e., ƒidi . | In this method we calculate step deviation(ui) of each class intervals and multiply them with corresponding frequency(ƒ), i.e., ƒiui . |
| The formula of this method is Mean = ∑ƒixi / ∑ƒi | Formula for assumed mean method is Mean = a + ∑ ƒidi / ∑ ƒi | Formula for step deviation method is Mean = a + h × ∑ ƒiui / ∑ ƒi |
Also Read,
Relation between Mean, Median and Mode
Mode of Grouped Data
Statistics Formula
Calculation of Median in Discrete Series
Solved Examples on Assumed Mean Method
Example 1: The following table gives information about the marks obtained by 120 students in an examination.
| | | | | | |
--- --- --- |
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 14 | 30 | 34 | 27 | 15 |
Find the mean marks of the students using the assumed mean method.
Solution:
Let assume mean of the given data be 25 i.e., a = 25.
| Class | Frequency | Class mark (xi) | di = (xi - a) | ƒidi |
--- ---
| 0-10 | 14 | 5 | 5 - 25 = -20 | -280 |
| 10-20 | 30 | 15 | 15 - 25 = -10 | -300 |
| 20-30 | 34 | 25 = a | 25 - 25 = 0 | 0 |
| 30-40 | 27 | 35 | 35 - 25 = 10 | 270 |
| 40-50 | 15 | 45 | 45 - 25 = 20 | 300 |
| | ∑ƒi = 120 | | | ∑ƒidi = -10 |
The formula,
x̄ = a + ∑ƒidi/∑ƒi
⇒ x̄ = 25 + (-10/120)
⇒ x̄ = 25 - 1/12
⇒ x̄ = (300-1)/12
⇒ x̄ = 299/12
⇒ x̄ = 24.91
Therefore, the mean marks of the students are 24.91 .
Example 2: A group of students surveyed as a part of their environmental awareness
| | | | | | | | |
--- --- --- --- |
| Number of plants | 0 - 2 | 2 - 4 | 4 - 6 | 6 - 8 | 8 - 10 | 10 - 12 | 12 - 14 |
| Number of houses | 1 | 2 | 2 | 4 | 3 | 2 | 6 |
The program in which they collected the following data of plants in 20 homes in a area. Find the mean number of plants per household using the assumed mean method.
Solution:
| No. of Plants | No. of houses/ Frequency (ƒi) | Class mark (xi) | di = (xi - a) | ƒidi |
--- ---
| 0-2 | 1 | 1 | -6 | -6 |
| 2-4 | 2 | 3 | -4 | -8 |
| 4-6 | 2 | 5 | -2 | -4 |
| 6-8 | 4 | 7 = a | 0 | 0 |
| 8-10 | 3 | 9 | 2 | 6 |
| 10-12 | 2 | 11 | 4 | 8 |
| 12-14 | 6 | 13 | 6 | 36 |
| | ∑ ƒi = 20 | | | ∑ƒidi = 32 |
x̄ = a+ (Σƒidi /Σƒi)
⇒ x̄ = 7+(32/20)
⇒ x̄ = 7+(8/5)
⇒ x̄ = 8.6
∴ Mean number of plants per household = 8.6
Practice Questions on Assumed Mean Method: Unsolved
Problem 1: The following table contains information on the grades received by 160 students in an examination. Find the mean marks of the students using the assumed mean method.
| | | | | | |
--- --- --- |
| Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
| Frequency | 20 | 30 | 45 | 35 | 30 |
Problem 2: Find the mean of the following data using the assumed mean method formula.
| Marks | Number of students |
--- |
| 0-10 | 5 |
| 10-20 | 3 |
| 20-30 | 4 |
| 30-40 | 3 |
| 40-50 | 3 |
| 50-60 | 4 |
| 60-70 | 7 |
| 70-80 | 9 |
| 80-90 | 7 |
Problem 3: Find the arithmetic mean using the assumed-mean method:
| | | | | | |
--- --- --- |
| Class interval | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |
| Frequency | 20 | 30 | 15 | 10 | 5 |
Problem 4: The following table contains, distribution of daily wages of 60 worker of a factory.
| | | | | | |
--- --- --- |
| Daily Wages | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |
| Number of Workers | 15 | 12 | 10 | 13 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Problem 5: Find the mean of the following data using assumed mean method.
| Class Interval | Frequency |
--- |
| 0-10 | 8 |
| 10-20 | 12 |
| 20-30 | 10 |
| 30-40 | 12 |
Conclusion
Assumed Mean Method is a valuable tool in statistics for estimating the mean of a dataset when exact values are unavailable. By leveraging an assumed mean, analysts can perform essential calculations and make informed decisions despite incomplete data. This method is particularly useful in fields such as quality control and financial analysis, where precise data may be hard to obtain. Whether you're dealing with preliminary data or refining your analysis, the assumed mean provides a practical approach to navigating uncertainty and improving decision-making.
R
rajaroy97
Improve
Article Tags :
Mathematics
School Learning
Class 10
Math-Statistics
Maths-Class-10
Explore
Maths
4 min read
Basic Arithmetic
What are Numbers?
15+ min readArithmetic Operations
9 min readFractions - Definition, Types and Examples
7 min readWhat are Decimals?
10 min readExponents
9 min readPercentage
4 min read
Algebra
Variable in Maths
5 min readPolynomials| Degree | Types | Properties and Examples
9 min readCoefficient
8 min readAlgebraic Identities
14 min readProperties of Algebraic Operations
3 min read
Geometry
Lines and Angles
9 min readGeometric Shapes in Maths
2 min readArea and Perimeter of Shapes | Formula and Examples
10 min readSurface Areas and Volumes
10 min readPoints, Lines and Planes
14 min readCoordinate Axes and Coordinate Planes in 3D space
6 min read
Trigonometry & Vector Algebra
Trigonometric Ratios
4 min readTrigonometric Equations | Definition, Examples & How to Solve
9 min readTrigonometric Identities
7 min readTrigonometric Functions
6 min readInverse Trigonometric Functions | Definition, Formula, Types and Examples
11 min readInverse Trigonometric Identities
9 min read
Calculus
Introduction to Differential Calculus
6 min readLimits in Calculus
12 min readContinuity of Functions
10 min readDifferentiation
2 min readDifferentiability of Functions
9 min readIntegration
3 min read
Probability and Statistics
Basic Concepts of Probability
7 min readBayes' Theorem
13 min readProbability Distribution - Function, Formula, Table
13 min readDescriptive Statistic
5 min readWhat is Inferential Statistics?
7 min readMeasures of Central Tendency in Statistics
11 min readSet Theory
3 min read
Practice
NCERT Solutions for Class 8 to 12
7 min readRD Sharma Class 8 Solutions for Maths: Chapter Wise PDF
5 min readRD Sharma Class 9 Solutions
10 min readRD Sharma Class 10 Solutions
9 min readRD Sharma Class 11 Solutions for Maths
13 min readRD Sharma Class 12 Solutions for Maths
13 min read |
189287 | https://math.stackexchange.com/questions/2863752/concept-multiplying-a-vector-by-a-scalar-quantity | Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Concept: Multiplying a vector by a scalar quantity
Ask Question
Asked
Modified 7 years, 1 month ago
Viewed 2k times
0
$\begingroup$
A scalar is a quantity which has the only magnitude $\|m\|$ in contrast to a vector which has both direction and magnitude $\|m\|\angle\theta$. Intuitively, multiplying a vector by a scalar scales a vector by the respective magnitude. Moreover, various texts claim, that multiplying a vector by scalar only changes the magnitude but direction remains unaffected.
An unexpected result is noticed when we multiply a negative scalar by a vector. Notably, a vector $\vec v = (x, y)$, when its multiplied by a negative scalar $-\|q\|$, it not only scales the magnitude by a quantity equals to $\|q\|$, but also flips the vector $-\|q\|\vec v = (-\|q\|x, -\|q\|y) = \|q(x^2+y^2)\|\angle{\left(\pi+\arctan\frac{y}{x}\right)}$.
Many notable texts, claim the behavior to be valid. But, I find a few contradictions
If by multiplying a vector $\vec v$ scales a vector, what is the intuition of negative scaling?
If the magnitude is absolute, and scalar has the only magnitude, how can scalar be negative?
If multiplying by a scalar does not change the direction, why is multiplying a vector by a negative quantity and thus flipping it, a valid behavior?
The way, I am trying to understand is, multiplying by a negative scalar is actually two operations.
Multiplying the vector by (-1) which flips a vector. This generates the negative inverse of a vector $\because \vec v + (-1)\times \vec v = 0$
Scale the resultant vector by the appropriate scalar quantity.
So, scalar -q is actually $\|q\| \times -1$ and thus multiplying with the vector $\vec v = \|v\|\angle\theta$
$$\Rightarrow -q \times \vec v = \|q\| \times -1 \vec v = \|q\| \times \|v\|\angle\left(\pi + \theta\right)$$
But, I cannot find a source to substantiate my reasoning and need the community support to help me with my understanding.
linear-algebra
vectors
Share
edited Jul 27, 2018 at 4:32
AbhijitAbhijit
asked Jul 26, 2018 at 19:52
AbhijitAbhijit
2,5691414 silver badges2323 bronze badges
$\endgroup$
7
$\begingroup$ I think when they say that multiplying by a scalar does not change the direction, what they mean is the span of the vector remains unchanged. $\endgroup$
gd1035
– gd1035
2018-07-26 20:01:34 +00:00
Commented Jul 26, 2018 at 20:01
$\begingroup$ Multiplying by a negative does "flip" and scale the vector, but the span of the vector remains unchanged. $\endgroup$
gd1035
– gd1035
2018-07-26 20:02:11 +00:00
Commented Jul 26, 2018 at 20:02
$\begingroup$ Multiplying by a negative number changes the magnitude and direction, as you say. I wouldn't worry too much about the words the text used to describe this. $\endgroup$
saulspatz
– saulspatz
2018-07-26 20:02:15 +00:00
Commented Jul 26, 2018 at 20:02
$\begingroup$ When they say a scalar is "a quantity that has magnitude but not direction", they are speaking roughly. The term "magnitude" in this context is not meant to imply that a scalar cannot be negative. A scalar can be negative, and that is fine. When they say that multiplying by a scalar "does not change direction", they are also speaking roughly, because multiplying by a negative scalar does reverse the direction. $\endgroup$
littleO
– littleO
2018-07-26 20:04:20 +00:00
Commented Jul 26, 2018 at 20:04
$\begingroup$ For a vector $v$: Magnitude: $\|v\|$, Direction: $O={rv:\ r\text{ scalar}}$, Orientation: Fixing a vector $w\in O$, with $w\neq0$, $v$ is positively oriented with respect to $w$ if there is $r>0$ such that $rv=w$, negatively oriented with respect to $w$ if there is $r<0$ such that $rv=w$. With these definitions, multiplying $v$ by $-2$, changes magnitude, and orientation with respect to a fixed $w$, and not direction. $\endgroup$
user577471
– user577471
2018-07-26 20:11:36 +00:00
Commented Jul 26, 2018 at 20:11
| Show 2 more comments
2 Answers 2
Reset to default
2
$\begingroup$
In a real vector space where the scalars are real numbers, a scalar has a magnitude and a sign.
Thus when you multiply a vector by a scalar you scale the magnitude of the vector and change or not change the direction of your vector based on the sign of the scalar.
( The vector stays on the same line, which sometimes is interpreted as not changing the direction )
The concept is clear but the terminology is sometimes confusing.
Share
answered Jul 26, 2018 at 20:12
Mohammad Riazi-KermaniMohammad Riazi-Kermani
70.1k44 gold badges4444 silver badges9393 bronze badges
$\endgroup$
Add a comment |
1
$\begingroup$
We have that for $\lambda\in \mathbb{R}$
for $\lambda>0$ we have that the operation $\lambda\vec v$ scale $\vec v$ of a factor $|\lambda|$ with the same orientation
for $\lambda<0$ we have that the operation $\lambda\vec v$ scale $\vec v$ of a factor $|\lambda|$ with the opposite orientation
Note that in both cases the direction doesn't change since both vectors belong to the same line.
Share
answered Jul 26, 2018 at 20:11
useruser
164k1414 gold badges8484 silver badges157157 bronze badges
$\endgroup$
Add a comment |
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
linear-algebra
vectors
See similar questions with these tags.
Featured on Meta
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Linked
2 Quick doubt about multiplying vectors by scalars
Related
What separates the dot product from the scalar projection?
1 Understanding components of a vector
1 Taking a magnitude of a vector consisting of other vectors
1 Terminology direction ratios and scalar components of vector
1 Gradients, Directional Derivatives and Change in Scalar Functions
2 Definition of Vector (Cross) Product
2 Is the component of a vector along another vector also a vector?
Are vectors in one dimension scalar?
Hot Network Questions
Where can a French citizen with Russian visa legally cross by bike/foot/car from China into Russia at a border point as close as possible to Dongning?
My pure mathematics PhD dissertation will likely be very short. I have until the last day of January 2026 to remedy this. What can I do?
Buffering line vectors that only buffer along the length of the line in QGIS
Is there any guideline to make it portable regarding filepath related code?
What is an application of Goursat's stronger version of Cauchy's theorem?
How are Ephesians 5:18 and Acts 2:13 related?
Does concatenating the digits of the k-th perfect powers always give a base-10 normal real number?
Installing firealpaca
What does the SH pin on HDMI do?
What is Sealord frequency used prior to entering restricted airspace?
Clearly allowed
What is the purpose of the option delete readonly = yes available for the configuration of a Samba service?
Beyond first place?
How to clear specific byte values inside a 64-bit value without looping
When can you stealth in terms of "other senses"?
Rendering tikz code as an image in a separate file, then using \includegraphics in the main file, whilst complying with PDF/UA
Short story about man deciding to end the stasis of eternal life on Earth
How to remove the empty star in Files (Nautilus)?
Pressure treated or regular wood for temporary ramp?
What are possible options to disembark for health issues if your airline keeps you on board at the tarmac for several hours?
Is electrolysis of alumina (Al2O3) without cryolite possible to produce pure aluminum?
What setting shows the objects as different colors by default?
Why do complex questions lead to simple elegant answers?
Did alignment charts of old use to have Law on the right (and when did that change)?
more hot questions
Question feed |
189288 | https://www2.chem.wisc.edu/deptfiles/genchem/lab/labdocs/modules/volflask/volflaskdisol.htm | | | | |
---
| | Volumetric Flask | |
| |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | Description | | | Preparing a Flask for Use | | | Dissolving a Solute | | | Filling a Volumetric Flask | | | Mixing the Solution in a Volumetric Flask | | | Using and Storing a Solution Prepared in a Volumetric Flask | | Additional Topics | | | Self Check Exercises | | Related Modules | | | Balance | | | Mohr Pipet | | | Volumetric Pipet | | | Hot Plate/Magnetic Stirrer | | | | Dissolving a Solute | | | | | --- --- | | | | | | Video. Dissolving a solid in a volumetric flask. | | Video. Self Check video. | When dissolving a solid or a liquid, remember to add the chemical to water, rather than vice versa, especially when diluting concentrated acids. When the solute is a solid, first dissolve the solid in a vessel other than the volumetric flask (a beaker or Erlenmeyer flask). Occasionally, a solid will not dissolve readily and the mixture must be heated to aid dissolution. NEVER heat a solution in a volumetric flask. After the solid dissolves, pour the solution into the volumetric flask and use the solvent to rinse all of the solution from the vessel into the flask (a "quantitative transfer"). However, if the solid is known to be very soluble, it may be added directly to the flask. The student in the video believes that he prepared a 0.5236 M NaCl solution. Is the actual molarity higher or lower than 0.5236 M? Why? Answer The actual molarity of the NaCl solution is lower than 0.5236 M. The student first attempted to dissolve the NaCl in a separate vessel, which is the correct procedure; however, he did not dissolve all of the NaCl, so the entire quantity of solute was not transferred to the volumetric flask. Since the number of moles transferred was less than anticipated, the actual molarity of the solution is lower than expected. | | |
189289 | https://www.youtube.com/watch?v=vkMXBvcImjA | OpenStax: Algebra and Trigonometry - Chapter 2, Section 4 | Complex Numbers
Scalar Learning
117000 subscribers
27 likes
Description
1622 views
Posted: 30 Jun 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 4, Complex Numbers. Watch Huzefa as he reviews exercises 1-55 odd.
Credit:
Link to Exercises:
To skip to a particular question, use the chapters below:
00:00 Introduction
00:15 Exercise 1
01:05 Exercise 3
01:41 Exercise 5
02:13 Exercise 7
03:16 Exercise 9
05:20 Exercise 11
06:00 Exercise 13
06:15 Exercise 15
06:39 Exercise 17
07:00 Exercise 19
07:18 Exercise 21
07:46 Exercise 23
08:01 Exercise 25
08:35 Exercise 27
09:09 Exercise 29
09:26 Exercise 31
10:03 Exercise 33
11:24 Exercise 35
12:06 Exercise 37
12:53 Exercise 39
13:43 Exercise 41
14:37 Exercise 43
18:08 Exercise 45
20:43 Exercise 47
21:23 Exercise 49
22:02 Exercise 51
23:19 Exercise 53
25:16 Exercise 55
CHECK OUT our Super Awesome SAT Math Video Course:
JOIN OUR TEST PREP DISCORD:
SUBSCRIBE NOW! And give us a thumbs up if you liked this video.
Need academic help? Learn more about the Scalar Learning Tutoring Team:
Get more tips and tricks by following us!
Check out our Math Music Videos!
