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[GAP Forum] Comparing real numbers
Frank Lübeck frank.luebeck at math.rwth-aachen.de
Tue Aug 22 17:36:37 BST 2017
On Mon, Aug 21, 2017 at 10:47:24AM -0400, Joey Iverson wrote:
> Dear GAP forum,
> I was recently caught off guard by the following:
> gap> Sqrt(2)<2;
> false
> Now that I've read the manual more carefully, I don't think this is a
> malfunction. Instead, it looks like GAP will always report irrational
> cyclotomics to be larger than rationals.
> Does anybody know of a workaround that compares real numbers with their
> usual ordering instead? Suppose we agree that E(n) = exp(2*pi*i/n).
> As a last resort, I suppose I could run GAP inside of SAGE and get
> numerical approximations of everything, but I would like to avoid that if
> possible.
> Thanks for any advice!
> Joey Iverson
> Research Associate
> University of Maryland, College Park
Dear Joey, dear Forum,
Comparisons of real cyclotomic numbers in the natural ordering of the reals
depend on the choice of embeddings of E(n) into the complex plane.
The only way I know to decide if a real cyclotomic number is positive is
indeed by using numerical approximations.
I have a small package
which can be used as a workaround.
It contains a function 'HasPositiveRealPartCyc' which decides what its name
suggests. It uses numerical approximations and a simple interval arithmetic
and automatically enlarges precision if needed. The function assumes the
suggested embedding E(n) = exp(2*pi*i/n).
Best regards,
/// Dr. Frank Lübeck, Lehrstuhl D für Mathematik, Pontdriesch 14/16,
\\\ 52062 Aachen, Germany
/// E-mail: Frank.Luebeck at Math.RWTH-Aachen.De
\\\ WWW: http://www.math.rwth-aachen.de/~Frank.Luebeck/
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Interactive physics simulator
Step is an interactive physical simulator. It allows you to explore the physical world through simulations. It works like this: you place some bodies on the scene, add some forces such as gravity or
springs, then click Simulate and Step shows you how your scene will evolve according to the laws of physics. You can change every property of the bodies/forces in your experiment (even during
simulation) and see how this will change evolution of the experiment. With Step you cannot only learn but feel how physics works!
* Classical mechanical simulation in two dimensions
* Particles, springs with damping, gravitational and coulomb forces
* Rigid bodies
* Collision detection (currently only discrete) and handling
* Soft (deformable) bodies simulated as user-editable particles-springs system, sound waves
* Molecular dynamics (currently using Lennard-Jones potential): gas and liquid, condensation and evaporation, calculation of macroscopic quantities and their variances
* Units conversion and expression calculation: you can enter something like "(2 days + 3 hours) * 80 km/h" and it will be accepted as distance value (requires libqalculate)
* Errors calculation and propagation: you can enter values like "1.3 ± 0.2" for any property and errors for all dependent properties will be calculated using statistical formulas
* Solver error estimation: errors introduced by the solver is calculated and added to user-entered errors
* Several different solvers: up to 8th order, explicit and implicit, with or without adaptive timestep (most of the solvers require GSL library)
* Controller tool to easily control properties during simulation (even with custom keyboard shortcuts)
* Tools to visualize results: graph, meter, tracer
* Context information for all objects, integrated wikipedia browser
* Collection of example experiments, more can be downloaded with KNewStuff
* Integrated tutorials. | {"url":"https://depot.haiku-os.org/__multipage/pkg/step/haikuports/haikuports_x86_64/22/04/0/-/1/x86_64?locale=pt","timestamp":"2024-11-06T05:48:36Z","content_type":"text/html","content_length":"6858","record_id":"<urn:uuid:698337d0-c99e-4968-b0a5-c4eb02063dd3>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00457.warc.gz"} |
M75 - Equatio (Premium Chrome extension)
What is it?
You may have heard of texthelp's literacy support tool Read&Write but did you know that they also have an education tool called Equatio which helps support the creation of accessible mathematical
content online? You may have already explored the free Google Chrome extension and if not, see our mini guide. For the purpose of this guide, we will focus on the Premium version of the Google Chrome
extension. Currently, UCL has a licence for the Premium version of Equatio for staff and students until July 2024.
While the free version of Equatio can help anyone create accessible maths, it lacks features such as prediction, Equatio Mobile integration, and the screenshot reader. Here is a handy overview of
what is included in the premium version compared to the free version.
Why use it?
Equatio can help you create maths expressions without having to use any code or programming languages. You can easily create formulas and equations through keyboard input, handwriting recognition or
voice recording. It is also compatible with LaTeX for more advanced users. In addition, it allows some graphing input.
Who can use it?
Teaching staff, students and anyone with a Google Chrome browser.
• Remember, the full Premium licence is only available until July 2024.
• You should always check your privacy and security settings for any Google Chrome extension.
• After you have installed it, click the Extensions icon and look for Equatio in the list of extensions, click the three dots to the right of it and choose Manage extension to check the various
settings and permissions.
• Google Chrome cannot prevent extensions from recording your browsing history. However, it is possible to browse in incognito mode in a different window. To disable this extension in Incognito
mode, make sure you unselect this option.
Getting started
1. Add the Equatio extension to Chrome.
2. Click the Extensions icon and look for Equatio in the list of extensions.
3. If it is there, click the pin to add it to your Chrome toolbar.
4. Log into UCL Moodle using the Chrome browser.
5. In order to gain access to the Premium Equatio features in Chrome to use with Moodle, you need to sign in using your ac.uk domain. Launch your Equatio extension and you will see the Equatio
toolbar appear at the bottom of your screen. Open the main Equatio menu in the far left, click on Switch User and you will then have the option to sign in with your microsoft email address (
ucl.ac.uk). You should now have full access to the Premium features.
6. Navigate to your Moodle course where you wish to use Equatio. You can create mathematical content in various ways with the Equatio extension then insert it into any Moodle text editor. Text
editors are available anywhere where text and media can be added on Moodle. This includes Text and Media areas (formerly Labels), Pages, Books, Discussion Forum messages, Assignment instructions,
Quiz questions etc.
When you launch the Equatio extension, a toolbar will appear at the bottom of the screen with 3 main sections:
1. Equatio menu - this menu enables you to customize your experience using Equatio’s options, access the Equatio Academy, open your mathspace dashboard, change users, and so on.
2. Equatio tools - these buttons are the main tools of Equatio, enabling you to create, insert, and read math, as well as inserting mathspaces and accessing Equatio’s STEM tools. When you click some
of these buttons, an input area opens above the toolbar. This is the Equation Editor, which you can use to build your math before inserting it into your document.
3. Action bar - there are up to three action buttons (Edit Math, Insert Math and Copy Math as) at the right of the Equation toolbar. Instead of using Insert Math to insert math into your document,
you can the Copy Math as button to export it to the clipboard in a variety of formats.
You can also explore the Equatio Interactive Toolbar. This interactive toolbar shows a summary for each button, along with video links so you can see each feature in action.
Setting Equatio Options
1. Open the Equatio menu in the bottom left and then click on Options
2. On the left of the window, click Math Options.
The Math Options page contains the following options:
1. Select your preferred Math font size. This applies to both Equatio’s editor and the size that math is inserted into your documents.
2. Select your preferred Language.
3. Equatio recommends that you leave the Speech Engine set to Automatic. While Clearspeak generally sounds more natural, some users may prefer Mathspeak, which is a widely-used standard for braille
4. Deselect any Prediction options that you do not use. As you type, Equatio offers the following types of suggestion:
• Math – for example, ^2 as you type squa…
• Chemistry – for example, Na as you type sodi…
• Formulas – for example, P = FA as you type pres…
If you don’t use a particular prediction type, disabling it will make Equatio’s prediction quicker and more likely to feature the suggestions that you want. For example, if you do not study
chemistry, you should disable the Chemistry prediction option. Equatio will no longer search for chemical names or formulas as you type.
On the left of the window, click Toolbar Options.
The Toolbar Options page contains entries for each toolbar button.To remove a button from your toolbar, click its toggle. Removing buttons that you don’t need makes the toolbar simpler and easier to
use. For example, if you do not use speech input, you should consider removing the Speech Input button.
Click Save to save changes and close the Options window.
Equation Editor
Click Equation Editor . The Equation Editor appears above the Equatio toolbar. It contains a single Math panel, which is a space for creating your math before adding it to your document.
To create simple math
1. Type into the equation editor
2. You can change change formatting such as the text colour
3. You can add to Favourites
4. You can clear the contents of the math pane
5. When ready, you can Copy Math As and choose HTML, go to source code view in the Moodle Text Editor, paste in the code then return to the editor view.
Five drawers are available under the More button:
1. General – buttons for bold, italic, underline, and strikethrough, as well as a shortcut for inserting a square root symbol.
2. Symbols – common symbols such as greater than and not equals. This drawer also contains most of the Greek alphabet.
3. Layout – layout forms such as vectors, matrices, column addition, and long division.
4. Formulas – library of scientific formulas.
5. My Favorites – any entries that you have added to your favorites are listed here.
You can use the Search bar below a drawer to search through its contents. If you know the name of the item you want, however, it is often quicker to use Equatio’s prediction; simply start typing the
item’s name in the Math panel and it will appear in the prediction list.
You can press the / key to create a fraction, for example 1 / 4. After typing a denominator, press the right arrow key ➜ to move the cursor out of the the fraction. Alternatively, you can start
typing fraction or over and accept the prediction to insert the framework for a fraction.
Creating multi-line math
If you enter a line of maths and press Ctrl+Shift+Enter, Equatio copies the current line to the line below. This makes it easy to work with more complex math without having to retype each line. Also,
the Math panel now has a new toolbar, containing buttons to align your math and to add new columns and rows. Click the Align (Relation) button to align the math using the equals sign on each line.
This can often be a neat way to align a solution.
Note: If you want to start a new line without copying the previous line, you can simply press Enter.
When you have finished typing, minimise the equation editor and put the cursor in the Moodle text editor then click Insert Math and then Save and return to course.
Using prediction
Open the equation editor:
• Type xsq and watch a predictive menu appear. To insert the first item in the prediction list, simply press Enter. If you want one of the other items, you can either click it with the mouse or use
the arrow keys to highlight it and then press Enter.
• Instead of using prediction, you could select the x^a (superscript) item from the More > Symbols This inserts an x^a, which you can use as the basis for x^2, x^3, and so on.
• To write the ÷ and √ symbols, start typing divide and square root (or just root) until you see the symbols appear in the prediction list.
• Alternatively, you can find them in the More > Symbols
Start typing the following into the Math panel and see what comes up in the prediction list:
• Quadratic
• velocity
• avogadro
• nitric
• magnesium
Equatio’s prediction tool knows a huge range of equations from math, physics, and other STEM subjects. It also predicts constants, expressions, chemical formulas, and more. For a list of Equatio’s
prediction items, go to http://text.help/EquatIOSpeech. If you want to type text without Equatio trying to make it into something else, you can use the Insert Text button. Remember that you can also
turn off prediction for Math, Formulas, or Chemistry in Equatio’s Math Options.
Handwriting recognition
Using handwriting recognition
Go to the text editor you wish to add content to, ensure the equation toolbar is visible and click the pen button for Handwriting Recognition:
• Handwriting panel – this is where you write your math, using a finger or stylus to write directly on your touchscreen. You can also use your mouse, graphics tablet, etc.
• Math panel – as you write math in the Handwriting panel, Equatio analyzes it and shows its results as “digital math” in the Math panel. This is the same Math panel that you’ve learned about
already, with all its usual functions remaining available.
For an overview of the main features for Handwriting Recognition, see the video tour: Equatio Feature Tour - Handwriting Recognition | Texthelp
Using context with handwritten math
Equatio will use positions, relative sizes, and whatever other information it can find to work out what you’re trying to write. For example, are you writing a cross or an x? A plus or a t?
Equatio continuously analyzes possibilities as you write and attempts to choose the correct one. If it picks the wrong one, it’s easy for you to correct it in the Math panel.
Speech input
Equatio’s speech input enables you to dictate maths into your computer’s microphone and have it converted automatically into digital math:
1. Click Speech Input
2. Speech panel – as you dictate maths, Equatio writes what it has heard here. If Equatio is not recognising your math correctly, this helps you to understand why.
3. Math panel – Equatio analyzes the maths in the Speech panel and shows its results as “digital math” in the Math panel. This is the same Math panel that you’ve learned about already, with all its
usual functions remaining available.
When you click Start Speech Input and dictate maths, note how the button changes to indicate that Equatio is recording, with three flashing green dots below it. Both the Speech panel and the Math
panel feature dynamic content, that is, their contents will change as you continue speaking. This is because Equatio builds up more context as you speak, so it has a better understanding of the math
that you are trying to dictate.
There are traffic light icons at the bottom right corner of the Equation Editor. When you are using Equatio’s speech input or handwriting recognition, these indicate how well Equatio can recognise
your math:
Green – Equatio can read and edit the math you are creating. No further action is required.
Yellow – Equatio can read the math that you are creating, but you may need to fine tune it using the LaTeX or Math panel.
Red – Equatio found an error in your math that is preventing it from rendering correctly. You need to fix it using the LaTeX or Math panel.
Hover your pointer over the icon to see more information.
For an overview of all the main features for Speech Input, see the video tour: Equatio Feature Tour - Speech Input | Texthelp
LaTeX Editor
LaTeX is a system that uses tags and markup to produce structured documents. It is widely used in academia for scientific documents and includes a powerful engine for writing math. Many students (and
teachers) learn LaTeX as part of their math, science, or engineering studies. Equatio includes a LaTeX editor so you can easily write and edit your math in LaTeX, with the results shown in real time.
LaTeX panel – this panel shows the LaTeX version of the selected math. You can use the LaTeX panel to edit any math created in Equatio.
Math panel – Equatio analyzes the math in the LaTeX panel and shows its results as “digital math” in the Math panel. This is the same Math panel that you’ve learned about already, with all its usual
functions remaining available.
You can write your math in the LaTeX panel, the Math panel, or a combination of the two – whichever is easier for you. You can even start your math using speech input or handwriting recognition, then
edit it using LaTeX.
For more details, see the LaTeX Editor feature tour.
Screenshot Reader
If you want to copy math that wasn’t created in Equatio, for example from a PDF worksheet or video, you can use the Screenshot Reader to “grab” it, read it aloud, and copy it into Equatio.
The Screenshot Reader works well for any source where the math is encoded as a picture or video, as long as it can be opened in Chrome.
When the Discoverability tool is active, Equatio highlights any math that it finds on your web pages. You can then use the Screenshot Reader to copy that math into Equatio.
The Discoverability button turns blue , to indicate that it is active, and Equatio draws blue boxes around any math it finds in the current web page. The Discoverability tool finds whatever math that
it can – this includes math that is encoded as text, for example, or as an image with a LaTeX alt tag. It may not recognise math within a larger image, or with graphical annotations.
While the Discoverability tool is active, Equatio will continue to find and highlight math in any web page you open until you click the button again to turn it off.
When you hover your mouse pointer over one of the equations in the web page, the follwoing buttons appear at the top right of the blue rectangle:
• What is this?
• Capture Math
• Disable Discoverability
For more details, see the Screenshot Reader feature tour.
Equatio Mobile
Equatio Mobile enables you to use your mobile device as an input tool for writing math. You may not have a touchscreen or camera on your laptop or desktop computer, but you’ll almost certainly have
them on your phone or tablet. Equatio Mobile’s OCR is a great tool for copying math from a worksheet, textbook, or whiteboard:
1. Click Equatio Mobile. A QR code appears above the Equatio toolbar which is compatible with Google Chrome on Android and Safari on IOS 11+.
2. Scan the QR code with your mobile device’s camera (or any other app that supports QR codes).
3. Open Chrome (Android) or Safari (iOS) and navigate directly to https://m.equat.io
4. Equatio Mobile opens in your mobile device’s browser.
5. You may be prompted to sign into the account, for example if you are using Equatio Mobile for the first time, or if you haven’t used that account before on your mobile device.
The Active Documents screen lists all your documents that are currently using Equatio. For a document to appear in the list, you must be editing it (on your computer) using the same account that
you’re using for Equatio Mobile, with the Equatio toolbar open. Equatio Mobile can insert math into Microsoft Word, Gmail, and many other websites such as Moodle that support text or image
Math OCR
1. Click Math OCR . If prompted, click Allow to give Equatio Mobile access to your mobile device’s camera.
2. Equatio Mobile takes control of your mobile device’s camera.
3. This may be useful if you’re writing a lot of maths – you can take photos of it all and upload each one in turn.
4. Tap and then select Save as math.
5. Equatio Mobile processes your image and shows it as digital math.
6. Instead of saving your image as digital maths, you can select Save as image to insert your math as a picture, without any processing. You can use the Screenshot Reader to turn it into digital
maths later. Check that the maths matches what you originally wrote down. If it doesn’t, either try again or make a note to correct it later using Equatio’s Math or LaTeX panel.
7. When you are happy with the math, click Upload.
8. Equatio inserts the math into your document and shows a confirmation message on your mobile device.
Using mobile handwriting recognition
1. After selecting your Moodle page in Equatio Mobile, tap the Handwriting Recognition button and then draw the math on your mobile device.
2. After you have drawn the math, continue with Equatio Mobile until it has been processed and inserted.
Using mobile speech input
1. After selecting your Moodle content in Equatio Mobile, tap the Speech Input button and then dictate the math into your mobile device.
2. Note that you may have to allow Equatio Mobile access to your mobile device’s microphone.
3. After you have dictated the math, continue with Equatio Mobile until it has been processed and inserted.
Speech input can be tricky, as you have to formulate math in your head and then dictate it without pausing for too long. It can be useful for simple math, however, especially if you have it written
down in front of you but cannot use OCR.
For more details on Equatio Mobile, see the Equatio Mobile feature tour.
Graph Editor
The Graph Editor is powered by the Desmos Graphing Calculator, a leading graphing calculator with a simple interface that makes it easy to set up graphs and investigate them in real-time.
For a video introduction to creating graphs in Equatio, see the Graph Editor feature tour.
For more in-depth knowledge of creating and working with graphs, visit the Desmos Help Center. This contains comprehensive instructions, tutorials, user guides, and more.
When using the full online version of the Desmos Graphing Calculator, you can save a graph under your Desmos account and export it using a simple URL. You then can paste that URL into Equatio’s Graph
At its simplest, Equatio mathspace is a digital whiteboard tool that helps you add diagrams alongside your math.
Mathspaces are also flexible collaborative workspaces that teachers can use to:
• Work on maths with their students
• Create and send out assignments
• Collate and grade assignments
• Send feedback to their students
Mathspaces make it easy for students to complete their assignments. Having all their assignments in one place, complete with feedback, makes it easy for them to review and keep track of their work.
Further info: Equatio mathspace learning resources on Texthelp Academy.
This video shows how you can send assignments to your students, receive their answers, and send feedback back to them: https://academy.texthelp.com/equatio/mathspace-assignments/
Using a standalone mathspace for an assignment gives you interactive options that simply aren’t possible with paper worksheets.
Your students can move items around in the mathspace – for example, you could provide a set of answers and ask students to drag them to the corresponding questions. Or you could include a stack of
coins in the mathspace and ask students to put a certain amount into a piggy bank.
This video shows how you can use Equatio mathspace’s Infinite Cloner tool to make it easy for students to use pre-set items in their answers: https://academy.texthelp.com/equatio/infinite-cloner/.
This video shows how you can lock items that you don’t want your students to be able to move: https://youtu.be/LRnSnCAVMio.
STEM Tools
To see Equatio’s STEM tools, click STEM Tools then one of the following buttons:
a. Periodic Table
Each element is displayed with its symbol and atomic number, as well as its full name if your window is large enough. The information window includes a picture (or diagram) of the element, its
properties, and a description, with a link to the element’s Wikipedia page.
b. Scientific Calculator
Equatio’s Scientific Calculator is the Desmos Scientific Calculator, which is used widely in US exams. The Scientific Calculator enables you to build complex calculations using scientific operators
and functions. It supports fractions and can show answers as either decimals or fractions.
c. Molecular Viewer
The Molecular Viewer shows the structure of a particular molecule.
For a video introduction to each of these tools, see the STEM Tools feature tour.
Further information | {"url":"https://ucldata.atlassian.net/wiki/spaces/MoodleResourceCentre/pages/31864255/M75+-+Equatio+Premium+Chrome+extension?atl_f=content-tree","timestamp":"2024-11-12T23:30:02Z","content_type":"text/html","content_length":"1050373","record_id":"<urn:uuid:61deaa13-bd05-43ef-83ec-2e7026d85697>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00746.warc.gz"} |
Non-tradability interval for heterogeneous rational players in the option markets
This study uses theoretical and empirical approaches to analyze a number of phenomena observed in trading floors, such as the changes in trading volumes and Bid–Ask spreads as a function of the
moneyness level and the remaining time until the option’s expiration. A mathematical model for pricing options is developed that assumes two players with heterogeneous beliefs, where the objective of
each player is to maximize their profit on the expiration day. By solving a system of algebraic equations, which takes into consideration the subjective beliefs of the players regarding the price of
the underlying asset on expiration day, a feasible price domain is constructed that defines the boundaries within which a transaction may be executed. The developed model is applied to the special
case in which the distribution of the underlying asset price on expiration day is uniform, and a sensitivity analysis for selected parameters is presented. An interesting theoretical result that
emerges from the proposed model is the existence of an interval under which there is no tradability near the expiration day. The existence of this interval offers an explanation for the decrease in
the apparent trading volumes of out-of-money (OTM) options, together with an increase in Bid–Ask spreads, as the expiration day approaches. The main parameters that affect the point of time after
which there will be no trading are those that represent the players’ subjective beliefs about the distribution of the expiration values, and the cost of trading.
Bibliographical note
Publisher Copyright:
© 2021, The Author(s), under exclusive licence to Springer-Verlag GmbH Germany, part of Springer Nature.
• Bid–Ask spreads
• Heterogeneous players
• Non-tradability interval
• Option pricing
Dive into the research topics of 'Non-tradability interval for heterogeneous rational players in the option markets'. Together they form a unique fingerprint. | {"url":"https://cris.biu.ac.il/en/publications/non-tradability-interval-for-heterogeneous-rational-players-in-th","timestamp":"2024-11-09T09:59:45Z","content_type":"text/html","content_length":"57901","record_id":"<urn:uuid:fbeca784-ba89-4f2c-a164-c5039da6b8f0>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00595.warc.gz"} |
(PDF) A RUL Estimation System from Clustered Run-to-Failure Degradation Signals
Citation: Cho, A.D.; Carrasco, R.A.;
Ruz, G.A. A RUL Estimation System
from Clustered Run-to-Failure
Degradation Signals. Sensors 2022,22,
5323. https://doi.org/10.3390/
Academic Editors: Ningyun Lu,
Hamed Badihi and Tao Chen
Received: 30 May 2022
Accepted: 14 July 2022
Published: 16 July 2022
Publisher’s Note: MDPI stays neutral
with regard to jurisdictional claims in
published maps and institutional affil-
Copyright: © 2022 by the authors.
Licensee MDPI, Basel, Switzerland.
This article is an open access article
distributed under the terms and
conditions of the Creative Commons
Attribution (CC BY) license (https://
A RUL Estimation System from Clustered Run-to-Failure
Degradation Signals
Anthony D. Cho 1,2 , Rodrigo A. Carrasco 1,3 and Gonzalo A. Ruz 1,4,5,*
1Faculty of Engineering and Sciences, Universidad Adolfo Ibáñez, Santiago 7941169, Chile;
acholo@alumnos.uai.cl (A.D.C.); rax@uai.cl (R.A.C.)
2Faculty of Sciences, Engineering and Technology, Universidad Mayor, Santiago 7500994, Chile
3School of Engineering, Pontificia Universidad Católica de Chile, Santiago 7820436, Chile
4Data Observatory Foundation, Santiago 7941169, Chile
5Center of Applied Ecology and Sustainability (CAPES), Santiago 8331150, Chile
*Correspondence: gonzalo.ruz@uai.cl
The prognostics and health management disciplines provide an efficient solution to improve
a system’s durability, taking advantage of its lifespan in functionality before a failure appears.
Prognostics are performed to estimate the system or subsystem’s remaining useful life (RUL). This
estimation can be used as a supply in decision-making within maintenance plans and procedures.
This work focuses on prognostics by developing a recurrent neural network and a forecasting method
called Prophet to measure the performance quality in RUL estimation. We apply this approach to
degradation signals, which do not need to be monotonical. Finally, we test our system using data
from new generation telescopes in real-world applications.
Keywords: prognostics; fault detection; recurrent neural networks; prophet
1. Introduction
Modern industry has evolved significantly in the past decades, building more complex
systems with greater functionality. This evolution has added many sensors for better control,
higher system reliability, and information availability. Given this improvement in data
availability, new adequate maintenance policies can be developed [
]. Thus, maintenance
policies have evolved from waiting to fix the system when a failure appears (known as
reactive maintenance) to predictive maintenance, where intervention is scheduled with the
information obtained from fault detection methods.
Various researchers confirm that sensors play a crucial role in preserving the proper
functioning of the system or subsystem [
] as they provide information about the oper-
ating status in real-time such as possible failure patterns, level of degradation, abnormal
states of operation, and others. Taking this into account, various methodologies have been
developed for fault detection [
], testability design for fault diagnosis [
], detection of
fault conditions malfunction using deep learning techniques [
], and test selection design
for fault detection and isolation [
], just to name a few. Most of them share the same goal of
being able to help increase the reliability, availability, and performance of a system.
The two main extensions of predictive maintenance are Condition Based Mainte-
nance (CBM) and Prognostics and Health Management (PHM); both terms have been
used as a substitute for predictive maintenance in the literature [
]. According to
Jimenez et al. [
], they aligned these terms by adopting predictive maintenance as the first
term to refer to a maintenance strategy, CBM as an extension of predictive maintenance,
and adding alarms to warn when there is a fault in the system. Later, Vachtsevanos and
Wang [
] introduced prognostics algorithms as tools for predicting the time-to-failure
of components; from this insight emerged PHM [
] as an extension of CBM to improve
the predictability and remaining useful life (RUL) estimation of a component in question
Sensors 2022,22, 5323. https://doi.org/10.3390/s22145323 https://www.mdpi.com/journal/sensors
Sensors 2022,22, 5323 2 of 29
after a fault appears. This information can then be used as a supply for decision-making in
maintenance scheduling [14].
It is necessary to highlight that fault detection and prognostics are not always exclusive.
Fault detection is usually an initial step in computing prognostics to estimate the future
behavior of the system or subsystem.
Generally, faults are generated by degradation of the components that make up the
system. Such degradation can be monitored through the signals collected from the sensors.
There are various types of degradations that have been addressed in the literature, one
of the most common are those signals that present degradation with slow decay that are
present in different components, such as, for example, an increase in resistivity of fuses,
reduction in currents on frequency processors, and the mean resolution of a telescope’s
camera, among others. Considering these similarities, it is possible that an automatic fault
detection framework that manages to detect the degradation in a frequency processor could
also effectively detect the degradation in the resolution of a camera or vice versa. Similarly,
it is possible that a good prediction of the RUL of a camera can be obtained using historical
fault information present in other components.
This work focuses on prognostics by developing recurrent neural networks (RNNs)
and a forecasting method called Prophet to measure the performance quality in RUL es-
timation. First, we apply this approach to degradation signals, which do not need to be
monotonical, using the fault detection framework proposed in [
] with some improve-
ments in the pre-processing and the cleaning data step. Later, we applied our approach to
similar degradation problems but with different statistical characteristics.
The difference between our research with the rest of the works is in the scalability of
the framework in fault detection towards other similar problems, showing its effectiveness
and robustness. On the other hand, the adjusted RNN models with historical data of one
type of fault to predict its RUL can also be used in other problems that have signals with
similar degradation, such as the resolution of a telescope’s camera, showing the power of
generalization and precision in the prediction of the RUL.
Our work has the following contributions:
We made improvements in cleaning spikes or possible outlines and smoothing time-
series in the pre-processing data step in the fault detection framework developed
in [
] to reduce the remaining noise level while maintaining its relevant characteristics
such as trends and stationarity.
We show that the fault detection framework in [
], together with our pre-processing
method, improves the robustness of the framework and can be transferable to another
problem with similar degradation, although with different statistical characteristics.
We built a strategy using clustering run-to-failure critical segments to define an
appropriate failure threshold that improves the RUL estimation. Moreover, using this
strategy, we predict the RUL of another problem with similar degradation.
The rest of this article is organized as follows. First, the background related to this
work is presented in Section 2. In Section 3, we present the proposed method for data
pre-processing for cleaning spikes or outlier points, the smoothing for time series, and the
process of prognostic for RUL estimation. In Section 4, the details of the application are
given, as well as the results. Section 5presents a discussion of results and performances
obtained for each application. Finally, the conclusion of the work is presented in Section 6
and future work in Section 7.
2. Background
The following subsections present a brief description of fault detection, prognostics,
performance measurements, and a method used for RUL estimation.
2.1. Fault Detection
Most modern industries are equipped with several sensors collecting process-related
data to monitor the status of the process and discover faults arising in the system. Fault
Sensors 2022,22, 5323 3 of 29
detection systems were developed around the 1970s [
], as an essential part of automatic
control systems to maintain desirable performance. Fault detection can be defined as a
process of determining if a system or subsystem has entered a mode different from the
normal operating condition [
], and a fault may appear at an unknown time, and the
speed of appearance may be different [17,18].
In the literature, a wide variety of methods used for fault detection can be classified
into signal processing approaches [
], model-based approaches [
], knowledge-
based approaches [
], and data-driven approaches [
]. With the arrival
of technology and the advancement of computing methods, data-driven approaches are
gaining attention in the last decades, where it is expected that the data will drive the
identification of normal and faulty modes of operation. See [
] for a general description of
fault detection and diagnosis systems.
Some recent developments have addressed this issue with deep learning to increase
accuracy in fault detection. For example, Yao Li [
] presented a branched Long-Short
Term Memory (LSTM) model with an attention mechanism to discriminate multiple states
of a system showing high performance in its prediction based on the F1-score metric. On
the other hand, Liu et al. [
] showed a strategy for failure prediction using the LSTM
model in a multi-stage regression model to predict the trend; this is then used to classify the
level of degradation by similarity with established failure profiles, achieving improvement
estimates with better precision.
Zhu et al. [
] addressed the problem of classifying multiple states of a system with a
convolutional network structure (CNN), specifically LeNet, optimized with Particle Swarm
Optimization (PSO). Their results showed that this strategy achieves better performance and
greater robustness compared to LeNet without PSO, VGG-11, VGG-13, VGG-16, AlexNet,
and GoogleNet. Another approach using CNN is presented in the work of Jana et al. [
which uses a suite of Convolutional Autoencoder (CAE) networks to detect each type of
failure. Its design allows addressing failures in multiple sensors with multiple failures,
obtaining an accuracy of around 99%.
Within the approaches not fully supervised, Long et al. [
] developed a Self-Adaptation
Graph Attention Network, one of the first models of this type of network to be able to use a
few-shot learning approach in which abundant data is available but very little is labeled
and also to be able to incorporate cases of failures that rarely occur. In their results, they
showed better performance at the level of accuracy compared to other models.
From an application perspective, fault detection systems have been developed in
many areas such as rolling bearing, machines, industrial systems, mechatronics sys-
tems, industrial cyber-physical systems, and industrial-scale telescopes, to name a
few [15,23–26,33–35,37,38,41,42].
Some of them describe some advantages and disadvantages over others in the ap-
plied methodology to obtain better results. However, there are still a lot of difficulties in
implementing fault detection methods for real industries due to the properties of the data.
2.2. Prognostic
The prognosis task is mainly focused on estimating or predicting the RUL of a degrad-
ing system and reducing the system’s downtime [
]. So, the development of effective
prognosis methods to anticipate the time of failure by estimating the RUL of a degrading
system or subsystem would be useful for decision-making in maintenance [
]. A failure
refers to the event or inoperable behavior in which the system or subsystem does not
perform correctly.
According to the literature, prognostics approaches can be classified into model-based
approaches [
], hybrid approaches [
], and data-driven approaches [
Data-driven approaches offer some advantages over the other approaches, especially when
obtaining large and reliable historical data is easier than constructing physical models that
require a deeper understanding of the system degradation. Also, they are increasingly
applied to industrial system prognostic [
]. Recently, these studies are also divided into
three branches: degradation state-based, regression-based, and pattern matching-based
Sensors 2022,22, 5323 4 of 29
prognostics methods [
]. The former usually estimates the RUL by estimating the
system’s health state and then using a failure threshold to compute the RUL. The second
method is dedicated to predicting the evolution behavior of a degradation signal, and the
estimation of the RUL can be obtained when the prediction reaches the failure threshold.
The last methods consist of characterizing the signal and comparing it in the run-to-failure
repository to compute the RUL by similarity.
In recent years, various deep learning models have been introduced to address fore-
casting problems in RUL prediction. For example, Kang et al. [53] developed a multilayer
perceptron neural network (MLP) model to predict the health index of a signal; this is used
in a polynomial interpolation model to estimate the RUL. They indicate that their strategy
outperforms direct prediction methods using SVR, Linear Regression, and Random Forest.
In an ensemble-type approach, Chen et al. [
] presented a hybrid method for RUL predic-
tion using Support Vector Regression (SVR) and LSTM in which the results are functionally
weighted, showing to be more robust as it takes advantage of the benefits provided by SVR
and LSTM.
Among the most innovative methods, Ding and Jia [
] designed a convolutional
Transformer network model that takes advantage of the attention mechanism and CNN to
capture global information and local dependence of a signal allowing to directly map the
raw signal to an estimated RUL, increasing its effectiveness and accuracy in prediction. On
the other hand, Zhang et al. [
] developed a model that allows evaluating health status and
predicting RUL simultaneously using a dual-task network model based on the bidirectional
gated recurrent unit (BiGRU) and multigate mixture-of-experts (MMoE), resulting in better
performance compared to traditional popular models such as ANN, RNN, LSTM, CNN,
GRU and Bi-GRU, and with satisfactory higher robustness.
Under the not fully supervised approach, He et al. [56] developed a semi-supervised
model based on a generative adversarial network (GAN) in regression mode, considering
historical data for prevention and scarce historical information for failures to predict
the RUL. This approach allows for avoiding overfitting, thus increasing its power of
generalization and manages to achieve satisfactory accuracy even when the amount of
historical data per failure is limited.
To measure the performance of the prognosis method, Saxena et al. [
] introduced
some standard evaluation metrics that were used to evaluate several algorithms compared
to other conventional metrics effectively. Such metrics can be used as a guideline for choos-
ing one model over another. A description of these metrics can be found in Appendix A;
they can be considered as a hierarchical validation approach for model selection described
in [
], where the first instance is to check out whether a model gives a sufficient prognostic
horizon, and if not, this method is not meant to compute the other metrics. If the model
passes PH’s criterion, it is followed by the computation of the
accuracy, which needs a
more strict requirement of staying within a converging cone of error margin as a system
reaches the End-of-Life (EoL). If this criterion is also met, we can quantify how well the
method does by computing the accuracy levels relative to the actual RUL and, finally,
measure how fast the method converges. This work will focus on the first two metrics since
they provide a meaningful level of accuracy of the model in the RUL estimation.
2.3. Recurrent Neural Networks (RNNs)
Among data-driven techniques used for prognostics, RNNs have been widely studied
in recent years and are one of the most powerful tools as they can model significant
nonlinear dynamical time series. A large dynamic memory is allowed to preserve temporal
dynamics of complex sequential information and has been used with success in several
prognostic applications [
]. Three types of RNN are chosen in this work: Echo State
Networks (ESNs), Long-Short Term Memory (LSTM), and Gated Recurrent Unit (GRU),
to measure the performance of RUL estimation applied in three problems described in
Section 4. A description of these RNNs appears in Appendix B.
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2.4. Prophet Model
The Prophet model was developed by Sean Taylor and Benjamin Letham [
] in 2018 to
produce more confident forecasts. Its methodology consists of the usage of a decomposable
time series model, consisting of three main components: trend, seasonality, and holidays.
It allows one to look at each component of the forecast separately. These components are
combined as an additive model in the following form:
y(t) = g(t) + s(t) + h(t) + e(t), (1)
is the trend function that represents the non-periodic changes of the time series,
describes the periodic changes (daily, weekly, and yearly seasonality),
the effects of holidays that occur on potentially irregular calendar schedules over one or
more days, and
represents the error term of any idiosyncratic changes which are not
accommodated by the model. This method has several advantages that allow the analyst to
make different assumptions about the trend, seasonality, and holidays if necessary, and the
parameters of the model are easy to interpret.
3. Methodology
3.1. Pre-Processing Data
The data or signals collected from a system, in most cases, are noisy, and some outliers
or spikes might be present. So, it is necessary to pre-process each signal before feeding
it to the forecasting model. This process is shown in Figure 1, and it consists of the
following steps:
Raw data
Smoothed data
Figure 1. Pre-processing flow chart.
1. Spikes cleaning
: it consists of clearing possible outliers and spikes points by compar-
ing time series values with the values of their preceding time window, identifying a
time point as anomalous if the change of value from its preceding average or median
is anomalously large.
An advantage of this outlier reduction strategy is that it considers the local dynamics
of the signal with time windows. Therefore, managing to identify as outliers the
samples that are outside the local range and thus reduce the number of samples
that are normal but that were identified as outliers, as could happen with traditional
methods that depend on the global mean and standard deviation. This method is
implemented in the ADTK library [59].
2. Double exponential smoothing
: this filter [
] is commonly used for fore-
casting in time series, but it can also be used for noise reduction. This method is
particularly useful in time series to smooth its behavior, preserving the trend and
without losing almost any information in the dynamics of the series. Also, the model is
simple to implement, depending on two main parameters. For more details, see [
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3. Convolutional smoothing
: this consist of applying the Fourier transform with a
fixed window size to smooth the signal maintaining the trend. In other words, this
method applies a central weighted moving average to the signal allowing short-term
fluctuations to be reduced and long-term trends to be highlighted. It is implemented
in the TSmoothie library [65].
Each of the methods that make up the pre-processing process offers some strengths
and weaknesses. To see its independent effect, each of the methods was applied to a signal
that presented outliers with a high level of noise, as shown in Figure 2.
The effect of the method that was mentioned in Step 1, shown in Figure 2a, can be
seen that it manages to reduce the large jumps that are considered outliers, but still, some
outliers remain with minor jumps. The noise reduction or smoothing methods that were
mentioned in Steps 2 and 3 present some artifacts in the signal dynamics due to outliers,
and their effects are unknown, as shown in Figure 2b,c.
It is for this reason that we combine the methods to use the advantages offered by
each one of them, allowing us to reduce large jump outliers, followed by a noise reduction
strategy and reduce minor jump outliers, and finally, reduce possible remaining artifacts
with smoothing procedure as presented in the designed pre-processing scheme, Figure 1.
The effect of this combination is shown in Figure 2d, where the resulting signal has smoother
dynamics and preserves the trend of the original signal.
No outliers
Figure 2.
Application of each method separately to the raw signal. (
) Outliers and spikes
cleaning. (
) Double Exponential Smoothing. (
) Convolutional smoothing. (
) Proposed pre-
processing method.
3.2. Run-to-Failures Critical Segments Clustering
The increase in processor speed, sensors monitoring, and the development of storage
technologies allow real-world applications to record changing data over time easily in
components of a system/subsystem [
]. It is necessary to highlight that the components
used in different environments lead to different degradation levels, even for one type of
component. Therefore, the failure threshold can be different in each situation. However,
from the historical run-to-failure signals, they can be clustered so that each signal in a
cluster behaves similarly; thus, it is possible to define a failure threshold based on the
Sensors 2022,22, 5323 7 of 29
signals that belong to a cluster. In other words, there is a failure threshold A that can be
defined as cluster A, a failure threshold B to cluster B, and so on.
Our scheme of clustering does not consider the entire signal since it starts running
until EoL; instead, we use the critical segment of the signal for clustering. Our definition of
a critical segment of a signal is the segment where the degradation begins until EoL. Under
these critical segments, we build clusters so that each cluster has signals with a degradation
level relatively similar.
The advantage of clustering by critical segments allows us to define, in an easy way,
the different failure thresholds. Therefore, we can define for each cluster an appropriate
failure threshold based on the critical segment signals that belong to a cluster. To increase
the effectivity, each critical segment is centered with its own standard normal condition
value before the clustering process, i.e, if
is the complete signal, and
is the critical
segment, then
is centered by
, where
is the standard normal condition
value and
is the first sample of
. Lastly, a threshold can be defined as the minimum
degradation level reached by critical signals in the cluster.
3.3. Prognostic Method
Two strategies are proposed to deal with the estimation of RUL in components. For all
strategies, we consider the fault date as the point in time
at which the fault prediction
starts [
]. We also assume that the recollected data consists of daily samples, which were
processed using the approach presented in Section 3.1. In what follows, a description of
these strategies is presented.
3.3.1. Strategy A
This strategy is based on a regression model, similar to the prognostic approach
proposed in [
]. In this strategy, we define a time window of
days in which we analyze
the data. Note that the number of samples in the time window can vary since data is not
assumed to be available every day. Figure 3a shows an example with missing data, whereas
Figure 3b shows an example where data is available through the whole time window.
(a) (b)
Figure 3.
Time-window examples. (
) Time-window with missing values. (
) Time-window without
missing values.
The data within the time window is used to train the model, which is then utilized
to predict a forecast for the next
days, following the structure shown in Figure 4. In this
represents the first
samples of
, the data used as input to train the
model. The model then estimates
, and the current window is updated by dropping
the oldest value and adding the newly calculated one:
. The forecasting
process is similar to the P-method developed in [48].
Using the previous forecast, we verify if the failure threshold is crossed within the
time window, calculating the RUL if this occurs. This procedure is applied in a rolling
window fashion whenever new data arrives.
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ModelData y(t+1)
Training Forecast
Figure 4. Model training and forecast structure.
Figure 5shows an application example using a time window of 365 days. The first
iteration result is shown in Figure 5a, with the time window between 18 November 2014
and 18 November 2015. Since some data is missing, we have 340 samples in this case.
In this step, our approach estimates the RUL to be 384 days. Next, Figure 5b shows the
results of the second iteration, where the time window lies between 14 September 2015 and
13 September 2016, containing 365 samples. In this step, the RUL is estimated to be 181 days.
The black line represents the ground truth in both figures, and the blue line represents the
obtained forecast. The green dashed line is
, the red dashed one is the failure threshold,
and the RUL value is computed as the difference between when the forecast crosses the
failure threshold and tP. Finally, the whole process is shown in the diagram in Figure 6.
1 January 2015
current [A]
1 May 2015
1 September 2015
1 January 2016
1 May 2016
1 September 2016
1 January 2017
1 May 2017
1 September 2017
1 January 2018
Ground truth
current [A]
1 January 2015
1 July 2015
1 January 2016
1 July 2016
1 January 2017
1 July 2017
1 January 2018
1 July 2018
Ground truth
Figure 5.
An example of RUL estimation using a time-window size of 365 days. (
) Time-window
samples until fault date tP. (b) Time-window shifted by 300 days.
Raw data
(input) Pre-processing
data Model
Is there new
Update raw
Wait for new
available data
Figure 6. Prognostic process: strategy A.
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3.3.2. Strategy B
Considering that one type of component could be in vastly different environments, it
is possible that their degradation level, and thus failure thresholds, could be very different.
Due to this, we need to adapt the previous strategy to account for this difference. We do
this by combining matching and regression-based methods. This technique consists of
two steps:
•Cluster-Model stage
: it consists of the usage of clustering described in Section 3.2, so
that, for each cluster we can fit a regression model. The train data is defined by the
critical signals limited by a defined failure threshold in the cluster with its residual
RUL, i.e., for each critical signal
with length
in cluster
such that
, and
l(S0)≈f ail ure_threshold
. Then, each sample
has a residual RUL
where l(S)is the length of the signal S,
Normalize(Si) = Si−min(S)
S0and S0
0are the first sample of Sand S0, respectively.
•Prediction stage
: it consists mainly in predicting the RUL of a component in the
signal that has been diagnosed as a fault, which means a degradation behavior has
started. In this step, we took a segment of the signal after a fault has been detected; it
is pre-processed and submitted to a classifier to identify to which cluster it belongs
and select the related regression model, already fitted in the Cluster-Model stage, to
predict the RUL. This procedure is executed when new samples are available.
The classifier works in matching segments to all run-to-failure critical segments using
Minimum Variance Matching (MVM) [
], which is a popular method for elastic
matching of two sequences of different lengths by mapping the problem of the best
matching subsequence to the problem of the shortest path in a directed acyclic graph
providing the minimum distance. The classification scope provides the assignment by
a voting criterion, i.e., the maximum number of signals of a cluster closer to a given
segment will be taken. A flow chart of this prognostic process is shown in Figure 7.
Figure 7. Prognostic process: strategy B.
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The principal models used in this work for training and computing forecasts or
RUL are mentioned in Sections 2.3 and 2.4: ESN, LTSM, GRU, and Prophet (only for
Prognostic Strategy A). To measure how well the model is for estimating RUL, we will use
the prognostic horizon and α–λaccuracy.
4. Application Setting
4.1. Crack Growth
The crack propagation description is one of the most important components in the
analysis of the life span of structural components, but it may require time and expense to
investigate experimentally [
]. Hence, the estimation of crack propagation and durability
of construction or structural component will be useful to estimate the remaining life of
the component.
4.1.1. Problem Description
As described in [
], components that are subjected to fluctuating loads are prac-
tically found everywhere: vehicles and other machinery that contain rotating axles and
gears, pressure vessels and piping may be subjected to pressure fluctuations or repeated
temperature changes, and structural members in bridges are subjected to traffic loads and
wind loads, and some other applications. If the components are subjected to a fluctuating
load of a certain magnitude for a sufficient amount of time, it may cause small cracks in
the material. Over time, the cracks will propagate up to the point where the remaining
cross-section of the component cannot carry the load, at which the component will be
subjected to sudden fracture. This process is called fatigue and is one of the main causes of
failures in structural and mechanical components.
The common Paris–Erdogan model is adopted [
] for describing the evolution of the
crack length
as a function of the load cycles
summarized by the following discrete-
time model
xt+1=xt+Ceωt(β√xt)n, (2)
ωt∼ N(
is a random variable depicting white Gaussian noise, and
are fixed constants. A generation of 30 crack growth trajectories using Equation (2) is
illustrated in Figure 8and consists of 900 days of samples per trajectory.
1 January 2000
Crack length [mm]
Crack growth
1 April 2000
1 July 2000
1 October 2000
1 January 2001
1 April 2001
1 July 2001
1 October 2001
1 January 2002
1 April 2002
Figure 8. 30 crack growth trajectories.
4.1.2. Prognostic
For practical purposes, we choose one trajectory from Figure 8to estimate RUL to
measure the performances of both strategies.
Sensors 2022,22, 5323 11 of 29
Strategy A: following the methodology in Section 3.3.1, we estimate RUL shifting the
time window by 15 days in every iteration, 1 year size of time-window, and 2 years
of forecast.
The results are shown in Figure 9. In the prognostic horizon, Figure 9b, we can see
that all the models underestimate RUL, with some exceptions like the Dense neural
network model. Neural network models had poor performances of RUL estimation
and mostly fall outside of the confidence interval. Only the Prophet model is relatively
close to the ground truth RUL. Concerning the
accuracy, only Prophet has a
segment close to the ground truth but then falls outside of the confidence interval,
underestimating the RUL.
2000 Apr Jul Oct Jan
2001 Apr Jul Oct Jan
2002 Apr
Crack length [mm]
Crack Growth (n_samples=900)
Fault date
RUL [days]
Prognostic horizon ( = 0.25)
Ground truth
RUL [days]
accuracy ( = 0.25)
Ground truth
RUL [days]
Prognostic horizon ( = 0.25)
Ground truth
RUL [days]
accuracy ( = 0.25)
Ground truth
Figure 9.
The crack growth prognostic. (
) Testing: a crack growth trajectory. (
) Strategy A: the
prognostic horizon metric. (
) Strategy A: the
accuracy metric. (
) Strategy B: the prognostic
horizon metric. (e) Strategy B: the α–λaccuracy metric.
Strategy B: using the technique proposed in Section 3.3.2 in this problem, we will
simplify some steps of this process. Given that all the degradation trajectories are
similar, we can assume only one cluster and the classifier will assign to it every time.
Hence, the Cluster-Model stage has only one model, which is used to predict the RUL.
Sensors 2022,22, 5323 12 of 29
Basically, this scheme becomes a simple regression model where it is fitted with all the
historical-critical segment trajectories limited by its failure threshold and its residual
RUL. We use 100 trajectories as run-to-failure signals generated from Equation (2) to
fit the model.
The performances can be seen in Figure 9d,e. All the models fall inside the confidence
interval in the prognostic horizon and are getting closer to the ground truth as they
reach the EoL, as illustrated in Figure 9d. Similar behavior is obtained for
as shown in Figure 9d. Only a few times, some methods go out and then go back into
the confidence interval, e.g., LSTM and GRU, but these behaviors are acceptable.
The results are shown to indicate a large difference in the estimation of the RUL
between the two strategies. This is due to the fact that the models that use strategy A are
more sensitive to small variations in the signal, making the EoL estimate highly variable
and, most critically, it is unaware of the possible variation that it may present in the future.
On the other hand, the models that use strategy B take advantage of historical information
to incorporate into the model information on how the signal could evolve, reducing the
sensitivity due to small disturbances and better mapping to a more precise RUL.
4.2. Intermediate Frequency Processor Degradation Problem
The Atacama Large Millimeter/submillimeter Array (ALMA) is a revolutionary instru-
ment operating in northern Chile’s Atacama desert’s very thin and dry air at an altitude of
5200 m above sea level. ALMA is one of the first industrial-scale new generation telescopes,
composed of an array of 66 high-precision antennas working together at the millimeter
and submillimeter wavelengths, corresponding to frequencies from about 30 to 950 GHz.
Adding to the observatory’s complexity, these 7 and 12-m parabolic antennas, with ex-
tremely precise surfaces, can be moved around at the high altitude of the Chajnantor
plateau to provide different array configurations, ranging in size from about 150 m to up
to 20 km. The ALMA Observatory is an international partnership between Europe, North
America, and Japan, in cooperation with the Republic of Chile [75].
4.2.1. Problem Description
The Intermediate Frequency Processor (IFP) of the antennas of the ALMA telescope, as
described in [
], is a critical component responsible for the second down-conversion, signal
filtering, and amplification of the total power measurement of sidebands and basebands.
This subsystem allows for effective communication of the captured data to the central
correlator for processing, thus making it a central and critical component of each antenna.
It is necessary to highlight that there are 2 IFPs per antenna, one for each polarization, and
each IFP has sensors measuring currents of three different voltage levels: 6.5, 8, and 10 volts.
For 6.5 and 8 volts, currents have four different basebands: A, B, C, and D, whereas, for
10 volts, sidebands USB and LSB, and switch matrices SW1 and SW2 currents are read.
Each current is sampled every 10 min.
One of the diagnosed degradation problems that occur in the IFP module is due to
hydrogen poisoning caused by hydrogen outgassing in tightly sealed packages [
], where
this degradation can be tracked by monitoring current signals collected from each module.
4.2.2. Prognostic
To measure the performance of both strategies, we selected one of the signals with a
fault detected in [15], and applied the data pre-processing. This is shown in Figure 10a.
Strategy A: the performances of this method are illustrated in Figure 10b,c, in which
we can see that none of these models give good predictions of RUL, nor when it
approaches the EoL.
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1 July 2012
1 October 2012
1 January 2013
1 April 2013
1 July 2013
1 October 2013
1 January 2014
1 April 2014
1 July 2014
1 October 2014
current [A]
Antenna 13 - Polarization 1 - 8 Volts (Channel BB-B)
Fault date
RUL [days]
Prognostic horizon ( = 0.25)
Ground truth
RUL [days]
accuracy ( = 0.25)
Ground truth
RUL [days]
Prognostic horizon ( = 0.25)
Ground truth
RUL [days]
accuracy ( = 0.25)
Ground truth
Figure 10.
The IFP prognostic. (
) Testing: a signal from an IFP. (
) Strategy A: the prognostic
horizon metric. (
) Strategy A: the
accuracy metric. (
) Strategy B: the prognostic horizon metric.
(e) Strategy B: the α–λaccuracy metric.
Strategy B: from the historical run-to-failure signals, different degradation levels
appears in each voltage’s current of the IFP. In this application, each voltage’s signals
are clustered into a few clusters so that signals in each cluster have similar degradation
levels making it easier to define an appropriate failure threshold in each cluster, just
as described in Section 3.2, defining a total of 5 clusters for this problem: 2 cluster for
6.5 volts, 1 cluster for 8 volts, and 2 clusters for 10 volts; they are shown in Figure 11,
in which, for each cluster has its corresponded failure threshold value, i.e., 0.566 is
the failure threshold for cluster 1, 0.2 for cluster 2, 0.127 for cluster 3, 0.246 for cluster
4, and 0.275 for cluster 5; or it can be explained as 5.7%, 2%, 36%, 18%, and 8.3% of
degradation levels for each cluster, respectively. These clusters are used to classify
the new arriving pre-processed signal to select the appropriate failure threshold and
predict the RUL.
The cluster generation criterion focuses mainly on the Minimum Variance Matching
(MVM) similarity metric, which is obtained by solving a shortest path (SP) problem
Sensors 2022,22, 5323 14 of 29
that measures the distance between pairs of signals. The principle is to fix a signal as a
centroid and compute the distances with the other signals; these distances are ordered,
and using the same fundamentals of the elbow method, a group of signals is selected
to form a cluster
and the rest in another group
. This process is repeated for the
to verify if the signals are similar or if another cluster is generated, and so
on. Repeated runs were made, resulting in most cases with 5 clusters being enough to
separate these signals.
The performances under both metrics, Figure 10d,e, show that almost all models have
relatively good predictions of RUL falling inside of the confidence interval. Only ESN
has some irregularities, but these underestimations are acceptable. The Dense neural
network model outperforms the others slightly when it gets close to the EoL.
Analyzing the results, the models that used strategy A showed a problem similar
to what occurs in the application of the Crack Growth in Section 4.1.2, in which
the models remain sensitive to small variations, generating a great variability in the
estimation of EoL and therefore, affects the prediction of the RUL.
Taking into account these effects that it could have on the models, if strategy B is used
and a set of historical run-to-failure signals is considered that have great variability in
the degradation behavior, different from that used in Section 4.1.2 in which the signals
are quite similar, could affect the models in predicting the RUL due to these variations
in the level of degradation of the historical signals.
To avoid this, it was decided to group the signals into groups that are similar in
degradation level and address them separately. As a consequence, the performance in
different models manages to predict the RUL close to the real value.
current [A]
6.5 Volts (type: 1) (Threshold: 0.566)
current [A]
6.5 Volts (type: 2) (Threshold: 0.588)
current [A]
8 Volts (type: 1) (Threshold: 0.127)
current [A]
10 Volts (type: 1) (Threshold: 0.246)
current [A]
10 Volts (type: 2) (Threshold: 0.275)
Figure 11.
The IFP signals clustering, the red dashed lines represent the failure threshold defined for
each cluster, and continuous lines are the critical segments segmented from the run-to-failure IFP
signals (
) Class 1: 6.5 Volts (Degradation type 1). (
) Class 2: 6.5 Volts (Degradation type 2). (
) Class
3: 8 Volts. (d) Class 4: 10 Volts (Degradation type 1). (e) Class 5: 10 Volts (Degradation type 2).
4.3. Validation in a Different Setting
To validate our approach, we considered testing this methodology in a very different
setting. In particular, we used measurements of camera resolution information from an
important optical telescope.
4.3.1. Problem Description
One of the problems presented in the studied instrument is the Teflon wear in the
lens support, increasing the humidity level, which affects the camera resolution. This
degradation can be tracked through measurements collected from the camera’s CCDs.
Sensors 2022,22, 5323 15 of 29
An example of degradation over 18 years is shown in Figure 12, where it can be seen
that this signal is noisy and has several spike points (large down jumps that may be possible
outliers). Some corrective or maintenance actions have been made (time indexes of up
jumps) are taken along these records. Therefore, a process of fault detection would be
excellent for anticipating an unacceptable deviation of the fault-free behavior and then a
prognostic process to compute the RUL of the component accurately.
Figure 12. Resolution media signal obtained from a CCD.
4.3.2. Fault Detection
Recently, Cho et al. [
] tackled similar degradation noisy signals using a fault detec-
tion framework based on ESNs applied to IFPs of the antennas of the ALMA observatory;
the authors highlighted the noise level in the data affected the performance of detection
significantly. In the case of the camera resolution, unlike the ALMA IFP data, it contains
larger spikes that distort the signal dynamics even after double exponential smoothing. For
this reason, it is necessary to adopt a mechanism that allows reducing spikes efficiently in
time series as a clean outlier method in the pre-processing stage of the framework proposed
in [
]. With this insight, the modified data pre-processing method was generated, and it is
described in Section 3.1. The results, applying the proposed data pre-processing method,
are shown in Figure 13, where the red signal represents the pre-processed signal, and the
trend is maintained from the raw signal.
Pre processed
Figure 13. Raw and pre-processed signal of the resolution media obtained from a CCD.
Once the pre-processing stage is done, the fault detection process is maintained almost
the same as in [
]. The result is shown in Figure 14. The vertical dashed red lines are fault
detected time indexes and the vertical dashed green lines are time indexes where corrective
or maintenance were made.
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31 March 2007
25 January 2014
11 April 2017
10 February 2017
04 October 2017
16 December 2004
07 December 2010
29 March 2015
15 June 2016
18 January 2018
Figure 14. Fault detection in the resolution media signal obtained from a CCD.
It is necessary to highlight that the framework designed in [
] deals with current
signals with a resolution of 10 min per sample, resulting in high performance on real
data. Now, with this modification in the pre-processing, the robustness of the framework
increases, and it is applied to the camera problem, which are signals coming from a
resolution camera with daily samples, resulting in the same effectiveness in fault detection;
this is justified in that the degradation characteristic is similar to the ones that were used
during the design of the method.
4.3.3. Prognostic
For the prognostic application to the camera resolution signal, we took the first segment
of the trajectory until the first maintenance, dated 2007-03-31, as the test signal for RUL
estimation, Figure 15a. The rest of the segment can be computed similarly by applying the
methodology described in Section 3.
Strategy A: applying this method, we can see Figure 15b,c, that neural networks
have a poor quality of predictions, whereas the Prophet model has some segments
that fall inside the confidence interval, but it is not good enough because of its
irregular behaviour.
Strategy B: in this problem, there are no historical run-to-failure signals. So, clustering
over this component is not possible. However, given that the degradation behavior
present in this component is similar to the IFP of ALMA, we can use these clusters and
try to transfer to this problem. To achieve this, it is necessary to transform the new
arriving pre-processed signal
and scale it to every cluster described in Section 3.2,
this means, for each cluster, we define a transformed signal of Qas follows
S=κ·Q, (3)
is the scaling constant,
are standard normal conditions and failure threshold
of the cluster
, respectively.
is the first sample of the signal in this problem, and
is its associated failure threshold.
The classifier result gives the final scope, which is used for model selection in the
prediction of RUL. In the prognostic horizon metric, Figure 15d, the GRU model
outperforms the other models. However, the other models fall inside the confidence
interval after 200 days. So, all the models in this metric are acceptable. From the
accuracy side, most of the time, these models are not inside the confidence interval,
underestimating the RUL on the first 300 days
. After that, they are around
the ground truth up to the EoL. In this case, the GRU model is close to the frontier of
Sensors 2022,22, 5323 17 of 29
the confidence interval, which is not as bad as an instance for RUL computation by
using a similar degradation signal developed from another system or component like
the IFP Problem.
Resolution media [R]
Ground truth
Fault date
RUL [days]
Prognostic horizon ( = 0.25)
Ground truth
RUL [days]
accuracy ( = 0.25)
Ground truth
RUL [days]
Prognostic horizon ( = 0.25)
Ground truth
RUL [days]
accuracy ( = 0.25)
Ground truth
Figure 15.
The Camera Resolution prognostic. (
) Testing: Resolution media trajectory. (
) Strategy A:
The prognostic horizon metric. (
) Strategy A: The
accuracy metric. (
) Strategy B: The prognostic
horizon metric. (e) Strategy B: The α–λaccuracy metric.
The way in which strategy B was approached in this application allows comparing
the critical segment of the new pre-processed and transformed incoming signal with the
clustered signals that have similar patterns at the level of degradation. In addition, this
helps to relate to possible trajectories of the signals of the cluster that is most assimilated
and, thus, to be able to approximate the RUL of this new signal when historical information
is not available. As the mean resolution signal has similar characteristics to some signals in
one of the clusters, this helps in obtaining a relatively good RUL prediction.
5. Discussion
Several frameworks of fault detection have been developed in the last decades, most
of them for a specific degradation present in an application of interest. In this work, we
Sensors 2022,22, 5323 18 of 29
are interested in a more general framework, transferable to many domains that present a
similar degradation problem. In Section 4.3.2, we show that the fault detection framework
developed in [
] can be transferable to other applications with similar degradation be-
havior as the one described in Section 4.3.1, without any adjustment to the structure but
only some improvement to the data pre-processing step. In particular, by adding other
properties of noise to get a better-smoothed signal, as the example shown in Figure 13. Such
improvement increases the performance of this framework slightly even when applied
to the IFP signals, which was the problem of interest in [
]. We obtained a smoothed
signal while maintaining the relevant characteristic of the raw data, such as the degradation
trend. This smoothed signal then was used as an input to verify if a fault was present and
returned the date where it was detected, as illustrated in Figure 14, where the red dashed
lines represent the dates of detected faults and the green ones represent the dates of the
performed maintenance.
The parameters used in the pre-processing steps were: Factor used to determine the
bound of the normal range based on the historical interquartile range was fixed as 3, and the
window size was fixed as 20 for both spike cleaner and convolutional smoothing methods.
It is necessary to highlight our meaning of transferable is not the same as transfer
learning used in the context of deep learning. The framework learns from the data auto-
matically but does not inherit the insights from another problem so that it can be scaled
and applied to other similar problems. Given that fault detection and prognostic are not
always exclusive to each other, in most of the cases, the former is considered as the previous
step of the prognostic process. Additionally, the pre-processing method that we designed
in Section 3.1 allows us to reduce as far as possible problems of outliers present in the
signal to be later used, either for fault prediction or forecasting. This allows to increase the
performance and reduce possible disturbances that affect the estimation.
For prognostic settings:
Strategy A: time-window size was 365 days, 2 years of forecasting, a lookback of
19 samples format (e.g., samples from time
19 until time
with a total of 20 samples)
as input, and 20 epochs for neural networks adjustments. For simplicity, we assume
for this method that new data is available every 15 days to update RUL estimation.
The model hyperparameters used for prognostics are summarized in Table 1.
Strategy B: a lookback of 9 samples format (e.g., samples from time
9 until time
with a total of 10 samples) as input, and 15 epochs for neural networks adjustments.
The model hyperparameters used for prognostics are summarized in Table 2.
All the algorithms were implemented in Python version 3.8.5 and ran on a computer
with an Intel
Processor i5-3230M of 2.6 GHz
4 cores, with 8 GB RAM, and using
Linux Mint 20.1 Ulyssa (64 bits) as OS.
Two prognostic strategies were tested in three problems:
Crack Growth in Section 4.1.2: is a classical problem in the literature in which the
degradation is a monotonical non-decreasing trajectory. The worst performances
are given by strategy A, where only the Prophet model was relatively close to the
ground truth RUL. Whereas, the strategy B, all prediction models are significantly
well performed on both metrics.
IFP Degradation in Section 4.2.2: the historical degradation signals are not totally
monotonous with different degradation levels and speeds, resulting in different failure
threshold values for a set of signals. With this insight, defining a unique failure
threshold for all the signals and forecasting the dynamic of the signal until reaching the
failure threshold as described by strategy A does not work well. Therefore, clustering
signals by degradation levels helps to define appropriately the failure threshold given
the characteristic of similarity to a set of historical run-to-failure signals from a cluster.
Therefore, using strategy B improves the prediction of RULs, in which ESN is the less
accurate model than the other models tested.
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Table 1. Models setting used for strategy A.
input_size: 20 input_shape: (20, 1) input_shape: (20, 1)
output_size: 1 units (GRU): 20 units (LSTM): 20
reservoir_size: 100 activation (GRU): reLU activation (LSTM): reLU
spectralRadius: 0.75 units (Dense): 20 units (Dense): 20
noise_scale: 0.001 activation (Dense): reLU activation (Dense): reLU
leaking_rate: 0.5 units (Dense): 1 units (Dense): 1
sparsity: 0.3 activation (Dense): linear activation (Dense): linear
activation: tanh optimizer: adam optimizer: adam
feedback: True
regularizationType: Ridge
regularizationParam: auto
changepoint_prior_scale: 0.05
seasonality_prior_scale 0.01
daily_seasonality: False
Table 2. Models setting used for strategy B.
ESN GRU Dense
input_size: 10 input_shape: (10, 1) input_shape: 10
output_size: 1 units (GRU): 15 units (Dense): 50
reservoir_size: 250 activation (GRU): reLU activation (Dense): reLU
spectralRadius: 1.0 recurrent_dropout (GRU): 0.5 dropout: 0.5
noise_scale: 0.001 units (GRU) 15 units (Dense): 25
leaking_rate: 0.7 activation (GRU): reLU activation (Dense): reLU
sparsity: 0.2 recurrent_dropout (GRU): 0.5 dropout: 0.5
activation: tanh units (Dense): 1 units (Dense): 1
feedback: False activation (Dense): linear activation (Dense): linear
regularizationType: Ridge optimizer: adam optimizer: adam
regularizationParam: 0.01
Camera Resolution Degradation in Section 4.3.3: the degradation trajectory showed
irregularities similar to the IFP signals, in which there is some segment increase
and then decrease, and vice versa. Therefore, the degradation trajectory is also not
completely monotonous. Addressing this problem with strategy A showed some
difficulties, particularly trying to forecast the dynamic or trend of the signal when the
trend of the segment changes in the opposite sense to the degradation, obtaining an
overestimation of the RUL. Working with this strategy showed that only the Prophet
approximates the ground truth, but it is still not good enough and acceptable. From
the strategy B perspective and using the RUL predictive model transferred from the
IFP setting provided better results compared to the previous strategy, converging to
the ground truth as it reaches the EoL with a few minor exceptions.
For the three problems addressed in this work, the degradation signals present ir-
regularities that affect the forecast of the dynamic of the signal by a fitted model; even
with Prophet, which is based on time series decomposition, it could not handle these
irregularities to allow a trustworthy RUL prediction to all the degradation problems.
In most of the cases, RNN models provided an underestimated RUL, opposite to the
results of the linear forecasting model such as Prophet. The time spent in the prognostic
Sensors 2022,22, 5323 20 of 29
process using strategy A are shown in Table 3, where we can see that ESN is the fastest
method because of its simplicity in training and forecast, followed by Prophet, and finally,
LSTM and GRU were similar in the time spent.
Table 3. Time performance measured in seconds.
Problem Prophet ESN LSTM GRU
Crack growth 252.40 109.49 2170.89 2197.84
Resolution Degradation 193.41 31.60 1995.64 1997.99
IFP Degradation 82.28 38.20 892.36 890.27
Concerning strategy B, the results showed that this strategy obtained better estimations
of RULs. It seems to be robust to irregularities present in the signal, and it is helpful for
problems with similar degradations and scarce historical run-to-failure signals. With this
method, it is only necessary to fit the models once and simply call the best representative
model by the classifier to predict the RUL, so the time spent using the fitted model to
calculate the RUL is almost negligible.
Finally, two main points must be highlighted. First, the fault detection framework
defined in our previous work [
] was designed from historical fault information of a pair
of IFPs out of the 132 available distributed in the 66 ALMA antennas and was validated on
other IFP data achieving good detection performance. Now by updating the pre-processing
module in this work, it was possible to improve the robustness by reducing the sensitivity
generated by the existing noise level. This was validated in other IFPs data preserving
the same performance and also found that the same effect applied to other signals similar
to those of IFPs can be obtained, such as the average resolution of the camera. Second,
the signals that are in the clusters do not fully represent the historical signals of the IFPs;
for validation purposes, some signals that were used to verify their effectiveness in the
prediction of the RUL were excluded; one of them is shown in Figure 10a, the other signals
showed very similar results, and most interestingly, that using the models fitted with
the IFPs data it is possible to obtain a good approximation in the RUL applied to other
components that have signals with similar degradations, in this case, applied to the camera
resolution signal. This indicates the power of generalization that the adjusted models have
against other similar problems.
6. Conclusions
This work shows a fault detection framework that can be transferable or scalable to
other applications with similar degradation behaviors but not necessarily with the same
statistical characteristics as the particular problem for which it was developed initially.
Hence, it is a helpful tool because it can be used in many applications to detect faults in the
system of interest without any changes in the method.
We also tested the performance of RNN models and a time series decomposition
model called Prophet to measure the precision of the RUL estimation using standard
metrics proposed in [
] that allow a systematic evaluation and a level of confidence for
model selection. Through this performance measurement scheme, one could eventually ask
which model is the best? We argue that the best would be one that has the largest PH value
and a lower
—additionally, an underestimation of the RUL close to the ground truth.
So, future works could use this as a guideline for model testing and the measurement of
quality of the model used for prognostic in RUL estimation.
One of the weaknesses of this proposal in forecasting is that it depends on a catas-
trophic failure threshold to estimate the RUL of a component. Furthermore, it considers
a deterministic threshold that could be a bit conservative if it is chosen as the worst
case scenario.
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7. Future Work
Our approach has shown to work effectively in different settings with slow degrada-
tion faults, adapting to each environment effectively. This method, together with several
others that have been developed in the literature, will help organizations transform data
into information. The challenge then becomes transforming this new vast information into
actionable decisions. Hence, as part of our future work, we will work in:
Improving the computation of uncertainty measurements of RUL predictions. This
computation will help develop new prescriptive maintenance approaches that help in
the decision-making process of maintenance procedures.
Test this approach on other problems with similar degradation faults to continue
evaluating the robustness of this run-to-failure critical segment clustering approach to
predict a component’s RUL value.
Author Contributions: Conceptualization and validation, A.D.C., R.A.C. and G.A.R.; methodology,
A.D.C. and G.A.R.; software, analysis, visualization, and writing—original draft preparation, A.D.C.;
supervision and writing—review and editing, R.A.C. and G.A.R.; funding acquisition, R.A.C. and
G.A.R. All authors have read and agreed to the published version of the manuscript.
This research was partially funded by FONDECYT 1180706, PIA/BASAL FB0002, and
ASTRO20-0058 grants from ANID, Chile.
Institutional Review Board Statement: Not applicable.
Informed Consent Statement: Not applicable.
Data Availability Statement:
The data presented in this study are available on request from the
corresponding author.