Music Videos for math concepts:
Synthetic Division -
Special Right Triangles -
Imaginary Numbers -
6 comments
Transcript:
Introduction [Music] what's up everybody and welcome back to open stacks algebra and trigonometry chapter 2 section 4 complex numbers let's do it so explain how to add Exercise 1 complex numbers so first what is a complex number that's the broadest category of numbers that we have so technically everything is going to be a complex number but what they mean here is the real component and the imaginary component so for example if you'd have a complex number as 5 plus 2i that's in the format of a plus bi so if i had another complex number and again we'll get to imaginary numbers more so later in this section but i is simply equal to the square root of negative 1 which we can't evaluate with real numbers there's no number times itself that's a real number that will equal negative 1. so we have 5 plus 2i negative 3 plus 7i so if i were to add these i would add the real parts 5 and negative 3 which is 2 2i and 7i which is 9i that's how you do it done Exercise 3 so a very simple example to show that the product of two imaginary numbers is not always imaginary is as follows if i do 2i times 3i the real parts will multiply to give us 6 and the imaginary parts will give us i squared so again if we recognize that i is the square root of negative 1 if i were to square this right square i squared it would get rid of the square root therefore i squared is negative 1. therefore we can replace i squared with negative 1 which means this would give a result of negative six which is real boom done here we're going to plug and Exercise 5 chug and we're going to solve for y when x equals 2i so we've got 2i squared plus 2i minus 4. so 2i squared becomes 4i squared plus 2i minus 4. once again i squared is negative 1 so that's negative 1 times 4 which is negative 4 plus 2i minus 4. we combine like terms negative 4 negative 4 make negative 8 plus 2 i is the winner for y done Exercise 7 here we're going to plug 2 plus i in for x so we've got 2 plus i squared plus 3 times two plus i plus five okay first i'm going to foil this so that's going to be 2 plus i times 2 plus i that's how we square a binomial this one we're going to distribute that 3 which is going to be 3 times 2 is 6 plus 3 i plus 5. and now let's simplify this so first we got 2 times 2 which is 4. then we've got a 2i and a 2i as they're multiplying so 2i plus 2i is i and then we got i times i which is i squared again i squared is just negative one so i'm gonna take a little shortcut there and then we've got this whole section now we can combine like terms we've got four and negative 1 and 6 and 5 so that's 3 9 plus 5 is 14 put in the real part first and then 4i and 3i makes 7i for the win done Exercise 9 we're going to plug in 5i for x so we got 5i plus 1 over 2 minus 5i now this is in and of itself fine but the general idea is when we got complex numbers we don't want to have an imaginary in the denominator as such we're going to multiply by something called the conjugate both on the top and bottom so it's the same as 2 minus 5i except we flip the minus 2a plus and then we have to do the same thing on top because that makes it so that we're multiplying technically by one which is allowed so now we're going to do our foiling so we've got 5 i times 2 which is 10 i plus 25 i squared again the i squared is negative 1 so it's minus 25 and then plus 1 times 2 which is 2 plus 1 times 5 i which is 5 i will combine like terms in a second then on the denominator we've got 2 times 2 which is 4. then we've got 10 i and negative 10i when they multiply so i'm not writing it out because this is the beautiful part about the difference uh or it's a difference of squares but it's multiplying by the conjugate the 10 i and the negative 10 i those are going to cancel out and that's how we get rid of the i on the denominator actually i will write it out why not because it's the first example so 10 i minus 10 i but those will cancel and then i've got negative 5i times 5i that's negative 25 i squared but again i squared is negative 1. so that cancels out with a minus so on the bottom it becomes 4 plus 25. so now let's combine like terms and finish this so i got 10 i and 5i that makes 15i negative 25 and 2 make minus 23 over 4 plus 25 which is 29. last but not least we're going to put this into a plus bi format so it's going to be the real part first so i'm going to split it up as negative 23 over 29 plus 15 i over 29 for the win Exercise 11 done but the following exercises plot the complex numbers on the complex plane so you might be like well how do i do that but it's really not so bad the real number aka this part is going to go on the x-axis or the real number axis in this case and then this complex part the negative 2i is going to go up here so we can just kind of number it out like this we say 1 2 3 just a couple numbers and then here we can say negative 1 negative 2 negative 3 again so negative 2 would represent negative 2i so this would be 1 on the real number axis negative 2 on the imaginary number axis so there's the point done Exercise 13 so in this case we've just got an i that we need to plot now technically that means the real number part is zero so we're not going left or right at all on the real axis we are going up because that's a one i so we're going right here for the win done Exercise 15 here we're simply going to simplify and the best way to think about this is to just treat i like a variable like x and since we got plus between these guys the parentheses aren't doing anything so i'm going to rewrite it like so and then i'm going to combine the 3 with the 5 and the 2i with the negative 3i so we have 8 minus 1i right because 2 and negative 3 is negative 1 so i'll just say 8 minus i boom done Exercise 17 here we have a minus and so i got to first distribute that minus so i got negative 5 plus 3i minus 6 and then minus and minus makes plus i now i'm going to combine like terms negative 5 and negative 6 makes a negative 11. 3i and i makes a 4i negative 11 plus 4i boom done here we're Exercise 19 going to start by distributing the negative and i got negative 4 plus 4 i plus 6 right negatives cancel out and then minus 9 i and then we're going to combine like terms negative 4 and 6 make 2 4i and negative 9i make negative 5i for the win done here we're going to Exercise 21 distribute distribute so 3i times 5 is 15 i and then minus 3 times 2 is 6 and then i times i is i squared so again i squared is just negative one so if i replace i squared with negative one it's going to cancel out with that minus so i've got 15 i plus six and then write it in a plus bi form it would just be 6 plus 15 i Exercise 23 done again we're going to distribute the 8 like so 8 times negative 2 is negative 16. 8 times 4 is 32 so you notice it's 32i just like if it was 4x it'd be 32x so this is the final answer done here Exercise 25 we're going to do some foiling so first negative 1 times negative 2 is 2 then we do the outer negative 1 times 3i is minus 3i then we do that which is negative 4i and then we got the last 2i times 3i which is 6i squared so again these guys combined negative 3i and negative 4i make negative 7i remember i squared is negative one so that's really just a negative six or six times negative one so two and negative six combine to become negative four for the win done once again we're going to foil here Exercise 27 three times three is nine 3 times negative 4 i is negative 12 i 4i times 3 is positive 12 i and 4i times negative 4i is negative 16 i squared now notice the negative 12i and positive 12i they're opposites so they cancel out and then that's a minus 16 i squared or minus 16 times negative 1. that's what i squared is so this just becomes a plus 16 and then 9 plus 16 is 25 boom done so here all we're doing is Exercise 29 we're simplifying we're dividing this entire expression by 3 so i can break it up as 6 divided by 3 which is 2 and then 2 divided by 3 which is 2 3 i and there we are in the format of a plus b i so this is the winner done here Exercise 31 we're going to divide both terms by i so i've got 6 over i plus 4 i over i just becomes 4. so now we've got this problem we've got the imaginary number in the denominator the way i'm going to get rid of that is i'm going to multiply it by i over i right so i times i is of course i squared so i've got 6 i over i squared plus four but i squared is the same as negative one so six divided six i divided by negative one is negative six i or minus six i then i'm gonna place the four in front because again we put the real part in front there's the winner done again we need to Exercise 33 get rid of this denominator having an imaginary so we're going to use the conjugate so check it out here's the complex conjugate instead of 2 minus i we're going to multiply by 2 plus i so on top we're going to do 3 times 2 which is 6 then 3 times i which is 3i then 4i times 2 which is 8i and then 4i times i which is 4i squared and then on the bottom we've got 2 times 2 which is 4. and again with this complex conjugate the nice thing is that you're going to get a 2i and a negative 2i which means the imaginaries are going to cancel out and then we got negative i times i which is negative i squared again since i squared is negative 1 negative i squared is negative negative 1 or 1 plus 1. okay now let's simplify so we've got like terms here that's 11i so we're going to put let's put it here 11i and then 4i squared that's 4 times negative 1. so that's a negative 4 right here 6 and negative 4 make a 2 and that's over 4 plus 1 which is 5. last but not least to put it into a plus bi form it would be two-fifths right two divided by five plus eleven fifths times i Exercise 35 done here we're gonna simplify as much as possible so first i've got i'm going to rewrite negative 9 as negative 1 times 9. i'm going to do the same with 16. so the reason why i'm doing this is because now i can take the square root of negative 1 which is i so i got i square root of 9 plus 3 i right square to negative 1 times or 3 i times square root of 16. then what is the square root of 9 it's 3 and what is the square root of 16 that's 4. so remember when the 4 comes out it multiplies that 3 in front so it becomes 12 i and then 3i plus 12i is 15 i done all right here i'm going to break Exercise 37 it up i'm going to first make it 2 over 2 plus square root of negative 12 over 2. so 2 over 2 is of course 1 and then for the 12 or negative 12 on top i'm going to break it up like so i'm going to make it negative 1 times 4 times 3. i could also broken it up and make it negative 1 times 2 times 2 times 3. but i did it this way in particular because i already know that 4 is a perfect square so check this out the square root of negative 1 is i the square root of 4 is 2 and that 3 is stuck inside we can't do anything to simplify that and that's over 2. so 1 plus this last but not least i can cross off the 2s on the top and the bottom so my final simplified expression is 1 plus i root three boom done so when we get Exercise 39 really big values of i like i to the eighth or i to the 20th or whatever there is a pattern the i values repeat every fourth so for example i equals just your regular square root of negative 1 which is just i uh i squared is negative 1. right we already have talked about that i cubed is that negative 1 right i squared times another i should be negative i and then i to the fourth is basically a negative i times another i which would be negative i squared since i squared is negative one it would come to positive one and then i to the fifth repeats because we're now just taking a one and multiplying it by another i so if we carry out this pattern we see that i to the eighth equals one boom done Exercise 41 so as i mentioned in question 39 these imaginary numbers when they're raised to exponents they follow a certain pattern so i to the first is one i squared is negative one i cubed is negative i oops let me put it make sure that it looks like an i and then i to the fourth equals one so to figure out like a big number like 22 is actually not that hard but imagine if it was 222 that'd be difficult so what you do is you take that exponent and you divide it by four so 4 goes into 20 5 times but that doesn't matter what matters is the remainder okay now if there was no remainder it would be 1. if there was a remainder of 1 it would be i if there was a remainder of three it would be negative i but we got a remainder of two so that means that i to the 22nd equals negative one Exercise 43 evaluate one minus i to the k power for k equals two six and ten predict the value of k equals 14. so we're gonna use this little binomial expansion calculator that i found and we're going to give it a shot and by the way this is the technology portion where they want you to use a calculator that's why i'm using one so i 1 minus i squared is i squared minus 2i plus 1. i squared since that's negative 1 it's just going to cancel out with the one but let's write it out so these guys will cancel and we get negative 2i that's for k equals 2. now for k equals six so i'm going to take that and i'm going to throw it up on the white board and copy it down and then we'll simplify it check it out so i to the sixth is again using our pattern right i to the fourth would be one out of the fifth will be i i to the sixth would be negative one and then i to the fifth is just i so that's minus six i i to the fourth is one so that's just plus fifteen cubed is just negative i so that's plus 20 i i squared is just negative 1 and that's minus 6 i plus 1 so we'll combine all the like terms so we got negative 1 that's 14 that's plus negative 15 is negative 1 and then that is 1. so there's no uh these guys all just cancel out it looks like and then we've got negative 6i negative 6i that's negative 12 i plus 20i which is 8 i now let's do k equals 10 so i'm going to take this we'll put it up and then we'll knock it out so my guess is that there's a pattern here where the real number part is going to cancel out again but we'll we'll see if that's true so i to the 10th that's 4 4 and then i squared which is just negative 1 right i to the 10th is negative 1. i to the ninth then it's probably just it's just i so that's minus 10 i i to the eighth is just one so that's plus 45. i to the seventh is negative i so then that's plus 120 i i to the six is like i squared so then it's like minus 210. i to the fifth is like i so that's minus 252 i i to the fourth is just one i to the third is negative i so that's plus 120 i i squared is negative 1 negative 45 minus 10 i and then it's plus 1. so once again the ones cancel the 45s cancel the two tens cancel positive and negative so it's true we're down to just i values so we got negative 10i that's plus 120 is 110 i that's negative 142 i um and then that's negative 1 52 i plus 120 negative 152 that's negative 32 i so the pattern i'm seeing here it seems like when this goes up by 4 we're multiplying this value by negative 4 because then we get positive 8i then we go up by 4 we multiply by negative 4 we get negative 32i so i believe that the pattern would be for k equals 14. we multiply this by negative 4 which would give a value of positive 128 i boom done Exercise 45 show that a solution of x to the 6 plus 1 is this well basically what that would mean is if we plugged this in for x we should get negative one because negative one plus one is zero so again i'm using my binomial expansion calculator the technology that they said we can use to get my binomial expansion where x is simply the i value so now let's go ahead and replace all these x's with i's and since the least common denominator is 64 i'm going to simultaneously convert everything to a denominator of 64. so i to the 6 is like i squared so that's negative 1 so we got negative 1 over 64 plus that's really i'm going to make that 64 so that would be 6 rad 3 and x to the 5th is i to the fifth which is just i so i'm gonna make it 6 i rad 3 over 64. right just doubling the top and bottom then x to the fourth is just 1 so that's plus 45 over 64 plus that's only out of 16 so i'd multiply that by 4 to get 64. so that's 4 times 15 is 60. i cubed is negative i so i'm actually going to change this to a minus so we've got minus 60 i root 3 over [Music] 64. and then that's minus 135 because i squared is negative 1 over 64. and then that's just i and since it's out of 32 i'm going to double it so that's plus 54 i rad 3 over 64. and then i'm going to leave that 27 over 64. so for this to work out this all needs to evaluate to negative 1. so my assumption is all the i's are going to cancel out let's see so 6 i root 3 54 i root 3 makes 60 i root 3 and we got a negative 60 i root 3 so those indeed cancel out like so and then we got to combine these hopefully again this will equal to negative 1. so negative 1 plus 45 is 44 over 64. negative 135 plus 27 is let's see one they take away 20 that's 115 negative 108 over 64. so minus 108 over 64. and negative 108 plus 44 taking 44 away basically that's 104 that's 64. boom which equals negative 64 over 64 which does equal negative one so we've proved our point done Exercise 47 here we are going to simplify by just getting rid of the i's in the denominator so for the first one i'm gonna multiply by i and i for the second one i'm gonna also multiply by i over i because i times i to the third is i to the fourth which evaluates to one so check this out this becomes i over i squared plus four i over i to the fourth i squared is negative one so that's just a negative i and i to the fourth is one so 4i over 1 is just 4i and then 4i plus negative i is 3i for the win Exercise 49 done here we're going to distribute like so that's i to the seventh right just times one is itself plus we add the exponents that's i to the ninth so if you remember our chart what we would do is we would take seven we divide it by four that's one with a remainder of three which means this is like i cubed i cubed is just negative i i times i times i i times i is negative 1 times another i is negative i now when we do 9 divided by 4 we get 2 with a remainder of 1. that means i to the ninth is same as i to the first which is just i negative i plus i is zero boom done so the first thing i'm Exercise 51 going to do is i'm going to foil this top so i've got 2 times 4 which is 8 and then 2 times negative 2i that's minus 4i and then i times 4 is positive 4i and then i times negative 2i is minus 2i squared these are going to cancel out nicely and then negative 2i squared that's negative 1 times negative 2 right i squared is negative 1 so it becomes a plus 2. so then we get 8 plus 2 which is 10 on top over 1 plus i last but not least we're going to multiply by the conjugate of 1 plus i because we don't want an imaginary in the denominator and that's 1 minus i 1 minus i so on top we're going to distribute distribute we got 10 minus 10 i on the bottom we have 1 times 1 which is 1 and then we have negative i and positive i which are going to cancel out and then we got i times negative i which is minus i squared which is the same as negative negative 1 or plus 1. last but not least we have 10 minus 10 i over 1 plus 1 which is 2 and then we can divide each of these by 2 and we get 5 minus 5 i for the win Exercise 53 done here let's go ahead and foil these out on top and bottom so on top we get 3 times 3 which is 9 plus 3i plus 3i plus i squared again i squared is just negative 1. so i'm going to shortcut that over 1 times 1 which is 1 plus 1 times 2i which is 2i again 2i times 1 which is 2i and then we got 4i squared but again i squared is negative 1. so i'm going to skip a step and just write minus 4. now i'm going to combine like terms and i got 9 and negative 1 is 8 plus 6i right 3i plus 3i is 6i over 1 and negative 4 make negative 3 and then 2i and 2i make 4i last but not least can't have an imaginary in the denominator so we got to multiply by that conjugate which will be negative 3 minus 4i so check this out so on top we're going to get 8 times negative 3 which is negative 24 8 times negative 4i which is negative 32i 6i times negative 3 which is negative 18i and then 6i times negative 4i is negative 24 i squared which again is just negative 24 times negative 1 which is positive 24. on the bottom we've got negative 3 times negative 3 which is 9. negative 3 times negative 4 which is 12i negative 3 times 4i which is negative 12i those cancel out so i just got to do negative 4i times 4i which is minus 16i squared since i squared is negative 1 this is actually a plus 16. all right now let's finish it off so negative 24 and 24 cancel out negative 32 i minus 18 i is negative 50 i over 9 plus 16 which is 25 negative 50 divided by 25 is negative 2 we keep the i there boom done Exercise 55 for 55 we again want to remove the imaginary number out of the denominator so 4 plus i over i we're going to multiply the top and bottom by i over i and then for this one we need to multiply by the conjugate of one minus i which is one plus i so check this out so up top we have four i mine plus i squared excuse me and i squared is negative one but we'll get to that in a second and on the bottom we have i squared as well so these are both gonna be minus one and minus one uh so that's pretty nice so if i want to simplify this a little bit more i can divide both by negative 1 so it's negative 4i and then negative 1 divided by negative 1 is plus 1. now over here we're going to foil i get 3 plus 3i minus 4i and then minus 4i squared and again i squared is negative 1 times negative 4 would make that a plus 4 over and then we got 1 times 1 which is 1 then we got 1i negative 1i cancels out and then minus i squared and again i squared is negative 1 so minus negative 1 is plus 1. so now this part simplifies to 3 and 4 becomes 7. 3i minus 4i is negative i over 2. and these two are being added together so now we can simplify further i can break this up into real and imaginary so i've got here let's make a little line there i got negative four i plus one plus seven over two and then minus i over two i'll call that one half i finally combining like terms one plus seven halves is the same as two over two plus seven halves which is nine over two and then negative four i minus one half i they'd be negative four and a half i but i'm gonna keep it as an improper so i'll make it negative nine over two nine over two is the same as four and a half negative nine over two i for the win 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 |
189290 | https://www.youtube.com/watch?v=R5U2GjpIXeM | Fatty Acid Biosynthesis | Chapter 25 - Lehninger Principles of Biochemistry
Last Minute Lecture
3020 subscribers
1 likes
Description
55 views
Posted: 4 Aug 2025
Chapter 25 of Lehninger Principles of Biochemistry (Eighth Edition) explores the biosynthesis of fatty acids, focusing on the enzymes, substrates, and regulation of this highly coordinated anabolic process. The chapter begins by contrasting fatty acid synthesis with β-oxidation, emphasizing that synthesis occurs in the cytosol (or plastids in plants) and uses NADPH as a reducing agent, while degradation occurs in mitochondria and generates NADH and FADH₂. Section 25.1 outlines the preparation of precursors, starting with the conversion of acetyl-CoA to malonyl-CoA by acetyl-CoA carboxylase (ACC)—the committed and highly regulated step in fatty acid synthesis. This reaction requires biotin and is activated by citrate and inhibited by palmitoyl-CoA and AMP-activated protein kinase (AMPK).
Section 25.2 details the multienzyme complex fatty acid synthase (FAS), which catalyzes a repeating four-step cycle: condensation, reduction, dehydration, and a second reduction. These steps extend the growing fatty acid chain by two carbon units per cycle, using malonyl-CoA as the donor. The process results in the formation of palmitate (C16:0) after seven cycles, which may then undergo further elongation or desaturation. The ACP (acyl carrier protein) plays a central role in shuttling intermediates between catalytic domains.
Section 25.3 explores the sources of NADPH required for fatty acid synthesis, primarily the pentose phosphate pathway and the malic enzyme reaction. The chapter also covers the elongation and desaturation of fatty acids in the endoplasmic reticulum, involving elongases and desaturases to produce unsaturated and longer-chain fatty acids, including essential ones like linoleate and linolenate, which must be obtained from the diet.
Section 25.4 focuses on the regulation of fatty acid biosynthesis. Key regulatory mechanisms include allosteric control of acetyl-CoA carboxylase, hormonal regulation via insulin and glucagon, and transcriptional regulation of lipogenic enzymes in response to nutritional status. The chapter concludes with clinical relevance, including obesity, metabolic syndrome, and fatty liver disease, all of which involve dysregulation of lipid synthesis pathways. The integration of fatty acid biosynthesis with carbohydrate and energy metabolism is emphasized as essential for metabolic homeostasis.
📘 Read full blog summaries for every chapter:
📘 Have a book recommendation? Submit your suggestion here:
Thank you for being a part of our little Last Minute Lecture family!
Lehninger Principles of Biochemistry Chapter 25 summary, fatty acid biosynthesis steps, acetyl-CoA carboxylase regulation, malonyl-CoA formation, fatty acid synthase complex FAS, acyl carrier protein ACP role, palmitate synthesis from acetyl-CoA, NADPH source in lipid biosynthesis, pentose phosphate pathway NADPH, malic enzyme function, elongation desaturation fatty acids, linoleate linolenate essential fatty acids, insulin and glucagon fatty acid control, obesity fatty liver disease lipogenesis
Transcript:
Welcome to the deep dive where we crack open complex topics and distill them into the most essential intriguing insights. Great to be diving in. Today we're plunging into the absolutely fascinating world of DNA metabolism. When you think of DNA, you probably picture it as this incredibly stable molecule, the static blueprint of life, right? It's the common picture. Yeah. But the truth is far, far more dynamic than that suggests. That's spot on. While DNA is the ultimate repository of genetic information, guiding everything from the size of a cell to the shape of an entire organism. Well, calling it merely stable storage really misses the bigger picture. How so? It just doesn't capture the constant intricate processes by which this precious genetic information is not only preserved uh kept uncorrupted, but also flawlessly transmitted from one generation to the next. Okay. So, our mission today is to unpack that complexity. We're talking DNA replication, repair, and recombination. Three huge areas. Exactly. And these aren't just biological footnotes. They're dynamic, relentless forces constantly at work. They ensure life's continuity battle daily damage and even, you know, fuel evolution itself. Absolutely. And the molecular mechanisms, the biochemical pathways involved, they're truly ingenious. We'll explore those often using landmark discoveries made with common bacteria like E.coli Poli is our guide. These are insights you find detailed in well foundational biochemistry understanding. So let's dive in. Let's begin with DNA replication. The fundamental process of making faithful copies. The core copying mechanism. For centuries, long before we even knew DNA structure, scientists marveled at how organisms seem to copy themselves perfectly. But uh once the double helis was discovered, the mystery truly began to unravel. Exactly. the very structure of DNA, you know, deciphered by Watson and Crick. It immediately hinted at the template concept, meaning each strand could serve as a guide, a template for building a new complimentary strand. This idea led straight to the discovery of semiconservative replication. Right? The measlesen install experiment beautifully demonstrated by them in 1957. Their elegant experiment showed that every new DNA molecule is a hybrid, one original parental strand and one brand new, newly synthesized strand. It's uh really a marvel of biological design. And how does this incredible copying process actually kick off and then proceed within a living cell? Where does it start? Replication typically begins at a unique specific point on the DNA called an origin. Okay. For circular bacterial chromosomes like ecolles, it usually proceeds in two directions simultaneously moving away from that origin. Two directions. Yeah. Forming two replication forks. Think of them as points where the DNA is unwinding and being copied as they move along. John Terren famously visualized this in E.coli back in the 60s. He had a structure image. That's the one. It showed this extra loop looked just like the Greek letter theta confirming that birectional movement. So DNA strands run in opposite directions, antiparallel as we call it. But new DNA can only be built in one direction, five prime to threep prime. How on earth do both strands get copied at the same time? It sounds like a well a biological paradox. It definitely did. It was a major puzzle. This paradox was brilliantly resolved by Raio Kazaki Okazaki fragments. Precisely. He discovered that while one new strand, the leading strand is synthesized continuously just following the replication fork. The other one, the lagging strand, is built discontinuously and pieces in short pieces now known as Okazaki fragments. These are maybe 1,000 to 2,000 bases long in bacteria shorter in us. They are then joined together later. It's like building one road smoothly and the other in short connected segments. Before we get deeper into building DNA, what about the flip side? The enzymes that actually break down DNA because that must play a role too, right? In managing all this. Absolutely. Those are crucial. They're called nucleuses or dasses if they're specific to DNA. And there are different types. Yeah. Two main types. Exonucleuses, which sort of chew away at DNA from its ends, either the five prime or threep prime end. And then endonucleuses, which can cut the DNA strand at specific internal sites, breaking it into smaller fragments. Both are critical for repair, recombination, and even parts of replication itself. Okay. Now, for the true workh horses of DNA synthesis, the DNA polymerises. Arthur Kornberg's pioneering work in 1955 really put them on the map. So, how do these molecular machines actually add new pieces to the DNA chain? What's the chemistry? At their core, all DNA polymerises perform a fundamental chemical reaction. They take a building block, a deoxyucleotide triphosphate or dntp A's, T's, C's, and G's. Exactly. And they attach it to the growing DNA strand. Think of it like snapping a new Lego brick onto an existing structure. The existing strand, provides a little hook chemically. It's a threep prime hydroxal group. Okay. And the polymerase brings in the new DNTP, aligns it with the template, releases some energy from it by breaking off two phosphate groups, and forms that new phosphodaster bond, linking it into the chain. And it needs help doing that. Well, the reaction is stabilized by magnesium ions at the enzyme's active site. And critically, every DNA polymerase needs two things to get started. But a template DNA strand to read that's the guide for which base comes next. And a primer. A primer like painting sort of. It's a short pre-existing piece of DNA or often RNA that provides that essential threep prime hydroxal hook for the polymerase to start adding onto. It can't start from scratch on a bare template. Ah, okay. And they stick around for a while. That's called processivity. It describes how many new nucleotides they can add before letting go of the DNA. Some are highly processive, adding thousands or even hundreds of thousands of bases. Replication needs to be incredibly accurate. Considering the billions of bases copied, how does the cell achieve such amazing precision? Fewer errors than you'd expect by chance. It's really a marvel of multiple layered mechanisms. It's not just one thing. Like what? First, the polymerase enzyme itself is quite picky. During that nucleotide addition step, it actively checks the shape of the incoming base. Yeah, the geometry of the correct aut base pairs is very specific. If an incorrect nucleotide tries to fit into the active site, it just doesn't have the right shape, the right geometry, and it's usually rejected before the chemical bond is even formed. So, a physical check first, right? But mistakes still happen occasionally. So layer two, many DNA polymerases have a built-in proofreading function like spellch check. Exactly like spellch check. It's a separate enzyatic activity, usually a threep prime to fivep prime exonucleus. If the polymerase accidentally adds the wrong nucleotide, yeah, this proofreading nucleus detects the mismatch, pauses, backs up, and specifically removes that incorrectly paired base. Then the polymerase gets another chance to add the right one. That must cost energy. It does. It's energetically costly for the cell, using up extra energy bonds, but it dramatically boosts accuracy, maybe 100 to a thousandfold. Maintaining perfect DNA copies is just that important. What's fascinating is that even a relatively simple organism like E.coli doesn't just have one DNA polymerase. It has several, right? Each with distinct roles like different tools in a toolkit. Indeed, it it's not a one-sizefits-all situation. DNA polymerase or pole one was the first one discovered by Kornberg. It's abundant and it does have a unique ability, a fivep prime to threep prime x and nucleus activity that's useful for removing RNA primers and filling in gaps later. But it's not the main replicator. No, it's actually too slow and its processivity is too low. It falls off the DNA too easily to copy the whole chromosome efficiently. That central heavy lifting role belongs to DNA polymerase the third or pole third. And pole third is different. How pull is the real speed demon. It's much much faster and incredibly processive. It can add hundreds of thousands even millions of nucleotides without dissociating from the DNA template. Wow. How does it manage that? It's because pole third is actually a huge complex hollow enzyme. Think of it less like a single tool and more like a sophisticated molecular machine made of many different protein subunits. It has the core polymerase parts, the proofreading exonucleus, but also other crucial components. One key part is a donut-shaped protein complex called the sliding clamp. a clamp. Yeah, the beta subunit in E.coli. It encircles the DNA like a ring and tethers the pullth through core enzyme to the template. This prevents the polymerase from falling off, which is what gives it that amazing processivity. There's also a clamp loader complex that uses ATP energy to open and close this clamp around the DNA at the right time. That makes perfect sense. So DNA replication isn't just a few isolated enzymes bumping around. It sounds like a precisely coordinated assembly line. A whole factory of molecular machines. Absolutely. That's a great analogy. The entire collection of proteins working together at the replication fork is called the reploum. The reposum. In E.coli, it's a massive team effort involving 20 or more different proteins beyond the DNA polymerasees we've mentioned. What else is in there? You have enzymes called helicases like DNAB and E.coli. Their job is to unwind the DNA double helix ahead of the polymerase like unzipping a zipper. And they use the energy currency of the cell ATP to do this. Unzipping must cause tangles, right? It's like twisting a rope. It does. That unwinding introduces torsional stress or super coiling into the DNA ahead of the fork. So you need to poison races like DNA gyra and E.coli which act like swivel points. They cut the DNA, let it unwind and then reseal it to relieve that stress. Clever. What else? Once the strands are separated, they're vulnerable. So singlestranded DNA binding proteins or SSB coat the exposed single strands to keep them from snapping back together or getting damaged, protecting the template. Exactly. And remember that primer requirement. An enzyme called primus dag in E.coli synthesizes those short RNA primers that DNA pulymerase needs to get started especially on the lagging strand. Okay. So polymerase extends, helicase unwinds, deposase relieves stress, SSB protects, primus starts. Anything else? Yes. Finally, after the RNA primers are removed and the gaps filled with DNA, there are still little nicks or breaks in the sugar phosphate backbone. DNA legacies are the molecular glue. They use energy often from ATP or NAD plus a day to form the final phosphodister bond and seal those nicks creating a continuous intact DNA strand. It's a beautifully orchestrated dance of molecules. Truly incredible coordination. It really is. Let's trace the actual steps of replication in E.coli then. Initiation, elongation, termination. How does this intricate process actually unfold from start to finish? Okay, let's break it down. Initiation is all about getting started at the right place in time. It begins at that specific origin site or C in E.coli, the starting line, right? Key initiator proteins like DMA and E.coli recognize and bind to specific sequences within orus. This binding fueled by ATP causes the DNA to wrap around the proteins and forces the strands to separate in a nearby region that's rich in ATS pairs which are easier to pull apart. The replication bubble. Exactly. Once that bubble is open, other proteins load the main helicase DNA B onto each of the separated strands. The helicase then starts unwinding the DNA in both directions, creating those two replication forks. SSB proteins jump on the single strands. Gyra gets busy relieving the super coiling ahead and the whole process is very tightly regulated. Regulated how? To make sure replication only happens once per cell cycle. Cells don't want to accidentally copy their DNA multiple times. Mechanisms involving protein binding, ATP hydrarolysis, and even chemical modifications like DNA methylation ensure it's a onceanddone event each generation. Okay, so initiation sets the stage. Then comes elongation, the main copying part. Right. This is where the bulk of the DNA synthesis happens. At the replication forks, the DNA telase continues to power ahead unwinding the DNA. On the leading strand, DNA pull third clamped on just follows right behind the helicase synthesizing DNA continuously. Smooth sailing there pretty much. But the lagging strand is the tricky one because it runs in the opposite direction. It has to be synthesized backwards relative to the fork movement. The Okazaki fragments again. Yep. Primus periodically hops onto the lagging strand template associated with the helicase and lays down a short RNA primer. Then a pulththeric complex clamps on and synthesizes an Okazaki fragment extending from that primer until it hits the primer of the previous fragment. How does a cell coordinate both strands being made at the same fork? That's where the trombone model comes in. It's thought that the lagging strand template is looped out so that the pole thread enzyme working on it can physically stay associated with the pole thread working on the leading strand even though they're moving on templates oriented in opposite directions. It keeps the whole reposum together as a single coordinated unit like a trombone slide adjusting the loop length. Exactly. As one Okazaki fragment finishes, the polymerase releases the loop. The clamp later puts a new clamp near the newly made primer. The pymerase re-engages and a new loop forms for the next fragment. It's quite elegant. And then clean up. Then clean up. The RNA primers have to go. In E.coli, DNA poly often does this using its 5 to 3 minute exonucleate activity, removing the RNA ahead of it while simultaneously filling the gap with DNA or another enzyme RNA's H can remove most of the RNA. Then PI fills any remaining gap and finally DNA liase seals the nick between the fragments. Okay. Initiation elongation leads to termination. How does it stop? In E.coli's circular chromosome, the two replication forks travel around the circle until they meet roughly opposite the origin. There's a specific terminus region containing special DNA sequences called tersites. Roadblocks sort of. A protein called tus binds to these ters. The tester complex acts like a one-way gate. It lets a replication fork pass through in one direction but blocks it if it approaches from the other direction. This ensures the two forks meet within the defined terminus region. And then the last few hundred base pairs are replicated and you're left with two complete newly synthesized circular DNA molecules, but they're often interlin like two rings of a chain. They're catenated, tangled up. Yep. So a final enzyme to poisonase the thief in E.oli, which is a type two to poisonase, makes a transient double strand break in one circle, passes the other circle through the break, and then receals it. This decatenates the chromosomes, separating them so they can be segregated into the two daughter cells during cell division. Phew, that's quite the process in bacteria. So if that's the bacterial story, what does this all mean for ukarotic cells like our own? How do these fundamental principles scale up? Our genomes are huge, right? Ukarotic replication shares many of the fundamental principles. Semiconservative origins, forks, leading lighting strands, polymerases, helicases, liguses. The basic chemistry is the same, but yes, the complexity ramps up significantly because of size and structure. Exactly. Our DNA molecules are vastly larger linear chromosomes, not circles, and they're packaged tightly into chromatin with histone proteins. This presents extra challenges. So, multiple starting points. Absolutely crucial. Instead of a single origin, human chromosomes have tens of thousands of origins of replication. Maybe 30,000 to 50,000 across the entire genome. Why so many? Because ukareotic replication forks move much slower than bacterial ones, only about 50 nucleotides per second compared to maybe 1,000 in E.coli. If we only had one origin per chromosome, it would take weeks or months to copy our DNA. So multiple origins are essential to replicate the entire genome within the Sphase of the cell cycle, which takes several hours. And controlling all those origins must be complex extremely. There's a very intricate licensing system involving proteins like cyclines and cycl dependent kinases CDKs. This ensures that each origin is activated or fires only once per cell cycle. It prevents disastrous re-replication. The key helicase the MCM complex is loaded onto origins early on but only activated later when the cell is ready to divide. Different tools to different polymerases. Yes, ukarotes have a larger cast of DNA polymerases for nuclear replication. The main workh horses are pole epsilon thought to synthesize the leading strand and pole delta for the lagging strand. Both of these have proofreading activity. There's also pole alpha which works with primus to make the RNA primer and then adds a short stretch of DNA but it lacks proof reading. Do we have a sliding clamp? We do. It's called PCNA proliferating cell nuclear antigen. It serves the same function as E.coli's Ki's beta clamp encircling the DNA and tethering the polymerases pole delta and pole epsilon to enhance their processivity looks structurally different but does the same job. So similar principles but scaled up and with more layers of regulation and specialized players. That's a good summary. More origins, slower forks, complex licensing, different polymerases, but the core logic remains. And on a more personal level, what's a fascinating application of this knowledge that directly impacts human health? You mentioned viruses earlier, right? Many DNA viruses like herpes simplex virus, which causes cold sores and genital herpes, are quite self-sufficient. They encode their own DNA polymerases to replicate their viral genomes inside our cells, making them a target. Exactly. Because the viral polymerase is different from our human polymerases, it makes an excellent drug target. A cycllover is a classic example. How does it work? It's clever. A cycllover is a modified version of a DNA building block. A viral enzyme, one that our cells don't have, adds the first phosphate group to a cycllover, activating it. Our own cellular enzymes then add more phosphates. This activated form inhibits the viral DNA polymerase much more strongly than it inhibits our own polymerases. So, it's selective, highly selective. And there's more. Because of its chemical structure, a cycllover lacks the proper threep prime hydroxal group needed to add the next nucleotide. So once it gets incorporated into the growing viral DNA chain, it stops everything. It acts as a chain terminator. It halts further elongation of that viral DNA strand. So a cycllover hits viral replication at multiple steps. Selective activation and chain termination. Very effective. Incredible. Okay, let's shift gears. DNA might be the blueprint, but as you said, it's constantly under attack. Thousands of tiny lesions, damages can accumulate daily in just one of our cells. How on earth does life manage to maintain genomic integrity against such a constant barrage? It's truly astonishing, isn't it? Despite this relentless assault from things like UV radiation, chemical mutagens, even just spontaneous chemical reactions within the cell, fewer than one in 1,000 of those initial DNA lesions actually become a permanent mutation. One in a thousand hugs to an elaborate, multi-layered, incredibly efficient set of DNA repair systems. These systems are constantly scanning the DNA, finding damage and fixing it. And their importance is huge, paramount. The link between unrepaired DNA damage, the accumulation of mutations, and the development of cancer is incredibly strong. Think about the Ames test, which measures a chemical's potential to cause mutations. Over 90% of known carcinogens test positive as mutagens. Wow. And genetic defects and DNA repair genes themselves are often devastating. conditions like zero dermma pigmentotosum or inherited mutations in BRCA1 or BRCA2 genes linked to breast and ovarian cancers. These dramatically increase cancer susceptibility because the cell's ability to fix DNA damage is compromised. It really shows the immense energy the cell invests in repair. Maintaining genetic integrity is a top priority. So what are some of these critical repair systems? Let's start with correcting those rare errors left behind after replication. The ones that even proofreading missed, right? Right. The mismatches that's handled by mismatch repair or MMR. This system acts like a final quality control check after replication, boosting the overall fidelity by another 100 to 1,000fold. How does it know which strand is wrong? The new one or the old template? That's the clever part, especially in E.coli. For a short time after replication, the newly synthesized strand is chemically distinct from the older template strand. The template strand has specific methylation tags added to certain DNA sequences, GATC sequences. The new strand hasn't been methylated yet, so it can tell them apart. Exactly. The MMR system involving proteins like MUTZ, MUTL, and MUT in E.coli recognizes the mismatch and specifically targets the unmethylated new strand for repair. MUT makes a cut in the new strand near a hemmethylated GATC site. Then an exonucleus chews away a segment of that new strand, including the mismatch. And then then DNA pole 3 comes back in to fill the gap correctly using the methylated template as a guide and DNA lius seals the final nick. It can be quite costly sometimes removing over a thousand base pairs just to fix one mismatch but accuracy demands it and in humans ukarotic cells have very similar mutess and miel homologues showing the system is highly conserved. Defects in these cause hereditary non-polyposis colon cancer H&PCC or Lynch syndrome. However, the exact mechanism for identifying the new strand in ukarotes is still being worked out. It doesn't seem to rely on the same GATC methylation system as E.oli. Okay. So, MMR fixes replication errors. What about more common types of damage that happen spontaneously like a base changing its chemical identity? Ah, yes. Like cytosine demination. Cytosine can spontaneously lose an amino group and turn into uricil, a base normally found only in RNA. This is a very common lesion. And the cell fixes it how? With base excision repair or BER. This pathway deals with damage to single bases. The first step is a specialized enzyme called a DNA glycosillase. There are different glycosillase specific for different types of damaged bases like a uricil glycos. Exactly. U glycosilase finds uricil in DNA recognizes it shouldn't be there and cleaves the bond linking the uricil base to the DNA sugar phosphate backbone. Simply removing the base itself, leaving a hole. Yes. It creates what's called an APSite opirinic or a pyramid. a sugar with no base attached. This AP site is then recognized by another enzyme, an AP endonucleus, which cuts the DNA backbone next to the site. Okay. Then a short patch of DNA around the cut is removed. DNA polymerase often pull I and E. coli or specialized polymerases like pole beta and ukarotes fills in the correct nucleotide using the opposite strand as a template and DNA liga seals the final nick. And that's why DNA uses thymine T instead of uricil U. Precisely. If DNA naturally contained uricil, the cell wouldn't be able to distinguish a correct uricil from one that arose from cytosine damination. By using symine, which is essentially methylated uricil, any uricil found in DNA is immediately recognized as damage arrived from cytosine and targeted for removal by beer. Very smart design. It really is. Now, what about bigger problems? Those bulky distortions in the DNA helix may be caused by UV light creating perramitine dimers or chemical adductions from things like cigarette smoke. Those are handled by a different more versatile system called nucleotide excision repair or NE. This system is absolutely vital for survival especially for damage caused by environmental mutagens. How does NE work? It recognizes distortions in the DNA helix rather than specific damaged bases. A large multi-UN enzyme complex called an exucleus is recruited. This complex makes two cuts in the damaged DNA strand, one on each side of the bulky lesion, cutting out a whole chunk. Exactly. In E.coli, the ABCucleus removes a fragment that's typically 12 or 13 nucleotides long. In humans, the NER machinery removes a larger chunk, usually 27 to 29 nucleotides long containing the damage and then fill and seal. Yep. The resulting gap is filled by a DNA polymerase, poli and e.coli, coli pole delta or epsilon in humans using the undamaged strand as a template and the remaining nick is sealed by DNA legus and defects here cause genetic defects in human ne genes are the cause of zerodermopigmentotosum XP individuals with XP are extremely sensitive to sunlight because they can't repair the pyramidine DRS caused by UV radiation they have a dramatically increased risk of skin cancer it highlights just how critical NE is for protecting us from environmental DNA damage Are there any repair mechanisms that are more direct that don't involve cutting out and reynthesizing a whole segment? That seems like a more efficient approach if possible. Yes, there are a few fascinating examples of direct repair where the enzyme directly reverses the damage in a single step. Like what? One example is enzymes called DNA photos. These enzymes can recognize UV induced pyramidine DR, bind to them, and then use the energy from visible light to directly break the abnormal bonds forming the dimer, restoring the original pyramidines. It's direct chemical reversal. But we don't have those. Strangely, no. Photosises are found in bacteria, fungi, plants, and many animals. But they seem to have been lost during the evolution of placental mammals, including humans. We rely entirely on ne to fix UV damage. Interesting. Any other direct repair examples? Another striking one is the repair of a highly mutagenic lean called 06 methyl guanine. This happens when guanine gets inappropriately methylated. 06 methyl granine tends to pair with thymine instead of cytosine during replication leading to mutations. How is it fixed? There is a protein called 06 methyl guanine DNA methyl transferase. It finds the 06 methyl guanine, binds to it, and directly transfers the harmful methyl group from the guanine base onto one of its own cyine amino acid residues within the protein. So, the protein takes the hit. Exactly. And here's the truly remarkable part. This transfer permanently inactivates the methyl transferase protein. It's a suicide mission. The protein sacrifices itself to repair just one lesion. Wow, that really underscores the priority. It absolutely does. that illustrates the immense cellular priority given to maintaining DNA integrity. The cell is willing to synthesize an entire protein just to fix one specific dangerous lesion and then the protein is gone. Incredible. What happens when the damage is really severe, so extensive that maybe the complimentary strand is also compromised or there's a break and the repair systems we've talked about can't use a template. How does the cell recover them? That's when things get really tricky and the cell might have to resort to what's often called a desperation strategy. This involves errorprone transion DNA synthesis or TLS. Transion meaning across the lesion. Precisely. Imagine a replication fork moving along the DNA and it suddenly encounters an unrepaired lesion like a pyramidine dimer or a bulky aduct that the main replicative polymerase like pull the third or pull delta epsilon simply cannot read past it stalls the fork which is bad very bad a stall replication fork can collapse and lead to chromosome breaks and cell death so under these conditions especially if the damage is widespread triggering what's called the SOS response in bacteria the cell can bring in specializ ized TLS DNA polymerases. The specialized how these TLS polymerases like pulv in ecol or petta iota kappa zeta in humans have much more open flexible active sites. They're essentially designed to tolerate weird shapes in the DNA template. They can synthesize DNA past the lesion effectively bypassing it and allowing replication to continue but at a cost. Yes, a significant cost. Fidelity because they're less picky and often lack proofreading activity. TLS polymerases frequently insert the wrong nucleotide opposite the lesion or even opposite undamaged bases nearby. They introduce mutations. So, it fixes the stall but causes mutations. Seems counterintuitive. It's a trade-off. From the cell's perspective, getting replication completed, even with a few errors, might be preferable to letting the fort collapse entirely, which could be lethal. It's a gamble allowing some cells in a population to survive an otherwise insurmountable barrier, even if it means introducing genetic variation. Some argue this provides fuel for evolution generating diversity under stress than necessary evil perhaps. In some situations, yes. Though it's worth noting that some TLS polymerases are better than others. For instance, human petta is actually quite good at inserting the correct adenine opposite UV induced thymine DR minimizing mutations from that specific leion. So, it's nuanced. Okay. So far, we've covered copying, DNA replication, fixing DNA repair. But DNA isn't always static. It also under goes incredible rearrangements, a process called genetic recombination. Barbara Mcccleintoch famously discovered jumping genes, transposons in maze way back in the 1940s, showing us just how dynamic genomes can be. Absolutely. Genetic re combination is the third major aspect of DNA metabolism involving the exchange or rearrangement of genetic information either between DNA molecules or within the same molecule. It broadly falls into three main classes. Homologous genetic recombination which involves exchange between DNA molecules with extensive sequence similarity. Then there's sightspecific recombination which occurs only at specific defined DNA sequences. And finally, DNA transposition, which involves those jumping genes or transposable elements moving to new locations. And these all have different jobs. They do. Recombination plays crucial roles in specialized DNA repair pathways, helps rescue stalled replication forks, can regulate gene expression, is vital for generating genetic diversity during sexual reproduction, and is even involved in programmed rearrangements during development like in our immune system. Let's tackle homologist recombination first. You mentioned repair. How does it work in bacteria? In bacteria, homologous recombination is primarily a really important DNA repair process, often called recombinational DNA repair. Its main job is to help reconstruct replication forks that have stalled or collapsed, often at sites of DNA damage like nicks or gaps. How does it fix a broken fork? When a fork collapses, it can often lead to a dangerous double strand break. The cell needs to fix this break and restart replication. Homologous recombination provides a way. Enzymes first process the broken DNA end, typically chewing back the fivep prime ending strand to create a threep prime singlestranded tail. This threep prime tail is then coated by a key bacterial recombinase protein called riia. Ray forms a helical filament on this singlestranded DNA. This reay filament is amazing. It can search the entire genome for a matching homologous sequence on an intact DNA molecule like the sister chromosome. Finds the backup copy. Exactly. Once it finds homology, the RK filament promotes strand invasion where the single stranded tail displaces one strand of the intact duplex and base pairs with the other. This creates a branched DNA structure. Sort of a crossover point. Sort of. It sets the stage. This branch point can then move along the DNA branch migration potentially forming a cross-shaped structure called a holiday intermediate or holiday junction. Specialized enzymes then cleave this junction in specific ways resolution and DNA legus seals any remaining nicks. The end result is that the broken DNA is repaired using the intact molecule as a template. The replication fork is effectively reconstructed and DNA synthesis can restart. So in bacteria homologous re combination is fundamentally intertwined with replication rescue and repair primarily yes it's a vital survival mechanism but in ukarotes like us it plays a critical role in a very different but equally fundamental process right meiosis indeed in ukarotes homologous recombination is absolutely essential for the precisely without these kaismada formed by recombination homologous chromosomes often fail to segregate properly leading into errors. Plus, this crossing over has another huge consequence. It shuffles the genetic alals between the maternal and paternal chromosomes, significantly increasing the genetic diversity in the resulting gametes. And if this process fails, the consequences can be severe. Failure of miotic recombination or improper chromosome segregation is a leading cause of an employee gametes and resulting embryos having the wrong number of chromosomes. Tricome 21 causing Down syndrome is a well-known example. Anipoloy is a major cause of pregnancy loss and developmental disabilities in humans. And this relates to maternal age. Yes. The increased risk of an employee with advancing maternal age is thought to be linked at least in part to the incredibly long time human eggs remain arrested in meiosis, the dictiat stage. Those kaismata formed early in fetal development have to remain intact for decades, potentially becoming less stable over time, increasing the chance of segregation errors later in life. Fascinating. You mentioned proteins detecting DNA damage earlier. Thinking about double strand breaks, how does the cell first sense that kind of serious damage in our complex chromosomes? That's a great question. One of the key first responders to DNA breaks, especially single strand breaks, but also involved near double strand breaks is a protein called PRP1. That stands for polyadpibos pulmerase 1. What does PAP1 do? PRP1 is constantly scanning the DNA. When it encounters a break, it binds tightly to the broken ends. Upon binding, it gets activated and starts synthesizing long branched chains of a molecule called polyadp ribos. Using NAD+ as a substrate, it attaches these polymers to itself and other nearby proteins, creating a signal flare. Exactly. This burst of polyadpose polymer acts like a molecular flag or signal flare at the site of damage. It's negatively charged and bulky and it serves to recruit a whole host of other DNA repair factors including proteins involved in both base excision repair and double strand break repair pathways to that specific location. It essentially shouts damage here needs fixing and this has therapeutic relevance huge relevance. Now PP inhibitors are a major class of cancer drugs. The logic is this. Many cancers, particularly those with mutations in the BRCA1 or BRCA2 genes, already have defects in their primary pathway for repairing double strand breaks, which relies on homologous recombination. These cancer cells become heavily dependent on other repair pathways, including those initiated by PRP-1, just to survive. So you target their backup plan. Precisely. By inhibiting PRP1, you take away that backup. The BRCA deficient cancer cells accumulate too much DNA damage, particularly double strand breaks during replication, and they die. Normal cells, which still have functional BRCA pathways, are much less affected. It's a great example of exploiting a cancer's specific vulnerabilities, synthetic lethality. Okay, so homologus recombination is great for repair if you have a template. But what if a double strand break occurs and there's no homologous chromosome nearby to use as a template, say outside of the S or G2 phases of the cell cycle? That's where the other major double strand break repair pathway comes in. Non-homologous N joining or NHEJ. This is actually the predominant pathway for fixing double strand breaks in mamalian sematic cells throughout most of the cell cycle. Non-homologous meaning it doesn't need a template. Correct. Instead of searching for a homologous sequence, NHJ simply takes the two broken ends of the chromosome and essentially sticks them back together. How does it do that? It involves a set of core proteins. First, a protein complex called coup7280 acts like a cap, quickly recognizing and binding to the broken DNA ends, protecting them from degradation. Coup then recruits other factors, including a large protein kynise called DNA PKCs. These proteins help bring the two ends together. Often the broken ends aren't clean. They might have damaged bases or overhangs. So, various nucleuses like Artemis and specialized DNA polymerises process the ends, trimming them or filling small gaps. Finally, a dedicated DNA legus complex legus store with XRCC4 joins the processed ends back together. But if you're trimming or adding bases randomly, Exactly. That's the catch. NHJ is considered an errorprone or mutagenic repair process. Because it doesn't use a template, the processing steps often result in small insertions or deletions of nucleotides induls at the site of the break. The original sequence isn't perfectly restored, but it fixes the break. It fixes the break, preventing catastrophic loss of chromosome fragments or cell death. For sematic cells in a large organism, losing a few base pairs might be an acceptable price to pay to maintain overall chromosome integrity. It highlights the cell's absolute need to fix double strand breaks even if the fix isn't perfect. Okay, let's move to the second type of recombination, sight specific recombination. How is this different? You said it happens only at specific sites, right? Unlike homologist recombination which can happen anywhere there's sufficient sequence similarity sight specific recombination is strictly limited to particular usually short DNA sequences the recombination sites and it uses different enzymes yes it relies on specific enzymes called recombinases these recombinases recognize their target DNA sequences bind to them and then catalyze the cutting and rejoining of the DNA strands they essentially act as both sequence specific endonucleuses and leguses all rolled into one. What kinds of rearrangements can it do? Depending on the location and orientation of the recombination sites, sight specific recombination can lead to different outcomes. If the sites are on the same DNA molecule and oriented inversely, recombination causes an inversion of the DNA segment between them. If they're oriented directly, it leads to a deletion of the segment as a circle. And if the sites are on two different DNA molecules, it can result in an insertion or integration of one molecule into the other. An example, a great bacterial example is the XCD system in E.coli. Sometimes due to errors in replication or repair, the circular bacterial chromosome can end up as a dimer, two circles linked together. This prevents proper cell division. The XCD recombinase acts at a specific site called diff present on the chromosome. Sight specific recombination at diff resolves the dimer back into two separate monomer circles, allowing the cell to divide successfully. It's essential for chromosome segregation. Very precise control. Finally, the third type, the jumping genes, transposons, DNA transposition. Yes, transposable elements or transposins. These are fascinating segments of DNA that possess the remarkable ability to jump or move from one location in the genome to another. This process is called transposition. Yeah, it's random. The movement itself doesn't usually require sequenceology between the transposent and its new insertion site. and the insertion site is often, though not always, relatively random. Because of this potential to disrupt genes or genome structure, transposition is generally a rare event and is often tightly regulated by the cell or the transpose itself. What do transposons contain? The simplest ones in bacteria called insertion sequences or is elements typically contain only the gene encoding the enzyme needed for transposition. the transposes flanked by short inverted repeat sequences that the transposes recognizes. More complex transposins carry additional genes besides the transpos such as genes conferring antibiotic resistance. Ah so that's how antibiotic resistance can spread. It's a major mechanism. When a complex transposing carrying an antibiotic resistance gene jumps onto a plasmid, a small circular DNA molecule that can be transferred between bacteria, it allows that resistance to spread rapidly through a bacterial population via plasma transfer. It's a huge driver of the antibiotic resistance crisis. How does the jumping actually work? Are there different ways? There are two main pathways. In direct transposition, sometimes called simple or cut and paste transposition, the transposase enzyme cuts the transposent completely out of its original location, often leaving behind a double strand break that needs repair and then inserts it into a new target site. So it moves, doesn't copy itself. Correct. The second mechanism is replicative transposition. In this case, the entire transposon is replicated during the transposition process. One copy remains at the original site and a new copy is inserted at the target site. So the number of transposons increases and when they insert they leave a footprint. Yes. A characteristic feature of most transposition events is the duplication of a short sequence typically 3 to 12 base pairs at the target site. This short target site duplication ends up flanking the inserted transposent element on both sides. Finding these flanking direct repeats is often a clue that a particular DNA segment is a transpos. Now, in vertebrates, you mentioned some DNA rearrangements are actually programmed essential parts of development. The immune system example. This is truly one of the most stunning examples of programmed recombination specifically related to sight specific recombination and potentially evolved from transposition. It's how our immune system achieves its incredible diversity. Diversity to fight off infections. Exactly. We need to be able to produce millions, maybe billions of different antibodies, immunogloabbulins and T- cell receptors to recognize and fight off the vast array of potential pathogens we might encounter. But we certainly don't have millions of separate genes encoding each one. So how do we do it? Through gene rearrangement. The genes encoding the variable regions of antibodies and T- cell receptors are assembled during the development of immune cells, B cells and T cells from separate gene segments. For antibbody light chains, for example, there are multiple variable V segments and joining J segments stored in the genome and they get pieced together precisely. During B cell development, specific enzymes recognize special recombination signal sequences, RSS that flank these V andJ segments. Proteins called ARG1 and ARG2 recombination activating gene products act like a sightspecific recombinase. They make precise double strand breaks at the RSS's next to one chosen V segment and one chosen J segment. Then the cell's general DNA repair machinery, particularly the NHJ pathway we discussed earlier, joins the ends of the chosen V andJ segments together. The DNA between them is deleted. This creates a unique functional VJ exxon that encodes part of the antibodies antigen binding site. Since there are many V&J segments to choose from and the joining process itself can be imprecise adding more variation. This combinatorial process generates a huge diversity of antibbody genes from a limited set of starting parts. And this looks like transposition. The RG enzymes and the RSS sequences bear striking similarities to transposes and the ends of certain transposons. The mechanism of cutting and rejoining is also very similar. This has led to the fascinating hypothesis that the entire VDJ recombination system, the foundation of our adaptive immune systems diversity, might have actually evolved from an ancient invasion of a vertebrate ancestors genome by a transposable element which was then harnessed and repurposed for this vital imunological function. Wow. Evolution repurposing molecular tools in amazing ways. Truly remarkable. What a journey. We've gone from the precise almost magical copying of our genetic code in replication. Yeah, the fidelity is amazing. to the vigilant repair systems that tirelessly guard against damage, constantly patching things up. And finally, to the deliberate, sometimes programmed rearrangement of DNA through recombination. It's abundantly clear that the idea of DNA is just stable storage is well, far too simple. Absolutely. It's anything but static. It's a continuous dynamic dance of incredibly sophisticated molecular machinery. This interplay underpins life itself. This constant interaction between replication, repair, and recombination, often using shared or repurposed molecular parts like NHJ and VDJ recombination or recombination proteins, helping restart replication forks. It really highlights how evolution builds complexity, doesn't it? Tinkering with existing tools. It absolutely does. It's a masterclass in molecular repurposing and integration. And it definitely raises an important, maybe profound question. What other fundamental biological processes are underpinned by such ancient repurposed molecular machinery? Machinery whose origins and full connections we are yet to fully appreciate. That is a great thought to ponder. What else is hiding in plain sight built from old parts? Thank you so much for taking us on this deep dive into the dynamic world of DNA metabolism. My pleasure. It's a fascinating subject. We hope you out there feel a little more well informed and perhaps like me, a lot more amazed by the intricate, relentless activity happening inside every single one of your cells right now. Keep exploring. Keep questioning. Exactly. Keep exploring. Keep questioning. And we'll see you next time on the deep dive. |
189291 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems?srsltid=AfmBOoo4hgwmUaCpHT7OKz_DXFuvBFpEcKd63bXC3JhhYOGp5drAoGQ4 | Art of Problem Solving
2022 AMC 12B Problems - AoPS Wiki
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical CampusVisit the Virtual Campus
Sign In
Register
online school
Class ScheduleRecommendationsOlympiad CoursesFree Sessions
books tore
AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates
community
ForumsContestsSearchHelp
resources
math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten
contests on aopsPractice Math ContestsUSABO
newsAoPS BlogWebinars
view all 0
Sign In
Register
AoPS Wiki
ResourcesAops Wiki 2022 AMC 12B Problems
Page
ArticleDiscussionView sourceHistory
Toolbox
Recent changesRandom pageHelpWhat links hereSpecial pages
Search
GET READY FOR THE AMC 12 WITH AoPS
Learn with outstanding instructors and top-scoring students from around the world in our AMC 12 Problem Series online course.