The Atacama Large Millimeter/submillimeter Array (ALMA), an international
astronomy facility, is a partnership of the European Organisation for Astronomical Research in the
Southern Hemisphere (ESO), the U.S. National Science Foundation (NSF), and the National Institutes
of Natural Sciences (NINS) of Japan in cooperation with the Republic of Chile. ALMA is funded by
ESO on behalf of its Member States, by NSF in cooperation with the National Research Council of
Canada (NRC) and the National Science Council of Taiwan (NSC) and by NINS in cooperation with
the Academia Sinica (AS) in Taiwan and the Korea Astronomy and Space Science Institute (KASI).
ALMA construction and operations are led by ESO on behalf of its Member States; by the National
Radio Astronomy Observatory (NRAO), managed by Associated Universities, Inc. (AUI), on behalf
of North America; and by the National Astronomical Observatory of Japan (NAOJ) on behalf of East
Asia. The Joint ALMA Observatory (JAO) provides the unified leadership and management of the
construction, commissioning, and operation of ALMA. The authors would like to thank José Luis
Ortiz, from ALMA, for his support with the relevant data.
Conflicts of Interest: The authors declare no conflict of interest.
The following abbreviations are used in this manuscript:
RUL Remaining Useful Life
RNN Recurrent Neural Network
ALMA Atacama Large Millimeter Array
CBM Condition-Based Maintenance
PHM Prognostic and Health Management
PH Prognostic Horizon
EoL End-of-Life
ESN Echo State Network
LSTM Long-Short Term Memory
GRU Gated Recurrent Unit
ADTK Anomaly Detection Toolkit
MVM Minimum Variance Matching
IFP Intermediate Frecuency Processor
Sensors 2022,22, 5323 22 of 29
LSB Lower Sideband
USB Upper Sideband
SW Switch matrix current
UT Unit Telescope
CCD Charge-Coupled Device
EoP End-of-Prediction
ANN Artificial Neural Network
SP Shortest Path
Appendix A. Evaluation Metrics
be the set of all time indexes when the prediction is made,
the ground
truth Remaining-Useful-Life (RUL),
is the allowable error bound,
time when the first
prediction is made, titime at the time index i, and EoP as End-of-Prediction of the RUL.
•Pronostic Horizon (PH)
: it identifies whether a method predicts within specified
limits around the ground truth End-of-Life (EoL) so that the predictions are considered
trustworthy. If it does, how much time does it allow for any maintenance action to be
taken. The longer PH better the model and more time to act based on the prediction
with some desired credibility. This metric is defined as:
PH =tEoL −ti, (A1)
i=minj|(j∈ J )∧$−
α=r∗−α·tEoL ,
α=r∗+α·tEoL .
: this metric quantifies the prediction quality by identifying whether
the prediction falls within specified limits at a particular time; this is a more stringent
requirement as compared to PH since it requires predictions to stay within a cone
of accuracy. Its output is binary since we need to evaluate whether the following
condition is met,
(1−α)·r∗(t)≤r(tλ)≤(1+α)·r∗(t), (A2)
tλ=tP+λ·(tEoL −tP).
•Relative Accuracy
: a similar notion as
accuracy where, instead of finding out
whether the predictions fall within given accuracy levels at a given time
, we also
quantitatively measure the accuracy by the following
r∗(tλ), (A3)
is defined previously at
accuracy. For measurement of the general
behavior of the algorithm over time, Cumulative Relative Accuracy (CRA) can be
used, and it is defined as
CR Aλ=1
is a weight factor as a function of the RUL at all time indices,
is the
set of all time indexes before
when a prediction is made, and
is the cardinality
operation of a set. The meaning of these metrics is that as more information becomes
Sensors 2022,22, 5323 23 of 29
available, the prognostic performance improvement will increase as it converges to
the ground truth RUL.
: it is a useful metric since we expect a prognostics algorithm to converge
to the true value as more information accumulates over time. Besides, it shows that the
distance between the origin and the centroid of the area under the curve for a metric
quantifies convergence, and a faster convergence is desired to achieve high confidence
in keeping the prediction horizon as large as possible. Lower distance means a faster
convergence. The computation of this metric is defined as, let
be the center of
mass of the area under the curve
. Then, the convergence
can be represented
by the Euclidean distance between the center of mass and (tP, 0), where
is a non-negative prediction error accuracy or precision metric. In other words,
this metric measures the fastness of convergence of a method.
Appendix B. Recurrent Neural Networks
Appendix B.1. Echo State Networks (ESNs)
The ESNs are a type of recurrent neural network developed by Herbert Jaeger [
that has a dynamical memory to preserve in its internal state a nonlinear transformation of
the input’s history. Hence, they have shown to be exceedingly good at modeling nonlinear
systems. Another advantage of ESNs is that they are easy to train because they do not need
to backpropagate gradients as classical ANNs do.
An ESN can be defined as follows: consider a discrete-time neural networks like
, with
input units,
internal units (also called reservoir units), and
units. Activations of input units at time step
, of internal units are
, and of output units
. The connection weight matrix
Win ∈IRNx×(1+Nu)
for the input weights,
for reservoir connections,
Wout ∈IRNy×(1+Nu+Nx)
connections to the output units, and
Wf b ∈IRNx×Ny
for the connections that are projected
back (also called feedback) from the output to the internal units. The connections go directly
from input to output units and connections between output units are allowed. Figure A1
shows the basic network architecture.
The activation of reservoir units are represented by
˜x(t+1) = tanhWin [1; u(t+1)]
+Wx(t) + Wfb y(t), (A5)
and are updated according to
x(t+1) = (1−δ)x(t) + δ˜x(t+1), (A6)
where δ∈(0, 1]is the leaky integrator rate. The output is calculated by
y(t+1) = Wout[1; u(t+1);x(t+1)], (A7)
denotes the vertical vector concatenation. The coefficients in
are computed
by using ridge regression, solving the following equation,
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Ytarget =WoutX, (A8)
with columns
[1; u(t);x(t)]
. . .
; and all
produced by presenting the reservoir with u(t)and Ytarget ∈IRNy×T.
Figure A1. The basic echo state network architecture.
Finally, the solution can be represented by
Wout =YtargetXTXXT+τI, (A9)
is the identity matrix and
is a regularization factor
(ridge constant). The ridge constant is estimated using grid search and time series cross-
validation methods.
Appendix B.2. Long-Short Term Memory (LSTM)
LSTM is another type of artificial recurrent neural network (RNN) architecture pro-
posed by Hochreiter and Schmidhuber [
] that deals with the vanishing gradient problem.
One LSTM unit is composed essentially of three gates: an input gate, an output gate, and a
forget gate; and a memory cell that remembers values over arbitrary time intervals, and the
three gates regulate the flow of information into and out of the cell. This type of RNN has
been found extremely successful in many applications [
] and was regarded as one of the
most popular and efficient RNN models using back-propagation as a training method. A
typical LSTM [82] is illustrated in Figure A2, and can be formulated as follow.
Let u(t)∈IRNuan input vector at time t, and consider Mof LSTM units, then
•Block input
: it consists of combining the input
and the previous output of LSTM
units h(t−1)for each time step t, and it is defined as
z(t) = φ(Wzu(t) + Rzh(t−1) + bz). (A10)
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•Input gate
: this gate decides which values needs to be updated with new information
to the cell state. It is computed as a combination of the input
, the previous output
of LSTM units h(t−1), and the previous cell state c(t−1)for each time step t,
i(t) = σ(Wiu(t) + Rih(t−1)
+pic(t−1) + bi). (A11)
•Forget gate
: it makes the decision of what information needs to be removed from the
LSTM memory, and it is calculated similarly to the input gate.
f(t) = σWfu(t) + Rfh(t−1)
+pfc(t−1) + bf. (A12)
•Cell state
: this step provides an update for the LSTM memory in which the current
value is given by the combination of block input
, input gate
, forget gate
and the previous cell state c(t−1).
c(t) = z(t)i(t) + c(t−1)f(t). (A13)
•Output gate
: this gate makes the decision of what part of the LSTM memory con-
tributes to the output and it is related to the current input vector
, the previous
output h(t−1), and the current cell state c(t).
o(t) = σ(Wou(t) + Roh(t−1)
+poc(t) + bo). (A14)
•Block output
: finally, this step computes the output
, which combines the current
cell state c(t)and the current output gate o(t).
h(t) = ψ(c(t)) o(t)(A15)
c(t − 1) c(t)
h(t − 1) h(t)
h(t − 1)
Figure A2. The basic LSTM architecture.
In the above description,
, for
k∈ {z
, are input weights, recurrent weights, peephole weights, and bias weights,
respectively. The operator
represent the point-wise multiplication of two vectors.
σ(x) =
1+exand φ(x) = ψ(x) = tanh(x).
Sensors 2022,22, 5323 26 of 29
Appendix B.3. Gated Recurrent Unit (GRU)
The GRU model was introduced by Cho et al. [
], which chose a new type of hidden
unit inspired by the LSTM unit. Basically, it combines the input gate and the forget gate
into a single update gate, and some operations are mixed with computing the update cell
state, making this model simpler, containing fewer variables than the basic LSTM model,
as shown in Figure A3. It can be formulated as follow,
h(t − 1) h(t)
Figure A3. The basic GRU architecture.
Let u(t)∈IRNuan input vector at time tand consider Mof GRU units, then,
•Update gate
: this gate determines how much previously learned information should
be passed on to the future,
z(t) = σ(Wzu(t) + Rzh(t−1) + bz). (A16)
•Reset gate: this gate decides how much previously learned information to forget.
r(t) = σ(Wru(t) + Rrh(t−1) + br). (A17)
•Cell state
: it consists of storing the relevant information from the past, using the reset
gate to affect the memory content.
c(t) = tanh(Wcu(t)
+Rch(t−1)r(t) + bc). (A18)
•Block output: finally, compute the output y(t)
h(t) = c(t)z(t) + h(t−1)(1−z(t))(A19)
In the above description,
, for
k∈ {z
, are
update gate weights, reset gate weights, cell state weigths, and bias weights, respectively.
The operator represent the point-wise multiplication of two vectors, and σ(x) = 1
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Math 7 Notes Volume & Surface Area of Prisms
Name: ______Date: ______
Math 7 Notes Volume & Surface Area of Prisms
A prism is a 3-dimensional figure with 2 congruent bases and rectangular lateral faces. The edges between the lateral faces are called lateral edges . All prisms are named by their bases, so the
prisms below is a pentagonal prism.
This particular prism is called a right prism because the lateral faces are perpendicular to the bases. Oblique prisms lean to one side or the other and the height is outside the prism.
Surface areais the sum of the areas of the faces of a solid.
Surface Area of a Right Prism: The surface area of a right prism is the sum of the area of the bases and the area of each rectangular lateral face. It is measured in square units.
Example AFind the surface area of the prism below.
Open up the prism and draw the net. Determine the measurements for each rectangle in the net.
Using the net, we have:
Because this is still area, the units are squared.
Example B Find the surface area of the prism below.
This is a right triangular prism. To find the surface area, we need to know the length of the hypotenuse of the base because it is the width of one of the lateral faces.
Looking at the net, the surface area is:
Volumeis the measure of how much space a three-dimensional figure occupies. The basic unit of volume is the cubic unit: cubic centimeter , cubic inch , cubic meter , cubic foot , etc. Each basic
cubic unit has a measure of one for each: length, width, and height.
Volume of a Rectangular Prism: If a rectangular prism is units high, units wide, and units long, then its volume is . If we further analyze the formula for the volume of a rectangular prism, we would
see that is equal to the area of the base of the prism, B, a rectangle. If the bases are not rectangles, this would still be true, however we would have to rewrite the equation a little.
Volume of a Prism: If the area of the base of a prism is and the height is , then the volume, .
Cavalieri’s Principle: If two solids have the same height and the same cross-sectional area at every level, then they will have the same volume.
Basically, if an oblique prism and a right prism have the same base area and height, then they will have the same volume.
Example C A typical shoe box is 8 in by 14 in by 6 in. What is the volume of the box?
We can assume that a shoe box is a rectangular prism. Therefore, we can use the formula above.
Example D You have a small, triangular prism shaped tent. How much volume does it have, once it is up?
First, we need to find the area of the base. That is going to be . Multiplying this by 7 we would get the entire volume. The volume is .
Even though the height in this problem does not look like a “height,” it is, when referencing the formula. Usually, the height of a prism is going to be the last length you need to use.
Example E What if your family were ready to fill the new pool with water and they didn't know how much water would be needed? The shallow end is 4 ft. and the deep end is 8 ft. The pool is 10 ft.
wide by 25 ft. long. How many gallons of water will it take to fill the pool? There are approximately 7.48 gallons in a cubic foot.
Even though it doesn’t look like it, the trapezoid is considered the base of this prism. The area of the trapezoidsare . Multiply this by the height, 10 ft, and we have that the volume is . To
determine the number of gallons that are needed, divide 1500 by 7.48. gallons are needed to fill the pool.
Watch this video for help with the Examples above. CK-12 Foundation: Chapter11PrismsB
Name: ______Date: ______
Math 7 Prep Task Volume & Surface Area of Prisms
1. Find the volume of the right rectangular prism below.
Use the right triangular prism to answer questions 2-6.
2. What shape are the bases of this prism? What are their areas?
3. What are the dimensions of each of the lateral faces? What are their areas?
4. Find the lateral surface area of the prism.
5. Find the total surface area of the prism.
6. Find the total volume of the prism.
7. Writing Describe the difference between lateral surface area and total surface area.
8. Fuzzy dice are cubes with 4 inch sides.
a. What is the surface area of one die?
b. Typically, the dice are sold in pairs. What is the surface area of two dice?
c. What is the volume of both dice?
9. Find the surface area and volume of the following solid. Bases are isosceles trapezoids
Find the value of , given the surface area.
Use the diagram below for questions 13-16. The barn is shaped like a pentagonal prism with dimensions shown in feet.
1. What is the area of the roof? (Both sides)
2. What is the floor area of the barn?
3. What is the area of the sides of the barn?
4. What is the total volume of the barn? | {"url":"https://docest.com/doc/88561/math-7-notes-volume-surface-area-of-prisms","timestamp":"2024-11-10T04:30:18Z","content_type":"text/html","content_length":"26819","record_id":"<urn:uuid:a358ac9c-d4e1-4cea-81e5-4a018189226d>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00636.warc.gz"} |
What is Fluid? & types of Fluid. - Fluid Mechanics
What is Fluid? & types of Fluid.
What is Fluid?
A Fluid is a substance that continually flows when an external force is applied. Fluids generally include liquids, gases and plasmas. To some extent, plastic solids are also considered fluids.
Fluids can be classified based on the following properties:
• Viscosity
• Conductivity
• density
• compressible or not.
Common types of Fluids based on Viscosity:
Fluids are separated in five basic types:
1. Ideal Fluid
2. Real Fluid
3. Newtonian Fluid
4. Non-Newtonian Fluid
5. Ideal Plastic Fluid
1. Ideal Fluid:
An Ideal Fluid is a fluid which is incompressible in nature and that has no viscosity. In practical, no fluid is ideal fluid because all the fluids have some viscosity. Thus, it is also called as
Imaginary Fluid.
2. Real Fluid:
Real fluids are the fluids which have some viscosity and are compressible in nature. All the fluids in actual are real fluids.
Examples: Kerosene, Petrol, Castor oil
3. Newtonian Fluid:
Newtonian fluids are the fluids that obey Newton’s law of viscosity. In other words, a real fluid whose shear stress is directly proportional to the rate of shear strain is known as Newtonian Fluid.
For a Newtonian fluid, viscosity totally depends upon the temperature and pressure of the fluid.
Examples: water, air, emulsions, Hydrogen
Main characteristics:
• Newtonian fluids do not have any elastic properties.
• They are incompressible, isotropic and unreal.
• Viscosity is temperature dependant,
• Viscosity further depends on the various pressures at which it is found.
• At a fixed temperature, their viscosity remains constant.
• With the increase in the temperature of a fluid, the viscosity decreases.
• The viscosity of this type of fluid is inversely proportional to the increase in its temperature.
• The Newtonian fluid was named after Sir Isaac Newton, who defined it as a viscous flow.
• They comply with Newton’s law of viscosity.
4. Non-Newtonian Fluid:
Non-Newtonian fluids are the fluids that do not obey Newton’s law of viscosity. In other words, a real fluid in which shear stress is not directly proportional to the rate of shear strain is known as
Non-Newtonian Fluid.
Examples: Flubber, Oobleck
Non-Newtonian fluids can further be classified as:
1. Time Dependent Fluids: These fluids are the ones for whose shear stress or viscosity decreases with time due to isothermal conditions and steady shear are known as Thixotropic and those that
increase with time under same circumstances are known as Rheopectic or Anti-thixotropic.
2. Time Independent Fluids: These fluids are the ones for which the rate of shear at a given point depends on the instantaneous shear stress at that point. These are also known as Non-Newtonian
viscous fluids or Purely viscous fluids.
These fluids are further of two types:
□ Fluids with yield stress
□ Fluids without yield stress: These also have two kinds-
☆ Pseudoplastic Fluids
☆ Dilatant Fluids
3. Visco-elastic Fluids
5. Ideal Plastic Fluid:
Ideal plastic fluid is a fluid, the shear stress is proportional to the rate of shear strain and in which shear stress is more than the yield value, is known as ideal plastic fluid.
Common types of Fluids based on density:
1. Gas
2. Liquid
There are two approaches to study the behaviour of fluid:
1. Lagrangian Approach: In this method, a single fluid particle is taken and studied carefully for its behaviour at varying size of sections.
2. Eulerian Approach: In this approach, fluid is passed through a section of fixed size and then behaviour of fluid particles are studied at different instants.
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What Is Unit Weight | What Is Density | What Is Unit Weight Material | Unit Weight Building Materials
What Is Unit Weight?
The ratio of the weight of a material to its volume is its unit weight, sometimes termed specific weight or weight density.
The unit weight of water, γw, is 9.81 kN/m^3 in the SI system and 62.4 lb/ft^3 in the English system.
Also, read: What Is a Field Dry Density Test | Different Type of Field Density Tests
What Is Density?
The term density is used herein to denote the mass-to-volume ratio of the material. However, some references, particularly older ones, use the term to describe unit weight.
Density is denoted by p. Because m = W/g, the unit weight terms defined above can be converted to mass densities as follows:
ρ = M/V
ρ = Density
M = Mass
V = Volume
In the SI system mass densities are commonly expressed in Mg/m^3, kg/m^3, or g/ml. The mass density of water can therefore be expressed ρw = 1000 kg/m^3 = 1 Mg/m^3 = 1 g/ml.
The mass density of soil solids typically ranges from 2640 to 2750 kg/m^3. Where mass or mass density values (g, kg, or kg/m^3) are given or measured, they must be multiplied by g(9.8 m/s^2) to
obtain weights or unit weights before performing stress calculations.
In the English system mass density values are virtually never used in geotechnical engineering and all work is performed in terms of unit weights (1b/ft^3).
Also, read: Tension Vs Compression | What Is Tension & Compression
What Is Unit Weight Material?
Unit weight material is also known as Specific material weight. Unit weight material is the weight of the material per unit volume.
As we know that the volume is identified in terms of liters or m^3 and weight is measured in terms of Kg or KN.
The unit weight of materials is the weight of material/unit volume which means the Unit weight is expressed in Kg/L or KG/m^3 or KN/m^3.
For easy reference, we organized all the building materials unit weights in a table. This list is a collective effort. Give a thumbs up if you liked it. Bookmark the page and use the search if
Also, read: Density of Cement Sand and Aggregate | Cement Density | Sand Density | Aggregate Density | list of Density
Unit Weight Building Materials
Sr.No. Material Unit Weight
1 A.C Sheet 17 Kg/m^3
2 Aerocon Bricks 551 to 600 Kg/m^3
3 Alcohol 780 Kg/m^3
4 Aluminum 2739 Kg/m^3
5 Anthracite Coal 1550 Kg/m^3
6 Ashes 650 Kg/m^3
7 Ballast 1720 Kg/m^3
8 Birch Wood 670 Kg/m^3
9 Bitumen 1040 Kg/m^3
10 Bituminous concrete 2243 Kg/m^3
11 Bituminous Macadum 2400 Kg/m^3
12 Brick 1600 – 1920 Kg/m^3
13 Brick Jelly 1420 Kg/m^3
14 Brick Masonry 1920 Kg/m^3
15 Cast iron 7203 Kg/m^3
16 Ceement Slurry 1442 Kg/m^3
17 Cement Concrete block 1800 Kg/m^3
18 Cement Grout 1500 to 1800 Kg/m^3
19 Cement Mortar 2000 Kg/m^3
20 Cement Plaster 2000 Kg/m^3
21 Cemrent 1400 Kg/m^3
22 Chalk 2220 Kg/m^3
23 Clay (Damp) 1760 Kg/m^3
24 Clay (dry) 1600 Kg/m^3
25 Clinker 750 Kg/m^3
26 Coal Tar 1200 Kg/m^3
27 Coarse Aggregate 1680-1750 Kg/m^3
28 Cobalt 8746 Kg/m^3
29 Copper 8940 Kg/m^3
30 Crude Oil 880 Kg/m^3
31 Cuddapa 2720 Kg/m^3
32 Disel 745 Kg/m^3
33 Dry Rubble Masonry 2080 Kg/m^3
34 Earth (Dry,loose) 1200 Kg/m^3
35 Fly Ash 1120 to 1500 Kg/m^3
36 Fly Ash Brick Masonry 2000 to 2100 Kg/m^3
37 Fly Ash Bricks 1468 to 1700 Kg/m^3
38 Galvanized Iron Steel (0.56 mm) 5 Kg/m^3
39 Galvanized Iron Steel (1.63 mm) 13 Kg/m^3
40 Gasoline 670 Kg/m^3
41 GeoPolimer Concrete 2400 Kg/m^3
42 Glass Reinforced Concrete 2000 to 2100 Kg/m^3
43 Granite Stone 2460-2800 Kg/m^3
44 Graphite 1200 Kg/m^3
45 Gravel Soil 2000 Kg/m^3
46 Green Concrete 2315 to 2499 Kg/m^3
47 Heavy Charcoal 530 Kg/m^3
48 Ice 910 Kg/m^3
49 Igneous rocks (Felsic) 2700 Kg/m^3
50 Igneous rocks (Mafic) 3000 Kg/m^3
51 Kerosene 800 Kg/m^3
52 Larch Wood 590 Kg/m^3
53 Laterite Stone 1019 g/m^3
54 Lead 11340 Kg/m^3
55 Light Charcoal 300 Kg/m^3
56 Light Weight Concrete 800 to 1000 Kg/m^3
57 Lime Concrete 1900 Kg/m^3
58 Lime Plaster 1700 Kg/m^3
59 Lime Stone 2400 – 2720 Kg/m^3
60 M Sand 1540 Kg/m^3
61 Magnesium 1738 Kg/m^3
62 Mahogany 545 Kg/m^3
63 Mangalore Tiles with Battens 65 Kg/m^3
64 Maple 755 Kg/m^3
65 Marble Stone 2620 Kg/m^3
66 Metamorphic rocks 2700 Kg/m^3
67 Mud 1600-1920 Kg/m^3
68 Nickel 8908 Kg/m^3
69 Nitric Acid (91 percent) 1510 Kg/m^3
70 Oak 730 Kg/m^3
71 Peat 750 Kg/m^3
72 Petrol 720 Kg/m^3
73 Pitch 1100 Kg/m^3
74 Plain Cement Concrete 2300 Kg/m^3
75 Plaster of Paris 881 Kg/m^3
76 Plastics 1250 Kg/m^3
77 Quarry Dust 1300 to 1450 Kg/m^3
78 Quartz 2320 Kg/m^3
79 Quick lime 33450 Kg/m^3
80 Rapid Hardening Cement 1250 Kg/m^3
81 Red Wood 450-510 Kg/m^3
82 Reinforced Cement Concrete 2400 Kg/m^3
83 Rubber 1300 Kg/m^3
84 Rubble stone 1600-1750 Kg/m^3
85 Sal Wood 990 Kg/m^3
86 Sand 1440-1700 Kg/m^3
87 Sandstone 2250 to 2400 Kg/m^3
88 Sedimentary rocks 2600 Kg/m^3
89 Shale Gas 2500 Kg/m^3
90 Silt 2100 Kg/m^3
91 Slag 1500 Kg/m^3
92 Stainless Steel 7480 Kg/m^3
93 Steel 7850 Kg/m^3
94 Sulphuric Acid (87 Percent) 1790 Kg/m^3
95 Teak 630-720 Kg/m^3
96 Tin 7280 Kg/m^3
97 Water 1000 Kg/m^3
98 Zinc 7135 Kg/m^3
Unit Weight of Materials
Unit weight or specific weight of any material is its weight per unit volume that means in a unit volume, how much weight of the material can be placed. Volume is measured in litres or cubic meters,
and weight is expressed in kg or kilo Newton.
Unit Weight
Unit Weight is the weight per unit volume of a material. Unit Weight = Weight of Material. Volume of Material. Example: U.W. = 103.2 lbs = 103.2 pcf.
What Is Unit Weight?
The unit weight of a substance is calculated by dividing its weight by its volume. In the International System of Units (SI), the unit weight is typically expressed in newtons per cubic meter (N/m³)
or kilograms per cubic meter (kg/m³). In the United States customary units, it is commonly expressed in pounds per cubic foot (lb/ft³).
Rubble Density Kg/m3
The bulk density ranges from 1,86 g/cm3 to 2,33 g/cm3.
Unit Weight Definition
Unit weight, also known as specific weight or weight density, is a measure of the weight of a substance per unit volume. It quantifies the amount of mass in a given volume of a material and is
expressed in units of force per unit volume, such as newtons per cubic meter (N/m³) or pounds per cubic foot (lb/ft³).
What Is Density?
Density is the substance’s mass per unit of volume. The symbol most often used for density is ρ, although the Latin letter D can also be used. Mathematically, density is defined as mass divided by
volume: where ρ is the density, m is the mass, and V is the volume.
Unit Weight Means
The specific weight, also known as the unit weight, is the weight per unit volume of a material. A commonly used value is the specific weight of water on Earth at 4 °C (39 °F), which is 9.807
kilonewtons per cubic metre or 62.43 pounds-force per cubic foot.
Cement Unit Weight
Normal weight concrete is in the range of 140 – 150 lbs./cu. ft. For normal weight concrete, a change in unit weight of 1.5 lbs./cu. ft.
Define Unit Weight
The specific weight, also known as the unit weight, is the weight per unit volume of a material. A commonly used value is the specific weight of water on Earth at 4 °C, which is 9.807 kilonewtons per
cubic metre or 62.43 pounds-force per cubic foot.
Density of Rubble Stone
The density of rubble stone can range from approximately 1500 to 2500 kilograms per cubic meter (kg/m³) or 94 to 156 pounds per cubic foot (lb/ft³). However, it’s important to note that these values
are approximate and can vary based on the specific type of rubble stone, its porosity, and other factors.
Density Vs Unit Weight
Density is mass per unit volume, whereas unit weight is force per unit volume. In this standard, density is given only in SI units. After the density has been determined, the unit weight is
calculated in SI or inch-pound units, or both.
What Is a Unit Weight?
The unit of measurement for weight is that of force, which in the International System of Units (SI) is the newton. For example, an object with a mass of one kilogram has a weight of about 9.8
newtons on the surface of the Earth, and about one-sixth as much on the Moon.
Density of Laterite Stone in Kn/m3
Laterite weighs 1.02 gm/cm³ or 1,019 kg/m³., i.e. density of Laterite is equal to 1,019 kg/m³.
The unit weight of cement can vary slightly depending on the specific type and brand of cement. However, as a general guideline, the typical unit weight of cement is around 1440 kilograms per cubic
meter (kg/m³) or 90 pounds per cubic foot (lb/ft³).
Unit Weight of Construction Materials
Unit weight of the material is simply defined as Mass per Volume (M/V) and also it is called Specific Weight or Density.
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Suggested Read –
1. Tshering Tobgyel says
□ Krunal Rajput says
Thanks, sir
Leave a Reply Cancel reply | {"url":"https://civiljungles.com/unit-weight-building-materials/","timestamp":"2024-11-12T13:40:20Z","content_type":"text/html","content_length":"169101","record_id":"<urn:uuid:c34dc114-a077-4ab4-91e8-4c77de1f4f5e>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00291.warc.gz"} |
Stefan Schwede: Global localization and equivariant Thom spectra | KTH
Stefan Schwede: Global localization and equivariant Thom spectra
Time: Tue 2022-03-15 14.15 - 16.00
Location: Institut Mittag-Leffler, Seminar Hall Kuskvillan and Zoom
Video link: Meeting ID: 921 756 1880
Participating: Stefan Schwede (University of Bonn)
Abstract: The aim of this talk is twofold. Firstly, I want to explain a systematic formalism to construct and manipulate Thom spectra in global equivariant homotopy theory. The upshot is a colimit
preserving symmetric monoidal global Thom spectrum functor from the infinity-category of global spaces over BOP to the infinity-category of global spectra. Here BOP is a particular
globally-equivariant refinement of the space Z x BO, which simultaneously represents equivariant K-theory for all compact Lie groups. Secondly, I want to use the formalism to derive certain universal
properties of real and complex bordism in the world of highly structured globally-equivariant spectra. For this we recall that equivariantly, the key features of the complex bordism spectrum are
embodied in two different objects: the spectrum mU is equivariantly connective and the natural target for the Thom–Pontryagin construction; the spectrum MU is equivariantly complex-oriented and
features in the theory of equivariant formal group laws. These two features are incompatible, and the morphism \(\mathrm{mU} \to \mathrm{MU}\) is not an equivalence. Both equivariant forms of complex
bordism assemble into multiplicative global homotopy types. I will explain why the morphism \(\mathrm{mU} \to \mathrm{MU}\) is a localization, in the infinity-category of commutative global ring
spectra, at the ‘inverse Thom classes’. The key principle underlying this result can be summarized by the slogan: ‘Thom spectra take group completion to localization at inverse Thom classes.’ | {"url":"https://www.kth.se/math/kalender/stefan-schwede-global-localization-and-equivariant-thom-spectra-1.1151262?date=2022-03-15&orgdate=2022-02-03&length=1&orglength=332","timestamp":"2024-11-06T11:29:47Z","content_type":"text/html","content_length":"56942","record_id":"<urn:uuid:bc58aa66-40ff-4483-a1fe-b47efcc4f841>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00768.warc.gz"} |
Fifth question of the practice exam – Q&A Hub – 365 Data Science
Resolved: Fifth question of the practice exam
can Lambda value be considered as the time taken for average number of events in that time interval?
as per the lecture Lambda is frequency of events in a specific time interval. So in that question - why is lambda not equal to avg events in time which is 2 (2 trees every four minutes).
2 answers ( 1 marked as helpful)
Hey Sandesh,
Thank you for reaching out!
Note that it is important to consider the same time intervals. In the Q&A example in the lecture, the time interval was 1 day:
Given that you receive 4 questions per day on average, what is the probability of receiving 7 questions today?
Therefore, lambda = 4.
In the practice exam question, we are told that the average number of trees per 4 minutes is 2. This makes for 4 trees per 8 minutes on average. Since the question asks for the probability of of less
that 5 trees appearing in 8 minutes, we need to use lambda = 4.
Hope this helps!
Kind regards,
365 Hristina | {"url":"https://365datascience.com/question/fifth-question-of-the-practice-exam/","timestamp":"2024-11-08T12:03:35Z","content_type":"text/html","content_length":"112342","record_id":"<urn:uuid:c3b52fb9-9416-4d33-be1c-f33f05e3f151>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00616.warc.gz"} |
I'm currently reading Understanding Distributed Systems. The author starts with few intro chapters and one of them touches security and cryptography. It briefly discusses ECC and gives very good
reference to learn more.
I won't be able to describe it better than Cloudflare's blog does, so here is a quick and inaccurate summary. ECC stands for Elliptic Curve Cryptography. The idea of the cryptographic systems is to
find so-called Trap Door Function. Trap Door Function is a function which is easy to do but very hard to undo. Most cryptography is based on that idea. RSA uses this as well and the "undo" function
in RSA is called factoring.
In ECC, we have an equation. When we create a graph with it, then selecting two points on the graph we can point to the third point (by drawing a line through the two selected points). When we arrive
to the third point we can either go straight up or straight down. Then we repeat such operation few times, and it becomes very hard to undo it.
This is very inaccurate description of ECC (this is not easy topic). The important facts are that ECC is a lot faster than RSA. This means that for the same number of bits, ECC is a lot stronger than
For more accurate description go here.
Thanks for reading! Read other posts? | {"url":"https://zbysiu.dev/til/ecc/","timestamp":"2024-11-07T19:10:17Z","content_type":"text/html","content_length":"5692","record_id":"<urn:uuid:770104cb-3379-4f58-b9ef-ca588ceec48b>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00460.warc.gz"} |
Article - ChannelFireball: Set Cubes
Might be one of your best articles to date. Loved the intersection of real math with cube design. I wished more articles were about decisions based on data analysis and less about "I think these
cards are powerful" (not that those aren't entertaining).
Terrific stuff, Jason. a few thoughts:
-did you know the 6
-i liked that this was mainly about the thought process behind building such a cube instead of the experience thereof, since that's what i was interested in and that's what we do best.
Not to go overboard on the praise, but I thought the article was stellar. These kind of recurring questions about common / uncommon / rare ratios pop up over and over again on various forums, so I
hope this article gets linked as an answer for years to come. I know you hate writing beginner articles, but this was easily an out-of-the-park home run.
Also, those stats and charts were sweet!
Thanks for the feedback guys.
I'm not really sure what you were saying. Were you being tautalogical? The rate at which a card is opened equates to the rate at which cards are in the packs?
I took it to mean that the ratio of cards within a rarity opened across First Big Set : Second Small Set: Third Small Set is roughly analogous to the ratio of commons / uncommons / rares opened
within a set itself.
Oh, uh, I see what he's saying. I didn't realize they were all same rarity / different set cards. | {"url":"https://riptidelab.com/forum/threads/channelfireball-set-cubes.191/","timestamp":"2024-11-12T06:59:38Z","content_type":"text/html","content_length":"58756","record_id":"<urn:uuid:beff15d2-cb8a-42ba-834b-c08003136e9e>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00164.warc.gz"} |
matematicasVisuales | The Diagonal of a Regular Pentagon and the Golden Ratio
We start with a regular pentagon and we want to know the ratio between the diagonal and the side of the pentagon.
We can think that the side of the regular pentagon is 1. And the diagonal is represented by
We have two homothetic isosceles triangles:
We can write a proportion:
Using the numbers 1 and
The value of
These are some basic properties of
We can say that the diagonal of a regular pentagon are in golden ratio to its sides.
The point of intersection of two diagonals of a regular pentagon are said to divide each other in the golden ratio (or "in extreme and mean ratio").
We suppose that Pythagoras and the Pythagoreans (around 500 BC) knew this ratio because their symbol was the pentagram (a pentagon with its diagonals) but the first written reference we have is in
Euclid´s Elements (around 300 BC):
Book VI, Definition 3: "A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less". (For example, in Euclid's
Elements at Clark University by D.E. Joyce)
The first reference to this kind of construction appeared in Book II, Proposition 11. That was before ratios were defined and then the proposition is in terms of areas: "To cut a given straight line
so that the rectangle contained by the whole and one of the segments equals the square of the remaining segment". (Euclid's Elements at Clark University by D.E. Joyce)
Using our notation:
Usign the golden ratio we can draw a regular pentagon, a golden triangle and a golden rectangle and is related with the icosahedron and the dodecahedron.
With a strip of paper we can make a knot and get a pentagon and a pentagram:
From Euclid's definition of the division of a segment into its extreme and mean ratio we introduce a property of golden rectangles and we deduce the equation and the value of the golden ratio.
A golden rectangle is made of an square and another golden rectangle.
A golden rectangle is made of an square an another golden rectangle. These rectangles are related through an dilative rotation.
The golden spiral is a good approximation of an equiangular spiral.
Two equiangular spirals contains all vertices of golden rectangles.
The twelve vertices of an icosahedron lie in three golden rectangles. Then we can calculate the volume of an icosahedron
With three golden rectangles you can build an icosahedron.
Some properties of this platonic solid and how it is related to the golden ratio. Constructing dodecahedra using different techniques.
The first drawing of a plane net of a regular dodecahedron was published by Dürer in his book 'Underweysung der Messung' ('Four Books of Measurement'), published in 1525 .
He studied transformations of images, for example, faces.
Two transformations of an equiangular spiral with the same general efect.
In an equiangular spiral the angle between the position vector and the tangent is constant.
Leonardo da Vinci made several drawings of polyhedra for Luca Pacioli's book 'De divina proportione'. Here we can see an adaptation of the dodecahedron.
A Dilative Rotation is a combination of a rotation an a dilatation from the same point. | {"url":"http://matematicasvisuales.com/english/html/geometry/goldenratio/pentagondiagonal.html","timestamp":"2024-11-05T10:38:38Z","content_type":"text/html","content_length":"24550","record_id":"<urn:uuid:2ee8d05e-db31-4e9e-b079-9bdd749a6708>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00459.warc.gz"} |
Data-driven modeling and control of an X-ray bimorph adaptive mirror
^aDepartment of Electrical Engineering and Computer Sciences, University of California, Berkeley, Berkeley, California, USA, ^bAdvanced Light Source, Lawrence Berkeley National Laboratory, Berkeley,
California, USA, and ^cAdvanced Photon Source, Argonne National Laboratory, Lemont, Illinois, USA
^*Correspondence e-mail: ggunjala@berkeley.edu, waller@berkeley.edu
Edited by A. Stevenson, Australian Synchrotron, Australia (Received 3 May 2022; accepted 18 November 2022)
Adaptive X-ray mirrors are being adopted on high-coherent-flux synchrotron and X-ray free-electron laser beamlines where dynamic phase control and aberration compensation are necessary to preserve
wavefront quality from source to sample, yet challenging to achieve. Additional difficulties arise from the inability to continuously probe the wavefront in this context, which demands methods of
control that require little to no feedback. In this work, a data-driven approach to the control of adaptive X-ray optics with piezo-bimorph actuators is demonstrated. This approach approximates the
non-linear system dynamics with a discrete-time model using random mirror shapes and interferometric measurements as training data. For mirrors of this type, prior states and voltage inputs affect
the shape-change trajectory, and therefore must be included in the model. Without the need for assumed physical models of the mirror's behavior, the generality of the neural network structure
accommodates drift, creep and hysteresis, and enables a control algorithm that achieves shape control and stability below 2nm RMS. Using a prototype mirror and ex situ metrology, it is shown that
the accuracy of our trained model enables open-loop shape control across a diverse set of states and that the control algorithm achieves shape error magnitudes that fall within diffraction-limited
1. Introduction
The next generation of light sources, including free-electron lasers and diffraction-limited storage rings, will produce X-ray beams of unprecedented brightness and coherent flux, enabling fast
experiments where wavefront phase information will be used to study matter in exquisite detail.
Reflective X-ray optics (e.g. mirrors and gratings) are illuminated at glancing angles of incidence and their surface shape tolerances are on the scale of nanometres. Achieving high Strehl ratios
from beamlines with several mirrors requires that individual-mirror height errors be limited to the nanometre scale, depending on the wavelength (Shi et al., 2016 ). To reach efficient
diffraction-limited X-ray optical performance in routine operation, it is necessary to correct residual aberrations arising from imperfect optical surfaces, misalignment, thermo-mechanical
deformations and dynamic mirror shape deformations caused by time-varying power loads and beam profiles (Sanchez del Rio et al., 2020 ; Cutler et al., 2020 ).
The development of X-ray adaptive optics started in the mid-1990s (Susini et al., 1996 ). Over the last decade, significant advances (Mimura et al., 2010 ; Sawhney et al., 2010 ) have led to the
commercial availability of piezo-bimorph mirrors (Alcock et al., 2019a ; Ichii et al., 2019 ) and their successful deployment on several beamlines (Matsuyama et al., 2016 ; Sutter et al., 2019 ). A
recent review by Cocco et al. (2022 ) summarizes alternative approaches to deformable mirrors. These mirrors have symmetrically placed bimorph elements attached to silicon mirror substrates, which
allow these systems to maintain thermal stability while providing one-dimensional shape actuation. Investigations of the mirrors' linear response demonstrate that their shape along the tangential
(longitudinal) direction can be controlled to a nanometre level in a predictive way (Vannoni et al., 2016 ; Alcock et al., 2019a ) as required for diffraction-limited performance. Studies of the
mirrors' dynamic response show that appreciable shape changes on the scale of 1s are possible (Alcock et al., 2019b ), as well as precise actuation relying on closed-loop feedback with accuracy
better than 1nm using arrays of laser interferometers (Alcock et al., 2019c ).
Measuring the performance of one such piezo-bimorph mirror using X-ray light and a wavefront sensor, we have observed time-dependent and history-dependent behaviors that defy a simple linear response
model. The use of piezo-electric materials to induce deformation of relatively thick substrates is associated with nonlinearities such as cross-talk between actuators, creep and hysteresis (Alcock et
al., 2015 ). The nanometre-scale magnitudes of these effects are relevant for our applications. Our work confronts the challenge of controlling the mirror shape in the presence of dynamic non-linear
behavior. For soft X-ray applications, in particular, wavefront monitoring interrupts the beam delivery. With the increased ability to predict the temporal behavior following actuation, fewer
wavefront measurements are required to achieve and maintain the desired shape, and systems progress toward the goal of open loop operation.
Current approaches for in situ mirror shape control rely on a linear model. Nonlinearities are usually compensated using closed-loop feedback from an X-ray wavefront sensor (Assoufid et al., 2016 ;
Liu et al., 2018 ; de La Rochefoucauld et al., 2018 ; Goldberg et al., 2021 ; Shi et al., 2020 ) or in situ monitoring (Badami et al., 2019 ). In practice, beam pick-up for feedback can be invasive
or interrupting, and systems that can operate in open-loop are desirable. Moreover, while linear models may be adequate for small surface changes, they have significant limitations for larger moves
and fail to capture the dynamic response at short (seconds) and long (minutes) time scales. Comprehensive physical modeling of the mirror (e.g. using finite-element analysis) is possible (Song et al.
, 2009 ; Jiang et al., 2019 ), but it requires highly specific system characterization, which cannot always be achieved in practice. In recent years, similar complex systems such as the storage ring
itself are now using techniques derived from machine learning to improve their stability (Leemann et al., 2019 ). Given their success, we aim to apply similar data-driven techniques to the operation
of adaptive X-ray optics and circumvent the limitations of linear methods.
2. Methods
In this paper we propose a two-part framework for the open-loop operation of an X-ray deformable mirror, involving (1) approximating the nonlinear system dynamics using a feedforward neural network,
and (2) control to a desired surface shape using nonlinear quadratic cost regulation over a finite time horizon. We developed and tested our approach using an ex situ visible-light Fizeau
interferometer to record the behavior of an adaptive mirror driven through various shape transitions. The test mirror is a PZT (lead zirconate titanate)-glued bimorph mirror fabricated by JTEC
Corporation (see Appendix A.1 for details). Our methodology is broadly applicable to X-ray or other optical systems utilizing an adaptive element, independent of the optical configurations and
wavelength ranges.
2.1. Predictive modeling
We want to find a discrete time model for the nonlinear dynamics of the bimorph mirror with the general form
where s[t] and v[t] represent the mirror surface and voltage input applied to each actuator at discrete times t, respectively. For a fixed time step of Δt, the model should predict the shape of the
mirror one time step in the future given its current shape, current input voltage, and a finite history of shapes and inputs [see Fig. 1 (a)]. The number of past states and inputs was chosen
empirically based on observed performance, and our time step of Δt – 2.0s was limited by interface latency with the prototype mirror.
We use a feedforward neural network (Bishop, 2006 ) as our discrete-time forward model for the nonlinear dynamics of the bimorph mirror, with five fully connected layers and exponential linear unit
activation. An additional skip (or shortcut) connection was introduced due to their effectiveness in modeling an identity mapping (Bishop, 2006 ; He et al., 2015 ); in our case, this greatly improved
the predictive performance when the mirror was at or close to rest. The network architecture is shown in Fig. 1 (b). The size of the input layer is determined by the dimension of our surface
representation, the number of actuators being controlled and the amount of history incorporated in making a prediction. Due to the limited field-of-view (FOV) of the Fizeau interferometer being used
to measure the mirror surface, we restrict actuation and analysis to the central 9 of 18 actuators. The mirror surface within the FOV is parametrized by heights at 14 equidistant points. A history of
three discrete-time inputs and measurements, in addition to the input at the current time, are concatenated and treated as the input to a neural network that predicts the surface shape at the next
time step.
To train our model, we collect a large ensemble of surface profiles occurring with sequences of random applied voltage inputs. The application of inputs and measurements of the mirror surface are
synchronized according to our discrete-time model. In this study, the voltage input to an individual actuator is limited to a range of [−100V,100V].
The surface measurements are acquired using a 4D Technology FizCam Fizeau interferometer, shown in Fig. 2 . Appendix A.1 contains more details about the setup. To minimize the contribution of small
vibrations, piston and tilt terms are removed from the 2D surface measurements. The surface is averaged in the narrow sagittal direction to produce 1D curves along the tangential dimension of
actuation. The sequential data are divided into sub-sequences which constitute input–output training examples for the supervised learning of our dynamics model. A total of 5964 examples were
collected and used to train the neural network. A detailed description of the data acquisition and learning process can be found in Appendix A.2 .
2.2. Control
Once the parameters of the nonlinear system dynamics model, f(·), are learned, the model is used to determine a sequence of voltage inputs that will drive the mirror from a measured initial state to
a desired final state in a given finite number of steps. Our algorithm solves a quadratic cost function, similar to the iterative linear quadratic regulator (Li & Todorov, 2004 ), that penalizes
state error at each simulated intermediate step. However, rather than linearizing the system dynamics around the initial state and finding an analytic solution to the reduced problem, we directly
minimize the non-convex objective function using the Adam algorithm (Kingma & Ba, 2014 ). While this does not theoretically guarantee optimality of the converged solution, the observed performance
meets our specifications in a majority of experiments.
The cost function is given by
subject to
with initial conditions
In this equation, N is the total number of computed steps, {v} represents the set of inputs {v[0],v[1],…v[N−1]}, s^* is the desired mirror shape, and the w[k] coefficients apply a weighted penalty
to the shape error at each time step. For the experiments discussed in this paper, we assume the first input, v[0], is applied at time step 0, and that the mirror begins at rest with some arbitrary
shape. Additionally, we use constant weights w[k] in these experiments.
3. Results and discussion
To test the predictive performance of our learned system dynamics model, we collect a sequential test dataset similar to our training data. The voltages applied to the mirror actuators are updated
and measurements of the surface are acquired at fixed time intervals (2,12,30s). The learned model is applied to sub-sequences of these data, and the prediction is compared with the next
measurement. Our test dataset consists of a total of 497 input–output examples. Here we evaluate the performance of our model when used with different time intervals, and in comparison with a
linear-response model.
3.1. Comparison with a linear-response model
We compare the performance of our model with that of a linear prediction based on `influence functions' (also called `actuator response functions' or `characteristic functions') (Hignette et al.,
1997 ; Goldberg & Yashchuk, 2016 ; Merthe et al., 2012 ). In that approach, voltage is supplied to individual actuators in isolation and the resulting surface is measured once the mirror has settled.
The set of shape measurements is treated as a basis, and a linear model predicts the resultant shape from an arbitrary set of inputs. To achieve a desired shape, the actuation matrix is then inverted
using least-squares in order to obtain the corresponding input voltages. In practice, a linear model such as this is applied iteratively, with measurement at each step, to overcome discrepancies with
the real-world response and converge to the target shape in several steps; this is known as feedback control.
Some examples of shape prediction are demonstrated in Fig. 3 (a), and the corresponding prediction errors for neural network (our method) and linear prediction are labeled. The aggregate performance
over the entire test dataset is shown in Fig. 3 (b). Overall, the mean prediction error for our method was 1.26nm root mean square (RMS), compared with 4.20nm RMS for linear prediction. Our neural
network model demonstrated lower prediction error than the linear model in 460/497 test examples. The cases for which linear prediction demonstrated lower prediction error generally involved very
small changes in input voltage.
It is well known that the linear model defined as a basis of influence functions is more appropriate for small changes in surface shape than for large changes. Moreover, since the influence functions
are measured after waiting for the mirror to settle, they fail to capture dynamics over small time scales. In Fig. 3 (c), the prediction errors for both models are plotted against the magnitude of
observed shape change for a variety of time scales.
3.2. Varying the time interval
While the training data are acquired using a fixed 2s time interval, we can apply the forward prediction process to a variety of time intervals, and compare with linear modeling. For all time
intervals tested, the linear prediction is computed using a basis of influence functions measured at steady state. Testing the model's predicted system response with a 2s time interval gives the
highest measured accuracy because the test data are acquired through a procedure identical to that of the training data. At this short time scale, errors from the linear model are essentially random
since transient effects are poorly characterized by influence functions measured at steady state.
For longer time scales, we apply our learned model iteratively, using predicted intermediate surface shapes as input for subsequent steps toward the goal shape. For 12s intervals (six prediction
steps), we see that the neural network prediction performs worse on a somewhat sparse set of examples, and linear prediction begins to exhibit a linear correlation between the shape change observed
(requested) and error. For 30s intervals (15 prediction steps), neural network predictive performance degrades over a larger set of examples, and linear prediction maintains its linear correlation,
albeit with lesser slope.
Unsurprisingly, the tests show that the predictive performance of our neural network model is best at the time interval used to acquire the training data; its performance degrades when used at longer
time intervals. This is likely due to the much more limited representation of repeated inputs in our training data set. As shown in Appendix A.2 , the training data set extracted from measurements
with repeated inputs (no voltage input change) is only one-third of the total training data set. Also, the maximum number of consecutive measurements with repeating voltage inputs is only five,
corresponding to a maximum of 10s mirror relaxation without varying voltages. While the network effectively predicts shape changes when all voltage inputs are updated, more data are required to
inform the network of convergence behaviors at different states when the inputs are held constant. We believe this can be improved with the acquisition of more finely sampled data. The results also
demonstrate that linear-model prediction can be effective for small shape changes (<20nm RMS) over long time scales. However, linear prediction, without repeated measurement and iteration, may be an
inappropriate choice for applications where speed and open-loop operation are important.
3.3. Directed shape control
In practice, mirror shape control will be used to compensate phase errors in the wavefront of a focused beam. Therefore, it is of central importance to be able to direct the mirror to achieve and
hold arbitrary shapes within its capabilities.
As a demonstration, we test the ability of our control algorithm to direct the mirror to a series of 50 random prescribed shapes. For each target shape, an initial surface measurement is acquired and
used to generate a ten-step sequence of voltage inputs, to achieve the shape and stabilize the mirror. We allow 150s to elapse between experiments so that the initial conditions [equation (4) ] are
approximately true.
The results of these experiments are shown in Fig. 4 . A selected sequence of transitions to three prescribed shapes is shown in Fig. 4 (a), where the measured surface profiles (colored, dashed)
closely approximate the desired shapes (black, solid). Figure 4 (b) shows the voltages applied to each of the nine actuators over the ten steps that were generated by the optimization algorithm.
We observe that these voltage sequences sometimes demonstrate oscillatory behavior, suggesting that the model is accommodating dynamic effects such as overshoot and creep. In Fig. 4 (c), we see that
the algorithm drives the mirror close to the goal after the first step, with the remaining nine steps being used to maintain the position. Some overshoot may still occur, as the error with respect to
the prescribed shape is often slightly larger after ten steps than after the first step. This may be caused by a combination of the non-convexity of the optimization problem and the lack of
guaranteed optimality, and by any residual errors in the predictive capability of our learned system dynamics model. The former can be somewhat addressed by changing the parameters of the Adam
algorithm (learning rate, iterations) or the weights w[k] in equation (2) , or by adding regularization to the objective function, e.g. penalizing RMS differences between time-adjacent voltage inputs
or predictions (`velocity'). Figure 4 (d) shows the aggregate performance of our control algorithm across the 50 test cases. The mean RMS errors between the measured and prescribed shapes are 1.70nm
after the first step and 1.91nm after ten steps.
Among the directed shape-control tests, we drove the mirror to a set of cylindrical shapes with prescribed radii of curvature from 2km to 6km. This emulates the case of an adaptive mirror used to
vary the focal distance, as in Sutter et al. (2019 ). In our applications, we consider these to be relatively large moves, with central surface height changes from 146.3nm in the 6km case to
429.0nm in the 2km case. Test results are shown in Fig. 5 . We observe that the mean RMS errors between the measured and prescribed shapes are 1.44nm after one step and 1.51nm after ten steps.
4. Conclusion
We have shown that the combination of a data-driven model for piezo-bimorph adaptive mirror shape dynamics and an optimization-based control strategy was able to reduce residual mirror figure errors
in open-loop operation below 2nm RMS, outperforming linear models and achieving the shape-control accuracy required to achieve diffraction-limited performance in the X-ray regime.
Our method effectively accounts for creep and hysteresis, nonlinear properties that currently limit the performance of such devices in open-loop operation. Accurate predictive modeling to achieve
stable arbitrary surface shapes is essential for effective deployment on high-coherent-flux X-ray beamlines where continuous feedback may be difficult to implement.
This calibration method is simple to implement and easily automated, requiring only a sequence of random shape commands and surface profile measurements. The data can be collected ex situ, as
presented in this paper, or even in situ with a wavefront sensor, where the phase of the beam can be mapped back into the mirror shape if required. The method is also robust, providing accurate
predictions and control across the full range of operation of the mirror. Other types of adaptive mirrors, such as resistive-element mirrors (Cocco et al., 2020 ), can also be characterized with this
The number of shape measurements required to build the training dataset is larger than what is required to acquire the characteristic functions in the linear model, but the training data can be
gathered during routine operation, over time. There is some flexiblity around the structure of the neural network itself (hyperparameters such as number of inputs, layers), but the performance level
we found is very close to the noise level of our sensor, and needed no further refinement despite being rather economical.
A1. Experimental setup
The bimorph mirror was fabricated by JTEC and measured with a metrology setup designed in-house at the Advanced Photon Source, using a FizCam 2000 Fizeau interferometer (4D Technology) with 100mm
aperture. The interferometer is oriented vertically to measure mirrors with the optical surface facing upwards. The setup includes a manual tip-tilt stage for high-precision mirror alignment. A Lexan
enclosure is constructed to enclose the X-ray mirror and the transmission flat of the interferometer to avoid air turbulence and reduce temperature fluctuation.
The bimorph mirror substrate (enclosed in the mirror box as shown in Fig. 6 ) has an overall dimension of 160mm (length) × 50mm (width) × 10mm (thickness). The Pt-coated optical surface is 150mm
(length) × 8mm (width), centered on the top side of the mirror. There are two piezo strips (one strip is visible in Fig. 6 ), each with 18 separate electrodes, that sandwich the optical surface. The
two electrodes at a corresponding position of the two strips comprise one electrode channel (referred to as the actuator in this paper), which locally perturbs the optical surface. There are a total
of 18 such channels (CH1 to CH18). There are two additional piezo strips glued to the bottom surface of the mirror, which act as a single electrode channel (CH20) that globally perturbs the optical
surface. The grounded channel (CH19) is on the back side of all piezo strips and adjacent to the mirror surface. The resting state of the mirror is defined as when all 20 channels are at the same
voltage. In this work, we kept CH19 and CH20 at the fixed voltage of 500V and only varied the top-surface channels in a relative range of [−100V,100V] around their nominal value of 500V. Since
the Fizeau interferometer aperture (100mm) is smaller than the mirror length, we only used the nine central channels (CH5 to CH13).
A2. Data acquisition and learning process
The data used to train our neural network model were acquired by applying voltages to the central nine mirror actuators and recording the resulting mirror shapes after a fixed, 2s, time interval. An
acquisition consists of ten images captured with an exposure time of 0.126ms and averaged together. Acquisitions are triggered such that this exposure time does not contribute delays to the 2s
interval. The voltages are uniform random values in the range [−100V,100V] and are rounded to the nearest tenth of a volt. A time interval of 2s between the application of voltage inputs and
surface measurement is maintained throughout the acquisition of training data.
We acquire eight separate sequences of surface measurements, starting with an initial measurement followed by 500 voltage changes. We acquire an additional four sequences of 501 measurements in which
voltage changes are held for five time steps (i.e. the same voltage input is applied for five consecutive measurements). We believe these data are vital for informing the model of system convergence
behavior. Since each training example requires four sequential measurements (one state to be predicted from three inputs), we can extract a maximum of 497 training examples from each of the
501-measurement sequences. In total, the 12 sequences give 5964 training examples.
We used the PyTorch library as the machine learning framework. To train our dynamics model, we perform a 15-fold cross validation on the training dataset with 3000 iterations of the Adam algorithm
per fold. To test the model's performance, we acquire another independent sequence of measurements and divide into sub-sequences, as described above, resulting in 497 test examples.
The authors appreciate the support of Elaine DiMasi, Simon Morton and Howard Padmore of Lawrence Berkeley National Laboratory, and Luca Rebuffi of Argonne National Laboratory. We are grateful for the
guidance and technical support of Yoshio Ichii of JTEC, and Greg Maksinchuk and Robert Mattern at 4D Technology.
Funding information
Funding for this research was provided by: US Department of Energy, Office of Science (contract No. DE-AC02-05CH11231; contract No. DE-AC02-06CH11357; contract No. DE-SC0014664 to GG). GG was
supported by the US Department of Energy, Office of Science, Office of Workforce Development for Teachers and Scientists, Office of Science Graduate Student Research (SCGSR) program. The SCGSR
program is administered by the Oak Ridge Institute for Science and Education for the DOE under contract number DE-SC0014664.
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P = iv - (Intro to Engineering) - Vocab, Definition, Explanations | Fiveable
P = iv
from class:
Intro to Engineering
The equation p = iv defines the relationship between power (p), current (i), and voltage (v) in an electrical circuit. This equation shows that power is the product of current flowing through a
circuit and the voltage across that circuit. Understanding this relationship is crucial for analyzing how electrical devices operate, as it allows engineers to calculate the power consumed or
generated by various components.
congrats on reading the definition of p = iv. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. In the equation p = iv, power is measured in watts, where 1 watt equals 1 joule per second.
2. If either current or voltage increases while keeping the other constant, the overall power will also increase.
3. This equation can be rearranged to find current or voltage if the power is known, which helps in designing circuits.
4. In practical applications, this relationship helps engineers determine the efficiency and performance of electrical devices.
5. Understanding p = iv is essential for troubleshooting circuits, as it helps identify whether issues arise from insufficient voltage or current.
Review Questions
• How does the equation p = iv help engineers analyze electrical circuits?
□ The equation p = iv provides a straightforward way for engineers to calculate the power consumption of devices within an electrical circuit. By knowing either the current or voltage along
with the resulting power, engineers can diagnose performance issues, ensure proper functioning, and optimize designs. It allows them to quickly evaluate how changes in voltage or current can
impact overall circuit performance.
• What happens to power consumption when both current and voltage are doubled in a circuit?
□ If both current and voltage are doubled in a circuit, the power consumption increases by a factor of four. This is because using the equation p = iv, if i and v each double, then p becomes 2i
× 2v, resulting in 4iv. Understanding this relationship highlights the exponential effect that changes in current and voltage have on overall power usage.
• Evaluate the importance of understanding the relationship defined by p = iv when designing efficient electrical systems.
□ Understanding p = iv is crucial for designing efficient electrical systems because it helps predict how changes in one variable—current or voltage—affect overall power consumption. By
analyzing these relationships, engineers can create systems that minimize energy waste and enhance performance. This knowledge allows for better component selection and ensures that devices
operate within optimal power ranges, which is essential for sustainability and cost-effectiveness in modern engineering practices.
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Dyson Mod 27 Identities -- from Wolfram MathWorld
The Dyson mod 27 identities are a set of four Rogers-Ramanujan-like identities given by
(OEIS A104501, A104502, A104503, and A104504).
Bailey (1947) systematically studied and generalized Rogers's work on Rogers-Ramanujan type identities in a paper submitted in late 1943. At the time, G. H. Hardy was the editor of the Proceedings of
the London Mathematical Society and Hardy had recently taught the young Freeman Dyson in one of his undergraduate classes at Cambridge. He was therefore aware of Dyson's interest in
Ramanujan-Rogers-type identities through his rediscovery of the Rogers-Selberg identities. Ignoring the usual convention of keeping the referee anonymous (since as far as Hardy knew, Bailey and Dyson
were the only two people in all of England who were interested in Rogers-Ramanujan type identities at the time) and thinking that they would like to be in contact with each other), Hardy asked Dyson
to referee Bailey's paper.
A correspondence between Bailey and Dyson ensued. Using the ideas in Bailey's paper, Dyson discovered a number of new Rogers-Ramanujan-type identities, including the four mod 27 identities above.
Bailey suggested that Dyson publish his results in a separate paper, but Dyson declined, instead asking Bailey to include these identities in his own paper (with proper attribution to Dyson of
course), which is what was done.
Due to the paper shortage caused by World War II, Bailey's paper wasn't published until 1947. Bailey's followup paper (Bailey 1949) was submitted about six months later and once again Dyson refereed
it as well as contributed some additional identities. | {"url":"https://mathworld.wolfram.com/DysonMod27Identities.html","timestamp":"2024-11-01T23:45:16Z","content_type":"text/html","content_length":"64571","record_id":"<urn:uuid:c15a1080-bd98-40c2-bc9c-ffa85cfdc5ec>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00467.warc.gz"} |
How to calculate the value of scrap gold?
How do you calculate gold scrap
How to Calculate the Cost of Scrap Gold Scrap Gold Calculator
Enter the price of gold per troy ounce in USD
Enter the weight of the scrap gold in grams
Select the gold carat value of the product. The price of a miraculous product is determined by its carat value (K).
How to calculate the value of scrap gold
The value of waste depends on two main factors: weight and purity. So, if you know something about the price of a unit of gold per unit of weight, and also know the current carat value of an old
piece of jewelry, you can evaluate its dollar value as scrap gold. For example, if you have an 18 carat disc weighing 100 grams, the actual gold content of the gift is 75% (18 carat
How much do gold buyers pay for scrap gold
Pure gold currently pays around $1,250 per bit. After going through the math, this means that 10 carat gold is definitely “scrap” at around $16.35 per gram. And 14 carat gold will also break $23 at
$0.50 per gram. The scrap fee is related to how much these big sellers get paid when they send that gold to be melted down and that’s what most of them write with your gold.
How much is 14K scrap gold worth today
First, this means that the available price per gram is $14.79 ($23 for $1.555 since 1dwt in stage 1 = 0.555g). However, since one gram of pure 14 carat gold contains 0.583 grams of pure gold, the
price per gram of pure gold is actually $25.37 ($14.79 for 0.583 grams).
How do you calculate the melt value of gold
The cost to melt coins and bars can be determined by multiplying the standard weight of most gold coins by the spot price of your old watch. For example, if you have 0.8 ounce (oz) necklaces and the
spot price of gold should be $1,600, then the melt value would be $1,280.
Are gold coins worth more than scrap gold
This suggests that scrap gold will almost certainly fetch valuable coins at a lower price, especially since the coins also typically have a higher gold value than most scrap coins. In terms of
accessories, a 14 carat piece of jewelry is 0.585 gold fineness, with most pieces being 0.900 gold fineness and above. | {"url":"https://www.vanessabenedict.com/gold-scrap-calculator/","timestamp":"2024-11-02T04:37:03Z","content_type":"text/html","content_length":"68802","record_id":"<urn:uuid:cf861acc-ae66-4a27-8ad4-67c3acfcbc1c>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00724.warc.gz"} |
Anna University - Electromagnetic Theory (EMT) - Question Bank - All Units
Electromagnetic Theory
Part A
1. State divergence theorem.
2. State Stoke’s theorem.
3. What is del operator? How is it used in density curl, gradient and divergence?
A = x ax + y ay+ y az
4. Define vector product of two vectors.
5. Write down expression for x, y, z in terms of spherical co-ordinates r,θ and φ.
6. Write down the expression for differential volume element in terms of spherical
7. What is the divergence of curl of a vector?
8. Write expression for differential length in cylindrical and spherical co-ordinates.
9. Find the divergence of F= x y ax+ y x ay + z x az
10. Define a vector and its value in Cartesian co-ordinate axis.
11. Verify that the vectors A= 4 ax - 2ay + 2az and B = -6ax + 3ay - 3az are parallel to each other.
12. List out the sources of electromagnetic fields.
13. When a vector field is solenoidal and irrotational.?
Part B
14. (i) State and prove Divergence theorem.
(ii) For a vector field A, show explicitly that ∆.∆ x A=0: that is the divergence of the curl of any vector field is zero.
15. (i) State and prove Stroke’s theorem.
(ii) Show that the vector H = (y2- z2+3yz-2x) ax + (3xz+2xy) ay + (3xy- 2xz+2z) az is both irrotational and solenoidal.
16. Using Divergence theorem, evaluate ∫∫ E.ds = 4xz ax - y2 ay + yz az over the cube bounded by x=0,x=1,y=0,y=1,z=0,z=1
17. What is the different co-ordinate systems used to represent field vectors? Discuss about them in brief.
18. (i) Given A= 5 ax and B= 4 ax + t ax Find t such that the angle between A and B is 45.
(ii) Using Divergence theorem evaluate ∫∫ A.ds where A = 2xy ax + y2 ay + 4yz az and S is the surface of the cube bounded by x =0 , x = 1; y = 0, y = 1; and z = 0, z = 1.
19. (i) Determine the divergence and curl of the vector A = x ax + y ay+ y az
(ii) Determine the gradient of the scalar field at P(√2, л/2, 5) defined in cylindrical co-ordination system as A = 25 r sin Ф.
20. Given point P(-2,6,3) and vector A= y ax + (x+z) ay Evaluate A and at P in the Cartesian, cylindrical and spherical systems
Part A
1. State coulomb’s law.
2. State Gauss’s law.
3. Define dipole moment.
4. Define electric flux and flux density.
5. Define electric field intensity or electric field.
6. What is a point charge?
7. Write the Poisson’s and Laplace equation.
8. Define potential and potential difference.
9. Give the relationship between potential gradient and electric field.
10. Define current density.
11. State point form of Ohm’s law.
12. Define polarization.
13. Express the value of capacitance for a coaxial cable.
14. What is meant by displacement current?
15. State the boundary conditions at the interface between two perfect dielectrics.
16. Write down the expression for the capacitance between (a) two parallel plates (b) two coaxial cylinders.
17. Calculate the capacitance of a parallel plate capacitor having an electrode area of 100 cm2. The distance between the electrodes is 3 mm and the dielectric used has a permittivity of 3.6 the
applied potential is 80 V. Also compute the charge on the plates.
18. An infinite line charge charged uniformly with a line charge density of 20 n C/m is located along z-axis. Find E at (6, 8, 3) m.
Part B
19. (i) Derive an expression for electric field due to an infinite long charge from its principles.
(ii)Derive the boundary conditions at the charge interfaced of two dielectric media.
20. Find the electric field intensity due to the presence of co-axial cable with inner conductors of ρs c/m2 and outer conductor of - ρs c/m2.
21. What is dipole? Derive the expression for potential and electric field intensity due to a dipole.
22. (i) Compare and explain conduction current and displacement current.
(ii) A circular of radius ‘a’ meter is charged uniformly with a charge density ρs c/m2. Find the electric field at a point ‘h’ meter from the disc along its axis.
23. A circular disc of 10cm radius is charged uniformly with a total charge of 10^-6 C. Find the electric intensity at a point 30cm away from the disc along the axis.
24. (i) Derive the expression for electric field intensity due to a circular surface charge.
(ii) Two parallel plates with uniform surface charge intensity equal and opposite to each other have an area of 2 m2 and distance of separation of 2.5 mm in free space. A steady potential of 200 V is
applied across the capacitor formed. If a dielectric of width 1 mm is inserted into this arrangement what is the new capacitance if the dielectric is a perfect non- conductor?
25. (i) State and prove Gauss’s law.
(ii) Derive an expression for energy density in electrostatic fields.
26. (i) Derive Poisson’s and Laplace equation.
(ii) Three concentrated charges of 0.25 µ C are located at the vertices of an equilateral triangle of 10 cm side. Find the magnitude and direction of the force on one charge due to other two charges.
27. (i) Using Laplace’s equation find the potential V between two concentric circular cylinders, if the potential on the inner cylinder of radius 0.1 cm is 0Vand that on the outer cylinder of radius
1 cm is 100 V.
(ii) A point charge of 5 n C is located at (-3, 4,0 ) while line y = 1, z = 1 carries uniform charge 2 n C/m. If V =0V at O (0, 0, 0) find V at A (5, 0, 1)
Part A
1. State Ampere’s circuital law.
2. State Biot-Savart law.
3. State Lorenz law of force.
4. Define magnetic scalar potential.
5. Write down the equation for general, Integral and point form of Ampere’s law.
6. What is field due to toroid and solenoid?
7. Define magnetic flux density.
8. Write down the magnetic boundary conditions.
9. Give the force on a current element.
10. Define magnetic moment.
11. Give torque on a solenoid.
12. State Gauss’s law for magnetic field.
13. Define magnetic dipole.
14. Define magnetization.
15. Define magnetic susceptibility.
16. What are the different types of magnetic materials?
17. What is the inductance per unit length of a long solenoid of N turns and having a length L meters?Assume that its carries a current of I amps.
18. A parallel plate capacitor with plate area of 5 cm2 plate separation of 3 mm has a voltage 50 sin 103 t applied to its plates. Calculate the displacement current assuming ξ= 2 ξ0
Part B
19. (i) Derive an expression for the force between two current carrying wires.Assume that the currents are in the same direction.
(ii) State and explain Biot-Savart’s law.
20. Obtain an expression for the magnetic field around long straight wire using magnetic vector potential.
21. (i) Obtain an expression for the magnetic flux density and field intensity due to finite long current carrying conductor.
(ii) Give a brief note on the magnetic materials.
22. Derive the expression for magnetic field intensity on the axis of solenoid at a) center and b) end point of the solenoid.
23. (i) State and explain Ampere’s circuital law.
(ii) State and prove boundary condition for magnetic field.
24. Derive an expression for the inductance of solenoid and toroid.
25. Derive an expression for the inductance per meter length of two transmission lines.
26. Obtain the expression for energy stored in magnetic field and also derive an expression for magnetic energy density.
27. (i) Derive and expression for self inductance of co-axial cable.of inner radius a and outer radius radius b.
(ii) A circular loop located on x2+y2 =9, z=0 carries a direct current of 10 A along aθ. Determine H at (0,0, 4) and (0,0,-4).
28. An air coaxial transmission line has a solid inner conductor of radius ‘a’ and very thin outer conductor of inner radius ‘b’. Determine the inductance per unit length of the line.
Part A
1. State Faraday’s law of electromagnetic induction.
2. Define self inductance.
3. Define mutual inductance.
4. Define coupling coefficient.
5. Define reluctance.
6. Give the expression for lifting force of an electromagnet.
7. Give the expression for inductance of a solenoid.
8. Give the expression for inductance of a toroid.
9. What is energy density in the magnetic field?
10. Define permeance.
11. Distinguish between solenoid and toroid.
12. Write down the general, integral and point form of Faraday’s law.
13. Distinguish between transformer emf and motional emf.
14. Compare the energy stored in inductor and capacitor.
15. State Lenz’s law.
16. Define magnetic flux.
17. Write the Maxwell’s equations from Ampere’s law both in integral and point forms.
18. Write the Maxwell’s equations from Faraday’s law both in integral and point forms.
19. Write the Maxwell’s equations for free space in point form.
20. Write the Maxwell’s equations for free space in integral form.
21. Determine the force per unit length between two long parallel wires separated by 5 cm in air and carrying currents of 40 A in the same direction.
Part B
22. (i) State and explain Faraday’s law.
(ii)Compare the field theory and circuit theory.
23. Develop an expression for induced emf of Faraday’s disc generator.
24. Derive the Maxwell’s equation for free space in integral and point forms explain.
25. Derive Maxwell’s equation from Faraday’s law and Gauss’s law and explain them.
26. Derive the Maxwell’s equation in phasor differential form.
27. Derive the Maxwell’s equation in phasor integral form.
28. Derive and explain the Maxwell’s equations in point form and integral form using Ampere’s circuital law and Faraday’s law.
Part A
1. Define a wave.
2. Mention the properties of uniform plane wave.
3. Define intrinsic impedance or characteristic impedance.
4. Calculate the characteristics impedance of free space.
5. Define propagation constant.
6. Define skin depth.
7. Define polarization.
8. Define linear polarization.
9. Define Elliptical polarization.
10. Define pointing vector.
11. What is complex pointing vector?
12. State Slepian vector.
13. State pointing theorem.
14. State Snell’s law.
15. What is Brewster angle?
16. Define surface impedance.
17. Write the wave equation in a conducting medium.
18. Compute the reflection and transmission coefficients of an electric field wave travelling in air and incident normally on a boundary between air and a dielectric having permittivity of 4.
19. Calculate the depth of penetration in copper at 10 MHZ given the conductivity of copper is 5.8 x 10 7 S/m and its permeability = 1.3=26 mH/m.
Part B
1. (i) Obtain the electromagnetic wave equation for free space in terms of electric field.
(ii)Derive an expression for pointing vector.
2. (i) Obtain the electromagnetic wave equation for free space in terms of magnetic field.
(ii) Calculate the intrinsic impedance, the propagation constant and wave velocity for a conducting medium in which σ = 58 ms/m, µ r = 1 at a frequency of f = 100 M Hz.
3. (i) Derive the expression for characteristic impedance from first principle.
(ii)Show that the intrinsic impedance for free space is 120π. Derive the necessary equation.
4. (i) Explain the wave propagation in good dielectric with necessary equation.
(ii)Define depth of penetration. Derive its expression.
5. (i) Derive the expressions for input impedance and standing wave ratio of transmission line.
(ii) Find the skin depth at a frequency of 1.6 MHz in aluminium σ= 38.2 ms/m and µr = 1.
6. (i) State and prove pointing theorem.
(ii) Define surface impedance and derive its expression.
7. Define Brewster angle and derive its expression. Also define loss tangent of a medium.
8. Determine the reflection coefficient of oblique incidence in perfect dielectric for parallel polarization.
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Optimization techniques for simulation must also account specifically for the randomness inherent in estimating the performance measure and satisfying the constraints of stochastic systems. We
described the most widely used optimization techniques that can be effectively integrated with a simulation model. We also described techniques for post-solution analysis with the aim of theoretical
unification of the existing techniques. However, a few studies rely on real computer simulations to compare different techniques in terms of accuracy and number of iterations.
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All factors must assume a finite number of values for this technique to be applicable. The analyst can attribute some degree of confidence to the determined optimal point when using this procedure.
Although the complete enumeration technique yields the optimal point, it has a serious drawback. If the number of factors or levels per factor is large, the number of simulation runs required to find
the optimal point can be exceedingly large. For example, suppose that an experiment is conducted with three factors having three, four, and five levels, respectively.
This procedure reduces the number of simulation runs required to yield an ‘optimal’ result; however, there is no guarantee that the point found is actually the optimal point. Of course, the more
points selected, the more likely the analyst is to achieve the true optimum. Note that the requirement that each factor assumes only a finite number of values is not a requirement in this scheme.
Replications can be made on the treatment combinations selected, to increase the confidence in the optimal point. Which strategy is better, replicating a few points or looking at a single observation
on more points, depends on the problem.
The pattern search technique is most suitable for small size problems with no constraint, and it requires fewer iterations than the genetic techniques. The most promising techniques are the
stochastic approximation, simultaneous perturbation, and the gradient surface methods. Stochastic approximation techniques using perturbation analysis, score function, or simultaneous perturbation
gradient estimators, optimize a simulation model in a single simulation run. This observing-updating sequence, done repeatedly, leads to an estimate of the optimum at the end of a single-run
simulation. With the growing incidence of computer modeling and simulation, the scope of simulation domain must be extended to include much more than traditional optimization techniques.
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Also suppose that five replications are desired to provide the proper degree of confidence. Then 300 runs of the simulator are required to find the optimal point. The simulated results based on the
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MU Computer Graphics & Virtual Reality - December 2012 Exam Question Paper | Stupidsid
MU Information Technology (Semester 5)
Computer Graphics & Virtual Reality
December 2012
Total marks: --
Total time: --
(1) Assume appropriate data and state your reasons
(2) Marks are given to the right of every question
(3) Draw neat diagrams wherever necessary
Solve any four:-
1(a) Draw and explain basic block diagram of Virtual Reality System.
5 M
1(b) Explain the significance of Homogenous co-ordinate System.
5 M
1(c) Rotate a triangle ABC by an angle 30^0 where the triangle has co-ordinates A (0, 0), B (10,2), C (7,4).
5 M
1(d) Compare DDA line algorithm with Bresenham's line algorithm.
5 M
1(e) List at least three Input and Output devices of VR system and explain any one in detail.
5 M
2(a) Prove that a shear transform can be expressed in terms of rotation and scaling operations.
7 M
2(b) Specify highlights and drawbacks of bezier curves. Construct the bezier curve of order three with control points P1(0,0), P2(1,3), P3(4,2) and P4(2,1). Generate at least five points on the
13 M
3(a) Describe any two VR architectures with neat diagrams.
10 M
3(b) What are fractals? Derive an equation D=(log N)/(log S). Outline the procedure of generating Koch curve or Hilbert curve
10 M
4(a) Develop a single transformation matrix which does the following on given object:
i. Reduces the size by 1/2
ii. Rotates about Y axis by (-300)
iii. Performs a single point perspective transformation projection z=0 and z=10
6 M
4(b) Derive a 3D inverse transformation for translating and scaling.
4 M
4(c) Explain with example Sutherland - Hodgeman Polygon clipping algorithm. List the shortcomings of this method, if any.
10 M
5(a) What are different types of projection? Derive the matrix representation for perspective transformation in XY plane and on negative Z axis.
10 M
5(b) Explain flood fill algorithm using four and eight connectivity method with suitable examples and diagrams. Compare the same with boundary fill algorithm.
10 M
6(a) Compare the capabilities and limitations of geometric and kinematic modelling techniques.
10 M
6(b) (i) Mesh and feature based warping.
5 M
6(b) (ii) 2D and 3D Morphing.
5 M
7 (a) Using Liang Barsky Algorithm, find the clipping co-ordinates of line segment with end co-ordinates A(-10,50), B(30,80) against the window (Xmin = -30, Ymin = 10) and (Xmax = 20, Ymax = 60).
10 M
7 (b) Write a detailed note on VR applications.
10 M
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PID Control in WPILib
PID Control in WPILib
This article focuses on in-code implementation of PID control in WPILib. For a conceptual explanation of the working of a PIDController, see Introduction to PID
WPILib supports PID control of mechanisms through the PIDController class (Java, C++, Python). This class handles the feedback loop calculation for the user, as well as offering methods for returning
the error, setting tolerances, and checking if the control loop has reached its setpoint within the specified tolerances.
Using the PIDController Class
Constructing a PIDController
While PIDController may be used asynchronously, it does not provide any thread safety features - ensuring threadsafe operation is left entirely to the user, and thus asynchronous usage is recommended
only for advanced teams.
In order to use WPILib’s PID control functionality, users must first construct a PIDController object with the desired gains:
// Creates a PIDController with gains kP, kI, and kD
PIDController pid = new PIDController(kP, kI, kD);
// Creates a PIDController with gains kP, kI, and kD
frc::PIDController pid{kP, kI, kD};
from wpimath.controller import PIDController
# Creates a PIDController with gains kP, kI, and kD
pid = PIDController(kP, kI, kD)
An optional fourth parameter can be provided to the constructor, specifying the period at which the controller will be run. The PIDController object is intended primarily for synchronous use from the
main robot loop, and so this value is defaulted to 20ms.
Using the Feedback Loop Output
The PIDController assumes that the calculate() method is being called regularly at an interval consistent with the configured period. Failure to do this will result in unintended loop behavior.
Using the constructed PIDController is simple: simply call the calculate() method from the robot’s main loop (e.g. the robot’s autonomousPeriodic() method):
// Calculates the output of the PID algorithm based on the sensor reading
// and sends it to a motor
motor.set(pid.calculate(encoder.getDistance(), setpoint));
// Calculates the output of the PID algorithm based on the sensor reading
// and sends it to a motor
motor.Set(pid.Calculate(encoder.GetDistance(), setpoint));
# Calculates the output of the PID algorithm based on the sensor reading
# and sends it to a motor
motor.set(pid.calculate(encoder.getDistance(), setpoint))
Checking Errors
getPositionError() and getVelocityError() are named assuming that the loop is controlling a position - for a loop that is controlling a velocity, these return the velocity error and the acceleration
error, respectively.
The current error of the measured process variable is returned by the getPositionError() function, while its derivative is returned by the getVelocityError() function:
Specifying and Checking Tolerances
If only a position tolerance is specified, the velocity tolerance defaults to infinity.
As above, “position” refers to the process variable measurement, and “velocity” to its derivative - thus, for a velocity loop, these are actually velocity and acceleration, respectively.
Occasionally, it is useful to know if a controller has tracked the setpoint to within a given tolerance - for example, to determine if a command should be ended, or (while following a motion profile)
if motion is being impeded and needs to be re-planned.
To do this, we first must specify the tolerances with the setTolerance() method; then, we can check it with the atSetpoint() method.
// Sets the error tolerance to 5, and the error derivative tolerance to 10 per second
pid.setTolerance(5, 10);
// Returns true if the error is less than 5 units, and the
// error derivative is less than 10 units
// Sets the error tolerance to 5, and the error derivative tolerance to 10 per second
pid.SetTolerance(5, 10);
// Returns true if the error is less than 5 units, and the
// error derivative is less than 10 units
# Sets the error tolerance to 5, and the error derivative tolerance to 10 per second
pid.setTolerance(5, 10)
# Returns true if the error is less than 5 units, and the
# error derivative is less than 10 units
Resetting the Controller
It is sometimes desirable to clear the internal state (most importantly, the integral accumulator) of a PIDController, as it may be no longer valid (e.g. when the PIDController has been disabled and
then re-enabled). This can be accomplished by calling the reset() method.
Setting a Max Integrator Value
Integrators introduce instability and hysteresis into feedback loop systems. It is strongly recommended that teams avoid using integral gain unless absolutely no other solution will do - very often,
problems that can be solved with an integrator can be better solved through use of a more-accurate feedforward.
A typical problem encountered when using integral feedback is excessive “wind-up” causing the system to wildly overshoot the setpoint. This can be alleviated in a number of ways - the WPILib
PIDController class enforces an integrator range limiter to help teams overcome this issue.
By default, the total output contribution from the integral gain is limited to be between -1.0 and 1.0.
The range limits may be increased or decreased using the setIntegratorRange() method.
// The integral gain term will never add or subtract more than 0.5 from
// the total loop output
pid.setIntegratorRange(-0.5, 0.5);
// The integral gain term will never add or subtract more than 0.5 from
// the total loop output
pid.SetIntegratorRange(-0.5, 0.5);
# The integral gain term will never add or subtract more than 0.5 from
# the total loop output
pid.setIntegratorRange(-0.5, 0.5)
Disabling Integral Gain if the Error is Too High
Another way integral “wind-up” can be alleviated is by limiting the error range where integral gain is active. This can be achieved by setting IZone. If the error is more than IZone, the total
accumulated error is reset, disabling integral gain. When the error is equal to or less than IZone, integral gain is enabled.
By default, IZone is disabled.
IZone may be set using the setIZone() method. To disable it, set it to infinity.
// Disable IZone
// Integral gain will not be applied if the absolute value of the error is
// more than 2
// Disable IZone
// Integral gain will not be applied if the absolute value of the error is
// more than 2
# Disable IZone
# Integral gain will not be applied if the absolute value of the error is
# more than 2
Setting Continuous Input
If your mechanism is not capable of fully continuous rotational motion (e.g. a turret without a slip ring, whose wires twist as it rotates), do not enable continuous input unless you have implemented
an additional safety feature to prevent the mechanism from moving past its limit!
Some process variables (such as the angle of a turret) are measured on a circular scale, rather than a linear one - that is, each “end” of the process variable range corresponds to the same point in
reality (e.g. 360 degrees and 0 degrees). In such a configuration, there are two possible values for any given error, corresponding to which way around the circle the error is measured. It is usually
best to use the smaller of these errors.
To configure a PIDController to automatically do this, use the enableContinuousInput() method:
// Enables continuous input on a range from -180 to 180
pid.enableContinuousInput(-180, 180);
// Enables continuous input on a range from -180 to 180
pid.EnableContinuousInput(-180, 180);
# Enables continuous input on a range from -180 to 180
pid.enableContinuousInput(-180, 180)
Clamping Controller Output
// Clamps the controller output to between -0.5 and 0.5
MathUtil.clamp(pid.calculate(encoder.getDistance(), setpoint), -0.5, 0.5);
// Clamps the controller output to between -0.5 and 0.5
std::clamp(pid.Calculate(encoder.GetDistance(), setpoint), -0.5, 0.5);
# Python doesn't have a builtin clamp function
def clamp(v, minval, maxval):
return max(min(v, maxval), minval)
# Clamps the controller output to between -0.5 and 0.5
clamp(pid.calculate(encoder.getDistance(), setpoint), -0.5, 0.5) | {"url":"https://docs.wpilib.org/en/latest/docs/software/advanced-controls/controllers/pidcontroller.html","timestamp":"2024-11-02T01:07:10Z","content_type":"text/html","content_length":"112337","record_id":"<urn:uuid:90537004-fa28-49ad-ba5a-b0b1407b0483>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00471.warc.gz"} |
Square Pyramid Calculator
Last updated:
Square Pyramid Calculator
This square pyramid calculator will help you find the total surface area and volume of any square pyramid, even if it's the famous Egyptian Pyramids (provided you know the length measurements 😉). You
can also find the lateral edge of a square pyramid by using this tool. Or if you're interested in only finding the lateral surface area, this square pyramid calculator will help with that, too!
What is a square pyramid?
A pyramid, in general, has a polygonal base with triangular lateral surfaces that meet at a point. In the case of a square pyramid, the base is surprise, surprise, a square!
Here are some of the key terminologies used for square pyramids:
• The side length of the base square is called base length;
• The edges of the triangular lateral surfaces are called lateral edges of the square pyramid;
• The area of the square at the base is called the base area;
• The combined area of all the triangular lateral surfaces is called the lateral surface area of the square pyramid;
• The combined area of all 5 surfaces of the pyramid, including the base, is called the total surface area of the square pyramid; and
• The volume of the pyramid is called the square pyramid volume.
How do I use the square pyramid calculator
To use the square pyramid calculator to find the surface area and volume, you may do the following:
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2. Enter the height of the square pyramid.
3. Tada! The square pyramid calculator will show you a host of output values, such as:
□ Square pyramid base area;
□ Lateral edge of the square pyramid;
□ Lateral surface area of the square pyramid;
□ Total surface area of the square pyramid; and
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Other pyramid and prism calculators
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May 22, 201922 May, 2019. The ball mill consists of a metal cylinder and a ball. The working principle is that when the cylinder is rotated, the grinding body (ball) and the object to be polished
(material) installed in the cylinder are rotated by the cylinder under the action of friction and centrifugal force. At a certain height, it will automatically | {"url":"https://www.toskana-reiseinformationen.de/talc/1702253270-Ball-Mill-Construction-Specs/3501/","timestamp":"2024-11-10T12:22:16Z","content_type":"text/html","content_length":"37577","record_id":"<urn:uuid:11b41627-bf72-4ed6-9051-667dbfe81548>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00515.warc.gz"} |
OpenStax College Physics for AP® Courses, Chapter 11, Problem 15 (Problems & Exercises)
The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure due to the ocean at the bottom of this trench, given its depth is 11.0 km and
assuming the density of seawater is constant all the way down.
Question by
is licensed under
CC BY 4.0
Final Answer
$1.10 \times 10^8 \textrm{ Pa}$
$1090 \textrm{ atm}$
Solution video
OpenStax College Physics for AP® Courses, Chapter 11, Problem 15 (Problems & Exercises)
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Video Transcript
This is College Physics Answers with Shaun Dychko. This question asks us to find the pressure at the bottom of the deepest part of the ocean, which is the Marian Trench in the Pacific near the
Philippines. So it’s going to be the gauge pressure anyway which you know is all that really matters here because the pressure due to the atmosphere will be negligible in comparison to the enormous
pressure of the high column of water. Now the pressure will be density of the sea water times g times the height. So density of sea water is 1.025 times ten to the three kilograms per cubic meter,
and we make the assumption here that this density is constant over this great height, which is probably not exactly true, it’s likely that density gets a bit bigger at the bottom but never mind it’s
close enough, times 9.8 newtons per kilogram, times 11 kilometers which is 11 times ten to the three meters, and this gives 1.10 times ten to the eight Pascals. Now that number is hard for us to
understand because it’s just some big number, but we can turn it into something, a unit that we can relate to a bit better by multiplying by one atmosphere for every 101 times ten to the three
Pascals, this works out to 1090 atmosphere so that’s really high pressure. | {"url":"https://collegephysicsanswers.com/openstax-solutions/greatest-ocean-depths-earth-are-found-marianas-trench-near-philippines-0","timestamp":"2024-11-04T02:48:08Z","content_type":"text/html","content_length":"200470","record_id":"<urn:uuid:3db575fd-1d9f-4269-9a3f-023997eaaf22>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00667.warc.gz"} |
Delta parameterization of ordinal veriables
Replied on Fri, 06/03/2022 - 14:09
I've attached a full script that uses the "theta" "parameterization". (I use quotes around these because they are both nonsense terms with no intrinsic meaning for the given context). The key bit is
copied below.
The big picture is you will generally use nonlinear constraints (mxConstraint()) to make the total mean and variance zero and one, respectively. To do this without writing the algebra for the
model-implied means and covariances yourself, you can refer to the model-implied (i.e., expected) means and covariance as outlined below.
thetaMod2 <- mxModel(thetaMod,
# Create 'empty' matrices to refer to means and covs
mxMatrix('Full', nrow=1, ncol=length(mv), name='emean'),
mxMatrix('Symm', nrow=length(mv), ncol=length(mv), name='ecov'),
# Refer to the 'empty' matrices in expectattion
mxExpectationRAM(A='A', S='S', F='F', M='M', thresholds='Thresholds',
expectedMean='emean', expectedCovariance='ecov'),
# Create matrices for constrained values
mxMatrix(type='Zero', nrow=1, ncol=length(mv), name='Z'),
mxMatrix(type='Unit', nrow=length(mv), ncol=1, name='I'),
# Constrain means and variances
mxConstraint(emean == Z, name='mcon'),
mxConstraint(diag2vec(ecov) == I, name='vcon'),
# Add data
mxData(type='raw', ds),
# Add fit function
Hopefully, this helps! | {"url":"https://openmx.ssri.psu.edu/forums/opensem-forums/general-sem-discussions/delta-parameterization-ordinal-veriables","timestamp":"2024-11-02T23:41:23Z","content_type":"text/html","content_length":"29925","record_id":"<urn:uuid:91806ea1-2acc-490a-9f33-693a416dd32d>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00348.warc.gz"} |
Dan has some marbles. Ellie has twice as many marbles as dan.frank has 15 marbles. Dan,Ellie and frank have a total of 63 marbles. How many does dan have
1. Home
2. General
3. Dan has some marbles. Ellie has twice as many marbles as dan.frank has 15 marbles. Dan,Ellie and fra... | {"url":"http://thibaultlanxade.com/general/dan-has-some-marbles-ellie-has-twice-as-many-marbles-as-dan-frank-has-15-marbles-dan-ellie-and-frank-have-a-total-of-63-marbles-how-many-does-dan-have","timestamp":"2024-11-04T14:19:20Z","content_type":"text/html","content_length":"29977","record_id":"<urn:uuid:5c63a579-5642-4704-8ea1-507e0a8c6cb1>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00625.warc.gz"} |
Felix Halim .NET
Last year contestants (or contestants who practiced with last year ICPC Jakarta 2012 problemset) should immediately know, only by looking at the problem title, that this problem is related to last
year Problem H - Alien Abduction. Also, they should have guessed that this problem is either as hard, or harder than last year :D.
It took me several weeks to come up with this problem (and the solution). My criterion for the problem are:
• The problem must be hard. This is necessary to avoid the embarrasement (to the problem-setters) that some teams sweept clean all the problems in 3 hours >.< (yes I'm talking about you, +1
ironwood branch).
• The problem must be a rare problem. That is, I want it to be the decider problem to separate the teams at the top. I do not want to pick the problem from the Competitive Programming Book Chapter
9 - Rare Topics, because it would be too obvious :P.
• The problem must have a secret message. Well... the secret message is meaningless for any contestants (i.e., the message will not help anyone to gain better insight in solving the problem).
Instead, the secret message was intended for one of the special guests in the closing ceremony :).
Given the criterion above, how should I write the problem? I search for inspiration by reading blogs, TopCoder forum posts, etc. After two weeks I decided to write a BIT (Binary Indexed Tree) with
range update. I always surprised by BIT that it can be used differently than its original design (prefix sums). For example, BIT can be used to compute range minimum query, and range updates too (see
Petr Mitrichev blog as well as the links in the comments section).
Here is the problem statement, you can try submit your solution below.
In short, the problem statement is like this (Click here to see the full problem statement):
Given lines/curves segments as a set of functions f[p](x) = ax^3 + bx^2 + cx + d, each on a range [x1[p], x2[p]], what is the total sum of the y-values of the points (generated by the functions)
with integral x-values in range [x1, x2] ?
Naively, this problem can be easily solved using (Lazy) Segment Tree + coordinate compression. I do not want this solution to pass (remember that I want to create a BIT range update problem).
Coordinate compression is a trick to make the input range smaller by pre-reading all input data and then make it dense. This will help the Segment Tree solution that it only needs to allocate 100K
instead of 1M tuples. Luckily, there is a trick to make coordinate compression to fail: make it as an interactive problem. However, ICPC style usually do not involve interactive problems, so I have
to uglify the input that the next input depends on the previous output of the program (i.e., to simulate an interactive problem). This is the reason that "the space distortion" was introduced every
time the transporter is used. With this, coordinate compression will no longer work, but I agree that this made the problem harder to read. In fact from the survey, almost half of the teams voted
problem J as the least liked problem :(. I guess that's the price I have to pay.
What about the (Lazy) Segment Tree? How to make it fail to work? I know that normally, in programming contests, two solutions with the same complexity should both get Accepted. But, this problem is
an exception. I made it clear in the problem statement that your solution must be very-very efficient (otherwise the device will be too late to disrupt the transport operation by the alien ship). So,
constant factor matters in this problem. I was hoping that the contestants realize that when reading the problem statement.
If you are familiar with Segment Tree, you should have an insight how slow it can be. Thus you should pick solutions with lower constant factor if any (e.g., Binary Indexed Tree). The Segment Tree
solutions run in 5+ seconds while the BIT solutions run in less than 2 seconds. Moreover, the memory consumption for Segment Tree will exceed 64 MB. If you recall, in the briefing, Suhendry mentioned
that all of the judges solutions use less than 32 MB. That should give you another hint that Segment Tree is not the way to go.
Unfortunately, due to PC2 inaccuracy in measuring the time limit, some teams got lucky to get it accepted with Segment Tree (albeit, they need to insanely optimize their code to get it run in 5
seconds). We set the time limit for this problem to be 4 seconds in the PC2, but somehow PC2 still accepts solutions with 5 seconds runtime! I didn't re-adjust the time limit to 3 seconds during the
contest and decided to let it be (otherwise I will be cursed by the accepted teams :P).
No team solved this problem using BIT range updates. It is not easy to convert a (Lazy) Segment Tree into BIT range updates. It probably deserve a problem on its own. To give an example, consider the
simplest case where f(x) = d. That is, the values for a, b, and c are all zero. In this case, the problem is equal to a very simple BIT range update. Here is a nice post on how to simulate a (Lazy)
Segment Tree using two BITs. We can generalize this for the other powers (a, b, and c) and we will need five BITs. The runtime for this approach is less than 2 seconds and it consumes only ~20MB
memory. This problem also requires the knowledge of mod-inverse. That is, you will need to do division with modulo somewhere in the calculation.
Well, the first two criterion have been fulfilled. The last criteria is the secret message. I hid the message in plain sight. No other judge (even the chief of judge) was aware that there was a
secret message. But no worries, the message is meaningless to anyone except me and the intended recipient :). I really had fun in setting this problem :D. | {"url":"https://blog.felix-halim.net/2013/","timestamp":"2024-11-04T23:04:58Z","content_type":"application/xhtml+xml","content_length":"65756","record_id":"<urn:uuid:3c1a8b83-9b63-48bc-8286-0808f26d77d1>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00757.warc.gz"} |
Notes for Number Theory
© 2019 Brian Heinold
Licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 Unported License
Here is a pdf version of the book.
Here are the notes I wrote up for a number theory course I taught. The notes cover elementary number theory but don't get into anything too advanced. My approach to things is fairly informal. I like
to explain the ideas behind everything without getting too formal and also without getting too wordy.
If you see anything wrong (including typos), please send me a note at heinold@msmary.edu.
Definition of divisibility
One of the most basic concepts in number theory is that of divisibility. The concept is familiar: 14 is divisible by 7, even numbers are divisible by 2, prime numbers are only divisible by themselves
and 1, etc. Here is the formal definition:
An integer d is a divisor of an integer n if there exists an integer k such that n = dk. We say that n is divisible by d, or that d divides n, and write d ∣ n.
For example, 20 is divisible by 4 because we can write 20 = 4 · 5; that is n = dk, with n = 20, d = 4, and k = 5. The equation n = dk is the key part of the definition. It gives us a formula that we
can associate with the concept of divisibility.
This formula is handy when it comes to proving things involving divisibility. If we are given that n is divisible by d, then we write that in equation form as n = dk for some integer k. If we need to
show that n is divisible by d, then we need to find some integer k such that n = dk.
Here are a few example proofs:
1. Suppose we want to prove the simple fact that if n is even, then n^2 is even as well.
Proof. Even numbers are divisible by 2, so we can write n = 2k for some integer k. Then n^2 = (2k)^2 = 4k^2, which we can write as n^2 = 2(2k^2). We have written n^2 as 2 times some integer, so
we see that n^2 is divisible by 2 (and hence even).◻
2. Prove that if a ∣ b and b ∣ c, then a ∣ c
Proof. Since a ∣ b and b ∣ c we can write b = aj and c = bk for some integers j and k.**Note that we must use different integers here since the integer j that works for a ∣ b does not
necessarily equal the integer k that works for b ∣ c. Plug the first equation into the second to get c = (aj)k, which we can rewrite as c = a(jk). So we see that a ∣ c, since we have
written c as a multiple of a. ◻
3. Prove that if a ∣ b, then ac ∣ bc, for any integer c.
Proof. Since a ∣ b, we can write b = ak for some integer k. Multiply both sides of this equation by c to get (ac)k = bc. This equation tells us that ac ∣ bc, which is what we want.◻
Here are a couple of other divisibility examples:
1. Disprove: If a ∣ (b+c), then a ∣ b or a ∣ c.
Solution. All we need is a single counterexample. Setting a = 5, b = 3, and c = 7 does the trick.
2. Find a divisor of 359951 besides 1 and itself.
Solution. The only answers are 593 and 607. It would be tedious to find them by checking divisors starting with 2, 3, etc. A better way is to use the algebraic fact x^2–y^2 = (x–y)(x+y). This
fact is very useful in number theory.
With a little cleverness, we might notice that 359951 is 360000–49, which is 600^2–7^2. We can factor this into (600–7)(600+7), or 593 × 607.
The division algorithm
The division algorithm, despite its name, it is not really an algorithm. It states that when you divide two numbers, there is a unique quotient and remainder. Specifically, it says the following:
Let a, b ∈ ℤ with b > 0. Then there exist unique q, r ∈ ℤ such that a = bq+r and 0 ≤ r < b.
The integers q and r are called the quotient and remainder. For example, if a = 27 and b = 7, then q = 3 and r = 6. That is, 27 ÷7 is 3 with a remainder of 6, or in equation form: 27 = 7 · 3+6. The
proof of the theorem is not difficult and can be found in number theory textbooks.
One of the keys here is that the remainder is less than b. Here are some consequences of the theorem:
• Taking b = 2, tells us all integers are of the form 2k or 2k+1 (i.e., every integer is either odd or even).**We use k instead of q here out of convention.
• Taking b = 3, all integers are of the form 3k, 3k+1, or 3k+2.
• Taking b = 4, all integers are of the form 4k, 4k+1, 4k+2, or 4k+3.
• For a general b, all integers are of the form bk, bk+1, …, or bk+(b–1).
These are useful for breaking things up into cases. Here are a few examples:
1. Suppose we want to show that n^3–n is always divisible by 3. We can break things up into cases n = 3k, n = 3k+1, and n = 3k+2, like below:
We see that in each case, n^3–n is divisible by 3. By the division algorithm, these are the only cases we need to check, since every integer must be of one of those three forms.
2. Prove that every perfect square is of the form 4k or 4k+1.
Proof. Every integer n is of the form 2k or 2k+1.
If n = 2k, then we have n^2 = 4k^2, which is of the form 4k.**We are being a bit informal here with the notation. When we say the number is of the form 4k, the k is different from the k used in n
= 2k. What we're really saying here is that the number is of the form 4 times some integer. A more rigorous way to approach this might be to let n = 2j, compute n^2 = 4j^2 and say n^2 = 4k, with
k = j^2.
If n = 2j+1, then we have n^2 = 4k^2+4k+1 = 4(k^2+k) + 1, which is of the form 4k+1. ◻
3. A prime number is a number greater than 1 whose only divisors are 1 and itself. Prove that every prime greater than 3 is of the form 6k+1 or 6k+5.
Proof. Every integer is of the form 6k, 6k+1, 6k+2, 6k+3, 6k+4, or 6k+5. An integer of form 6k is divisible by 6. A integer of the form 6k+2 is divisible by 2 as it can be written as 2(3k+1).
Similarly, an integer of the form 6k+3 is divisible by 3 and an integer of the form 6k+4 is divisible by 2. None of these forms can be prime (except for the integers 2 and 3, which we exclude),
so the only forms left that could be prime are 6k+1 and 6k+5.◻
4. Prove that 16 ∣ a^4+b^4–2 for any odd integers a and b.
Proof. We will start by writing a^4+b^4–2 as (a^4–1) + (b^4–1). Let's take a look at the a^4–1 term. We can factor that into (a^2–1)(a^2+1) and further into (a–1)(a+1)(a^2+1). Since we know a is
odd, we can write a = 2k+1 and we have k(k+1) part of the expression. We know that one of k and k+1 is even, since k and k+1 are consecutive integers, and that gives us our factor of 16.
So we have that a^4–1 is divisible by 16. A similar argument tells us that b^4–1 is divisible by 16, and from there we get that a^4+b^4–2 is divisible by 16, since it is the sum of two multiples
of 16. ◻
5. One of the oldest and most famous proofs in math is that √2 is irrational. That is, √2 cannot be written as a ratio of integers. Here is the proof:
Proof. First, note that the square of an even is even and the square of an odd is odd since (2k)^2 = 2(2k^2) is even and (2k+1)^2 = 2(2k^2+2k)+1 is odd. In particular, if an integer a^2 is even,
then a is even as well.
Suppose √2 = p/q, with p and q positive integers. By clearing common factors, we can assume the fraction is in lowest terms. Multiply both sides by q and square both sides to get 2q^2 = p^2. This
tells us that 2 ∣ p^2. Thus 2 ∣ p by the statement above, and we can write p = 2k for some integer k. So we have 2q^2 = (2k)^2, which simplifies to q^2 = 2k^2. This tells us that 2 ∣
q^2 and hence 2 ∣ q. But this is a problem because p and q both have a factor of 2 and p/q is supposed to already be in lowest terms. So we have a contradiction, which shows that it must not
be possible to write √2 as a ratio of integers.
It is not to hard to extend this result to show that ^n√m is irrational unless m is a perfect nth power.
Perfect squares
The second example above is sometimes useful when working with perfect squares. We record it as a theorem:
Every perfect square is of the form 4k or 4k+1.
For example, 3999 is not a perfect square since it is 4000–1, which is of the form 4k–1 (same as a 4k+3 form). On the other hand, just because something is of the form 4k+1 does not mean it is a
perfect square. For instance, 41 is of the form 4k+1, but isn't a perfect square.
As another example, 3n^2–1 is not a perfect square for any integer n. To see this, break the problem into two cases: n = 2k and n = 2k+1. If n = 2k, then 3n^2–1 = 3(4k^2)–1 = 4(3k^2)–1. If n = 2k+1,
then 3n^2–1 = 3(4k^2+4k+1)–1 = 4(3k^2+3k)+2. Neither of these are of the form 4k or 4k+1, so they are not perfect squares.
Similar results to the theorem above can be proved for other integers. For instance, every perfect square is of the form 5k, 5k+1, or 5k+4.
More about remainders
Note that we can also take the remainders to be in other ranges besides from 0 to b–1. The most useful range is from –b/2 and b/2. For instance, with b = 3, we can also write every integer as being
of the form 3k–1, 3k, or 3k+1. An integer of the form 3k–1 is also of the form 3k+2. As another example, we can write every integer in one of the forms 6k–2, 6k–1, 6k, 6k+1, 6k+2, or 6k+3.
The modulo operation
In grammar school, the remainder always seemed to me to be an afterthought, but in higher math, it is quite useful and important. It is built into most programming languages, usually with the symbol
mod or %
The remainder when an integer a is divided by b is denoted by a mod b. It is the integer r in the division algorithm expression a = bq+r, with 0 ≤ r < b. We have
For example, suppose we want to find 68 mod 7. The definition above tells us to find the nearest multiple of 7 less than n and subtract. The closest multiple of 7 less than 68 is 63, and 68–63 = 5.
So 68 mod 7 = 5.
This procedure applies to negatives as well. For instance, to compute –31 mod 5, the closest multiple of 5 less than or equal to -31 is -35, which is 4 away from -31, so –31 mod 5 = 4, or –31 ≡ 4
(mod 5).
As one further example, suppose we want 179 mod 18. 179 is one less than 180, a multiple of 18, so it leaves a remainder of 18-1 = 17. So 179 mod 18 = 17.
A nice way to compute mods mentally or by hand is to use a streamlined version of the grade school long division algorithm. For example, suppose we want to compute 34529 mod 7. Here is the procedure:
The end result is 34529 mod 7 = 5.
The greatest common divisor
The greatest common divisor, or gcd, of two integers a and b is the largest integer that divides both a and b. We denote it by gcd(a, b).**In some texts, the notation (a, b) is used instead of gcd(a,
For example, the gcd of 24 and 96 is 12, since 12 is the largest integer that divides both 24 and 96. As another example, the gcd of 18 and 25 is 1. Those numbers have no divisors in common besides
The Euclidean algorithm
To find the gcd of two integers, the Euclidean algorithm is used. We'll start with an example, finding gcd(21, 78):
In general, to find gcd(a, b), assume a ≤ b, and compute **Note that the gcd is already built into Python's fractions module.:
def gcd(a, b):
while b != 0:
b, a = a, b % a
return a
Why the Euclidean algorithm works
The reason it works is that the common divisors of a and b are exactly the same as the common divisors as a and b mod a, so their gcds must be the same. Because of this, when we apply the Euclidean
algorithm, the gcd of the two numbers on the left side stays constant all the way through the algorithm. For example, when we compute gcd(21, 78), we get the following: gcd(a, b) = gcd(a, b mod a)
tells us that gcd(21, 78), gcd(15, 21), gcd(6, 15), and gcd(3, 6) are all the same, and since the last step gives us 6 mod 3 = 0, we know that 3 is a divisor of 6, and hence gcd(15, 21) = 3, so we
could stop there.
It is worth showing why the common divisors of a and b are the same as the common divisors as a and b mod a. First, when we apply the division algorithm to a and b, we get b = aq + r, where r = b mod
a. If a and b are both divisible by some common divisor d, then r = b–aq will be as well, since we can factor d out of the right side. On the other hand, if a and r are both divisible by some common
divisor d, then b = r–aq will be as well, since we can factor d out of the right side.
Gcds and linear combinations
In number theory, a linear combination of the integers a and b is an expression of the form ax+by for some integers x and y. For example, a linear combination of a = 6 and b = 15 is an expression of
the form 6x+15y, like 6(1)+15(2) = 36 or 6(4)+15(–1) = 9. Linear combinations are important in a variety of contexts.
Linear combinations have a close connection with the gcd. Suppose we want to know if it is possible to write 4 as a linear combination of 6 and 15. That is, can we find integers x and y such that 6x
+15y = 4? The answer is no, since the left side is a multiple of 3 (namely 3(2x+5y)), but the right side is not a multiple of 3. By the same reasoning, in general, if c is not a multiple of gcd(a, b)
, then it is impossible to write c as a linear combination of a and b.
What about multiples of the gcd? Is it always possible to write ax+by = c if c is a multiple of the gcd? The answer is yes. We just have to show how to write the gcd as a linear combination of a and
b. Once we have this (see the proof below for how), we can then multiply through to get c. For instance, suppose we want to find x and y such that 6x+15y = 21. We have gcd(6, 15) = 3 and it is not
hard to find that 6(3)+15(–1) = 3. If we multiply through by 7, we get 6(21)+15(–7) = 21. So it just comes down to writing the gcd as a linear combination.
Here is a formal statement of the above along with a proof:
a, b, c ∈ ℤ
d = gcd(a, b)
. There exist integers
such that
ax+by = c
if and only if
d ∣ c
First, since
d ∣ a
d ∣ b
, we have
d ∣ c
. Thus, we cannot write
as a linear combination of
if it is not divisible by
We now show that it is possible to write d as a linear combination of a and b. Start by letting e be the smallest positive linear combination of a and b. We need to show that e = d.
By the division algorithm, we can write a = eq+r for some integers q and r. Then we have r is a linear combination of a and b. But, by the division algorithm, 0 ≤ r < e. Since e is the smallest
positive linear combination of a and b, we must have r = 0. Thus, a = eq+r with r = 0 tells us that e ∣ a. A similar argument shows that e ∣ b. So e is a common divisor of a and b. But, as we
mentioned earlier, any linear combination of a and b is a multiple of d. So e is both a multiple of d and a common divisor of a and b, which means e = d.
Finally, if c is a multiple of d (say c = dk for some integer k), and we have integers x and y such that ax+by = d, then we can multiply through by k to get a(kx)+b(ky) = c.
In particular, we have the following important special case:
Let a, b ∈ ℤ with d = gcd(a, b). Then there exist integers x and y such that ax+by = d.
This fact is useful when working with gcds because it gives us an equation to work with. On the other hand, be careful. Just because we can write ax+by = c, that does not mean c = gcd(a, b). All we
are guaranteed is that c is a multiple of gcd(a, b).
The extended Euclidean algorithm
Theorem 3 tells us that the gcd is a linear combination, but it doesn't tell us how to find that linear combination. Being able to find that linear combination is important in a number of contexts.
The trick is to use the Euclidean algorithm in a particular way.
In the earlier example when we found gcd(21, 78) using the Euclidean algorithm, we used the modulo operation. Written out fully using the division algorithm, the Euclidean algorithm on this example
is as follows:
We can use this sequence of steps to find the linear combination by working backwards. Start with the second-to-last equation from the Euclidean algorithm and work back up in the following way: 3 =
78(3) + 21(–11). Here's what happens: In the Euclidean algorithm, we generate the sequence 78, 21, 15, 6, 3 of quotient/remainders. We start with the last equation from the Euclidean algorithm that
contains those numbers and solve it for the gcd, 3, in terms of the next two terms of the sequence, 6 and 15 (namely, 3 = 15–6(2)). At the next stage, solve the next equation up for the remainder (6
= 21–15(2)) and use it to eliminate 6. We then simplify to write things in terms of 15 and 21 and then solve the next equation for the remainder (15 = 78–21(3)). We use this to eliminate 15, simplify
to write things in terms of 21 and 78, and then we are done because the equation is in terms of 21 and 78, which is what we want.
Note that at each step we can check our work by making sure the expression equals the gcd. For instance, in the third line above, 15(3)–21(2) = 45–42 = 3.
Here is another example. Suppose we want to find integers such that 11x+41y = 1. Start with the Euclidean algorithm:
Thus we have x = 15 and y = –4 that give us 11x+41y = 1.
This algorithm is called the extended Euclidean algorithm. It turns out to have a number of important uses, as we will see. Here is a short Python program implementing a version of it. This is a
streamlined version of the algebra above, based on the algorithm given at http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm.
def extended_euclid(a, b):
s, old_s, t, old_t, r, old_r = 0, 1, 1, 0, b, a
while r != 0:
q = old_r // r
old_r, r = r, old_r - q * r
old_s, s = s, old_s - q * s
old_t, t = t, old_t - q * t
return (old_s, old_t)
A few example proofs
The keys to many proofs involving gcds are as follows:
1. Rewrite divisibility statements as equations. For instance, a ∣ b becomes b = ak for some integer k.
2. Rewrite gcd(a, b) = d as a linear combination equation, like ax+by = d for some integers x and y.
3. Algebraically manipulate the equations from (1) and (2).
4. If, at some point, you get a linear combination au+bv = e, you can conclude that gcd(a, b) divides e, but not necessarily that gcd(a, b) = e.
Here are a few example proofs involving gcds:
1. If d = gcd(a, b), then gcd(a/d, b/d) = 1.
Proof. The basic idea here is intuitively clear: If we divide through by the gcd, then what's left should not have factors in common, since all the common factors should be in the gcd.
Here is a more algebraic approach: Start by writing ax+by = d for some integers x and y. We can then divide through by d to get (a/d)x + (b/d)y = 1. Theorem 3 tell us that gcd(a/d, b/d) divides
any linear combination of a/d and b/d. So gcd(a/d, b/d) is a divisor of 1, meaning it must equal 1.◻
2. If d = gcd(a, b), a ∣ c, and b ∣ c, then ab ∣ cd.
Proof. Write c = aj, c = bk, and d = ax+by for some integers j, k, x, and y. Multiply the last equation through by c to get cd = acx+bcy. Then plug in c = bk into the acx term and c = aj into the
bcy term to get cd = abkx + bajy. So we have cd = ab(kx+jy), showing that ab ∣ cd.◻
3. If k > 0, then gcd(ka, kb) = k gcd(a, b).
Proof. Let d = gcd(a, b) and d′ = gcd(ka, kb). We can write d = ax+by for some integers x and y. Multiplying through by k gives kd = kax+kby. This is a linear combination of ka and kb, so we know
that d′ ∣ kd. On the other hand, we can write d′ = kax′+kby′ for some integers x′ and y′. Divide through by k to get d′/k = ax′+by′. This is a linear combination of a and b, so d ∣ d′/k,
or kd ∣ d′. Since d′ ∣ kd and kd ∣ d′, and k > 0, we must have d′ = kd.◻
4. If c ∣ (a–b), then gcd(a, c) = gcd(b, c).
Proof. Let d[1] = gcd(a, c) and d[2] = gcd(b, c). From these, we can write ax[1]+cy[1] = d[1] and bx[2]+cy[2] = d[2] for some integers x[1], x[2], y[1], and y[2]. Also, since c ∣ (a–b) we can
write a–b = ck for some integer k, which we can solve to get a = ck+b and b = a–ck. Plugging the former into the equation for d[1] gives (ck+b)x[1]+cy[1] = d[1], which we can write as c(kx[1]+y
[1])+bx[1] = d[1]. The left hand side is a linear combination of b and c, so it is a multiple of d[2] = gcd(b, c). Thus d[2] ∣ d[1]. Plugging b = a–ck into the equation for d[2] and doing a
similar computation gives d[1] ∣ d[2]. Thus d[1] = d[2]. ◻
The least common multiple
A relative of the gcd is the least common multiple.
The least common multiple (lcm) of two integers a and b, denoted lcm(a, b), is the smallest positive integer which is divisible by both a and b.
The following gives an easy way to find the lcm:
Here an example: If a = 14 and b = 16, then gcd(a, b) = 2, ab = 224 and lcm(a, b) = 224/2 = 112. A simple way to think of this theorem is that ab is a multiple of both a and b, but it has some
redundant factors in it. Dividing out by gcd(a, b) removes all of the redundancies, leaving the smallest possible common multiple. Here is a formal proof of the theorem:
d = gcd(a, b)
. We need to show that
ab/d = lcm(a, b)
. We can do this by first showing that
is a multiple of
and then showing that no other multiple is smaller than it.
First, since d is divisor of a and b, we can write a = dk and b = dj. Then ab/d = aj and ab/d = bk, which shows that ab/d is a multiple of both a and b.
Next, let m be any common multiple of a and b. So we have m = as and m = bt for some integers r and s, and we can write 1/a = s/m and 1/b = t/m.
We can also write d as a linear combination d = ax+by for some integers x and y. Dividing both sides of this equation by ab, we get dab = xb + ya. Plugging in our earlier equations for 1/a and 1/b
gives dab = xtm + ysm, which we can rewrite as m = (xt+ys)abd. This tells us that ad/b is a divisor of m.
Since m is an arbitrary multiple of a and b, and ab/d divides it, that means that ab/d is the smallest possible multiple, which is what we wanted to prove.
There are other ways to find lcm(a, b). One way would be to list all the multiples of a and all the multiples of b and find the first one they have in common. This, however, is slow unless a and b
are small. Another way uses the prime factorization. See Section 2.1 for an example.
Here is a simple example where the lcm is useful. Suppose one thing happens every 28 days and other happens every 30 days, and that both things happened today. When will they both happen again? The
answer is lcm(28, 30) = 420 days from now.**A more general approach to handling these kinds of cyclical problems is covered in Section 3.10.
As another example, some people theorize that the timing of periodic cicadas has to do with the lcm. Some species of cicadas only emerge every 13 years, while others emerge every 17 years. People
have noticed that these are values are both prime. This means that the lcm of these numbers with other numbers is relatively large. The things that eat cicadas are often on a boom-bust cycle, and it
would be bad for cicadas to emerge in a year when there are a lot of predators. Suppose a certain predator is on a 4-year cycle. How often would a boom year coincide with a 13-year cicada emergence?
The answer is every lcm(4, 13) = 52 years. On the other hand, if cicadas were on say a 14-year cycle, then it would happen every 28 years. It would also be bad for both the 13-year and the 17-year
cicadas to emerge at once, since they would be competing for resources. But this will only happen every lcm(13, 17) = 191 years.
Relatively prime integers
Integers a and b are called relatively prime (or coprime) if gcd(a, b) = 1.
In other words, a and b are relatively prime provided they have no divisors in common besides 1 (or maybe -1 if they are negative). There are a number of facts in number theory that are only true for
relatively prime integers. Here is one useful fact that follows quickly from Theorem 3:
One of the most useful tools in number theory is the following result:
For example, if c ∣ (5 × 12), and c has no factors in common with 5, then in order for it to divide 5 × 12 = 60, it must divide 12. On the other hand, if a and c do have factors in common besides
1, then the result might not hold. For instance, if a = 10, then 10 ∣ (5 × 12), but 10 ∤ 12.
Since gcd(c, a) = 1, we can write cx+ay = 1 for some integers x and y. We can solve this to get ay = 1–cx. Further, since c ∣ ab, we can write ck = ab for some integer k. Multiply both sides by y
and plug in ay = 1–cx to get cky = (1–cx)b. We can rearrange this to get c(ky+bx) = b, which tells us that c ∣ b, as desired.
The gcd and lcm of more than two integers
The concept of gcd can be applied to more than two integers. Namely, gcd(a[1], a[2], …, a[n]) is the largest integer that divides each of the a[i]. For instance, gcd(24, 36, 60) = 12. The gcd can be
computed from the Euclidean algorithm and the following fact: gcd(14, 28, 50, 77), we can compute gcd(14, 28) = 14, then gcd(14, 50) = 2 and finally, gcd(2, 77) = 1. So gcd(14, 28, 50, 77) = 1.
It is also possible to compute the gcd by extending the ideas of the Euclidean algorithm. We perform several modulos at each step, always modding by the smallest value. Here is some (slightly tricky)
Python code implementing these ideas:
def gcd(*A):
while A[0] != 0:
A = sorted([A[0]] + [x % A[0] for x in A[1:]])
return A[-1]
One thing to be careful of is that it is possible to have gcd(a, b, c) = 1, but not have gcd(a, b) and gcd(b, c) both equal to 1. For instance, gcd(2, 4, 5) = 1 since 1 is the largest integer
dividing 2, 4, and 5, but gcd(2, 4)≠ 1.
For many theorems that require a bunch of integers, a[1], a[2], …, a[n] to not have any factors in common, instead of requiring gcd(a[1], a[2], …, a[n]) = 1, often the following notion is used:
Integers a[1], a[2], …, a[n] are said to be pairwise relatively prime if gcd(a[i], a[j]) = 1 for all i, j = 1, 2, …, n with i≠ j.
The lcm of integers a[1], a[2], …, a[n] is the smallest integer that is a multiple of each of the a[i]. Similarly to the gcd, the lcm can be computed by using the rule below to break things down.
def lcm(*X):
m = 1
for a in X:
m = a*m // gcd(a, m)
return m
Some useful facts about divisibility and gcds
Here is a list of facts that might come in handy from time to time. It's not worth memorizing this list, but it may be useful to refer back to if you need a certain fact for something you are working
Most important facts
1. Euclid's lemma: If a ∣ bc with a and b relatively prime, then a ∣ c.
2. If d = gcd(a, b), then d = ax+by for some integers x and y.
3. Any linear combination of a and b is a multiple of gcd(a, b).
4. Integers a and b are relatively prime if and only ax+by = 1 for some integers x and y.
6. If a ∣ b and b ∣ c, then a ∣ c.
7. If a ∣ b and c ∣ d, then ac ∣ bd.
8. If a ∣ b and a ∣ c, then a ∣ (bx+cy) for any integers x and y.
9. a ∣ b and b ∣ a if and only if a = b or a = –b.
10. If c ∣ a and c ∣ b, then c ∣ gcd(a, b).
11. If d|a and d|b, then d = gcd(a, b) if and only if gcd(a/d, b/d) = 1.
12. gcd(ka, kb) = |k| gcd(a, b) for any integer k≠ 0.
13. Let d = gcd(a, b). If a ∣ c and b ∣ c, then ab ∣ cd.
14. gcd(a+bc, b) = gcd(a, b) for any integer c.
15. If gcd(a, b) = 1, then gcd(c, ab) = gcd(c, a) gcd(c, b).
16. If a ∣ bc, then a/ gcd(a, b) is a divisor of c.
17. gcd(a, a+n) ∣ n. In particular, gcd(a, a+1) = 1.
18. gcd(a, b) lcm(a, b) = ab.
Prime numbers are one of the main focuses of number theory.
An integer greater than 1 is called prime if its only divisors are 1 and itself. It is called composite otherwise.
Notice that 1 is not considered to be prime. The reason is that primes are thought of as fundamental building blocks of numbers. As we will soon see, every number is a product of primes, each prime
helping to build up the number. However, 1 doesn't do any building, as multiplying by 1 doesn't accomplish anything. There are other ways in which 1 behaves differently from prime numbers, and for
these reasons 1 is not considered prime.
The first few primes are 2, 3, 5, 7, 11, 13, 17, 19. Much of number theory is concerned with the structure of the primes—how frequent they are, gaps between them, whether there is any sort of pattern
to them, etc.
Euclid's lemma
Recall Euclid's lemma from Section 1.9. It states that if c ∣ ab and gcd(c, a) = 1, then c ∣ b. If a is prime, then gcd(c, a) = 1, so Euclid's lemma holds whenever p is prime. Here is
Euclid's lemma restated for primes:
(Euclid's lemma) If p is prime and p ∣ ab, then p ∣ a or p ∣ b.
Euclid's lemma could be used as an alternate definition for prime numbers as it is not too hard to show that no other number besides 1 has this property. In fact, Euclid's lemma is used to define
analogs of prime numbers (like prime ideals) in abstract algebra.
Using induction, Euclid's lemma can be extended as follows:
If p is prime and p ∣ a[1]a[2]… a[n], then p ∣ a[i] for some i = 1, 2, … n.
We proceed by induction. The base case, n = 2, is Euclid's lemma. Now assume the statement holds for n and suppose p ∣ a[1]a[2]… a[n]a[n+1]. We can write a[1]a[2]… a[n]a[n+1] as (a[1]a[2]… a[n])(
a[n+1]) and by Euclid's lemma, either p ∣ a[n+1] or p ∣ a[1]a[2]… a[n]. In the latter case, by the induction hypothesis, p ∣ a[i] for some i = 1, 2, …, n. So overall, p ∣ a[i] for
some i = 1, 2, …, n+1. Thus the result is true by induction.
A direct consequence of this is the following:
is prime and
p ∣ q[1]q[2]… q[n]
, where
q[1], q[2], …, q[n]
are all prime, then
p = q[i]
for some
i = 1, 2, …, n
Euclid's lemma is one of the most important tools in elementary number theory and we will see it appear again and again.
The fundamental theorem of arithmetic
In math there are a number of “fundamental theorems.” There is the fundamental theorem of algebra which states that every nonconstant polynomial has a root, the fundamental theorem of calculus that
relates integration to differentiation, and in number theory, there is the fundamental theorem of arithmetic, which states that every integer greater than 1 can be factored uniquely into primes. For
instance, we can factor 60 into 2 × 2 × 3 × 5 and there is no other product of primes equal to 60, other than changing the order of 2 × 2 × 3 × 5. Here is the formal statement of the theorem:
Every integer n > 1 can be written uniquely as a product of primes.
Here is intuitively why we can write n as a product of primes: Either n itself is prime (in which case we are done) or else n can be factored into a product ab. These integers are either prime or
they themselves can be factored. These new factors are in turn either prime or they can be factored. We can continue this process, but eventually it must stop since the factors of a number are
smaller than the number itself, and things can't keep getting smaller forever. This can be made formal using induction.
We use strong induction to show that each number can be written as a product of primes. The base case
n = 2
is clear. Now assume that each integer
1, 2, …, n–1
can be written as a product of primes. Now either
is prime, in which case
is trivially a product of primes, or else we can write
n = ab
for some integers
in the range from 1 to
. By the induction hypothesis,
can be written as products of primes, so
n = ab
is a product of primes. Thus the result is true by induction.
To show the representation is unique, suppose n = p[1]p[2]… p[k] and n = q[1]q[2]… q[m] are two different representations. From the first representation, we have p[1] ∣ n, and so p[1] ∣ q[1]q
[2]… q[m]. By Corollary 10, we must have p[1] = q[i] for some i. By rearranging terms, we can assume i = 1, so p[1] = q[1]. By the same argument, we can similarly conclude that p[2] = q[2], p[3] = q
[3], etc. Thus the two representations are the same.
In the factorization, some of the primes may be the same, like in 720 = 2 × 2 × 2 × 2 × 3 × 3 × 5. We can gather those factors up and write the factorization as 2^3 · 3^2 · 5. In general, we can
always write an integer n > 1 as a unique product of the form pe[1]1pe[2]2… pe[k]k.
Applications of the fundamental theorem
1. The gcd and lcm can be computed easily from the prime factorization. For example, suppose we have a = 168 and b = 180. We have 168 = 2^3 · 3 · 7 and 180 = 2^2 · 3^2 · 5.
To get the gcd, we go prime-by-prime through the two representations, always taking the lesser of the two amounts. For instance, both 168 and 180 have a factor of 2: 168 has 2^3 and 180 has 2^2,
and so we use the lesser factor, 2^2, in the gcd. Moving on to the factor 3, 168 has 3^1 and 180 has 3^2, so we take 3^1. The next factors are 5 and 7, but they are not common to both 168 and
180, so we ignore them. We end up with gcd(168, 180) = 2^2 · 3 = 12.
The lcm is done similarly, except that we always take the larger amount. We get lcm(168, 180) = 2^3 · 3^2 · 5 · 7 = 2520.
Using the prime factorization to find the gcd and lcm is fast if we have the factorization is available. However, finding the prime factorization is a slow process for large numbers. The
Euclidean algorithm is orders of magnitude faster.**A little more formally, the Euclidean algorithm's running time grows linearly with the number of digits in the number, whereas the running
times of the fastest known factoring algorithms grow exponentially with the number of digits.
2. The fundamental theorem is a useful tool in proofs. For instance, let's prove that if n is a perfect square with n = ab and gcd(a, b) = 1, then a and b are perfect squares.
We can write n = m^2 for some m and assume m has the prime factorization m = pe[1]1pe[2]2… pe[k]k. Then n = ab, the prime factorization of a includes some of these primes and the prime
factorization of b includes the rest of them (or possibly a or b equals 1, in which case the result is trivial). But if the prime factorization of a includes a prime p[i], then the factorization
of b cannot include p[i] since gcd(a, b) = 1. Thus, after possibly reordering, there exists some integer j such that we can break the prime factorization of n up into a = p2e[1]1p2e[2]2… p2e[j]j
and b = p2e[j+1]j+1p2e[j+2]j+2… p2e[k]k. Thus a = (pe[1]1pe[2]2… pe[j]j)^2 and b = (pe[j+1]j+1pe[j+2]j+2… pe[k]k)^2 are perfect squares.
3. As another example, let's prove that if a ∣ c and b ∣ c with gcd(a, b) = 1, then ab ∣ c.
Since a ∣ c, every term, pe[i]i, in the factorization of a occurs in the factorization of c. Similarly, since b ∣ c, every term in the factorization of b occurs in the factorization of c.
Since gcd(a, b) = 1, the primes in the factorization of a must be different from the primes in the factorization of b. Thus every term in the factorization of ab occurs in the factorization of c.
So ab ∣ c.
There are infinitely many primes
One of the first things we might wonder about prime numbers is how many there are. The ancient Greeks answered this question with a proof somewhat like the one below.
There are infinitely many primes.
Let p[1], p[2], … p[n] be primes, and define P = p[1]p[2]… p[n] + 1. By the fundamental theorem, P must be divisible by some prime, and that prime must be different from p[1], p[2], …, p[n] since for
any i = 1, 2, …, n the fact that p[i] divides P means that p[i] cannot divide P+1. Thus, given any list of primes, we can use the list to generate a new prime, meaning the number of primes is
An alternate way to do the above proof would be to assume that there were only finitely many primes and to use the process above to derive a contradiction. A quick web search will turn up dozens of
other interesting proofs of the infinitude of primes.
It is worth noting that the integer P in the proof above need not be prime itself. It only needs to be divisible by a new prime. Numbers of the form given in the proof above are sometimes called
Euclid numbers or primorials. The first few are E[6] = 59 × 509 is not. It is not currently known whether infinitely many Euclid numbers are prime. There are not many Euclid numbers that are known to
be prime. The next few that are prime are E[7], E[11], E[31], and E[379]. Note that E[379] is already a large number, having about 4300 digits.
Finding primes
One of the simplest ways to check if a number is prime is trial division. Just check to see if it is divisible by any of the integers 2, 3, 4, etc. When considering divisors of n, they come in pairs,
one less than or equal to √n and the other greater than or equal to √n. For instance, if n = 30, we have √30 ≈ 5.47 and we can write 30 as 2 × 15, 3 × 10, and 5 × 6, with 2, 3, and 5 less than √30
and 6, 10, 15 greater than √30. So, in general, we can stop checking for divisors at √n.
This process can be made more efficient by just checking for divisibility by 2 and by odd numbers, or better yet, checking for divisibility only by primes (provided the number is small or we have a
list of primes). For example, to check if 617 is prime, we have √617 = 24.84 and we check to see if it is divisible by 2, 3, 5, 7, 11, 13, 17, 19, and 23. It is not divisible by any of those, so it
is prime.
This approach is reasonably fast for small numbers, but for checking the primality of larger numbers, like ones with several hundred digits, there are faster techniques, which we will see later.
To find all the primes less than an integer n, we can use a technique called the sieve of Eratosthenes. Start by listing the integers from 2 to n and cross out all the multiples of 2 after 2. Then
cross out all the multiples of 3 after 3. Then cross out all the multiples of 5 after 5. Note that we don't have to cross out the multiples of 4 since they have all already been crossed out as they
are multiples of 2. We keep going, crossing out multiples of 7, 11, 13, etc., until we get to √n. At the end, only the primes will be left. Here is what we would get for n = 100:
Here is how we might code it in Python. We create a list of zeroes and ones, with a zero at index i meaning i is not prime and a one meaning i is prime. The list starts off initially with all ones
and we gradually cross off all the composites.
def sieve(n):
L = [0,0] + [1]*(n-1)
p = 2
while p <= n**.5:
while L[p]==0:
p = p + 1
for j in range(2*p,n+1,p):
L[j] = 0
p += 1
return [i for i in range(len(L)) if L[i]==1]
The sieve works relatively well for finding small primes. The code above, inefficient though it may be, takes 55 seconds to find all the primes less than 10^8 on my laptop.
The prime number theorem
A natural question to ask is how common prime numbers are. An answer to this question and some other questions is given by the prime number theorem, which states that the number of primes less than n
is roughly n/ ln n . Formally, we can state it as follows:
(Prime number theorem) Let π(n) denote the number of primes less than or equal to n. Then
For example, for n = 1,000,000, we have π(n) = 78498 and n/ ln n = 72382. The prime number theorem's estimate is off by about 8% here. A more accurate estimate is n/( ln(n)–1), which in this case
gives 78030.
The theorem tells us that roughly 100/ ln n percent of the numbers less than n are prime. For n = 1,000,000, the theorem tells us that roughly 7-8% of the numbers less than 1,000,000 are prime, and
that around n = 1,000,000 the average gap between primes is roughly ln(1000000) ≈ 14.
The prime number theorem is not easy to prove. The first proofs used sophisticated techniques from complex analysis. There is a proof using only elementary methods, but it is fairly complicated.
A more accurate estimate
The prime number theorem tells us that the probability an integer near x is prime is roughly 1/ ln x. Summing up all these probabilities for all real x from 2 through n gives us another estimate for
the number of primes less than n. Such a sum is a continuous sum (an integral), and we get the following estimate for π(n): logarithmic integral and is denoted Li(n). This integral predicts 78,628
primes less than 1,000,000, which is only 130 off from the correct value.
An interesting note about this is that for n up through at least 10^14 it has been shown that Li(n) < π(n). But it turns out not to be true for all n. In fact, it was proved in 1933 that Li(n)–π(n)
changes sign infinitely often. This illustrates an important lesson in number theory: just because something is true for the first trillion (or more) integers, does not mean it is true in general.
The proof is interesting in that it included one of the largest numbers to ever be used in a proof, namely that π(n) < Li(n) for some value less than e^e^e^79. More recent research has brought the
bound down to e^728.
Twin primes
Twin primes are pairs (p, p+2) with both p and p+2 prime. Examples include (5, 7), (11, 13) and (41, 43).
One of the most famous open problems in math is the twin primes conjecture, which asks if there are infinitely many twin primes, Most mathematicians think the conjecture is true, though it is
considered to be a very difficult problem.
Referring back to the prime number theorem, we know that the probability an integer near x is prime is roughly 1/ ln x. Assuming independence, the probability that both x and x+2 would be prime is
then 1/( ln x)^2 and summing up these probabilities gives ∫n2 1( ln x)^2 dx as an estimate for the number of twin primes pairs less than n. Independence is not quite a valid assumption here, but it
is not too far off. It is currently conjectured that the number of twin prime pairs less than n is approximately 2C[2]∫n2 1( ln x)^2 dx, where C[2] ≈ .66016 is something called the twin prime
constant.**See Section 1.2 of Prime Numbers: A Computational Perspective, 2nd edition by Crandall and Pomerance for more on this approach.
So it seems reasonable that there are infinitely many twin primes, but it has turned out to be very difficult to prove. The best result so far is that there are infinitely many pairs (p, p+2) where p
is prime and p+2 is either prime or the product of two primes, proved by Chen Jingrun in 1973.
There are a number of analogous conjectures. For instance, it is conjectured that there are infinitely many pairs of primes of the form (p, p+4) or infinitely many triples of the form (p, p+2, p+6).
Recent work has shown that there are infinitely many primes p such that one of p+2, p+4, … p+246 is also prime.
Prime gaps
It is interesting to look at the gaps between consecutive primes. Here are the gaps between the first 100 primes, from 2 to 541:
1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4, 6, 2, 10, 6, 6, 6, 2, 6, 4, 2, 10, 14, 4, 2,
4, 14, 6, 10, 2, 4, 6, 8, 6, 6, 4, 6, 8, 4, 8, 10, 2, 10, 2, 6, 4, 6, 8, 4, 2, 4, 12, 8, 4, 8, 4, 6, 12, 2, 18
We see gaps of 2 quite often. These correspond to twin primes. There are a few larger gaps, like a gap of 14 from 113 to 127 and a gap of 18 from 523 to 541. It is not too hard to find gaps that are
arbitrarily large. Just find an integer n divisible by 2, 3, 4, 5, …, k and then n+2, n+3, … n+k will all be composite.
The prime number theorem tells us that the average gap between a prime p and the next prime is approximately ln p. Thus for p near 1,000,000, we would expect an average gap of about 14, and for p =
10^200, we would expect an average gap of around 460.
One important result is relating to prime gaps is Bertrand's postulate.
(Bertrand's postulate) For any integer n > 1 there exists a prime p such that n < p < 2n.
For example, for n = 1000, we are guaranteed that there is a prime between 1000 and 2000. This is not news, but it is useful in some cases to have an interval on which you are guaranteed to have a
prime, even if that interval is rather large.
The proof of Bertrand's postulate is actually not too difficult, but we won't cover it here. There are a number of improvements on Bertrand's postulate. For instance, in 1952 it was proved that for n
≥ 25 there exists a prime between n and (1+15)n. The range can be narrowed further for larger n.**In general, it has been proved that for any ε > 0 there is an integer N such that for n ≥ N, there is
always a prime between n and (1+ε)n. In fact, as n → ∞, the number of primes in that range approaches ∞ as well. In math, there are many results concerning how things behave as n approaches ∞.
Results of this sort are called asymptotic results.
A related, but unsolved, question is if there is always a prime between consecutive perfect squares, n^2 and (n+1)^2.
Finding large primes
A popular sport among math enthusiasts is finding large prime numbers. The largest primes known are all Mersenne primes, which are primes of the form 2^n–1. There is a relatively fast algorithm for
checking if a number of the form 2^n–1 is prime, and that is why people searching for large primes use these numbers. As of early 2016, the largest known prime is 2^74207281–1, a number over 22
million digits long.
Most of the largest primes found recently were found by the Great Internet Mersenne Prime Search (or GIMPS), where volunteers from around the world donate their spare CPU cycles towards checking for
primes. Finding large primes usually involves either a combination of sophisticated algorithms and finely-tuned hardware or a distributed computer search like GIMPS.
People also look for large primes of special forms. For instance, as of early 2016, the largest known twin primes are 3756801695685 · 2^666669± 1, about 200,000 digits long. See http://primes.utm.edu
/largest.html for a nice list of large primes.
Prime-generating formulas
One of the most remarkable polynomials in all of math is p(n) = n^2+n+41. It has the property that for each n = 0, 1, … 39, p(n) is prime. However, p(40) and p(41) are not prime as p(40) = 41^2 and p
(41) = 41^2+41+41 is clearly divisible by 41. Still, the polynomial keeps on generating primes at a pretty high rate as p(n) is prime for 34 of the next 39 values of n. In total, 156 of the first 200
values of p(n) are prime, and 581 of the first 1000.
There are a number of other polynomials that are good at generating primes, like n^2+n+11 and n^2+n+17, though neither of these is quite as good as n^2+n+41. The formula n^2–79n+1601 generates primes
for each integer from n = 0 through n = 79. It is actually a modification of n^2+n+41: namely, n^2–79n+1601 = (n–40)^2+(n–40)+41. The 80 primes are the same as the 40 primes from n^2+n+41, each
appearing twice.
The site http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html has a nice list of some other prime-generating polynomials.
Prime spirals
There is a nice way to visualize prime numbers known as a prime spiral or Ulam spiral. Start with 1 in the middle, and spiral out from there, like in the figure below on the left. Then highlight the
primes, like on the right.
If we expand the view to the first 20,000 primes, we get the figure below.
Notice the primes tend to cluster along certain diagonal lines. The dots highlighted in red correspond to Euler's polynomial, n^2+n+41.
An interesting twist on this is something called the Sacks spiral. Instead of the rectangular spiral we used above, we instead spiral along an Archimedean spiral, where both the angular and radial
velocity are constant, with those constants chosen so that the perfect squares lie along the horizontal axis. Here is a typical Archimedean spiral and the Sacks spiral for the first few primes:
And here it is for a wider range. Euler's polynomial, n^2+n+41, is highlighted in red.
More with primes and polynomials
It is not too difficult to show that there is no nonconstant polynomial that can give only primes. Just plug in the constant term. For instance, if p(x) = a[n]x^n+a[n–1]x^n–1+…+a[1]x+a[0], then every
term of p(a[0]) is divisible by a[0]. Thus p(a[0]) will be composite, except possibly when a[0] = 0 or ± 1. If a[0] = ± 1, then p(0) is not prime, and if a[0] = 0, factor out an x and try plugging in
the new constant term.
An interesting open problem is whether or not there are infinitely many primes of the form n^2+1.
Other formulas
There are some other interesting formulas for generating primes. For instance, it turns out that there exists a real number r such that floor(r^3^n) is prime for any integer n. The exact value of r
is unknown, but it is thought to be approximately 1.30637788. Another example is the recurrence x[n] = x[n–1] + gcd(n, x[n–1]), x[1] = 7. The difference between consecutive terms is always either 1
or a prime.
Primes and arithmetic progressions
One interesting result is Dirichlet's theorem, stated below:
(Dirichlet's theorem) If gcd(a, b) = 1, then there are infinitely many primes of the form ak+b.
For example, with a = 4 and b = 3, the theorem tells us that there are infinitely many primes of the form 4k+3. If a = 100 and b = 1, the theorem tells us there are infinitely many primes of the form
100k+1 (i.e., primes that end in 01). The first few are 101, 401, 601, 701, 1201, …. Like the prime number theorem, Dirichlet's theorem is difficult to prove, relying on techniques from analytic
number theory.