CHECK SCHEDULE
2022 AMC 12B Problems
2022 AMC 12B (Answer Key)
Printable versions: Wiki • AoPS Resources • PDF
Instructions
This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
Figures are not necessarily drawn to scale.
You will have 75 minutes working time to complete the test.
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
Contents
1 Problem 1
2 Problem 2
3 Problem 3
4 Problem 4
5 Problem 5
6 Problem 6
7 Problem 7
8 Problem 8
9 Problem 9
10 Problem 10
11 Problem 11
12 Problem 12
13 Problem 13
14 Problem 14
15 Problem 15
16 Problem 16
17 Problem 17
18 Problem 18
19 Problem 19
20 Problem 20
21 Problem 21
22 Problem 22
23 Problem 23
24 Problem 24
25 Problem 25
26 See also
Problem 1
Define to be for all real numbers and What is the value of
Solution
Problem 2
In rhombus , point lies on segment so that , , and . What is the area of ? (Note: The figure is not drawn to scale.)
Solution
Problem 3
How many of the first ten numbers of the sequence are prime numbers?
Solution
Problem 4
For how many values of the constant will the polynomial have two distinct integer roots?
Solution
Problem 5
The point is rotated counterclockwise about the point . What are the coordinates of its new position?
Solution
Problem 6
Consider the following sets of elements each: How many of these sets contain exactly two multiples of ?
Solution
Problem 7
Camila writes down five positive integers. The unique mode of these integers is greater than their median, and the median is greater than their arithmetic mean. What is the least possible value for the mode?
Solution
Problem 8
What is the graph of in the coordinate plane?
Solution
Problem 9
The sequence is a strictly increasing arithmetic sequence of positive integers such that What is the minimum possible value of ?
Solution
Problem 10
Regular hexagon has side length . Let be the midpoint of , and let be the midpoint of . What is the perimeter of ?
Solution
Problem 11
Let , where . What is ?
Solution
Problem 12
Kayla rolls four fair -sided dice. What is the probability that at least one of the numbers Kayla rolls is greater than and at least two of the numbers she rolls are greater than ?
Solution
Problem 13
The diagram below shows a rectangle with side lengths and and a square with side length . Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?
Solution
Problem 14
The graph of intersects the -axis at points and and the -axis at point . What is ?
Solution
Problem 15
One of the following numbers is not divisible by any prime number less than Which is it?
Solution
Problem 16
Suppose and are positive real numbers such that What is the greatest possible value of ?
Solution
Problem 17
How many arrays whose entries are s and s are there such that the row sums (the sum of the entries in each row) are and in some order, and the column sums (the sum of the entries in each column) are also and in some order? For example, the array satisfies the condition.
Solution
Problem 18
Each square in a grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
Any filled square with two or three filled neighbors remains filled.
Any empty square with exactly three filled neighbors becomes a filled square.
All other squares remain empty or become empty.
A sample transformation is shown in the figure below. Suppose the grid has a border of empty squares surrounding a subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)
Solution
Problem 19
In medians and intersect at and is equilateral. Then can be written as , where and are relatively prime positive integers and is a positive integer not divisible by the square of any prime. What is
Solution
Problem 20
Let be a polynomial with rational coefficients such that when is divided by the polynomial , the remainder is , and when is divided by the polynomial , the remainder is . There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?
Solution
Problem 21
Let be the set of circles in the coordinate plane that are tangent to each of the three circles with equations , , and . What is the sum of the areas of all circles in ?
Solution
Problem 22
Ant Amelia starts on the number line at and crawls in the following manner. For Amelia chooses a time duration and an increment independently and uniformly at random from the interval During the th step of the process, Amelia moves units in the positive direction, using up minutes. If the total elapsed time has exceeded minute during the th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most steps in all. What is the probability that Amelia’s position when she stops will be greater than ?
Solution
Problem 23
Let be a sequence of numbers, where each is either or . For each positive integer , define Suppose for all . What is the value of the sum
Solution
Problem 24
The figure below depicts a regular -gon inscribed in a unit circle. What is the sum of the th powers of the lengths of all of its edges and diagonals?
Solution
Problem 25
Four regular hexagons surround a square with a side length , each one sharing an edge with the square, as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be written as , where , , and are integers and is not divisible by the square of any prime. What is ?
Solution
See also
2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
2022 AMC 12A ProblemsFollowed by
2023 AMC 12A Problems
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 12 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
Retrieved from "
Category:
AMC 12 Problems
Art of Problem Solving is an
ACS WASC Accredited School
aops programs
AoPS Online
Beast Academy
AoPS Academy
About
About AoPS
Our Team
Our History
Jobs
AoPS Blog
Site Info
Terms
Privacy
Contact Us
follow us
Subscribe for news and updates
© 2025 AoPS Incorporated
© 2025 Art of Problem Solving
About Us•Contact Us•Terms•Privacy
Copyright © 2025 Art of Problem Solving
Something appears to not have loaded correctly.
Click to refresh. |
189292 | https://math.stackexchange.com/questions/1506706/how-to-calculate-the-distance-between-two-points-with-polar-coordinates | calculus - How to calculate the distance between two points with polar coordinates? - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
How to calculate the distance between two points with polar coordinates?
Ask Question
Asked 9 years, 11 months ago
Modified1 year, 3 months ago
Viewed 31k times
This question shows research effort; it is useful and clear
3
Save this question.
Show activity on this post.
I want to calculate the distance between p 1 p 1 and p 2 p 2 as shown in the image. These points are in a polar coordinate space. The red arrows in the image show in which direction ρ ρ and θ θ "grow".
p 1 p 1 = (ρ 1,θ 1)(ρ 1,θ 1) and p 2 p 2 = (ρ 2,θ 2)(ρ 2,θ 2).
In both cases, θ∈[0,180 º]θ∈[0,180 º] and ρ∈[0,ρ m a x]ρ∈[0,ρ m a x].
ρ m a x ρ m a x is the size of the diagonal of the black square.
I've tried to use the euclidean distance formula for polar coordinates:
d(a,b)=ρ 2 1+ρ 2 1−2 ρ 1 ρ 2 cos(θ 1−θ 2)−−−−−−−−−−−−−−−−−−−−−−√d(a,b)=ρ 1 2+ρ 1 2−2 ρ 1 ρ 2 cos(θ 1−θ 2).
However, the value calculated does not seem to be correct.
I've tried to calculate this distance using this formula:
d(a,b)=(ρ 1−ρ 2)2+sin(θ 1−θ 2)2−−−−−−−−−−−−−−−−−−−−−√d(a,b)=(ρ 1−ρ 2)2+sin(θ 1−θ 2)2
Then it occured to me that I might have to normalize ρ ρ, so it can only take values between zero and one (just like the sin sin). Thus, both coordinates have the same weight.
Is this a correct way to calculate the distance between these two points?
calculus
coordinate-systems
polar-coordinates
Share
Share a link to this question
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this question to receive notifications
edited Jun 15, 2024 at 4:41
qwr
11.4k 4 4 gold badges 49 49 silver badges 84 84 bronze badges
asked Oct 31, 2015 at 18:43
J. HarrisJ. Harris
133 1 1 gold badge 1 1 silver badge 3 3 bronze badges
4
2 Try converting to Cartesian coordinates and then calculating the distance. That'll tell you if you're getting the right answer or not.user137731 –user137731 2015-10-31 18:46:39 +00:00 Commented Oct 31, 2015 at 18:46
Why didn't d(a,b)=ρ 2 1+ρ 2 1−2 ρ 1 ρ 2 cos(θ 1−θ 2)−−−−−−−−−−−−−−−−−−−−−−√d(a,b)=ρ 1 2+ρ 1 2−2 ρ 1 ρ 2 cos(θ 1−θ 2) work?John Douma –John Douma 2015-10-31 18:50:28 +00:00 Commented Oct 31, 2015 at 18:50
@JohnDouma Using that formula messed up the results of the algorithm I'm implementing. I've checked several times, but always got the same results. That's why I decided to try something different.J. Harris –J. Harris 2015-10-31 22:31:37 +00:00 Commented Oct 31, 2015 at 22:31
Wouldn't it be d(a,b)=ρ 2 1+ρ 2 2−2 ρ 1 ρ 2 cos(θ 1−θ 2)−−−−−−−−−−−−−−−−−−−−−−√d(a,b)=ρ 1 2+ρ 2 2−2 ρ 1 ρ 2 cos(θ 1−θ 2). ?Mark –Mark 2016-09-01 16:39:13 +00:00 Commented Sep 1, 2016 at 16:39
Add a comment|
3 Answers 3
Sorted by: Reset to default
This answer is useful
9
Save this answer.
Show activity on this post.
Consider the cosine law:
c 2=a 2+b 2−2⋅a⋅b⋅c o s(θ)c 2=a 2+b 2−2⋅a⋅b⋅c o s(θ)
which gives the length of 3rd side c c in a triangle with legs a a and b b that are separated by angle θ θ. The legs a a and b b correspond to points (r 1,θ 1)(r 1,θ 1) and (r 2,θ 2)(r 2,θ 2) drawn as vectors from the coordinate origin, with respective lengths r 1 r 1 and r 2 r 2 and separated by angle θ=θ 2−θ 1 θ=θ 2−θ 1. The cosine law then becomes an expression for the distance d d between the points:
d 2=r 1 2+r 2 2−2 r 1 r 2 c o s(θ 2−θ 1)d 2=r 1 2+r 2 2−2 r 1 r 2 c o s(θ 2−θ 1)
Notice that difference θ 1−θ 2 θ 1−θ 2 would give identical results since c o s(θ)=c o s(−θ)c o s(θ)=c o s(−θ). While for triangles the separation angle is always θ<180∘θ<180∘, the symmetry of the cosine function holds for larger angles as well. I would ask the original poster to give more information about the code that uses this correct equation, verify its implementation, and the values of equation inputs.
The polar distance equation can also be derived from the Euclidean distance in Cartesian coordinates:
(x 2−x 1)2+(y 2−y 1)2−−−−−−−−−−−−−−−−−−√(x 2−x 1)2+(y 2−y 1)2
By substituting x=r⋅c o s(θ)x=r⋅c o s(θ) and y=r⋅s i n(θ)y=r⋅s i n(θ) for the coordinates (x 1,y 1)(x 1,y 1) and (x 2,y 2)(x 2,y 2) we get
d 2=[r 2 c o s(θ 2)−r 1 c o s(θ 1)]2+[r 2 s i n(θ 2)−r 1 s i n(θ 1)]2 d 2=[r 2 c o s(θ 2)−r 1 c o s(θ 1)]2+[r 2 s i n(θ 2)−r 1 s i n(θ 1)]2
expand and simplify using c o s 2(θ)+s i n 2(θ)=1 c o s 2(θ)+s i n 2(θ)=1
d 2=r 1 2+r 2 2−2 r 1 r 2[c o s(θ 1)c o s(θ 2)+s i n(θ 1)s i n(θ 2)]d 2=r 1 2+r 2 2−2 r 1 r 2[c o s(θ 1)c o s(θ 2)+s i n(θ 1)s i n(θ 2)]
simplify further using c o s(α−β)=c o s(α)c o s(β)+s i n(α)s i n(β)c o s(α−β)=c o s(α)c o s(β)+s i n(α)s i n(β)
d 2=r 1 2+r 2 2−2 r 1 r 2 c o s(θ 2−θ 1)d 2=r 1 2+r 2 2−2 r 1 r 2 c o s(θ 2−θ 1)
which is the same as before. The original expression inside the cosine term confirms that θ 1 θ 1 and θ 2 θ 2 are interchangeable.
EngrStudent's answer is misleading because d 2=(Δ r)2+(r Δ θ)2 d 2=(Δ r)2+(r Δ θ)2 so the terms Δ r=r 1 2+r 2 2 Δ r=r 1 2+r 2 2 should not cancel out. The result for (r Δ θ)2(r Δ θ)2 is accurate but not the Euclidean distance, unless both points happen to have the same radius as in the example on the unit circle (Δ r=0 Δ r=0). The distance r Δ θ r Δ θ also vanishes when r 1 r 1 or r 2 r 2 is zero, unlike the cosine law.
SchrodingersCat's answer is correct as long as infinitesimal distances are used, but the extension to finite distance is wrong as indicated by the ambiguous definition of r r. Extending the increment d l d l to a finite distance l l requires solving ∫B A d l∫A B d l along a straight line between points A and B, which should give the same result.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
edited Oct 8, 2018 at 20:49
USERID_UNK
3 3 3 bronze badges
answered Jan 11, 2018 at 19:19
Bart Van HoveBart Van Hove
91 1 1 silver badge 2 2 bronze badges
1
1 This should be marked as the answer, both of the ones posted below are wrong.F. Remonato –F. Remonato 2022-02-25 12:21:49 +00:00 Commented Feb 25, 2022 at 12:21
Add a comment|
This answer is useful
5
Save this answer.
Show activity on this post.
In Euclidean geometry, we have where d l d l is the distance, we have
d l 2=d x 2+d y 2 d l 2=d x 2+d y 2
In polar co-ordinates, we have
x=r cos θ x=r cosθ
y=r sin θ y=r sinθ
So
d x=d r cos θ−r sin θ d θ d x=d r cosθ−r sinθ d θ
and
d y=d r sin θ+r cos θ d θ d y=d r sinθ+r cosθ d θ
Put these values in the first equation and we have
d l 2=(d r cos θ−r sin θ d θ)2+(d r sin θ+r cos θ d θ)2 d l 2=(d r cosθ−r sinθ d θ)2+(d r sinθ+r cosθ d θ)2
=(d r)2+(r d θ)2=(d r)2+(r d θ)2
So the distance between 2 points P(r 1,θ 1 r 1,θ 1) and Q(r 2,θ 2 r 2,θ 2) is given by
l=(r 2−r 1)2+r 2(θ 1−θ 2)2−−−−−−−−−−−−−−−−−−−−√l=(r 2−r 1)2+r 2(θ 1−θ 2)2
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Oct 31, 2015 at 18:54
SchrodingersCatSchrodingersCat
24.9k 7 7 gold badges 47 47 silver badges 90 90 bronze badges
5
In this case, don't we have to include a normalization in r 1 r 1 and r 2 r 2? Also, since we're talking about angles and distances, shouldn't we be working in the same ranges?J. Harris –J. Harris 2015-10-31 22:27:56 +00:00 Commented Oct 31, 2015 at 22:27
Can you be a bit more simple? I didn't get what you meant...SchrodingersCat –SchrodingersCat 2015-11-01 14:38:53 +00:00 Commented Nov 1, 2015 at 14:38
Since r r and θ θ have different ranges, isn't it necessary to normalize r r and θ θ so both can only take values between zero and one?J. Harris –J. Harris 2015-11-01 17:58:45 +00:00 Commented Nov 1, 2015 at 17:58
Not necessarily. You will get the values of r r and θ θ from the co-ordinates itself. So I don't think you need any kind of normalization.SchrodingersCat –SchrodingersCat 2015-11-02 12:55:10 +00:00 Commented Nov 2, 2015 at 12:55
18 What is r 2 r 2 in your final formula?Daniel –Daniel 2016-03-23 14:49:49 +00:00 Commented Mar 23, 2016 at 14:49
Add a comment|
This answer is useful
3
Save this answer.
Show activity on this post.
SchrodingersCat made a nice answer, but it has not been updated in a year and a half and is missing a term, and an actual answer. (The answer is one we should know both from Algebra and from Trigonometry.)
d l 2=(d r)2+(r d θ)2 d l 2=(d r)2+(r d θ)2
Δ l 2=(Δ r)2+(r Δ θ)2 Δ l 2=(Δ r)2+(r Δ θ)2
now we evaluate this between 1 and 2
Δ l 2|2 1=Δ r 2|2 1+(r Δ θ)2|2 1 Δ l 2|1 2=Δ r 2|1 2+(r Δ θ)2|1 2
so how is r r in the (r 2 Δ θ 2)|2 1(r 2 Δ θ 2)|1 2 handled? Do we evaluate r at the same place we evaluate theta? Should the answer look like this?
Δ l 2|2 1=(r 2−r 1)2+(r 2 θ 2−r 1 θ 1)2 Δ l 2|1 2=(r 2−r 1)2+(r 2 θ 2−r 1 θ 1)2
if we think of two points on a unit circle at θ=[0,π]θ=[0,π] then the distance between the points is 2 and the square of that is 4. When we use this expression what do we get? We get a value we know is not correct.
Though they had their misconceptions about 3.0, even 1800's era politicians were sure that π π was not equal to 2.
Why not resort to the law of cosines, from algebra?
c 2=a 2+b 2−2⋅a⋅b⋅c o s(C)c 2=a 2+b 2−2⋅a⋅b⋅c o s(C)
where ABC is the triangle with side AC of length b, side CB of length a, and angle AC to CB of size C. When we try it on the previously given points we get this:
It seems to work. I'm pretty sure that 2=2 2=2.
Inspection tells us that, if derivation was correct, we can say this
[(Δ r)2+(r Δ θ)2]|2 1=r 2 2+r 2 1−2⋅r 1⋅r 2⋅c o s(θ)[(Δ r)2+(r Δ θ)2]|1 2=r 2 2+r 1 2−2⋅r 1⋅r 2⋅c o s(θ)
expanding out the Δ r Δ r and isolating the r Δ θ r Δ θ yields
(r Δ θ)2|2 1=r 2 2+r 2 1−2⋅r 1⋅r 2⋅c o s(θ)−(r 2−r 1)2(r Δ θ)2|1 2=r 2 2+r 1 2−2⋅r 1⋅r 2⋅c o s(θ)−(r 2−r 1)2
We then expand :
(r Δ θ)2|2 1=r 2 2+r 2 1−2⋅r 1⋅r 2⋅c o s(θ)−(r 2 2−2 r 2 r 1+r 2 1)(r Δ θ)2|1 2=r 2 2+r 1 2−2⋅r 1⋅r 2⋅c o s(θ)−(r 2 2−2 r 2 r 1+r 1 2)
then simplify
(r Δ θ)2|2 1=−2 r 1 r 2 c o s(θ)+2 r 2 r 1(r Δ θ)2|1 2=−2 r 1 r 2 c o s(θ)+2 r 2 r 1
We can then gether terms of r r
(r Δ θ)2|2 1=2 r 1 r 2⋅(1−c o s(θ))(r Δ θ)2|1 2=2 r 1 r 2⋅(1−c o s(θ))
We also know that "C" is really the interior angle θ 2−θ 1 θ 2−θ 1 so this becomes:
(r Δ θ)2|2 1=2 r 1 r 2⋅(1−c o s(θ 2−θ 1))(r Δ θ)2|1 2=2 r 1 r 2⋅(1−c o s(θ 2−θ 1))
The left side of the expression is the unknown and the right contains only our known values.