A related theorem is the Green-Tao theorem, proved in 2004. It concerns arithmetic progressions of primes, sequences of the form p, p+a, p+2a, … p+(n–1)a, all of which are prime. In other words, we
are looking at sequences of equally spaced primes, like (3, 5, 7) or (5, 11, 17, 23, 29). The Green-Tao theorem states that it is possible to find prime arithmetic progressions of any length. The
proof, like many in math, is an existence proof. It shows that these progressions exist but doesn't tell how to find them. According to the Wikipedia page on the Green-Tao theorem, as of 2010 the
longest known arithmetic progression was 25 terms long, starting at the integer 43142746595714191.
Fermat numbers
In the 1600s, Pierre de Fermat studied primes of the form 2^2^n+1. Numbers of this form are now called Fermat numbers. The first one is 2^2^0+1 = 3, and the next few are 5, 17, 257, and 65537. These
are all prime, and Fermat conjectured that every Fermat number is prime. However, the next one, F[5] = 4294967297, turns out to have a factor of 641.
The remarkable fact is that there are no other Fermat numbers that are known to be prime. In fact, it is an open question as to whether any other Fermat numbers are prime.
Considerable effort has gone into trying to factor Fermat numbers. This is difficult because of the shear size of the numbers. As of early 2014, F[5] through F[11] have been completely factored.
Partial factorizations have been found for many other Fermat numbers. The smallest Fermat number which is not known to be composite is F[33]. See www.prothsearch.net/fermat.html for a comprehensive
list of results.
Sophie Germain primes
A Sophie Germain prime is a prime p such that 2p+1 is also prime. For instance, 11 is a Sophie Germain prime because 2 · 11 + 1 = 23 is also prime. The first few Sophie Germain primes are 2, 3, 5,
11, 23, 29, 41, 53, 83, 89. It is not currently known if there are infinitely many, though it is thought that there are.**In fact, it is suspected that there are about as many Sophie Germain primes
as there are twin primes pairs. They are named for the 19th century mathematician Sophie Germain, who used them in her work on Fermat's Last Theorem.**Fermat's last theorem states that there are no
integer solutions to x^n+y^n = z^n if n > 2. It was one of the most famous problems in math for a few hundred years.
If p is a Sophie Germain prime, then 2p+1 is also prime by definition, and is called a safe prime. Safe primes are important in modern cryptography. See Section 4.1.
It is also interesting to create chains where p, 2p+1, 2(2p+1)+1, etc. are all prime. Such chains are called Cunningham chains. One such chain is 2, 5, 11, 23, 47. It can't be extended any further as
2 · 47 +1 = 95 is not prime. It is thought that there are infinitely many chains of all lengths, but no one knows for sure. According to Wikipedia, the longest chain so far found is 17 numbers long,
starting at 2,759,832,934,171,386,593,519.
Goldbach's conjecture
Goldbach's conjecture is one of the most famous open problems in math. It simply states that any even number greater than two is the sum of two primes. For instance, we can write 4 = 2+2, 6 = 3+3, 8
= 5+3, and 10 = 5+5 or 7+3.
Goldbach's conjecture has been verified numerically by computer searches up through about 10^18. The number of ways to write an even number as a sum of two primes seems to increase quite rapidly. For
instance, numbers between 2 and 100 have an average of about four ways to be written as sums of primes. This increases to 18 for numbers between 100 and 1000, 93 for numbers between 1000 and 10,000,
and 554 for numbers between 10,000 and 100,000. Moreover, in the range from 10,000 to 100,000 no number can be written in less than 92 ways.
Here is a graph showing how the number of possible ways to write a number as a sum of two primes increases with n. The horizontal axis runs from n = 4 to n = 100,000 and the vertical axis runs to
about 2000.
Despite the overwhelming numerical evidence, the Goldbach conjecture is still far from being proved. However, there are a number of partial results. For instance, in the early 1970s Chen Jingrun
proved that every sufficiently large even number can be written as a sum p+q, where p is prime and q is either prime or a product of two primes.
There is also the weak Goldbach conjecture that states that every odd number greater than 7 is the sum of three primes. In the 1930s, I.M. Vinogradov proved that it was true for all sufficiently
large integers. It seems that the weak Goldbach conjecture may have been proved in 2013 by Harald Helfgott, though as of this writing, the proof has not been fully checked. Like, Vinogradov's result,
Helfgott proved the result true for all sufficiently large integers, but in this case “sufficiently large” was small enough that everything less than it could be checked by computer.
Some sums
One of the most important functions in analytic number theory is the zeta function, ζ(1), the well-known harmonic series. We also have ζ(2) and ζ(4), whose sums were famously determined by Euler. In
general, the even values, ζ(2n), always sum to some constant times π^2n, whereas not too much is known about any of the odd values, except for the harmonic series.
It has been known since at least the 14th century that the harmonic series is divergent. A particularly nice proof of this fact is shown below: Nth partial sum of the harmonic series is nearly equal
to ln(N). For instance, we have the following:
N ∑Nn = 11n ln(N) Difference
100 5.187378 4.605170 .582207
10,000 9.787606 9.210340 .577266
1,000,000 14.392727 13.815511 .577216
100,000,000 18.997896 18.420681 .577216
In general, we have that ∑Nn = 1 1n – ln(N) converges to γ ≈ .5772156649, the Euler-Mascheroni constant. This is one of the most famous constants in math, showing up in a number of places in higher
mathematics. It is actually not known whether γ is rational or irrational.
Euler also discovered a beautiful relationship between the harmonic series and prime numbers:
It is interesting to note that a similar sum involving twin primes is actually convergent, namely **A 1994 calculation of the constant was responsible for finding a bug in the Pentium processor. See
the article How Number Theory Got the Best of the Pentium Chip in the January 13, 1995 issue of Science magazine.
The Riemann hypothesis
Perhaps the most famous unsolved problem in mathematics is the Riemann hypothesis, first stated by Bernhard Riemann in 1859. It is a statement involving the zeta function. A process called analytic
continuation is used to find a function defined for most real and complex values that agrees with ζ(n) wherever ζ(n) is defined. This new function is called the Riemann zeta function.
The Riemann zeta function has zeroes at –2, –4, –6, …. These are called its trivial zeros. It has many other complex zeros that have real part 1/2. The line with real part 1/2 is called the critical
line. The Riemann hypothesis states that the only nontrivial zeros of the Riemann zeta function lie on the critical line.
It might not be clear at this point what the Riemann hypothesis has to do with primes. Here is the connection, first shown by Euler:
If true, the Riemann hypothesis would imply that the prime numbers are distributed fairly regularly, whereas if it were false, then it would mean that prime numbers are distributed considerably more
wildly. There are a number of potential theorems in number theory that start “If the Riemann hypothesis is true, then….” So a solution to the Riemann hypothesis would tell us a lot about primes and
other things in number theory.
There have been a variety of different approaches to proving the Riemann hypothesis, though none have thus far been successful. Many mathematicians believe it is likely true, but there are some that
are not so sure. Numerical computations have shown that the first 10^13 nontrivial zeroes all lie on the critical line. The Riemann hypothesis is one of the Clay Mathematics Institute's seven
Millennium Prize problems, with $1,000,000 offered for its solution.
Number-theoretic functions
There are a few functions that show up a lot in number theory.
be positive integer.
1. The number of positive divisors of n is denoted by τ(n).
2. The sum of the positive divisors of n is denoted by σ(n).
3. The number of positive integers less than n relatively prime to n is denoted by φ(n).
For example, n = 12 has six divisors: 1, 2, 3, 4, 6, and 12. Thus τ(12) = 6 and σ(12) = 1+2+3+4+6+12 = 28. The only positive integers less than 12 that are relatively prime to 12 are 1, 5, 7, and 11,
so φ(12) = 4.
As another example, suppose p is prime. Then the divisors of p are just 1 and p, so τ(p) = 2, σ(p) = p+1, and every positive integer less than p is relatively prime to it, so φ(p) = p–1.
The most important of the three functions is φ(n), called the totient function or simply the Euler phi function.
Computing τ(n)
Let's start by computing τ(1400). We have 1400 = 2^3 · 5^2 · 7. Any divisor will include 0, 1, 2, or 3 twos, 0, 1, or 2 fives, and 0 or 1 seven. So we have 4 choices for the twos, 3 choices for the
fives, and 2 choices for the sevens. There are thus 4 · 3 · 2 = 24 possible divisors in total.
This reasoning works in general. Given the prime factorization n = pe[1]1pe[2]2 ··· pe[k]k, we have
Computing σ(n)
Let's compute σ(1400). Again, we have 1400 = 2^3 · 5^2 · 7. Consider the following product: σ(1400); each divisor appears exactly once in the sum. On the other hand, each term on the left side can be
rewritten using the geometric series formula. So we end up with n = pe[1]1pe[2]2 ··· pe[k]k, we have
Computing φ(n)
Finding a formula for φ(n) is considerably more involved. First, if p is prime, then φ(p) = p–1 since every positive integer less than p is relatively prime to p.
Next, consider φ(p^i), where p is prime. The positive integers less than p^i that are not relatively prime to p^i are precisely the multiples of p. There are p^i–1 such multiples, namely 1, p, 2p, 3p
, …, p^i–1p. So φ(p^i) = p^i–p^i–1, which we can rewrite as p^i(1–1p).
To extend this to the factorization n = pe[1]1pe[2]2 ··· pe[k]k, we will show (in a minute) that φ(mn) = φ(m)φ(n), whenever m and n are relatively prime. Since each of the terms pe[i]i are relatively
prime, this will give us φ(1400), we note 1400 = 2^3 · 5^2 · 7 and compute φ(164934), we have 164934 = 2 · 3^2 · 7^2 · 11 · 17 and so
We now show that φ(mn) = φ(m)φ(n) whenever gcd(m, n) = 1.
First, consider an example: φ(30) = 8. We have 30 = 5 × 6, φ(5) = 4, and φ(6) = 2. See the figure below. Notice there are 2 columns with 4 integers relatively prime to 30 in each.
As another example, consider φ(33) = 20. We have 33 = 3 × 11, φ(3) = 2, and φ(11) = 10. Notice in the figure below there are 10 columns with 2 integers relatively prime to 33 in each.
This sort of thing happens in general. Suppose we have relatively prime integers m and n, and consider φ(nm).
1 2 … m
m+1 m+2 … 2m
2m+1 2m+2 … 3m
⋮ ⋮ ⋱ ⋮
(n–1)m+1 (n–1)m+2 … nm
We claim that there are φ(m) columns that each contain φ(n) integers relatively prime to nm. This tells us that φ(nm) = φ(n)φ(m). The following three steps are enough to prove the claim:
1. If an entry x in row 1 is not relatively prime to m, then none of the entries in the same column as x are relatively prime to m. Thus only φ(m) columns can contain entries that are relatively
prime to mn.
This is true because all the entries in the column are of the form x+mk. Since gcd(x, m) ≠ 1 that means some integer d divides both x and m and hence x+mk as well.
2. Each column is a permutation of 0, 1, 2, …, n–1 modulo n.
To show this, suppose two entries in the column were congruent modulo n, say x+mk ≡ x+mj (mod n) for some i and j. Then mk ≡ mj (mod n) and since gcd(m, n) = 1, we can cancel to get k ≡ j (mod n)
, which is to say the entries must have come from the same row. In other words, entries in the column from different rows can't be the same.
3. If we can show that whether an integer is relatively prime to n or not depends only on its congruence class modulo n, then we would be done, since the column is a permutation of 0, 1, 2, …, n–1
modulo n, and there are φ(n) integers in that range relatively prime to n.
To do this, suppose s ≡ t (mod n) and s is relatively prime to n. We want to show that t is relatively prime to n. We can write s–t = nk and sx+ny = 1 for some integers k, x, and y. Solve the
former for s and plug it into the latter to get (nk+t)x+ny = 1. Rearrange to get tx+n(kx+y) = 1 from which we get that t and n are relatively prime.
Here is an interesting result about the Euler phi function:
That is, if we sum φ(d) over the divisors of n, the result is n. For an idea as to why this is true, take a look at the following example:
d Integers a with gcd(a, 10) = d φ(10/d)
1 1, 3, 7, 9 φ(10) = 4
2 2, 4, 6, 8 φ(5) = 4
5 5 φ(2) = 2
10 10 φ(1) = 1
Recall from Section 1.7 that gcd(a, n) = d if and only if gcd(a/d, n/d) = 1. That is, if we divide a and n by their gcd, then the resulting integers have nothing in common (and the converse holds as
well). So for instance, if we want to find all the integers a such that gcd(a, 10) = 2, we find all the integers whose gcd with 10/2 is 1; there are φ(5) such integers. And this works in both
directions. In general then n as each integer from 1 through n must fall into exactly one of the sets |{a : gcd(a, n) = d}|.
Multiplicative functions
A little earlier we showed that φ(mn) = φ(m)φ(n) provided gcd(m, n) = 1. This way of breaking up a function is important in number theory. Here is a definition of the concept:
A function f defined on the positive integers is called multiplicative provided f(mn) = f(m)f(n) whenever gcd(m, n) = 1.
We have the following:
The functions τ, σ, and φ are multiplicative.
We already showed φ is multiplicative, and it is easy to show τ and σ are multiplicative using the formulas we have for computing them.
The Möbius function
The Möbius function, defined below is important in higher number theory, though we won't use it much in this text.
The Möbius function, denoted μ(n), is 1 if n = 1, (–1)^k if n = p[1]p[2] ··· p[k] is a product of k distinct primes, and is 0 otherwise.
For instance, 12 = 2 × 2 × 3 is not a product of distinct primes, so μ(12) = 0. On the other hand, 105 = 3 × 5 × 7 is a product of 3 distinct primes, so μ(105) = (–1)^3 = –1.
It easy to show μ is multiplicative from its definition. The following fact, called Möbius inversion is an important tool in analytic number theory:
(Möbius inversion) If f and g are two number-theoretic functions such that g(n) = ∑[d ∣ n] f(d) for every integer n ≥ 1, then for every integer n ≥ 1,
The proof is not difficult and can be found in many textbooks.
Modular arithmetic
Modular arithmetic is a kind of “wrap around” arithmetic, like arithmetic with time. In 24-hour time, after 23:59, we wrap back around to the start 00:00. After the 7th day of the week (Saturday), we
wrap back around to the start (Sunday). After the 365th or 366th day of the year, we wrap back around to the first day of the year. Many things in math and real-life are cyclical and a special kind
of math, known as modular arithmetic, is used to model these situations.
Let's look at some examples of arithmetic modulo (mod) 7, where we use the integers 0 through 6. We can think of the integers as arranged on a circle, like below:
We have the following:
1. 7 is the same as 0, 8 is the same as 1, 9 is the same as 2, etc.
2. In general, any multiple of 7 is the same as 0, any number of the form 7k+1 is the same as 1, any number of the form 7k+2 is the same as 2, etc.
3. 4+5 is the same as 2. Adding 5 corresponds to moving around the circle 5 units clockwise.
4. 4–5 is the same as 6. Subtracting 5 corresponds to moving 5 units counterclockwise.
5. 4+21 is the same as 4. Adding 21 corresponds to going around the circle 3 times and ending up where you started.
Instead of saying something like “8 is the same as 1,” we use the notation 8 ≡ 1 (mod 7). This is read as “8 is congruent to 1 mod 7.” Such an expression is called a congruence. Here is the formal
Let a, b, and n be integers. We say a ≡ b (mod n) if a and b both leave the same remainders when divided by n. Equivalently, a ≡ b (mod n) provided n ∣ (a–b).**We can also use n ∣ (b–a) in
place of n ∣ (a–b) in the definition.
It is not too hard to show that the two definitions are equivalent. We can use whichever one suits us best for a particular situation. The latter definition is useful in that it gives us an equation
to work with, namely a–b = nk for some integer k. For example, 29 ≡ 15 (mod 7) since both 29 and 15 leave the same remainder when divided by 7. Equivalently, 29 ≡ 15 (mod 7) because 29–15 is
divisible by 7.
Modular arithmetic is usually defined by noting that ≡ is an equivalence relation. See Section 3.6 for more on this approach. For now we will just approach things informally.
A few examples
Here are a few examples to get us some practice with congruences:
1. Find the remainder when 1!+2!+3!+… 100! is divided by 12.
Solution: Notice that 4! = 4 · 3 · 2 · 1 contains 4 · 3, so it is divisible by 12. Similarly, 5!, 6!, etc. all contain 4 × 3, so they are all divisible by 12 and hence congruent to 0 modulo 12.
Thus 1!+2!+3!+… 100! ≡ 1! + 2! + 3! ≡ 9 (mod 12)
2. A useful fact is that a ≡ 0 (mod n) if and only if n ∣ a. This is useful in computer programs. For instance, to check if an integer a is even in a computer program, we check if a % 2 == 0.
3. Mods give an easy way of finding the last digits of a number. The last digit of an integer n is n mod 10. The last two digits are n mod 100, and in general, the last k digits are n mod 10^k.
For example, suppose we want the last digit of 2^1000. To find it, we compute 2^1000 modulo 10. Notice that 2^5 ≡ 2 (mod 10). Then 2^25 = (2^5)^5 ≡ 2^5 ≡ 2 (mod 10). Similarly, 2^125 ≡ 2 (mod 10)
, and 2^1000 = 2^125 · 8 = (2^125)^8 ≡ 2^8 (mod 10). Since 2^8 = 256 ≡ 6 (mod 10), our answer is 6.
Working with congruences
Modular arithmetic is like a whole new way of doing math. It's good to have a list of some of the common rules for working with it.
Relationship to the mod operator
Modular arithmetic is related to the mod operation from Section 1.3. For instance, in arithmetic mod 7 we have 1, 8, 15, 22, … as well as -6, -13, -20, … all corresponding to the same value. Often,
but not always, the most convenient value to use to represent all of them is the smallest positive integer value, in this case 1. To find that value for, we use the mod operation from Section 1.3.
For instance, to find the smallest positive integer that 67 is congruent to modulo 3, we can compute 67 mod 3 to get 1.
In general, we have the following:
Given an integer m, if we want to find the smallest positive integer k such that m ≡ k (mod n), we have k = m mod n.
A similar and useful rule is the following:
An integer n is of the form ak+b if and only if n ≡ b (mod a).
For instance, if a number n is of the form 3k+1, then n ≡ 1 (mod 3). And conversely, any number congruent to 1 modulo 3 is of the form 3k+1.
Algebraic rules
Here are a few rules for working with congruences:
1. The ≡ symbol satisfies three properties that make it an equivalence relation:
1. Reflexive property: For all x, x ≡ x (mod n).
2. Symmetric property: If x ≡ y (mod n), then y ≡ x (mod n).
3. Transitive property: If x ≡ y (mod n), and y ≡ z (mod n), then x ≡ z (mod n).
These are simple rules that are easy to prove. We will use them without referring to them.
2. a + cn ≡ a (mod n).
In other words, adding a multiple of the modulus n is the same as adding 0. For instance, 2 + 45 ≡ 2 (mod 9) and 17 + 1000 ≡ 17 (mod 100).
3. n–a ≡ –a (mod n).
This rule is a special case of the previous rule and is often useful in computations. For instance, –1 ≡ 99 (mod 100). So we can replace 99 in computations mod 100 with -1, which is a lot easier
to work with. If we needed to compute 99^50 modulo 100, we could replace 99 with -1 and note that (–1)^50 is 1, so 99^50 ≡ 1 (mod 100).
4. We can work with congruences in similar (but not identical) ways to how we work with algebraic equations. For example, we can add or subtract a term on both sides of a congruence, multiply both
sides by something, or raise both sides to the same power. That is, if a ≡ b (mod n), then for any c
5. We can add two congruences, just like adding two equations. In particular, if a ≡ b (mod n) and c ≡ d (mod n), then
Canceling terms
We have to be careful canceling terms. For example, 2 × 5 ≡ 2 × 11 (mod 12) but 5 ≢ 11 (mod 12). We see that we can't cancel the 2. In general, we have the following theorem:
Suppose ca ≡ cb (mod n). Then a ≡ b (mod n/ gcd(c, n)). In particular, if gcd(c, n) = 1, then we can cancel c to get a ≡ b (mod n).
In the example above, since gcd(2, 12)≠ 1, we can't cancel out the 2. However, using the theorem we can say that 7 ≡ 9 (mod 6). On the other hand, given 2 × 3 ≡ 2 × 10 (mod 7), since gcd(2, 7) = 1,
we can cancel out the 2 to get 3 ≡ 10 (mod 7).
Breaking things into cases
Modular arithmetic is useful in that it can break things down into several cases to check. Here are a few examples:
1. Suppose we want to show that n^4 can only end in 0, 1, 5 or 6.
To solve this, we find the possible values of n^4 modulo 10. There are 10 cases to check, 0^4, 1^4, …, 9^4, since every integer is congruent to some integer from 0 to 9. We end up with 0^4 ≡ 0
(mod 10), 5^4 ≡ 5 (mod 10), 1^4 ≡ 3^4 ≡ 7^4 ≡ 9^4 ≡ 1 (mod 10), and 2^4 ≡ 4^4 ≡ 6^4 ≡ 8^4 ≡ 1 (mod 10).
2. In Section 1.2 we showed that any perfect square n^2 is of the form 4k or 4k+1.
To show this using modular arithmetic, we just consider cases modulo 4. Squaring 0, 1, 2, and 3 modulo 4 gives 0, 1, 0, and 1, so we see that n^2 ≡ 0 (mod 4) or n^2 ≡ 1 (mod 4), which is the same
as saying that n^2 is of the form 4k or 4k+1.
3. Show that if p > 3 is prime, then p^2–1 is divisible by 24.
First, note that if p > 3 is prime, then p must be of the form 24k+r, where r = 1, 5, 7, 11, 13, 17, 19, or 23. Any other form is composite (for example 24k+3 is divisible by 3 and 24k+10 is
divisible by 2). Thus p ≡ r (mod 24) for one of the above values of r, and it is not hard to check that p^2–1 ≡ r^2–1 ≡ 0 (mod 24) for each of those values.
In general, if an integer n is of the form ak+b, then we can write the congruence n ≡ b (mod a). The converse holds as well. For instance, all numbers of the form 5k+1 are congruent to 1 modulo 5 and
Breaking up a mod
We have the following useful fact:
If a ≡ b (mod m) and a ≡ b (mod n), with gcd(m, n) = 1, then a ≡ b (mod mn).
This follows from a fact proved in Section 2.1.
We often use this to break up a large mod into smaller mods. For example, suppose we want to show that n^5 and n always end in the same digit. That is, we want to show that n^5 ≡ n (mod 10). Using
the above fact, we can do this by showing n^5 ≡ n (mod 2) and n^5 ≡ n (mod 5). The first congruence is easily seen to be true. For the second, we check the five cases 0^5, 1^5, 2^5, 3^5, and 4^5. A
short calculation verifies that the congruence holds for each of them.
The theorem above can be generalized to the following:
Consider the system of congruences
a ≡ b (mod m[1])
a ≡ b (mod m[2])
, …,
a ≡ b (mod m[k])
. If
, …
are all pairwise relatively prime, then
a ≡ b (mod m[1]m[2]… m[2])
If the moduli are not necessarily pairwise relatively prime, we still have
Working with the definition
One of the most important parts of working with congruences is using the definition. In particular, we have that x ≡ y (mod n) provided n ∣ (x–y) or equivalently that x–y = nk for some integer k.
Here are several examples:
1. Prove that n^3–n is divisible by 3 for any n ∈ ℤ.
We can turn this into a statement about congruences, namely n^3 ≡ n (mod 3). Modulo 3 there are only three cases to check: n = 0, 1, and 2. And we have 0^3 ≡ 0 (mod 3), 1^3 ≡ 1 (mod 3), and 2^3 ≡
2 (mod 3). Compare this argument to the longer one using the division algorithm in Example 1 of Section 1.2.
2. Prove if a ≡ b (mod n), then a+c ≡ b+c (mod n).
We can start by writing the first congruence as an equation: a–b = nk for some integer k. Add and subtract c to the left side to get a–b+(c–c) = nk. Then rearrange terms to get (a+c)–(b+c) = nk.
Then rewrite this equation as the congruence a+c ≡ b+c (mod n), which is what we needed to show.
3. Prove if a ≡ b (mod n), then a^c ≡ b^c (mod n).
This is a little trickier. The definition tells us we need to show n ∣ (a^c–b^c). The trick is to factor the left side into (a–b)(a^c–1+a^c–2b+a^c–3b^2 + … + ab^c–2 + b^c–1). Since a ≡ b (mod
n), we have n ∣ (a–b). Since a–b is a factor of a^c–b^c, we get n ∣ (a^c–b^c).
4. Prove if ca ≡ cb (mod n), then a ≡ b (mod n/d), where d = gcd(c, n).
Start with nk = ca–cb for some integer k. Also, since d is gcd(a, b), we have dx = n and dy = c for some integers x and y.
Plugging in, we get dxk = dya – dyb. We can cancel d to get xk = y(a–b). We know gcd(x, y) = 1 as otherwise any common factor of x and y could be included in d to get a larger common divisor of n
and c. So we can use Euclid's lemma to conclude that x ∣ a–b. Note that x = n/d, so we have a ≡ b (mod n/d).
Powers turn out to be interesting in modular arithmetic. For instance, here is a table of powers modulo 7:
a^0 a^1 a^2 a^3 a^4 a^5 a^6
a = 1 1 1 1 1 1 1 1
a = 2 1 2 4 1 2 4 1
a = 3 1 3 2 6 4 5 1
a = 4 1 4 2 1 4 2 1
a = 5 1 5 4 6 2 3 1
a = 6 1 6 1 6 1 6 1
There are a number of interesting things here, some of which we will look at in detail later. We see that a^6 ≡ 1 (mod 7) for all values of a. We can also see that some of the powers run through all
the integers 1 through 6, while others don't. In fact, all the powers cycle with period 1, 2, 3, or 6, all of which are divisors of 6.
Note: To create the table, it is not necessary to compute large powers. For instance, instead of computing 5^3 = 125 and reducing modulo 7, we can instead write 5^3 = 5^2 · 5. Since 5^2 is 4 modulo
7, we get that 5^3 = 4 · 5, which is 6 modulo 7.
Below are the tables for modulo 23 and 24. Smaller integers are colored red and larger values are green.**The program that produced these images can be found at http://www.brianheinold.net/mods.html.
Computing numbers to rather large powers turns out to be pretty important in number theory and cryptography. Here are a couple of examples:
1. Compute 6^100 modulo 7.
We have 6 ≡ –1 (mod 7) so 6^100 ≡ (–1)^100 ≡ 1 (mod 7).
2. Compute 2^100 modulo 7.
Note that 2^3 ≡ 1 (mod 7). Further, we can write 100 = 3 × 33 + 1. Thus we have 2^100 = 2 × (2^3)^33. Then
In general, if we can spot a power that is simple, like that 6^1 ≡ –1 (mod 7) or 2^3 ≡ 1 (mod 7), then we can leverage that to work out large powers. Otherwise, we can use the technique demonstrated
Suppose we want to compute 5^100 (mod 11). Compute the following: 5^100 as 5^64+32+4, which is 5^64 · 5^32 · 5^4 or 9 · 3 · 9. This reduces to 1 modulo 11.
This process is called exponentiation by squaring. In general, to compute a^b (mod n), we compute a^1, a^2, a^4, a^8, etc., up until the exponent is the largest power of 2 less than b. We then write
b in terms of those powers and use the rules of exponents to compute a^b. Writing b in terms of those powers is the same process as converting b to binary. For instance, 100 in binary is 1100100,
which corresponds to 64 · 1 + 32 · 1 + 16 · 0 + 8 · 0 + 4 · 1 + 2 · 0 + 1 · 0, or 64+32+4.
Here is how we might code this algorithm in Python:
def mpow(b,e,n):
prod = 1
while e > 0:
if e % 2 == 1:
prod = (prod * b) % n
e = e // 2
b = (b*b) % n
return prod % n
Note, however, that this algorithm is already built into Python with the built-in function pow. In particular, pow(a, n, m) will compute a^n mod m. It can handle quite large powers.
Some further examples of modular arithmetic
1. Unlike with ordinary arithmetic, it is possible for the product of two nonzero integers to be 0 in modular arithmetic. For example, 2 × 5 ≡ 0 (mod 10).
Note that this cannot happen if the modulus is prime. This is because if ab ≡ 0 (mod n), then we have p ∣ ab. By Euclid's lemma, since p is prime, either p ∣ a or p ∣ b, which would
imply at least one of a and b is congruent to 0 modulo p.**By the more general version of Euclid's lemma (Theorem 7), if n is relatively prime to both a and b, then we can't have ab ≡ 0 (mod n).
2. An easy way to tell if a number is divisible by 3 is if the sum of its digits is divisible by 3. We can use modular arithmetic to show that this is true. Suppose we have a number n with ones
digit d[0], tens digit d[1], etc. We can write that number as S = d[k]+d[k–1]+… + d[1] + d[0]. We can see that n ≡ S (mod 3) because n and S are congruent modulo 3, whenever one is divisible by
3, the other is, too.
The key here is that when we compute n–S, each of the coefficients is a multiple of 3. It is possible to use this idea to develop tests for divisibility by other integers. For instance, for
divisibility by 11, we use S = d[0]–d[1]+d[2]–d[3]+… ± d[k], where the last sign is + or -, depending on whether k is even or odd. When we compute n–S, the coefficients become 11, 99, 1001, 9999,
etc., which are all divisible by 11.
As another example, suppose we want a test for whether a four-digit number is divisible by 7. The ideas above can be streamlined into the following procedure:
– 994 98 7 0
The numbers in the second row are the closest multiples of 7 less than the powers of 10 directly above.
Our divisibility test for the four-digit number abcd is to check if 6a+2b+3c+d is divisible by 7. Or, since 6 ≡ –1 (mod 7), we can check if –a+2b+3c+d is divisible by 7.
3. Modular arithmetic can be used to find the day of the week of any date. For example, here is how to compute the date of Christmas is any given year Y:
Reduce (b+c–2a+1) modulo 7. A 0 corresponds to Sunday, 1 to Monday, etc.
For example, if y = 1977, then a = 19 mod 4 = 3, b = 77, and c = floor(77/4) mod 7 = 5. Then we get b+c–2a+1 ≡ 0 (mod 7), so Christmas was on a Sunday in 1977.
A more general process can be used to find the day of the week of any date in history. See Secrets of Mental Math by Art Benjamin and Michael Shermer. Modular arithmetic can also be used to
determine the date of Easter in a given year.
Fermat's little theorem
Fermat's Little theorem is a useful rule that is simple to state:
An equivalent way to state the theorem is: If p is prime, then a^p ≡ a (mod p) for any integer a.
To get from the this statement to the original, we can divide both sides through by a, which works as long as gcd(p, a) = 1. To get from the original to this statement, just multiply through by a.
We will now prove Fermat's little theorem. To do so, we will need the following lemma:
If p is prime, then (a+b)^p ≡ a^p + b^p (mod p).
Use the binomial theorem to write (pk) with 1 < k < p, can be written as below: p is prime, no term in the denominator will cancel with p, meaning that each binomial coefficient is divisible by p and
hence congruent to 0 modulo p. Thus only the a^p and b^p terms survive.
The lemma above is sometimes called the freshman's dream since many freshman calculus and algebra students want to say (a+b)^2 = a^2+b^2, forgetting the 2ab term. You can't forget the 2ab term in
ordinary arithmetic, but you can when working mod 2.
We can now prove the alternate statement of Fermat's little theorem (a^p ≡ a (mod p)) using the lemma.
The proof is by induction on a. The base case a = 1 is simple. Now assume a^p ≡ a (mod p). We need to show (a+1)^p ≡ a (mod p). Using the previous lemma (a+1)^p ≡ a^p + 1 (mod p) and by the induction
hypothesis a^p ≡ a (mod p). Thus we have (a+1)^p ≡ a+1 (mod p), as desired.
Here are some examples of Fermat's little theorem in action:
1. If a is not a multiple of 7, then a^6 ≡ 1 (mod 7). We saw this in the last row of the table of powers we computed in Section 3.2.
2. Find the remainder when 5^38 is divided by 11.
By Fermat's little theorem, 5^10 ≡ 1 (mod 11). Thus 5^30 ≡ 1 (mod 11) and so
3. The inverse of an integer a modulo a prime p is an integer a^–1 such that aa^–1 ≡ 1 (mod p). Show that a^p–2 is the inverse of a, provided gcd(a, p) = 1.
By Fermat's little theorem, we have a · a^p–2 ≡ a^p–1 ≡ 1 (mod p). From this we see that a^p–2 fits the definition of a^–1. As an example, the inverse of 3 modulo 7 is 3^5 ≡ 4 (mod 7).
4. Show that if a not divisible by 7, then a^3+1 or a^3–1 is divisible by 7.
By Fermat's little theorem, a^6 ≡ 1 (mod 7). From this we get that 7 ∣ (a^6–1). We can factor a^6–1 into (a^3–1)(a^3+1) and use Euclid's lemma to conclude that 7 ∣ (a^3–1) or 7 ∣ (a^
5. Show that p^q–1+q^p–1 ≡ 1 (mod pq).
By Fermat's little theorem, we have p^q–1 ≡ 1 (mod q). Also, q^p–1 ≡ 0 (mod q). Adding these gives p^q–1+q^p–1 ≡ 1 (mod q). A similar argument shows p^q–1+q^p–1 ≡ 1 (mod p). Since the same
congruence holds modulo p and modulo q (and gcd(p, q) = 1), it holds modulo pq.
6. Show that any prime other than 2 and 5 divides infinitely many of the numbers 1, 11, 111, 1111, ….
These numbers are of the form 1+10+10^2+10^3+…+10^k, which can be rewritten as 10^k+1–19 using the geometric series formula. By Fermat's little theorem 10^p–1 ≡ 1 (mod p) for any prime p such
that gcd(p, 10) = 1 (i.e. for any prime besides 2 and 5). Thus also 10^2(p–1), 10^3(p–1), etc. are congruent to 1 modulo p. Thus 10^k+1–19 will be congruent to 0 modulo p for infinitely many
values of k.
For example, the integers 111111 (6 ones), 111111111111 (12 ones), 111111111111111111 (18 ones) etc. are all divisible by 7 since 10^6 ≡ 1 (mod 7). As another example, numbers that consist of 16,
32, 48, etc. ones are all divisible by 17 since 10^16 ≡ 1 (mod 17).
7. Fermat's little theorem and a generalization of it called Euler's theorem (see the next section) are a key part of the RSA algorithm, which is of fundamental importance in modern cryptography.
See Section 4.2.
Euler's theorem
Euler's theorem is a generalization of Fermat's little theorem to nonprimes.
(Euler's theorem) If gcd(a, n) = 1, then a^φ(n) ≡ 1 (mod n).
Recall that φ(n) is the Euler phi function from Section 2.15. Since φ(p) = p–1 for any prime p, we see that Euler's theorem reduces to Fermat's little theorem when n = p is prime.
As an example of Euler's theorem, 3^4 ≡ 1 (mod 10) since gcd(3, 10) = 1 and φ(10) = 4.
To understand why this works, recall that 1, 3, 7, and 9 are the φ(10) = 4 integers relatively prime to 10. Multiply each of these by a = 3 to get 3, 9, 21, and 27, which reduce modulo 10 to 3, 9, 1,
and 7, respectively. We see these are a rearrangement of the originals. So 3^4 ≡ 1 (mod 10). We can formalize this example into a proof of Euler's theorem.
, …
be the integers from 1 to
that are relatively prime to
. Multiply each by
to get
, …
. We claim this is just a rearrangement of the original values. To show this, we need to show that the
are all distinct and relatively prime to
The ax[i] are distinct because if ax[i] ≡ ax[j] (mod n), then since gcd(a, n), we can cancel a to get x[i] ≡ x[j] (mod n).
We have ax[i] is relatively prime to n because if some prime p divides ax[i], then by Euclid's lemma, p ∣ a or p ∣ x[i], and as gcd(a, n) = gcd(x[i], n) = 1, p cannot divide n. Thus ax[i] and
n cannot have any prime factors (and hence any factors besides 1) in common.
Thus we have x[i] is relatively prime to n, we can cancel it from both sides, leaving us with a^φ(n) ≡ 1 (mod n).
Formal definition and inverses
The notation ℤ[n] refers to the set {0, 1, 2, …, n–1} with all arithmetic done modulo n.**In many texts the notation ℤ/nℤ is used. More formally, the way ℤ[n] is defined usually goes as follows:
The relation ≡ is an equivalence relation (it is reflexive, symmetric, and transitive). As such, it partitions ℤ into disjoint sets called equivalence classes, where every integer in a given set is
congruent to everything else in that set and nothing else.
For instance, here are the sets we get modulo 5: ℤ[n] to be the set {[0], [1], …, [n–1]}, with the addition and multiplication defined by [a]+[b] = [a+b] and [a] × [b] = [a × b].
The formal definition is used to make sure everything is on a firm mathematical footing. There is more to show to make sure that everything works out mathematically, but we will skip that here and
just think of ℤ[n] as the set {0, 1, …, n–1} with arithmetic done modulo n.
Some integers have an inverse in ℤ[n]. That is, for some integers a, there exists an integer a^–1 such that aa^–1 ≡ 1 (mod n). For instance, in ℤ[7], 3 · 5 ≡ 1 (mod 7), so we can say that the inverse
of 3 is 5 (and also that the inverse of 5 is 3). Here is a useful fact about inverses.
Finding an inverse of
is the same as solving
ax ≡ 1 (mod n)
, which is the same as finding
such that
ax–ny = 1
. Theorem
guarantees that this has a solution if and only if
gcd(a, n) = 1
To see that the inverse is unique, suppose ax ≡ 1 (mod p) and ay ≡ 1 (mod p). Then ax ≡ ay (mod p) and since gcd(a, p) = 1, we can cancel a to get x ≡ y (mod p).
The particular case of interest is ℤ[p] where p is prime:
If p is prime, then each element of ℤ[p] has a unique inverse.
For instance, in ℤ[7], we have 1^–1 = 1, 2^–1 = 4, 3^–1 = 5, 4^–1 = 2, 5^–1 = 3, and 6^–1 = 6. Notice that 1 and 6 (which is -1 mod 7) are their own inverses. We have the following:
For any integer
, 1 and -1 are their own inverses in
. If
is prime, then no other integers are their own inverses.
1 · 1 ≡ 1 (mod n)
, 1 is its own inverse. Similarly,
–1 · –1 ≡ 1 (mod n)
, so
is its own inverse.
Now assume n = p is prime. If a is its own inverse, then a · a ≡ 1 (mod p). From this we get that p ∣ (a^2–1). We can factor a^2–1 into (a–1)(a+1). By Euclid's lemma, p ∣ a–1 or p ∣ a+1,
which tells us that a ≡ 1 (mod p) or a ≡ –1 (mod p).
Modulo a composite, there can be integers besides ± 1 that are their own inverses. For instance, 5 · 5 ≡ 1 (mod 8), so 5 is its own inverse mod 8.
Wilson's theorem
Wilson's theorem is a nice theorem that it gives a simple characterization of prime numbers in terms of modular arithmetic.
(Wilson's theorem) An integer p is prime if and only if (p–1)! ≡ –1 (mod p).
This gives us a way to check if a number is prime: just compute (p–1)! modulo p. Its fatal flaw is that (p–1)! is huge and difficult to compute even for relatively small values of p. So Wilson's
theorem is not a practical primality test, unless someone were to find an easy way to compute factorials modulo a prime. Still, here is an example with p = 11:
The proof of Wilson's theorem is interesting. Let's look at some examples to help understand it.
Take p = 14, a composite. We have (14–1)! ≡ 0 (mod 14) since 13! contains 2 · 7 = 14. This will work in general for a composite number—factor it and find its factors in (p–1)!. We have to be a little
careful if p is the square of a prime, though. But as long as p > 4, we can still find its factors in (p–1)!. For instance, if p = 25, then (25–1)! ≡ 0 (mod 25) since 24! is divisible by both 5 and
10 (which contains a factor of 5), so we do get 25 ∣ 24!.
Now take p = 7, a prime. In ℤ[7], 2 and 4 are inverses of each other, as are 3 and 5. Then we have p is prime, then in ℤ[p], we can pair off everything except 1 and p–1 into inverse pairs. The
numbers in each pair will cancel each other out in (p–1)!, leaving just p–1.
Here is a formal write-up of the proof:
It is easy to check that the theorem holds for
p < 5
, so suppose
p ≥ 5
If p is not prime, then (p–1)! ≡ 0 (mod p) since p ∣ (p–1)!. We have this because we can write p = ab for some positive integers a and b less than p (since p is not prime) and those integers both
show up in (p–1)!, except possibly if the square of a prime, specifically p = q^2. But in that case as long as p > 4, we know that both q and 2q will show up in (p–1)!.
On the other hand, suppose p is prime. By Theorems 25 and 27, if p is prime, then in ℤ[p] each integer 1, 2, … , p–1 has a unique inverse, with 1 and p–1 being the only integers that are their own
inverses. This means that all the other integers come in inverse pairs. Thus, looking at (p–1)! = (p–1)(p–2)… 3 · 2 · 1, we know that all the terms in the middle, p–2, p–3, …, 3, 2 pair off into
inverse pairs and cancel each other out, leaving us with (p–1)! ≡ p–1 ≡ –1 (mod p).
Even though Wilson's theorem is not practical for checking primality, it is a useful tool in proofs. There is also an interesting twin-prime analogue of Wilson's Theorem proved by P. Clement in 1949:
(p, p+2) are a twin-prime pair if and only if 4(p+1)! ≡ –(p+4) (mod p(p+2)).
Solving congruences
In ordinary algebra, solving the linear equation ax = b is very useful and also very easy. Solving the congruence ax ≡ b (mod n) is also useful, but a little trickier than solving ax = b.
For example, suppose we want to solve 2x ≡ 5 (mod 11). To solve the ordinary equation 2x = 5, we would divide by 2 to get x = 2.5, but in modular arithmetic, we don't quite have division, so we have
to find other approaches. A simple approach is trial and error, as there are only 11 values of x to try. The solution ends up being x = 8. We will see some faster approaches shortly.
The algebraic equation ax = b always has a single solution (unless a = 0), but with ax ≡ b (mod n), it often happens that there is no solution or multiple solutions. For example, 2x ≡ 5 (mod 10) has
no solution, while 2x ≡ 4 (mod 10) has two solutions modulo 10, namely x = 2 and x = 7.
Procedure for solving ax ≡ b (mod n)
Let d = gcd(a, n).
1. If d ∤ b, then there is no solution. Otherwise there are d solutions.
2. If there is a solution, we first find one solution, x[0], and use it to find all the other solutions. To find x[0], a variety of techniques can be used, such as the extended Euclidean algorithm,
properties of congruences, and systematic checking of all possibilities (if n is small).
3. Once a solution, x[0], has been found, all solutions are of the form x[0] + ndt for t = 0, 1, … d–1.
The idea for why this works is we can write ax ≡ b (mod n) as ax–nk = b for some integer k, Thus, solving the congruence is the same as finding integer solutions to the equation ax–nk = b. We know
from Theorem 3 that such an equation only has a solution provided d = gcd(a, n) divides b. Further, the extended Euclidean algorithm can be used to find x and k in the equation.
To see where the formula for all the solutions comes from, note that we can divide ax ≡ b (mod n) through by d to get ad x ≡ bd (mod nd). By a similar argument to the one used to prove Theorem 25
(about the existence of inverses), this equation must have a unique solution: x ≡ x[0] (mod nd). So x[0] will also be a solution modulo n, as will x[0]+nd, x[0]+2nd, x[0]+3nd, …, x[0]+(d–1)nd, which
are all less than n and congruent to x[0] modulo nd. This gets us d different solutions. It is not hard to show that these are the only solutions.
A few notes on finding an initial solution
Below are a few different ways to find an initial solution x[0].
1. Extended Euclidean algorithm — Probably the most effective way in general to find an initial solution is to use the extended Euclidean algorithm, which was covered in Section 1.6. As an example,
suppose we want to solve 8x ≡ 11 (mod 23). Write it as 8x–23k = 11. Using the Euclidean algorithm on 8 and 23, we get 8(3) – 23 (1) = 1 and multiplying through by 11 gives 8(33) + 23(–11) = 11.
So x[0] = 33 is a solution, which we can reduce mod 23 to get x[0] = 10.
2. Trial and error — If the modulus is small enough, we can just try things systematically until something works. For example, to solve 5x ≡ 7 (mod 11), we can list all the multiples of 5 until we
get to one that is 7 more than a multiple of 11. Doing this, we get 5, 10, 15, 20, 25, 30, 35, 40. We stop here as 40 is a 7 away from 33, which is a multiple of 11. Thus x[0] = 8 is a solution.
Another way of doing this is to list all the integers of the form 11k+7 (18, 29, …) until we get a multiple of 5.
3. Congruence rules — We can use rules for working with congruences to manipulate the congruence into giving us an initial solution. For example, to find a solution to 11x ≡ 31 (mod 45), we can add
90 (which is congruent to 0 mod 45) to both sides to get 11x ≡ 121 (mod 45). Then, since gcd(11, 45) = 1, we can divide both sides by 11 to get x ≡ 11 (mod 45).
4. “Division” — One final method is to use “division.” To solve the ordinary linear equation ax = b, we would divide by a to get x = b/a. We can't divide by a in modular arithmetic, but we can do
its equivalent, which is to multiply by a^–1 (provided that a^–1 exists). A solution to ax ≡ n (mod b) is x ≡ a^–1n (mod b). For instance, to solve 3x ≡ 4 (mod 7), we note that 3^–1 = 5 and 5 · 4
≡ 6 (mod 7), so that x[0] = 6 is a solution.
One problem with this is that finding a^–1 itself takes some work. However, if you have a lot of equations of the form ax ≡ n (mod b), where a is fixed, but n varies, then it makes sense to use
this method, since all the work goes into finding a^–1 and once we have it, it is short work to solve all those equations.
One way to find a^–1 is to note that if p is prime and gcd(a, p) = 1, then by Fermat's little theorem, a^–1 ≡ a^p–2 (mod p). In general, by Euler's theorem, a^–1 ≡ a^φ(n)–1 (mod n) as long as gcd
(a, n) = 1.
1. Solve 3x ≡ 11 (mod 36).
Solution: Since gcd(3, 36) = 3 and 3 ∤ 11, there is no solution.
2. Solve 14x ≡ 4 (mod 37).
Solution: Since gcd(14, 37) = 1, there will be exactly 1 solution.
Using the Euclidean algorithm on 14 and 37, we get 14(8) – 37(3) = 1 and multiplying through by 4 gives 14(32) + 37(–12) = 4. From this, we get x[0] = 32 is the solution.
3. Solve 12x ≡ 18 (mod 30).
Solution: Since gcd(12, 30) = 6 and 6 ∣ 18, there 6 different solutions mod 30. We can divide the whole equation through by gcd(12, 30) = 6 to get 2x ≡ 3 (mod 5). By trial and error, we get x
[0] = 4 is a solution. Then the solutions of the original are
Solving linear Diophantine equations
Diophantine equations are algebraic problems where we are looking for integer solutions. They are among some of the trickiest problems in mathematics. For instance, Fermat's Last Theorem, which took
400 years and some seriously high-powered math to solve, is about showing that x^n+y^n = z^n has no integer solutions if n > 2. However, linear Diophantine equations of the form ax+by = c can be
easily solved since they closely are related to the congruence ax ≡ c (mod b).
Using the formula for the solutions to that congruence gives us the following formula for solutions to ax+by = c: t ∈ ℤ. Here we need some solution (x[0], y[0]) to ax+by = c to get us started. Such a
solution can be found using the extended Euclidean algorithm, trial and error, or by working with congruences. Note that there is no solution if c is not divisible by gcd(a, b).
Here are a few example problems:
1. Find all the integer solutions to 37x+14y = 11.
Solution: Notice that this is equivalent to the congruence 37x ≡ 11 (mod 14), which we did earlier. In that example, we found 14(32) + 37(–12) = 4. From this, we get x[0] = 32 and y[0] = –12.
From here, all the solutions are of the form t ∈ ℤ. Each value of t gives a different solution. For instance t = 1 gives (69, –26) and t = –1 gives (–5, 2). And we can check our work: (–5, 2) is
a solution because 14(–5)+37(2) = 4.
2. Find all integer solutions of 12x+30y = 18.
Solution: This equation is also equivalent to a congruence we solved earlier. In that example, we got x[0] = 4 and from 12x[0]+30y[0] = 18, we get y[0] = –1. Then all solutions are of the form t
∈ ℤ. For example, t = –2, -1, 0, 1, and 2 give (–6, 3), (–1, 1), (4, –1), (9, –3), and (14, –5) .
3. Linear Diophantine equations are the key to many famous old word problems. As a simple example, suppose apples are 69 cents and oranges are 75 cents. We spend a total of $12.09. How many of each
did we buy?
This reduces to the equation 69x+75y = 1209. The extended Euclidean algorithm gives 69(12)+75(–11) = 3 (we'll skip the details here; a quick way to get this would be to use the program of Section
1.6). Multiply through by 403 to get 69(4836)+75(–4433) = 1209. All the solutions are of the form t ∈ ℤ. However, not all solutions will make sense since neither x nor y can be negative as they
represent the numbers of apples and oranges bought. So we must have 4836 + 25t ≥ 0 and –4433 – 23t ≥ 0. The first inequality can be solved to give us t ≥ –193.44. The second gives us t ≤ –192.74.
So only t = –193 will give us positive values for x and y, which turn out to be x = 11 and y = 6.
4. Oystein Ore's Number Theory and Its History has a number of interesting old problems from various cultures. One such problem comes from a 12th century Hindu manuscript. Here is a modification of
A person has 5 rubies, 8 sapphires, 7 pearls, and 92 coins, while the other has 19 rubies, 14 sapphires, 2 pearls, and 4 coins. The combined worth is the same for both people. How many coins
must rubies, sapphires, and pearls each be worth?
Letting r, s, and p denote the values of rubies, sapphires, and pearls, we have 5r+8s+7p+92 = 19r + 14s + 2p + 4, which becomes 14r+6s–5p = 88. This has three variables, as opposed to our
previous examples which all have two.
To handle this we start with the first two terms, 14r+6s. We have gcd(14, 6) = 2 and we can write 14(1)+6(–2) = 2. Thus, the number of rubies and sapphires is always a multiple of 2, say 2n for
some integer n. Then consider 2n–5p = 88. We have gcd(2, 5) = 1 and we can write 2(3)–5(1) = 1. Multiply through by 88 to get 2(264)–5(88) = 88. Thus all solutions of 2n–5p = 88 can be written in
the form t ∈ ℤ. Going back to the equation 14(1)+6(–2) = 2 and multiplying through by n = 264–5t gives 14(264–5t)+6(–528+10t) = 2(264–5t). Thus all solutions of 14r+6s–5p = 88 are of the form t,
u ∈ ℤ. We are looking for positive integers, so we know that 88–2t > 0 and hence t < 44. From this and the equation for s, we get that u < –14. Note that we can make things look a little nicer by
setting t = 43–x and u = –14–y. Then we have x, y ≥ 0. For instance, x = y = 0 produces r = 4, s = 7, and p = 2, while x = y = 1 produces r = 6, s = 4, p = 4. Not all values of x and y will
produce positive solutions, but there are infinitely many that do.
Equations with 3 or more unknowns
The example above is an equation of the form ax+by+cz = d. The procedure used in that example can be streamlined and used in general:
1. Let e = gcd(a, b) and f = gcd(e, c).
2. Find a solution (x[0], y[0]) to ax+by = e.
3. Find a solution (w[0], z[0]) to ew+cz = d.
4. Then all solutions are given by s, t ∈ ℤ. This works provided gcd(a, b, c) ∣ d.
In general a[1]x[1] + a[2]x[2] + … + a[n]x[n] = b has a solution provided gcd(a[1], a[2], …, a[n]) ∣ b. The equation can be solved by an iterative process like above.
The Chinese remainder theorem
Just like we can solve systems of algebraic equations, we can solve systems of congruences. The technique used is called the Chinese remainder theorem. The name comes from its appearance in a third
century Chinese manuscript.
(Chinese remainder theorem) Suppose we have a system like the one below: n[i] are pairwise relatively prime. Then we can find a solution that is unique modulo the product n[1]n[2] ··· n[k].
To solve such a system, let N = n[1]n[2] ··· n[k]. Then for each i = 1, 2, …, k, let N[i] = N/n[i] (the product of all the moduli except n[i]), and solve the congruence N[i]x[i] ≡ 1 (mod n[i]). The
solution to the system is given by a[1]N[1]x[1] + a[2]N[2]x[2]+… + a[k]N[k]x[k], which is unique modulo N.
1. Suppose we have the system gcd(5, 7) = gcd(5, 8) = gcd(7, 8) = 1, so we can use the Chinese remainder theorem. We have N = 5 · 7 · 8 = 280 along with N[1] = 7 · 8 = 56, N[2] = 5 · 8 = 40, and N
[3] = 5 · 7 = 35. We then solve the following three congruences: x[1] ≡ 1 (mod 5), so x[1] = 1. The second reduces to 5x[2] ≡ 1 (mod 7), from which we get x[2] = 3. The last reduces to 3x[3] ≡ 1
(mod 8), from which we get x[3] = 3. Then the solution is
2. The Chinese remainder theorem is useful for finding when several cyclical events will all line up. For instance, suppose there are three salesmen that visit a town on different cycles. Salesman A
visits every 10 days, B visits every 7 days, and C visits every 3 days. Suppose A was last there 8 days ago, B was last there yesterday, and C is there today. When will all three be in town on
the same day?
We can describe this problem with a system of three congruences: N = 10 · 7 · 3 = 210 and N[1] = 7 · 3 = 21 and N[2] = 10 · 3 = 30. Because of the 0 in the last congruence, there is no need to
worry about N[3] and its associated congruence. We then solve the following two congruences: x[1] = 1 and x[2] = 4. The solution to the problem is then
Moduli that aren't relatively prime
The Chinese remainder theorem can't be used if the moduli are not relatively prime, but there are things that can be done:
1. A useful trick is to replace the moduli m[i] with new moduli c[i]. The rules for the new moduli are that c[i] must be a divisor of m[i] for each i, and the lcm Of the new moduli must be the same
as the lcm of the originals. For example, suppose we have the system
The problem is that the moduli 4 and 6 share a factor of 2. We can set c[1] = m[1], c[2] = m[2], and remove a factor of 2 from m[3] = 6 to get c[3] = 3. This doesn't change the lcm, as lcm(4, 5,
6) = 60 and lcm(4, 5, 3) = 60. We then solve the reduced system
2. Sometimes the above trick is not enough. In that case, it might be possible to combine or eliminate some congruences. For instance, if x ≡ 1 (mod 2) and x ≡ 3 (mod 4), the first congruence is
redundant. It tells us that x must be odd, but the second congruence also implies that.
3. Another way of approaching these problems is shown in the following example. Suppose we have x ≡ 1 (mod 4) and x ≡ 3 (mod 6). From the first congruence, we have x–1 = 4k for some integer k. Plug
this into the second to get 4k ≡ 2 (mod 6). We can divide through by 2, though this changes the modulus to 3. So we get 2k ≡ 1 (mod 3), which we can solve to get k ≡ 2 (mod 3). We can write this
as k = 3j+2 for some integer j. Thus we have x = 4k+1 = 12j+9. In other words, x ≡ 9 (mod 12) solves the two congruences.
One way to think about this is if a number is of the form 4k+1 and 6k+3, then it is of the form 12k+9.
It is not too hard to generalize the procedure above to solve x ≡ a[1] (mod n[1]) and x ≡ a[2] (mod n[2]). What we do is set d = gcd(n[1], n[2]), solve m[1]dk ≡ a[2]–a[1]d (mod m[2])d for k, and
then the solution to both congruences is x ≡ a[1]+m[1]k (mod lcm(a[1], a[2])). Note that this works provide d ∣ (a[2]–a[1]).
In general, some combination of these techniques can be used for tricky problems. In fact, we have the following theorem:
Suppose we have a system like the one below: lcm(n[1], n[2], ··· , n[k]) if and only if a[i] ≡ a[j] (mod gcd(n[i], n[j])) for all i and j.
A classic Chinese remainder theorem problem is the following: There are some eggs in a basket. When they are removed in pairs, there is one left over. When three at a time are removed, there are two
left over. When four, five, six, or seven at a time are removed, there remain three, four, five, or zero respectively. How many eggs are in the basket?
This corresponds to the following system: 4k+3, which is odd. So we can drop the first congruence.
Then the remaining moduli 3, 4, 5, 6, 7 can be reduced to 3, 4, 5, 6, 7 using the first trick given above. This turns x ≡ 5 (mod 6) into x ≡ 5 (mod 3), which is the same as x ≡ 2 (mod 3), which we
already have, so we can drop it. We are thus left with the following: N = 3 · 4 · 5 · 7 = 420, N[1] = 4 · 5 · 7 = 140, N[2] = 3 · 5 · 7 = 105, N[3] = 3 · 4 · 7 = 84. We don't need N[4] because of the
0 in the last congruence. We then solve x[1] = 2, x[2] = 1, and x[3] = 4. The solution is then
A few notes
All of the congruences we considered above are of the form x ≡ a (mod m). It is possible to have more general cases, where we have cx ≡ a (mod m). To handle these, we first have to solve them for x.
The Chinese remainder theorem has a number of applications. As seen above, it is useful any time we need to know when several cyclical events will line up. The Chinese remainder theorem is also an
important part of modern cryptography, and it shows up here and there in higher math.
One important use of the Chinese remainder theorem is breaking up composite moduli into smaller pieces that are easier to work with. For instance, if we need to solve a congruence f(x) ≡ a (mod mn)
with gcd(m, n) = 1, we can solve the congruences f(x) ≡ a (mod m) and f(x) ≡ a (mod n) and combine them by the Chinese remainder theorem to get a solution modulo mn. Note the similarity between this
and the fact mentioned on in Section 3.1.
It is interesting to look at powers modulo an integer. For example, if we look at the powers of 2 modulo 9, we get the repeating sequence 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, …. If we look at the
powers of 7 modulo 9, we get the repeating sequence 7, 4, 1, 7, 4, 1, …. The powers of 8 give the repeating sequence 8, 1, 8, 1, …. The goal of this section is to understand a little about these
repeating sequences.
In particular, we are interested in the first power that turns out to be 1. This power is called the order. We have the following definition:
Let a and n be relatively prime positive integers. The order of a modulo n is the least positive integer k such that a^k ≡ 1 (mod n).
For instance, the order of 7 modulo 9 is 3, since 7^3 ≡ 1 (mod 9) and no smaller positive power of 7 (7^1 or 7^2) is congruent to 1.
If a is not relatively prime to n, the order is undefined, as no power of a beside a^0 can ever be 1. So we will only concern ourselves with values of a that are relatively prime to n.
Euler's theorem tells us that a^φ(n) ≡ 1 (mod n), so the order must always be no greater than φ(n). But there is an even closer connection between the order and φ(n), namely that the order must
always be a divisor of φ(n).
For example, take n = 13. We have φ(13) = 12. Suppose some integer a had an order that is not a divisor of 12, say order 7. Then we would have a^7 ≡ 1 (mod 13) and a^14 ≡ 1 (mod 13). But since a^12 ≡
1 (mod 13) (by Euler's theorem), we would have a was assumed to be 7, it is not possible to have a power less than the seventh power be congruent to 1. Here is a formal statement and proof of the
The order of an integer a modulo n is a divisor of φ(n).
Let k be the order of a. By the division algorithm, we can write φ(n) = kq+r for some integers q and r with 0 ≤ r < k. Our goal is to show that k is a divisor of φ(n), which means we need to show
that r = 0. Euler's theorem tells us a^φ(n) ≡ 1 (mod n), so we have a^r ≡ 1 (mod n), But since 0 ≤ r < k and the order of a is k, this is only possible if r = 0.
This theorem is a special case of Lagrange's Theorem, an important result in group theory.
Primitive roots
Below is a table of orders of integers modulo 13. We have φ(13) = 12 and the possible orders are the divisors of 12, namely 1, 2, 3, 4, 6, and 12.
Order Integers with that order modulo 13
3 3, 9
4 5, 8
6 4, 10
12 2, 6, 7, 11
Of particular interest are 2, 6, 7, and 11, which have order 12, the highest possible order. These values are called primitive roots. In general, we have the following:
Let n be a positive integer. An integer a is said to be a primitive root of n if the order of a modulo n is φ(n).
Not every integer has primitive roots. For instance, 8 doesn't have any. We have φ(8) = 4, but 1, 3, 5, and 7 all have order 1 or 2. We have the following theorem:
If n = 2, 4, p^k ,or 2p^k for some odd prime p and positive integer k, then n has a primitive root. Otherwise it doesn't.
We won't prove this theorem here as it is a bit involved. However, you can find a complete development of the theorem in many number theory texts. The integers between 2 and 50 that have primitive
roots are shown below:
The powers of a primitive root a of n are all unique and run through all of the integers relatively prime to n. For instance, consider a = 2 and n = 13. The powers of 2 are 2, 4, 8, 3, 6, 12, 11, 9,
5, 10, 7, 1. We see that they run through all the integers relatively prime to 13.**Using the terminology of abstract algebra, we can say a is a generator of the multiplicative group of integers
relatively prime to n since every integer is a power of a. The group is thus cyclic provided n has a primitive root.
If a is a primitive root of n, the integers a, a^2, …, a^φ(n) are distinct and hence run through all the integers relatively prime to φ(n).
Suppose i and j are exponents in the range from 1 to φ(n) with i ≤ j and a^i ≡ a^j (mod n). Since gcd(a, n) = 1, we must also have gcd(a^j, n) = 1 and we can divide through by a^i to get a^j–i ≡ 1
(mod n). But j–i < φ(n) and the order of a is φ(n), so we must have j–i = 0, meaning i = j.
Phrased another way, if a is a primitive root of n, then every integer relatively prime to n is of the form a^k for some k. This gives us a way to determine the orders of other elements modulo n.
As an example, consider an integer n with φ(n) = 30 and suppose we want the order of some integer b that turns out to equal a^8. Since a is a primitive root, its order must be 30. That is a^30 ≡ 1
(mod n), and for that matter a^60, a^90, a^120, etc. are all congruent to 1 modulo n as well. We are looking for the order of b, so we want to find a power of a^8 that is congruent to 1. So we go
through a^16, a^24, a^32, etc. until we get to one that matches up with one of the powers 30, 60, 90, etc.
In other words, we are looking for when multiples of 8 match up with multiples of 30. Thus we just need to find lcm(8, 30), which is 120. Then b is thus 15, which is lcm(8, 30)/8. In general, we have
the following:
Let n be a positive integer with a primitive root a. If b = a^k, then the order of b is lcm(k, φ(n))/k; equivalently, the order is φ(n)/ gcd(k, φ(n)).
By Euler's theorem, the only powers of
that are congruent to 1 are multiples of
. Thus for
to be congruent to 1, we must have
be a multiple of
. The smallest such multiple is
m = lcm(k, φ(n))
. Thus we have
1 ≡ a^m ≡ (b^k)^m/k
So the order of
. By Theorem
, we can also write the order of
φ(n)/ gcd(k, φ(n))
The theorem above actually allows us to determine how many primitive roots there are for a given integer n:
If an integer n has a primitive root, then it has φ(φ(n)) of them.
Let a be a primitive root of n, and consider a^k for some integer k. If k is relatively prime to φ(n), then the order of a[k] is φ(n) by the previous theorem, making a^k a primitive root. There are φ
(φ(n)) such integers k relatively prime to φ(n).
We can actually say something a little more general:
If n has a primitive root, then there are φ(m) integers with order m in ℤ[n].
be a primitive root of
. We know the order of
φ(n)/ gcd(k, φ(n))
. Setting this equal to
and rewriting gives
gcd(k, φ(n)) = φ(n)/m
. From the proof of Theorem
, any integer
whose gcd with
must satisfy
gcd(k, φ(n)/(φ(n)/m)) = 1
. In other words,
must be relatively prime to
. There are
such integers.
Finding primitive roots
So we know how many primitive roots any integer must have, but what are they? That is a much harder question to answer. Even simple questions such as which integers have 2 as a primitive root or
finding the smallest primitive root for a given integer don't have easy answers. It is suspected that every positive integer that is not a perfect square is a primitive root of infinitely many
primes. This has not been proved. It is known as Artin's conjecture. It is further suspected that the number of primes less than x for which a non-perfect square a is a primitive root is
approximately some constant times x/ ln(x). In terms of partial results, it has been shown that Artin's conjecture is true for almost all primes, specifically that there at most two primes for which
Artin's conjecture fails.
For relatively small integers, it is not too difficult to write a program to find the primitive roots. Here is some Python code to do that:
from fractions import gcd
def phi(n):
return len([x for x in range(1,n) if gcd(x,n)==1])
def order(a, n):
if gcd(a,n) != 1:
return -1
c = a % n
p = 1
while c != 1:
c = c*a % n
p += 1
return p
def prim_roots(n):
return [a for a in range(1,n) if order(a, n)==phi(n)]
The code above mostly brute-forces things. There are more efficient approaches, like using the formula for calculation φ(n) from its prime factorization. And for finding the primitive roots, we know
that possible orders are divisors of φ(n), so to check if a is a primitive root, we could just compute a^d for each divisor d of φ(n) less than φ(n). If none of those come out to 1, then we know a is
a primitive root.
Decimal expansions
Decimal expansions have interesting connections to modular arithmetic. Here are a few decimal expansions:
1/3 .3
1/5 .2
1/6 .16
1/7 .142856
1/11 .09
1/13 .076923
1/17 .0588235294117647
1/19 .052631578947368421
1/37 .027
The overbar indicates a repeating decimal. For instance, 1/11 = .09 is shorthand for the decimal expansion .09090909…. To find the decimal expansion of a/b, repeatedly apply the division algorithm as
in the example below that computes the decimal expansion of 1/7: b–1, eventually a remainder will be repeated, causing the decimal expansion to repeat. It is interesting to note that changing the 10
to another integer d will give the base-d decimal expansion of a/b.
Below is a short Python implementation of the algorithm above. It will print the first n digits of the decimal expansion of a/b. With a little work it could be modified to find how long it takes
before the expansion repeats itself.
for i in range(n):
print(10 * a // b)
a = 10 * a % b
Every fraction has a decimal expansion that will repeat or terminate. A terminating expansion is a special case of a repeating expansion, with a 0 repeating, like in 1/4 = .250000 ··· . The expansion
of a/b is terminating if and only if b is of form 2^j5^k for some integers j and k. This is not too hard to show using the division algorithm and noting that 2 and 5 are the prime divisors of 10, the
base of the decimal system.
Further, only rational numbers have repeating or terminating expansions. The decimal expansion of irrational numbers cannot have an endlessly repeating pattern of this sort. To see why, consider the
process below that will find the fraction corresponding to the repeating decimal x = .345634563456….
Multiply both sides by 10000 to get 10000x = 3456.34563456…. We can rewrite this as 10000x = 3456+x, which we can solve to get 3456/9999. We can reduce this in lowest terms to 384/1111. A similar
process works in general. For example, .123 would be 123/999, and .235711 would be 235711/999999. A variation of the process can be used for other numbers where the pattern does not start right away,
like .2345 or .18.
This technique can help us find the length of the repeating pattern in the expansion of 1/n. Suppose we have 1/n = .d[1]d[2]… d[k]. Write 10^k/n = d[1]d[2]… d[k] + 1/n. From here we get 10^k–1 = (d
[1]d[2]… d[k])n, which we can write as 10^k ≡ 1 (mod n). Thus k, the length of the cycle, is given by the order of 10 modulo n.
For example, 10 has order 1 modulo 3, so 1/3 has a decimal expansion with a repeating cycle of length 1. Also, 10 has order 6 modulo 7, so 1/7 has a decimal expansion with a repeating cycle of length
6. In general, since the maximum order of 10 modulo n is φ(n), that is the maximum length of a repeating cycle.
An interesting note is that the factors of 10^k–1 tell us for what primes p that 1/p might have a repeating decimal expansion of length k. For instance, 10^3–1 = 999, which factors into 3^3 × 37, so
only 1/3 and 1/37 can have length 3, and 1/3 has length 1, so only 1/37 has a length of 3.
Quadratic Reciprocity
This section is about what positive integers are perfect squares in ℤ[n]. For example, in ℤ[7], squaring 1, 2, 3, 4, 5, and 6 gives 1, 4, 2, 2, 4, and 1. So the only perfect squares are 1, 4, and 9.
As another example, in ℤ[11], the squares of the integers 1 through 10 are 1, 4, 9, 5, 3, 3, 5, 9, 4, and 1, in that order. So the perfect squares are 1, 3, 4, 5, and 9.
Perfect squares modulo n have long been studied and are usually referred to as quadratic residues. Here is the formal definition:
Let a and n be relatively prime integers. Then a is called a quadratic residue of n if there exists a b such that b^2 ≡ a (mod n). Otherwise, a is called a quadratic nonresidue of n.
To denote whether an integer is a quadratic residue or not, a special notation called the Legendre symbol is used. Here is the formal definition:
Let p be an odd prime and let a ∈ ℤ. The Legendre symbol, (ap), is defined to be 1 if a is a quadratic residue of p, –1 if a is a quadratic non-residue of p, and 0 if a is 0.
Basic properties
Here are the squares of the nonzero elements of ℤ[13] in order from 1^2 to 12^2: 1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4, 1. Notice the symmetry about the middle. This always happens. Notice also that
each square appears exactly twice in the list above and that exactly half of the integers from 1 through 12 are squares. This always happens modulo a prime.
The following hold in
, where
is prime.
1. Every quadratic residue is the square of exactly two elements of ℤ[p], one of which is less than p/2 and the other of which is greater than p/2.
2. Exactly half of the elements of ℤ[p] are quadratic residues.
b^2 ≡ a (mod p)
, then
(–b)^2 ≡ a (mod p)
as well. So every quadratic residue is the square of at least two things. Note that if
b < p/2
, then
, which is congruent to
, is greater than
. And if
b > p/2
, then
n–b < p/2
Now suppose b^2 ≡ c^2 (mod p). Then p ∣ (b^2–c^2) = (b–c)(b+c). By Euclid's lemma, p ∣ (b–c) or p ∣ (b+c). Hence b ≡ ± c (mod p). So each quadratic residue is the square of exactly two
integers modulo p. From this, we also get that half of the integers modulo a prime are quadratic residues and the other half are not.
This doesn't necessarily work for composite moduli. For instance, in ℤ[8], the integer 1 has four square roots, namely 1, 3, 5, and 7. So in what follows, we will usually be working modulo a prime.
Euler's identity
In ℤ[7], if we raise the integers from 1 to 6 to the 3rd power, we get 1, 1, -1, 1, -1, 1. In ℤ[11], if we raise the integers from 1 to 10 to the 5th power, we get 1, -1, 1, 1, 1, -1, -1, -1, 1, -1.
Do we always get ± 1 when raising an integer to the (p–1)/2 power modulo a prime p? The answer is yes. Fermat's little theorem tells us that a^p–1 ≡ 1 (mod p), and the two square roots of 1 are ± 1,
so a^(p–1)/2 ≡ ± 1 (mod p).
The interesting fact, known as Euler's identity, is that whether a^(p–1)/2 is 1 or -1 tells us if a is a quadratic residue or not.
(Euler's identity) Let p be an odd prime and let a ∈ ℤ. Then (ap) ≡ a^(p–1)/2 (mod p).
As an example, to tell if 2 is a quadratic residue of 17, we just compute 2^8 modulo 17. Doing so, we get 1, so 2 is a quadratic residue of 17.
To understand how the proof works, consider p = 11 and the quadratic residue a = 5. We have 5 ≡ 4^2 (mod 11) so 5^(10–1)/2 ≡ 4^10–1 ≡ 1 (mod 11) by Fermat's little theorem.
On the other hand, take a = 2, which is not a quadratic residue of 11. The integers 1 through 10 pair up into products that equal 3, namely 2 = 1 · 2, 3 · 8, 4 · 6, 5 · 7, and 9 · 10. Since a is not
a quadratic residue, none of those pairs could have a repeat (like 3 · 3). Thus we have 10! ≡ 2^5 (mod 11). Notice that 5 is (p–1)/2 here and by Wilson's theorem, 10! (which is (p–1)!) will be -1.
Here is a formal proof:
Suppose a is a quadratic residue. Then a ≡ b^2 (mod p) for some integer b and we have by Fermat's little theorem that a is a quadratic nonresidue. For any b = 1, 2, …, p–1, the congruence bx ≡ a (mod
p) has a unique solution since p is prime, and since a is a quadratic nonresidue, we can't have x = b. As a consequence, the integers 1, 2, …, p–1 can be broken up into (p–1)/2 pairs whose products
all equal a. Therefore, we have a^(p–1)/2 ≡ (p–1)! (mod p) and this is congruent to -1 by Wilson's theorem.
One nice consequence of Euler's identity is the following:
If p is prime, then -1 is a quadratic residue of p if and only if p is of the form 4k+1.
If p is of the form 4k+1, then using Euler's identity, we get 4k+1 prime. On the other hand, for a 4k+3 prime, a similar computation results in -1, showing -1 is a quadratic nonresidue for 4k+3
One can use this fact to show that there are infinitely many primes of the form 4k+1. The proof is reminiscent of Euclid's proof that there are infinitely many primes, but with a twist. Suppose p[1],
p[2], …, p[n] are all 4k+1 primes. Consider (2p[1]p[2] ··· p[n])^2+1. It is odd, so it is divisible by some odd prime p, and that prime cannot be any of the p[i]. We can then write (2p[1]p[2] ··· p[n
])^2 ≡ –1 (mod p). We have written -1 as the square of an element of ℤ[p], so -1 is a quadratic residue of p, which means p is a 4k+1 prime by the theorem above. Thus, given any list of 4k+1 primes,
we can generate another one, giving us infinitely many.
While we're at it, we can also show that there are infinitely many 4k+3 primes. This proof is also reminiscent of Euclid's proof, but it doesn't require anything about quadratic residues. First note
that the product of integers of the form 4k+1 is also of the form 4k+1, so any number of the form 4k+3 can't consist of only 4k+1 primes; it must be divisible by some prime of the form 4k+3. Let p[1]
, p[2], …, p[n] be 4k+3 primes, none of which equal 3. Consider P = 3p[1]p[2] ··· p[n]+2 and P′ = 3p[1]p[2] ··· p[n]+4. When multiplying numbers of the form 4k+3 together, the product will either be
of the form 4k+1 or 4k+3, depending on whether there are an odd or even amount of numbers in the product. So one of P and P′ must be of the form 4k+3. None of the p[i] can divide P or P′ because they
are all divisors of 3p[1]p[2] ··· p[n] and are greater than 4. But, as mentioned, one of P and P′ must be divisible some 4k+3 prime, and prime cannot be one of the p[i]. Thus, given any list of 4k+3
primes, we can generate another one, giving us infinitely many.
Another easy consequence of Euler's identity is that the Legendre symbol is multiplicative.
The Legendre symbol is multiplicative. That is, (abp) = (ap)(bp) whenever gcd(a, b) = 1.
In other words, if a and b are both quadratic residues or both quadratic nonresidues of p, then their product is a quadratic residue of p. Otherwise their product is a quadratic nonresidue.
By Euler's formula, we have
Gauss's lemma
Another way to tell if something is a quadratic residue is Gauss's lemma:
(Gauss's lemma) Let p be an odd prime with a relatively prime to p. Consider the following multiples of a: a, 2a, … (p–1)/2 · a. Let n be the number of them that are greater than p/2. Then (ap) =
To help us understand why this is true, look at the multiples of a = 2 in ℤ[11]: 2, 4, 6, 8, 10, 1, 3, 5, 7, 9. Replacing the elements greater than 5 with their equivalent negative forms, we get 2,
4, -5, -3, -1, 1, 3, 5, -4, -2. Notice that 1, 2, 3, 4, and 5 or their negatives appear exactly once in the first five multiples and once in the last five. This always happens. If we multiply the
first five multiples together, one the one hand we get 2^5 · 5!, and on the other hand we get (–1)^3 · 5!. This gives 2^5 ≡ (–1)^3 (mod 11). By Euler's identity, 2^5 tells us whether 2 is a quadratic
residue or not, so here we have another way of computing (25), by looking at how many negatives (numbers greater than 11/2) appear in the first 5 multiples of 2. Here is a formal proof:
If two multiples ja and ka are congruent mod p, then we have ja ≡ ka (mod p), and we can cancel a to conclude j ≡ k (mod p). If we have ja ≡ –ka (mod p), then we have j ≡ –k (mod p). Thus the
absolute values of the multiples |a|, |2a|, …, |(p–1)/2 · a| are distinct. Therefore, multiplying all of the multiples together gives us (–1)^n · ((p–1)/2)!. On the other hand, it gives us a^(p–1)/2
((p–1)/2)!. Equating these and canceling out the common terms gives us a^(p–1)/2 ≡ (–1)^n (mod p). By Euler's identity, (ap) = a^(p–1)/2, so the result follows.
A straightforward calculation using Gauss's lemma shows the following:
Let p be prime, then 2 is a quadratic residue of p if and only if p is of the form 8k± 1.
Looking at the multiples 2, 4, …, 2(p–1)/2, there are (p–1)/2 in total, and floor((p–1)/4) that are less than p/2. So there are (p–1)/2 – floor((p–1)/4) that are greater than p/2. If p is of the form
8k+1, that expression simplifies to 2p and since (–1)^2p = 1, we get that 2 is a quadratic residue of p by Gauss's lemma. The other cases 8k+3, 8k–3, and 8k–1, can be similarly verified.
Eisenstein's lemma
Eisenstein's lemma builds on Gauss's lemma to give us another way to compute (ap).
(Eisenstein's lemma) If p is prime and a is odd, then
To better understand the statement and its proof, consider an example with a = 5 and p = 13. The terms floor(kap) are the quotients in the division algorithm when ka is divided by p. For instance,
below is the division algorithm written out for all the multiples, a, 2a, … 6a. p/2 in a particular way. If we add up all these equations, we get (1+2+3+4+5+6) and 5–3+2–6–1+4 are congruent. The
right side of the congruence, 3, is the number of multiples greater than p/2, which we know from Gauss's lemma is equal to (ap).
Here is the formal proof:
For each
k = 1, 2, …, (p–1)/2
, we use the division algorithm to write
ka = q[k]p + r[k]
, where the quotient
and the remainder
0 ≤ r[k] < p
. For each
, define the modified remainder
to be
r[k] < p/2
r[k] > p/2
. By Gauss's lemma there are
remainders greater than
. So we can rewrite
r[1]+r[2]+… r[k]
s[1]+s[2]+…+s[k] + p(ap)
Add up all of the equations from the division algorithm to get a and p are both congruent to 1 mod 2. By the argument used in the proof of Gauss's lemma, {s[1], s[2], …, s[k]} = {1, 2, …, (p–1)/2}.
Since x ≡ –x (mod 2) for any integer x, we must then have s[1]+s[2]+…+s[k] ≡ 1+2+…+(p–1)/2 (mod 2). Thus the equation above, when written as a congruence mod 2, reduce to
The law of quadratic reciprocity
We have been building up to one of the most famous results in number theory, the law of quadratic reciprocity:
(Law of quadratic reciprocity) Let p and q be odd primes. Then
In short, if either p or q is a 4k+1 prime, then (pq) = (qp), and if both p and q are 4k+3 primes, then (pq) = –(qp). This is a bit of a surprising connection between prime numbers.
We can use Eisenstein's lemma to prove it. The sum from Eisenstein's lemma, ∑(p–1)/2k = 1 floor(kap) has a nice geometric interpretation. It is the number of lattice points (points with integer
coordinates) under the line y = ap x and above the x-axis, between x = 0 and x = p/2. See the figure below for an example with a = 5 and p = 13.
For example, the height of the line y = 513x at x = 6 is about 2.3 and so there are 2 (that is, floor(6 × 513)) lattice points below that point and above the axis. Eisenstein's lemma for this example
can be rephrased to say that (513) = (–1)^n, where n is the number of lattice points under y = 513 x and above the axis between x = 0 and x = 13/2. By symmetry, (135) is (–1)^m, where m is the number
of lattice points to the left of x = 135y and right of the y-axis between y = 0 and y = 5/2.
We can think of (513) as coming from a count of the lattice points in the interior of the bottom green triangle shaded in the figure above and (135) as coming from a count of the lattice points in
the top pink triangle. Those two triangles fit together to give a rectangle. Inside that rectangle there are exactly ((5–1)/2) · ((13–1)/2) = 12 lattice points. Therefore, we have
Here is a proof of the law of quadratic reciprocity:
are prime. Then
m = ∑(p–1)/2k = 1 floor(kqp)
is the number of lattice points below the line
y = qp x
and above the
x = 0
x = p/2
. Similarly,
n = ∑(q–1)/2k = 1 floor(kpq)
is the number of lattice points to left of the line
x = qp y
and right of the
-axis, between
y = 0
y = q/2
. None of the lattice points are the same since none of them can lie on the line
y = qp x
(which is the same as
x = pq y
), since if some lattice point
(x[0], y[0])
lies on that line, then we would have
py[0] = qx[0]
, which can only happen for
a multiple of
a multiple of
are prime.
The lattice points are all the lattice points of the interior of the rectangle running from x = 0 to p/2 and y = 0 to q/2. There are p–12q–12 of them, so m+n = p–12q–12. By Eisenstein's lemma, we
have (pq) = (–1)^m and (qp) = (–1)^n. Thus we have
The law of quadratic reciprocity can be rephrased as follows:
If either p or q is of the form 4k+1, then (pq) = (qp). Otherwise, (pq) = –(qp).
It's quite surprising in that there is no a priori reason why the congruence x^2 ≡ p (mod q) should have anything at all to do with the congruence x^2 ≡ q (mod p). It is what mathematicians call a
deep theorem, in that it is not at all obvious. The proof we gave requires Fermat's little theorem, Euler's identity, Gauss's lemma, and some other little parts. Gauss, who came up with the first
proof, said he struggled for a year trying to figure out a proof. Euler, who conjectured the result, was unable to prove it. There are well over 100 proofs known for quadratic reciprocity, many of
them quite different from the others.
Quadratic reciprocity is responsible for much of modern number theory, as generalizations of it have occupied a lot of number theorists' time. One such generalization was a key part of the proof of
Fermat's last theorem.
Here are a couple of applications of quadratic reciprocity.
1. Suppose we want to compute (747). Both of these are 4k+3 primes, so by quadratic reciprocity, we have (747) = –(477). We are trying to see if 47 is a quadratic residue of 7, so we can reduce 47
mod 7 to get 5. So we can compute (57), which by quadratic reciprocity is the same as (75). We can reduce this to (25) and our rule from earlier tells us that 2 is not a quadratic residue of 5
since 5 is not of the form 8k± 1. In short we have
2. As another example, we have (1343) = –(4313) = –(413). We can't use quadratic reciprocity to compute (413) since 4 is not prime, but we can use the fact that the Legendre symbol is multiplicative
to write (413) = (213)(213). Then, since 13 is not of the form 8k± 1, 2 is not a quadratic residue of 13, and so overall we get (1343) = –1. So 13 is not a quadratic residue of 43.
3. In general, we have (3p) = 1 if and only if p is of the form 12k± 1. We can write (3p) = (p3) if p is of the form 4k+1 and (3p) = –(p3) if p is of the form 4k+3. The only quadratic residue of 3
is 1, so (p3) = 1 if and only if p is of the form 3k+1. So one way for (3p) to equal 1 is if p ≡ 1 (mod 4) and p ≡ 1 (mod 3). Combining these congruences, we get that this will happen when p ≡ 1
(mod 12). The other way for (3p) to equal 1 is if p ≡ 1 (mod 4) and p ≡ 2 (mod 3), which happens when p ≡ –1 (mod 12).
4. A nice application of quadratic reciprocity is Pepin's test, which is used to tell if a Fermat number, F[n] = 2^2^n+1 is prime. Fermat numbers are covered in Section 2.10. Pepin's test is as
Let n > 0. Then F[n] is prime if and only if 3^(F[n]–1)/2 ≡ –1 (mod F[n]).
Here is how we can prove this fact. First suppose F[n] is prime. By Euler's identity, 3^(F[n]–1)/2 tells us whether 3 is a quadratic residue of F[n]. We know that 3 is a quadratic residue of a
prime p if and only if p is of the form 12k± 1. However, all Fermat numbers besides F[0] are of the form 12k+5. This is because 2^2^1 ≡ 4 (mod 12) and 2^2^n+1 ≡ (2^2^n)^2 ≡ 4^2 ≡ 4 (mod 12) So we
must have 3^(F[n]–1)/2 ≡ –1 (mod F[n]).
On the other hand, suppose 3^(F[n]–1)/2 ≡ –1 (mod F[n]). Squaring this tells us that 3^F[n]–1 ≡ 1 (mod F[n]), so the order of 3 modulo F[n] must be a divisor of F[n]–1 = 2^2^n, a power of 2. Its
proper divisors are 2, 4, 8, 16, …, (F[n]–1)/2. But since 3^(F[n]–1)/2 ≡ –1 (mod F[n]), 3^d ≢ 1 (mod F[n]) for any proper divisor d of F[n]–1. So the order of 3 is F[n]–1. Since the order
is a divisor of φ(F[n]), we have φ(F[n]) = F[n]–1, which means that F[n] is prime.
We can implement Pepin's test in Python like below:
def pepin(n):
f = 2**(2**n)+1
return pow(3,(f-1)//2,f) == f-1
My laptop was able to verify that F[14] was not prime in about 10 seconds. It took 80 seconds to verify F[15] is not prime and about 10 minutes to show F[16] is not prime. The largest one ever
done, according to Wikipedia, was F[24] in 1999. Note that this requires raising 3 to a humongous power, as F[24] is over 5 million digits long.
The Jacobi symbol
Just a quick note: There is an important generalization of the Legendre symbol to all positive integers, called the Jacobi symbol. If the prime factorization of n is pe[1]1pe[2]2 ··· pe[k]k, then the
Jacobi symbol is defined by (ap[i]) are Legendre symbols. The Jacobi symbol has many of the same properties as the Legendre symbol. The one notable exception is that it doesn't tell us wither a is a
quadratic residue or not. However, it turns out to have a number of important applications in math and cryptography.
Since the 1970s number theory has been a critical part of cryptography. The following sections detail two important modern cryptographic algorithms: Diffie-Hellman and RSA.
One thing that should be noted before proceeding is that it is usually recommended that you not try to write your own cryptographic routines in any important program as there are many details to get
right (including things that you would probably never think of). Any lapse can make it easy for attackers to break your system.
Diffie-Hellman key exchange
Most forms of cryptography require a key. For example, one of the simplest methods is the substitution cipher. Under the substitution cipher, each letter of the alphabet is replaced with another
letter of the alphabet. For instance, maybe A is replaced by Q, B is replaced by W, and C is replaced by B, like in the figure below:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Q W B R S E T P U M F X H V Z O K A G Y C N D J I L
The message SECRET is encoded as GSBASY. The substitution cipher's key is the replacement alphabet, QWBRSETPUMFXHVZOKAGYCNDJIL. A major obstacle to the substitution cipher and many more sophisticated
methods is that both the person sending the message and the person decrypting the message need to have a copy of the key. Safely transferring that key can be a serious challenge since you would not
want your key to be intercepted.
The Diffie-Hellman key exchange algorithm is a way for two people, usually called Alice and Bob, to publicly create a shared secret number (a key) known only to them. Imagine they are shouting some
numbers back and forth to each other across a crowded room. Anyone can hear the numbers that Alice and Bob are yelling to each other, but at the end of the process Alice and Bob will have a shared
secret number that only they will know.
It starts with Alice and Bob picking a large prime p and a generator g. The generator g is usually a primitive root modulo p, or at least an element of high order. These values are not secret.
Once p and g have been chosen, Alice picks a secret random number a and Bob picks a secret random number b. Alice sends g^a mod p to Bob and Bob sends g^b mod p to Alice. Alice computes (g^b)^a = g^
ab mod p and Bob computes (g^a)^b = g^ab mod p. Alice and Bob now have the value g^ab in common and this is the secret key. Even though g^a and g^b were sent publicly, only someone who knows a or b
can easily compute g^ab.
We say easily compute g^ab because, in theory, someone could compute g^ab just knowing p, g, g^a, and g^b (which were all sent in public), but if p is sufficiently large, there is no known way to do
this efficiently.
Here is an example with p = 23. It turns out that g = 5 is a primitive root. Alice and Bob agree on these values and don't need to keep them secret. Then Alice and Bob pick their secret numbers, say
Alice picks a = 6 and Bob picks b = 7. Alice then sends Bob g^a mod p, which is 5^6 mod 23, or 8. Bob sends Alice g^b mod p, which is 5^7 mod 23, or 17. These values, 8 and 17, are sent publicly.
Alice then computes (g^b)^a mod p, which is 17^6 mod 23, and Bob computes (g^a)^b mod p, which is 8^7 mod 23. Both of these values work out to g^ab mod p, which is 12. This is the shared key.
Someone monitoring Alice and Bob's exchanges would see the values 23, 5, 8, and 17. For this small example, they could solve 5^a ≡ 8 (mod 23) or 5^b ≡ 17 (mod 23) by brute-force to get a and b, as
there are only 22 possible values for each of a and b. Once they have a and b, they can easily compute the shared key, g^ab. However, if p is very large, then this brute-force search would be
infeasible. There are other techniques that are better than a brute-force search, but if p is large enough, these techniques are still computationally infeasible. The problem of determining g^ab from
the public information is known as the Diffie-Hellman problem. More generally, given g, p, and a value c, solving g^x ≡ c (mod p) is called the discrete logarithm problem. Both are currently
considered intractable for large p.
The shared secret number created by this algorithm can be used as a key for some other cryptographic algorithm, like DES or a more modern variant of it.
Possible problems with Diffie-Hellman
One major problem is that this method is subject to a so-called man-in-the-middle attack. This is where a third party, usually called Eve (for eavesdropper) sits in the middle, pretending to be Bob
to Alice and pretending to be Alice to Bob. Alice and Bob think they are communicating with each other, but they are really communicating with Eve. Eve ends up with a key in common with Alice and a
key in common with Bob. Alice and Bob will only be able to discover that it was Eve they were communicating with once Eve leaves and they find that they can't communicate with each other (because
they have different keys). The solution to this problem is to add some form of authentication so that Alice and Bob can be sure that they are communicating with each other.
Another problem is a poor choice of p and g. For example, suppose p = 61 and g = 13. The problem is that g^3 = 1, so that the powers of g get caught in the repeating cycle 1, 13, 47, 1, 13, 47, ….
With only three values to choose from, it will be pretty easy to guess what g^ab is. This is the reason that g is supposed to be a primitive root; we want the powers of g to take on a wide range of
values to make brute-forcing infeasible. Actually, because of this, it is not necessary that g be a primitive root as long as the order of g is relatively large.
In practice, p is chosen to be what's called a safe prime, a prime of the form 2q+1, where q is also prime.**Note that q is a Sophie Germain prime. Since φ(p) = p–1 = 2q and q is prime, the only
possible orders mod p are 1, 2, q, and 2q (recall that the order is always a divisor of φ(n)). Only 1 has order 1 and only p–1 has order 2 (see Theorem 27). The other elements have order q or 2q. So
any value except 1 and p–1 would be fine to use for g.
One further problem is not choosing a large enough value of p. Big primes are usually measured in bits. A 256-bit prime is a prime on the order of 2^256. This corresponds to log[10](2^256) ≈ 77
digits. Despite being a very large number, the discrete logarithm problem with primes of this size can be solved using a technique called a number field sieve. 1024-bit primes are a bit safer, but
not recommended. Primes on the order of 2048 bits or larger are recommended. Of course, the question arises, why not use a truly huge prime far beyond what technology could ever break? The reason is
using such huge numbers slows down the Diffie-Hellman algorithm too much and may require more computational power than a device (like a phone) can deliver. There is a tradeoff between speed and
Finally, it is worth noting that Diffie-Hellman can be used with other algebraic structures. The positive integers modulo p are a type of algebraic structure known as a group. There are many other
types of groups whose elements are numbers or other types of objects, called groups, for which a kind of arithmetic works. Diffie-Hellman can be extended to these more general groups. In particular,
elliptic curve cryptography involves using groups whose elements are rational points on elliptic curves. See Section 4.3.
RSA cryptography
RSA is similar to Diffie-Hellman in that both methods rely on the intractability of a number-theoretic problem. Diffie-Hellman relies on the difficulty of the discrete logarithm problem, while RSA
relies on the fact that it is difficult to factor very large numbers.
The scenario is this: Alice wants other people to be able to send her secret messages. She posts a public key that anyone can use to encrypt messages. She maintains a private key (related to, but
different from, the public key) that only she can use to decrypt the messages.
Alice creates the keys as follows: She picks two large prime numbers, p and q, and computes n = pq. The primes p and q must be kept secret, but n is part of the public key. Then Alice picks an
integer e between 1 and (p–1)(q–1) that shares no divisors with (p–1)(q–1). That integer is also part of the public key. She then finds a value d such that de ≡ 1 (mod (p–1)(q–1)). In other words, d
is the inverse of e modulo (p–1)(q–1). This value is found with the extended Euclidean algorithm. And it is kept secret. In summary, n and e are public, while p, q, and d are kept private.
Here is how Bob can encrypt a message using Alice's public key: We'll assume the message is an integer a (text can be encoded as integers in a variety of different ways). Bob computes a^e mod n and
sends it to Alice. Alice can then decrypt the message using d. In particular, Alice computes (a^e)^d ≡ a^ed (mod n).
We know that ed ≡ 1 (mod (p–1)(q–1)) but not necessarily that ed ≡ 1 (mod n). However, we have φ(n) = (p–1)(q–1), and since ed ≡ 1 (mod φ(n)), we can write ed = 1+kφ(n). By Euler's theorem a^φ(n) ≡ 1
(mod n), so p = 13 and q = 17. Then n = pq = 221. Note also that (p–1)(q–1) = 192. Alice then chooses an e with no factors in common with 192, say e = 11. She then computes d such that de ≡ 1 (mod (p
–1)(q–1)), which in our case becomes 11d ≡ 1 (mod 192). We get it by using the extended Euclidean algorithm, as follows (starting with the Euclidean algorithm): 11 · 35 – 192 · 2 = 1, so d = 35.
Now suppose Bob encrypts a message a = 65. He computes a^e mod n , or 65^11 mod 221 to get 78. He sends this to Alice. Alice can decrypt it by computing 78^d mod n , which is 78^35 mod 221 or 65.
Someone observing this communication would see n = 221, e = 11, as well as a^e = 78 pass by. In order to decrypt the message, they would need to solve a^11 ≡ 78 (mod 221), factor n, or find the
decryption exponent d. These are easy tasks for n = 221, but for large values of n there are no known efficient ways to do them.
Possible problems with RSA
There are a number of attacks possible on RSA, some of which are quite devious, so that anyone implementing RSA needs to be careful. Here is a short, incomplete list:
1. The first thing is that p and q need to be large primes in order to make it computationally infeasible to factor n = pq.
2. We also need to make sure p and q should not themselves be predictable. If the random number generator used to generate p and q is not completely random, then that leaves an opening for an
attacker (and such openings have been exploited in the past).
3. Because of the way some factoring algorithms work, not just any large primes p and q will work. If p–1 or q–1 have a lot of small prime factors, those algorithms have an easier time factoring n.
If p–1 and q–1 have small prime factors it also makes it more likely that e has a small order (since the order of e divides φ(n) = (p–1)(q–1)).
4. It can be shown that if an attacker is able to figure out just 1/4 of the bits of n (specifically the 1/4 least significant or the 1/4 most significant bits), it is possible to use those bits to
efficiently find d.
5. The value of n should not be reused among different people. Suppose Alice has n, d, and e and Bob also uses n but with a different d and e. Bob can use his d to factor n. Then if Bob intercepts a
message encrypted with Alice's public key, he can easily decrypt it.
6. If the value of d is too small, there is an efficient way for an attacker to figure out d. The problem with using a large value of d is it might make the RSA algorithm too computationally intense
for some devices. One way around this is the Chinese remainder theorem. To compute a^d modulo n = pq, compute a^d modulo p and q separately and then use the Chinese remainder theorem to combine
the two parts. This delivers a nice speedup and is used in practice.
7. If the message a^e to be sent is less than n, then solving a^e ≡ b (mod n) reduces to just finding the ordinary eth root of b, which can be done very easily.
8. RSA must be implemented with a padding scheme, where letters of the message are permuted and random stuff is added to the message (this process can be undone by the receiver). If a padding scheme
is not used, then RSA is susceptible to the following attacks.
1. There can be a problem if e is too small. Suppose Alice sends the same message a to three different people using e = 3 and three values of n, say n[1], n[2], and n[3]. We have three
congruences: x ≡ a^3 (mod n[1]), x ≡ a^3 (mod n[2]), and x ≡ a^3 (mod n[3]). We can combine those into one congruence: x ≡ a^3 (mod n[1]n[2]n[3]). The problem with this is since a < n[1], a <
n[2], and a < n[3], we have a < n[1]n[2]n[3], and so solving x ≡ a^3 (mod n[1]n[2]n[3]) reduces to just computing an ordinary (not a modular) cube root.
In general, if we send e or more messages using the same e, then this attack can be applied, unless a padding scheme is used. This is important because small values of e are often used in
practice to speed up the RSA algorithm. This is important for devices without a lot of power that can't don't handle serious computations well.**Actually, e is often chosen to be a Fermat
prime (a prime of the form 2^2^n+1) since exponentiation by repeated squaring on Fermat primes is much faster than with other exponents.
2. Another attack, called a chosen plaintext attack, involves intercepting some encrypted text and then trying to encrypt likely messages using the public key until you get something that
matches the text that you intercepted. Many messages start with something predictable, like “Dear so and so” or maybe some information about the sender or something like that. Once an
attacker knows a few different inputs and their corresponding outputs, they will have an easier time breaking the encryption.
3. There is a related attack, called a chosen ciphertext attack. One way this attack works is if an attacker has access to the decryption algorithm, though not the actual values of d, p, or q.
The attacker decrypts a bunch of messages in an attempt to learn something about those values. Another way this can work is as follows: Suppose Eve intercepts a message a^e that Bob is
sending to Alice. Eve wants to know what a is. She picks some integer b, and sends (ab)^e to Alice. Alice decrypts the message, which should come out as garbage because of the extra factor
that Eve added. If Eve can somehow convince Alice to send her the decrypted message (possibly by pretending to be Bob), then Eve can figure out Bob's original message. This is because Alice
decrypts (ab)^e into ab and if she sends that to Eve, then Eve can just divide by b to find a.
9. One particularly interesting class of attacks on RSA is what are known as side-channel attacks. In these, an attacker observes the state of a computer's CPU while it is encrypting and decrypting,
in particular, when it is raising numbers to powers. The CPU works harder at some points of the process than others, depending on the bit pattern of the key. An attacker can look at where the CPU
is working harder and where it isn't in order to determine the bit pattern of the key (and hence the key itself). They can do this, for instance, by placing a cell phone nearby the computer. The
CPU makes a high frequency noise that varies based on how hard it is working, and there are programs that can pick up and decode the changes in the sound to figure out the key. Another approach
involves putting one hand on the computer and holding a voltmeter in the other to detect small changes in power output of the CPU.**See http://www.cs.tau.ac.il/~tromer/acoustic/. Still another
approach simply relies on timing how long it takes the CPU to perform various steps of the encryption process. In order to stop this, if you implement RSA, you would need to disguise things so
that the CPU is working at the same rate at all times.
10. It is also theoretically possible to factor n = pq, even if p and q are large, if you have a quantum computer. A regular computer is based on bits that have one of two states: on or off (0 or 1).
A quantum computer, using the properties of quantum physics, has qubits instead of bits, which can exist in a variety of states from 0 to 1. It has been proved that a quantum computer could
quickly factor pq even for very large values of p and q. However, the largest quantum computers that have been built consist of only a few qubits and haven't been able to factor numbers larger
than 100.
Elliptic curve cryptography
This section provides a brief introduction to elliptic curve cryptography. We will leave many mathematical details out and oversimplify some things.
Elliptic curves are curves with equations of the form y^2 = x^3+ax+b. A typical elliptic curve looks a lot like the curve shown below:
Mathematicians have been interested for a while in rational points on such curves; namely, points (x, y) whose coordinates are both rational numbers. For instance, the curve y^2 = x^3–x+1 contains
the points (1/4, 7/8) and (3, 5), both of which have rational coordinates. See Section 5.2 for a nice application of using rational points on the unit circle to find Pythagorean triples.
We can define a way to combine rational points on the curve to get new points. Many lines will intersect an elliptic curve at three points. If we connect two rational points on an elliptic curve by a
line, that line will meet the curve in one other point, and it's not too hard to show that point will also have rational coordinates. We then reflect that point across the x-axis, using the fact that
elliptic curves are symmetric about the x-axis, to get a new point. So given two different points, P and Q, on the curve, P+Q is defined to be this new point. See the figure below on the left.
A line that is tangent to the curve may only intersect the curve in two points. We can use this to define a rule for P+P. Namely, we follow the tangent line from P until it hits the curve and then
reflect across the x-axis, like in the figure above on the right.
The only other possibilities for lines intersecting the curve are vertical lines, which can meet the curve in one or two points. The key to understanding them is there one other point, called the
point at infinity, that we need to add to our curve. We can sort of think of it sort of sitting out at infinity. We use the symbol 0 for it. It acts as the additive identity. The vertical lines are
important for showing that P+0 = P and P–P = 0, where –P is the point obtained by reflecting P across the x-axis.
We are omitting a lot of technical details here. But the important point is that this addition operation makes the rational points into a mathematical object called an abelian group. Basically, this
means that the addition operation behaves nicely, obeying many of the rules that ordinary addition on integers satisfies.**An abelian group, roughly speaking, is a set along with an operation that is
commutative, associative, has an additive identity akin to the number 0, and every element of the set has an additive inverse.
It is possible to work out formulas for P+Q and P+P. If P has coordinates (x, y), and Q has coordinates (x′, y′), then P+Q has coordinates (λ^2–x–x′, λ(x–x′)–y), where λ = y′–yx′–x. And P+P has
coordinates (μ^2–2x, μ(x–(μ^2–2x))–y), where μ = 3x^2+a2y.
For cryptography, we use modular arithmetic with elliptic curves. For instance, say we use arithmetic in ℤ[7] on y^2 = x^3–x+1. In that case, the point (5, 2) is on the curve since 2^2 ≡ 5^3–5+1 (mod
7). We define addition of points on the curve using the formulas given above, but in place of division we use the modular inverse. For instance, instead of doing 4/3 in ℤ[7], we would do 4 · 3^–1,
which is 4 · 5 ≡ 6 (mod 7).
Finally, to actually do cryptography, we pick a curve (that is, we pick values of a and b in y^2 = x^3+ax+b). There are certain values that people suggest to use. Then we pick a large prime p so that
all the arithmetic will be done modulo p. Then we pick a random point G on the curve and a random integer a. We then compute aG, which denotes G added to itself a times. The public key is aG and the
private key is a. Note the similarity with Diffie-Hellman, where we have a generator g and a random integer a and g^a is sent publicly. If the group of points on the curve is large, it is thought to
be very difficult for someone to recover a from aG (which will appear as just a random point on the curve).
Once we have a way of generating a public and private key like this, we can do all sorts of cryptographic things, including analogs of Diffie-Hellman key exchange, secure communication of messages,
and more. The benefits of elliptic curve cryptography over cryptography with ordinary modular arithmetic are that arithmetic on elliptic curves is less computationally intensive than raising numbers
to large powers and the best known algorithms for brute-force breaking the private key in elliptic curve cryptography are not as good as the best-known algorithms for brute-force breaking the private
key in ordinary modular arithmetic. In short, you can get more security for less computational effort.
For more information, do a web search for A Tutorial on Elliptic Curve Cryptography by Fuwen Liu or A (Relatively Easy To Understand) Primer on Elliptic Curve Cryptography by Nick Sullivan.
Special numbers
Perfect numbers and Mersenne primes
A number is called perfect if it equals the sum of its proper divisors. For example, 6 is perfect because 6 is the sum of its proper divisors, 1, 2, and 3. Recall that σ(n) denotes the sum of the
divisors of n. So we can define perfect numbers as below:
A positive integer n is called perfect if σ(n) = 2n.
Perfect numbers have been of interest to mathematicians since at least the time of the ancient Greeks. They knew of four perfect numbers: 6, 28, 496, and 8128. The next perfect number, 33,550,336,
was not found until possibly as late as the 15th century. The next one after that, 8,589,869,056 was discovered in the late 16th century.
Perfect numbers that are even are associated with a particular type of number called a Mersenne number, named for the 17th century monk and mathematician, Marin Mersenne.
A positive integer of the form 2^n–1 is called a Mersenne number. If it is prime, it is called a Mersenne prime.
The relationship between perfect numbers and Mersenne primes is that whenever 2^n–1 is a Mersenne prime, then 2^n–1(2^n–1) is perfect. For instance, the first Mersenne prime is 2^2–1 = 3 and 2^1(2^
2–1) = 6 is perfect. The next Mersenne prime is 2^3–1 = 7 and 2^2(2^3–1) = 28 is perfect. The next two Mersenne primes are 2^5–1 and 2^7–1, which correspond to the perfect numbers 496 and 8128.
The next several exponents of Mersenne primes are 13, 17, 19, 31, 61, 89, 107, 127. The next one, 521, was found by a computer search in 1952. As of 2019, there were only 51 known Mersenne primes,
the largest of which has exponent 82,589,933. It corresponds to a number with close to 25 million digits. It is also the largest prime number that had been found. It is suspected that there are
infinitely many Mersenne primes (and hence perfect numbers), but no one has been able to prove it.
One thing to note is that if 2^n–1 is prime, then n itself must be prime. This is because if n = ab is not prime, then 2^ab–1 can be factored into (2^a–1)(1+2^a+2^2a+2^3a+ ··· + 2^(b–1)a). A similar
factorization shows that there are no primes of the form 3^n–1, 4^n–1, etc. as m^n–1 is divisible by m–1.
We record the relationship between Mersenne primes and perfect numbers in the following theorem. A proof of it appears in Euclid.
If 2^n–1 is prime, then 2^n–1(2^n–1) is perfect.
Let k = 2^n–1(2^n–1). We have to show that σ(k) = 2k. Since σ is multiplicative and gcd(2^n–1, 2^n–1) = 1, we have σ(k) = σ(2^n–1)σ(2^n–1).
The formula for computing σ tells us that σ(2^n–1) = 2^n–1, and we have σ(2^n–1) = 2^n since 2^n–1 is prime. Putting this together gives us σ(k) = 2k.
We can take things one step further.
Every even perfect number is of the form 2^n–1(2^n–1), where 2^n–1 is a Mersenne prime.
be an even perfect number. Factor as many twos as possible out to write
k = 2^n–1m
, where
is odd and
n ≥ 2
. We have
is perfect,
σ(k) = 2k = 2^nm
, and so we have
is a divisor of
. By Euclid's lemma, since
gcd(2^n–1, 2^n) = 1
, we must have
2^n–1 ∣ m
. So we can write
(2^n–1)j = m
for some integer
. Plugging this in to the equation above and simplifying gives
2^nj = σ(m)
We know that both j and m are divisors of m, so σ(m) ≥ j + m = j + (2^n–1)j = 2^nj = σ(m). But 1 is also a divisor of m and wasn't included in that sum, so we have a contradiction unless j = 1. In
that case, the only divisors of m are 1 and itself, meaning m is prime and further that m = 2^n–1 and k = 2^n–1(2^n–1).
The above theorems tell us about even perfect numbers but not about odd perfect numbers. In fact, no odd perfect numbers have ever been found. Mathematicians have not been able to prove that they
don't exist, but if they do, it would come as a surprise. There are a number of things that have been proved must be true of an odd perfect number, should one exist:
• It must be larger than 10^1500.
• It must be congruent to 1 mod 12, 117 mod 468, or 81 mod 324.
• It must have at least 9 distinct prime factors, the largest of which is greater than 10^8
There are quite a few others. See the Wikipedia page on perfect numbers for more.
An interesting fact about the even perfect numbers we've seen (6, 28, 496, 8128, 33,550,336, and 8,589,869,056) is that they all end in 6 or 8. This is in fact true for all even perfect numbers.
Every even perfect number is of the form 2^n–1(2^n–1), where n is prime. Since n is prime, either n = 2 or n is of the form 4k± 1. If n = 2, we get the perfect number 6. If n is of the form 4k+1,
then a simple calculation allows us to reduce 2^n–1(2^n–1) to 6 mod 10. A similar calculation for n = 4k–1 reduces 2^n–1(2^n–1) to 8 mod 10.
Finding Mersenne primes
There is a reason why the largest known primes are all Mersenne primes: there is an easy test to tell if a Mersenne number is prime:
(Lucas-Lehmer test for Mersenne primes) Let S[1] = 4 and S[k+1] = S2k–2 for k ≥ 1. Then 2^n–1 is prime if and only if 2^n–1 ∣ S[n–1].
For example, let's use it to show that 2^7–1 = 127 is prime. We have to show that 127 ∣ S[6], or equivalently that S[6] ≡ 0 (mod 127). Working mod 127, we start with S[1] = 4. We then get the
following: S[6] is congruent to 0 mod 127, we conclude that 2^7–1 is a Mersenne prime.
There are various optimizations that can be made to make this process more efficient. The Lucas-Lehmer test is used by GIMPS, the Great Internet Mersenne Prime Search, which uses the idle time of
computers around the world to search for new Mersenne primes. GIMPS has found most of the recent record prime numbers.
As mentioned earlier, it has not been proved that there are infinitely many Mersenne primes, but it is suspected that there are. A conjecture of Lenstra, Pomerance, and Wagstaff is that there are
roughly e^γ log[2]( log[2] x) Mersenne prime numbers less than x, where γ is the Euler-Mascheroni constant. Evidence for this conjecture is shown in the graph below, which shows log[2]( log[2] M[n]),
where the M[n] run through the first 43 Mersenne primes. Note the very nearly linear relationship.
See http://primes.utm.edu/mersenne/heuristic.html for a simple heuristic argument in favor of the conjecture.
Pythagorean triples
We've all seen a 3-4-5 triangle, like the one below, where all the sides are integers.
A 3-4-5 triangle is not the only one with integer sides. There are many others, like 6-8-10 and 5-12-13. Each of these triangles has integer sides (a, b, c) that satisfy the Pythagorean theorem a^2+b
^2 = c^2. We make the following definition:
A triple of positive integers, (a, b, c), is called a Pythagorean triple if a^2+b^2 = c^2.
Given any triple (a, b, c), we can generate infinitely many other triples by multiplying through by a constant. That is, for any integer k ∈ ℤ, (ka, kb, kc) is also a Pythagorean triple. This is
because (ka)^2+(kb)^2 = k^2(a^2+b^2) = k^2c^2. For instance, (3, 4, 5) leads to (6, 8, 10), (9, 12, 15), (12, 16, 20), etc.
We are not too interested in Pythagorean triples that are multiples of other ones, so we make the following definition.
A Pythagorean triple (x, y, z) is called primitive if gcd(x, y, z) = 1.
So (3, 4, 5) and (5, 12, 13) are primitive. Can we find a primitive Pythagorean triple that includes the integer 7? The answer is yes. We want to find b and c such that 7^2+b^2 = c^2. We can rewrite
this as c^2–b^2 = 7^2 or (c–b)(c+b) = 49. Since c and b are integers, c–b will be a divisor of 49 and c+b will be its complement. We have 49 equal to 1 × 49 or 7 × 7. The latter doesn't give a
solution, but the former does. We have c–b = 1 and c+b = 49. Solving this system gives b = 24 and c = 25, so (7, 24, 25) is a new primitive Pythagorean triple.
In general, one way to find Pythagorean triples involving the integer a is to write a^2 = (c–b)(c+b) and assign factors of a^2 to c–b and c+b. For instance, with a = 15, one way to factor a^2 = 225
is as 9 × 25. Setting c–b = 9 and c+b = 25 gives c = 17 and b = 8. So we get the triple (8, 15, 17).
There is a nice formula describing all primitive Pythagorean triples.
(Euclid's formula) Every primitive Pythagorean triple is of the form (2mn, m^2–n^2, m^2+n^2) for some positive integers m and n, where gcd(m, n) = 1 and m and n are not both odd.
For example, with m = 5 and n = 2, we get the triple (20, 21, 29). If we remove the conditions on m and n, we still get Pythagorean triples, just not primitive ones. This formula was first proved by
One way to prove it is to use the technique we used in the examples preceding the theorem. Here is a different argument that has connections to higher mathematics. Suppose we have a^2+b^2 = c^2.
Dividing through by c^2 gives (ac)^2+(bc)^2 = 1. This process can be reversed, so we see that each Pythagorean triple corresponds to a point on the unit circle with rational coordinates and
It is not too hard to show with a little algebra that if a line with rational slope intersects the unit circle at a rational point, then the other point of intersection must also be rational.
Conversely, if we draw a line with rational slope from a rational point on the unit circle, then the other point of intersection with the circle will also be rational. Therefore, if we pick a
convenient rational point on the unit circle (like (–1, 0) and draw lines with rational slope from that point, we will hit all of the other rational points on the unit circle (and hence find all the
Pythagorean triples). See the figure below for a few example slopes and the rational points (and Pythagorean triples) they generate.
A line through (–1, 0) with rational slope r has equation y = r(x+1). Plugging this into the unit circle equation x^2+y^2 = 1 gives x^2+(r(x+1))^2 = 1. After a little algebra, we can write this as x
= 1–r^21+r^2 and plugging back into the line equation gives y = 2r1+r^2. Each value of r gives a different rational point (x, y) on the curve. If we write r = m/n for some integers m and n, and
convert the rational point into a triple, we get the desired formulas a = n^2–m^2, b = 2mn, c = n^2+m^2.
Studying rational points on other curves, especially elliptic curves like y^2 = x^3+ax+b, is a major focus of modern mathematics.
There are a number of interesting properties that a Pythagorean triple (a, b, c) must satisfy. Here are a few of them:
• One of a and b is even and the other is odd.
• Exactly one of a and b is divisible by 3.
• Exactly one of a and b is divisible by 4.
• Exactly one of a, b, and c is divisible by 5.
See the Wikipedia page on Pythagorean triples for more properties.
Fermat's last theorem
So we have seen that there are infinitely many integer solutions to x^2+y^2 = z^2. What about x^3+y^3 = z^3 or x^4+y^4 = z^4? One of the most famous stories from math concerns these equations. Fermat
wrote in the margin of a copy of Diophantus' Arithmetica that he had a proof that x^n+y^n = z^n has no integer solutions and said he couldn't include it because the book's margin was too small to
hold the proof. People tried for the next 350 years to prove it before it was finally resolved by Andrew Wiles in the 1990s.
Primality testing and factoring
We will look at two important problems in number theory: determining if a number is prime and factoring a number. The former can be done relatively quickly, even for pretty large numbers, while there
is no known fast way to do the latter.
Primality testing
As mentioned in Section 2.3, one way to tell if a number n is prime be to test if it is divisible by 2, 3, 4, …, √n. One improvement is to just check for divisibility by primes, but we might not know
all the primes from 2 to √n. One thing we can do is to check if n is divisible by 2 or 3 and then check divisibility by all the integers of the form 6k± 1 up through √n. These are the integers 5, 7,
11, 13, 17, 19, 23, 25, 29, 31, …. There is no need to check divisibility by integers of the form 6k, 6k+2, 6k+3, or 6k+4, since if a number is divisible by one of those numbers then it must already
be divisible by 2 or 3.
Here is some Python code implementing this method.
def is_prime(n):
if n in [2,3,5,7,11,13,17,19,23]:
return True
if n==0 or n==1 or n%2==0 or n%3==0:
return False
d = 5
step = 2
stop = n**.5
while d <= stop:
if n % d == 0:
return False
d += step
step = 4 if step == 2 else 2
return True
On my laptop here is some data for how long it took this program to verify that various integers are prime:
prime running time in seconds
100000000003 0.08
1000000000039 0.22
10000000000037 0.91
100000000000031 2.47
1000000000000037 8.14
10000000000000061 24.54
We see that adding a digit multiplies the running time roughly by 3. Modern cryptography needs primes that are hundreds of digits long. This algorithm would take around 10^40 seconds to verify that a
100-digit number is prime, which is simply unreasonable. The running time here is exponential in the number of digits.
Probabilistic primality tests
Recall Fermat's little theorem, which states that if p is prime, then a^p–1 ≡ 1 (mod p) for any a relatively prime to p. The contrapositive of this statement gives us a way to show a number n is not
prime: just find an a such that a^n–1 ≢ 1 (mod n). For instance, 10 is not prime because 2^9 ≡ 2 (mod 10). Similarly, 2^11 ≡ 8 (mod 12), so 12 is not prime. Can we always use 2^n–1 to show
that n is not prime? No, though it usually does work. The smallest value for which this fails is n = 341. We have 2^340 ≡ 1 (mod 341) but 341 is not prime. Because of this, 341 is called a
pseudoprime to base 2.
Note that even though we couldn't use a = 2 to show that 341 is not prime, we can use a = 3 since 3^340 ≡ 56 (mod 341). The question then becomes: if a^n ≡ a (mod n) for every a relatively prime to n
, is n prime? The answer, perhaps surprisingly, is still no. There are numbers, called Carmichael numbers, that are not prime and yet a^n ≡ a (mod n) for every a relatively prime to n. The first
several Carmichael numbers are 561, 1105, 1729, 2465, 2821, 6601, and 8911. Carmichael numbers are considerably more rare than primes, there being only 43 less than 1,000,000 and 105,122 less than 10
^15 (versus about 29 trillion primes less than 10^15). Despite their relative rarity, it was proved in 1994 that there are infinitely many of them.
Because they are so rare, one way to test if a number n is prime, is for several values of a to test if a^n ≡ a (mod n). If it fails any test, then the number is composite. If it passes every test,
then it might not be prime, but there is a good chance that it is.
This is an example of a probabilistic primality test. There is a small, but not negligible, chance of it being wrong. We can modify this test, however, to create a test whose probability of being
wrong can be made vanishingly small. This new test is based on the following fact:
Let n be an odd prime, with n–1 factored into 2^st for some integer s and odd integer t. Let a be relatively prime to n. Then one of the following congruences is true: a^t ≡ 1 (mod n), a^t ≡ –1 (mod
n), a^2t ≡ –1 (mod n), a^4t ≡ –1 (mod n), …, a^2^s–1t ≡ –1 (mod n).
Since n–1 = 2^st, we can factor a^n–1–1 into (a^2^s–1t+1)(a^2^s–1t–1). We can then factor a^2^s–1t–1 into (a^2^s–2t+1)(a^2^s–2t–1). Continuing this way, we get a^n–1–1 = 0, so one of the factors of
the right side must be 0. Hence once of those congruences must hold.
We can use this theorem as probabilistic primality test in a similar way that we use Fermat's little theorem. Here is the algorithm to test if n is prime. It is called the Miller-Rabin probabilistic
primality test.
1. Factor as many twos as possible out of n–1 to write it as n–1 = 2^st with t odd.
2. Repeat the following steps several times:
1. Choose a random integer a in the range from 2 through n–2.
2. Consider the following congruences: a^t ≡ 1 (mod n), a^t ≡ –1 (mod n), a^2t ≡ –1 (mod n), a^4t ≡ –1 (mod n), …, a^2^s–1t ≡ –1 (mod n).
3. If none of those congruences are true, then n is composite and we stop the algorithm. If at least one of those congruences is true, then go back to step (a).
Here is a Python implementation of this algorithm:
def miller_rabin(n, tries=10):
s = 0
t = n-1
while t%2 == 0:
t //= 2
s += 1
for i in range(tries):
a = random.randint(2,n-2)
b = pow(a,t,n)
if b!=1 and b!=n-1:
for j in range(s):
b = pow(b,2,n)
if b == n-1:
return False
return True
The basic idea of the algorithm is that we apply the theorem to several random values of a. It can be shown that the probability that at least one of the congruences in the theorem is true and yet
the number is still composite is at most 1/4 (and actually often quite a bit less). Assuming independence, if we repeat the process with k values of a, and some of the congruences are true each time,
the probability that a composite will pass through undetected will be less than (1/4)^k.**Note that since composites are so much more common than primes, the probability that the number is prime is
not quite (1/4)^k. But actually the 1/4 probability is an extremely conservative estimate. For large numbers, the probability is actually much lower, so in fact the probability will always turn out
to be less than (1/4)^k. For instance, Pomerance and Crandall in Number Theory: A Computational Perspective, 2nd edition, report that for a 150-digit prime, the probability is actually less than 1/4^
28, not 1/4.
The upshot of all of this is that we just perform Step 2 of the algorithm several times (for large enough values even once or twice is enough), and if the algorithm doesn't tell us the number is
composite, then we can be nearly certain that the number is prime.
It took my laptop a little over 7 seconds with one step of the test to verify (with high probability) that the 4000-digit number 1477!+1 is prime. It took about 16 minutes to show that 6380!+1 (a
21000-digit number) is prime.
There are efficient primality tests that are not probabilistic, but they are also not as fast as the Miller-Rabin test. If you can live with a very small amount of uncertainty, the Miller-Rabin test
is a good way to test primality.**Note that if an important conjecture known as the extended Riemann hypothesis is true, then the Miller-Rabin test could be turned into a true primality test by
running the test for all a less than 2( ln n)^2.
The simple way to factor a number is to check the possible divisors one-by-one. Just like with primality testing, there are more efficient ways to do things than the simple approach.
Fermat's method
Suppose we want to factor 9919. We might notice that it is 10000–81, which is 100^2–9^2 or (100–9)(100+9). Thus we have written 9919 as 91 × 109. We could further factor this into 7 × 13 × 109 if we
This leads to an approach known as Fermat's method. We systematically try to write our integer n as a difference of two squares, which we can then easily factor. This process will always work. Given
n = ab, we can write n = (a+b2)^2–(a–b2)^2.
Here is the general process: We want to write n = x^2–y^2. We can rewrite this as x^2–n = y^2. We start with x equal to the smallest perfect square greater than n and continually increment x by 1
unit until x^2–n is a perfect square. Here is a step-by-step description:
1. Let x = ceiling( √n).
2. If x^2–n is a perfect square, then we can factor n into (x–y)(x+y).
3. Otherwise, increase x by 1 and go to step 2.
For example, let n = 119, 143. We start with x = ceiling(√119143) = 346. We then compute
346^2–119143 = 573
347^2–119143 = 1266
348^2–119143 = 1961
349^2–119143 = 2658
350^2–119143 = 3357
351^2–119143 = 4058
352^2–119143 = 4761 = 69^2.
We stop at x = 352, since we get a perfect square at that step. We then factor n into 352^2–69^2, which is (352–69)(352+69) or 283 × 421, both of which are prime. Notice that this is considerably
less steps than trial division. Fermat's method is good for finding factors close to √n. It is bad at finding small factors. Thus it is a good complement for trial division. Trial division can be
used to find small factors and Fermat's method can be used to find the others.
Note of course that Fermat's method just finds one factor, d. To find more factors, we can run the algorithm or another on n/d. Here is a Python implementation of Fermat's method:
from math import floor, ceil
def is_perfect_square(n):
return abs(n**.5 - floor(n**.5)) < 1e-14
def fermat_factor(n):
x = ceil(n**.5)
y = x*x - n
while not is_perfect_square(y):
x += 1
y = x*x - n
return (x-floor(y**.5), x+floor(y**.5))
Note the way we check if an integer n is a perfect square is if |√n–floor(√n)| is less than some (small) tolerance. One way to speed this up a bit would be to save the value of √y instead of
computing it three separate times.
The Pollard rho method
We will consider one more factoring technique to give a sense for what factoring techniques are out there. This method is called the Pollard rho method.
Say we need to factor 221, which is 13 × 17. Consider iterating the function f(x) = x^2+1 starting with x[0] = 1. We get the following sequence of iterates: x[4]–x[0] = 26–1 is divisible by 13. Also,
x[5]–x[1] = 195 and x[6]–x[2] = 130 are divisible by 13, and in general, x[m+4]–x[m] is divisible by 13 for any m ≥ 4. Further, notice that x[7]–x[1] = 102, x[8]–x[2] = 204, x[9]–x[1] = 119, etc. are
all divisible by 17. This sort of thing will always happen for the divisors of an integer. This suggests a way to find factors of an integer n: Look at the sequence of iterates of x^2+1 and look at
gcd(n, x[k]–x[j]). Eventually this should (but not always) lead to a factor of n.
To see why this works, consider iterating f(x) = (x^2+1) mod 13 starting with x = 1. We get the repeating sequence 1, 2, 5, 0, 1, 2, 5, 0, …. Notice the period of the repeat is 4, which corresponds
to differences of the form x[m+4]–x[m] being divisible by 13 in the iteration mod 221. Similarly, iterating f(x) = (x^2+1) mod 17 starting with x = 1, gives the sequence, 1, 2, 5, 9, 14, 10, 16, 2,
5, 9, which has a repeating cycle of length 6 starting at the second element. This corresponds to differences of the form x[m+6]–x[m] being divisible by 17 in the iteration mod 221. Note that x[6]–x
[0] above is not divisible by 17 as the repeating pattern mod 17 doesn't start until the second term of the sequence.
In general, if we iterate f(x) = (x^2+c) mod n for any integers c and n, and any starting value, we will eventually**The sequence might not right away start repeating. For instance, for f(x) = (x^
2+1) mod 11 starting at 1, we get the sequence 1, 2, 5, 4, 6, 4, 6, 4, 6, …. Notice that the sequence ends up in a repeating cycle. This is where the ρ in the name comes from—the starting values 1,
2, 5 are the tail of the ρ and then the values end up in a cycle, which is the circular part of the ρ. end up in a repeating cycle. This is because the sequence can only take on a finite number of
values, so eventually we must get a repeat, and since each term in the sequence is completely determined by the term before it, that means the next term and all subsequent terms must fall into that
How many iterations do we have to do before we see get a repeated value? If we think of the iteration as generating random numbers between 0 and m–1, then we are looking at how many numbers in that
range we can randomly generate before running into a repeat. This is just like the birthday problem from probability, where we want to know how many people we have to have in a room before there is a
50/50 chance that some two people in the room share a birthday. It turns out that we need just 23 people. We can think of the birthday problem as just like our problem with m = 365. An analysis of
the birthday problem, which we omit here, shows that after roughly √n terms, we should have a good chance of seeing a repeat.
So to find a factor of n, we can iterate x^2+1 and look at the gcd of various differences, x[j]–x[k]. If one of those is not relatively prime to n, then we have found a divisor of n. The problem is
we don't know the length of the cycle we are trying to find or where it starts. The Pollard rho method uses a technique called Floyd's cycle-finding algorithm to find a cycle. It searches for cycles
by reading through the sequence at two different rates, one unit at a time and two units at a time. So we will be looking at the differences x[1]–x[2], x[2]–x[4], x[3]–x[6], x[4]–x[8], etc. This will
eventually find a cycle, even if it isn't the shortest possible cycle.
Note It might happen that we have n = ab and the cycle length for f(x) = (x^2+1) (mod a) is a divisor of the cycle length for f(x) = (x^2+1) (mod b). This would mean that the gcd we calculate would
come out to n and we wouldn't find a nontrivial factor. In this case, we can switch to another function, f(x) = x^2+c, for some other value of c, and try again. Also note that we don't have to start
the iteration at x[0] = 1. It might be better to start with a random value.
If n = ab, with a < b, the time it takes to find a factor a should be on the order of √a. Here is the Pollard rho algorithm to find a nontrivial factor of n:
1. Pick a random c in the range from 1 to n–3 and a random s in the range from 0 to n–1.
2. Set u and v to s and define a function f(x) = (x^2+c) mod n.
3. Set g = 1 and compute u = f(u), v = f(f(v)), and g = gcd(u–v, n). Keep repeating the computation until g≠ 1. This value of g will be a factor of n. However, if g = n, then go back to step 1, as
we want a nontrivial factor.
Here is some Python code implementing this algorithm:
from random import randint
from fractions import gcd
def pollard_rho(n):
g = n
while g == n:
c = randint(1,n-3)
s = randint(0,n-1)
u = v = s
f = lambda x:(x*x+c) % n
g = 1
while g == 1:
u = f(u)
v = f(f(v))
g = gcd(u-v,n)
return g
On my laptop, this program was able to factor a 20-digit number into two 10-digit primes in a few seconds. It took about six minutes to factor a 30-digit number into two 15-digit primes. Compare this
to the Miller-Rabin probabilistic primality test, where my laptop was able to determine (with high probability) that a 21,000-digit number was prime in about 16 minutes. In short, factoring seems to
be a lot harder than primality testing.
Note that because of the random choices in the algorithm, if we run the algorithm on the same number multiple times, we might find different factors. Also, just Fermat's method, this method will just
return a single factor, a of n. We can repeat the algorithm on n/a to find more factors. Don't try to run this algorithm on a prime, however, as it will end up in an infinite loop.
Appendix of useful algebra
Here are a few tricks that are occasionally useful in number theory:
1. The geometric series 1+a+a^2+… + a^n can be rewritten as a^n+1–1a–1.
2. We can factor x^2–y^2 into (x–y)(x+y) and more generally we can factor x^n–y^n as y = 1.
3. The binomial theorem states that (nk) is called a binomial coefficient and is often read as “n choose k”. We have (nk) is the entry in row n, column k of the triangle (where we start counting
rows and columns at index 0 instead of 1).
There are a number of books I used in preparing these notes. Here they are listed, roughly in order of how much I used them.
1. Burton. Elementary Number Theory, 5th edition. McGraw-Hill, 2002.
2. Pommersheim, Marks, and Flapan. Number Theory: A Lively Introduction with Proofs, Applications, and Stories. Wiley, 2010.
3. Crandall and Pomerance. Prime Numbers: A Computational Perspective, 2nd edition. Springer, 2010.
4. Tattersall. Elementary Number Theory in Nine Chapters, 2nd edition. Cambridge, 2005.
5. De Koninck and Mercier. 10001 Problems in Classical Number Theory. American Mathematical Society, 2007.
6. Ore. Number Theory and Its History. McGraw-Hill, 1948.
7. Bressoud and Wagon. A Course in Computational Number Theory. Key College Publishing, 2000.
8. Niven and Zuckerman. An Introduction to The Theory of Numbers. Wiley, 1972.
In addition, I used Wikipedia quite a bit, as well as http://primes.utm.edu. | {"url":"https://brianheinold.net/number_theory/number_theory_book.html","timestamp":"2024-11-09T11:06:27Z","content_type":"text/html","content_length":"400974","record_id":"<urn:uuid:e6577879-16b9-48db-8499-79d306936857>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00803.warc.gz"} |
Blackjack Betting Strategy Based On The True Count
I informative post discuss the dangers that may come from naively trusting the Kelly criterion’s suggestions and propose a practical solution. This practical solution is fully in line with how I
invest at Junto. In the first section, I introduce the mental model of gambler’s ruin and the mental model of leverage.
Proportional Staking
We have verified how betting with Kelly formula is doing when the outcome probabilities are not exactly the ones we believe in. We’ve applied the Kelly formula to the simple betting game. Let’s do
the same analysis as above for when the coin has odds 65/35, and we bet either 20% or 30% of our capital. We can clearly see that having 20% of https://tchingame.com/
wagering-guidelines-forecasts-posts/ capital “invested” in each bet is riskier – with each bet we put more money at stake, but do we receive more reward for that? Blue lines are significantly more
dispersed than the red ones, some of the blue runs finish with capital larger than the red ones, but also a prominent group ends with much smaller winnings. We could calculate easily Kelly ratios for
each of above scenarios if we knew the real probabilities.
How Kelly Criterion Works Exactly
For obvious reason, you don’t want to bet in any game where the expected payout is 0 or negative. How do serious sports bettors approach their wagering? What are the key differences between casual
punters and professional bettors? If you’re serious about sharpening your betting intelligence, the bettingexpert Academy guide to Advanced Betting Theory provides you with the tools and techniques
to truly take your sports betting to the next level. To take your reasoning seriously, the reason why you might not want to bet 1 cent at a time is because the Kelly bet is guaranteed to eventually
overtake your 1-cent-bet-strategy. Furthermore, it is completely incorrect to say that the Kelly bet has a 1/4 chance of losing all your money in the given situation.
A positive percentage implies an edge in favour of your bankroll, so your funds grow exponentially. You can also test the criterion for different values in this online sheet by using the code below.
We also showed that, in the long term, such strategies increase the probability of ruin. Note that we do not imply that the player should invariably use , but rather just in the general scenario
studied here. Given further information, using may actually be advantageous.
While these are technically “up” rounds, that company is going sideways. And, of course, if you think the probability of success is only 10% but exit is imminent it’s probably your thought process
you should worry about, not your portfolio allocation. The good news is that this issue can be easily overcome if you just reduce the amount of the stake suggested by the formula a little. Many
bettors go with the so-called fractional Kelly strategy, according to which they need to use a fraction of the stake suggested by the formula. The two examples express a negative value and show that
it is not worth placing a bet on either opportunity because it has a negative expected value. On the surface of it, the second situation seems to offer a good chance of success, but once you do the
calculations, you realise that it is not a wise idea to risk your money, as the odds are not high enough.
Optimizing Investment Sizing With The Kelly Criterion
The latter is also valued more than the tangent portfolio but, consistent with the literature, it presents more pronounced drawdowns. This strategy is compared with the Half Kelly, the Triple Kelly
and the buy-and-hold strategies considering Banca Intesa as a unique asset. For sake of simplicity, we assume there are no transaction costs, and that the risk-free rate corresponds to the borrowing
rate. In the previous sections we have described the Kelly criterion and its properties. All the computations presented in the following are performed using the R software for statistical computing
installed on a PC equipped with a Intel i9 5.3 GHz processor. Beside investing in a single financial asset it is also possible to compose portfolios optimized under the Kelly criterion.
By applying the Kelly criterion method properly, using completely accurate inputs, you are certain to see your bankroll grow exponentially. Inject those numbers into the Kelly Criterion formula with
the available odds, and the result will be as accurate as reasonably possible. Now, we look at the straight bet on this game, and we see the lines set at Team A 1.65, and Team B 2.70. Sports
predictions are base on probability and need consistent wagering as per the advice of the winning betting system to succeed. If there was a universal winning formula at our disposal, the bookies
would not be working and would most likely change something about their approach.
In the following we use the term Kelly portfolio to refer to such a kind of portfolio. Growth rate function G for different proportions of wealth . For instance, let’s assume that Arsenal have a 50%
probability of winning. | {"url":"https://listexlojavirtual.com.br/blackjack-betting-strategy-based-on-the-true-count/","timestamp":"2024-11-07T16:09:15Z","content_type":"text/html","content_length":"131840","record_id":"<urn:uuid:a9e42b6d-e8c9-457b-a55d-a7bbd3170c49>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00432.warc.gz"} |
How do you convert seconds to days ? | HIX Tutor
How do you convert seconds to days ?