Note: while I like the test case of θ⃗=[0,π]θ→=[0,π], I think that a decent test would include cases such as θ⃗=[0,π 2]θ→=[0,π 2] or r⃗=[1,2]r→=[1,2] as well. Multiple cases with different "physics" should be tested.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Apr 21, 2017 at 13:41
EngrStudentEngrStudent
181 1 1 silver badge 8 8 bronze badges
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
calculus
coordinate-systems
polar-coordinates
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Linked
3How this guy goes from a equation to another equation
Related
3Distance between 2 points in 3D space (in spherical polar coordinates)
1How to calculate the distance between two points when x and y have different ranges?
0Distance between two Polar-Coordinates
0Calculate length of curve created by interpolation of polar coordinates
4Graphically representing vectors with polar unit vectors without converting to Cartesian coordinates
0How to find the distance between two points on a sphere in polar coordinates with fixed ϕ ϕ
1How to integrate over all space in polar coordinates?
2The polar coordinates change of variables
1Equation for the development of an oblique cone in polar coordinates
Hot Network Questions
Interpret G-code
What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left?
ConTeXt: Unnecessary space in \setupheadertext
Weird utility function
For every second-order formula, is there a first-order formula equivalent to it by reification?
Drawing the structure of a matrix
Does the Mishna or Gemara ever explicitly mention the second day of Shavuot?
Why include unadjusted estimates in a study when reporting adjusted estimates?
Origin of Australian slang exclamation "struth" meaning greatly surprised
Stress in "agentic"
Bypassing C64's PETSCII to screen code mapping
What's the expectation around asking to be invited to invitation-only workshops?
A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man
Another way to draw RegionDifference of a cylinder and Cuboid
How to fix my object in animation
Storing a session token in localstorage
How many color maps are there in PBR texturing besides Color Map, Roughness Map, Displacement Map, and Ambient Occlusion Map in Blender?
Calculating the node voltage
Passengers on a flight vote on the destination, "It's democracy!"
Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation
Countable and uncountable "flavour": chocolate-flavoured protein is protein with chocolate flavour or protein has chocolate flavour
How to rsync a large file by comparing earlier versions on the sending end?
Making sense of perturbation theory in many-body physics
Where is the first repetition in the cumulative hierarchy up to elementary equivalence?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
189293 | https://airlinecareer.com/tests/flight-attendant-dictionary/ | Flight Attendant Dictionary and Airline Terminology - AirlineCareer.com
Skip to content
The #1 Flight Attendant Job Site Since 1998!
The #1 Flight Attendant Job Site Since 1998!
Facebook page opens in new windowX page opens in new window
Facebook page opens in new windowX page opens in new window
AirlineCareer.com
Flight Attendant Job Resources and Hiring Updates
Home
Products
Flight Attendant Resumes and Cover Letters
Flight Attendant Career Books
Training
Pre-Qualification Test
Airport Codes Test
24-Hour Clock Test
Hours and Minutes Test
PA Announcements
Flight Attendant Dictionary
About
About Us
FAQ
Privacy Policy
Legal Terms
Blog
The #1 Flight Attendant Job Site Since 1998!
Home
Products
Flight Attendant Resumes and Cover Letters
Flight Attendant Career Books
Training
Pre-Qualification Test
Airport Codes Test
24-Hour Clock Test
Hours and Minutes Test
PA Announcements
Flight Attendant Dictionary
About
About Us
FAQ
Privacy Policy
Legal Terms
Blog
Facebook page opens in new windowX page opens in new window
Flight Attendant Dictionary
Terminology for Flight Attendants
Flight attendants and other airline employees all speak a different “language.” Many of the everyday terms they use are unique to the airline business. Knowing these terms before you attend new-hire training will be very helpful. You will not be tested on this material here, but you may be tested by your airline so try to familiarize yourself with the following terms. There are over 250 of them!
24-Hour Clock Time – See “Military Time.”
“A” Flight Attendant –The flight attendant assigned to work the senior position on a flight. Also referred to as “Lead” or “Senior” flight attendant or “Purser” (on international flights). Warrants additional pay.
Aborted Takeoff –See “Rejected Takeoff.”
Aerophobia –Fear of flying.
AFA – CWA –Abbreviation for “Association of Flight Attendants,” the largest flight attendant labor union in the U.S.
Agent –See “PSR.”
Air Rage – A phenomenon whereby passengers vent their frustrations (sometimes violently) on crewmembers during a flight due to perceived inadequacies in airline service.
Air Traffic Control – See “ATC.”
Air Turbulence –See “Turbulence.”
Aircraft Aft –Section in the rear of an aircraft. Also referred to as “Aircraft Rear.”
Aircraft Forward –Section in the forward part of an aircraft.
Aircraft Fuselage –The complete central structure of an aircraft to which wings, tail surfaces, and engines are attached. Includes entire cabin and cockpit areas.
Aircraft Left –Refers to left interior portion of an aircraft.
Aircraft Rear –See “Aircraft Aft.”
Aircraft Right –Refers to right interior portion of an aircraft.
Airline Deregulation –Refers to the Airline Deregulation Act of1978, which removed governmental authority to regulate airfares and airline route structures. In the current deregulated environment, airlines can fly anywhere and charge any price they desire. Also referred to as “Deregulation.”
Airport Code –A unique, 3-letter airport identification code (e.g., BOS = Boston’s Logan Airport). Every airport in the world has an airport code. Also referred to as an “Airport Identifier.”
Airport Identifier –See “Airport Code.”
Alternate Airport –The airport to which an aircraft diverts when it is unable to land at its destination airport due to weather or other problems. See also “Divert.”
APFA – Abbreviation for “Association of Professional Flight Attendants,” the Flight Attendant Union for American Airlines.
Arbitration –A hearing by an independent arbitrator to reach a settlement with relation to a dispute between a company and labor union.
Arrival Time –See “ETA.”
A-Scale –An airline’s maximum contractual pay scale.
ATC –Abbreviation for “Air Traffic Control,” the FAA agency that controls the movement of aircraft both on the ground and in-flight.
Background Check –A 5 to 10-year check by an employer of an applicant’s past history. Can include checks on work, education, family, medical and criminal or civil offense history.
Base Rate –The basic hourly rate a crewmember is paid. Excludes per diem and other forms of incentive pay.
Base –See “Domicile.”
Beverage Cart –A heavy, wheeled cart onboard an aircraft, which flight attendants use for beverage and meal service. Also called a “meal cart.” Collapses for quick stowage in the galley area.
Bid Closing Date –A published date and time marking the end of the bidding period.
Bid Opening Date –A published date and time marking the beginning of the bidding period. Flight attendants may bid on monthly lines, vacations, and domiciles.
Bid Package –A published listing of all trip pairings and lines of time available for a specific time period, usually 1 month.
Bid Sharing –A practice offered at some airlines in which 2 flight attendants may share or split the same line of time.
Bid –Flight attendants bid on monthly lines, vacations, and domiciles. Bid results are awarded based on relative seniority.
Block –See “Regular Line of Time.” Also a wheel chock placed under the airplane to prevent movement.
Blockholder –See “Lineholder.”
Block Time –Also known as “Block-to-Block” time or “Hard Time,” this is the actual time an aircraft leaves the blocks and pushes back from the gate to the time it arrives (and is blocked in) at the gate at its destination. Most airlines have an onboard computer system known as ACARS (ARINC Communication Addressing and Reporting System) which, along with simplifying engine readings and other routine flight data, logs departure times when all doors are closed on the aircraft and pushback commences. Flight attendant pay is based largely on block time.
Block-to-Block time –See “Block Time.”
Boarding –The term used to describe the process of passengers getting on an aircraft, stowing their carry-on articles, and taking their assigned seats.
Briefing –A procedure initiated by an individual in charge of a group, during which specific items of responsibility are reviewed and delegated. Captains and senior flight attendants initiate crew briefings following every crew change. Briefings can also cover irregular operations, emergency procedures, medical emergencies, etc.
B-Scale – A second-tier pay scale within an airline.
Buddy Bidding –A system that allows 2 flight attendants to bid the identical schedule.
Buddy Pass –A reduced rate, space-available pass that allows a friend or family member to travel on an airline. Most airlines allow a specified number of buddy passes for each employee per year. Typical cost is 50 to 90% off the full coach fare. Also referred to as a “Companion Pass.”
Bulkhead Seat –The seat or seats located directly behind the partition separating various cabins of an aircraft.
Business Class –A premium service offered on most international flights with 2 or 3 classes of service. Amenities may include larger seats, complimentary cocktails, and upgraded meal service.
Cabin Crew –The flight attendants assigned to work a specific flight, trip or series of trips.
Captain –The pilot in command of an aircraft who is responsible for the safe operation of the flight.
Cardiopulmonary Resuscitation –See “CPR.”
CHAOS –An acronym for “Create Havoc Around Our System.” An accepted legal practice used by flight attendant labor unions as an alternative to a strike with the goal of expediting an agreement between the union and the airline. The practice involves flight attendants walking off certain flights in order to create havoc at the airline.
Charter Flight –A non-scheduled flight, often exclusively booked by a sports team, band, or vacation group. Routes may be on the airline’s regular route system or “offline” to a city not served by the airline.
Check-inTime– Time you are required to check-in at the airport.
Civil Rights Act of 1964 –Legislation passed by Congress that prevents an airline from discriminating on the basis of race, sex, age or marital status.
Claim Time –This is time paid in excess of hard time. Examples include pay for deadheading and excessive on-duty time.
Cleaners –The ground personnel that clean an aircraft between legs. At some airlines, these individuals are members of the same union as mechanics and are also used for other duties such as aircraft pushback. Also called “Utility Crew.”
Coach –The largest and normally most basic class of service on an aircraft. Some aircraft have only a coach class of service, while on others, coach may be a part of a 2 or 3 class configuration.
Cockpit Crew –The individuals responsible for piloting an aircraft. Normally a 2 or 3 pilot crew consisting of a captain, first officer, and second officer (if applicable). Long international flights may carry 2 complete cockpit crews.
Commute –The process of commuting by air to a given domicile.
Commuter Airline –Small airlines that are part of the regional airline category. Some are affiliated with major airlines and act as feeders to the mainline routes.
Commuter Apartment – See “Crash Pad.”
Commuter –An individual who commutes by air to a given domicile.
Companion Pass –See “Buddy Pass.”
Configuration –Refers to the class(es) of service onboard an aircraft: First class, Business class, Coach class, etc.
Contract Negotiations –The formal talks between an airline and elected representatives of a labor group that allow for the establishment (or renewal) of a labor contract. The working agreement covers pay, benefits, union security, seniority, scheduling, work rules, vacations, sick time, the handling of grievances (complaints), etc.
Contractual Work Rules –The airline-specific rules published in a labor contract that govern maximum flight hours, on-duty time, days worked, scheduling procedures, etc. Also referred to as “Work Rules.”
Copilot –See “First Officer.”
CPR –Abbreviation for “Cardiopulmonary Resuscitation.” A first-aid procedure that attempts to restore normal operation of the heart and lungs when an individual has stopped breathing.
Crash Pad –Term used by commuters to refer to their “home away from home.” Also referred to as a “Commuter Apartment.” Usually shared by several commuters to reduce costs.
Crew Base – See “Domicile.”
Crew Legalities –A measure of whether or not a crewmember is abiding by the section of the labor contract that governs hours flown, rest periods, and on-duty times.
Crew Schedulers –The group of individuals who are responsible for making sure routine and non-routine flights depart on time with the sufficient number of crewmembers.
CRM – Abbreviation for “Crew Resource Management” or “Cockpit Resource Management.” A set of training procedures that reduces the likelihood of error.
Date of Hire –The date a flight attendant begins or graduates from training. Date of hire establishes a seniority number within an airline.
Deadhead Time –The pay time associated with a deadhead flight. See “Claim Time.”
Deadhead – The process of traveling on an aircraft as a passenger while on-duty (often in uniform). Flight attendants frequently deadhead to reposition for flight segments originating from other cities. Usage: “I had to deadhead to Boston.”
Defibrillator –A medical device carried on an aircraft that allows flight attendants to treat victims of cardiac arrest.
Delay –Time period during which an aircraft is held before it is allowed to depart. Delays may be caused by maintenance (mechanical problems), air traffic, weather, connecting passengers, weight and balance, etc.
Demo –The FAA-mandated demonstration of aircraft equipment and emergency procedures. The demo must be performed before takeoff for all passengers.
Department Head –The head of a specific department within an airline.
Departure Time –See “ETD.”
Deplaning –The term used to describe the process of passengers getting off an aircraft once it has arrived at the gate and the seat belt sign has been turned off by the captain.
Deregulation –See “Airline Deregulation.”
Direct Flight– A single flight number that makes 1 or more stops before reaching its final destination. Many people confuse “direct” flights with “non-stop” flights.
Divert –A procedure where pilots elect to land at an airport other than the airport of intended destination. This can be due to adverse weather, a medical emergency, a closed runway, etc.
Domicile –The city and associated airport a flight attendant is assigned to. The domicile is where all trips begin and end. Also referred to as “Base” or “Crew Base.”
Door Arming –The procedure for preparing an aircraft door so that its associated escape slide will deploy upon opening.
Door Slide –A device used to quickly escape from an aircraft during an emergency evacuation.
Duty Time –The period of time a crewmember is on-duty. Includes the period from check-in to off-duty time.
Emergency Exits – Specified doors and windows on an aircraft used during an emergency evacuation.
ETA –Abbreviation for “Estimated Time of Arrival.” Also referred to as “Arrival Time.”
ETD –Abbreviation for “Estimated Time of Departure.” Also referred to as “Departure Time.”
Evacuation –An emergency procedure that allows for the rapid removal of passengers and crew from an aircraft, using all suitable exits – including slides and over-wing exits. Also referred to as a “Passenger Evacuation.”
Extended Sick Leave –An extended period of time during which a flight attendant is unable to work due to serious illness.
Extra Section –An additional flight added to a scheduled destination in order to accommodate additional passengers. Extra sections are common during peak travel periods (especially during the holidays to popular destinations).
F/A – Abbreviation for flight attendant.
F/A Emergency Manual –A manual required by the FAA to be personally carried at all times by every on-duty flight attendant. Also called an “Emergency Procedures Manual,” this manual governs all normal and emergency flight attendant procedures.
F/A Jumpseat –A fold-down seat that is used by flight attendants during taxi, take-off, and turbulence. Non-working flight attendants can ride the jump seat on heavily booked flights.
FAA –Abbreviation for “Federal Aviation Agency,” the government agency that oversees all airline operations.
FARs –Abbreviation for “Federal Aviation Regulations,” the specific government regulations instituted by the FAA.
FAM –Abbreviation for “Federal Air Marshall.” An armed individual who rides as a passenger on the aircraft to protect the passengers and crew against terrorist acts. The captain is always notified when FAMs are present on the aircraft.
Federal Aviation Agency –See “FAA.”
Federal Aviation Regulations – See “FARs.”
Ferry Flight –A repositioning flight flown with no revenue passengers onboard. Usually flown by pilots only, but flight attendants are sometimes onboard as well.
FFDO – Abbreviation for “Federal Flight Deck Officer.” A designated pilot who is armed in order to defend the cockpit against terrorist attacks and air piracy. An FFDO may be working or deadheading.
Firearm Authorization –A notification slip presented to the senior flight attendant that identifies passengers carrying authorized firearms (e.g., federal marshals and secret service members). The firearm authorization slip identifies the armed passenger (and accompanied prisoner, if applicable) by seat number, and must be presented to the captain.
First Class –The highest class of service onboard an aircraft. Amenities may include larger seats, complimentary cocktails, and upgraded meal service.
First Officer –Second-in-command pilot on an aircraft. Also referred to as a “Copilot.”
Flight Attendant–A person responsible for the safety and comfort of all passengers during a flight.
Flight Engineer –See “Second Officer.”
Flight Hour Option –A bid position offered by some airlines that allows a flight attendant to fly less or more than a normal monthly line of time.
Flight Hours –Total number of hours flown in any given period of time used for pay computation.
Flight Miles –Miles flown in any given period of time used for pay computation at some airlines. 1 trip equals 243 miles.
Flight Time– Sometimes known as “Airtime,” this is the time that is allowed from takeoff to touchdown for a specific flight. Flight time is block time less taxi time and gives flight attendants an idea how much time they will have to provide in-flight service.
FOB –Fuel on board.
Full Month –A term used to refer to the maximization of flight hours for any given month.
Furlough –A layoff of an airline employee due to financial difficulties, labor union strikes, etc.
G.E.D. –Abbreviation for “Government Equivalency Degree,” which is the minimum education requirement for most airlines. This applies only to individuals without a high school diploma.
Galley –The area inside an aircraft where meals and beverages are prepared by flight attendants.
Gate –The airport boarding area at an airport terminal for a specific flight number.
Get your time in –Refers to the practice of maximizing flight hours in order to get a “Full Month.”
Government Equivalency Degree – See “G.E.D.”
Green Card – See “Registered Alien.”
Greenwich Mean Time –The exact time in Greenwich, England. Also known as “Zulu” or “Z” time. Pilot weather packets, clearances, dispatch releases, and other paperwork all use Greenwich Mean Time as a worldwide airline standard.
Grievances –Complaints issued by a labor union to an airline when it appears that portions of the labor contract are being violated by the company. Grievances are most common in scheduling scenarios.
Hard Time –See “Block Time.”
Hijacking –A militaristic act of aggression by a passenger onboard an aircraft whereby crewmembers and other passengers are taken hostage until the hijacker’s demands are met.
Holding –A procedure used by air traffic controllers to delay arrivals of aircraft due to traffic saturation or bad weather. Holding patterns are areas along a route system that allow aircraft to fly around a specified track until released.
Hub –An airport through which the majority of an airline’s traffic passes in order to meet connecting flights. Many airlines have multiple hubs.
IAM – Abbreviation for “International Association of Machinists.”
IBT – Abbreviation for “International Brotherhood of Teamsters.”
Illegal –A term used to describe a crewmember who can no longer legally continue to work due to a violation of scheduling restrictions in the labor contract. For example, a crewmember who exceeds the 16-hour daily maximum on-duty period becomes illegal.
Initial Training –Training that must be completed by all new-hires. Also referred to as “New-Hire Training.”
Instructor –An individual who conducts initial and recurrent flight attendant training classes. Also referred to as a “Trainer.”
Interline Discounts –Travel industry discounts available to airline crewmembers.
International Flight –A flight that departs a domestic location and lands at an international destination.
Interviewer –An individual who conducts interviews with flight attendant applicants.
Jetway –The passageway between the gate area and an aircraft that allows passengers to board and deplane.
Jumpseat – See “F/A Jumpseat.
Junior– A term used as a measure of one’s relative seniority at an airline or at any given base (e.g., “How junior are you?”). Also used to describe the seniority of a specific flight attendant domicile (“Is that a junior base?”).
Labor Contract –An official document negotiated between labor and management. Also called a “working agreement,” covering pay, benefits, union security, seniority, scheduling, work rules, vacation, sick time, handling of grievances, etc.
Labor Union –A body of members and elected representatives that seek to negotiate labor contracts and handle job disputes. They also actively represent members in the media and fight battles in the political arena.
Language of Destination or Origin – See “LOD/O.”
Layover –A period of time a crewmember is scheduled to remain at a specified location (e.g., a 2-hour layover in PIT). Extended layovers are termed “RON,” for “Remain Overnight.”
Lead Flight Attendant –See “‘A Flight Attendant.”
Leave of Absence –A period of time granted by an airline in order for a flight attendant to pursue other interests or needs before returning to work. Types of leaves include education, military and maternity leave. Seniority continues to accrue during most leaves.
Leg –A single flight from 1 departure point to 1 destination point. A leg includes 1 takeoff and 1 landing.
LEO – Abbreviation for “Law Enforcement Officer.” An armed law enforcement individual on the aircraft who may or may not be on duty. LEOs are informed when other LEOs and FAMs are onboard and the captain is always advised of their presence as well.
Lifeguard Flight –A flight responsible for transporting human organs to a medical facility. Flight is given priority routing to expedite arrival.
Limo time– See “Van Time.”
Limo –Generic term used by crewmembers to refer to the vehicle that transports a crew to and from the hotel (even if it is only a van).
Line of Time – A monthly schedule.
Lineholder –A crewmember with sufficient seniority to fly (or hold) a regular line of time. Also referred to as a “Lineholder” or “Blockholder.”
Line Sharing – A system that allows 2 flight attendants to split a monthly line of time.
LOD/O –Acronym for“Language Of Destination/ Origin.” A flight attendant who is assigned to an international trip as a second language speaker.
Mainline –A term used when referring to the larger entity of an airline that provides both commuter and large jet service.
Major Airline –An airline with over $1 billion in annual operating revenue.
Maternity Leave- See “Leave of Absence.”
Meal Cart –See “Beverage Cart.”
Mechanical –Refers to a problem with an aircraft that must be remedied by maintenance before departure. Often boarding is delayed until the mechanical is fixed. Usage: “This plane has a mechanical.”
Mechanic –A member of an airline’s ground support crew who is responsible for maintenance and repair of aircraft.
Mediation –An individual hired from outside an airline that assists labor and management in reaching terms to a working agreement.
Medical Emergency –An emergency onboard an aircraft that requires first aid, medical attention, and possible aircraft diversion.
Medlink –A medical agency used by some airlines that provides in-flight personnel direct communication with a physician during medical emergencies.
MEL –Abbreviation for“Minimum Equipment List,” an FAA-approved document that spells out what equipment is allowed to be inoperative at the time of dispatch and for how long.
Military Time– The time on a 24-hour clock, used by airlines and crewmembers industry-wide in reading trip pairings. Also referred to as “24-hour clock time.”
Minimum Guarantee –Minimum number of hours to be paid in a given month regardless of number of hours flown. Applies to reserves only.
Monthly Projection –Projected number of flight hours at the end of the current month based on projected flying activity.
Narrow Body –An aircraft with a single aisle with seats on either side of the aisle.
National Airline – An airline with between $100 million and $1 billion annual operating revenue.
New-Hire Training – See “Initial Training.”
New-Hire –A flight attendant who has just completed initial training. Flight attendants are usually classified as new-hires during the first six months of employment.
No Contact –A flight attendant infraction resulting from a crew scheduler being unable to reach a flight attendant for a trip assignment. Usually applies to reserves only.
Non-Rev –Abbreviation for “Non-Revenue,” a term used to describe airline employees who travel using space-available passes. Also referred to as “Space-A travel.” Usage: “Are you a non-rev passenger?” or “Are you flying non-rev?”
Non-Stop Flight –A single flight number with 1 takeoff and 1 landing to reach its destination. See also “Direct Flight.”
On-Call –A period of time during which a reserve flight attendant may be assigned a trip.
On-Duty –The period of time a flight attendant is “working,” from check-in time until going off-duty. Reserves also use this term in reference to being on call.