Answer 1
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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File: Q1-Q6_ReTest27.R.CorticalAreas_dil_Final_Final_Individual.32k_fs_LR.dlabel.nii
A Multi-modal Parcellation of Human Cerebral Cortex
CIFTI : Dense Label
HCP_ReTest / Q1-Q6_ReTest27 / MNINonLinear / fsaverage_LR32k / Q1-Q6_ReTest27.R.CorticalAreas_dil_Final_Final_Individual.32k_fs_LR.dlabel.nii
3 MB
NIFTI 2, CIFTI 2
• TABLE NAME: INDEXMAX
click to show/hide
Maps to Surface Maps to Volume Maps with LabelTable Maps with Palette
true false true false
Number of Rows:
Number of Columns:
Palette Type:
ALONG_COLUMN Map:
• BRAIN_MODELS
• Has Volume Data: false
• CortexRight: 29716 out of 32492 vertices
Surface Mesh:32k fs LR, Species:Human | {"url":"https://balsa.wustl.edu/file/644r","timestamp":"2024-11-15T02:41:00Z","content_type":"text/html","content_length":"75858","record_id":"<urn:uuid:31dd8f50-a5ba-4f4e-b503-2d4165dda806>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00806.warc.gz"} |
EMU of Charge/Square Kilometer to Ampere Hour/Square Decimeter Converter (EMU/km2 to A·h/dm2) | Kody Tools
Conversion Description
1 EMU of Charge/Square Kilometer = 0.00001 Coulomb/Square Meter 1 EMU of Charge/Square Kilometer in Coulomb/Square Meter is equal to 0.00001
1 EMU of Charge/Square Kilometer = 1e-7 Coulomb/Square Decimeter 1 EMU of Charge/Square Kilometer in Coulomb/Square Decimeter is equal to 1e-7
1 EMU of Charge/Square Kilometer = 1e-9 Coulomb/Square Centimeter 1 EMU of Charge/Square Kilometer in Coulomb/Square Centimeter is equal to 1e-9
1 EMU of Charge/Square Kilometer = 1e-11 Coulomb/Square Millimeter 1 EMU of Charge/Square Kilometer in Coulomb/Square Millimeter is equal to 1e-11
1 EMU of Charge/Square Kilometer = 1e-17 Coulomb/Square Micrometer 1 EMU of Charge/Square Kilometer in Coulomb/Square Micrometer is equal to 1e-17
1 EMU of Charge/Square Kilometer = 1e-23 Coulomb/Square Nanometer 1 EMU of Charge/Square Kilometer in Coulomb/Square Nanometer is equal to 1e-23
1 EMU of Charge/Square Kilometer = 10 Coulomb/Square Kilometer 1 EMU of Charge/Square Kilometer in Coulomb/Square Kilometer is equal to 10
1 EMU of Charge/Square Kilometer = 0.0000083612736 Coulomb/Square Yard 1 EMU of Charge/Square Kilometer in Coulomb/Square Yard is equal to 0.0000083612736
1 EMU of Charge/Square Kilometer = 9.290304e-7 Coulomb/Square Foot 1 EMU of Charge/Square Kilometer in Coulomb/Square Foot is equal to 9.290304e-7
1 EMU of Charge/Square Kilometer = 6.4516e-9 Coulomb/Square Inch 1 EMU of Charge/Square Kilometer in Coulomb/Square Inch is equal to 6.4516e-9
1 EMU of Charge/Square Kilometer = 25.9 Coulomb/Square Mile 1 EMU of Charge/Square Kilometer in Coulomb/Square Mile is equal to 25.9
1 EMU of Charge/Square Kilometer = 1e-8 Kilocoulomb/Square Meter 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Meter is equal to 1e-8
1 EMU of Charge/Square Kilometer = 1e-10 Kilocoulomb/Square Decimeter 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Decimeter is equal to 1e-10
1 EMU of Charge/Square Kilometer = 1e-12 Kilocoulomb/Square Centimeter 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Centimeter is equal to 1e-12
1 EMU of Charge/Square Kilometer = 1e-14 Kilocoulomb/Square Millimeter 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Millimeter is equal to 1e-14
1 EMU of Charge/Square Kilometer = 1e-20 Kilocoulomb/Square Micrometer 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Micrometer is equal to 1e-20
1 EMU of Charge/Square Kilometer = 1e-26 Kilocoulomb/Square Nanometer 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Nanometer is equal to 1e-26
1 EMU of Charge/Square Kilometer = 0.01 Kilocoulomb/Square Kilometer 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Kilometer is equal to 0.01
1 EMU of Charge/Square Kilometer = 8.3612736e-9 Kilocoulomb/Square Yard 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Yard is equal to 8.3612736e-9
1 EMU of Charge/Square Kilometer = 9.290304e-10 Kilocoulomb/Square Foot 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Foot is equal to 9.290304e-10
1 EMU of Charge/Square Kilometer = 6.4516e-12 Kilocoulomb/Square Inch 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Inch is equal to 6.4516e-12
1 EMU of Charge/Square Kilometer = 0.02589988110336 Kilocoulomb/Square Mile 1 EMU of Charge/Square Kilometer in Kilocoulomb/Square Mile is equal to 0.02589988110336
1 EMU of Charge/Square Kilometer = 0.01 Millicoulomb/Square Meter 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Meter is equal to 0.01
1 EMU of Charge/Square Kilometer = 0.0001 Millicoulomb/Square Decimeter 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Decimeter is equal to 0.0001
1 EMU of Charge/Square Kilometer = 0.000001 Millicoulomb/Square Centimeter 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Centimeter is equal to 0.000001
1 EMU of Charge/Square Kilometer = 1e-8 Millicoulomb/Square Millimeter 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Millimeter is equal to 1e-8
1 EMU of Charge/Square Kilometer = 1e-14 Millicoulomb/Square Micrometer 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Micrometer is equal to 1e-14
1 EMU of Charge/Square Kilometer = 1e-20 Millicoulomb/Square Nanometer 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Nanometer is equal to 1e-20
1 EMU of Charge/Square Kilometer = 10000 Millicoulomb/Square Kilometer 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Kilometer is equal to 10000
1 EMU of Charge/Square Kilometer = 0.0083612736 Millicoulomb/Square Yard 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Yard is equal to 0.0083612736
1 EMU of Charge/Square Kilometer = 0.0009290304 Millicoulomb/Square Foot 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Foot is equal to 0.0009290304
1 EMU of Charge/Square Kilometer = 0.0000064516 Millicoulomb/Square Inch 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Inch is equal to 0.0000064516
1 EMU of Charge/Square Kilometer = 25899.88 Millicoulomb/Square Mile 1 EMU of Charge/Square Kilometer in Millicoulomb/Square Mile is equal to 25899.88
1 EMU of Charge/Square Kilometer = 10 Microcoulomb/Square Meter 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Meter is equal to 10
1 EMU of Charge/Square Kilometer = 0.1 Microcoulomb/Square Decimeter 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Decimeter is equal to 0.1
1 EMU of Charge/Square Kilometer = 0.001 Microcoulomb/Square Centimeter 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Centimeter is equal to 0.001
1 EMU of Charge/Square Kilometer = 0.00001 Microcoulomb/Square Millimeter 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Millimeter is equal to 0.00001
1 EMU of Charge/Square Kilometer = 1e-11 Microcoulomb/Square Micrometer 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Micrometer is equal to 1e-11
1 EMU of Charge/Square Kilometer = 1e-17 Microcoulomb/Square Nanometer 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Nanometer is equal to 1e-17
1 EMU of Charge/Square Kilometer = 10000000 Microcoulomb/Square Kilometer 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Kilometer is equal to 10000000
1 EMU of Charge/Square Kilometer = 8.36 Microcoulomb/Square Yard 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Yard is equal to 8.36
1 EMU of Charge/Square Kilometer = 0.9290304 Microcoulomb/Square Foot 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Foot is equal to 0.9290304
1 EMU of Charge/Square Kilometer = 0.0064516 Microcoulomb/Square Inch 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Inch is equal to 0.0064516
1 EMU of Charge/Square Kilometer = 25899881.1 Microcoulomb/Square Mile 1 EMU of Charge/Square Kilometer in Microcoulomb/Square Mile is equal to 25899881.1
1 EMU of Charge/Square Kilometer = 10000 Nanocoulomb/Square Meter 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Meter is equal to 10000
1 EMU of Charge/Square Kilometer = 100 Nanocoulomb/Square Decimeter 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Decimeter is equal to 100
1 EMU of Charge/Square Kilometer = 1 Nanocoulomb/Square Centimeter 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Centimeter is equal to 1
1 EMU of Charge/Square Kilometer = 0.01 Nanocoulomb/Square Millimeter 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Millimeter is equal to 0.01
1 EMU of Charge/Square Kilometer = 1e-8 Nanocoulomb/Square Micrometer 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Micrometer is equal to 1e-8
1 EMU of Charge/Square Kilometer = 1e-14 Nanocoulomb/Square Nanometer 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Nanometer is equal to 1e-14
1 EMU of Charge/Square Kilometer = 10000000000 Nanocoulomb/Square Kilometer 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Kilometer is equal to 10000000000
1 EMU of Charge/Square Kilometer = 8361.27 Nanocoulomb/Square Yard 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Yard is equal to 8361.27
1 EMU of Charge/Square Kilometer = 929.03 Nanocoulomb/Square Foot 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Foot is equal to 929.03
1 EMU of Charge/Square Kilometer = 6.45 Nanocoulomb/Square Inch 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Inch is equal to 6.45
1 EMU of Charge/Square Kilometer = 25899881103.36 Nanocoulomb/Square Mile 1 EMU of Charge/Square Kilometer in Nanocoulomb/Square Mile is equal to 25899881103.36
1 EMU of Charge/Square Kilometer = 10000000 Picocoulomb/Square Meter 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Meter is equal to 10000000
1 EMU of Charge/Square Kilometer = 100000 Picocoulomb/Square Decimeter 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Decimeter is equal to 100000
1 EMU of Charge/Square Kilometer = 1000 Picocoulomb/Square Centimeter 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Centimeter is equal to 1000
1 EMU of Charge/Square Kilometer = 10 Picocoulomb/Square Millimeter 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Millimeter is equal to 10
1 EMU of Charge/Square Kilometer = 0.00001 Picocoulomb/Square Micrometer 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Micrometer is equal to 0.00001
1 EMU of Charge/Square Kilometer = 1e-11 Picocoulomb/Square Nanometer 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Nanometer is equal to 1e-11
1 EMU of Charge/Square Kilometer = 10000000000000 Picocoulomb/Square Kilometer 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Kilometer is equal to 10000000000000
1 EMU of Charge/Square Kilometer = 8361273.6 Picocoulomb/Square Yard 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Yard is equal to 8361273.6
1 EMU of Charge/Square Kilometer = 929030.4 Picocoulomb/Square Foot 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Foot is equal to 929030.4
1 EMU of Charge/Square Kilometer = 6451.6 Picocoulomb/Square Inch 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Inch is equal to 6451.6
1 EMU of Charge/Square Kilometer = 25899881103360 Picocoulomb/Square Mile 1 EMU of Charge/Square Kilometer in Picocoulomb/Square Mile is equal to 25899881103360
1 EMU of Charge/Square Kilometer = 62415090744608 Elementary Charge/Square Meter 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Meter is equal to 62415090744608
1 EMU of Charge/Square Kilometer = 624150907446.08 Elementary Charge/Square Decimeter 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Decimeter is equal to 624150907446.08
1 EMU of Charge/Square Kilometer = 6241509074.46 Elementary Charge/Square Centimeter 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Centimeter is equal to 6241509074.46
1 EMU of Charge/Square Kilometer = 62415090.74 Elementary Charge/Square Millimeter 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Millimeter is equal to 62415090.74
1 EMU of Charge/Square Kilometer = 62.42 Elementary Charge/Square Micrometer 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Micrometer is equal to 62.42
1 EMU of Charge/Square Kilometer = 0.000062415090744608 Elementary Charge/Square Nanometer 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Nanometer is equal to 0.000062415090744608
1 EMU of Charge/Square Kilometer = 62415090744608000000 Elementary Charge/Square Kilometer 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Kilometer is equal to 62415090744608000000
1 EMU of Charge/Square Kilometer = 52186965048449 Elementary Charge/Square Yard 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Yard is equal to 52186965048449
1 EMU of Charge/Square Kilometer = 5798551672049.9 Elementary Charge/Square Foot 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Foot is equal to 5798551672049.9
1 EMU of Charge/Square Kilometer = 40267719944.79 Elementary Charge/Square Inch 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Inch is equal to 40267719944.79
1 EMU of Charge/Square Kilometer = 161654342934080000000 Elementary Charge/Square Mile 1 EMU of Charge/Square Kilometer in Elementary Charge/Square Mile is equal to 161654342934080000000
1 EMU of Charge/Square Kilometer = 1.0364272140124e-10 Farady (C12)/Square Meter 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Meter is equal to 1.0364272140124e-10
1 EMU of Charge/Square Kilometer = 1.0364272140124e-12 Farady (C12)/Square Decimeter 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Decimeter is equal to 1.0364272140124e-12
1 EMU of Charge/Square Kilometer = 1.0364272140124e-14 Farady (C12)/Square Centimeter 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Centimeter is equal to 1.0364272140124e-14
1 EMU of Charge/Square Kilometer = 1.0364272140124e-16 Farady (C12)/Square Millimeter 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Millimeter is equal to 1.0364272140124e-16
1 EMU of Charge/Square Kilometer = 1.0364272140124e-22 Farady (C12)/Square Micrometer 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Micrometer is equal to 1.0364272140124e-22
1 EMU of Charge/Square Kilometer = 1.0364272140124e-28 Farady (C12)/Square Nanometer 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Nanometer is equal to 1.0364272140124e-28
1 EMU of Charge/Square Kilometer = 0.00010364272140124 Farady (C12)/Square Kilometer 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Kilometer is equal to 0.00010364272140124
1 EMU of Charge/Square Kilometer = 8.6658515028438e-11 Farady (C12)/Square Yard 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Yard is equal to 8.6658515028438e-11
1 EMU of Charge/Square Kilometer = 9.6287238920487e-12 Farady (C12)/Square Foot 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Foot is equal to 9.6287238920487e-12
1 EMU of Charge/Square Kilometer = 6.6866138139227e-14 Farady (C12)/Square Inch 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Inch is equal to 6.6866138139227e-14
1 EMU of Charge/Square Kilometer = 0.00026843341615209 Farady (C12)/Square Mile 1 EMU of Charge/Square Kilometer in Farady (C12)/Square Mile is equal to 0.00026843341615209
1 EMU of Charge/Square Kilometer = 0.000001 EMU of Charge/Square Meter 1 EMU of Charge/Square Kilometer in EMU of Charge/Square Meter is equal to 0.000001
1 EMU of Charge/Square Kilometer = 1e-8 EMU of Charge/Square Decimeter 1 EMU of Charge/Square Kilometer in EMU of Charge/Square Decimeter is equal to 1e-8
1 EMU of Charge/Square Kilometer = 1e-10 EMU of Charge/Square Centimeter 1 EMU of Charge/Square Kilometer in EMU of Charge/Square Centimeter is equal to 1e-10
1 EMU of Charge/Square Kilometer = 1e-12 EMU of Charge/Square Millimeter 1 EMU of Charge/Square Kilometer in EMU of Charge/Square Millimeter is equal to 1e-12
1 EMU of Charge/Square Kilometer = 1e-18 EMU of Charge/Square Micrometer 1 EMU of Charge/Square Kilometer in EMU of Charge/Square Micrometer is equal to 1e-18
1 EMU of Charge/Square Kilometer = 1e-24 EMU of Charge/Square Nanometer 1 EMU of Charge/Square Kilometer in EMU of Charge/Square Nanometer is equal to 1e-24
1 EMU of Charge/Square Kilometer = 8.3612736e-7 EMU of Charge/Square Yard 1 EMU of Charge/Square Kilometer in EMU of Charge/Square Yard is equal to 8.3612736e-7
1 EMU of Charge/Square Kilometer = 9.290304e-8 EMU of Charge/Square Foot 1 EMU of Charge/Square Kilometer in EMU of Charge/Square Foot is equal to 9.290304e-8
1 EMU of Charge/Square Kilometer = 6.4516e-10 EMU of Charge/Square Inch 1 EMU of Charge/Square Kilometer in EMU of Charge/Square Inch is equal to 6.4516e-10
1 EMU of Charge/Square Kilometer = 2.59 EMU of Charge/Square Mile 1 EMU of Charge/Square Kilometer in EMU of Charge/Square Mile is equal to 2.59
1 EMU of Charge/Square Kilometer = 29940.12 ESU of Charge/Square Meter 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Meter is equal to 29940.12
1 EMU of Charge/Square Kilometer = 299.4 ESU of Charge/Square Decimeter 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Decimeter is equal to 299.4
1 EMU of Charge/Square Kilometer = 2.99 ESU of Charge/Square Centimeter 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Centimeter is equal to 2.99
1 EMU of Charge/Square Kilometer = 0.029940119760479 ESU of Charge/Square Millimeter 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Millimeter is equal to 0.029940119760479
1 EMU of Charge/Square Kilometer = 2.9940119760479e-8 ESU of Charge/Square Micrometer 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Micrometer is equal to 2.9940119760479e-8
1 EMU of Charge/Square Kilometer = 2.9940119760479e-14 ESU of Charge/Square Nanometer 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Nanometer is equal to 2.9940119760479e-14
1 EMU of Charge/Square Kilometer = 29940119760.48 ESU of Charge/Square Kilometer 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Kilometer is equal to 29940119760.48
1 EMU of Charge/Square Kilometer = 25033.75 ESU of Charge/Square Yard 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Yard is equal to 25033.75
1 EMU of Charge/Square Kilometer = 2781.53 ESU of Charge/Square Foot 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Foot is equal to 2781.53
1 EMU of Charge/Square Kilometer = 19.32 ESU of Charge/Square Inch 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Inch is equal to 19.32
1 EMU of Charge/Square Kilometer = 77544554201.68 ESU of Charge/Square Mile 1 EMU of Charge/Square Kilometer in ESU of Charge/Square Mile is equal to 77544554201.68
1 EMU of Charge/Square Kilometer = 29940.12 Franklin/Square Meter 1 EMU of Charge/Square Kilometer in Franklin/Square Meter is equal to 29940.12
1 EMU of Charge/Square Kilometer = 299.4 Franklin/Square Decimeter 1 EMU of Charge/Square Kilometer in Franklin/Square Decimeter is equal to 299.4
1 EMU of Charge/Square Kilometer = 2.99 Franklin/Square Centimeter 1 EMU of Charge/Square Kilometer in Franklin/Square Centimeter is equal to 2.99
1 EMU of Charge/Square Kilometer = 0.029940119760479 Franklin/Square Millimeter 1 EMU of Charge/Square Kilometer in Franklin/Square Millimeter is equal to 0.029940119760479
1 EMU of Charge/Square Kilometer = 2.9940119760479e-8 Franklin/Square Micrometer 1 EMU of Charge/Square Kilometer in Franklin/Square Micrometer is equal to 2.9940119760479e-8
1 EMU of Charge/Square Kilometer = 2.9940119760479e-14 Franklin/Square Nanometer 1 EMU of Charge/Square Kilometer in Franklin/Square Nanometer is equal to 2.9940119760479e-14
1 EMU of Charge/Square Kilometer = 29940119760.48 Franklin/Square Kilometer 1 EMU of Charge/Square Kilometer in Franklin/Square Kilometer is equal to 29940119760.48
1 EMU of Charge/Square Kilometer = 25033.75 Franklin/Square Yard 1 EMU of Charge/Square Kilometer in Franklin/Square Yard is equal to 25033.75
1 EMU of Charge/Square Kilometer = 2781.53 Franklin/Square Foot 1 EMU of Charge/Square Kilometer in Franklin/Square Foot is equal to 2781.53
1 EMU of Charge/Square Kilometer = 19.32 Franklin/Square Inch 1 EMU of Charge/Square Kilometer in Franklin/Square Inch is equal to 19.32
1 EMU of Charge/Square Kilometer = 77544554201.68 Franklin/Square Mile 1 EMU of Charge/Square Kilometer in Franklin/Square Mile is equal to 77544554201.68
1 EMU of Charge/Square Kilometer = 1e-11 Megacoulomb/Square Meter 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Meter is equal to 1e-11
1 EMU of Charge/Square Kilometer = 1e-13 Megacoulomb/Square Decimeter 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Decimeter is equal to 1e-13
1 EMU of Charge/Square Kilometer = 1e-15 Megacoulomb/Square Centimeter 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Centimeter is equal to 1e-15
1 EMU of Charge/Square Kilometer = 1e-17 Megacoulomb/Square Millimeter 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Millimeter is equal to 1e-17
1 EMU of Charge/Square Kilometer = 1e-23 Megacoulomb/Square Micrometer 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Micrometer is equal to 1e-23
1 EMU of Charge/Square Kilometer = 1e-29 Megacoulomb/Square Nanometer 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Nanometer is equal to 1e-29
1 EMU of Charge/Square Kilometer = 0.00001 Megacoulomb/Square Kilometer 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Kilometer is equal to 0.00001
1 EMU of Charge/Square Kilometer = 8.3612736e-12 Megacoulomb/Square Yard 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Yard is equal to 8.3612736e-12
1 EMU of Charge/Square Kilometer = 9.290304e-13 Megacoulomb/Square Foot 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Foot is equal to 9.290304e-13
1 EMU of Charge/Square Kilometer = 6.4516e-15 Megacoulomb/Square Inch 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Inch is equal to 6.4516e-15
1 EMU of Charge/Square Kilometer = 0.00002589988110336 Megacoulomb/Square Mile 1 EMU of Charge/Square Kilometer in Megacoulomb/Square Mile is equal to 0.00002589988110336
1 EMU of Charge/Square Kilometer = 29940.12 Statcoulomb/Square Meter 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Meter is equal to 29940.12
1 EMU of Charge/Square Kilometer = 299.4 Statcoulomb/Square Decimeter 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Decimeter is equal to 299.4
1 EMU of Charge/Square Kilometer = 2.99 Statcoulomb/Square Centimeter 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Centimeter is equal to 2.99
1 EMU of Charge/Square Kilometer = 0.029940119760479 Statcoulomb/Square Millimeter 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Millimeter is equal to 0.029940119760479
1 EMU of Charge/Square Kilometer = 2.9940119760479e-8 Statcoulomb/Square Micrometer 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Micrometer is equal to 2.9940119760479e-8
1 EMU of Charge/Square Kilometer = 2.9940119760479e-14 Statcoulomb/Square Nanometer 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Nanometer is equal to 2.9940119760479e-14
1 EMU of Charge/Square Kilometer = 29940119760.48 Statcoulomb/Square Kilometer 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Kilometer is equal to 29940119760.48
1 EMU of Charge/Square Kilometer = 25033.75 Statcoulomb/Square Yard 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Yard is equal to 25033.75
1 EMU of Charge/Square Kilometer = 2781.53 Statcoulomb/Square Foot 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Foot is equal to 2781.53
1 EMU of Charge/Square Kilometer = 19.32 Statcoulomb/Square Inch 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Inch is equal to 19.32
1 EMU of Charge/Square Kilometer = 77544554201.68 Statcoulomb/Square Mile 1 EMU of Charge/Square Kilometer in Statcoulomb/Square Mile is equal to 77544554201.68
1 EMU of Charge/Square Kilometer = 0.000001 Abcoulomb/Square Meter 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Meter is equal to 0.000001
1 EMU of Charge/Square Kilometer = 1e-8 Abcoulomb/Square Decimeter 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Decimeter is equal to 1e-8
1 EMU of Charge/Square Kilometer = 1e-10 Abcoulomb/Square Centimeter 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Centimeter is equal to 1e-10
1 EMU of Charge/Square Kilometer = 1e-12 Abcoulomb/Square Millimeter 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Millimeter is equal to 1e-12
1 EMU of Charge/Square Kilometer = 1e-18 Abcoulomb/Square Micrometer 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Micrometer is equal to 1e-18
1 EMU of Charge/Square Kilometer = 1e-24 Abcoulomb/Square Nanometer 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Nanometer is equal to 1e-24
1 EMU of Charge/Square Kilometer = 1 Abcoulomb/Square Kilometer 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Kilometer is equal to 1
1 EMU of Charge/Square Kilometer = 8.3612736e-7 Abcoulomb/Square Yard 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Yard is equal to 8.3612736e-7
1 EMU of Charge/Square Kilometer = 9.290304e-8 Abcoulomb/Square Foot 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Foot is equal to 9.290304e-8
1 EMU of Charge/Square Kilometer = 6.4516e-10 Abcoulomb/Square Inch 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Inch is equal to 6.4516e-10
1 EMU of Charge/Square Kilometer = 2.59 Abcoulomb/Square Mile 1 EMU of Charge/Square Kilometer in Abcoulomb/Square Mile is equal to 2.59
1 EMU of Charge/Square Kilometer = 2.7777777777778e-9 Ampere Hour/Square Meter 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Meter is equal to 2.7777777777778e-9
1 EMU of Charge/Square Kilometer = 2.7777777777778e-11 Ampere Hour/Square Decimeter 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Decimeter is equal to 2.7777777777778e-11
1 EMU of Charge/Square Kilometer = 2.7777777777778e-13 Ampere Hour/Square Centimeter 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Centimeter is equal to 2.7777777777778e-13
1 EMU of Charge/Square Kilometer = 2.7777777777778e-15 Ampere Hour/Square Millimeter 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Millimeter is equal to 2.7777777777778e-15
1 EMU of Charge/Square Kilometer = 2.7777777777778e-21 Ampere Hour/Square Micrometer 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Micrometer is equal to 2.7777777777778e-21
1 EMU of Charge/Square Kilometer = 2.7777777777778e-27 Ampere Hour/Square Nanometer 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Nanometer is equal to 2.7777777777778e-27
1 EMU of Charge/Square Kilometer = 0.0027777777777778 Ampere Hour/Square Kilometer 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Kilometer is equal to 0.0027777777777778
1 EMU of Charge/Square Kilometer = 2.322576e-9 Ampere Hour/Square Yard 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Yard is equal to 2.322576e-9
1 EMU of Charge/Square Kilometer = 2.58064e-10 Ampere Hour/Square Foot 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Foot is equal to 2.58064e-10
1 EMU of Charge/Square Kilometer = 1.7921111111111e-12 Ampere Hour/Square Inch 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Inch is equal to 1.7921111111111e-12
1 EMU of Charge/Square Kilometer = 0.0071944114176 Ampere Hour/Square Mile 1 EMU of Charge/Square Kilometer in Ampere Hour/Square Mile is equal to 0.0071944114176
1 EMU of Charge/Square Kilometer = 0.00001 Ampere Second/Square Meter 1 EMU of Charge/Square Kilometer in Ampere Second/Square Meter is equal to 0.00001
1 EMU of Charge/Square Kilometer = 1e-7 Ampere Second/Square Decimeter 1 EMU of Charge/Square Kilometer in Ampere Second/Square Decimeter is equal to 1e-7
1 EMU of Charge/Square Kilometer = 1e-9 Ampere Second/Square Centimeter 1 EMU of Charge/Square Kilometer in Ampere Second/Square Centimeter is equal to 1e-9
1 EMU of Charge/Square Kilometer = 1e-11 Ampere Second/Square Millimeter 1 EMU of Charge/Square Kilometer in Ampere Second/Square Millimeter is equal to 1e-11
1 EMU of Charge/Square Kilometer = 1e-17 Ampere Second/Square Micrometer 1 EMU of Charge/Square Kilometer in Ampere Second/Square Micrometer is equal to 1e-17
1 EMU of Charge/Square Kilometer = 1e-23 Ampere Second/Square Nanometer 1 EMU of Charge/Square Kilometer in Ampere Second/Square Nanometer is equal to 1e-23
1 EMU of Charge/Square Kilometer = 10 Ampere Second/Square Kilometer 1 EMU of Charge/Square Kilometer in Ampere Second/Square Kilometer is equal to 10
1 EMU of Charge/Square Kilometer = 0.0000083612736 Ampere Second/Square Yard 1 EMU of Charge/Square Kilometer in Ampere Second/Square Yard is equal to 0.0000083612736
1 EMU of Charge/Square Kilometer = 9.290304e-7 Ampere Second/Square Foot 1 EMU of Charge/Square Kilometer in Ampere Second/Square Foot is equal to 9.290304e-7
1 EMU of Charge/Square Kilometer = 6.4516e-9 Ampere Second/Square Inch 1 EMU of Charge/Square Kilometer in Ampere Second/Square Inch is equal to 6.4516e-9
1 EMU of Charge/Square Kilometer = 25.9 Ampere Second/Square Mile 1 EMU of Charge/Square Kilometer in Ampere Second/Square Mile is equal to 25.9
1 EMU of Charge/Square Kilometer = 1.6666666666667e-7 Ampere Minute/Square Meter 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Meter is equal to 1.6666666666667e-7
1 EMU of Charge/Square Kilometer = 1.6666666666667e-9 Ampere Minute/Square Decimeter 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Decimeter is equal to 1.6666666666667e-9
1 EMU of Charge/Square Kilometer = 1.6666666666667e-11 Ampere Minute/Square Centimeter 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Centimeter is equal to 1.6666666666667e-11
1 EMU of Charge/Square Kilometer = 1.6666666666667e-13 Ampere Minute/Square Millimeter 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Millimeter is equal to 1.6666666666667e-13
1 EMU of Charge/Square Kilometer = 1.6666666666667e-19 Ampere Minute/Square Micrometer 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Micrometer is equal to 1.6666666666667e-19
1 EMU of Charge/Square Kilometer = 1.6666666666667e-25 Ampere Minute/Square Nanometer 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Nanometer is equal to 1.6666666666667e-25
1 EMU of Charge/Square Kilometer = 0.16666666666667 Ampere Minute/Square Kilometer 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Kilometer is equal to 0.16666666666667
1 EMU of Charge/Square Kilometer = 1.3935456e-7 Ampere Minute/Square Yard 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Yard is equal to 1.3935456e-7
1 EMU of Charge/Square Kilometer = 1.548384e-8 Ampere Minute/Square Foot 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Foot is equal to 1.548384e-8
1 EMU of Charge/Square Kilometer = 1.0752666666667e-10 Ampere Minute/Square Inch 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Inch is equal to 1.0752666666667e-10
1 EMU of Charge/Square Kilometer = 0.431664685056 Ampere Minute/Square Mile 1 EMU of Charge/Square Kilometer in Ampere Minute/Square Mile is equal to 0.431664685056 | {"url":"https://www.kodytools.com/units/scharge/from/emuchargepkm2/to/amphpdm2","timestamp":"2024-11-10T19:01:58Z","content_type":"text/html","content_length":"157661","record_id":"<urn:uuid:e3b89c73-3015-4534-b84e-7b4e9fa08fee>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00643.warc.gz"} |
Session 7 - Duration and Modified Duration
At the end of this session, you should be able to understand the difference between duration and modified duration and how they are related as measures of the interest rate risk of a bond. Examples
used throughout this session are based on fixed rate or nominal bonds for ease of calculation but floating rate note duration and modified duration are calculated using the same principles.
Duration and modified duration - useful tools for comparing bonds
Duration measures the average term to maturity of a bond, weighted by the present value of each cashflow. It is also commonly referred to as “Macaulay Duration”. It is a more meaningful measure of a
bond than simply looking at the term to maturity of the bond, as it takes into account the coupon payments that are made. If a bond is a zero-coupon bond, then the duration is equal to the term to
Mathematically, duration is represented as follows
t = term in years to the coupon payment date
n = term in years to the maturity date
disct = discount factor for term t based on the yield to maturity
Modified duration gives you the ability to be able to quickly determine how the price of a bond will change in response to interest rate movements. For those mathematically inclined, it is the first
derivative of price with respect to yield. That is, it represents the ‘rate of change’ of the price to changes in yields. Fortunately, the mathematics of this calculation means that modified duration
can be calculated simply once you know the duration.
i = yield to maturity
f = frequency (eg monthly, quarterly, semi-annually etc)
Calculating the modified duration of a bond or portfolio allows you to quickly estimate how much the value with rise/fall when there are changes in interest rates. For example, if a bond has a
modified duration of 3 years, this means that the value of the bond will increase by around 3% if interest rates decrease by 1%. Conversely, the value of the bond will decrease by around 3% if
interest rates increase by 1%.
The longer the duration, the more sensitive the security is to interest rate movements. Therefore, in a rising interest rate environment, investors would prefer a security with a lower duration as
the value of the security would fall by less than one with a higher duration. Conversely, in a falling interest rate environment, investors would prefer a security with a higher duration as the value
of the security would rise by more than one with a lower duration.
If an investor holds a security until maturity then the changes in prices along the way due to interest rate movements are not relevant as the yield is locked in from the start. While a bond’s value
may change as prices change, gains and losses are only realised when the security is sold. The changes in prices, and therefore duration, are relevant to those investors who may choose to sell
investments prior to maturity.
One way to compare bonds is by measuring the average life of a bond including its cash flows and this is known as duration. It measures when, on average, the investor will receive cash back equal to
their initial investment. It can also be described as measuring the present value of the weighted average cash flows of the bond. The coupons and the face value can all be treated as cash flows from
an investment. Duration is typically expressed in years, as is the term to maturity of a security.
When calculating this average term of cash flow, the size of each cash flow is not used to weight the averages, rather the present value of each cash flow is used as the weighting factor. This is
very important because it ensures the cash flows are weighted by their importance and value in today’s terms.
It is known that a dollar today is worth more than a dollar in ten years. If given the choice of having a dollar today or a dollar in ten years, an investor would naturally pick today due to the time
value of money, where you can reinvest your dollar, and in ten years from now it will be worth more than one dollar. Similarly, in order to compare cash flows in the future, each term needs to be
weighted by its present value.
As mentioned earlier, Duration is calculated as follows:
This may appear to be time consuming, however it is quite simple once you set up a spreadsheet to do the calculations for you.
Example 1
Face Value = $100,000
Coupon = 7%
Yield = 8%
Duration graphically
Figure 7.1 below shows a graphical example of duration. There are five cash flows of equal size, which represent the coupon payments, and a larger cash flow at the end, which is the Face Value plus
the final coupon.
The coupon amounts are all the same, however the present values of those cash flows are smaller the more distant in the future they are. The final cashflow, being the coupon and face value, is the
main contributor to the duration, however duration is shortened due to the coupon payments.
Look at the present value rectangles or the dark blue sections in the chart, and imagine the X- axis represents the top of a seesaw and that the triangle below the X-axis is the fulcrum of the
seesaw. The value (or duration) is the average of the cash flows. In this case it would be 2.75 years for this three year semi-annual coupon bond.
The effect of coupon rate and yield
How does the coupon rate and yield affect duration? The coupon rate and yield are important inputs into pricing and duration calculation formulas. Let’s see how duration is affected by different
coupon rates and yields.
Earlier above we calculated the duration of a three year, 7% coupon bond with a yield of 8%. What if the bond had a 6% coupon instead of 7% coupon? How would the duration be affected?
Example 2
Face Value = $100,000
Coupon = 6%
Yield = 8%
The duration in this example is 2.7831 (coupon = 6%) compared to a duration of 2.7537 (coupon = 7%). Therefore you can see that there is an inverse relationship between duration and coupon rate.
Example 3
We discovered above that there is an inverse relationship between duration and coupon. What is the relationship between duration and yield?
Face Value = $100,000
Coupon = 7%
Yield = 7%
The duration in this example is 2.7575 (yield = 7%) compared to a duration of 2.7537 (yield = 8%). Therefore you can see that there is an inverse relationship between duration and yield.
Relationships between price, duration, interest rates and coupon
Therefore, if you want a security with a higher duration, you should purchase in a security with a lower yield or coupon. Conversely, if you want a security with a lower duration, you should purchase
a security with a higher yield or coupon.
Duration over time
Duration of a bond decreases over time, however it does increase slightly on each coupon payment date. Remember that the duration can be thought of as the balancing point of a see-saw. When a coupon
payment is made, the fulcrum needs to move a little to the right to rebalance it.
Here is a graphical representation of the duration of a 3 year bond.
Portfolio duration and modified duration
Duration and modified duration can also be calculated for a portfolio, not just for each individual security.
For duration, weight the duration of each bond in the portfolio by its size, in comparison to the portfolio, and likewise for modified duration. It is important to remember when doing these
weightings and calculations, that the market value of the bond and the market value of the portfolio are used. The market value is the present value and it is essential to use this value to weight
portfolios to ensure that like is being compared with like.
For a portfolio of bonds, duration of the portfolio is the present value weighted term to cash flow for every coupon flow and face value in the portfolio. It represents when, on average, the
portfolio will return its cash flows to the investor.
In the same way that modified duration impacts a single bond, if our portfolio had a modified duration of four, and interest rates were to rise by 1% on every security in the portfolio, the portfolio
would lose about 4% in value. Or if interest rates were to fall by 1%, the portfolio would gain 4% across the entire portfolio.
Investors can use this duration indicator to position their portfolios in accordance with their view on the underlying interest rate cycle.
Example 4
Here is a portfolio of bonds. The duration and modified duration are calculated as the weighted average based on the market value.
By now you should have an understanding of the terms duration and modified duration, and that they are very important tools for fixed income investors and are often used in the description of managed
bond funds. The duration measures the average term of all cash flows of the bond and is the present value of all that bond’s weighted average cash flows. The modified duration measures the price
sensitivity of a bond to changes in interest rates.
Review questions
1. What does duration measure?
1. The time to maturity of a bond
2. The face value of the bond
3. The value of coupon payments you can expect over the life of a bond
4. The average term to maturity of a bond weighted by the present value of each cashflow
2. Modified duration allows investors
1. To estimate the change in price of a bond for a given change in interest payments
2. The change in price of a bond given payment of a coupon
3. If I have a zero coupon bond, then
1. Duration will be shorter than the term to maturity
2. Duration will be equivalent to the term to maturity
3. Duration will be longer than the term to maturity | {"url":"https://fiig.com.au/news/2016/11/23/session-7---duration-and-modified-duration","timestamp":"2024-11-02T17:56:34Z","content_type":"text/html","content_length":"263917","record_id":"<urn:uuid:4e56acf8-bdea-4c3f-9a83-4f6634655534>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00830.warc.gz"} |
Algorithms to Live By: The Computer Science of Human Decisions
Algorithms to Live By
Imagine you’re searching for an apartment in San Francisco—arguably the most harrowing American city in which to do so. The booming tech sector and tight zoning laws limiting new construction have
conspired to make the city just as expensive as New York, and by many accounts more competitive. New listings go up and come down within minutes, open houses are mobbed, and often the keys end up in
the hands of whoever can physically foist a deposit check on the landlord first.
Such a savage market leaves little room for the kind of fact-finding and deliberation that is theoretically supposed to characterize the doings of the rational consumer. Unlike, say, a mall patron or
an online shopper, who can compare options before making a decision, the would-be San Franciscan has to decide instantly either way: you can take the apartment you are currently looking at, forsaking
all others, or you can walk away, never to return.
Let’s assume for a moment, for the sake of simplicity, that you care only about maximizing your chance of getting the very best apartment available. Your goal is reducing the twin,
Scylla-and-Charybdis regrets of the “one that got away” and the “stone left unturned” to the absolute minimum. You run into a dilemma right off the bat: How are you to know that an apartment is
indeed the best unless you have a baseline to judge it by? And how are you to establish that baseline unless you look at (and lose) a number of apartments? The more information you gather, the better
you’ll know the right opportunity when you see it—but the more likely you are to have already passed it by.
So what do you do? How do you make an informed decision when the very act of informing it jeopardizes the outcome? It’s a cruel situation, bordering on paradox.
When presented with this kind of problem, most people will intuitively say something to the effect that it requires some sort of balance between looking and leaping—that you must look at enough
apartments to establish a standard, then take whatever satisfies the standard you’ve established. This notion of balance is, in fact, precisely correct. What most people don’t say with any certainty
is what that balance is. Fortunately, there’s an answer.
Thirty-seven percent.
If you want the best odds of getting the best apartment, spend 37% of your apartment hunt (eleven days, if you’ve given yourself a month for the search) noncommittally exploring options. Leave the
checkbook at home; you’re just calibrating. But after that point, be prepared to immediately commit—deposit and all—to the very first place you see that beats whatever you’ve already seen. This is
not merely an intuitively satisfying compromise between looking and leaping. It is the provably optimal solution.
We know this because finding an apartment belongs to a class of mathematical problems known as “optimal stopping” problems. The 37% rule defines a simple series of steps—what computer scientists call
an “algorithm”—for solving these problems. And as it turns out, apartment hunting is just one of the ways that optimal stopping rears its head in daily life. Committing to or forgoing a succession of
options is a structure that appears in life again and again, in slightly different incarnations. How many times to circle the block before pulling into a parking space? How far to push your luck with
a risky business venture before cashing out? How long to hold out for a better offer on that house or car?
The same challenge also appears in an even more fraught setting: dating. Optimal stopping is the science of serial monogamy.
Simple algorithms offer solutions not only to an apartment hunt but to all such situations in life where we confront the question of optimal stopping. People grapple with these issues every
day—although surely poets have spilled more ink on the tribulations of courtship than of parking—and they do so with, in some cases, considerable anguish. But the anguish is unnecessary.
Mathematically, at least, these are solved problems.
Every harried renter, driver, and suitor you see around you as you go through a typical week is essentially reinventing the wheel. They don’t need a therapist; they need an algorithm. The therapist
tells them to find the right, comfortable balance between impulsivity and overthinking.
The algorithm tells them the balance is thirty-seven percent.
There is a particular set of problems that all people face, problems that are a direct result of the fact that our lives are carried out in finite space and time. What should we do, and leave undone,
in a day or in a decade? What degree of mess should we embrace—and how much order is excessive? What balance between new experiences and favored ones makes for the most fulfilling life?
These might seem like problems unique to humans; they’re not. For more than half a century, computer scientists have been grappling with, and in many cases solving, the equivalents of these everyday
dilemmas. How should a processor allocate its “attention” to perform all that the user asks of it, with the minimum overhead and in the least amount of time? When should it switch between different
tasks, and how many tasks should it take on in the first place? What is the best way for it to use its limited memory resources? Should it collect more data, or take an action based on the data it
already has? Seizing the day might be a challenge for humans, but computers all around us are seizing milliseconds with ease. And there’s much we can learn from how they do it.
Talking about algorithms for human lives might seem like an odd juxtaposition. For many people, the word “algorithm” evokes the arcane and inscrutable machinations of big data, big government, and
big business: increasingly part of the infrastructure of the modern world, but hardly a source of practical wisdom or guidance for human affairs. But an algorithm is just a finite sequence of steps
used to solve a problem, and algorithms are much broader—and older by far—than the computer. Long before algorithms were ever used by machines, they were used by people.
The word “algorithm” comes from the name of Persian mathematician al-Khwãrizmī, author of a ninth-century book of techniques for doing mathematics by hand. (His book was called al-Jabr wa’l-Muqãbala
—and the “al-jabr” of the title in turn provides the source of our word “algebra.”) The earliest known mathematical algorithms, however, predate even al-Khwãrizmī’s work: a four-thousand-year-old
Sumerian clay tablet found near Baghdad describes a scheme for long division.
But algorithms are not confined to mathematics alone. When you cook bread from a recipe, you’re following an algorithm. When you knit a sweater from a pattern, you’re following an algorithm. When you
put a sharp edge on a piece of flint by executing a precise sequence of strikes with the end of an antler—a key step in making fine stone tools—you’re following an algorithm. Algorithms have been a
part of human technology ever since the Stone Age.
In this book, we explore the idea of human algorithm design—searching for better solutions to the challenges people encounter every day. Applying the lens of computer science to everyday life has
consequences at many scales. Most immediately, it offers us practical, concrete suggestions for how to solve specific problems. Optimal stopping tells us when to look and when to leap. The explore/
exploit tradeoff tells us how to find the balance between trying new things and enjoying our favorites. Sorting theory tells us how (and whether) to arrange our offices. Caching theory tells us how
to fill our closets. Scheduling theory tells us how to fill our time.
At the next level, computer science gives us a vocabulary for understanding the deeper principles at play in each of these domains. As Carl Sagan put it, “Science is a way of thinking much more than
it is a body of knowledge.” Even in cases where life is too messy for us to expect a strict numerical analysis or a ready answer, using intuitions and concepts honed on the simpler forms of these
problems offers us a way to understand the key issues and make progress.
Most broadly, looking through the lens of computer science can teach us about the nature of the human mind, the meaning of rationality, and the oldest question of all: how to live. Examining
cognition as a means of solving the fundamentally computational problems posed by our environment can utterly change the way we think about human rationality.
The notion that studying the inner workings of computers might reveal how to think and decide, what to believe and how to behave, might strike many people as not only wildly reductive, but in fact
misguided. Even if computer science did have things to say about how to think and how to act, would we want to listen? We look at the AIs and robots of science fiction, and it seems like theirs is
not a life any of us would want to live.
In part, that’s because when we think about computers, we think about coldly mechanical, deterministic systems: machines applying rigid deductive logic, making decisions by exhaustively enumerating
the options, and grinding out the exact right answer no matter how long and hard they have to think. Indeed, the person who first imagined computers had something essentially like this in mind. Alan
Turing defined the very notion of computation by an analogy to a human mathematician who carefully works through the steps of a lengthy calculation, yielding an unmistakably right answer.
So it might come as a surprise that this is not what modern computers are actually doing when they face a difficult problem. Straightforward arithmetic, of course, isn’t particularly challenging for
a modern computer. Rather, it’s tasks like conversing with people, fixing a corrupted file, or winning a game of Go—problems where the rules aren’t clear, some of the required information is missing,
or finding exactly the right answer would require considering an astronomical number of possibilities—that now pose the biggest challenges in computer science. And the algorithms that researchers
have developed to solve the hardest classes of problems have moved computers away from an extreme reliance on exhaustive calculation. Instead, tackling real-world tasks requires being comfortable
with chance, trading off time with accuracy, and using approximations.
As computers become better tuned to real-world problems, they provide not only algorithms that people can borrow for their own lives, but a better standard against which to compare human cognition
itself. Over the past decade or two, behavioral economics has told a very particular story about human beings: that we are irrational and error-prone, owing in large part to the buggy, idiosyncratic
hardware of the brain. This self-deprecating story has become increasingly familiar, but certain questions remain vexing. Why are four-year-olds, for instance, still better than million-dollar
supercomputers at a host of cognitive tasks, including vision, language, and causal reasoning?
The solutions to everyday problems that come from computer science tell a different story about the human mind. Life is full of problems that are, quite simply, hard. And the mistakes made by people
often say more about the intrinsic difficulties of the problem than about the fallibility of human brains. Thinking algorithmically about the world, learning about the fundamental structures of the
problems we face and about the properties of their solutions, can help us see how good we actually are, and better understand the errors that we make.
In fact, human beings turn out to consistently confront some of the hardest cases of the problems studied by computer scientists. Often, people need to make decisions while dealing with uncertainty,
time constraints, partial information, and a rapidly changing world. In some of those cases, even cutting-edge computer science has not yet come up with efficient, always-right algorithms. For
certain situations it appears that such algorithms might not exist at all.
Even where perfect algorithms haven’t been found, however, the battle between generations of computer scientists and the most intractable real-world problems has yielded a series of insights. These
hard-won precepts are at odds with our intuitions about rationality, and they don’t sound anything like the narrow prescriptions of a mathematician trying to force the world into clean, formal lines.
They say: Don’t always consider all your options. Don’t necessarily go for the outcome that seems best every time. Make a mess on occasion. Travel light. Let things wait. Trust your instincts and
don’t think too long. Relax. Toss a coin. Forgive, but don’t forget. To thine own self be true.
Living by the wisdom of computer science doesn’t sound so bad after all. And unlike most advice, it’s backed up by proofs.
Just as designing algorithms for computers was originally a subject that fell into the cracks between disciplines—an odd hybrid of mathematics and engineering—so, too, designing algorithms for humans
is a topic that doesn’t have a natural disciplinary home. Today, algorithm design draws not only on computer science, math, and engineering but on kindred fields like statistics and operations
research. And as we consider how algorithms designed for machines might relate to human minds, we also need to look to cognitive science, psychology, economics, and beyond.
We, your authors, are familiar with this interdisciplinary territory. Brian studied computer science and philosophy before going on to graduate work in English and a career at the intersection of the
three. Tom studied psychology and statistics before becoming a professor at UC Berkeley, where he spends most of his time thinking about the relationship between human cognition and computation. But
nobody can be an expert in all of the fields that are relevant to designing better algorithms for humans. So as part of our quest for algorithms to live by, we talked to the people who came up with
some of the most famous algorithms of the last fifty years. And we asked them, some of the smartest people in the world, how their research influenced the way they approached their own lives—from
finding their spouses to sorting their socks.
The next pages begin our journey through some of the biggest challenges faced by computers and human minds alike: how to manage finite space, finite time, limited attention, unknown unknowns,
incomplete information, and an unforeseeable future; how to do so with grace and confidence; and how to do so in a community with others who are all simultaneously trying to do the same. We will
learn about the fundamental mathematical structure of these challenges and about how computers are engineered—sometimes counter to what we imagine—to make the most of them. And we will learn about
how the mind works, about its distinct but deeply related ways of tackling the same set of issues and coping with the same constraints. Ultimately, what we can gain is not only a set of concrete
takeaways for the problems around us, not only a new way to see the elegant structures behind even the hairiest human dilemmas, not only a recognition of the travails of humans and computers as
deeply conjoined, but something even more profound: a new vocabulary for the world around us, and a chance to learn something truly new about ourselves.
Copyright © 2016 by Brian Christian and Tom Griffiths | {"url":"https://algorithmstoliveby.com/excerpt?mc_cid=76acee3efa&mc_eid=490dd6709c","timestamp":"2024-11-08T05:43:28Z","content_type":"text/html","content_length":"38681","record_id":"<urn:uuid:46be1a44-6df6-4308-ad1b-c443a31ac901>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00276.warc.gz"} |
zlartg: generates a plane rotation so that [ CS SN ] [ F ] [ R ] [ __ ] - Linux Manuals (l)
zlartg (l) - Linux Manuals
zlartg: generates a plane rotation so that [ CS SN ] [ F ] [ R ] [ __ ]
ZLARTG - generates a plane rotation so that [ CS SN ] [ F ] [ R ] [ __ ]
F, G, CS, SN, R )
DOUBLE PRECISION CS
COMPLEX*16 F, G, R, SN
ZLARTG generates a plane rotation so that
-SN CS ] [ G ] [ 0 ]
This is a faster version of the BLAS1 routine ZROTG, except for the following differences:
F and G are unchanged on return.
If G=0, then CS=1 and SN=0.
If F=0, then CS=0 and SN is chosen so that R is real.
F (input) COMPLEX*16
The first component of vector to be rotated.
G (input) COMPLEX*16
The second component of vector to be rotated.
CS (output) DOUBLE PRECISION
The cosine of the rotation.
SN (output) COMPLEX*16
The sine of the rotation.
R (output) COMPLEX*16
The nonzero component of the rotated vector.
3-5-96 - Modified with a new algorithm by W. Kahan and J. Demmel This version has a few statements commented out for thread safety (machine parameters are computed on each entry). 10 feb 03, SJH. | {"url":"https://www.systutorials.com/docs/linux/man/l-zlartg/","timestamp":"2024-11-12T22:16:51Z","content_type":"text/html","content_length":"8972","record_id":"<urn:uuid:dde5e00b-37b1-4afe-97fc-6f042ae5fe8c>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00545.warc.gz"} |
Topological characterization of fractional quantum hall ground states from microscopic hamiltonians
We show how to numerically calculate several quantities that characterize topological order starting from a microscopic fractional quantum Hall Hamiltonian. To find the set of degenerate ground
states, we employ the infinite density matrix renormalization group method based on the matrix-product state representation of fractional quantum Hall states on an infinite cylinder. To study
localized quasiparticles of a chosen topological charge, we use pairs of degenerate ground states as boundary conditions for the infinite density matrix renormalization group. We then show that the
wave function obtained on the infinite cylinder geometry can be adapted to a torus of arbitrary modular parameter, which allows us to explicitly calculate the non-Abelian Berry connection associated
with the modular T transformation. As a result, the quantum dimensions, topological spins, quasiparticle charges, chiral central charge, and Hall viscosity of the phase can be obtained using data
contained entirely in the entanglement spectrum of an infinite cylinder.
All Science Journal Classification (ASJC) codes
• General Physics and Astronomy
Dive into the research topics of 'Topological characterization of fractional quantum hall ground states from microscopic hamiltonians'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/topological-characterization-of-fractional-quantum-hall-ground-st","timestamp":"2024-11-10T03:37:27Z","content_type":"text/html","content_length":"47969","record_id":"<urn:uuid:2a9c0afa-a9e6-4fce-b3be-0adb77614107>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00164.warc.gz"} |
How do you find two solutions (in degree and radians) for cscx = (2sqrt3)/3? | HIX Tutor
How do you find two solutions (in degree and radians) for cscx = (2sqrt3)/3?
Answer 1
Solve $\csc x = \frac{2 \sqrt{3}}{3}$
#csc x = 1/(sin x) = (2sqrt3)/3.# Find sin x. #sin x = 3/(2sqrt3) = sqrt3/2.#
Trig Table of Special Arcs gives --> #sin x = sqrt3/2# ---> arc #x = pi/3 (or 60^@)#, and #x = (2pi)/3 (or 120^@)#
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Answer 2
To find the solutions for the equation csc(x) = (2√3)/3, you can follow these steps:
1. Identify the reference angle θ in the first quadrant using the given value: csc(θ) = (2√3)/3. The reference angle is the angle whose sine is equal to the reciprocal of the given value. In this
case, θ = 30 degrees or π/6 radians.
2. Recognize that csc(x) = 1/sin(x), so if csc(x) = (2√3)/3, then sin(x) = 3/(2√3).
3. Since sin(x) = 3/(2√3), we can find the angle x by taking the inverse sine (arcsin) of 3/(2√3).
4. Calculate the values of x using the inverse sine function. Remember that sine is positive in the first and second quadrants.
5. Once you find the value of x in radians, convert it to degrees if necessary.
So, the solutions in degrees and radians are:
1. ( x = 30^\circ ) (or ( x = \frac{\pi}{6} ) radians)
2. ( x = 180^\circ - 30^\circ ) (or ( x = \pi - \frac{\pi}{6} ) radians)
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Why You Should Pay Attention to RC-STARKs
This article provides a friendly exposition to our new paper: “Really Complex Codes with Application to STARKs” by @Yuval_Domb
Intro: My Secret Playbook for Consuming Research
For over a decade I’ve been diving into cryptography research papers. For the longest time, I had this ritual: to truly grasp a paper, I’d read it three times. The first pass was just to get the lay
of the land — skimming the main results and understanding the structure. The second read was more thorough; I’d dig into the introduction and make sure I understood all the key constructions. The
third time reading was the most time-consuming — it was all about learning the proof techniques, the primitives, and their security definitions.
But there are always exceptions. Some papers are masterpieces; others are just plain cool. There are some I even know by heart (shoutout to L17). Lately, though, my attention span isn’t what it used
to be. Usually I only have time for a quick pass before deciding on any action items. Weekends are my time for those deeper dives — the second and third reads — but the papers I choose then are more
for the soul. I guess I’ve gained enough experience to zero in on the parts that matter most to me and extract the knowledge I need.
In recent years, I’ve added a new twist to my reading habit that makes it a lot more fun. With every paper I read, I try to reverse-engineer the author’s journey: How did this paper come to be?
Thinking of it like a mystery novel, I ask myself questions like: What’s the real discovery here? Why hasn’t anyone else thought of this before — why now? Why this specific author and not someone
else? Why are the results presented in this specific way? What results are missing? And so on..
Answering these questions helps me navigate those hefty 40- or 50-page papers with much more ease. Usually, the “core” of a paper boils down to a very specific idea. Often, that idea is simple, and
the rest of the work is about proving it or wrapping it up in a way that delivers a more complete system or story. For example, a paper about running a zero-knowledge prover with multi-party
computation (MPC) might boil down to a single clever trick on how to efficiently divide the multi-scalar multiplication (MSM) primitive between two parties. The rest is mostly repeating known results
to ensure a full ZK protocol can run in the same MPC setup.
Getting into the author’s mind is an art, but when it works, it turns a paper into a story — a reflection of the author’s struggles and victories on their journey to a finished work. I believe that
good papers should be written with this narrative in mind.
Here’s My Take on How the Circle STARK Paper Came to Be
STARKs are incredibly powerful.
The Mersenne prime with 31 bits, often referred to as M31, has unique properties that make it exceptionally attractive. However, using M31 as the base field for STARKs isn’t efficient because M31 has
a low 2-adicity, meaning we can’t use the Fast Fourier Transform (FFT) directly — and FFTs are crucial for STARK performance in areas like low-degree extensions (LDEs) and Reed-Solomon (RS) code
The authors of the Circle STARK paper were trying to tackle this problem. They wanted a transform similar to FFT that takes elements from the M31 field and outputs elements in the same field. Their
first key observation was that while M31 doesn’t have a high enough order of 2, its first field extension does. The rest of the paper is dedicated to developing the required FFT-like transform. The
end result is a new transform with the essential 2:1 folding property, enabling STARKs to work over M31 and achieve significant efficiency gains. The paper involves some pretty complex mathematical
gymnastics to achieve this novel result.
So Far, So Good, So What
But here’s the thing: the new transform is FFT-like, but it’s not the FFT we know and love. For example, we don’t yet know if there’s a way to support radices larger than 2. When we implement FFTs in
hardware, we rely on higher-order radices because they help us save on memory — which is the bottleneck we’re always trying to avoid with FFTs.
Yuval’s paper tells a different story. Yuval is the lead author of our textbook on NTTs, and his work feels like he pulled a few pages straight from that book. His paper doesn’t present a novel idea;
instead, it constructs an FFT based on concepts that have been known for decades. Essentially, the paper presents an FFT over the complex field. The main breakthrough is identifying the right
symmetry in the problem, allowing the evaluations (or coefficients) to come from the real numbers — in this case, the elements of M31.
This leads to a straightforward integration into a STARK system, replacing the Circle STARK transform almost one-to-one. The complexity is roughly the same as in the Circle STARK approach, but now we
can immediately leverage existing hardware optimizations for efficient memory management, thanks to the massive amount of work that’s been put into FFTs over the past 50 years.
To add some color, the table below, taken from the paper, shows that the number of multiplications for Circle STARK (left column) is comparable to the number of multiplications with our FFT
(rightmost column). Refer to section 6 in the paper for a full analysis.
Important note: I oversimplified what it takes to go from our new RC-Code to a full production ready STARK like Stwo. I also downplayed the genius in Yuval’s work. My goal is just to make a point ❤️.
In his book Skunk Works: A Personal Memoir, Ben Rich quotes his predecessor, Kelly Johnson, saying, “Planes that look beautiful fly beautiful.” This sentence captures a profound truth: aerodynamics
obey the laws of physics. Physics describes nature and is encoded through mathematics. Math is elegant, and so are airplanes. The same goes for science. Just think about the elegance of Maxwell’s
equations. The FFT is another example — it’s often the first algorithm taught in any algorithms class or book. It has a certain beauty and is incredibly powerful. Our work is not groundbreaking,
Circle STARK is. Our work is effective: we leverage symmetries in the problem to keep STARK classy, even with the challenges introduced by M31.
When I say this is worth your attention, I mean it embodies in the most perfect way what we are here to do: ZK. Effective.
Needless to say, this work was not possible without the support of @StarkWareLtd and @StarknetFndn. | {"url":"https://www.ingonyama.com/blog/why-you-should-pay-attention-to-rc-starks","timestamp":"2024-11-05T18:48:54Z","content_type":"text/html","content_length":"87210","record_id":"<urn:uuid:68d58a61-61a9-4652-bffd-9f821303b3ef>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00768.warc.gz"} |
GDP Formula - Calculation of GDP Using 3 Formulas
GDP Formula
Last Updated :
21 Aug, 2024
Formula to Calculate GDP
GDP is Gross Domestic Product and is an indicator to measure economic health. The formula to calculate GDP is of three types: Expenditure Approach, Income Approach, and Production Approach.
#1 - Expenditure Approach -
There are three main groups of expenditure household, business, and the government. By adding all-expense, we get the below equation.
GDP = C + I + G +NX
• C = All private consumption/ consumer spending in the economy. It includes durable goods, nondurable goods, and services.
• I = All of a country’s investment in capital equipment, housing, etc.
• G = All of the country’s government spending. It includes the salaries of government employees, construction, maintenance, etc.
• NX= Net country export – Net country import
We can also write this as:-
GDP = Consumption + Investment + Government Spending + Net Export
The Expenditure Approach is a commonly used method for calculating GDP.
#2 - Income Approach -
The Income Approach is a way to calculate GDP by total income generated by goods and services.
GDP = Total National Income + Sales Taxes + Depreciation + Net Foreign Factor Income
• Total National Income = Sum of rent, salaries profit.
• Sales Taxes = Tax imposed by a government on sales of goods and services.
• Depreciation = the decrease in the value of an asset.
• Net Foreign Factor Income = Income earned by a foreign factor like the amount a foreign company or foreign person earns from the country. It is also the difference between a country's citizens
and country's earn.
#3 - Production or Value-Added Approach -
From the name, it is clear that value is added at the time of production. It is also known as the reverse of the expenditure approach. Estimating the gross value-added total cost of economic output
is reduced by the cost of intermediate goods used to produce final goods.
Gross Value Added = Gross Value of Output – Value of Intermediate Consumption
GDP = Sum of all value-added to products during the production of a process
• GDP is Gross Domestic Product. It is an indicator to calculate economic health.
• The formula to calculate GDP is of three types: Expenditure Approach, Income Approach, and Production Approach.
• The industries included in the GDP are manufacturing, mining, banking and finance, construction, real estate, agriculture, electricity, gas, petroleum, and trade.
• The two ways to calculate the GDP in India are economic activity or factor cost and expenditure or market price.
GDP Calculation
Let us see how to use these formulas to calculate GDP.
• GDP can be calculated by considering various sector net changed values.
• GDP is defined as the market value of all goods and services produced within a country in a given period. It can be calculated on an annual or quarterly basis.
• GDP includes every expense in a country like government or private costs, investment, etc. Apart from this, export is also added, and import is excluded.
The industries are as follows:
• Manufacturing
• Mining
• Banking & Finance
• Construction
• Real Estate
• Agriculture
• Electricity, gas, and petroleum
• Trade
GDP vs GNP Explanation in Video
Examples of GDP Formula (with Excel Template)
Here, we are taking a sample report for Q2 of 2018.
Below are two ways to calculate the GDP in India:
• Economic Activity or Factor Cost
• Expenditure or Market Price
Example #1
Let us take an example where one wants to compare multiple industries' GDP with the previous year's GDP.
In the below-given figure, we have shown the calculation of total GDP for Quarter 2 of 2017.
Similarly, we have done the calculation of GDP for Quarter 2 of 2018.
And then, changes between two quarters are calculated in percentage, i.e., GDP of industry upon a sum of total GDP multiple by 100.
At the bottom, it provides an overall change in GDP between two quarters. Again, this is an economic activity-based method.
It helps the government and investors decide on investment and allows the government for policy formation and implementation.
Example #2
Now, let's see an example of an expenditure method that considers expenditure from different means. It is inclusive of expenditure and investment.
Below are the different expenditures, gross capital, export, import, etc. These will help calculate GDP.
For Quarter 2 of 2017, total GDP at market price is calculated in the below-given figure.
Similarly, we have calculated GDP for Quarter 2 of 2018.
Here, first, the sum of expenditure is taken along with gross capital, change in stocks, valuables, and discrepancies which are an export minus import.
A rate of GDP at Market Price:
Similarly, we can calculate the rate of GDP for Quarter2 of 2018.
GDP at market price is a sum of all expenditures. The GDP market price percentage rate is calculated when expenditure is divided by total GDP at market price multiplied by 100.
Through this, one can compare and get a market situation. For example, in a country like India, the global slowdown does not have any major impact only affected factor is export. But on the other
hand, if a country has high export, it will get affected by the global recession.
Frequently Asked Questions (FAQs)
What are GDP formula components?
The components of the GDP formula are consumption, investment, government spending, exports, and imports
What is the PPP-adjusted GDP formula?
The PPP adjusted GDP formula refers to the PPP GDP (Gross Domestic Product), in which GDP is converted into international dollars utilizing the purchasing power parity rates.
What is savings in the GDP formula?
The savings in the GDP formula is the disposable income minus the consumption function. Therefore, analyst expresses it as a GDP percentage. Saving in GDP is defined as the portion of household
income not consumed and reserved for future use. Since savings are not directly spent on goods and services during the current period, they are not included in the calculation of GDP.
Recommended Articles
This article has been a guide to GDP Formula. We discuss the calculation of GDP using 3 types of formulas (Expenditure, Income & Production Approach) with examples and a downloadable Excel template.
You can learn more about Economics from the following articles: -
• Nominal GDP Formula
• Calculate Real GDP
• Differences Between GDP and GNP | {"url":"https://www.wallstreetmojo.com/gdp-formula/","timestamp":"2024-11-08T21:55:51Z","content_type":"text/html","content_length":"344912","record_id":"<urn:uuid:f3c182f1-bb6e-4e02-8d1e-0595401e289e>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00176.warc.gz"} |
converts a compressed adjacency representation
integer vector of length (n+1).
integer vector of length n+1 (pointers).
integer vector
integer vector
Utility function is used to convert a compressed adjacency
representation into standard adjacency representation.
// A is the sparse matrix:
//For this matrix, the standard adjacency representation is given by:
adjncy=[1, 2, 3,4,5,6,7, 4,5,6,7, 5, 6,7, 7];
//(see sp2adj).
// increments in vector xadj give the number of non zero entries in each column
// ie there is 2-1=1 entry in the column 1
// there is 3-2=1 entry in the column 2
// there are 8-3=5 entries in the column 3
// 12-8=4 4
//The row index of these entries is given by the adjncy vector
// for instance,
// adjncy (3:7)=adjncy(xadj(3):xadj(4)-1)=[3,4,5,6,7]
// says that the 5=xadj(4)-xadj(3) entries in column 3 have row
// indices 3,4,5,6,7.
//In the compact representation, the repeated sequences in adjncy
//are eliminated.
//Here in adjncy the sequences 4,5,6,7 and 7 are eliminated.
//The standard structure (xadj,adjncy) takes the compressed form (lindx,xlindx)
lindx=[1, 2, 3,4,5,6,7, 5, 6,7];
//(Columns 4 and 7 of A are eliminated).
//A can be reconstructed from (xadj,xlindx,lindx).
[xadj,adjncy,anz]= sp2adj(A);
See also
• sp2adj — converts sparse matrix into adjacency form
• adj2sp — converts adjacency form into sparse matrix.
• spget — retrieves entries of sparse matrix | {"url":"https://help.scilab.org/docs/2023.0.0/ru_RU/spcompack.html","timestamp":"2024-11-04T05:09:03Z","content_type":"text/html","content_length":"18320","record_id":"<urn:uuid:f1216710-56c8-41d5-a997-b8d186a79e82>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00892.warc.gz"} |
The ratio of shear to elastic modulus of in-plane loaded masonry (vol 53, 40, 2020)
Nicolas Candau, Oguzhan Oguz, Adrien Julien Demongeot
There has been a considerable interest in developing stiff, strong, tough, and highly stretchable hydrogels in various fields of science and technology including biomedical and sensing applications.
However, simultaneous optimization of stiffness, strength ... | {"url":"https://graphsearch.epfl.ch/en/publication/284044","timestamp":"2024-11-06T05:57:34Z","content_type":"text/html","content_length":"102316","record_id":"<urn:uuid:0c1c220e-092a-4ecd-ab96-58f57e338558>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00878.warc.gz"} |
Shock Testing & Analysis
SECTION 1
Shock Isolator Photo Gallery
Figure 1.1. Titan II Missile Silo, Launch Control Room
The control room is mounted underground via huge isolation springs. A typical spring is shown in the background. The purpose is to isolate the control room from mechanical shock and vibration in the
event of a nuclear strike above the launch site. The springs allow 18 inches of relative displacement. The control room could thus carry out a retaliatory strike, as ordered by the U.S. president.
This site is located south of Tucson, Arizona. It has been decommissioned and is now a museum.
SECTION 2
Simple Drop Shock
Figure 2.1. iPhone 6 Accidental Drop
The first iPhone 6 was purchased in Perth, Australia on September 19, 2014. The event was covered by a live TV report. The buyer mishandled the phone as he unboxed it. The phone survived the drop
onto the ground but may have had some fatigue or fracture damage.
Portable electronic devices (PEDs) are expected to survive multiple drops. Most original equipment suppliers specifying between 30 and 50 drops. Recommended test methods are given in Reference [28].
Figure 2.2. Drop Shock Analytical Model
The accidental drop shock of a component onto a hard surface is difficult to model accurately. The item may undergo rigid-body translation and rotation in each of its six degrees-of-freedom during
freefall. The item may have a nonlinear response with plastic deformation during impact. It may or may not bounce. Furthermore, a box-shaped object may strike the ground on any of its corners, edges
or faces.
A very simple method, as a first approximation, is to assume that the object is a linear, undamped, single-degree-of-freedom subjected to initial velocity excitation as it strikes the ground and
remains attached to it via its spring. The object then undergoes free vibration in his configuration. The initial velocity is calculated using a familiar physics formula where the change in kinetic
energy is equal to the change in potential energy due to gravity.
Assume that the object is dropped from rest. The initial velocity as it strikes the ground is
The equation of motion is
The peak velocity is equal to the initial velocity in equation (1.1).
An example is shown in the following table for three natural frequency cases.
Table 2.1. Peak Response Values for 36 inch Drop Height
Natural Freq (Hz) Displacement (in) Velocity (in/sec) Acceleration (G)
200 0.133 167 543
600 0.044 167 1630
1000 0.027 167 2710
SECTION 3
Classical Shock
Figure 3.1. Idealized Classical Pulse Examples
Classical pulses are the simplest base excitation pulses. They are deterministic and can be represented by simple mathematical functions. They are typically one-sided. An SDOF system’s response to a
classical pulse can be solved for exactly using Laplace transforms.
Four classical pulse types are shown in Figure 18.4. Other types include rectangular and trapezoidal pulses. These pulses do not necessarily represent real field environments, but they are still used
throughout industry to test equipment ruggedness for convenience.
Shock tests are performed on military equipment [26] to:
• a. Provide a degree of confidence that materiel can physically and functionally withstand the relatively infrequent, non-repetitive shocks encountered in handling, transportation, and service
environments. This may include an assessment of the overall materiel system integrity for safety purposes in any one or of the handling, transportation, and service environments.
• b. Determine the materiel's fragility level, in order that packaging may be designed to protect the materiel's physical and functional integrity.
• c. Test the strength of devices that attach materiel to platforms that can crash.
Potential equipment failure modes due to shock excitation include:
• a. Materiel failure resulting from increased or decreased friction between parts, or general interference between parts.
• b. Changes in materiel dielectric strength, loss of insulation resistance, variations in magnetic and electrostatic field strength.
• c. Materiel electronic circuit card malfunction, electronic circuit card damage, and electronic connector failure. (On occasion, circuit card contaminants having the potential to cause short
circuit may be dislodged under materiel response to shock.)
• d. Permanent mechanical deformation of the materiel resulting from overstress of materiel structural and nonstructural members.
• e. Collapse of mechanical elements of the materiel resulting from the ultimate strength of the component being exceeded.
• f. Accelerated fatiguing of materials (low cycle fatigue).
• g. Potential piezoelectric activity of materials, and materiel failure resulting from cracks in fracturing crystals, ceramics, epoxies, or glass envelopes.
Figure 3.2. Drop Shock Test Machine, Initial Velocity Excitation
(Courtesy of Lansmont)
Classical pulse shock testing has traditionally been performed on a drop tower. The component is mounted on a platform which is raised to a certain height. The platform is then released and travels
downward to the base, which has pneumatic pistons to control the impact of the platform against the base. In addition, the platform and base both have cushions for the model shown. The pulse type,
amplitude, and duration are determined by the initial height, cushions, and the pressure in the pistons. This is a textbook example of case where the initial potential energy of the raised platform
and test item are converted to kinetic energy. The final velocity of the freefall becomes the initial velocity of the shock excitation.
Figure 3.3. A 50 G, 11 msec, Terminal Sawtooth Pulse for Shaker Shock Test
Classical pulse shock testing can sometimes be performed on shaker tables, but there some constraints. The net velocity and net displacement must each be zero. Also, the acceleration, velocity and
displacement peaks must each be within the shaker table stroke limits. Pre and post pulses are added to classical pulses to meet these requirements. A hypothetical terminal sawtooth pulse suitable
for shaker shock testing is shown in Figure 1.6.
SECTION 4
Half-Sine Shock Example
Consider a single-of-freedom is subjected to a 50 G, 11 msec half-sine pulse applied as base excitation per Figure 12.12. Set the amplification factor to Q=10. Allow the natural frequency to be an
independent variable. Solve for the absolute response acceleration. The equation of motion is for the relative displacement z is
The primary response occurs during the half-sine pulse input. The residual response occurs during the quiet period thereafter. The total response is the combination of primary and residual. The exact
response for a given time can be calculated via a Laplace transform. Note that the quiet period solution is free vibration with initial velocity and displacement excitation.
Figure 4.1. Fourier Transform Half-Sine Shock
A common misunderstanding is to regard the half-sine shock pulse as having a discrete frequency which would be the case if it were extended to a full-sine pulse. The Fourier transform in Figure 18.7
shows that the half-sine pulse has a continuum of spectral content beginning at zero and then rolling-off as the frequency increases. There are also certain frequencies where the magnitude drops to
zero. The magnitude represents the acceleration, but the absolute magnitude depends on the total duration include the quiet period after the pulse is finished. The post-pulse duration was 10 seconds
in the above example, for a total of 10.011 seconds.
Figure 4.2. SDOF System Response to Half-Sine Shock, 10 Hz
The response in the above figure has an absolute peak value less than the peak input. This is an isolation case. The response positive and negative peaks occur after the base input pulse is over.
Figure 4.3. SDOF System Response to Half-Sine Shock, 75 Hz
The response in Figure 1.9 has an absolute peak value that is 1.65 times the peak input. This is resonant amplification case. The absolute peak response occurs during the base input pulse.
Figure 4.4. SDOF System Response to Half-Sine Shock, 400 Hz
The response converges to the base input as the natural frequency becomes increasingly high. This becomes a unity gain case. The system is considered as hard-mounted.
The peak results from the three cases are shown in Table 1.2. The calculation can be repeated for a family of natural frequencies. The peak acceleration results can then be plotted as a shock
response spectrum (SRS) as shown in Figure 1.11. The peak relative displacement values can likewise be plotted as an SRS as shown in Figure 1.12.
Figure 4.5. Acceleration SRS, 50 G, 11 msec Half-Sine Pulse
The two curves in Figure 1.11 contain the coordinates in Table 1.2.
The initial slope of each SRS curve is 6 dB/octave indicating a constant velocity line. The curves also indicate that the peak response can be lowered by decreasing the natural frequency. A low
natural frequency could be achieved for a piece of equipment by mounting it via soft, elastomeric isolator bushings or grommets. But a lower natural frequency leads to a higher relative displacement
as shown in Figure 1.12.
Table 4.1. Summary of Peak Response, 50 G, 11 msec, Half-Sine Base Input
Natural Frequency (Hz) Peak Positive (G) Absolute Value of Peak Negative (G)
10 20.3 17.3
75 82.5 65
400 51.8 4.38
Figure 4.6. Relative Displacement SRS, 50 G, 11 msec Half-Sine Pulse
The curves in the above figure could be used for designing isolator mounts for a component. The isolators must be able to take up the relative displacement without bottoming or topping out. There
must also be enough clearance and sway space around the component.
Figure 4.7. Acceleration SRS, 50 G, 11 msec Terminal Sawtooth Pulse
The positive and negative SRS curves are reasonably close for the terminal sawtooth pulse shown in Figure 1.13. In contrast, the positive and negative SRS curves for the half-sine pulse in Figure
1.11 diverge as the natural frequency increases above 80 Hz. The terminal sawtooth pulse is thus usually preferred over the half-sine pulse for classical shock testing.
Another means of visualizing the SRS concept is given in Figure 1.14.
Figure 4.8. Half-Sine Shock Applied as Base Input to Independent SDOF Systems
The systems are arranged in order of ascending natural frequency from left to right and subjected to a common half-sine base input. The Soft-mounted system on the left has high spring relative
deflection, but its mass remains nearly stationary. The Hard-mounted system on the right has low spring relative deflection, and its mass tracks the input with near unity gain. The Middle system
ultimately has high deflection for both its mass and spring. The peak positive and negative responses of each system are plotted as a function of natural frequency in the shock response spectrum.
SECTION 5
Response to Arbitrary Excitation
Recall from Section 12.3 that the response of a single-degree-of-freedom system to base excitation can be expressed in terms of a second order, ordinary differential equation for the relative
Equation (1.7) can be solved via Laplace transforms if the base acceleration is deterministic such as a half-sine pulse. A convolution integral is needed if the excitation varies arbitrary with time.
The convolution integral is computationally inefficient, however. An alternative is to use the Smallwood ramp invariant digital recursive filtering relationship [17], [19].
Again, the recursive filtering algorithm is fast and is the numerical engine used in almost all shock response spectrum software. It is also accurate assuming that the data has a sufficiently high
sample rate and is free from aliasing. One limitation is that it requires a constant time step.
The equation for the absolute acceleration is
The damped natural frequency is
The digital recursive filtering relationship for relative displacement is omitted for brevity but is available in Reference [19].
The relationship in equation (1.8) is recursive because the response at the current time depends on the two previous responses, which are the first two terms on the righthand side of the equation.
This is a feedback loop in terms of control theory. The relationship is filtering because the energy at and near the natural frequency is amplified whereas higher frequency energy above √2 times the
natural frequency is attenuated. See Figure 12.14.
Shock Response Spectrum Test Specification Objective
Figure 5.1. Mid-Field Pyrotechnic Shock Time History
Consider the measured mid-field shock time history in the above figure as taken from Reference [29]. The time history would be essentially impossible to reproduce in a test lab given that it is a
high-frequency, high-amplitude complex, oscillation pulse. The aerospace practice instead is to derive an SRS to represent the damage potential of the shock event. The test conductor may then use an
alternate pulse to satisfy the SRS specification within reasonable tolerance bands. This is an indirect method of achieving the test goal. There are some limitations to this approach. One is that the
test item is assumed to be linear. Another is that it behaves as a single-degree-of-freedom system. Nevertheless, this method is used in aerospace, military and earthquake engineering fields, for
both analysis and testing.
Figure 5.2. Mid-Field Shock Response Spectrum & Envelope
The customary approach is to draw a conservative envelope over the measured SRS. The ramp-plateau format is the most common, although there are variations. The enveloping process shown in the above
vibration is very conservative in the mid frequency domain. An additional dB uncertainty factor may be needed to develop the envelope into a test specification, given that the SRS envelope is derived
from a single time history. Industry standards, such as Reference [30], give guidelines for the dB factor.
Seismic Waveforms
Figure 6.1. Mother Earth
The Earth experiences seismic vibration. The fundamental natural frequency of the Earth is 309.286 micro Hertz, equivalent to a period of approximately 54 minutes [1]. The structure of Earth's deep
interior cannot be studied directly. But geologists use seismic waves to determine the depths of layers of molten and semi-molten material within Earth.
Figure 6.2. P-Wave
The primary wave, or P-wave, is a body wave that can propagate through the Earth’s core. This wave can also travel through water. The P-wave is also a sound wave. It thus has longitudinal motion.
Note that the P-wave is the fastest of the four waveforms.
Figure 6.3. S-Wave
The secondary wave, or S-wave, is a shear wave. It is a type of body wave. The S-wave produces an amplitude disturbance that is at right angles to the direction of propagation. Note that water cannot
withstand a shear force. S-waves thus do not propagate in water.
Figure 6.4. Love Wave
Love waves are shearing horizontal waves. The motion of a Love wave is similar to the motion of a secondary wave except that Love wave only travel along the surface of the Earth. Love waves do not
propagate in water.
Figure 6.5. Rayleigh Wave, Retrograde
Rayleigh waves produce retrograde elliptical motion. The ground motion is thus both horizontal and vertical. The motion of Rayleigh waves is similar to the motion of ocean waves in Figure 1.22 except
that ocean waves are prograde. Rayleigh waves resulting from airborne acoustical sources may either be prograde or retrograde per Reference [31]. In some cases, the motion may begin as prograde and
then switch to retrograde. Airborne acoustical sources include above ground explosions and rocket liftoff events.
Figure 6.6. Ocean Surface Wave Particle Motion, Prograde
The Love and Rayleigh waves are both surface waves. These are the two seismic waveforms which can cause the most damage to building, bridges and other structures.
As an aside, seismic and volcanic activity at the ocean floor generates a water-borne longitudinal wave called a T-wave, or tertiary wave. These waves propagate in the ocean’s SOFAR channel, which is
centered on the depth where the cumulative effect of temperature and water pressure combine to create the region of minimum sound speed in the water column. SOFAR is short for “Sound Fixing and
Ranging channel.” These T-waves may be converted to ground-borne P or S-waves when they reach the shore. Acoustic waves travel at 1500 m/s in the ocean whereas seismic P and S-waves travel at
velocities from 2000 to 7000 m/s in the crust.
Seismic Response Spectrum Method
Professors Theodore von Kármán and Maurice Biot were very active in the early 1930s in the theoretical dynamics aspects of what would later become known as the response spectrum method in earthquake
engineering. Biot proposed that rather than being concerned with the shape of the input time history, engineers should instead use a method describing the response of systems to those shock pulses.
The emphasis should instead be on the effect, as represented by the response of a series of single-degree-of-freedom oscillators, similar to that previously shown for the case of a half-sine input in
Figure 18.14.
Practical use of the response spectrum method had to wait until the 1970s due to the intricacy of the response calculation for complex, oscillating pulses which required digital computers. Time was
also needed to establish and publicize databases of strong motion acceleration time histories.
The response spectrum method was adopted for pyrotechnic shock in the aerospace industry and renamed as shock response spectrum.
El Centro Earthquake
Figure 6.7. El Centro, Imperial Valley Earthquake Damage
Nine people were killed by the May 1940 Imperial Valley earthquake. At Imperial, 80 percent of the buildings were damaged to some degree. In the business district of Brawley, all structures were
damaged, and about 50 percent had to be condemned. The shock caused 40 miles of surface faulting on the Imperial Fault, part of the San Andreas system in southern California. Total damage has been
estimated at about $6 million. The magnitude was 7.1. The was the first major earthquake for which strong motion acceleration data was obtained that could be used for engineering purposes.
Figure 6.8. El Centro Earthquake, Triaxial Time History
The highest excitation is in the North-South axis, parallel to the ground.
Figure 6.9. El Centro Earthquake North-South SRS, Three Damping Cases
The acceleration levels reached 1.5 G for the 1% damping curve. Recall that large civil engineering structures can have nonlinear damping. The damping values tend to increase as the base excitation
levels increase as shown for the Transamerica Title Building in Section 10.7.
Figure 6.10. El Centro Earthquake Tripartite SRS
Seismic SRS curves are often plotted in tripartite format which displays relative displacement, pseudo velocity and acceleration responses all on the same graph. The pseudo velocity
Stress can be calculated from pseudo velocity using the methods in Section 19. The acceleration curve might be the most important design metric for equipment mounted inside a building. The relative
displacement might be the most important concern for analyzing the foundational strength. The curves also show design tradeoffs. Lowering the building’s natural frequency below 1 Hz reduces the
acceleration response but increases the relative displacement. Note that Q=10 is the same as 5% damping.
Golden Gate Bridge
Figure 6.11. Golden Gate Suspension Bridge, San Francisco, California
In addition to traffic loading, the Golden Gate Bridge must withstand the following environments:
• 1. Earthquakes, primarily originating on the San Andreas and Hayward faults
• 2. Winds of up to 70 miles per hour
• 3. Strong ocean currents
The Golden Gate Bridge has performed well in all earthquakes to date, including the 1989 Loma Prieta Earthquake. Several phases of seismic retrofitting have been performed since the initial
construction. The bridge’s fundamental mode is a transverse mode with a natural frequency of 0.055 Hz, with a period of 18.2 seconds
Note that California Department of Transportation (CALTRANS) standards require bridges to withstand an equivalent static earthquake force (EQ) of 2.0 G. This level is plausibly derived as a
conservative envelope of the El Centro SRS curves in Figure 1.25.
Vandenberg AFB
Figure 6.12. Rocket Launch from Vandenberg AFB, California
The Vandenberg launch site is near the San Andreas fault system. The vehicle is mounted on the pad as a tall cantilever beam. The vehicle must be analyzed to verify that it can withstand a major
seismic event. The vehicle may be mounted to the pad for only two weeks prior to launch. The odds of an earthquake occurring to that time window are miniscule. But the launch vehicle and payload
together may cost well over $1 billion. The risk thus necessitates the analysis. Areas of concern are the loads imparted at the launch vehicle’s joints and to the payload.
Figure 6.13. NASA SRS Curves for Launch Vehicles at Vandenberg
SRS curves are given for three damping cases. The curves are taken from Reference [29]. The vehicle would typically be analyzed as a multi-degree-of-freedom system via a finite element model. Each
SRS curve could be applied to the model using a modal combination method. An alternative is to synthesize a time history to satisfy a selected SRS curve. The time history could then be applied to the
model via a modal transient analysis.
Seismic Testing
Figure 6.14. Electrical Power Generator Seismic Shock Test
The diesel generator is mounted onto a platform at the top of a shaker table which is located below the ground floor. This could be an emergency power generator for a hospital in an active seismic
zone. A video clip of the test is available on YouTube at: https://youtu.be/5uSqI7kSYE4
SECTION 7
Pyrotechnic Shock
Flight Events
Figure 7.1. Stage Separation Ground Test, Linear Shaped Charge
The plasma jet cuts the metal inducing severe mechanical shock energy, but the smoke and fire would not occur in the near-vacuum of space.
Launch vehicle avionics components must be designed and tested to withstand pyrotechnic shock from various stage, fairing and payload separation events that are initiated by explosive devices. Solid
rocket motor ignition is another source of pyrotechnic shock. The source shock energy can reach thousands of acceleration Gs at frequencies up to and beyond 100 kHz. The corresponding velocities can
reach a few hundred in/sec, well above the severity thresholds in the empirical rules-of-thumb in Section 19.4.
Empirical source shock levels for a variety of devices are given in References [29], [32], [33]. These levels are intended only as preliminary estimates for new launch vehicle designs. The estimates
should be replaced by ground test data once the launch vehicle hardware is built and tested.
Pyrotechnic Device Photo Gallery
Figure 7.2. Metal Clad Linear Shaped Charge
The chevron focuses a pyrotechnic plasma jet at the launch vehicle’s separation plane material. Severe shock levels are generated as a byproduct.
Figure 7.3. Frangible Joint
A frangible joint may be used for stage or fairing separation. The key components of a frangible joint:
• Mild Detonating Fuse (MDF)
• Explosive confinement tube
• Separable structural element
• Initiation manifolds
• Attachment hardware
The hot gas pressure from the MDF detonation cause the internal tube to expand and fracture the joint.
Figure 7.4. Frangible Nuts, Hold Down Posts
The purpose of the nuts was to hold the SRBs in place against wind and ground-borne excitation. The nuts were separated just before liftoff.
Figure 7.5. Clamp Band with Pyrotechnic Bolt-Cutters
(image courtesy of European space agency)
Clamp bands are often used for payload separation from launch vehicle adapters. They are also commonly used for stage separation in suborbital launch vehicles, similar to the one in Figure 4.1. A
pyrotechnic bolt-cutter uses an explosive charge to drive a chisel blade to cut the band segments’ connecting bolt. The cutters produce some shock energy, but much of the shock is due to the sudden
release of strain energy in the preloaded clamp band. This action can excite a ring mode in the radial axis. Recall Section 7.3.
The total clamp band release shock is significantly less than linear shaped charge and frangible joint. Note that an analysis must be performed to verify that no gapping will occur in the between the
band and the joint as the vehicle undergoes bending mode vibration during powered flight.
SECTION 8
Pyrotechnic Shock Data
Initial SRS Slopes
Figure 8.1. Expected Pyrotechnic SRS Initial Slope Limits
Near-field pyrotechnic shock can be difficult to measure accurately. The accelerometer data may have a baseline shift or spurious low frequency transient. This error could be a result of the
accelerometer’s own natural frequency being excited or to some other problem. Aerospace pyrotechnic shock SRS specifications usually begin at 100 Hz due to the difficult in accurately measuring
low-amplitude, low-frequency shock, while simultaneously measuring high-amplitude, high-frequency shock.
There are several methods for checking whether the data is acceptable. One is to the check the initial slopes of both the positive and negative SRS curves. Each should have an overall trend of 6 to
12 dB/octave. The actual slopes may have local peaks and dips due to low frequency resonances. A 12 dB/octave slope represents constant displacement and zero net velocity change. A 6 dB/octave slope
indicates constant velocity. Recall the slope formulas in Section 16.1.2.
A second method for checking data accuracy is to verify that the positive and negative SRS curves are within about 3 dB of each other across the entire natural frequency domain. A third method is to
integrate the acceleration time history to velocity. The velocity time history should oscillate in a stable manner about the zero baselines.
These verification goals are challenging to meet with near-field shock measurements of high-energy source shock, such as that from linear shaped charge and frangible joints. In practice, some
high-pass filtering or spurious trend removal may be necessary. There is no one right way to perform this “data surgery.” It is a matter of engineering judgment.
Re-entry Vehicle Separation, Flight Data
Figure 8.2. Re-Entry Vehicle, Separation Shock, Near-Field Measurement, Time History
The source device was linear shaped charge.
Figure 8.3. Re-Entry Vehicle, Separation Shock, Near-Field Measurement, SRS
The SRS reached 20,040 G at 2400 Hz, which is an extreme level per the rule-of-thumb Section 19.4.
SECTION 9
Water Impact Shock
Figure 9.1. Solid Rocket Booster Water Impact Shock
Each of the two Space Shuttle Boosters was recovered and refurbished after every flight. Each booster contained sensitive avionics components which underwent shock testing for the water impact event.
Figure 9.2. Solid Rocket Booster Water Impact Shock, Time History
The data is from the STS-6 Mission. The accelerometer was mounted at the forward end of the booster adjacent to a large IEA avionics box. This was the worst-case shock event for this component.
Figure 9.3. Solid Rocket Booster Water Impact Shock, Tripartite SRS
The maximum acceleration response is 257 G at 85 Hz. The maximum pseudo velocity response is 201 in/sec at 76 Hz, which is severe per the rule-of-thumb Section 1.4.
SECTION 10
Shock Response Spectrum Synthesis
Synthesis Objectives
Consider an SRS specification for a component, subsystem or a large structure. The article should be tested, if possible, to verify that it can withstand the shock environment. Certain shock tests
can be performed on a shaker table, like the generator test in Figure 18.30. This requires synthesizing an acceleration time history to satisfy the SRS. The net velocity and net displacement must
each be zero for this test. These requirements can be met with a certain type of wavelet series, where the individual wavelets may be nonorthogonal. The resulting wavelet time history should meet the
SRS within reasonable tolerance bands, but it may not “resemble” the expected time history which is a limitation of this method. Another requirement is that the peak acceleration, velocity and
displacement values must be within the shaker table’s capabilities.
An innovative method for meeting the SRS with a wavelet series that resembles one or more measured shock time histories is given in Reference [34].
A synthesized time history can also be used for modal transient analysis. This could be done for articles which are too large or heavy for shaker tables. This analysis can also be done on small
components prior to shock testing to determine whether they will pass the test. Or the analysis could be done in support of isolator mounting design. There is again a need for the synthesized time
history to have net velocity and net displacement which are each zero, to maintain numerical stability in stress calculations from relative displacement values.
Damped-sines can be used for modal transient analysis where the goal is to meet the SRS with a time history that plausibly resembles the expected field shock event. But damped-sines do not meet the
desired zero net velocity and displacement goals. The workaround is to first synthesize a damped-sine series to meet the SRS and then reconstruct it via a wavelet series, in Rube Goldberg fashion.
Wavelet and damped-sine synthesis are shown in the following examples.
Wavelet Synthesis
Wavelet Equation
A wavelet is a sine function modulated by another sine function. The equation for an individual wavelet is
The total acceleration at time
Figure 10.1. Sample Wavelet
A sample, individual wavelet is shown in the above figure. This wavelet was a component of a previous analysis for an aerospace project.
Wavelet Synthesis Example
Consider the specification: MIL-STD-810E, Method 516.4, Crash Hazard for Ground Equipment.
Table 10.1. SRS Q=10, Crash Hazard
Natural Frequency (Hz) Peak Accel (G)
10 9.4
Synthesize a series of wavelets as a base input time history for a shaker shock test to meet the Crash Hazard SRS. The goals are:
• Satisfy the SRS specification
• Minimize the displacement, velocity and acceleration of the base input
The synthesis steps are shown in the following table.
Table 10.2. Wavelet Synthesis Steps
Step Description
1 Generate a random amplitude, delay, and half-sine number for each wavelet. Constrain the half-sine number to be odd. These parameters form a wavelet table.
2 Synthesize an acceleration time history from the wavelet table.
3 Calculate the shock response spectrum of the synthesis.
4 Compare the shock response spectrum of the synthesis to the specification. Form a scale factor for each frequency.
5 Scale the wavelet amplitudes.
6 Generate a revised acceleration time history.
7 Repeat steps 3 through 6 until the SRS error is minimized or an iteration limit is reached.
Calculate the final shock response spectrum error.
8 Also calculate the peak acceleration values.
Integrate the signal to obtain velocity, and then again to obtain displacement. Calculate the peak velocity and displacement values.
9 Repeat steps 1 through 8 many times.
10 Choose the waveform which gives the lowest combination of SRS error, acceleration, velocity and displacement.
The resulting time history and SRS are shown in Figure 1.43 and Figure 1.44, respectively.
Figure 10.2. Crash Hazard Time History Synthesis
The acceleration time history has a reverse sine sweep character. It is an efficient and optimized waveform for a shaker shock test, and it satisfies the SRS as shown in the next figure. A drawback
is that it does not resemble an actual crash shock time history.
Figure 10.3. Crash Hazard SRS
The positive and negative curves are from the synthesized waveform. The tolerance bands are set at +3 dB.
Damped-Sine Synthesis
Damped-Sine Equation
The equation for an individual damped sinusoid is
The total acceleration at time
Damped-Sine Example
Consider the following specification which could represent a stage separation shock level as some location in a launch vehicle. A modal transient finite element analysis is to be performed on a
component to verify that the component will pass its eventual shock test. The immediate task is to synthesize a time history to satisfy the SRS. The time history should “resemble” an actual
pyrotechnic shock pulse.
Note that pyrotechnic SRS specifications typically begin at 100 Hz. The author’s rule-of-thumb is to extrapolate the specification down to 10 Hz in case there are any component modes between 10 and
100 Hz. This guideline is also to approximate the actual shock event which should have an initial ramp somewhere between 6 and 12 dB/octave.
Table 10.3. SRS Q=10, Stage Separation
Natural Frequency (Hz) Peak Accel (G)
10,000 2000
The specification has an initial slope of 6 dB/octave. The synthesis steps are shown in the following table.
Table 10.4. Damped-Sine Synthesis Steps
Step Description
1 Generate random values for the following for each damped sinusoid: amplitude, damping ratio and delay. The natural frequencies are taken in one-twelfth octave steps.
2 Synthesize an acceleration time history from the randomly generated parameters.
3 Calculate the shock response spectrum of the synthesis.
4 Compare the shock response spectrum of the synthesis to the specification. Form a scale factor for each frequency.
5 Scale the amplitudes of the damped sine components.
6 Generate a revised acceleration time history.
7 Repeat steps 3 through 6 as the inner loop until the SRS error diverges.
8 Repeat steps 1 through 7 as the outer loop until an iteration limit is reached.
9 Choose the waveform which meets the specified SRS with the least error.
10 Perform wavelet reconstruction of the acceleration time history so that velocity and displacement will each have net values of zero.
The resulting time history and SRS are shown in Figure 1.45 and Figure 1.46, respectively.
Figure 10.4. Stage Separation Time History Synthesis
The acceleration time history somewhat resembles a mid or far-field pyrotechnic shock. The velocity and displacement time histories each have a stable oscillation about their respective baselines.
Figure 10.5. Stage Separation SRS
The positive and negative curves are from the damped-sine synthesis.
Modal Transient Finite Element Analysis for Uniform Base Excitation
Consider a rectangular plate mounted to a base at each of its four corners. The plate is to be subject to uniform seismic excitation. There are two methods to apply the base excitation in a finite
element analysis, as shown in Figure 1.47 and Figure 1.48.
Figure 10.6. Direct Enforce Acceleration Method
The direct enforcement method is computationally intensive, requiring matrix transformations and a matrix inversion as shown in Reference [35].
Figure 10.7. Seismic Mass Method
The seismic mass is chosen to be several orders of magnitude higher in mass than the plate. An equivalent force is calculated and apply to the seismic mass to excite the desired acceleration at each
of the plate’s corners. This method adds a degree-of-freedom to the plate system resulting in a rigid-body mode at zero frequency. But the remaining natural frequencies and mode shapes should be the
same as if the plate were mounted normally to its joining structure. The author’s experience is that the seismic mass method is faster and more accurate and reliable than the direct enforced method.
This method was introduced in Section 12.3.3.
Shock Fields
The near-field environment is dominated by direct stress wave propagation from the source. Peak accelerations in excess of 5000 G occur in the time domain with a frequency content extending beyond
100 kHz. The near-field usually includes structural locations within approximately 15 cm of the source for severe devices such as linear shaped charge and frangible joint. No shock-sensitive hardware
should be mounted where it would be exposed to a near-field environment.
The mid-field environment is characterized by a combination of wave propagation and structural resonances. Peak accelerations may occur between 1000 and 5000 G, with substantial spectral content
above 10 kHz. The mid-field for intense sources usually includes structural locations between approximately 15 and 60 cm of the source, unless there are intervening structural discontinuities.
The far-field environment is dominated by structural resonances. The peak accelerations tend to fall below 2000 G, and most of the spectral content below 10 kHz. The far-field distances occur outside
the mid-field. The typical far-field SRS has a knee frequency corresponding to the dominant frequency response. The knee frequency is the frequency at which the initial ramp reaches the plateau in
the log-log SRS plot.
Joint & Distance Attenuation
The source shock energy is attenuated by intervening material and joints as it propagates from the near-field to the far-field. Empirical distance and joint attenuation factors for the SRS reduction
are given in References [29], [32], [33]. A typical attenuation curve from [32] is given in Figure 18.49, which assumes an input source shock SRS consisting of a ramp and plateau in log-log format.
Such curves should be used with caution given that the attenuation is highly dependent on damping and structural details. The curves can be used for preliminary estimates for new launch vehicle
designs, but ground separation tests are still needed for the actual launch vehicle hardware. These tests are needed to measure the source shock as well as the levels at key component mounting
Figure 10.8. Shock Response Spectrum Versus Distance from Pyrotechnic Shock Source
Shock Mitigation
Figure 10.9. Sensor Electronics, Wire Rope Isolators
(image courtesy of NASA/JPL)
The source shock energy is attenuated as it propagates to avionics mounting locations through the launch vehicle’s material and joints. The input shock to a component can be mitigated by mounting the
component as far away from the source device as possible. Another effective attenuation technique is to mount the component via elastomeric bushings or wire rope isolators. The NASA Mars Science
Laboratory Sensor Support Electronics unit is mounted on vibration isolators in Figure 1.50.
Figure 10.10. SCUD-B Avionics Component Isolation
The bushings are made from some type of rubber or elastomeric compound. The bushings provide damping, but their main benefit is to lower the natural frequency of the system. The isolators thus
attenuate the shock and vibration energy which flows from the instrument shelf into the avionics component.
Pyrotechnic Shock Testing Methods
Figure 10.11. Near-Field Shock Simulation using a Plate
The test component is mounted on other side of plate. The source device is a textile explosive cord with a core load of 50 gr/ft (PETN explosive). Up to 50 ft of Detonating Cord has been used for
some high G tests. The maximum frequency of shock energy is unknown so analog anti-aliasing filters are needed for the accelerometer measurements per the guidelines in Section 14.
Note that a component may be mounted in the mid-field shock region of a launch vehicle. The SRS test level derivation for this zone may include a significant dB factor for uncertainty or as a
qualification margin. This conservatism may require a near-field-type shock test for a component that is actually located in a mid-field zone. This situation can also occur for components mounted in
Figure 10.12. NASA JPL Tunable Shock Beam
The NASA/JPL Environmental Test Laboratory developed and built a tunable beam shock test bench based on a design from Sandia National Laboratory many years ago. The excitation is provided by a
projectile driven by gas pressure. The beam is used to achieve SRS specifications, typically consisting of a ramp and a plateau in log-log format. The intersection between these two lines is referred
to as the “knee frequency.” The beam span can be varied to meet a given knee frequency. The high frequency shock response is controlled by damping material.
Shock Failure Modes
Figure 10.13. Sensitive Electronic Parts
Pyrotechnic shock can cause crystal oscillators to shatter. Large components such as DC-DC converters can detached from circuit boards. In addition, mechanical relays can experience chatter or
Figure 10.14. Shock Test Case History, Adhesive & Solder Joint Failure
The image shows adhesive failure and rupture of solder joint after a stringent shock test. A large deflection of the PCB resulting from an insufficient support/reinforcement of the PCB combined with
high shock loads can lead to these failures. Staking is needed for parts weighing more than 5 grams.
Figure 10.15. Shock Test Case History, Lead Wire Failure
The image shows a sheared lead between solder joint and winding of coil. The lacing cord was insufficient by itself. The lacing should be augmented by staking with an adhesive. | {"url":"https://endaq.com/pages/shock-testing-analysis","timestamp":"2024-11-14T11:53:28Z","content_type":"text/html","content_length":"212262","record_id":"<urn:uuid:0222b1a5-2960-4dd4-abbb-94d9a23028fd>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00032.warc.gz"} |
How to Calculate Degrees of Freedom in Excel?
How to Calculate Degrees of Freedom in Excel?
Are you trying to understand how to calculate degrees of freedom in Excel? If so, you’re in the right place. In this article, we’ll walk you through the steps necessary to calculate degrees of
freedom in Excel, providing detailed instructions and examples along the way. Whether you’re a student, a business owner, or a data analyst, understanding how to calculate degrees of freedom in Excel
can help you save time and make the most of your data. So let’s get started!
How to Calculate Degrees of Freedom in Excel?
• Open Microsoft Excel and enter the data into the spreadsheet.
• In the cell under the last entry, enter the formula =SQRT(COUNT(A2:A8))-1.
• Press the Enter key and the degrees of freedom will be calculated.
• The result should be the same as the number of entries minus one.
Measure Degrees of Freedom with Excel
Degrees of freedom (DF) is an important concept in statistics that measures the number of values that can vary in a given population. It is used to determine the reliability of a statistical model or
test. In Excel, you can calculate DF using an equation and the data that you have entered into the spreadsheet. This guide will explain how to calculate DF in Excel and provide some examples.
Understanding Degrees of Freedom
DF is the number of values that can vary in a given population. For example, if you have a population of 10 people and want to calculate DF, the number of values that can vary is 10, so DF = 10. In
statistical models, DF is used to calculate the reliability of the model or test. The more DF a model has, the more reliable it is.
Calculating Degrees of Freedom in Excel
To calculate DF in Excel, you need to use an equation. The equation for calculating DF is: DF = N – 1, where N is the number of values in the population. For example, if you have a population of 10
people, DF = 10 – 1 = 9.
Examples of Calculating Degrees of Freedom in Excel
Now that you understand the equation for calculating DF in Excel, let’s look at some examples. In the first example, you have a population of 10 people and you want to calculate DF. Using the
equation above, DF = 10 – 1 = 9.
Using Degrees of Freedom in Excel
Once you have calculated DF in Excel, you can use it to determine the reliability of your statistical model or test. The higher the DF, the more reliable the model or test is. For example, if you
have a population of 10 people and you calculate DF to be 9, the model or test is more reliable than if you had calculated DF to be 8.
Using the Results of Degrees of Freedom in Excel
Once you have calculated DF in Excel and determined the reliability of the model or test, you can use the results to make decisions about the validity of the model or test. If the model or test is
reliable, you can use it to make decisions or draw conclusions. If the model or test is not reliable, you should not use it to make decisions or draw conclusions.
Degrees of freedom (DF) is an important concept in statistics that measures the number of values that can vary in a given population. In Excel, you can calculate DF using an equation and the data
that you have entered into the spreadsheet. Once you have calculated DF, you can use it to determine the reliability of your statistical model or test. If the model or test is reliable, you can use
it to make decisions or draw conclusions. If the model or test is not reliable, you should not use it to make decisions or draw conclusions.
Frequently Asked Questions
What is Degrees of Freedom?
Degrees of freedom is a statistical concept used in hypothesis testing and in analysis of variance. It is the number of observations that are free to vary, once the number of parameters required to
estimate them have been set. For example, if you calculate the average of 5 numbers, the degrees of freedom is 4 since one parameter (the average) has already been set.
What is the Formula for Degrees of Freedom?
The formula for degrees of freedom is DF = n – k, where n is the number of observations and k is the number of parameters required to estimate them. For example, if you have 10 observations and 2
parameters, the degrees of freedom would be 8 (10 – 2).
How to Calculate Degrees of Freedom in Excel?
To calculate degrees of freedom in Excel, you can use the formula DF = n – k, where n is the number of observations and k is the number of parameters required to estimate them. For example, if you
have 10 observations and 2 parameters, the formula in Excel would be =10-2, which would return 8 as the result.
What are Some Tips to Calculate Degrees of Freedom in Excel?
Some tips to calculate degrees of freedom in Excel include:
1. Make sure that the number of parameters is greater than the number of observations.
2. Use the formula DF = n – k, where n is the number of observations and k is the number of parameters.
3. Make sure that all the parameters are correctly entered in the formula.
4. Check the result of the calculation to make sure it is correct.
What is the Significance of Calculating Degrees of Freedom in Excel?
Calculating degrees of freedom in Excel is important as it helps in understanding the accuracy of the data. It is used to determine the reliability of a statistical analysis. It can also be used to
identify the number of observations that are free to vary, once the number of parameters required to estimate them have been set.
Are there any Limitations to Calculating Degrees of Freedom in Excel?
Yes, there are some limitations to calculating degrees of freedom in Excel. Since the formula used is DF = n – k, where n is the number of observations and k is the number of parameters, there is a
limit to the number of observations that can be used. Additionally, the parameters used in the formula must be correctly entered in order to get an accurate result.
Degrees of Freedom Video
Calculating degrees of freedom in Excel can be a useful tool for many different types of projects. With the help of Excel’s built-in functions and formulas, you can easily calculate degrees of
freedom in a matter of seconds. Not only that, but you can also use Excel to help you find the probability of a given event occurring and to analyze data trends over time. With the help of the many
tools available in Excel, you can make sure your data is accurate and up to date. | {"url":"https://keys.direct/blogs/blog/how-to-calculate-degrees-of-freedom-in-excel","timestamp":"2024-11-12T20:38:10Z","content_type":"text/html","content_length":"351104","record_id":"<urn:uuid:8aa0afb5-816f-46fe-8ecd-fcf615c454d6>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00289.warc.gz"} |
WG21 N1567=04-0007
CRITIQUE OF WG14/N1016
P.J. Plauger
Dinkumware, Ltd.
Both WG14 and WG21 have accepted WG14/N1016 as the basis for parallel
non-normative Technical Reports, adding decimal floating-point arithmetic
to C and C++. The decimal formats are based on work done at IBM plus
current standardization work within IEEE -- a revision of IEEE 754, the
widely adopted standard for binary floating-point arithmetic. The revised
standard IEEE 754R will describe both binary and decimal formats.
N1016 proposes adding three more basic types to C (and C++), for
decimal floating-point to coexist alongside whatever an implementation
currently uses for float, double, and long double. That also involves:
-- adding literal formats for the new types, such as 1.0DF
-- adding promotion and conversion rules between the new types and
existing basic types
-- adding macros to <fenv.h> to describe new rounding modes and exceptions
-- adding a new header <decfloat.h> to describe properties of the new types
a la <float.h>
-- adding macros to <math.h> to describe huge values and NaN for the new types
-- adding versions of all the math functions in <math.h> for the new types
-- adding half a dozen new functions to <math.h> to perform operations
particular to decimal floating-point on the new types
-- adding new conversion specifications, such as %GLD, to the formatted
input/output conversions in <stdio.h> and <wchar.h>
-- adding strto* functions to <stdlib.h> for the new types
-- adding wcsto* functions to <wchar.h> for the new types
-- adding the relevant macros to <tgmath.h> for the new functions added
to <math.h>
There is no provision for new complex types in C based on the decimal
floating-point types.
Adding three new types is clearly a major change to C. C++ could avoid
the proliferation of names by overloading existing names, but it shares
all the other problems. It is not at all clear to me that much need
exists for having two sets of floating-point types in the same program.
It is certainly not clear to me that whatever need exists is worth the
high cost of adding all these types.
Lower-cost alternatives exist. Probably the cheapest is simply to define
a binding to IEEE 754R, much like the current C Annex F for IEC 60559 (the
international version of IEEE 754). If we do this, most of the above
list evaporates. But I believe we should still add a few items to the
C library, and the obvious analogs to C++, independent of whether the
binding is provided by the compiler. These involve:
-- adding macros to <fenv.h> to describe *some optional* new rounding
modes but *no* new exceptions
-- adding half a dozen new functions to <math.h> to perform operations
demonstrably useful for decimal floating-point but not unreasonable
even for other floating-point formats
-- adding the relevant macros to <tgmath.h> for just the half a dozen
new functions added to <math.h>
An implementation that chooses this option would also get C complex decimal
in the bargain.
Defining a binding does not oblige compiler vendors to switch to
decimal floating-point, however. For programmers to get access to this
new technology, they would have to wait for a vendor to feel motivated
to make major changes to both compiler and library. And it is not just
the vendor who pays a price:
-- The major cost for the unconcerned user is a slight reduction in
average precision, due mainly to the greater "wobble" inherent in decimal
vs. binary format. The major payoff is fewer surprises of the form
(10.0 * 0.1 != 1.0), and perhaps faster floating-point input/output.
-- A bigger cost falls on those programmers who need to convert often between
decimal floating-point and existing formats. These could be a rare and
special breed, but there might also be a distributed cost in performance
and complexity among users who have to access databases that store
floating-point results in non-decimal encoded formats. Experts have to
write the converters; many non-experts might have to use them.
So if all we do is define a binding, it could take a long time for
decimal floating-point to appear in the marketplace. But a reasonably
cheap alternative can mitigate this problem. Simply define a way to
add decimal floating-point as a pure bolt-on to C and C++ -- a library-only
package that can work with exsiting C and C++ compilers. For C, this
means adding one or more new headers that define three structured types
and a slew of functions for manipulating them. For C++, the solution
can look much like the existing standard header <complex> -- a template
class plus operators and functions that manipulate it, with the three
IEEE 754R decimal formats as explicit instantiations.
The major compromise in a bolt-on solution is the weaker integration
of decimal floating-point with the rest of the language and library.
C suffers most because it doesn't permit operator overloading for
user-defined types. (C++ seems to be doing just fine with complex as
a library-defined type.) The payoff is a much greater chance that vendors
will supply implementations sooner rather than later.
I believe the best thing is to do both of these lightweight things,
instead of adding three more floating-point types to the C and C++
languages. Implementing a TR of this form assures programmers that they
can reap the benefits of decimal floating-point one way or the other.
And such a TR provides a road map for how best to supply decimal
floating-point for both the short and long term.
A FEW DETAILED CRITICISMS OF N1016
The header <decfloat.h> should not define names that differ arbitrarily
from existing names in <float.h> (e.g. DEC32_COEFF_DIG).
The rounding modes in <fenv.h> have even more confusing differences
in naming. In C99, for example, "down" means "toward -infinity",
while in N1016 it means "toward zero". Here's a Rosetta Stone:
N1016 C99 (meaning)
FE_DEC_ROUND_DOWN FE_TOWARDZERO (toward zero)
FE_DEC_ROUND_HALF_EVEN FE_TONEAREST (ties to even)
FE_DEC_ROUND_CEILING FE_UPWARD (toward +inf)
FE_DEC_ROUND_FLOOR FE_DOWNWARD (toward -inf)
FE_DEC_ROUND_HALF_UP (ties away from zero)
FE_DEC_ROUND_HALF_DOWN (ties toward zero)
FE_DEC_ROUND_UP (away from zero)
Only the last two modes are optional in N1016.
Similarly, for floating-point exceptions, we have:
N1016 C99
FE_DEC_DIVISION_BY_ZERO FE_DIVBYZERO
FE_DEC_INVALID_OPERATION FE_INVALID
FE_DEC_INEXACT FE_INEXACT
FE_DEC_OVERFLOW FE_OVERFLOW
FE_DEC_UNDERFLOW FE_UNDERFLOW
There's no good reason for the differences in the first two lines.
Many of the functions cited in N1016 are *not* present in the
latest draft of IEEE 754R, as advertised. I had to hunt down
specifics at:
There's no reason for <math.h> to have HUGE_VALF, etc. followed
by DEC32_HUGE, etc. Once again, the names should not differ
It's also not clear why there should be a DEC_NAN and not a
DEC_INF (or DEC_INFINITY). Either both are easily generated
as inline expressions (0D/0D, 1D/0D, etc.) or neother is.
I favor defining both as macros (perhaps involving compiler
N1016 calls for the interesting function:
T divide_integerxx(T x, T y);
(where xx stands for d32, d64, or d128).
This generates an integer quotient only if it's exactly representable.
But N1016 doesn't require the corresponding remainder function. I suggest
loosely following the pattern of remquo and adding an optional pointer to
where to return the remainder:
T divide_integerxx(T x, T y, T *prem);
Or we could follow the pattern of remquo even closer and replace this
function with remainder_integerxx that returns the quotient on the side.
N1016 calls for the function:
T remainder_nearxx(T x, T y);
This has the same specification as the C99 remainder function.
It should share the same root name (e.g. remainderd32, after remainderf).
N1016 calls for the function:
T round_to_integerxx(T x, T y);
This has the same specification as the C99 rint function.
It should share the same root name.
N1016 calls for the interesting function:
T normalizexx(T x);
This shifts the coefficient right until the least-significant decimal
digit is nonzero, or it changes a zero value to canonical form. It's not
clear what should happen if such a shift would cause an overflow,
but that behavior must be specified. (It's also not clear what the
purpose of this function is in the best of circumstances, but maybe
I haven't read and played enough to understand.)
Finally, this function does exactly the opposite of what "normalize"
has meant as a term of art in floating-point for many decades. The
name suggests that the coefficient is shifted *left* until the
*most-significant" decimal digit is nonzero. (And I can even think
of uses for that operation.) Either the spec should change or the
N1016 calls for the function:
bool check_quantumxx(double x, double y);
This returns true only if x and y have the same exponent. There
is no spec for this function in N1016, but it is clearly the same
as the function same_quantum in the defining document from IBM.
I see no good reason for changing the name.
N1016 calls for the interesting function:
T quantizexx(T x, T y);
This changes x, as need be, to have the same exponent as y. It is,
in effect, a "round to N decimal places" function. In conjunction
with the rounding mode, it provides the much-touted "proper" decimal
rounding rules to match various government rules and commercial
practices. It's not entirely clear to me (yet) why a similar
function couldn't reap much the same benefit even when used with
binary floating-point. (I hasten to add that there are still other
good reasons for using decimal floating-point instead.) In any
event, I believe that it can and should be generalized so that
its application to binary floating-point makes equal sense.
Unfortunately, the function in its current form gets its parameter
N from the *decimal exponent* of y. That supports cute notation such
price = quantized32(price * (1DF + tax_rate), 1.00DF);
assuming that you find "1.00DF" more revealing than "2". But it
thus relies heavily on the ability to write literals (or generate
values) with a known decimal exponent. An earlier version evidently
required/permitted you to write the number of decimal digits instead.
I've long used an internal library function that truncates binary
floating-point values to a specified number of binary places (plus or
minus). I could see a real benefit in having both binary and
decimal versions of "quantize" that apply to floating-point
values of either base. But this particular form does not generalize
at all well.
N1016 does *not* call for several other functions that suggest themselves.
These include decimal equivalents of frexp and ldexp, and possibly an
exp10. I know from past experience that some of these are highly useful
in writing math functions; but I need more experience writing IEEE 754R
decimal floating-point math functions before I can make really
informed recommendations.
Nevertheless, the absence of such functions, and the lapses in the
functions presented in N1016, suggests to me the need for more
experience in using the decimal stuff from IEEE 754R as a general
floating-point data type before we freeze a TR of any sort, for either
C or C++. | {"url":"https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2004/n1567.htm","timestamp":"2024-11-14T17:48:55Z","content_type":"text/html","content_length":"12570","record_id":"<urn:uuid:77b08d04-3cfe-42a5-b1e7-c49930b0c887>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00483.warc.gz"} |
statistical inference hypothesis testing
Photo by Siora Photography on Unsplash. Example: You want to examine whether "brain gym" (a mixture of small mental and physical exercises) will improve your pupils' scores. 1.0 HYPOTHESIS TESTING.
Hypothesis testing addresses this random sampling “error” (i.e. The two branches of statistical inference are estimation and testing of hypothesis. Hypothesis testing is very important part of
statistical analysis. One of the main applications of frequentist statistics is the comparison of sample means and variances between one or more groups, known as statistical hypothesis testing. So
statistics helps us in arriving at the criterion for such decision is known as Testing … The confidence interval and hypothesis tests are carried out as the applications of the statistical
inference.It is used to make decisions of a population’s parameters, which are based on random sampling. In most cases, it may be easier to disprove a hypothesis than to verify it. STATISTICAL
INFERENCE . 4.2 (4,139 ratings) 5 stars. Reviews. In short: If the other side is not important or not possible. Statistics, Statistical Inference, Statistical Hypothesis Testing. In this blog post, I
explain why you need to use statistical hypothesis testing and help you navigate the essential terminology. Hypothesis Testing & Confidence Intervals are the main statistical methods by which we do
this but they are not the only methods. Estimation versus Hypothesis Testing Lead Author(s): George Howard, DrPH Inference; ESTIMATION. It helps to assess the relationship between the dependent and
independent variables. 56.97%. The aim of statistical inference is to predict the parameters of a population, based on a sample of data. Statistical Hypothesis Testing. Your null hypothesis …
Inferential Statistics is the process of examining the observed data (sample) in order to make conclusions about properties/parameter of a Population. 3 stars. Hypothesis testing is also referred to
as “Statistical Decision Making”. Unlike many introductory Statistics students, they had excellent math and computer skills and went on to master probability, random variables and the Central Limit
Theorem. Statistics Statistical Inference Overview Hypothesis Testing. These tests are also helpful in getting admission in different colleges and Universities. Whenever we observe data, we are
usually observing one or a few samples from a much larger population. Testing a Mean Value (µ) with σ 2 Known Testing A Mean Value (µ) with σ 2 Unknown Hypothesis Testing: Single Variance. An
important and time-saving skill is to ALWAYS do exploratory data analysis using dplyr and ggplot2 before thinking about running a hypothesis test. The purpose of statistical inference to estimate the
uncertainty or sample to sample variation. 23.43%. In statistical inference, there are three techniques in estimating the population parameter by utilizing sample information (statistics) as follows:
1) Point estimation 2) Confidence interval The basis of statistical inference is to determine (infer) an unknown parameter for a given population, based on a sample or subset of individuals belonging
to the mentioned population, and fundamented upon the frequency interpretation concept of probability. 2 stars. Statistical inference is a technique by which you can analyze the result and make
conclusions from the given data to the random variations. Statistical hypothesis testing plays an important role in the whole of statistics and in statistical inference. These tests are also helpful
in getting admission in different colleges and Universities. Statistical Inference - Confidence Interval & Hypothesis Testing 13 minute read Introduction. Learning statistics should be fun and
intuitive, at least that’s what I think. The present article describes the hypothesis tests or statistical significance tests most commonly used in … Chapter 9 Hypothesis Testing. In addition, the
concept of statistical significance was defined. Basically, the aspects studied by inference statistics are divided into estimation and hypothesis testing. What is an estimate? User Preferences ×
Font size. Hypothesis testing is a crucial procedure to perform when you want to make inferences about a population using a random sample. For example, you might be asked to test the hypothesis that
the mean weight gain of an women was more than 30 pounds. Question 2. 4.61%. Before we delve into hypothesis testing, it’s good to remember that there are cases where you need not perform a rigorous
statistical inference. 10.29%. Hypothesis Testing with Two Means: Population Variances Unknown but Assumed Equal Hypothesis testing provides a useful alternative. In Section 8.4, we showed you how to
construct confidence intervals.We first illustrated how to do this using dplyr data wrangling verbs and the rep_sample_n() function from Subsection 7.2.3 which we used as a virtual shovel. The
strategy for model selection in multivariate environment should have been explained with an example. Hypothesis testing and confidence intervals are the applications of the statistical inference.
Statistical Inference. Photo by Rana Sawalha on Unsplash. Multiple Choice Questions from Statistical Inference for the preparation of exams and different statistical job tests in Government/
Semi-Government or Private Organization sectors. Step 1: Null hypothesis is one of the common stumbling blocks–in order to make sense of your sample and have the one sample z test give you the right
information it must make sure written the null hypothesis and alternate hypothesis correctly. Null Hypothesis \(H_0\): The status quo that is assumed to be true. E. Inference Inference comes from the
verb “to infer” and is about the drawing of conclusions (both strong and weak) from data. The estimator is a function of the data arid so it is also a random variable. A hypothesis test is a
statistical test that assists in the decision to prove or disprove the statement. Statistics Inferences Based on Two Samples: Confidence Intervals & Tests of Statistical inference is a method of
making decisions about the parameters of a population, based on random sampling. Hypothesis Testing: Two Population Means with Variances Known. on descriptive statistics and interpreting graphs. 12
min read. View Hypothesis Testing ----- Two Sample Test 2.pptx from STAT 106 at University of the Fraser Valley. The present article describes the hypothesis tests or statistical significance tests …
AP. Statistics 101 – Inference and Hypothesis Testing (Part 1 of 3) Post author By Jason Oh; Post date June 15, 2019; As a generalist consultant you are unlikely to need any statistics for day-to-day
project work (there are specialists to call on for situations where it’s needed). Statistical Inference: Hypothesis Testing for Single Populations. Reset. Font family. 6b.5 - Statistical Inference -
Hypothesis Testing . Statistical Inference and Hypothesis Testing. In particular, we constructed confidence intervals by resampling with replacement by setting the replace = TRUE argument to the … 8
min read. Forecasting and Risk Modelling are two other options available among many. Hypothesis testing is a statistical procedure for testing whether chance is a plausible explanation of an
experimental finding. Statistics in Estimation; Repeated Estimates; Uncertainty in Estimation In statistics, we may divide statistical inference into two major part: one is estimation and another is
hypothesis testing.Before hypothesis testing we must know about hypothesis. With respect to hypothesis testing, there was a discussion of the null and alternative hypotheses, one- and two-tailed
hypothesis tests, and Type I and Type II errors in hypothesis testing. Cards. Inferential statistics encompasses the estimation of parameters and model predictions. The aim of statistical inference
is to predict the parameters of a population, based on a sample of data. In such cases, confidence interval estimation may not be the most suitable form in which to present the statistical
information. Key Questions. Introduction. In hypothesis testing, one form of statistical inference, a claim about a population is evaluated using data observed from a sample of the population. Now
that we’ve studied confidence intervals in Chapter 8, let’s study another commonly used method for statistical inference: hypothesis testing.Hypothesis tests allow us to take a sample of data from a
population and infer about the plausibility of competing hypotheses. In Chapter 15 we considered inference procedures that relied on estimation. Multiple Choice Questions from Statistical Inference
for the preparation of exams and different statistical job tests in Government/ Semi-Government or Private Organization sectors. 9.3 Conducting hypothesis tests. The conclusion of a statistical
inference is called a statistical proposition. It employs statistical techniques to arrive at decisions in certain situations where there is an element of uncertainty on the basis of sample, whose
size is fixed in advance. The researcher has a proposed hypothesis about a population characteristic and conducts a study to discover if it is reasonable, or, acceptable. 1 Mar 21, 2017. 4 stars.
4.68%. The data one observes will be different depending on which individuals of the population the sample captures. A hypothesis is a statement, inference or tentative explanation about a population
that can be tested by further investigation. Question 3. 1 star. That was the fourth part of the series, that explained hypothesis testing and hopefully it clarified your notion of the same by … This
book is a mathematically accessible and up-to-date introduction to the tools needed to address modern inference problems in engineering and data science, ideal for graduate students taking courses on
statistical inference and detection and estimation, … When would you use a one-sided alternative hypothesis? Answer: An estimator is a statistic that is used to infer the value of an unknown
population parameter in a statistical model. Statistics 101; by Karl - December 9, 2018 December 31, 2018 0. A A Mode. Introduction. In some situations, however, we want our statistical methods to
provide a more direct guide for decision making. For example, if we are looking at daily stock market returns for AAPL for last year, we are looking at only a small portion of the overall daily
returns. Inferential statistics encompasses the estimation of parameters and model predictions.. What is hypothesis testing? By the help of hypothesis testing many business problem can be solved
accurately. so we can define hypothesi as below-A statistical hypothesis is a statement about a population which we want to verify on the basis of information which contained in a sample. What is an
estimator? Is called statistical inference hypothesis testing statistical inference, you might be asked to test the hypothesis tests statistical. Inference procedures that relied on estimation
reasonable, or, acceptable plausible explanation of an experimental finding using dplyr ggplot2., statistical inference hypothesis testing test is a statistic that is used to infer the value of an
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\ ): George Howard, DrPH inference ; estimation situations, however we. Basically, the concept of statistical analysis do exploratory data analysis using dplyr ggplot2. ; by Karl - December 9, 2018
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And intuitive, at least that ’ s what I think on random “. | {"url":"https://www.marymorrissey.com/canada-goose-rhbnem/statistical-inference-hypothesis-testing-0bec39","timestamp":"2024-11-06T01:46:41Z","content_type":"text/html","content_length":"34794","record_id":"<urn:uuid:380a2583-b797-4139-b8df-1d218875be83>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00873.warc.gz"} |
The Rule of 72
The Rule of 72
The rule of 72 teaches that money can work for you or against you.
Albert Einstein said: “Compound interest is the greatest mathematical discovery of all time. It’s the eighth wonder of the world. He who understands it, earns it – he who doesn’t, pays it.”
The rule of 72 states that if you divide the interest rate by 72 you will get approximately how long it will take for your money to double. When you save or invest, money can work for you. Let’s look
at how this works and how your interest rate can make all the difference in what your return will be.
If on the day you were born your parents put $10,000 into a savings account, and that lump sum yielded a 1% fixed interest rate, you’d have around $20,000 waiting for you when you turned 72. If the
interest rate was 4%, you’d have $168,423. What do you think the value would be at 8%? Would it surprise you that you’d have $2,549,825. That’s 15 times more money by simply doubling the rate from 4%
to 8% over your lifetime.
The rule of 72 is the kind of principles that enable you to take advantage of The Wealth Wave.
When you borrow money, it works against you! Credit card debt, mortgages, bank loans for your business, student loans, and car loans are all examples of compound interest working against you and
working for someone else like the banks.
Here’s how the rule of 72 works:
At 1% rate of return, it takes 72 years for $1 to turn into $2.
At 4%, it takes 18 years for money to double. It’s a simple formula. Now, instead of your money doubling once over your lifetime, you could experience two or three doubles.
At double the rate of return – 8% – it takes half the time, 9 years for money to double. What if your money doubled four or five times in your life?
So if you think that a difference of 1% or 2% won’t amount to much, you’re seriously underestimating the power of compounding, and you’ll pay a huge price.
Do you have any idea how much money in this country is earning less than 1% today? Would you be surprised to hear there is over $11 trillion sitting in savings, money market and cash equivalent
accounts as of June 2, 2014… all earning less than 1%? Savings account rates are currently averaging .11% while 1 year CDs average .24% and 5 year CDs average .79%. That means your money will never
double in your lifetime. Sucks Ha?
Let’s look at the rule of seventy two this way. At 29 years old, if you had $10,000 earning a 4% rate of return, your money would double in 18 years. You would have $20,000 at age 47.
If you earned 8% rate of return, your money would double in only 9 years. At age 38 you would have $20,000. Lets double it one more time in 9 more years you would have $40,000 at age 47 and one more
9 years $80,000 at age 56. Now you can see the power of the rule of 72 and compounding interest. The more rate of return you earn your money doubles faster.
If you earned 6% on a $10,000 investment, after 36 years, you’ll have $80,000. That’s three doubles in your working lifetime.
If you double your return from 6% to 12% you double every 6 years, it’s not just double the money, it’s actually EIGHT times the money. Your money could double 7 times in your lifetime. At age 65 you
will have $640,000 and at age 71 you will have $1,280,000. That’s a lot of doubles you could get over your working lifetime.
The factors that make the Rule of 72 work for you are: time – the interest rate you earn – and how much money you put into the account.
When people don’t have time on their side, they’re faced with doing one of two things – either adding more money or earning a higher interest rate. Generally aiming for higher returns often means
increasing risk. So if you don’t want to add risk and you don’t have all the time in the world, what do they have to do? You have to save more money. And reduce the risk and taxation. Take a look at
the IUL it might be a perfect fit for you. Read about is an IUL right for you.
Also the rule of seventy two works against you with inflation also. Today we have one of the lowest inflation rates in last 20 years, but if we take the average for the last 20 years of 3.33%, lets
round it to just 3% divided by 72 you have 24. So you need to double your income every 24 years to keep up with inflation. Has your income doubled in the last 24 years? If so do you think it will
double in the next 24 years? If not you better find a way to add income or you are losing money. That is why I started a business in the financial industry. If you are interested in a part time or
full time business in the financial industry watch this video then call me http://wealthwaveinfo.com
When I looked into starting my own business as a financial professional, learning about the Rule of 72 was the BIG A-HA! for me. I can still remember how mesmerized I was – like I had just been
handed the skeleton key to the halls of wealth.
It’s a tremendous mental math shortcut to estimate the effect of any growth rate.
The Rule of 72 is so simple and powerful.
Once I’d learned these concepts my financial thinking was changed forever. As I shared this new knowledge with others, I found that it had the same effect on them. I learned that most people you know
have never heard of these truths. By giving a financial education, you have the opportunity to help unlock the doors of wealth and prosperity for people you know and care about. This is what drew me
into this business and why I love what I do more every day.
Chief Inspiration Officer
Vincent St.Louis
Fighting the forces of Mediocrity
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Journal Assignment 3
Math 203 Contemporary Mathematics
Due on Oct 14
This assignment is about making estimates, estimates with reason. For each
of the four questions below you must come up with an answer. To do so,
you may have to make extra assumptions and look up info. Your solution
must explain how you arrived at the answer, i.e. which extra assumptions
you made, what data your calculations are based on etc.
1. How big is India? Would the country fit inside one of the 50 United
2. My password is 7 characters long. How many different possible 7character passwords are there? How long would it take a hacker to
guess my password by just trying all possible combinations? (I.e using
a ”Brute Force” method.)
3. How thick is a sheet of paper? How many do you need to stack to make
the height of the Empire State Building?
4. The price of gas in Italy is currently about 1.15 euros per liter; how
does that compare to gas prices here?
Submit your journal by e-mail to [email protected] by midnight on the due date. | {"url":"https://studyres.com/doc/18431014/-journal3.pdf-","timestamp":"2024-11-06T01:49:26Z","content_type":"text/html","content_length":"56170","record_id":"<urn:uuid:2b39a35d-70a0-44fa-8cfa-a528283742aa>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00523.warc.gz"} |
Finding Kinetic Energy from graph of Power
• Thread starter mintsnapple
• Start date
In summary, the tutor attempted to solve two problems using the Work Energy Theorem, but was not sure if potential energy played a role. They suggest approximating the curves as either sine or
negative quadratic.
Homework Statement
Homework Equations
P = dW/dt
Change in work = Change in Kinetic Energy
The Attempt at a Solution
Since the integral of the power graph is work done in the system, and since it starts at 0, does this mean kinetic energy is the same thing? So I can probably make a triangle with a height of 20 and
base 1 to get an area of 10 for the first problem? And then 30 for the next one?
Though I feel this is much more complicated than that...
Science Advisor
Homework Helper
Gold Member
mintsnapple said:
I can probably make a triangle with a height of 20 and base 1 to get an area of 10 for the first problem? And then 30 for the next one?
Yes, except that they are clearly not triangles. Can you think of a nonlinear equation that might better represent those curves?
haruspex said:
Yes, except that they are clearly not triangles. Can you think of a nonlinear equation that might better represent those curves?
The question says estimate, so would an equation for the curve be necessary?
Also, one of the tutors explicitly said there were two things to the Work Energy Theorem:
1. The net work on a system is equal to the change in total energy of
that system.
2. The net work on a structureless element of a system is equal to the
change in kinetic energy of that element.
My concern: Does potential energy play any part in this problem?
Science Advisor
Homework Helper
Gold Member
mintsnapple said:
The question says estimate, so would an equation for the curve be necessary?
Also, one of the tutors explicitly said there were two things to the Work Energy Theorem:
1. The net work on a system is equal to the change in total energy of
that system.
2. The net work on a structureless element of a system is equal to the
change in kinetic energy of that element.
My concern: Does potential energy play any part in this problem?
The question is really much too vague. If you allow for the possibility that some of the power has gone into potential energy (e.g. pushing it against a strong electric field) then all you can hope
to do is provide an upper bound on the KE.
Since it doesn't say how accurate the estimate is to be, yes, you could just treat the curve as a sawtooth, but by the same argument you could just estimate 0.
So we are left to guess what is wanted. My guess would be to approximate the curves as either sine or negative quadratic.
Your approach is on the right track, but there are a few things to consider when using a graph to find kinetic energy. First, the power graph represents the rate of change of work, not the actual
work done. So to find the work done, you would need to integrate the power graph over the given time period. Then, to find the change in kinetic energy, you would need to use the equation Change in
work = Change in Kinetic Energy. This equation takes into account any other sources of work or energy that may be present in the system. So while the triangle approach may work in some cases, it may
not be accurate in all cases. It's important to consider all factors and equations when using a graph to find kinetic energy.
FAQ: Finding Kinetic Energy from graph of Power
1. How is kinetic energy measured from a graph of power?
Kinetic energy can be measured from a graph of power by finding the area under the power curve. This area represents the work done over time, which is equal to the change in kinetic energy.
2. What is the relationship between power and kinetic energy?
The relationship between power and kinetic energy is that power is the rate at which energy is transferred, while kinetic energy is the energy an object possesses due to its motion. In other words,
power is the derivative of kinetic energy with respect to time.
3. Can kinetic energy be negative on a graph of power?
Yes, kinetic energy can be negative on a graph of power. This can happen if the object is losing kinetic energy, such as when it is slowing down or changing direction.
4. How does the shape of the power curve affect the kinetic energy on a graph?
The shape of the power curve can affect the kinetic energy on a graph by determining the rate at which energy is transferred. A steeper slope on the power curve indicates a higher rate of energy
transfer, resulting in a larger change in kinetic energy.
5. Is it possible to find the kinetic energy from a graph of power if the power is not constant?
Yes, it is still possible to find the kinetic energy from a graph of power if the power is not constant. This can be done by breaking the graph into smaller sections where the power is constant,
finding the area under each section, and then adding them together to get the total change in kinetic energy. | {"url":"https://www.physicsforums.com/threads/finding-kinetic-energy-from-graph-of-power.739940/","timestamp":"2024-11-05T10:37:13Z","content_type":"text/html","content_length":"99274","record_id":"<urn:uuid:1903c7d7-b962-45c4-b80e-e26f62fabd3a>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00069.warc.gz"} |
Choosing image quality metrics for better results. - BEIJING OPTICS
Image Quality Criteria –Which Metric Should I Use?
Image quality criteria define the expected imaging performance of an optical system. It may also be described as the set of requirements that characterizes how well the imaging system works. It is
imperative to have image quality criteria when designing an optical imaging system, this is how one knows when to be done designing and begin building the system. The image quality criteria are
usually defined by a set of image quality metrics such as wave front error (WFE), point spread function (PSF), and modulation transfer function (MTF).
During the design of the optical system the metrics chosen to define the image quality are usually modeled. During this modeling process an error analysis is performed in order to quantify the system
performance as a function of planned and unplanned errors. Once the system is built, those metrics also need to be tested to verify that the imaging system will be able to meet its objectives and
that the engineering practices used to design the system are valid and can be used again with confidence. Selection of metrics that can be accurately modeled and tested (on schedule and within cost)
is important to the success of the imaging program.
The optical engineer must understand how the different physical parts and functions of an imaging system design will influence the image quality metrics. Ultimately, the image quality metrics used to
quantify the performance are affected by almost all aspects of engineering disciplines required to build an optical system (electrical, mechanical, thermal, contamination, etc…) For example, material
selection of the metering rods in a traditional Cassegrain telescope determines the primary-secondary mirror separation as a function of temperature. This may significantly impact the telescope WFE.
Communicating engineering requirements that flow down from the system image quality metrics to other engineers is also critical to the success of the program.
Understanding the image quality of an optical system can be a complicated process. Imaging systems are built for almost any number of applications. Image quality criteria should be selected that is
appropriate for the application. Also, one metric is usually not enough of a descriptor to guarantee that the system will be designed appropriately. It is not uncommon for image quality criteria to
be explained using multiple metrics. Selecting the appropriate image quality metrics for a particular application is an important first task.
Image Quality
There are many kinds of image quality metrics. They can be categorized into two distinct groups, geometric and diffraction. Some metrics are both geometric and diffraction based, these metrics can be
especially useful for setting the image quality criteria of certain imaging applications.
Geometric image quality metrics can be computed or solved using simple ray tracing through an optical system by following the practices of geometric optics. The geometric optics methods are the most
valid methods for applications where the expected wavefront error is greater than 1 wave. Geometric methods are usually very easy to compute with modern computers and especially with the use of ray
trace programs.
One of the most common geometric image quality metrics is spot size. Geometric spot size is a useful metric for systems investigating the image of points in different wavelengths and from different
field locations as illustrated in Figure 1. The geometric spot size is estimated by tracing rays though the optical system and plotting their coordinates at the image plane. The spot size is usually
quantified by the geometric and/or RMS radius.
Figure 1. A plot of spot diagrams for 3 different field points.
Spot diagrams can be used to compute the ensquared or encircled energy. Ensquared energy is used as a metric because it enables a geometric estimate of how many of the geometric rays of an imaged
spot will land inside a square that represents a pixel. Once again, in the geometric regime, this is usually a good metric for systems with WFE’s of greater than 1 wave. Figure 2 illustrates a
simulated spot with a square overlayed.
Figure 2. The ensquared energy calculated with geometric methods.
Perhaps one of the most widely used geometric methods for measuring image quality is the WFE. The WFE is the optical path difference measured from the reference wavefront in the exit pupil of the
imaging system. As can be seen in Figure 3, the reference wave is perfectly spherical and the aberrated wave is not. The optical path difference is shown on the figure by ΔW(x,y) where x and y are
the exit pupil coordinates. This is a geometric measurement that can be used for imaging systems that expect less than 1 wave of error. Some advantages of using the WFE metric is that the types of
system aberrations can be learned by studying the wave fan plots, lens manufacturing tolerances such as surface figure can be converted to WFE using rules of thumb, and finally the WFE can be related
to other metrics, even diffraction metrics such as the Strehl ratio. It is often required to convert between image quality metrics using rules of thumb rather than detailed models.
Figure 3. An illustration of the wave front error optical path.
The modulation transfer function MTF is both a geometric and diffraction metric for image quality. The MTF is the modulus of the Fourier Transform of the point spread function (PSF) and the complex
auto-correlation of the complex pupil function. Therefore the diffraction effects from the imaging system are shown in the MTF. The MTF is plotted as modulation (contrast) versus spatial frequency.
This metric is most useful for application intended to image extended scenes. It gives you the ability to understand how other aspects of the imaging system like the detector will affect image
Figure 4 shows the utility of the MTF by plotting the optics MTF and the detector MTF on the same graph. The optics imaging ability is cutoff by the system F# and wavelength. This particular curve
shows that the pixel is small enough that the detector cutoff occurs at a larger spatial frequency than the optics. The blue line shows a diffraction limited optics MTF and the red line shows how
aberrations will affect the contrast performance at certain spatial frequencies. To obtain a system MTF the pixel curve is multiplied by the optics curve point by point. To obtain even more accurate
modeling of the imaging application the MTF of the atmosphere, jitter environment, and other imperfections can be included in the MTF plot and multiplied to obtain a system estimate including all the
prevalent errors.
Figure 4. The modulation transfer function.
The PSF of the optical system describes the effects of diffraction and all aberrations. A picture of a typical PSF is shown in Figure 5. The width of the central core for a diffraction limited system
is a function of the system F# and wavelength. It should be noted that ensquared energy or encircled energy can also be a computed using the PSF when systems have a WFE less than about ¼ of a wave.
Therefore, the ensquared energy can be both a diffraction and geometric image quality metric.
Figure 5. A 3D and 2D view of the point spread function.
The Strehl ratio is a common way of measuring image quality using the PSF. The Strehl ratio is computed by taking the ratio of the height of the central core of the actual system PSF with respect to
the central core of an aberration free PSF. Figure 6 illustrates this ratio.
Figure 6. The Strehl ratio.
The RMS WFE of the system can be related to the Strehl ratio by the following relationship,
The W[rms] is RMS WFE of the system in units of λ. This is a convenient way to relate the WFE to the PSF or ensquared energy metric. In summary, Table 1 lists the image quality metrics that have been
discussed. MTF and ensquared energy are metrics that are both geometric and diffraction based. The RMS WFE, while best used when the errors are expected to be less than 1 wave is a geometric image
quality metric. A spot diagram is a geometric method used to describe the systems ability to image a point when the expected system errors are greater than 1 wave. The PSF is a diffraction metric
used to describe the systems ability to image a point when the expected errors are less than 1 wave and more ideally closer to ¼ of a wave.
Table 1. A comparison of the geometric and diffraction image quality metrics.
Conclusion –Select Wisely
In conclusion, a simple example is given to express the need to carefully select the image quality metric or metrics used to establish the image quality criteria of an imaging system. Figure 7 shows
the MTF for several aberration free optical systems. However, each optical system has a varying size of central obscuration. It is noticed in the plot that the MTF drastically decreases as the
obscuration size grows. The MTF analysis clearly shows this phenomenon. Because of the obscuration, the central core of the PSF gets smaller, but more of the energy in the PSF is diffracted outside
of the central core causing a blurred or less sharp image. In other words there is a reduction of contrast in the mid spatial frequency range of the images produced by this system.
The image quality of a telescope system intended for imaging extended scenes is affected by the size of the central obscuration. If WFE was the only metric chosen to establish the image quality
criteria of the system, the obscuration affect could have been overlooked resulting in poorer than expected imagery. For a case such as this, the WFE metric might be used to set the optical element,
mechanical, thermal, and other pertinent tolerancing requirements that influence the image quality. However, diffraction metrics such as the MTF or PSF must also be used to understand the affects of
the central obscuration to the systems image quality.
Figure 7. MTF degradation as a function of obscuration size.
Lightsey, Paul A., “Image for large segmented space telescopes”, Proceedings of SPIE, Vol 4850, 2003.
Grievenkamp, John, “Field Guide to Geometric Optics”, SPIE Vol. FG01.
ZEMAX Lens Design Manual
Burge, Jim, “Opti 521 course notes”, University of Arizona. | {"url":"https://www.beijing-optics.com/2023/06/30/image-quality-criteria-which-metric-should-i-use/","timestamp":"2024-11-13T03:03:14Z","content_type":"text/html","content_length":"70150","record_id":"<urn:uuid:cebe807b-8f93-4525-bb99-964f8f7b96a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00446.warc.gz"} |
Simplifying Trigonometric Expressions Using Double-Angle Identities
Question Video: Simplifying Trigonometric Expressions Using Double-Angle Identities Mathematics • Second Year of Secondary School
Simplify (sin ๐ ผ)/(1 + tan ๐ ผ) โ (sin 2๐ ผ)/(2 cos ๐ ผ + 2 sin ๐ ผ).
Video Transcript
Simplify sin ๐ ผ over one plus tan ๐ ผ minus sin two ๐ ผ over two cos ๐ ผ plus two sin ๐ ผ.
So Iโ ve copies down the expression to simplify, and my goal first of all is to get everything in terms of sin ๐ ผ and cos ๐ ผ. That means rewriting tan ๐ ผ as sin ๐ ผ over cos ๐ ผ and sin two
๐ ผ as two times sin ๐ ผ times cos ๐ ผ, where here we used the double angle identity for sine.
Okay so now we have something in terms of only sin ๐ ผ and cos ๐ ผ. Letโ s simplify. We multiply the first fraction by cos ๐ ผ over cos ๐ ผ in an attempt to simplify the denominator. And of
course because cos ๐ ผ over cos ๐ ผ is just one, this doesnโ t change the value of the fraction.
So now the first fraction is sin ๐ ผ cos ๐ ผ over cos ๐ ผ plus sin ๐ ผ. Is there anything we can do to simplify the second fraction before we perform the subtraction? Yes, the numerator and
denominator have a common factor of two, which we can cancel out. So we are left with sin ๐ ผ cos ๐ ผ over cos ๐ ผ plus sin ๐ ผ.
We can notice at least two terms of the same, and so when we subtract one from the other, we get zero. So sin ๐ ผ over one plus tan ๐ ผ minus sin two ๐ ผ over two cos ๐ ผ plus two sin ๐ ผ is
simply equal to zero, and you canโ t get much simpler than that. | {"url":"https://www.nagwa.com/en/videos/934174159192/","timestamp":"2024-11-13T22:34:30Z","content_type":"text/html","content_length":"249374","record_id":"<urn:uuid:a9807cf6-677a-4a82-a55e-ee0ced927b7f>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00776.warc.gz"} |
Which Equation Represents A Linear FunctionXy=5y=x-3y= 2/x + 73x + 2y = 6
3x + 2y = 6
Step-by-step explanation:
When you solve for y in this equation, it is in the same format as the slope-intercept equation.
3x + 2y = 6
2y = -3x + 6
y = -3/2 x + 3
The slope intercept equation is ...
y = mx + b
(m can be any number)
The slope-intercept equation is for linear functions!
Three integers, a, b, and c, where c is a positive integer.
The product of a and b is 6.
The product of a and c is -4.
The product of b and c is -6.
To find:
The values of a,b and c.
According to the given information:
[tex]ab=6[/tex] ...(i)
[tex]ac=-4[/tex] ...(ii)
[tex]bc=-6[/tex] ...(iii)
From (ii), we get
[tex]a=-\dfrsc{4}{c}[/tex] ...(iv)
From (iii), we get
[tex]b=-\dfrsc{6}{c}[/tex] ...(v)
Putting [tex]a=-\dfrsc{4}{c}[/tex] and [tex]b=-\dfrsc{6}{c}[/tex] in (i), we get
[tex]\dfrac{-4}{c}\times \dfrac{-6}{c}=6[/tex]
Taking square root on both sides, we get
[tex]\pm \sqrt{4}=c[/tex]
[tex]\pm 2=c[/tex]
It is given that c is a positive integer. So, it cannot be negative and the only value of c is [tex]c=2[/tex].
Putting [tex]c=2[/tex] in (iv), we get
Putting [tex]c=2[/tex] in (v), we get
Therefore, the values of a,b,c are [tex]a=-2,b=-3,c=2[/tex]. | {"url":"https://www.cairokee.com/homework-solutions/which-equation-represents-a-linear-functionbr-xy5br-yx2-3br-mvah","timestamp":"2024-11-05T10:30:11Z","content_type":"text/html","content_length":"68827","record_id":"<urn:uuid:60e626a7-2916-4852-b2db-9a1df470ce33>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00367.warc.gz"} |
Announcing Maple 2017
Maple 2017 has launched!
Maple 2017 is the result of hard work by an enthusiastic team of developers and mathematicians.
As ever, we’re guided by you, our users. Many of the new features are of a result of your feedback, while others are passion projects that we feel you will find value in.
Here’s a few of my favourite enhancements. There’s far more that’s new - see What’s New in Maple 2017 to learn more.
MapleCloud Package Manager
Since it was first introduced in Maple 14, the MapleCloud has made thousands of Maple documents and interactive applications available through a web interface.
Maple 2017 completely refreshes the MapleCloud experience. Allied with a new, crisp, interface, you can now download and install user-created packages.
Simply open the MapleCloud interface from within Maple, and a mouse click later, you see a list of user-created packages, continuously updated via the Internet. Two clicks later, you’ve downloaded
and installed a package.
This completely bypasses the traditional process of searching for and downloading a package, copying to the right folder, and then modifying libname in Maple. That was a laborious process, and,
unless I was motivated, stopped me from installing packages.
The MapleCloud hosts a growing number of packages.
Many regular visitors to MaplePrimes are already familiar with Sergey Moiseev’s DirectSearch package for optimization, equation solving and curve fitting.