Open Time –Uncovered trips available for bid by lineholders (can also be assigned to reserves).
Origination –Usually refers to the first flight of the day for a specific aircraft, but can also refer to a crew’s first flight.
Out-and-In –A 1-day trip that flies to 1 destination and returns. Also referred to as a “Turnaround.”
Out-of-Time –A situation in which a flight attendant reaches the maximum time ceiling (i.e., flight hour limit) and is no longer legal to fly. Usually applies to being “Out of Time” for the month.
Over Water Flight –A flight that exceeds 50 nautical miles from the coastline requiring an emergency life vest and raft demo. See also “Transoceanic Flight.”
Overhead Bin –Stowage area above the passenger seats. Carry-on bags are required to fit into these bins. Also referred to as an “Overhead Compartment.”
Overhead Compartment – See “Overhead Bin.”
Pass Privileges –Rules and regulations published by each airline that provide specific guidance on non-revenue travel.
Pass –A standby ticket issued to a “Non-Rev.”
Passenger Count –The final count of passengers delivered by the senior flight attendant to the cockpit crew for weight and balance considerations.
Passenger Evacuation –See “Evacuation.”
Passenger Service Agent –See “PSR.”
Passenger Service Unit –See “PSU.”
Pattern – See “Regular Line of Time.”
Pay Time –This is the time used for pay purposes.
Per Diem –Latin for “by the day.” Refers to hourly rate paid to flight attendants for meal expenses while on-duty.
Picket Line –A line of workers carrying signs during labor negotiations expressing dissatisfaction with their airline. Picket lines may be formed for information purposes only. Picket lines may also be formed during a work stoppage or just prior to a union’s implementation of “CHAOS.” See “Strike,” “CHAOS,” “Labor Contract, ” Labor Union.”
Pickup time –See “Van Time.”
Pre-board –A procedure during which passengers with small children and those needing assistance (e.g., wheelchair passengers) are boarded before regular passengers. These individuals are also referred to as “Pre-boards.”
Pre-flight –The period of time before passenger boarding, during which emergency equipment is checked, briefings are completed, catering supplies are checked, etc.
Probation –See “Probationary Period.”
Probationary Period –A period of time, usually lasting from 6 to 12 months, during which a new-hire’s performance is evaluated by an airline. Also referred to as “probation” or “being on probation.”
PSR – Abbreviation for “Passenger Service Agent,” the person responsible for passengers from the time they check-in (at the gate) until the aircraft’s cabin door is closed. Duties include passenger check-in, baggage checking, and assistance with passengers in wheelchairs. Also responsible for confirming that fuel quantity and passenger count is correct before entry door is closed. Also referred to as a “Ticket Agent” or “Agent.”
PSU –Abbreviation for “Passenger Service Unit,” a unit above each row of passenger seats that houses individual passenger oxygen units, reading lights, flight attendant call buttons, etc.
Purser –See “‘A’ Flight Attendant.”
Pushback –The process of moving an aircraft backwards from the gate, which is accomplished by coordination between the pilots and ground maintenance crew. The crew connects a towbar between the nosewheel of the aircraft and a ground tractor known as a “tug.” Once the captain releases the aircraft’s parking brake and deems it safe to pushback (doors closed, jetway clear of aircraft, clearance from the ramp or ground controller, and notification from the senior flight attendant that all passengers are seated and all bags are stowed), the tug slowly pushes the aircraft off the gate, while the aircraft engines are simultaneously started. When pushback is complete, the pilot sets the brakes, the tug is disconnected, and the maintenance crew waves off the aircraft so that it may proceed under its own power.
Quick Call –A trip assigned to a reserve lineholder, which requires reporting to the flight as soon as possible.
Ramp Workers –Maintenance personnel, cleaners, fuelers, caterers, etc.
Ramp –The area around an aircraft where ground personnel perform their duties. This includes maintenance, baggage handling, catering, fueling, etc.
Reach Test –A test instituted by several airlines to determine whether a flight attendant applicant will be tall enough to perform the duties required for the job.
Recall –The action of calling back a furloughed worker to the job.
Recurrent Training –Annual refresher training required by the FAA to be completed by all flight attendants.
Red-eye –A flight, typically from the West Coast that departs late in the evening and flies eastbound all night to the destination airport.
Regional Airline – An airline with less than $100 million in annual operating revenue. Includes commuter airlines.
Registered Alien – A person who has the legal right to accept employment in the United States. Registered aliens must possess what is called a “green card.”
Regular Line of Time –A unique schedule that features specific trip pairings typically over a 4-week period. Also referred to as a “Block” or “Pattern.”
Rejected Takeoff –A sudden, unexpected stop of an aircraft (on the runway) following the takeoff roll, due to a mechanical or other type of problem. Also referred to as an “Aborted Takeoff.”
Report Time- Time you are required to be at the airport for check-in.
Reservationist – An airline representative who books flights and assists passengers with reservation and/or ticket problems.
Reserve – A crewmember with insufficient seniority to hold a regular line of time. A reserve must fly a reserve line of time, which features no assigned trips and no set schedule other than days off.
Reserve Line of Time – A schedule which features no assigned trips and no set schedule other than days off.
RON –Abbreviation for “Remain Overnight.” A layover, which includes a hotel stay. See also “Layover.”
Scab –A crewmember who crosses a picket line and continues to work during a strike or other type of work stoppage.
Second Officer –Third-in-command of an aircraft. Normally a non-flying position. Also referred to as a “Flight Engineer.”
Senior – A term used as a measure of one’s relative seniority at an airline or at any given base (e.g., “How senior are you?”). Also used to describe the seniority of a specific flight attendant domicile (“Is that a senior base?”).
Seniority List –A list published by an airline listing the seniority number of every employed flight attendant.
Seniority Number –A unique number assigned to each flight attendant based on date of hire. See also “Date of Hire.”
Seniority –A numerical ranking system (based on date of hire) used by the airlines to determine awards of line positions, vacations, domiciles, etc.
Show Time –See “Check-in Time.”
Sick Time –Accrued time (in a sick bank) that is required to receive paid sick days.
Space-A Travel –See “Non-Rev.”
Standby –See “Reserve.”
Sterile Cockpit –The period of time during the critical phase of flight when the cockpit door must be locked and flight attendants are restricted from entering (except in an emergency). A sterile cockpit is required when an aircraft is below 10,000 feet and includes taxi, takeoff, and landing.
Stewardess –Original term used to describe a flight attendant before the 70s.
Strike –Legal work stoppage by labor due to an inability to reach a labor contract agreement with management.
Supervisor –A flight attendant manager who is directly responsible for flight attendants at a given domicile.
Taxi – The act of moving an aircraft on the ground under its own power.
Termination –Usually refers to the last flight of the day for a specific aircraft, but can also be used to refer to the last flight of the day for a crew.
Terminator – An aircraft that is finished flying on a given day upon arrival at its destination.
Ticket Agent –See “PSR.”
Trainer – See “Instructor.”
Transcontinental Flight –A flight that travels non-stop across the country, usually from coast-to-coast.
Transoceanic Flight –A flight that travels across the Atlantic or Pacific Ocean to reach its destination.
Trip Check-in –The time a flight attendant is required to check-in for an assigned trip, usually at least 1 hour before departure, depending upon the airline.
Trip Pairing –A series of flight numbers that comprise a “trip.” Also simply called a “Trip.”
Trip – See “Trip Pairing.”
Tuff Cuffs – A handcuff restraining device carried onboard an aircraft used by flight attendants for restraining unruly passengers.
Turbulence –Irregular motion of the atmosphere, causing a rough ride on an aircraft. Also referred to as “Air Turbulence.”
Turnaround –See “Outandin.”
Two-Tiered Wage System– See “B-Scale“.
TWU – Abbreviation for “Transport Workers Union of America.”
Unacc –See “Unaccompanied Minor.”
Unaccompanied Minor –An underage child traveling alone on an airplane. Also referred to as a “unacc.”
Union Dues –Monthly payment required to maintain individual labor union membership in good standing.
Utility Crew –See “Cleaners.”
Van Time –The time the crew is expected to meet in the hotel lobby after a RON. Also called “Limo Time” or “Pickup Time.”
Weight and Balance –Series of computations, normally automated and sent via computer to the cockpit. Includes aircraft gross weight, passenger and cargo weight, optimum runway, wing flap settings, etc.
Wide Body –An aircraft with two aisles with rows of seats in the center of the two aisles and on each side.
Work Rules – See “Contractual work rules.”
Working Agreement –See “Labor Contract.”
Working Conditions –See “Contractual Work Rules.”
Write-up –A logbook entry that describes a defect or discrepancy on an aircraft that needs maintenance. Usually these items are entered by pilots in the cockpit logbook, but in some instances, there is also a cabin logbook for flight attendant write-ups.
Zulu Time –See “Greenwich Mean Time.”
Join Our Email List - Breaking Flight Attendant News & Info
Email:
Select list(s) to subscribe to- [x] AirlineCareer.com ezine
[x] Example: Yes, I would like to receive emails from AirlineCareer.com. (You can unsubscribe anytime)
Constant Contact Use. Please leave this field blank.
AirlineCareer.com 1998-2017
Go to Top |
189294 | https://www.physics.umd.edu/courses/Phys621/gradlab/glhb/muon.html | Graduate Physics Laboratory Handbook: Lifetime of the Cosmic Ray Muon
Graduate Physics Laboratory Handbook: Lifetime of the Cosmic Ray Muon
Guide · Table of Contents · List of Experiments · Calendar · Student Roster
2.4.1Lifetime of the Cosmic Ray Muon
The muon is considered by many to be a "heavy" electron. (The mass of the muon is 205 times that of the electron). It is unstable and is known to have an intrinsic lifetime of several microseconds. Its decay is explained in the weak interaction theory in which it is used as a basic parameter.
Pions, with a mass 273 times that of the electron, are created in the upper atmosphere in nuclear interactions produced by high-energy cosmic-ray protons. These decay into a muon and a neutrino with a relatively shorter lifetime of a few tens of nano seco nds. The muons that are so produced, due to their longer lifetime and almost complete absence of nuclear interaction, are the principle components of penetrating particles produced by cosmic rays that are observed at sea level. Both µ+ and µ- are observed at sea level. On coming to rest both µ+ and µ- decay but the lifetime of the µ- is altered due to weak interaction with the atomic nucleus when captured to form a mesic atom.
The purpose of this experiment is to measure the intrinsic lifetime of the muon. The equipment for this purpose is available in the Graduate Laboratory. A description of one method of performing the experiment is described in a Graduate Laboratory Note. T he difficulty with this method lies in the small volume of the described detector and the limited solid angle possible of the cosmic-ray telescope. Another method is described below but the experimenter may choose to devise his own.
The principal problem is the dearth of muons that can actually be made to come to the end of their ranges in any finite detector. This number should be estimated to understand the problem. There is, however, a large activated mineral-oil scintillation cou nter (8" x 8" x 24") available in the lab. It has no wide angle telescope to define the cosmic-ray beam but can be used without it by using a coincidence between two photomultiplier tubes which look at the scintillator volume. In this way, a muon is detec ted by a coincidence pulse and, if it stops and decays, a second coincidence pulse detects its decay. The time interval between these two pulses is measured with a time-to-pulse height converter which, when converted with a pulse height analyzer, can be u sed to measure the frequency of decay times. From this frequency distribution the mean life for the cosmic ray muon can be determined and from this the intrinsic muon lifetime.
In this brief description a number of statements of a general nature have been made. They are meant to serve only as a guide for a more detailed and thorough study which will climax, with diligence, in an experiment (and paper) which quantitatively measur es and discusses the muon lifetime.
References
E. U. Condon and H. Odishaw, Handbook of Physics, New York: McGraw Hill (1967). See page 9-159 for overview of muon nuclear capture.
R. E. Hall, D. A. Lind, and R. A. Ristinen, "A Simplified Muon Lifetime Experiment for the Instructional Laboratory", Am. J. Phys. 38, 1196 (1970).
A. A. Bartlett, "Student Experiment on the Observation of a Cosmic-Ray Shower Transition Curve", Am. J. Phys. 23, 286 (1955).
A. Owens and A. E. MacGregor, "Simple Technique for Determining the Mean Lifetime of the Cosmic Ray Mu-Meson", Am. J. Phys. 46, 8 (1978).
D.H. Frisch and J.H. Smith, "Measurement of the Relativistic Time Relation Using µ-Mesons", Am. J. Phys. 31, 342 (1963).
G. C. Kyker, "Resolving Time Effect on Counting Statistics", Am. J. Phys. 49, 561 (1981).
E. A. Bogomolov, et al., "Investigation of the Cosmic Ray East-West Asymmetry", Can. J. Phys. 46, 805 (1968).
P. S. Freier and C. J. Waddington, "Electrons, Hydrogen Nuclei, and Helium Nuclei observed in the Primary Cosmic Radiation during 1963", J. Geophys. Res. 70, 5753 (1965).
J. R. Winckler and K. Anderson, "Geomagnetic and Albedo Studies with a Cerenkov Detector at 40 Degrees Geomagnetic Latitude", Phys. Rev. 93, 596 (1954).
M. A. Shea and D. F. Smart, "A Study of Vertical Cut-Off Rigidities Using Sixth Degree Simulations of the Geomagnetic Field", J. Geophys. Res. 70, 4117 (1965).
J. R. Winckler, et al., "A Directional and Latitude Survey of Cosmic Rays at High Altitudes", Phys. Rev. 79, 656 (1950).
J. B. Birks, The Theory and Practice of Scintillation Counting, New York: Pergamon Press Ltd. (1964). QC787.C6B53.
Glenn F. Knoll, Radiation Detection and Measurement, New York: John Wiley (2000). To be added to GL library 10/00.
William R. Leo, Techniques for Nuclear and Particle Physics Experiments, New York: Springer Verlag (1996). QC793.L46 Modern treatment of devices and techniques.
B. Rossi, High Energy Particles, New York: Prentice Hall, (1952). QC721.R828.
E. Segre, Nuclei and Particles, New York: W. A. Benjamin, (1964). Grad Lab #25. QC721.S4475.
A. Wolfendale, Cosmic Rays, New York: Philosophical Library (1963). QC485.W6.
A. E. Sandstrom, Cosmic Ray Physics, Amsterdam: North Holland Publishing Co. (1965). Grad Lab #88. QC485.S3.
B. Rossi and N. Neison, Phys. Rev. 62, 417 (1933), 64, 199 (1935).
Photomultiplier Handbook, (Burle Industries, Inc.,Lancaster, Pa., 1989). This is a complete guide to the understanding of photomultiplier tubes. Note the temperature dependence of photomultiplier tubes.
Go to Top
Laboratory CoordinatorGo to the Handbook Table of Contents
© Dept. of Physics, Univ. of MD
This page was maintained by the Laboratory Staff.
For queries regarding:
Content contact the Laboratory Coordinator.
Technical Questions contact the Laboratory Coordinator.
Last updated on 08/31/2004 11:58:37 |
189295 | https://goopenva.org/courseware/lesson/6451/overview | Lesson E: Understanding the Binary Number System | #GoOpenVA
Virginia Department of EducationAn official website of the Commonwealth of Virginia Here's how you knowAn official websiteHere's how you know
Find a Commonwealth Resource
Discover
Resources
Collections
Providers
Hubs
Sign in to see your Hubs
Login Featured HubsComputer Science Advancing Computer Science Education (ACSE) Internet Safety Resources Public Media Partners Health Education Resource Hub CodeVA See all Hubs
Groups
Sign in to see your Groups
Login Featured GroupsVirginia Society for Technology in Education See all Groups
Learn More
About
Help Center
Add OER
Open Author
Create a standalone learning module, lesson, assignment, assessment or activity
Create Resource
Submit from Web
Submit OER from the web for review by our librarians
Add Link
Learn more about creating OER
Add OER
Add Link
Create Resource
About creating OER
Notifications
Sign In/Register
Search
Advanced Search
Sign In/Register
Discover
Resources
Collections
Providers
Hubs
Sign in to see your Hubs
Login Featured HubsComputer Science Advancing Computer Science Education (ACSE) Internet Safety Resources Public Media Partners Health Education Resource Hub CodeVA See all Hubs
Groups
Sign in to see your Groups
Login Featured GroupsVirginia Society for Technology in Education See all Groups
Learn More
About
Help Center
Add OER
Open Author
Create a standalone learning module, lesson, assignment, assessment or activity
Create Resource
Submit from Web
Submit OER from the web for review by our librarians
Add Link
Learn more about creating OER
Add OER
Add Link
Create Resource
About creating OER
Student View
Preview
Copy
Save
Please log in to save materials. Log in
Report
Details
Standards
Author:VCU CISTEME .Subject:Computer Science Material Type:Lesson Level:Middle School, High School Tags: VCU CS Lessons
Log in to add tags to this item.
License:Creative Commons Attribution Non-Commercial Share AlikeLanguage:English Media Formats:Downloadable docs
Show MoreShow Less
Education Standards
1. 1
2. 2
VA.CS.MSCSE.25
Virginia Computer Science Standards of Learning 2017
Grades 6-8
Learning Domain: Computing Systems
Standard: The student will recommend improvements to the design of computing devices, based on an analysis of how users interact with the devices.
VA.8.CS.CSY.1
Virginia Computer Science Standards of Learning 2024
Learning Domain: Computing Systems
Standard: The student will recommend and design improvements to computing devices based on the needs of various users.
Lesson E: Understanding the Binary Number System
Overview
Students will use binary numbers to count and send messages.
Context of the Activity
The Big Idea:Students will use binary numbers to count and send messages.
Students will
learn to convert between binary and decimal numbers in the context of topics that are important to computer science.
count forward and backward in binary.
explain why binary numbers are important in computer science.
use binary numbers to encode and decode messages.
Theteacher will
model how to count in binary.
monitor teamwork skills during activities.
practice with students and reteach binary as needed.
explain how to use binary to code and decode a message.
Prerequisite Knowledge and Skills
Some background knowledge of what a computer is used for and basic parts such as the keyboard, monitor, and mouse
Ability to count and add and subtract are necessary
Recommended but not required:
Students have completed an Algebra course prior to enrolling
Vocabulary
binary numbers
message
forward
backward
decimal numbers
encode
decode
Narrative / Summary
Students will
complete a journal entry to introduce the topic.
practice binary countingorally in class.
complete the activities in groups or individually.
complete the decoding assignment.
The Teacher will
show students how to count in binary using binary cards.
supply students with 2 practice activities to practice binary skills.
monitor student ability and teamwork during activities.
review how to count in binary as needed with groups.
assign students a decoding activity using their binary counting skills.
prompt discussion about why encoding might be used.
Lesson Objectives (learning targets)
Students will be able to:
demonstrate counting in binary.
talk about the significance of binary in Computer Science.
encode and decode binary messages.
understand the foundational reasons for encoding.
Essential Questions (CRT & CS)
What impact does computer science and counting in binary have on you?
Why are binary numbers important to computer science?
Materials
Pens/Pencils/Marker for creating your own code
Card stock/paper
Rapid Tables: Binary to Text Translator:
Note: To ensure all students have equitable materials, the teacher will supply all materials for activity.
Lesson Structure and Activities (use 5E structure)
(5-10 min) Bell Ringer Activity (Engage):
Students willcreate a journal entry to answer the following question:How high can you count with your ten fingers?
Theteacher will
review the topic for today’s journal entry.
allow students 3-5 minutes to complete their journal entry.
ask students about their answers and why they chose them.
(5-10 min ) Launching Activity:
Students will
answer questions about the binary cards displayed in class.
participate in the discussion about binary numbers.
flip the cards over to make different numbers.
Theteacher will
d isplay the binary cards (from can be displayed on the wall, board, or students can hold them..
arrange the cards in binary order. How would you make the number 6? Add additional questions as appropriate with time and students.
have students use the cards to make different numbers. This is a good time to allow students to come to the front and change the cards to make the numbers. Explain how some cards can be flipped over and not counted to make zero and how cards facing up show as a 1.
(25 min) Exploring Activity:
Students will
complete the Learning How To Count activity on page 5,sing the cards provided on page 6 or some that they make if time allows.
make the numbers the sheet tells them.
make additional numbers assigned by the teacher.
answer questions from the teacher orally about binary numbers.
complete the Working with Binary activity.
review the Working with Binary activity with the class.
answer the journal entry question again?How has your answer changed? Where else could you use binary numbers in your life?
NOTE:can be done as a class, small group, or pair share
Theteacher will
determine if students will use the preprinted cards or students can make their own binary cards.
show students how binary digits add and count up.
ask students to use the cards to count in order starting at 1 to whatever number feesl appropriate and there is time for.
ask students: Are there any numbers that you cannot make?
walk around and monitor students using the cards and completing the numbers on the worksheet.Model again for students that are having trouble.
EXTRA: Review things in your house that might use binary code.
explain the Working with Binary worksheet to students.
review the activity with the class.
have students answer the journal entry question again.Then ask students to explain how their answers have changed?Are there other places in their lives that they can relate to binary numbers?Where do they use computer science?
NOTE: this can be done as a whole class discussion, small group, or pair share
(10-15 min) Concluding Activity:
Students will
complete the Sending Secret Messages activity.
use the chart at the bottom of the page to turn the binary number of each row into a letter.
Theteacher will
explain that students will be decoding the message using binary numbers from the rows of Christmas trees/other symbols.
provide numbers for students to convert to binary.
have students turn in assignments at the end of class or review together. If you are going to use it as an assessment, collect and review at the beginning of the next class.
Assessment Strategies (formative):
The teacher will use formative and summative assessment strategies to access students’ mastery of the concepts of data and data analysis.
Formative assessment strategies:
review original journal entries.
take note of students changes in thoughts related to journal entry.
evaluate student ability to count in binary with the Binary Dots activity.
monitor teamwork during assignments.
check for understanding during the Working with Binary activity
Summative Assessment Strategy(student work might be assessed individually or cumulatively): A ssess Sending Secret Messages for accuracy.
Both formative and summative assessment strategies enable the teacher to hold students accountable in a manner that attends to their unique needs (e.g., cultural backgrounds, learning styles, IEPs, 504s, accommodations, etc.), ensuring each student reaches their learning targets and demonstrates mastery.
Online resource for formative and summative assessments:
Computer Science information
Extension Ideas
What opportunities do students have for more investigation?You may provide additional extension activities such as:
CS Unplugged has a variety of computer science activities that can be done anywhere and without a computer.This is a good alternative for students with connectivity issues at home:
Have students make a code and then give it to another student and have them decode it on their own.Do not provide a key but let the student figure it out on their own.
Put students in groups of 5. You may adjust group size based on your class.Give each group a set of binary cards.Then the teacher will call out numbers for them to make using the binary cards. See which groups make the numbers the fastest.This can be a class game or just small group work.
Secret Codes. Have students research and present information on how secret codes are used in the military at different times during history.This can be a general history of assignment or students could focus on a certain time in history (Civil War, WWI, WWII, etc.) based on the current topic in history course.
Alignment
Students should have a basic understanding of computers and basic computer skills.Teachers should make connections between how a student uses the computer and how the computer uses the binary code.
The lesson should align with basic computer education standards across the CTE curriculum.
Lesson also aligns with basic mathematical skills.
Students use creativity and innovation skills to produce their own code.
Modifications (resources are linked in headings):
American with Disabilities Act (ADA) and Individuals With Disabilities Education Act (IDEA):
The teacher will ensure that the appropriate accommodations (504 plan) and/or special education services (IEP) are put in place to accommodate the student(s).
The teacher will ensure that students with disabilities (e.g., vision and hearing impaired, mobility, speech, etc.) are provided the appropriate tools (e.g., braille, microphone, wheelchair accessible desk, etc.) to improve and provide an equitable learning experience and outcomes.
English Language Learners (ELL):
The teacher will ensure that directions and materials are scaffolded and translated to ELLs preferred language.
Note:ELLs’ background knowledge might be lacking, especially with things unique to American or Westernized culture.
Learning Modalities (i.e., learning styles):
The teacher will e nsure, when appropriate, that students preferred learning modality is used to maximized all learning situations.
Note:Teachers will need to observe and document students’ preferred learning modalities to maximize student learning and outcomes.
Resources
If desired, additional background, lessons and worksheets for Binary Numbers may be accessed through the Exploring Computer Science Curriculum:
Discover
Resources
Collections
Providers
Advanced Search
Create
Open Author
Submit a Resource
Connect
Groups
Hubs
Learn More
About
Help Center
Virginia Department of Education
Contact Us
Connect with #GoOpenVA
Powered By
Terms of Agreement
The Virginia Department of Education (VDOE) is a collaborative partner in the #GoOpenVA initiative. The VDOE has not evaluated these resources for content nor accessibility but they have generally been reviewed by educational peers. Use your professional and subject expertise in evaluating and using these resources. For more details, see About #GoOpenVA Resources.
GoOpenVA was last updated: 12/20/2019
×
Sign in / Register
Your email or username:
Did you mean ?
Password:
Forgot password?- [x] Show password
or
Sign In through your Institution
Sign in with Clever
Create an account Register
or
Sign In through your Institution |
189296 | https://www.youtube.com/watch?v=r0lOBWmldcY | Let's solve this question,here it is →Q.If arcTanA+arcTanB+arcTanC=π,prove that A+B+C=ABC.
LUNJAPAO BAITE
3340 subscribers
3 likes
Description
52 views
Posted: 8 Feb 2019
You can check out my video on proof of arcTanA+arcTanB=arcTan [(A+B)/(1-AB)] here and arcTan(1/A)=arcCotA here and also arcCot(-A)=π-arcCotA here
If you're watching for the first time do like my video and please subscribe to my channel, so that you'll not miss any new video which is about to come in the next video and comment down below for the next video...
Thanks for watching my video....
Transcript:
France today I get one question here if tan inverse of X plus 10 were so y plus the inverse function is called a PI then prove that X plus y plus 10 is equal to XY Z so let's prove this we are given that tan inverse of x plus tan inverse of y plus tan inverse of the musical de pie we know tangent of X plus 10 was always turn reverse off X plus y over 1 minus X Y which is if you transfer you turn this off so it's turned worse off X plus y over 1 minus X Y is equal to PI minus so in thumb of khorinis go inverse off good inverse of one over Z so tan inverse of X plus y by 1 minus XY and we know by minus cot inverse of 1 by Z news just cold inverse off may not 1 over Z and so tan inverse of X plus y minus over 1 minus XY is equal to in thermo tan in this tan inverse off may not serve so this cooking a cancer so X plus 1 over 1 minus X Y Z colder may not serve so if you cross multiply X plus y is equal to may not Z minus minus plus XY Z so lastly X plus y plus Z squared X Y Z |
189297 | https://math.ucr.edu/home/baez/spin/node10.html | Selection Rules
Next:Magnetic MomentsUp:SpinPrevious:Angular Momentum Quantum Numbers
Selection Rules
How do the normal and anomalous Zeeman formulas account for the details of the Zeeman splitting?
I omitted many transition lines from figures1 and2 to avoid cluttering the diagram. Nature seems to share this taste for simplicity. Of the multitude of transition lines one might draw, most are forbidden by selection rules.
For figures1 and2, the appropriate selection rules assert that changes by one unit, and changes by at most one unit: , . Conservation of angular momentum accounts for these rules, as hinted earlier. But before getting into that, let us explore the role of selection rules in the Zeeman effect.
As it happens, the normal and anomalous Zeeman effects call for different selection rules. Four selection rules apply to the normal Zeeman effect:
.
.
.
.
Plugging this into the formula , we get:
(3)
Spin has dropped out completely! has only three possible values, thus the magnetic perturbation splits each spectral line into three.
For example, consider transitions from states to states. Without the magnetic field, each state belongs to a degenerate quintuplet of states (distinguished by the five possible values of ) - where ``degenerate'', you may recall, simply means that all states in the quintuplet have the same energy. Likewise, each state belongs to a degenerate triplet. This degeneracy stems from the fact that the energy does not depend on , when there is no field.
Had we no selection rules at all, we would have 15 possible transitions from the quintuplet to the triplet. The selection rule cuts this down to 9 (we can draw three lines to each state in the triplet). Since the quintuplet and the triplet are each degenerate, is the same for all 9 quintuplet-triplet transitions. We see a single spectral line from 9 transitions.
Now turn on the magnetic field. The degeneracy is lifted - the energy levels in the quintuplet separate slightly, as do those in the triplet (see figures2). Group the 9 quintuplet-triplet transitions into three groups of three, according to the value of . Each group has the same , and so 9 transitions give rise to three spectral lines.
For the anomalous Zeeman effect, different selection rules apply:
.
.
and fade from the picture. We will search for them again when we come to the Paschen-Back effect. ( and are still with us, though.)
We want to plug this into formula2 for the anomalous Zeeman . We don't need the full complexity of formula2; the following is enough:
(4)
The function contains the complicated fraction we had before, plus constants (like the magnetic field strength ). Consider a transition from a state to a state. Plugging in the selection rules, we get:
(5)
As before, we first turn the field off, then on. Without the field, the energy of a state does not depend on , and so each state belongs to a degenerate multiplet. We label the multiplet with the quantum numbers . Without the field, all transitions from one multiplet to another have the same -- shining out one single spectral line.
Turn the field on, and this overloaded spectral line splits according to the different values of . In contrast to normal Zeeman splitting, transitions with the same generally don't have the same . Count the spectral lines, and you've counted the transitions (apart from accidental coincidences, which are rare).
Figure 3: Sodium D-lines
Figure 4: Zeeman Effect for Sodium
For example, consider sodium, a historically important case. Sodium has two bright yellow spectral lines, known as D-lines (D and D, close together).11 In a weak magnetic field, the D line splits in four, and the D line splits in six.
``Scarcely credible!'' wrote Lorentz of this discovery. Each D-line splits into an even number of lines. It would seem that some multiplet must contain an even number of states. But there are different 's for each value of ! Heisenberg and Landé unflinchingly ascribed half-integer values to .
Here are the details. The D-lines come from three multiplets: a quartet and two doublets (see figure3). The pair of quantum numbers specify a multiplet, as can be seen from the figure- is for all eight states. The D-lines are indicated by the slanted lines. The D-line actually encompasses all transitions from the higher to the lower doublet; the D-line encompasses all transitions from the quartet to the lower doublet.
Applying the selection rule for , we find that all four conceivable D transitions are permitted; six out of the eight conceivable D transitions are permitted. Turn on the field, and these all separate. (See figure4. To avoid cluttering the figure, the D-transitions are only partly drawn.) Hence the observed splitting.
I note finally that sodium is a hydrogenic'' atom. It has 11 electrons, one more than the noble gas neon. Ten electrons form a relatively inertcore''-- they make up two complete ``shells'', which contribute collectively zero spin and zero orbital angular momentum (all electron spins are paired in the core.) The remaining electron is called a valence electron, and is far more loosely bound. A good approximation treats sodium as a system with a positively charged, immobile core, and a single orbiting electron-- rather like hydrogen. The eight states we just discussed are (to a high degree of approximation) eight different states of the valence electron (which is why they all have ).
Where do the selection rules come from? Here is one way to view a transition:
atom atom + photon
where atomImage 69: $_{\rm initial}$'' andatom'' refer of course to the initial and final states of the atom. , so our rules for adding angular momenta say that
in steps of 1, or in other words, for the atom.
We must supplement the catalog of spin facts to get the rule for . Our first ``spin fact'' told us that if we have a composite system:
then , in steps of 1. We now add a new requirment:
in steps of 1.
You may be wondering how to interpret these equations. What does it mean to say ? Modern quantum mechanics gives a clear answer: tensor product. (So would be a better notation than .) But a muddled view hews closer to history. The old quantum theory did offer derivations of the selection rules, via a strange brew of Maxwell's equations and other ingredients. On the other hand, the Paschen-Back effect presented a severe conceptual stumbling block. How does one set of selection rules fade away, and the other take over? Heisenberg discarded the derivations when it suited his purpose, a matter of some controversy at the time (see Cassidy's biography). So I leave a coherent discussion to the textbooks, and continue with cruder arguments.
To get the selection rules for and , we shift gears, and treat light as an electromagnetic wave, not photons. Also, we consider the reverse process, where an atomic state absorbs energy from incident light. An electron is so small (point-like, so far as we know today) that it feels at any moment a uniform electric field, and a uniform magnetic field. Classical physics says that magnetic effects are much smaller than electric effects, so we will ignore the former. (In modern parlance, we are ``deriving'' the selection rule for electric dipole transitions, and ignoring magnetic dipole transitions.)
The uniform electric field can shove the electron this way or that, but cannot flip it over. So S is unaffected, and hence . On the other hand, if the wavelength of light is short enough, the electric field will be shoving the electron one way in one part of its orbit, and another way for another part, and so may change its orbital angular momentum L.
. We've just seen that S doesn't change, so L and J must change the same way. So the selection rules for and must be the same as those for and .
Next:Magnetic MomentsUp:SpinPrevious:Angular Momentum Quantum Numbers
© 2001 Michael Weiss
home |
189298 | https://www.ajnr.org/content/ajnr/30/8/1623.full.pdf | of September 28, 2025.
This information is current as Vertigo and Hearing Loss F.J. Wippold II and P.A. Turski 2009, 30 (8) 1623-1625 AJNR Am J Neuroradiol ACR APPROPRIATENESS CRITERIA Vertigo and Hearing Loss F.J. Wippold II P.A. Turski Dizziness and Vertigo Dizziness and vertigo (Table 1) are common clinical com-plaints. Vertigo is caused by a disturbed vestibular system and is subdivided into peripheral vertigo (due to failure of the end organs) or central vertigo (due to failure of the vestibular nerves or central connections to the brainstem and cerebellum).1-5 Benign Positional Vertigo, Me ´nie re Disease, and Peripheral Vestibular Disorders Patients with benign positional vertigo rarely demonstrate im-aging findings.2,4 Me ´niere disease manifests as paroxysmal at-tacks of whirling vertigo due to failure of regulation of en-dolymph. CT or MR imaging, or both, may be used to evaluate the vestibular aqueduct, endolymphatic duct, and sac and to rule out associated infectious or neoplastic disease.4-12 Vestibular neuritis and labyrinthitis may also cause vertigo.
Labyrinthitis is usually viral in origin with few sequelae; how-ever, bacterial labyrinthitis may progress to partial or com-plete occlusion of the lumen of the affected labyrinth, detect-able on MR imaging as loss of the signal intensity of the fluid contents.3,4 Progressive labyrinthitis obliterans may be diag-nosed on high-resolution CT.13 Gadolinium enhancement of the labyrinthine structures or vestibular nerves may also occur and should not be mistaken for hemorrhage.14-16 Superior semicircular canal dehiscence, another cause of vertigo, can be diagnosed by high-resolution coronal CT im-aging of the temporal bones.17-19 Diseases of the internal au-ditory canal and cerebellopontine angle, such as tumors, are readily evaluated with CT and MR imaging techniques.
Central Vestibular Disorders Central lesions of the brainstem or cerebellum that result in central vertigo can be readily diagnosed by MR imaging. Pos-terior fossa vascular disorders may be evaluated with MR an-giography or conventional angiography of the posterior fossa vasculature.3,20,21 Cervical spondylosis, which causes vertigo by compressive osteophyte formation, may be evaluated with CT.3,20,21 Sensorineural Hearing Loss Sensorineural hearing loss (SNHL) results from the pathologic changes of inner ear structures such as the cochlea or the au-ditory nerve1 and is best evaluated with gadolinium-enhanced MR imaging.22-25 Patients with fluctuating SNHL may have congenitally en-larged vestibular aqueducts (apertures greater than 4 mm) de-tected by either CT or MR imaging.26-29 The imaging findings must be correlated with audiometry.27,28 Initial evaluation of symmetric or unilateral SNHL re-quires determination of whether the site of the lesion is cochlear30 or retrocochlear.31 Following preliminary au-diometric or auditory brain response testing, patients with retrocochlear localization should have a complete MR im-aging study of the head to include the internal auditory canal, temporal bones, central nuclei in the brainstem, and the auditory pathways extending upward into the cerebral hemispheres.22,23,32-34 Gadolinium contrast enhancement may be used. CT is sometimes diagnostic in lesions 1.5 cm or greater in diameter when dedicated techniques are used, but it does not readily detect small brainstem lesions such as infarctions or demyelination.33-40 In general, most cochlear disorders such as otosclerosis are evaluated by high-resolution CT imaging. Similarly, preoper-ative assessment for cochlear implants is usually best accom-plished by using thin-section CT with reformatted multipla-nar images. In patients with congenital etiologies for hearing loss, recent reports suggest that high-resolution MR imaging is more useful for surgical planning.41,42 Conductive Hearing Loss Conductive hearing loss results from pathologic changes of either the external or middle ear structures and is best eval-uated with CT. Indications include suspected complica-tions of acute and chronic otomastoiditis, such as cho-lesteatoma, and the assessment of congenital or vascular anomalies. Fistulization through the tegmen tympani of the temporal bone is usually detected by CT, though the actual involvement of the meninges and veins is better assessed by MR imaging. MR imaging is also indicated when compli-cated inflammatory lesions are suspected to extend into the inner ear or toward the sigmoid sinus or jugular vein. Neo-plasms arising from or extending into the middle ear re-quire the use of both techniques, as their combined data provide essential information. Vascular imaging should be performed when there is suspicion of a paraganglioma ex-tending into the middle ear.43 Trauma CT is used extensively to delineate fractures, ossicular disloca-tions, fistulous communications, and facial nerve injury and to evaluate post-traumatic hearing loss.44 This article is a summary of the complete version of this topic, which is available on the ACR Website at www.acr.org/ac. Practitioners are encouraged to refer to the complete version.
Reprinted with permission of the American College of Radiology.
Please address correspondence to Franz J Wippold II, MD, FACR, Neuroradiology Section, Mallinckrodt Institute of Radiology, 510 S Kingshighway Blvd, St. Louis, MO 63110-1076; e-mail: wippold@mir.wustl.edu; or Patrick A Turski, MD, FACR, Department of Radiology, University of Wisconsin Hospital, E1/398, 600 Highland Ave, Madison, WI 53792-0001; e-mail: pturski@uwhealth.org ACR CRITERIA AJNR Am J Neuroradiol 30:1623–25 Sep 2009 www.ajnr.org 1623 Congenital and Childhood Hearing Loss The ideal imaging method for children with unilateral or asymmetric sensory neural hearing loss is still controversial.
Most reports suggest that children with unilateral or asym-metric sensory neural hearing loss should have a high-resolu-tion temporal bone CT scan and that brain and temporal bone MR imaging be obtained in select cases. In general high-reso-lution CT has been shown to be efficacious for the preopera-tive work-up for congenital hearing loss due to aural dysplasia, congenital ossicular anomalies, large vestibular aqueduct syn-drome, congenital absence of cochlear nerve, and labyrinthitis ossificans.45-54 Review Information This guideline was originally developed in 1996. The last re-view and update was completed in 2008.
Appendix Expert Panel on Neurologic Imaging: Franz J. Wippold II, MD, Co-Author and Panel Chair; Patrick A. Turski, MD, Co-Author; Rebecca S. Cornelius, MD; James A. Brunberg, MD; Patricia C. Davis, MD; Robert L. De La Paz, MD; Pr. Didier Dormont; Linda Gray, MD; John E. Jordan, MD; Suresh Ku-mar Mukherji, MD; David J. Seidenwurm, MD; Robert D.
Zimmerman, MD; Brian Nussenbaum, MD, American Acad-emy of Otolaryngology; Michael A. Sloan, MD, MS, American Academy of Neurology.
References 1. Bagai A, Thavendiranathan P, Detsky AS. Does this patient have hearing im-pairment? JAMA 2006;295:416–28 2. McGee SR.
Dizzy patients: diagnosis and treatment.
West J Med 1995;162:37–42 3. Phelps PD, Lloyd GA. Radiology of vertigo. In: Phelps PD, Lloyd GA, eds.
Radiology of the Ear. St. Louis: Blackwell Scientific Publications; 1983:137–41 4. Dickins JR, Graham SS. Evaluation of the dizzy patient. Ear Hear 1986;7:133–37 5. Macleod D, McAuley D. Vertigo: clinical assessment and diagnosis. Br J Hosp Med (Lond) 2008;69:330–34 6. Albers FW, Van Weissenbruch R, Casselman JW. 3DFT-magnetic resonance imaging of the inner ear in Meniere’s disease.
Acta Otolaryngol 1994;114:595–600 7. Kraus EM, Dubois PJ. Tomography of the vestibular aqueduct in ear disease.
Arch Otolaryngol 1979;105:91–98 8. Lorenzi MC, Bento RF, Daniel MM, et al. Magnetic resonance imaging of the temporal bone in patients with Meniere’s disease. Acta Otolaryngol 2000;120:615–59 9. Mateijsen DJ, Van Hengel PW, Krikke AP, et al. Three-dimensional Fourier transformation constructive interference in steady state magnetic resonance imaging of the inner ear in patients with unilateral and bilateral Meniere’s disease. Otol Neurotol 2002;23:208–13 10. Nakashima T, Naganawa S, Sugiura M, et al. Visualization of endolymphatic hydrops in patients with Meniere’s disease. Laryngoscope 2007;117:415–20 11. Sajjadi H, Paparella MM. Meniere’s disease. Lancet 2008;372:406–14 12. Xenellis J, Vlahos L, Papadopoulos A, et al. Role of the new imaging modalities in the investigation of Meniere’s disease. Otolaryngol Head Neck Surg 2000;123:114–19 13. Hasso AN, Ledington JA. Imaging modalities for the study of the temporal bone. Otolaryngol Clin North Am 1988;21:219–44 14. Mark AS, Seltzer S, Nelson-Drake J, et al. Labyrinthine enhancement on gado-linium-enhanced magnetic resonance imaging in sudden deafness and vertigo:correlationwithaudiologicandelectronystagmographicstudies.Ann Otol Rhinol Laryngol 1992;101:459–64 15. Seltzer S, Mark AS. Contrast enhancement of the labyrinth on MR scans in patients with sudden hearing loss and vertigo: evidence of labyrinthine dis-ease. AJNR Am J Neuroradiol 1991;12:13–16 16. Weissman JL, Curtin HD, Hirsch BE, et al. High signal from the otic labyrinth on unenhanced magnetic resonance imaging. AJNR Am J Neuroradiol 1992;13:1183–87 17. Belden CJ, Weg N, Minor LB, et al. CT evaluation of bone dehiscence of the superior semicircular canal as a cause of sound- and/or pressure-induced ver-tigo. Radiology 2003;226:337–43 18. Curtin HD. Superior semicircular canal dehiscence syndrome and multi-de-tector row CT. Radiology 2003;226:312–14 19. Mong A, Loevner LA, Solomon D, et al. Sound- and pressure-induced vertigo associated with dehiscence of the roof of the superior semicircular canal.
AJNR Am J Neuroradiol 1999;20:1973–75 20. Kikuchi S, Kaga K, Yamasoba T, et al. Slow blood flow of the vertebrobasilar system in patients with dizziness and vertigo.
Acta Otolaryngol 1993;113:257–60 21. Norrving B, Magnusson M, Holtas S. Isolated acute vertigo in the elderly: ves-tibular or vascular disease? Acta Neurol Scand 1995;91:43–48 22. Busaba NY, Rauch SD. Significance of auditory brain stem response and gad-olinium-enhanced magnetic resonance imaging for idiopathic sudden senso-rineural hearing loss. Otolaryngol Head Neck Surg 1995;113:271–75 23. Hendrix RA, DeDio RM, Sclafani AP. The use of diagnostic testing in asym-metric sensorineural hearing loss. Otolaryngol Head Neck Surg 1990;103: 593–98 24. Kano K, Tono T, Ushisako Y, et al. Magnetic resonance imaging in patients with sudden deafness. Acta Otolaryngol Suppl 1994;514:32–36 25. Weber PC, Zbar RI, Gantz BJ. Appropriateness of magnetic resonance imaging in sudden sensorineural hearing loss.
Otolaryngol Head Neck Surg 1997;116:153–56 26. Davidson HC, Harnsberger HR, Lemmerling MM, et al. MR evaluation of ves-Rating of Techniques: Clinical condition—vertigo and hearing loss MRI Head and Internal Auditory Canal without and with Contrast MRI Head and Internal Auditory Canal without Contrast CT Temporal Bone without Contrast CT Head without and with Contrast MRA Head with or without Contrast CTA Head Sensorineural hearing loss, acute and intermittent vertigo 8 7 6† 3 N/A N/A Sensorineural hearing loss, no vertigo 8 7 5 4 N/A N/A Conductive hearing loss, rule out petrous bone abnormality 3 3‡ 8 3 N/A N/A Total deafness, cochlear implant candidate, surgical planning 5 5 9 3 N/A N/A Fluctuating hearing loss, history of meningitis or to rule out congenital anomaly 7 7 8 4 N/A N/A Episodic vertigo, new onset (hours to days) 7 6 4 5 6 5 Vertigo, no hearing loss, normal findings on neurologic examination 8 7 5 4 N/A N/A Note:—MRI indicates MR imaging; MRA, MR angiography; CTA, CT angiography; N/A, not rated.
Appropriateness criteria scale from 1 to 9: 1 indicates least appropriate; 9, most appropriate.
† For possible cholesteatoma with labyrinthine fistula.
‡ MR imaging is superior to CT for the detection of dural invasion and extradural extension.
1624 Wippold AJNR 30 Sep 2009 www.ajnr.org tibulocochlear anomalies associated with large endolymphatic duct and sac.
AJNR Am J Neuroradiol 1999;20:1435–41 27. Mafee MF, Charletta D, Kumar A, et al. Large vestibular aqueduct and congen-ital sensorineural hearing loss. AJNR Am J Neuroradiol 1992;13:805–19 28. Okumura T, Takahashi H, Honjo I, et al. Sensorineural hearing loss in patients with large vestibular aqueduct. Laryngoscope 1995;105:289–93, discussion 93–44 29. Valvassori GE, Clemis JD. The large vestibular aqueduct syndrome. Laryngo-scope 1978;88:723–28 30. Hegarty JL, Patel S, Fischbein N, et al. The value of enhanced magnetic reso-nance imaging in the evaluation of endocochlear disease. Laryngoscope 2002;112:8–17 31. Davidson HC. Imaging evaluation of sensorineural hearing loss. Semin Ultra-sound CT MR 2001;22:229–49 32. Cueva RA. Auditory brainstem response versus magnetic resonance imaging for the evaluation of asymmetric sensorineural hearing loss. Laryngoscope 2004;114:1686–92 33. Selesnick SH, Jackler RK, Pitts LW. The changing clinical presentation of acoustic tumors in the MRI era. Laryngoscope 1993;103:431–36 34. Gebarski SS, Tucci DL, Telian SA. The cochlear nuclear complex: MR location and abnormalities. AJNR Am J Neuroradiol 1993;14:1311–18 35. Daniels RL, Swallow C, Shelton C, et al. Causes of unilateral sensorineural hearing loss screened by high-resolution fast spin echo magnetic resonance imaging: review of 1,070 consecutive cases. Am J Otol 2000;21:173–80 36. Kocaoglu M, Bulakbasi N, Ucoz T, et al. Comparison of contrast-enhanced T1-weighted and 3D constructive interference in steady state images for pre-dicting outcome after hearing-preservation surgery for vestibular schwan-noma. Neuroradiology 2003;45:476–81 37. Kwan TL, Tang KW, Pak KK, et al. Screening for vestibular schwannoma by magnetic resonance imaging: analysis of 1821 patients. Hong Kong Med J 2004;10:38–43 38. Somers T, Casselman J, de Ceulaer G, et al. Prognostic value of magnetic reso-nance imaging findings in hearing preservation surgery for vestibular schwannoma. Otol Neurotol 2001;22:87–94 39. Swartz JD. Lesions of the cerebellopontine angle and internal auditory canal: diagnosis and differential diagnosis. Semin Ultrasound CT MR 2004;25:332–52 40. ZealleyIA,CooperRC,CliffordKM,etal.MRIscreeningforacousticneuroma: a comparison of fast spin echo and contrast enhanced imaging in 1233 pa-tients. Br J Radiol 2000;73:242–47 41. Parry DA, Booth T, Roland PS. Advantages of magnetic resonance imaging over computed tomography in preoperative evaluation of pediatric cochlear implant candidates. Otol Neurotol 2005;26:976–82 42. Rauch SD. Clinical practice: idiopathic sudden sensorineural hearing loss.
N Engl J Med 2008;359:833–40 43. Maroldi R, Farina D, Palvarini L, et al. Computed tomography and magnetic resonance imaging of pathologic conditions of the middle ear. Eur J Radiol 2001;40:78–93 44. Swartz JD. Temporal bone trauma. Semin Ultrasound CT MR 2001; 22:219–28 45. Bamiou DE, Phelps P, Sirimanna T. Temporal bone computed tomography findings in bilateral sensorineural hearing loss.
Arch Dis Child 2000;82:257–60 46. Glastonbury CM, Davidson HC, Harnsberger HR, et al. Imaging findings of cochlear nerve deficiency. AJNR Am J Neuroradiol 2002;23:635–43 47. Mafee MF. Congenital sensorineural hearing loss and enlarged endolym-phatic sac and duct: role of magnetic resonance imaging and computed to-mography. Top Magn Reson Imaging 2000;11:10–24 48. McClay JE, Tandy R, Grundfast K, et al. Major and minor temporal bone ab-normalities in children with and without congenital sensorineural hearing loss. Arch Otolaryngol Head Neck Surg 2002;128:664–71 49. Morzaria S, Westerberg BD, Kozak FK. Evidence-based algorithm for the eval-uation of a child with bilateral sensorineural hearing loss. J Otolaryngol 2005;34:297–303 50. Robson CD. Congenital hearing impairment. Pediatr Radiol 2006;36:309–24 51. Simons JP, Mandell DL, Arjmand EM. Computed tomography and magnetic resonance imaging in pediatric unilateral and asymmetric sensorineural hearing loss. Arch Otolaryngol Head Neck Surg 2006;132:186–92 52. Tan TY, Goh JP. Imaging of congenital middle ear deafness. Ann Acad Med Singapore 2003;32:495–99 53. Westerhof JP, Rademaker J, Weber BP, et al. Congenital malformations of the inner ear and the vestibulocochlear nerve in children with sensorineural hearing loss: evaluation with CT and MRI. J Comput Assist Tomogr 2001;25:719–26 54. Yuen HY, Ahuja AT, Wong KT, et al. Computed tomography of common congenital lesions of the temporal bone. Clin Radiol 2003;58:687–93 AJNR Am J Neuroradiol 30:1623–25 Sep 2009 www.ajnr.org 1625 |
189299 | https://www.uobabylon.edu.iq/eprints/publication_10_10772_6130.pdf | 3333333333333333311 HARMONIC FUNCTIONS 2 HARMONIC FUNCTIONS 1.1 INTRODUCTION Harmonic function is a mathematical function of two variables having the property that its value at any point is equal to the average of its values along any circle around that point, provided the function is defined within the circle. An infinite number of points are involved in this average, so that it must be found by means of an integral, which represents an infinite sum. In physical situations, harmonic functions describe those conditions of equilibrium such as the temperature or electrical charge distribution over a region in which the value at each point remains constant. Harmonic functions can also be defined as functions that satisfy Laplace’s equation, a condition that can be shown to be equivalent to the first definition. The surface defined by a harmonic function has zero convexity, and these functions thus have the important property that they have no maximum or minimum values inside the region in which they are defined. 1.2 HARMONIC FUNCTIONS Areal-valued functions H of two real variables x and y is said to be harmonic in a given domain of the xy plane if, through that the domain, it has continuous partial derivatives of the first and second order and satisfies the partial differential equations, ( , ) ( , ) 0 xx yy H x y H x y knows as Laplace's equation. Theorem 1.2.1 If a function , , f z u x y iv x y is analytic in a domain , then its component functions u and v are harmonic in . 3 Proof: Assuming that is analytic in , we start with the observation that the first order partial derivatives of its component functions must satisfy the Cauchy-Riemann equations throughout . x y u v and .
y x u v Differentiating both sides of these equations with respect to x we have: xx xy u v , .
xy xx u v The continuity of the partial derivatives of u and v ensures that yx xy u u and .
yx xy v v Therefore 0 xx yy u u and 0.
xx yy v v Example 1.2.2 Prove that ( sin cos ) x u e x y y y is harmonic. Proof: (sin ) ( )( sin cos ) x x u e y e x y y y x sin x e y sin cos x x xe y ye y 2 2 ( sin sin cos ) x x x u e y xe y ye y x x 2 sin sin cos x x x e y xe y ye y ---------------------------- (1) ( cos sin cos ) x u e x y y y y y cos sin cos x x x xe y ye y e y 2 2 ( cos sin cos ) x x x u xe y ye y e y y y = sin 2 sin cos x x x xe y e y ye y ------------------------------ (2). Adding (1) and (2) yield 2 2 u x + 2 2 u y = 0. Hence u is harmonic. 4 1.3 HARMONIC CONJUGATE A function ( , ) u x y defined on some open domain 2 R is said to have as a conjugate function ( , ) v x y if and only if they are respectively real and imaginary part of a holomorophic function of the complex variable z x iy That is, is conjugated to if = ( , ) u x y + ( , ) iv x y is holomorphic on Ω. As a first consequence of the definition, they are both harmonic real-valued functions on Ω. Moreover, the conjugate of , if it exists, is unique up to an additive constant. Also, is conjugate to if and only if is conjugate to Equivalently, is conjugate to in Ω if and only if and satisfy the Cauchy–Riemann equations in Ω. As an immediate consequence of the latter equivalent definition, if is any harmonic function on 2, the function uy is conjugate to − ux, for then the Cauchy– Riemann equations are just Δ and the symmetry of the mixed second order derivatives, uxy = uyx. Therefore an harmonic function admits a conjugated harmonic function if and only if the holomorphic function g(z): = ux(x,y) − iuy(x,y) has a primitive in Ω, in which case a conjugate of is, of course, ( ).
imf x iy So any harmonic function always admits a conjugate function whenever its domain is simply connected, and in any case it admits a conjugate locally at any point of its domain. There is an operator taking a harmonic function on a simply connected region in 2 to its harmonic conjugate (putting e.g. v( 0 x ) = 0 on a given 0 x in order to fix the indeterminacy of the conjugate up to constants). Conjugate harmonic functions (and the transform between them) are also one of the simplest examples of a Bäcklund transform (two PDEs and a transform relating their solutions), in this case linear; more complex transforms are of interest in solution and integral systems. Geometrically and are related as having orthogonal trajectories, away from the zeroes of the underlying holomorphic function; the contours on which and are constant cross at right angles. In this regard, would be the complex potential, where is the potential function and is the stream function. 5 Example 1.3.1 Consider the function ( , ) sin x u x y e y Solution: Since sin x u e y x , 2 2 sin x u e y x and cos x u e y y , 2 2 sin x u e y y It satisfies 0.
u And thus is harmonic. Now suppose we have a v( , ) x y such that the Cauchy–Riemann equations are satisfied: sin , x u v e y x y and cos x u v e y y x Simplifying, sin , x v e y y and cos x v e y x which when solved gives cos .
x v e y Observe that if the functions related to and v was interchanged, the functions would not be harmonic conjugates, since the minus sign in the Cauchy–Riemann equations makes the relationship asymmetric. The conformal mapping property of analytic functions (at points where the derivative is not zero) gives rise to a geometric property of harmonic conjugates. Clearly the harmonic conjugate of x is y, and the lines of constant x and constant y are orthogonal. Co formality says that equally contours of constant ( , ) u x y and ( , ) v x y will also be orthogonal where they cross (away from the zeroes of That means that is a specific solution of the orthogonal trajectory problem for the family of contours given by (not the only solution, naturally, since we can take also functions of ): the question, going back to the mathematics of the seventeenth century, of finding the curves that cross a given family of non-intersecting curves at right angles. There is an additional occurrence of the term harmonic conjugate in mathematics, and more specifically in geometry. Two points and are said to be 6 harmonic conjugates of each other with respect to another pair of points , if , where is the cross-ratio of points 1.4 The Maxim Principle And The Mean Value Property The Mean Value Property: suppose that if u :U R is harmonic functions on an open set U C and that is a point 0 max ( ) z U p u z ,then u is a constant (0,1) D U for some 0.
r Then 2 0 1 ( ) ( ) 2 i u p u p re d . To understand why this result is true, let us simplify matters by assuming (with a simple translation of coordinates) that 0 p . Notice that if 0 k and ( ) k u z z then 2 0 1 ( ) ( ) 2 i u p u p re d ( ) p 2 0 1 2 k ik r e d = 0 = (0).
u The same hold when 0 k by a similar calculation. So that is the mean value property for power of . The Maximum Principle For Harmonic Functions: If : u U R is a real –valued, harmonic function on a connected open set U and if there is a point 0 , p U with the property that (0,1) a D = min ( ) Q U u Q , then u is a constant on .
U The Minimum Principle Of Harmonic Functions: If : u U R is a real-valued, harmonic function on a connected open set U C and if there is point 0 p U such that 0 ( ) U p min ( ) Q U u Q , then u is consistent on .
U 1.5 The Poisson Integral Formula The next result shows how to calculate a harmonic function on the disc from its “boundary values” that is its values on the circle that bounds the disc. Let u be a harmonic functions on a neighborhood of (0,1) D then, for any point (0,1) a D .Then 2 2 2 0 1 1 ( ) ( ).
.
2 i i a u p u re d a e 7 1.6 HARMONIC FUNCTION ON DISK Before studying harmonic functions in the large it is necessary to study then locally. That is we must study these functions on disk. The plane is to study harmonic functions on the open unit disk : 1 .
z z What means unit disk? We learning the meaning unit disk, the open unit disk around (where is a given point in the plane) is the set of a point whose distance from is less then 1. 1( ) D p : 1 .
Q p Q The closed unit disk around is the set of points whose distant from is less then or equal to one . ( ) D p : 1 .
Q p Q Poisson Kernel: Is important in a complex analysis because its integral against a function defined on a unit circle. The Poisson kernel commonly finds applications in control theory and two – dimensional problems in electrostatics. The definition of Poisson kernels are often extended to n - dimension . Definition 1.6.1 let be the unit disc In the complex plane the Poisson kernel for the unit disc is given by This can be thought of in two ways: either as a function of r and , or as a family of functions of indexed by If is the open unit disc in , is the boundary of the disc, and a function on T that lies in L1 then the function u given by 8 is harmonic in and has a radial limit that agrees with almost everywhere on the boundary of the disc. Theorem 1.6.2 The function for 1 o r and is a poisson kernal. Proof: Let i z re , 1, o r then 2 1 (1 )(1 ...) 1 i i re z z z re = 1 1 2 n n z Hence 1 Re 1 i i re re = 1 1 2 cos n n r n ( ) p 2 2 1 1 1 1 i i i i i re re re r re re so that also ( ) r p 2 2 1 1 2 cos r r r = 1 Re .
1 i i re re Theorem 1.6.3 Let : 1 D z z and suppose that D is continuous function. Then there is a continuous function : u D R such that: (a) ( ) ( ) u z f z for z in , D (b) u is harmonic in D more over u is unique and is defined by the formula: 1 ( ) ( ) ( ) 2 i i r u re p t f e dt for 0 1 r and 0 2 .
Proof: Define : u D R by letting ( ) i u re in the above. If 0 1 r and letting 1 1 ( ) n in in n r e e ( ) n in r n p r e 9 ( ) = ( ) i i u re f re (i)u is harmonic in d. if 0 1 r then 1 1 ( ) Re ( ) 2 1 i i i i re u re f e dt re 2 ( ) ( ) 2 1 1 ( )Re 2 1 i t i i t re R f e dt re 2 2 1 ( ) 2 i i i i i e re R f e dt e re 2 2 1 ( ) .
2 it i it e z f e dt e z Hence u is continuous on .
D Harnacks inequality: In mathematics, Harnacks inequality is an in equality relating the values of positive harmonic function at two points introduced by A. Harnack (1887) generalized Harnacks inequaliy to solution of elliptic or parabolic partial differential Equation. Harnacks inequality is used to prove Harnacks theorem about the convergence of sequences of harmonic functions. Let = 0 ( , ) D z R be an open disk in the plane and let be a harmonic functions on , such that ( ) f z is non-negative for all .
z D Then the following inequality holds for all , z D 2 0 0 0 ( ) ( ), R f z f z R z z for general domain in n R the inequality can be stated as follows: if is abounded domain with , then there is a constant such that:sup ( ) inf ( ) x x u x C u x . For every twice differentiable, harmonic and non-negative functions ( ).
u x The constant is independent of .
u In complex analysis, Harnacks principle or Harnacks theorem about the convergent of sequences of harmonic functions that follow from Harnacks inequality. 10 If the function 1 2 ( ), ( ),...
u z u z are harmonic in an 1 2 ( ) ( ),...
u z u z every point of , then the limit, lim ( ) n n u z either is an infinite in every point of the domain or it is infinite in every point of the domain, in both cases uniformly in each compact subset of . Weierstrass approximation theorem: states that every continues functions defined on an interval , a b can be uniformly approximated as closely as desired by a polynomial functions. The statement of the approximation theorem as originally discovered by the Weierstrass is as follows: suppose is a continuous complex–valued function defined on the real interval , , a b for every 0 there exists a polynomial functions over such that for all in , , a b we have ( ) ( ) f x p x or equivalent, the supermom norm ( ) .
f x p If f is real valued, the polynomial function can be taken over . Definition 1.6.4 Any continuous functions on abounded interval can be uniformly approximated by polynomial functions. Theorem 1.6.5 Let U C be any open set. let 1 2 , ,...
a a be a finite or infinite sequence in U (possibly with repetitions) that has no accumulation point in .
U Then there exists a holomorphic function f onU whose zero set is precisely( ).
j a The function f is constructed by taking an infinite product. The proof converts the problem to a situation on the entire plan, and then uses the Weierstrass product. We next want to formulate a result about maximal domains of existence (or domains of definition) of holomorphic functions. But first we need a geometric fact about open subset of the plane. 11 1.7 SUBHARMONIC FUNCTIONS AND SUPERHARMONIC FUNCTIONS What is a sub harmonic function?. Subharmonic functions Areal- valued function u whose domains a two-dimensional domain D is subharmonic in a D provided u satisfies the following conditions in : D 1- ( , ) u x y where ( , ) u x y . 2- u is upper semicontinuouse in . 3- For any sub domain ' D included together with its boundary ', B in D and for any function h harmonic in ' D continuous in ' ' D B , and satisfying ( , ) ( , ) h x y u x y on ' B we have ( , ) ( , ) h x y u x y in '.
D Subharmonic functions can be defined by the same description, only replacing“ no larger ” with“ no smaller”. Alternatively, a subharmonic function is just negative of a sub harmonic function, and for these reason and property of subharmonic functions can be easily transferred to super harmonic function. Formally, the definition can be stated as follows. Let be a subset of the n R and let : ( ) G R be an upper semi-continuous function. Then, is called sub harmonic if for any closed ball ( , ) B x R of centre x and radius contained in and every real-valued continuous function h on ( , ) B x R that is harmonic in ( , ) B x r and satisfies ( ) ( ) x h x for all x on the boundary ( , ) B x r of ( , ) B x r we have ( ) ( ) x h x for all ( , ) x B x r . Note that by the above, the function which is identically −∞ is subharmonic, but some authors exclude this function by definition. Definition 1.7.1 Let G C be a region and let :G R be a continuous function is said to be subharmonic if whenever ( , ) D z r C (where ( , ) D z r is closed disc around z of radius r ) we have 2 0 1 ( ) ( ) 2 i z z re d and is said to be superharmonic if whenever ( , ) D z r G we have 12 2 0 1 ( ) ( ) .
2 i z z re d Intuitively what this means is that a subharmonic function is at any point no greater than the average of the values in a circle around that point. This implies that a non-constant subharmonic function does not achieve its maximum in a region (it would achieve it at the boundary if it is continuous). Similarly for a superharmonic function, but then a non-constant superharmonic function does not achieve its minimum in . It is also easy to see that is subharmonic if and only if − is superharmonic. Note that when equality always holds in the above equation then would in fact be a harmonic function. That is, when is both subharmonic and superharmonic, then is harmonic. A function ( ) ( , ) h z h x y is said to be subharmonic on adomain if it has the following properties: 1-( , ) h x y is defined continuous at every point of except possibly at a finite number of points or at points of a sequence ( , ) n n a b which has no limit points in , while every exceptional point ( , ).
n n a b Satisfies the relation ) ( , ) ( , lim ( , ) n n x y a b h x y which is used to define ( , ) n n h a b 2- The integral 2 0 0 1 ( ) 2 i h z e d exists for every point 0 z G and all less then 0 ( ) D z the distance between 0 z and the boundary of G. 3- The inequality 2 0 0 0 1 ( ) ( ) 2 i h z h z e d holds for every point 0 z G and all sufficiently small 0.
Theorem 1.7.2 Let be open inC and let u be use on , suppose that u is not identically on any connected component of . 13 1- If u is subharmonic on , and if a and 0 R are such that u then , ( ) ( )( ) a r u z P u z for all z D where , ( ) a r P u is the Poisson integral of u 2 0 1 ( ) ( Re ) 2 it u a u a dt 2- if, for every athere is 0 a R such that 2 0 1 ( ) ( Re ) 2 it u a u a dt for 0 a R R . Then u is subharmonic on . Proof: part1. Let 1 ( ) n n be a sequence of continuous functions on ( , ) D a R such that n u . Let , ( ) n a R n h p then n h is continuous on ( , ) D a R and harmonic ( , ) D a R and n n h u on ( , ).
D a R Hence, if u is harmonic, we have 2 , 0 1 ( ) ( ) ( , ) ( Re ) 2 it n a R n u z h z P z t a dt . Now , ( , ) 0 a r P z t and n u . Hence by the monotone convergence. Part2. Let U let h be continuous on U and harmonic on .
U Suppose that u h on U . Given 0 , every boundary point a U has neighborhood a D such that u h on a D . Let , z U and let V be an open set such that , z V V U and .
n a a U V D Hence ( ) ( ) sup ( ( ) ( )) .
w V u z h z u w h w Since z U and 0 are arbitrary, we have 0 u h onU and u is subharmonic. The Dirichlet Problem: Definition: A region is called a Dirichlet Region if the Dirichlet problem can be solved by , that is is a Dirichlet Region if for each continuous function : , f G R there is continuous function : u G R such that u is harmonic inG and ( ) ( ) u z f z for all z in G . Theorem 1.7.3 Let f be a continuous function on define 2 2 2 0 1 1 ( ).
, if (0,1) 2 ( ) e ( ), if (0,1) i i z f e d z D u z z z z D 14 Then u is continuous on (0,1) D and harmonic on (0,1) D . closely related to this result is the reproducing proper of the Poisson kernel: let u be continuous on (0,1) D and harmonic on Then, for (0,1) z D 2 2 0 1 1 ( ) ( ) .
2 i i z u z u e d z e Some of the basic questions in regard harmonic functions are related to the so-called Dirichlet problem, which we frame in the following ways: Given a region Z let p be a continuous real-valued function whose domain is .
D Determine a function (or all function) that is continuous in D Harmonic in , and whose restriction to D is .
p The Dirichlet Problem On A General Disc: A change variable shows that the result of remain true on a general disc. Let f be a continuous on ( , ).
D P R Define 2 2 2 0 1 ( ).
, if ( , ) 2 ( ) ( ) e ( ) , if ( , ) i i R z P f e d z D P R u z z P R z z D P R Then u is a continuous on ( , ) D P R and harmonic on ( , ).
D P R If instead u is harmonic on a neighborhood of ( , ) D P R then, for ( , ).
z D P R Application Of Conformal Mapping To The Dirichlet Problem: Let C be a domain whose boundary consists of finitely many smooth curves. The dirichlet problem which mathematical problem of interest in its own right, is the boundary value problem 0 u on u f on .
The way to think about this problem is a follows: a data a function. on the boundary of the domain is given to solve the corresponding Dirichlet problem, one seeks a continuous function on the closure of (that is, the union of andits boundary) such that u is harmonic on, and agrees with, on the boundary. We shall 15 now describe three distinct physical situations that are mathematically modeled by the Dirichlet problem. Heat Diffusion: I imagine that Is a thin plat of heat –conducting metal. The shape of is arbitrary (not necessarily a rectangle) see figure I A function ( , ) u x y describes the temperature at each point ( , ) x y in .
It is standard situation in engineering or physics to consider idealized heat sources or sinks that maintain specified (fixed) clause of u on certain parts of the boundary, other parts of the boundary are to be thermally insulated. One wants to find the steady states heat distribution on (that is, as to t ) that is determined by the given boundary condition. Figure I Electrostatic Potential: Now we describe a situation in electrostatics that is modeled by the boundary value. Imagine a long, bellow the cylinder made of a thin sheet of some conducting material, such as copper. split the cylinder lengthwise into two equal pieces figure II separate the two pieces with strips of insulating material. Now ground the upper of the semi-cylinder pieces to potential zero, and keep the lower piece at some non zero fixed potential. for simplicity in the present discussion, let us say that this last fixed potential is 1. 16 Figure II 1.8 CONCLUSION The title of this project is harmonic functions it has very interesting and help full for solve many problems are important in the areas of applied mathematics, engineering, and mathematical physics. Harmonic functions are used to solve problems involving steady state temperatures, two-dimensional electrostatics, and ideal fluid flow. We get some difficult from this title, so this title to facilitate for the students of the degree and graduate of the university to concept from this title of this project. Because this research consist basic, definition, examples, demonstrated a among this title of this project which Can make the reader to get some concepts and advantages from this title of this project. Finally we will encourage students to read this title from various books, because the concept of harmonic function, subharmonic function and subharmonic functions of the fundamental ideas. 17 REFERENCES A.I.Markushevich, 1965. Theory of function of a complex variable VolumII, 4 th Editions published in USA. James Ward Brown, 1976. Complex Variables and Applications, 6 th Edition published in USA. John B. Conway, 1973. functions of one Complex Variable, 2ndEdition published in USA by Springer. Verlagin C. Johnd D. Depre, 1969. Elements of complex analysis, 3rdEdition published in USA. Maurica Heins, 1968. Complex function of theory, 2nd Edition, published in the uK 1968 by academic press. Murray R. Spiegel, 1988. Complex variables, 5th Editions. Raghavan Nara simhan,1985. complex analysis is one Variable 2ndEdition, published in USA by Birkhauser Boston. Steven G. Krautz, 2008. A Guid to complex Variable, 6th Edition published in U.S.A by association of America. Steven G. Krautz, 2008. Complex Variables A physical applications, 5th Edition, published in U.S.A by taylor and Francic Group. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.