My fellow product manager, @DSkoog has written a package for grouping data into similar clusters (called ClusterAnalysis on the Package Manager)
Here’s a sample from a package I hacked together for downloading maps images using the Google Maps API (it’s called Google Maps and Geocoding on the Package Manager).
You’ll also find user-developed packages for exploring AES-based encryption, orthogonal series expansions, building Maple shell scripts and more.
Simply by making the process of finding and installing packages trivially easy, we’ve opened up a new world of functionality to users.
Maple 2017 also offers a simple method for package authors to upload workbook-based packages to the MapleCloud.
We’re engaging with many package authors to add to the growing list of packages on the MapleCloud. We’d be interested in seeing your packages, too!
Advanced Math
We’re committed to continually improving the core symbolic math routines. Here area few examples of what to expect in Maple 2017.
Resulting from enhancements to the Risch algorithm, Maple 2017 now computes symbolic integrals that were previously intractable
Groeber:-Basis uses a new implementation of the FGLM algorithm. The example below runs about 200 times faster in Maple 2017.
gcdex now uses a sparse primitive polynomial remainder sequence together. For sparse structured problems the new routine is orders of magnitude faster. The example below was previously intractable.
The asympt and limit commands can now handle asymptotic cases of the incomplete Γ function where both arguments tend to infinity and their quotient remains finite.
Among several improvements in mathematical functions, you can now calculate and manipulate the four multi-parameter Appell functions.
Appel functions are of increasing importance in quantum mechanics, molecular physics, and general relativity.
pdsolve has seen many enhancements. For example, you can tell Maple that a dependent variable is bounded. This has the potential of simplifying the form of a solution.
Plot Builder
Plotting is probably the most common application of Maple, and for many years, you’ve been able to create these plots without using commands, if you want to. Now, the re-designed interactive Plot
Builder makes this process easier and better.
When invoked by a context menu or command on an expression or function, a panel slides out from the right-hand side of the interface.
Generating and customizing plots takes a single mouse click. You alter plot types, change formatting options on the fly and more.
To help you better learn Maple syntax, you can also display the actual plot command.
Password Protected Content
You can distribute password-protected executable content. This feature uses the workbook file format introduced with Maple 2016.
You can lock down any worksheet in a Workbook. But from any other worksheet, you can send (author-specified) parameters into the locked worksheet, and extract (author-specified) results.
Plot Annotations
You can now get information to pop up when you hover over a point or a curve on a plot.
In this application, you see the location and magnitude of an earthquake when you hover over a point
Here’s a ternary diagram of the color of gold-silver-copper alloys. If you let your mouse hover over the points, you see the composition of the points
Plot annotations may seem like a small feature, but they add an extra layer of depth to your visualizations. I’ve started using them all the time!
Engineering Portal
In my experience, if you ask an engineer how they prefer to learn, the vast majority of them will say “show me an example”. The significantly updated Maple Portal for Engineers does just that,
incorporating many more examples and sample applications. In fact, it has a whole new Application Gallery containing dozens of applications that solve concrete problems from different branches of
engineering while illustrating important Maple techniques.
Designed as a starting point for engineers using Maple, the Portal also includes information on math and programming, interface features for managing your projects, data analysis and visualization
tools, working with physical and scientific data, and a variety of specialized topics.
Geographic Data
You can now generate and customize world maps. This for example, is a choropleth of European fertility rates (lighter colors indicate lower fertility rates)
You can plot great circles that show the shortest path between two locations, show varying levels of detail on the map, and even experiment with map projections.
A new geographic database contains over one million locations, cross-referenced with their longitude, latitude, political designation and population.
The database is tightly linked to the mapping tools. Here, we ask Maple to plot the location of country capitals with a population of greater than 8 million and a longitude lower than 30.
There’s much more to Maple 2017. It’s a deep, rich release that has something for everyone.
Visit What’s New in Maple 2017 to learn more.
Tags are words are used to describe and categorize your content. Combine multiple words with dashes(-), and seperate tags with spaces. | {"url":"https://wamp.mapleprimes.com/maplesoftblog/208276-Announcing-Maple-2017","timestamp":"2024-11-05T17:06:07Z","content_type":"text/html","content_length":"125589","record_id":"<urn:uuid:7bca2dd4-f905-4e34-bcfb-a480f1745fdf>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00815.warc.gz"} |
Ultimate GMAT Questions pack
Ultimate GMAT Past Questions and Answers Pack. Updated PDF with Study Guide
This ebook will get you ready for your best score on the GMAT with expert tips from our experienced seasoned tutors.
This pack will certainly guarantee your success.
This is a downloadable product and is accessible as soon as your purchase is complete. The study pack is in .PDF format.
Sample Questions for Ultimate GMAT Aptitude Test
Question:Problem: If a train travels 240 miles in 4 hours, what is its average speed in miles per hour?
A) 40 mph
B) 60 mph
C) 80 mph
D) 100 mph
Answer:B 60 mph
Question: Problem: If x + 3 = 7, what is the value of x?
A) 2
B) 3
C) 4
D) 5
Answer:A 2
Question: Problem: What is the next number in the sequence: 2, 4, 6, 8, __?
A) 10
B) 11
C) 12
D) 14
Answer:A 10
Question: Problem: If a rectangle has a length of 10 units and a width of 5 units, what is its area?
A) 15 sq units
B) 20 sq units
C) 25 sq units
D) 30 sq units
Answer:C 25 sq units
Question:Problem: If 3 apples cost $6, how much would 5 apples cost?
A) $10
B) $12
C) $15
D) $18
Answer:B $12
Question:Problem: If a book costs $20 before tax and the tax rate is 8%, what is the total cost after tax?
A) $20.80
B) $21.60
C) $22.40
D) $23.20
Answer:A $20.80
Question: Problem: If the average of three numbers is 15, and two of the numbers are 12 and 18, what is the third number?
A) 8
B) 10
C) 15
D) 20
Answer:A 8
Question: Problem: If a triangle has a base of 10 units and a height of 8 units, what is its area?
A) 20 sq units
B) 30 sq units
C) 40 sq units
D) 50 sq units
Answer:B 30 sq units
Question: Problem: If the ratio of boys to girls in a class is 3:5 and there are 24 students, how many girls are there?
A) 6
B) 10
C) 12
D) 15
Answer:C 12
Question: Problem: If a car travels 360 miles in 6 hours, what is its average speed in miles per hour?
A) 45 mph
B) 60 mph
C) 75 mph
D) 90 mph
Answer:C 75 mph
Question: Problem: If the price of a shirt is reduced by 20% in a sale, and the sale price is $40, what was the original price?
A) $32
B) $40
C) $45
D) $50
Answer:D $50
Question: Problem: What is the missing number in the sequence: 3, 7, 11, __, 19?
A) 12
B) 14
C) 15
D) 17
Answer:B 14
Question:Problem: If the area of a square is 64 sq units, what is the length of one side?
A) 4 units
B) 6 units
C) 8 units
D) 10 units
Answer:C 8 units
Question: Problem: If 2 pencils cost $1, how much would 5 pencils cost?
A) $1
B) $2
C) $2.50
D) $3
Answer:B $2
Question:Problem: If the sum of two numbers is 30 and one of the numbers is 18, what is the other number?
A) 8
B) 12
C) 15
D) 20
Answer:B 12
Question:If a jacket costs $120 and its price is increased by 25%, what will be the new price?
A $130
B $140
C $150
D $160
Answer:C $150
Question: A company’s revenue for the first quarter was $800,000. If this represents a 20% increase from the previous quarter, what was the revenue for the previous quarter?
(A) $600,000
(B) $640,000
(C) $680,000
(D) $720,000
Answer:B $640,000
Question:A container holds 500 liters of water. If 30% of the water is removed, how many liters of water remain?
A 150 liters
B 250 liters
C 350 liters
D 450 liters
Answer:C 350 liters
Question:The price of a smartphone decreased by 15% from $400. What is the new price?
A $320
B $340
C $360
D $380
Answer:A $320
Question:A car travels 240 miles on 12 gallons of gasoline. What is the car’s miles-per-gallon (MPG) rate?
A 15 MPG
B 18 MPG
C 20 MPG
D 24 MPG
Answer:D 24 MPG
Question: If the cost of 5 books is $75, what is the cost of 12 books at the same price?
A $140
B $180
C $200
D $220
Answer:B $180
Question:A shop sells a shirt at a 30% discount. If the original price was $50, what is the discounted price?
A $15
B $25
C $30
D $35
Answer:A $25
Question:If a train travels at a speed of 80 miles per hour, how far will it travel in 2.5 hours?
A 180 miles
B 200 miles
C 220 miles
D 250 miles
Answer:C 220 miles
Question:The price of a stock increased by 10% on Monday and decreased by 5% on Tuesday. If the initial price was $100, what is the price after Tuesday?
A $95
B $97.50
C $99
D $100.50
Answer:B $97.50
Question: If the area of a square garden is 144 square meters, what is the length of one side of the square?
A 9 meters
B 10 meters
C 12 meters
D 14 meters
Answer:C 12 meters
Question:A store offers a 20% discount on all items. If a customer buys a $60 item, how much will they pay after the discount?
A $40
B $44
C $48
D $52
Answer:B $44
Question: If a recipe requires 2/3 cup of sugar and the entire recipe is doubled, how much sugar will be needed?
A 1/3 cup
B 2/3 cup
C 1 cup
D 1 1/3 cups
Answer:D 1 1/3 cups
Question:A swimming pool is 40 meters long and 20 meters wide. What is its total perimeter?
A 80 meters
B 100 meters
C 120 meters
D 160 meters
Answer:C 120 meters
Question: If a package of pens contains 24 pens and costs $12, what is the cost per pen?
A $0.25
B $0.40
C $0.50
D $0.75
Answer:A $0.25
Question: The population of a city increased from 80,000 to 100,000 in a year. What is the percentage increase?
A 10%
B 20%
C 25%
D 30%
Answer:B 20%
Question:Choose the word that is most similar in meaning to “Eloquent”:
A Inarticulate
B Fluent
C Reserved
D Vague
Answer:B Fluent
Question:Select the word that is the opposite in meaning to “Cautious”:
A Bold
B Careful
C Reckless
D Prudent
Answer:C Reckless
Question:Identify the synonym for the word “Meticulous”:
A Careless
B Sloppy
C Precise
D Inaccurate
Answer:C Precise
Question:Choose the word that best completes the sentence: “Her singing was so beautiful that it __________ the entire audience.”
A Captivated
B Annoyed
C Bored
D Frustrated
Answer:A Captivated
Question: Which word does not belong with the others?
A Apple
B Banana
C Orange
D Carrot
Answer:D Carrot
Question: Identify the antonym for the word “Confident”:
A Unsure
B Arrogant
C Assertive
D Positive
Answer: A Unsure
Question: Choose the word that is most similar in meaning to “Diligent”:
A Lazy
B Neglectful
C Industrious
D Careless
Answer: C Industrious
Question: Which word is the odd one out?
A Square
B Circle
C Triangle
D Rectangle
Answer:B Circle
Question:Complete the sentence: “The novel’s plot was so __________ that it kept me engaged until the very end.”
A Predictable
B Boring
C Compelling
D Dull
Answer:C Compelling
Question: Identify the word that correctly completes the analogy: Cat is to Meow as Dog is to __________.
A Bark
B Purr
C Hiss
D Roar
Answer: A Bark
Question: Choose the word that does not belong with the others:
A Violin
B Trumpet
C Flute
D Carrot
Answer: D Carrot
Question: Identify the word that is most opposite in meaning to “Generous”:
A Stingy
B Giving
C Altruistic
D Kind
Answer:A Stingy
Question: Complete the sentence: “Her speech was so __________ that the audience was left speechless.”
A Boring
B Unimpressive
C Engaging
D Mediocre
Answer:C Engaging
Question: Which word is an example of an animal?
A Table
B Tree
C Cat
D Book
Answer:C Cat
Question: Identify the synonym for the word “Voracious”:
A Indifferent
B Hungry
C Full
D Small
Answer: B Hungry
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trademark holders are affiliated with Test Marshal International; or this website. [/message_box]
Sample of Ultimate GMAT Questions pack
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cRi3D POWERUP Game Theory | DENKBOTS
cRi3D POWERUP Game Theory
Game Theory is a tool used in several fields of Science including Political Science and Economics. For this article a very simplified version will be used where the model consists of a rational
agent (it always chooses to perform the action with the optimal expected outcome for itself from among all feasible actions) competing in a mathematically defined competition.
To put it more plainly, we are going to try and find the theoretical scoring maximums for each part of the game.
Using a game theory based evaluation of the scoring system accomplishes three things:
1. Makes sure the students understand the scoring rules
2. Positions optimized scoring strategies at the front of the discussion leading into system strategy brainstorming
3. Provides the team with an objective reference that can be used as an unbiased evaluation of strategies in later phases of design
To start this article, it is assumed that the reader is familiar with this season’s game and corresponding game manual.
To begin, we will break the game into three sections for evaluation:
• Autonomous Mode (15 Seconds)
• Teleoperation Mode (1 Minute 45 Seconds)
• End Game (30 Seconds).
In each section we ask: What are all of the ways our robot can score points? Are any of these scoring methods mutually exclusive (for example, any given CUBE can only be put on the SWITCH or the
SCALE)? Do any of the scoring methods share casual relationships (for example, a POWER UP becomes available when a set number of CUBES have been installed)?
For reference we will add a copy of “Figure 3.2 – Zones and Markings” from the game manual to clarify location terminology.
Figure 3-2: Zones and Markings
Autonomous Mode
In Autonomous Mode, the robot is placed in contact with its ALLIANCE WALL with no more than one (1) CUBE. The game manual (Table 2-1: Auto Point Values) details that a robot can score points in this
mode by:
• Cross the Auto Line, a.k.a Auto-Run (5 points)
• Switch Ownership (2, + 2 points per second)
• Scale Ownership (2, + 2 points per second)
If the robot deposits their CUBE on the SWITCH it can not deposit it on the SCALE, and vice-versa.
Treating the robot as a rational agent, it chooses to perform the actions with the optimal expected outcomes and Cross the Auto Line (5 pts), then place the CUBE on the SWITCH..
Why does the robot place the CUBE on the SWITCH and not the SCALE? If we assume an average speed of 10 ft/s for the robot, then evaluate the distance to the SWITCH (~12ft) and the distance to the
SCALE (~27ft), we find that since the scoring is the same 2 points per second for both goals the rational agent will choose the option that allows a CUBE to be placed the soonest and provide optimal
Since it takes ~1.2s to reach the SWITCH and ~1s to place the CUBE (we will make the assumption here that placing a CUBE on the SWITCH takes ~1s), we will round up to the nearest whole number and say
we begin scoring points after 3s. This provides 12s of scoring for 2 + 24pts.
This MAX* Autonomous Mode achieves a score of 31 pts.
*NOTE: If we assume our robot can do everything, the rational agent would use the remaining 12s to drive forward and collect a CUBE from the edge of the PLATFORM ZONE (we will make the assumption
here that picking a CUBE off the ground takes ~2s), then drive forward to the SCALE (1.5s), then place the CUBE on the SCALE (we will make the assumption here that placing a CUBE on the SCALE takes
~2s). Since it takes ~2s to get the CUBE, ~1.5s to get to the SCALE, and ~2s to place the CUBE on the SCALE, we will round up to the nearest whole number and say it takes us 6s to begin scoring
points, which is 9s into AUTO MODE. This provides 6s of scoring for 2 + 12pts.
This MAXX Autonomous Mode Achieves a score of 45 pts.
Teleop Mode
In Teleoperation Mode, the robot is located in the NULL TERRITORY from the end of Autonomous Mode. The game manual (Table 2-2: Teleop Point Values) details that a robot can score points in this mode
• Switch Ownership 1, + 1 point per second
• Scale Ownership 1, + 1 point per second
• Power Cube in Vault 5 points
• Boost Power Up Bonus 2 points per second
• Parked on Platform 5 points
• Successful Climb 30 points
Also, the game manual (Table 4-1: FIRST POWER UP rewards) details that a robot (or robots) can also score Ranking Points (RP) by:
• FACE THE BOSS (1 RP)
• AUTO QUEST (1 RP)
In order to attempt to model this game, we will make the assumption that an unattended SWITCH and SCALE will change ownership to the opponent every fifteen (15) seconds.
Treating the robot as a rational agent, it chooses to perform the actions with the optimal expected outcomes and turns and travels to the stacked CUBES (2s), then picks up a CUBE (2s), then travels
to the VAULT (2.7s), and deposits the CUBE in the VAULT (1s).
TIME ELAPSED: 7.7s
Because it has only been 7.7s the SWITCH and SCALE have not switched possession yet, so the robot repeats the steps above to put another CUBE in the VAULT.
TIME ELAPSED: 15.4s
Now that the possession has changed, the robot moves to begin scoring points again as quickly as possible on the SWITCH, so it travels to the stacked CUBES (1.2s), picks up a CUBE (2s), travels to
the SWITCH (1s), and deposits the CUBE on the SWITCH (1s).
TIME ELAPSED: 20.6s
Continuing this attempt to resuming scoring as quickly as possible, the robot travles to the CUBES behind the SWITCH (1.2), picks up a CUBE (2s), then turns and deposits the CUBE on the SCALE (2s).
TIME ELAPSED: 26.1s
With both the SWITCH and the SCALE now scoring points again, the robot moves to score the maximum number of points at the moment and travels to the stacked CUBES (2.7s), picks up a CUBE (2s), travels
to the VAULT (1.2s), and deposits the CUBE in the VAULT (1s).
TIME ELAPSED: 33s
Now that the possession has changed again, the robot repeats the SWITCH and SCALE cycle detailed above to regain scoring.
TIME ELAPSED: 43.7s
Once possession is secured, the robot repeats the VAULT cycle.
TIME ELAPSED: 50.6s
Now that the possession has changed again, the robot repeats the SWITCH and SCALE cycle detailed above to regain scoring.
TIME ELAPSED: 61.3s
Since the robot was in the middle of the VAULT cycle when possession changed, as soon as the robot finishes reclaiming the SCALE, possession changes again, so the robot repeats the SWITCH and SCALE
TIME ELAPSED: 72s
Once possession is secured, the robot repeats the VAULT cycle.
TIME ELAPSED: 78.9s
The robot then repeats the double SWITCH and SCALE cycles to regain scoring.
TIME ELAPSED: 100.3s
Once possession is secured, the robot repeats the VAULT cycle.
TIME ELAPSED: 107.2s
We assume the possession of the SWITCH will not change in the final 30s of the match so the robot retakes the SCALE by repeating the SCALE cycle.
TIME ELAPSED: 112.7s
Since we are assuming a 10s hang time, the robot optimizes its scoring by completing one final VAULT cycle.
TIME ELAPSED: 119.6s
If we tally the points we scored during TELEOP, we find we scored six (6) CUBES in the VAULT for 30pts, we gained possession of the SWITCH seven (7) times for 7pts, we held the SWITCH for a total of
67s, we gained possession of the SCALE (7) times for 7pts, and we held the SCALE for a total of 32s.
If we optimized the BOOST until we had both the SWITCH and the SCALE, we could add an additional 13pts to our score in TELEOP. This would give us a total of 126pts in TELEOP.
Therefore, the MAX TELEOP Mode using this model achieves a score of 126 pts.
End Game
We choose to CLIMB during the final 10s of the match to gain 30pts. We also choose to use the LEVITATE at the end of the match for an additional 30pts.
This MAX End Game achieves a score of 60 pts.
Final Score
• MAX Autonomous Mode: 45 pts
• MAX TELEOP Mode: 126 pts
• MAX End Game: 60 pts
MAX TOTAL: 233 pts
So what good has this exercise done? By walking through the match as a rational agent, the flow of a match can be better understood. By picking the optimal actions, necessary robot functions begin
to take shape. By approaching the game systematically, rules and strategic advantages provided by those rules can be uncovered.
This exercise has also provided a baseline to evaluate other game strategies against and to help define robot functions from.
Head on over to Part 2 – Strategy and Research for more fun!
If you want to learn more about this process, check out our presentation from the 2017 Purdue FIRST Forums on Robot Requirements!
Please feel free to join the conversation on our Facebook or Twitter with your questions, thoughts, and feedback on these articles! | {"url":"https://denkbots.com/2018/01/08/cri3d-powerup-game-theory/","timestamp":"2024-11-02T23:28:47Z","content_type":"text/html","content_length":"72939","record_id":"<urn:uuid:d16ba14c-15a5-4504-ab67-379b45acf7ce>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00043.warc.gz"} |
coef.lmList: Extract lmList Coefficients in nlme: Linear and Nonlinear Mixed Effects Models
The coefficients of each lm object in the object list are extracted and organized into a data frame, with rows corresponding to the lm components and columns corresponding to the coefficients.
Optionally, the returned data frame may be augmented with covariates summarized over the groups associated with the lm components.
## S3 method for class 'lmList' coef(object, augFrame, data, which, FUN, omitGroupingFactor, ...)
object an object inheriting from class "lmList", representing a list of lm objects with a common model.
an optional logical value. If TRUE, the returned data frame is augmented with variables defined in the data frame used to produce object; else, if FALSE, only the coefficients are
augFrame returned. Defaults to FALSE.
data an optional data frame with the variables to be used for augmenting the returned data frame when augFrame = TRUE. Defaults to the data frame used to fit object.
an optional positive integer or character vector specifying which columns of the data frame used to produce object should be used in the augmentation of the returned data frame.
which Defaults to all variables in the data.
an optional summary function or a list of summary functions to be applied to group-varying variables, when collapsing the data by groups. Group-invariant variables are always
summarized by the unique value that they assume within that group. If FUN is a single function it will be applied to each non-invariant variable by group to produce the summary for
that variable. If FUN is a list of functions, the names in the list should designate classes of variables in the frame such as ordered, factor, or numeric. The indicated function
FUN will be applied to any group-varying variables of that class. The default functions to be used are mean for numeric factors, and Mode for both factor and ordered. The Mode
function, defined internally in gsummary, returns the modal or most popular value of the variable. It is different from the mode function that returns the S-language mode of the
an optional logical value. When TRUE the grouping factor itself will be omitted from the group-wise summary of data but the levels of the grouping factor will continue to be used
omitGroupingFactor as the row names for the returned data frame. Defaults to FALSE.
... some methods for this generic require additional arguments. None are used in this method.
an object inheriting from class "lmList", representing a list of lm objects with a common model.
an optional logical value. If TRUE, the returned data frame is augmented with variables defined in the data frame used to produce object; else, if FALSE, only the coefficients are returned. Defaults
to FALSE.
an optional data frame with the variables to be used for augmenting the returned data frame when augFrame = TRUE. Defaults to the data frame used to fit object.
an optional positive integer or character vector specifying which columns of the data frame used to produce object should be used in the augmentation of the returned data frame. Defaults to all
variables in the data.
an optional summary function or a list of summary functions to be applied to group-varying variables, when collapsing the data by groups. Group-invariant variables are always summarized by the unique
value that they assume within that group. If FUN is a single function it will be applied to each non-invariant variable by group to produce the summary for that variable. If FUN is a list of
functions, the names in the list should designate classes of variables in the frame such as ordered, factor, or numeric. The indicated function will be applied to any group-varying variables of that
class. The default functions to be used are mean for numeric factors, and Mode for both factor and ordered. The Mode function, defined internally in gsummary, returns the modal or most popular value
of the variable. It is different from the mode function that returns the S-language mode of the variable.
an optional logical value. When TRUE the grouping factor itself will be omitted from the group-wise summary of data but the levels of the grouping factor will continue to be used as the row names for
the returned data frame. Defaults to FALSE.
some methods for this generic require additional arguments. None are used in this method.
a data frame inheriting from class "coef.lmList" with the estimated coefficients for each "lm" component of object and, optionally, other covariates summarized over the groups corresponding to the
"lm" components. The returned object also inherits from classes "ranef.lmList" and "data.frame".
Pinheiro, J. C. and Bates, D. M. (2000), Mixed-Effects Models in S and S-PLUS, Springer, New York, esp. pp. 457-458.
fm1 <- lmList(distance ~ age|Subject, data = Orthodont) coef(fm1) coef(fm1, augFrame = TRUE)
For more information on customizing the embed code, read Embedding Snippets. | {"url":"https://rdrr.io/cran/nlme/man/coef.lmList.html","timestamp":"2024-11-09T19:13:05Z","content_type":"text/html","content_length":"33214","record_id":"<urn:uuid:8d5818c2-fc64-4ce6-8f0f-6765b337681e>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00321.warc.gz"} |
Mace, Kristina / PreCalculus
• PRE-CALCULUS
Ms. Grossmann Phone: (503)673-7815 ext 4816
Pre-Calculus E-mail: grossmak@wlwv.k12.or.us
Room: B102 http://www.wlhs.wlwv.k12.or.us/Page/3400
Office Hours: 7:30am-8:15am
Course Description: This course is the analysis of polynomial, rational, power, exponential, logarithmic, trigonometric, and piecewise functions and their general characteristics. In addition
logic, probability, statistics, matrices, transformations, composition, inverses, and the binomial theorem will be covered. Students will be exposed to some beginning calculus topics.
Applications are emphasized throughout the material and algebraic, graphical, numerical, and verbal methods will be used to analyze and interpret problems.
Major Units:
│Chapter 1 Algebraic Fundamentals │Chapter 7 Analytic Trigonometry │
│ │ │
│Chapter 2 Transformations of Functions │Chapter 8 Polar Coordinates and Vectors│
│ │ │
│Chapter 3 Polynomial and Rational Functions │Chapter 9 Systems of Equations and │
│ │ │
│Chapter 4 Exponential and Logarithmic │Inequalities │
│ │ │
│Functions │Chapter 10 Analytic Geometry │
│ │ │
│Chapter 5 Trigonometric Functions of Real │Chapter 11 Sequences and Series │
│ │ │
│Numbers │Chapter 12 Limits and Derivatives │
│ │ │
│Chapter 6 Trig Functions of Angles │ │
Learning Objectives: Using a graphing calculator to investigate and solve problems, by engaging students in critical thinking tasks. Students will be required to communicate mathematical ideas
verbally, graphically, algebraically, and numerically. Describe general properties of functions as they relate to calculus, using the concept of limit as it pertains to sequences and functions,
analyzing the graphs of polynomial, rational, radical, and transcendental functions, using the Pythagorean Theorem to develop and understand both circular and right triangle trigonometry.
Learning Goals:
Find and interpret average rate of change and communicate how it applies to a relative application
Find and interpret the difference quotient and explore its application in real life
Find and interpret properties of piecewise-defined, polynomial, rational, power, radical, exponential and logarithmic functions
Evaluate and graph piecewise-defined, polynomial, rational, power, radical, exponential and logarithmic functions
Solve equations involving piecewise-defined, polynomial, rational, power, radical, exponential and logarithmic functions
Apply the solving of piecewise-defined, polynomial, rational, power, radical, exponential and logarithmic functions within real life applications and effectively communicate the results in the
proper context
Analyze and communicate differences in behaviors of different types of functions both graphically and numerically
Data will be modeled using the appropriate regression ad the model will be used to answer real-life questions and make predictions
Apply transformations to functions
Factor polynomial functions from a graphical perspective and write equations of polynomials given a graph
Find and interpret composition of functions and use the composition function to answer questions pertaining to a real-life application
Find and interpret inverse functions
Utilize proper notation to define and evaluate sequences and series
Solve applications involving sequences and series
Apply Pascal’s Triangle and the Binomial Theorem
Define and identify trigonometric functions
Convert between radian measure and degrees
Use radian measure to compute the length of an arc
Find trigonometric values for particular angles in a right triangle
Evaluate the sine and cosine functions for particular angles on the unit circle from memory
Define sine and cosine functions based on the unit circle
Graph, transform, and analyze the graphs of sine and cosine functions
Rewrite tangent, secant, cosecant, and cotangent functions in terms of sine and cosine functions
Use the trigonometric identities and inverse trigonometric functions appropriately to solve mathematical problems
Verify trigonometric identities
Use the laws of sine and cosine to solve mathematical problems
Recognize, model, and solve applications using trigonometry
Perform vector arithmetic
Use vectors to model applications and solve mathematical problems
Use parametric equations to describe curves
Convert between Cartesian and polar coordinates
Use polar equations to describe curves
Recognize, and solve mathematical problems with polar equations
Graph and translate graphs of conic sections (parabolas, ellipses, hyperbolas, and circles)
Demonstrate an appropriate use of technology to solve problems
Standards of Mathematical Practice:
The student will:
Make sense of problems and persevere in solving them
Reason abstractly and quantitatively
Construct viable arguments and critique the reasoning of others
Model with mathematics
Use appropriate tools strategically
Attend to precision
Look for and make sense of structure
Look for and express regularity in repeated reasoning.
all math courses are designed to meet the requirements of the WLWV Mathematics Curriculum and the Common Core State Standards.
Grading: Grading Breakdown:
A: 90 and above Tests 45%
B: 80.0-89.9 Quizzes 20%
C: 70.0-79.9 Final Exam 20%
D: 60.0-69.9 Homework 15%
F: 59.9 and below
Class Preparedness/Supplies: Students must come to class prepared to learn.
Graphing calculators are required for daily use, no cell phone calculators will be allowed.
Notebook paper, pencil, and a red pen are required to complete assignments. All assignments must be done in pencil and corrected in red pen for full credit.
Books are required for daily use.
Three-ring binder for organizational use of notes, assignments, quizzes, and tests.
Homework: Homework is assigned on a daily basis and will be corrected the next class period. In order to be successful, it is critical to practice new ideas and concepts. Homework is graded on a
scale of 1 to 10 and will be graded on completion, effort, and demonstration of techniques rather than correctness. Homework assignment must be marked with red pen while answers are read in
class. Any problems that are demonstrated during this time must show correct work in order to earn full credit. If an excused absence occurs, students have one day per missed class to turn in
Policies: In order to have a successful and safe learning environment the following policies will be enforced:
Act with kindness
o Be respectful to each other, the teacher and to the things in the room. This includes putting cell phones away when entering the room and not taking them out until teacher allows.
o Do not write or doodle on desks.
Work together
o This course emphasizes collaboration: class discussions, group work, and pair-sharing. There is a lot to be learned from your peers and having the opportunities to articulate, challenge, and
defend ideas will strengthen every individuals mathematical understanding. We are all in this together, so let’s work together.
No cheating
o Cheating is not tolerated. If you are caught cheating, you will be given a zero and disciplinary actions will follow. This applies to the person(s) caught cheating as well as any individual(s)
caught contributing.
Electronic Devices
o Cell phones and other electronic devices will NOT be tolerated during class time. It is expected that all electronic devices be put AWAY by at the beginning of class and not taken out until
teacher approval is given. Cell phones may NOT be used as calculators.
Website: If you miss class, please check the website calendar to see what you have missed. You should always check the class webpage before emailing the teacher to see what you missed.
Additional Support: Further academic support is offered through the academic center , appointments are not needed. If at any point you wish to get a private tutor your guidance counselor can
provide you with a district approved list. | {"url":"https://www.wlwv.k12.or.us/Page/3952","timestamp":"2024-11-12T20:19:18Z","content_type":"text/html","content_length":"465429","record_id":"<urn:uuid:d61b0706-d269-4f5e-8cd2-4f04b6c3cf50>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00408.warc.gz"} |
Bits To Yobibits Converter
Enter the number of bits:
Yobibits (Yibit)
Navigating the Digital Data Seas: A Comprehensive Bits to Yobibits Guide
Deciphering Digital Dimensions
In the expansive realm of digital data, the journey from the microscopic “bits” to the colossal “yobibits” represents a pivotal shift in comprehending and managing vast amounts of information. This
exploration aims to demystify the fundamental concepts of bits and yobibits while introducing an efficient converter designed to ease the transition.
Bits: The Foundation of Digital Language
At the core of digital information lie “bits” — the elemental components that make up the language of computers. These binary digits, denoted as 0s and 1s, act as the groundwork for encoding a
plethora of digital content. Whether it’s textual data, images, or videos, bits are the fundamental building blocks that define and represent every facet of the digital landscape.
Yobibits: Navigating the Seas of Immense Data
On the far end of the data spectrum, we encounter “yobibits” (Yibit), a term synonymous with extraordinary data magnitudes. A single yobibit is an astonishing 2^80 bits, making it the go-to unit for
measuring and grappling with the immense capacities of data storage. The realm of yobibits becomes prominent in discussions involving supercomputing, global data traffic, and the overwhelming expanse
of information characterizing our modern digital era.
Bridging the Gap: Converting Bits to Yobibits
The journey from bits to yobibits is akin to a transformation from the infinitesimal to the astronomical, requiring a bridge to navigate this vast expanse of digital dimensions. To facilitate this
transition, behold the “Bits to Yobibits Converter” provided above. The conversion process is elegantly simple:
javascriptCopy code
Number of Yobibits (Yibit) = Number of Bits / 2^80
Executing this conversion is a breeze — input the quantity of bits you aim to convert, click the “Convert” button, and witness the rapid display of the equivalent yobibits. This tool is a
cornerstone for professionals dealing with extensive datasets, streamlining the management of colossal data volumes in our data-centric world.
Navigating the Digital Data Landscape
In conclusion, understanding the transition from bits to yobibits is pivotal for anyone navigating the vast seas of digital information. Whether you’re a tech enthusiast, a data scientist, or an IT
professional, this guide and converter are invaluable resources in deciphering the language of digital data and managing its ever-expanding dimensions. | {"url":"https://www.calculators24.com/bits-to-yobibits-converter/","timestamp":"2024-11-06T04:14:00Z","content_type":"text/html","content_length":"182167","record_id":"<urn:uuid:11608c84-eed6-49a1-bbc0-7114a9e2ec0f>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00746.warc.gz"} |
In triangle ABC, how do you solve the right triangle given 12, 12 & 20 as sides? | Socratic
In triangle ABC, how do you solve the right triangle given 12, 12 & 20 as sides?
1 Answer
Largest angle $\cong 1.970 \left(\text{radians}\right) \cong {112.9}^{\circ}$
Other two angles $\cong 0.586 \left(\text{radians}\right) \cong {33.6}^{\circ}$
Note: despite the question statement, a triangle with sides $12 , 12 , 20$ is not a right triangle (the sides do not satisfy the Pythagorean Theorem).
The Cosine Law tells us:
$\cos \left(c\right) = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}$
Setting $c$ as the longest side ($20$)
we can calculate
$\textcolor{w h i t e}{\text{XXX}} \cos \left(c\right) = \frac{7}{18}$
Then use a calculator (or similar) determine
$\textcolor{w h i t e}{\text{XXX}} c = \arcsin \left(\frac{7}{18}\right) = 1.970$ (radians)
Similar calculations can be made for $a$ and $b$
noting that $a = b$ and $a + b + c = \pi$ we can calculate $a$ and #b" that way.
Impact of this question
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Managing a Derivative Portfolio & Strategies Thereof
Kiran Kumar K V
The risk of a derivative product depends on what an investor does with it. It is not the fork’s fault if someone sticks it in an electric outlets and gets shocked. People should think of derivatives
as neutral products that can be combined with other assets to create a more preferred risk-return trade-off. If used widely, they can help an investor or portfolio manager quickly adapt to changing
market conditions or client needs. In this article, we discuss certain dimensions of building and managing the portfolio of derivative products.
Setting Investment Objective
Every trade should begin with an opinion of the underlying market. Only then can an informed trading decision be made. With stocks and most assets, one thinks about the direction of the market: Is it
going up or down? With options, it is not enough to think about only the market direction: it is also necessary to think about the volatility. In other words, what matters to users of options is not
just the direction the underlying is headed, but the volatility of the underlying. For example, in a simple call option purchase, just having the underlying go up is not sufficient to make money. The
underlying must go up enough that the option reaches breakeven. The gain in the value of the stock must be sufficient that the call overcomes the loss of its time value. A long call must have some
upside volatility, and a long put must have some downside volatility.
Suppose an investor is neutral on market direction. Again, he should consider the volatility. A straddle buy is neither bullish nor bearish, but expects a sharp increase in volatility. Someone who is
neutral directionally, but does not anticipate a sudden price change may want to write the straddle. Spreads tend to be middle-of-the-road strategies that fit best in neutral, flat markets. A spread
trade is often appropriate when the outlook for an underlying market is either neutral or non-trending, meaning that it does not have a strong bullish or bearish outlook. In table-1 below, one way of
looking at the interplay of direction and volatility, is presented:
│ │ │ Direction │
│ │ │ Bearish │Neutral / No Bias│ Bullish │
│ │ High │ Buy Puts │ Buy Straddle │ Buy Calls │
│ ├───────┼─────────────────────────┼─────────────────┼─────────────────────────┤
│Volatility│Average│Write Calls and Buy Puts │ Spreads │Buy Calls and Write Puts │
│ ├───────┼─────────────────────────┼─────────────────┼─────────────────────────┤
│ │ Low │ Write Calls │ Write Straddle │ Write Puts │
Some option traders refer to the concept of volatility as speed, with high volatility market being a fast market and a low volatility market being a slow market. Market speed can, however, also refer
to the rate at which a market moves, so investors should take care to be aware of the distinction in the uses of these characterisations.
Market Risk
Derivatives enable market participants to take position that is extremely bearish, extremely bullish, or somewhere in between and to quickly and efficiently shift along this continuum as desired.
Suppose a pension fund owns one million shares of HSBC holdings. For some reason, the portfolio manager would like to temporarily reduce the position by 10%, meaning to reduce the market exposure and
convert the equivalent funds to cash. There are a variety of ways in which this might be done. The portfolio manager could do any of the below:
1. Sell 100000 shares, which is 10% of the holding
2. Enter into a futures or forward contract to sell 100000 shares
3. Write call contracts sufficient to generate minus 100000 delta points
4. Buy put contracts sufficient to generate minus 100000 delta points
5. Enter into a collar sufficient to generate minus 100000 delta points
Each of these alternatives has its own strengths and weaknesses. The first alternative, i.e., selling shares, has the advantage of clearly accomplishing the goal of a 10% reduction. For some
investors, though this solution could create tax problem or result in inadvertently putting downward pressure on the stock price. Forward contracts are simple and effective, but they involve
counterparty risk and are not easily cancel of later there is a desire to unwind the trade (It is possible to enter into an offsetting trade with the same counterparty, to net the position to zero.
This approach is essentially the same as covering a futures contract position in which the counterparty is the clearing house. Offsetting a forward with a new forward is relatively easy to do).
Writing calls brings in a cash premium, but leaves the writer subject to exercise risk. Buying the puts requires a cash outlay but leaves the investor in control with respect to exercise risk. In
short, all the outcomes have advantages and disadvantages, but with derivatives, the investor has multiple choices instead of just one.
Analytics of the Break Even Price
Investors often construct a profit and loss diagram for an option strategy that is under consideration. These diagrams are helpful in understanding the range of possible outcomes. It may also make
sense to use option pricing theory to learn more about what the break even points on the profit and loss diagram really mean. The current option price is based on an outlook that takes into account
an assumed future volatility of the prices of the underlying asset. Volatility is measured by standard deviation of percentage changes in the spot price of the underlying asset, and with an
understanding of some basic statistical principles, this volatility can be used to estimate the likelihood of achieving a particular price target.
There are a variety of factors that come into play when determining the value of an option. The underlying market price, the exercise price of the option contract, the time left until option
expiration, the current risk-free rate, and any dividends paid before expiration are all taken into account by the market when valuing an option. The exercise price is known, and the underlying
market price and risk-free rate are easily accessible. The time remaining until expiration is consistently changing, but we always are aware of how much time is left until expiration. Dividends paid
by companies are fairly stable as well.
30 days until March option expiration
A and B underlying stock price = Rs. 50
A Mar 50 Call= Re. 1.00
A Mar 50 Put = Re. 0.95
B Mar 50 Call= Re. 2.50
B Mar 50 Put = Re. 2.45
There is another pricing factor that is taken into account, and that is the expected price movement or volatility of the underlying stock. The more volatility that is expected from the underlying
over the life of an option, the higher the option premium. Consider two stocks, A and B that are both trading at Rs. 50. A is a utility company that is expected to not have much volatility over the
next month. B, however, is a biotech company that often experiences 5% price moves in a single day. All else being same, options on A would have lower premiums than those on B. The price data may
look approximately like this:
A trader buying a A Mar 50 Call and B Mar 50 put would be purchasing a A Mar 50 straddle for 1.95. Buying a B Mar 50 straddle would involve purchasing the B Mar 50 call at 2.50 and B Mar 50 put at
2.45 for a net cost of 4.95. Breakeven at expiration for the A straddle occurs if the stock moves up or down 1.95 whereas breakeven for the B straddle would require a move of 4.95. In percentage
terms, this means that breakeven for the A straddle is 3.90% (1.95/50) whereas B needs to rise or fall 9.90% (4.95/50) for the straddle to just breakeven.
We say that B options have a higher implied volatility than comparable A option contracts. Implied volatility is the standard deviation that causes an option pricing model to give the current option
price. Option users, in fact, use implied volatility as a form of option currency, meaning that prices are quoted as implied volatilities. For instance, knowing that an option sells for 2.00 reveals
nothing about its relative price because that depends on the moneyness (i.e., the extent to which the option is in or out of the money) and the remaining life of the option. If, instead, someone says
an option sells for an annualised implied volatility of 45% that is a standalone statistic that can be directly compared with other options. B option premiums are higher than A option premiums
because the market expects greater potential price moves out of B than from A.
There are a few different perspectives on how we should measure the time periods with option volatility. There are usually 365 calendar days in a year, but the markets are not open on Saturdays and
Sundays or official holidays, which vary by country. Because the stock price does not have the opportunity to change when the market is closed, most experts believe that those days should not count.
For this reason, some people use 252 day or some other appropriate number for the number of trading days in a year. Dispersion is a function of both the size of the “jumps” in the variable and the
number of those jumps. We convert an annual variance (σ^2) to a daily variance by dividing by 252, and we convert an annual standard deviation (σ) to a daily standard deviation by dividing by √252.
With options, volatility is measured by the annual standard deviation. | {"url":"https://www.isme.in/managing-a-derivative-portfolio-strategies-thereof/","timestamp":"2024-11-13T20:52:57Z","content_type":"text/html","content_length":"181395","record_id":"<urn:uuid:6d83e8f3-4b9f-4a99-8edf-5d5c16a47d7c>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00143.warc.gz"} |
Advanced Algorithms
Graduate Course - Summer Term 2019
Fabian Kuhn
Course description
In the course, we discuss modern algorithmic techniques. The course covers a variety of topics, such as for example:
• approximation algorithms
• randomized algorithms
• graph embeddings
• graph sparsification
• theory of learning
• sketching and streaming algorithms
• continuous methods in combinatorial optimization
There is no formal requirement, however some background in algorithm design/analysis and probability theory is expected. Having passed the algorithm theory course (or a similar course) prior to
taking the advanced algorithms lecture is highly recommended.
• Lecture: Friday 10:00 - 12:00, Building 106 SR 00 007
• Exercise Class: Friday 12:00 - 14:00, Building 106 SR 00 007
There will be a weekly exercise sheet, which is published here. While exercise sheets are not graded, you can do it yourself and submit your solution to us and we will give you some feedback on it.
This is recommended, but not mandatory (in order to take the exam). Solutions of the exercise sheet will be discussed in the following exercise class.
Exercise sheet Assigned Solution
Problem Set 01 (Set Cover, Integrality Gap) 26.04.2019 Sample Solution 01
Problem Set 02 (Chernoff Bounds) 03.05.2019 Sample Solution 02
Problem Set 03 (Probabilistic Tree Embeddings) 10.05.2019 Sample Solution 03
Problem Set 04 (Cut Based Tree Decompositions) 17.05.2019 Sample Solution 04
Problem Set 05 (Multiplicative Weight Updates) 27.05.2019 Sample Solution 05
Problem Set 06 (Learning Linear Classifiers with MWU) 31.05.2019 Sample Solution 06
Problem Set 07 (Max Flow, Zero Sum Games, MWU) 14.06.2019 Sample Solution 07
Problem Set 08 (Additive & Multipl. Spanners, APSP) 28.06.2019 Sample Solution 08
Problem Set 09 (Cuts, Sparsifiers) 05.07.2019 Sample Solution 09
Problem Set 10 (Max Flow, Cong. Approximators) 15.07.2019 Sample Solution 10
Problem Set 11 (Aggregation in the MPC Model) 23.07.2019 Sample Solution 11
Lecture Material
All slides and recordings can be found on our Webserver. | {"url":"https://ac.informatik.uni-freiburg.de/teaching/ss_19/advanced_algorithms.php","timestamp":"2024-11-08T17:28:42Z","content_type":"application/xhtml+xml","content_length":"12787","record_id":"<urn:uuid:e0f7d34a-3042-4a9d-b2f6-e202c0780d72>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00634.warc.gz"} |
Online math problem solver
online math problem solver
Author Message Author Message
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Vofj Timidrov Posted: Friday 05th of Jan 12:08 Vnode Posted: Tuesday 09th of Jan 13:45
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Koem Posted: Thursday 11th of Jan 07:44
Voumdaim of Obpnis Posted: Saturday 06th of Jan 07:09
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How do you solve equations in order?
How do you solve equations in order?
In mathematics, the order of operations define the priority in which complex equations are solved. The top priority is your parenthesis, then exponents, followed by multiplication and division, and
finally addition and subtraction (PEMDAS).
How do you put brackets in a quote?
When writers insert or alter words in a direct quotation, square brackets—[ ]—are placed around the change….Bracket Use: Quick Summary.
Do Don’t
Use brackets to enclose a change in letter case or verb tense when integrating a quote into your paper. Use bracketed material in a way that twists the author’s meaning.
How do you use brackets in math?
Brackets are used to provide clarity in the order of operations, the order in which several operations should be done in a mathematical expression. For example, suppose you have the following
expression: 2 + 4 * 6 – 1.
What are square brackets used for in quotations?
Use square brackets to include words within a quote that are not part of the original quote. For example, if a quoted passage is not entirely clear, words enclosed in square brackets can be added to
clarify the meaning.
What are brackets used for?
Brackets (parentheses) are punctuation marks used within a sentence to include information that is not essential to the main point. Information within parentheses is usually supplementary; were it
removed, the meaning of the sentence would remain unchanged.
Do you multiply or add first without brackets?
Order of operations tells you to perform multiplication and division first, working from left to right, before doing addition and subtraction. Continue to perform multiplication and division from
left to right. Next, add and subtract from left to right.
What is the difference between parentheses and brackets in math?
Parentheses are smooth and curved ( ), brackets are square [ ], and braces are curly { }. In mathematics, they are mostly used for order of operations.
What do brackets mean in MLA?
Academic style guides such as the MLA Handbook for Writers of Research Papers go into such matters at great length. The most common use of brackets is to enclose explanatory matter that one adds in
editing the work of another writer. They indicate that some kind of alteration has been made in the original text.
Which bracket should be solved first?
Ans: According to BODMAS rule, the brackets have to be solved first followed by powers or roots (i.e. of), then Division, Multiplication, Addition and at the end Subtraction. Solving any expression
is considered correct only if the BODMAS rule or the PEMDAS rule is followed to solve it.
In mathematics, the order of operations define the priority in which complex equations are solved. The top priority is your parenthesis, then exponents, followed by multiplication and division, and
finally addition and subtraction (PEMDAS)….
What are the most important math skills?
Key Math Skills for School
• Number Sense. This is the ability to count accurately—first forward.
• Representation. Making mathematical ideas “real” by using words, pictures, symbols, and objects (like blocks).
• Spatial sense.
• Measurement.
• Estimation.
• Patterns.
• Problem-solving.
What is the correct order of operations in math?
The order of operations is a rule that tells the correct sequence of steps for evaluating a math expression. We can remember the order using PEMDAS: Parentheses, Exponents, Multiplication and
Division (from left to right), Addition and Subtraction (from left to right).
How can I teach myself basic math?
How to Teach Yourself Math
1. Step One: Start with an Explanation. The first step to learning any math is to get a first-pass explanation of the topic.
2. Step Two: Do Practice Problems.
3. Step Three: Know Why The Math Works.
4. Step Four: Play with the Math.
5. Step Five: Apply the Math Outside the Classroom.
What are the four fundamentals of mathematics?
Definition: The four fundamental operations in mathematics are addition, subtraction, multiplication and division.
How do you simplify fractions step by step?
How to Reduce Fractions
1. Write down the factors for the numerator and the denominator.
2. Determine the largest factor that is common between the two.
3. Divide the numerator and denominator by the greatest common factor.
4. Write down the reduced fraction.
How do you brush up on basic math skills?
How to Improve Math Skills
1. Focus on Understanding Concepts. You can memorize formulas and rules to complete many math problems, but this doesn’t mean that you understand the underlying concepts behind what you’re doing.
2. Go Over New Concepts and Practice Problems.
3. Solve Extra Problems.
4. Change Word Problems Up.
5. Apply Math to Real Life.
6. Study Online.
Is 10th easier than 9th?
Most of say that 10th class is easy than 9th. It’s only because, the chapters you study in 10th class are extension of what you study in 9th. Most of say that 10th class is easy than 9th. It’s only
because, the chapters you study in 10th class are extension of what you study in 9th.
What is basic and standard maths?
The Standard exam is meant for students who wish to study Mathematics in higher classes, while the Mathematics Basic is taken by students who do not wish to pursue advanced Mathematics in higher
classes. As a result, the level of difficulty of the Standard exam was way higher than the Basic exam….
How do you simplify powers?
To simplify a power of a power, you multiply the exponents, keeping the base the same. For example, (23)5 = 215. For any positive number x and integers a and b: (xa)b= xa· b. Simplify.
What are the four basic rules of algebra?
The Basic Laws of Algebra are the associative, commutative and distributive laws. They help explain the relationship between number operations and lend towards simplifying equations or solving them.
The arrangement of addends does not affect the sum….
What is the easiest topic in math?
Which topic is so easy in mathematics? If it were so easy, then it would have been no fun….
• Arithmetic.
• Algebra.
• Geometry.
• Trigonometry.
• Analysis – Calculus, Differential Equations, Vector Analysis, etc.
• Linear algebra/matrices.
• Finite math – Probability, Combinatorics, Set theory, etc.
• Statistics.
How do you simplify Surds?
In general: To simplify a surd, write the number under the root sign as the product of two factors, one of which is the largest perfect square. Note that the factor 16 is the largest perfect square.
Recall that the numbers 1, 4, 9, 16, 25, 36, 49, are perfect squares.
What is basic of maths?
Basic math is nothing but the simple or basic concept related with mathematics. Generally, counting, addition, subtraction, multiplication and division are called the basic math operation. One can
understand the application of these concept and their use in practical life through the word problems. …
What comes first in order of operations multiplication or division?
Order of operations tells you to perform multiplication and division first, working from left to right, before doing addition and subtraction.
How is maths used in everyday life?
Math Matters in Everyday Life
1. Managing money $$$
2. Balancing the checkbook.
3. Shopping for the best price.
4. Preparing food.
5. Figuring out distance, time and cost for travel.
6. Understanding loans for cars, trucks, homes, schooling or other purposes.
7. Understanding sports (being a player and team statistics)
8. Playing music.
How do you simplify brackets?
Removing brackets
1. To remove brackets, multiply the term on the outside of the bracket with each term inside the brackets.
2. Here, we combine multiplying out brackets and collecting like terms, to simplify algebraic expressions.
3. Have a look at the example questions below.
How do I write a basic maths application?
I am writing this letter with a request for choosing Basic Math in Class ten. I want to take the subject as I have a keen interest in studying the subject. I hope I’ll be allotted with the same after
promoting to standard tenth. For this, I will be highly obliged to you….
Is basic math easy?
In simple terms, Basic Mathematics is supposed to be easier than Standard Mathematics. But that is not the only case. The difference in the two papers is not limited to the difficulty-level of the
What are maths skills?
What are math skills? Math skills help individuals deal with basic, everyday tasks—from getting to work on time to paying bills. Students learn these skills in school, and as they get older and
obtain a job, they often use them more frequently. Math skills are important for both work and personal life….
What are the 4 basic operations?
The four operations are addition, subtraction, multiplication and division. | {"url":"https://gowanusballroom.com/how-do-you-solve-equations-in-order/","timestamp":"2024-11-04T15:32:09Z","content_type":"text/html","content_length":"62267","record_id":"<urn:uuid:f753e613-c3a9-4698-82be-a85568c0b3dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00830.warc.gz"} |
Reducing probability of decision error using stochastic resonance
The problem of reducing the probability of decision error of an existing binary receiver that is suboptimal using the ideas of stochastic resonance is solved. The optimal probability density function
of the random variable that should be added to the input is found to be a Dirac delta function, and hence, the optimal random variable is a constant. The constant to be added depends upon the
decision regions and the probability density functions under the two hypotheses and is illustrated with an example. Also, an approximate procedure for the constant determination is derived for the
mean-shifted binary hypothesis testing problem.
• Modeling
• Pattern classification
• Signal detection
ASJC Scopus subject areas
• Signal Processing
• Electrical and Electronic Engineering
• Applied Mathematics
Dive into the research topics of 'Reducing probability of decision error using stochastic resonance'. Together they form a unique fingerprint. | {"url":"https://experts.syr.edu/en/publications/reducing-probability-of-decision-error-using-stochastic-resonance","timestamp":"2024-11-14T04:22:49Z","content_type":"text/html","content_length":"51000","record_id":"<urn:uuid:f7722a1c-ac66-4a42-a021-2f319faf7b48>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00315.warc.gz"} |
Classification: Accuracy, recall, precision, and related metrics | Machine Learning | Google for Developers
True and false positives and negatives are used to calculate several useful metrics for evaluating models. Which evaluation metrics are most meaningful depends on the specific model and the specific
task, the cost of different misclassifications, and whether the dataset is balanced or imbalanced.
All of the metrics in this section are calculated at a single fixed threshold, and change when the threshold changes. Very often, the user tunes the threshold to optimize one of these metrics.
Accuracy is the proportion of all classifications that were correct, whether positive or negative. It is mathematically defined as:
\[\text{Accuracy} = \frac{\text{correct classifications}}{\text{total classifications}} = \frac{TP+TN}{TP+TN+FP+FN}\]
In the spam classification example, accuracy measures the fraction of all emails correctly classified.
A perfect model would have zero false positives and zero false negatives and therefore an accuracy of 1.0, or 100%.
Because it incorporates all four outcomes from the confusion matrix (TP, FP, TN, FN), given a balanced dataset, with similar numbers of examples in both classes, accuracy can serve as a
coarse-grained measure of model quality. For this reason, it is often the default evaluation metric used for generic or unspecified models carrying out generic or unspecified tasks.
However, when the dataset is imbalanced, or where one kind of mistake (FN or FP) is more costly than the other, which is the case in most real-world applications, it's better to optimize for one of
the other metrics instead.
For heavily imbalanced datasets, where one class appears very rarely, say 1% of the time, a model that predicts negative 100% of the time would score 99% on accuracy, despite being useless.
Recall, or true positive rate
The true positive rate (TPR), or the proportion of all actual positives that were classified correctly as positives, is also known as recall.
Recall is mathematically defined as:
\[\text{Recall (or TPR)} = \frac{\text{correctly classified actual positives}}{\text{all actual positives}} = \frac{TP}{TP+FN}\]
False negatives are actual positives that were misclassified as negatives, which is why they appear in the denominator. In the spam classification example, recall measures the fraction of spam emails
that were correctly classified as spam. This is why another name for recall is probability of detection: it answers the question "What fraction of spam emails are detected by this model?"
A hypothetical perfect model would have zero false negatives and therefore a recall (TPR) of 1.0, which is to say, a 100% detection rate.
In an imbalanced dataset where the number of actual positives is very, very low, say 1-2 examples in total, recall is less meaningful and less useful as a metric.
False positive rate
The false positive rate (FPR) is the proportion of all actual negatives that were classified incorrectly as positives, also known as the probability of false alarm. It is mathematically defined as:
\[\text{FPR} = \frac{\text{incorrectly classified actual negatives}} {\text{all actual negatives}} = \frac{FP}{FP+TN}\]
False positives are actual negatives that were misclassified, which is why they appear in the denominator. In the spam classification example, FPR measures the fraction of legitimate emails that were
incorrectly classified as spam, or the model's rate of false alarms.
A perfect model would have zero false positives and therefore a FPR of 0.0, which is to say, a 0% false alarm rate.
In an imbalanced dataset where the number of actual negatives is very, very low, say 1-2 examples in total, FPR is less meaningful and less useful as a metric.
Precision is the proportion of all the model's positive classifications that are actually positive. It is mathematically defined as:
\[\text{Precision} = \frac{\text{correctly classified actual positives}} {\text{everything classified as positive}} = \frac{TP}{TP+FP}\]
In the spam classification example, precision measures the fraction of emails classified as spam that were actually spam.
A hypothetical perfect model would have zero false positives and therefore a precision of 1.0.
In an imbalanced dataset where the number of actual positives is very, very low, say 1-2 examples in total, precision is less meaningful and less useful as a metric.
Precision improves as false positives decrease, while recall improves when false negatives decrease. But as seen in the previous section, increasing the classification threshold tends to decrease the
number of false positives and increase the number of false negatives, while decreasing the threshold has the opposite effects. As a result, precision and recall often show an inverse relationship,
where improving one of them worsens the other.
Try it yourself:
What does NaN mean in the metrics?
NaN, or "not a number," appears when dividing by 0, which can happen with any of these metrics. When TP and FP are both 0, for example, the formula for precision has 0 in the denominator, resulting
in NaN. While in some cases NaN can indicate perfect performance and could be replaced by a score of 1.0, it can also come from a model that is practically useless. A model that never predicts
positive, for example, would have 0 TPs and 0 FPs and thus a calculation of its precision would result in NaN.
Choice of metric and tradeoffs
The metric(s) you choose to prioritize when evaluating the model and choosing a threshold depend on the costs, benefits, and risks of the specific problem. In the spam classification example, it
often makes sense to prioritize recall, nabbing all the spam emails, or precision, trying to ensure that spam-labeled emails are in fact spam, or some balance of the two, above some minimum accuracy
Metric Guidance
Use as a rough indicator of model training progress/convergence for balanced datasets.
Accuracy For model performance, use only in combination with other metrics.
Avoid for imbalanced datasets. Consider using another metric.
Recall Use when false negatives are more expensive than false positives.
(True positive rate)
False positive rate Use when false positives are more expensive than false negatives.
Precision Use when it's very important for positive predictions to be accurate.
(Optional, advanced) F1 score
The F1 score is the harmonic mean (a kind of average) of precision and recall.
Mathematically, it is given by:
\[\text{F1}=2*\frac{\text{precision * recall}}{\text{precision + recall}} = \frac{2\text{TP}}{2\text{TP + FP + FN}}\]
This metric balances the importance of precision and recall, and is preferable to accuracy for class-imbalanced datasets. When precision and recall both have perfect scores of 1.0, F1 will also have
a perfect score of 1.0. More broadly, when precision and recall are close in value, F1 will be close to their value. When precision and recall are far apart, F1 will be similar to whichever metric is
Exercise: Check your understanding
A model outputs 5 TP, 6 TN, 3 FP, and 2 FN. Calculate the recall.
Recall is calculated as [\frac{TP}{TP+FN}=\frac{5}{7}].
Recall considers all actual positives, not all correct classifications. The formula for recall is [\frac{TP}{TP+FN}].
Recall considers all actual positives, not all positive classifications. The formula for recall is [\frac{TP}{TP+FN}]
A model outputs 3 TP, 4 TN, 2 FP, and 1 FN. Calculate the precision.
Precision is calculated as [\frac{TP}{TP+FP}=\frac{3}{5}].
Precision considers all positive classifications, not all actual positives. The formula for precision is [\frac{TP}{TP+FP}].
Precision considers all positive classifications, not all correct classifications. The formula for precision is [\frac{TP}{TP+FP}]
You're building a binary classifier that checks photos of insect traps for whether a dangerous invasive species is present. If the model detects the species, the entomologist (insect scientist) on
duty is notified. Early detection of this insect is critical to preventing an infestation. A false alarm (false positive) is easy to handle: the entomologist sees that the photo was misclassified and
marks it as such. Assuming an acceptable accuracy level, which metric should this model be optimized for?
In this scenario, false alarms (FP) are low-cost, and false negatives are highly costly, so it makes sense to maximize recall, or the probability of detection.
False positive rate (FPR)
In this scenario, false alarms (FP) are low-cost. Trying to minimize them at the risk of missing actual positives doesn't make sense.
In this scenario, false alarms (FP) aren't particularly harmful, so trying to improve the correctness of positive classifications doesn't make sense. | {"url":"https://developers.google.com/machine-learning/crash-course/classification/accuracy-precision-recall?ref=broadbandbreakfast.com","timestamp":"2024-11-06T12:41:45Z","content_type":"text/html","content_length":"185802","record_id":"<urn:uuid:6eb8c24d-779c-447d-a5fa-efea1b7cfc4a>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00116.warc.gz"} |
How many shovels in a yard of dirt - Civil Sir
How many shovels in a yard of dirt
The number of shovels needed to move a cubic yard of dirt depends on the volume of each shovel and how densely the dirt is packed. If, for example, a shovel holds 0.0069 cubic yards of dirt, then you
would need 145 shovels to move one cubic yard. The actual number would depend on the specific shovel size and the dirt’s density.
It is not a standard unit for measurement, it is just calculated on average and depends on how shovel is big in size and how heaped up.
How many shovels in a yard of dirt
It all depends on size of shovel, how is big or small and how “heaped up” the shovel is each time. It would also depend on weight of dirt or top soil.
Dirt or soil is made up of varying amounts of minerals, decaying organic matter, nutrients and living organisms. How much 1 cubic yard of dirt weighs depends on the amounts of the components it’s
made up of.
Weight of dirt depends on, dry, moist, loose and compact condition, one cubic yard of dry sandy dirt weighs about 2,000 pounds, while 1 cubic yard of dry clay soil weighs in around 1,700 pounds. If
you want an exact weight, the supplier can tell you when you purchase the soil.
But roughly measurement with shovel make estimate, shovel are required in brick work, concrete work, removing a lot of sand and cement, top soil or dirt so here is my estimate.
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How many shovels in a yard of dirt
A cubic yard of dirt is 3 feet long by 3 feet wide and 3 feet height of heap, such that 1 cubic yard = 27 cubic feet, generally 5 to 6 shovels full need to heaped up 1 cubic feet of dirt. So my
estimate is, there are 135 to 162 shovels full needs to heaped up one yard of dirt or top soil. But your mileage will vary depending on the shovel size and how “heaped” each shovel is.
2D and 3D Ghar ka Naksha banane ke liye sampark kare
How many shovels in a bulk bag of sand and ballast
How many shovels in a yard of sand
How to figure yards, cubic feet or bag of dirt
How many shovels of dirt in a cubic yard.
A cubic yard of dirt is 3 feet long by 3 feet wide and 3 feet height of heap, such that 1 cubic yard = 27 cubic feet, generally 5 to 6 shovels full need to heaped up 1 cubic feet of dirt. So my
estimate is, there are 135 to 162 shovels full needs to heaped up one cubic yard of dirt or top soil. But your mileage will vary depending on the shovel size and how “heaped” each shovel is. | {"url":"https://civilsir.com/how-many-shovels-in-a-yard-of-dirt/","timestamp":"2024-11-09T16:31:18Z","content_type":"text/html","content_length":"85616","record_id":"<urn:uuid:219c7f4e-8f7d-4c23-ae2b-27a369eced36>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00384.warc.gz"} |
Probability Theory: Examples and Applications
Are you an EPFL student looking for a semester project?
Work with us on data science and visualisation projects, and deploy your project as an app on top of Graph Search.
This lecture covers various probability theory concepts and applications, such as calculating the probability of specific events like consecutive zeros in bit strings, independence in Bernoulli
trials, and the Monty Hall problem. It also delves into the generalized Bayes' theorem and the distribution of random variables.
This page is automatically generated and may contain information that is not correct, complete, up-to-date, or relevant to your search query. The same applies to every other page on this website.
Please make sure to verify the information with EPFL's official sources. | {"url":"https://graphsearch.epfl.ch/en/lecture/0_7bow1z1s","timestamp":"2024-11-05T18:18:54Z","content_type":"text/html","content_length":"115972","record_id":"<urn:uuid:43bac6fe-d069-4b48-8249-28f707556b1d>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00363.warc.gz"} |
Combination Calculator: An In-Depth Exploration
In the realm of mathematics and probability, combinations play a critical role in understanding how elements can be grouped or selected from a larger set. A combination calculator is a tool designed
to facilitate these calculations, making it easier for users to determine the number of possible ways to select items from a set. This article provides a thorough examination of combination
calculators, their importance, applications, and how they function.
Understanding Combinations
What Are Combinations?
Combinations refer to the different ways in which a subset of items can be selected from a larger set, where the order of selection does not matter. For example, if you are selecting team members
from a group, the order in which you select them is irrelevant—what matters is the group itself.
The Role of Combinations in Mathematics
In mathematics, combinations are fundamental to combinatorial analysis, a branch concerned with counting, arrangement, and combination of objects. They are crucial in probability theory, where
determining the likelihood of various outcomes often involves combinations.
Difference Between Combinations and Permutations
While combinations focus on the selection of items without regard to order, permutations involve arranging items in a specific sequence. The distinction between these two concepts is vital in various
mathematical applications, but both are interrelated.
The Functionality of a Combination Calculator
Purpose of a Combination Calculator
A combination calculator is designed to simplify the process of determining the number of ways items can be selected from a larger set. This tool is especially useful in fields such as statistics,
probability, and computer science, where understanding possible outcomes is crucial.
How a Combination Calculator Works
Combination calculators use algorithms to compute the number of possible combinations based on the input parameters. While the specific mechanics of these algorithms can vary, they generally follow
principles derived from combinatorial mathematics.
User Input and Output
Users typically input two main parameters into a combination calculator: the total number of items in the set and the number of items to be selected. The calculator then processes these inputs to
provide the number of possible combinations. The output is presented in a straightforward manner, often as a single numerical value representing the total number of combinations.
Applications of Combination Calculators
In Probability and Statistics
In probability and statistics, combination calculators are used to determine the likelihood of various outcomes. For instance, they are employed in calculating probabilities in card games, lottery
draws, and other scenarios where combinations of items are relevant.
In Game Theory
Game theory often involves scenarios where players must make decisions based on possible combinations of strategies or moves. Combination calculators help analyze these possibilities and develop
optimal strategies.
In Cryptography
Cryptography relies on combinatorial mathematics to secure information. Combination calculators assist in evaluating the strength of encryption methods by determining the number of possible keys or
In Research and Data Analysis
Researchers use combination calculators to analyze data and design experiments. By understanding the number of possible combinations, researchers can ensure that their studies are robust and
Choosing the Right Combination Calculator
Features to Consider
When selecting a combination calculator, users should consider several factors, including:
• Accuracy: Ensure that the calculator provides accurate results based on the inputs.
• User Interface: A user-friendly interface can make it easier to input data and interpret results.
• Additional Functions: Some calculators offer extra features, such as the ability to perform permutation calculations or integrate with other mathematical tools.
Online vs. Offline Calculators
Combination calculators are available in both online and offline formats. Online calculators are easily accessible through web browsers, while offline calculators can be downloaded as software or
applications. The choice between online and offline options depends on the user's preferences and needs.
Practical Tips for Using Combination Calculators
Input Precision
Ensure that the values entered into the calculator are precise and accurately represent the problem at hand. Even small errors in input can lead to incorrect results.
Understanding the Output
Interpret the results provided by the calculator carefully. The output typically represents the total number of combinations, which can be used to draw conclusions or make decisions.
Integrating with Other Tools
Consider using combination calculators in conjunction with other mathematical tools or software to enhance your analysis and decision-making processes.
Challenges and Limitations
Handling Large Numbers
Combination calculations involving very large numbers can be challenging due to computational limits. Some calculators may struggle with extremely large values, leading to potential inaccuracies.
Software Limitations
Not all combination calculators offer the same level of functionality. Some may lack advanced features or customization options, which can limit their usefulness for specific applications.
Future Trends in Combination Calculators
Advancements in Technology
As technology advances, combination calculators are likely to become more sophisticated, incorporating features such as artificial intelligence and machine learning to enhance their capabilities.
Integration with Data Analysis Tools
Future combination calculators may offer better integration with data analysis and visualization tools, providing users with more comprehensive insights and capabilities.
Combination calculators are essential tools in the fields of mathematics, statistics, probability, and beyond. They simplify the process of determining the number of possible ways to select items
from a set, providing valuable insights for various applications. By understanding how these calculators work and their potential applications, users can leverage them effectively to make informed
decisions and analyze complex scenarios. As technology continues to evolve, combination calculators will undoubtedly become even more powerful and versatile, offering enhanced capabilities and
A tool for finding the number of ways to select items from a set without regard to order.
Enter the total number of items and the number to be selected.
A combination calculator simplifies the process of computing combinations by automating the factorial calculations and applying the combination formula, providing quick and accurate results.
Some may have limitations with very large numbers.
Yes, combination calculators are available both online and as downloadable software. | {"url":"https://hightechtools.online/combination-calculator","timestamp":"2024-11-12T08:48:16Z","content_type":"text/html","content_length":"79204","record_id":"<urn:uuid:81da7b5f-87b9-4d29-89d1-e9698ba9d87e>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00105.warc.gz"} |
Theory of Combinatorial Algorithms
Mittagsseminar (in cooperation with J. Lengler, A. Steger, and D. Steurer)
Mittagsseminar Talk Information
Date and Time: Thursday, January 26, 2006, 12:15 pm
Duration: This information is not available in the database
Location: This information is not available in the database
Speaker: Thomas Bruderer
Comparing Top k Lists
Motivated by several applications, we introduce various distance measures between "top k lists." Some of these distance measures are metrics, while others are not. For each of these latter distance
measures: we show that it is "almost" a metric in the following two seemingly unrelated aspects: step (i) it satisfies a relaxed version of the polygonal (hence, triangle) inequality, and step (ii)
there is a metric with positive constant multiples that bounds our measure above and below.This is not a coincidence---we show that these two notions of almost being a metric are the same. Based on
the second notion, we define two distance measures to be equivalent if they are bounded above and below by constant multiples of each other. We thereby identify a large and robust equivalence class
of distance measures. Besides the applications to the task of identifying good notions of (dis-)similarity between two top k lists, our results imply polynomial-time constant-factor approximation
algorithms for the rank aggregation problem with respect to a large class of distance measures.
By Ronald Fagin, Ravi Kumar, D. Sivakumar (SODA 2003).
Upcoming talks | All previous talks | Talks by speaker | Upcoming talks in iCal format (beta version!)
Previous talks by year: 2024 2023 2022 2021 2020 2019 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2004 2003 2002 2001 2000 1999 1998 1997 1996
Information for students and suggested topics for student talks
Automatic MiSe System Software Version 1.4803M | admin login | {"url":"https://ti.inf.ethz.ch/ew/mise/mittagssem.html?action=show&what=abstract&id=7d1f877feebb46f8d67c35e9d53650a6c70519c0","timestamp":"2024-11-04T09:00:54Z","content_type":"text/html","content_length":"14008","record_id":"<urn:uuid:ea9e82bc-ab9d-476b-acd3-3a98ad618dd4>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00308.warc.gz"} |
Return on Total Assets
The return on assets ratio (ROA) is found by dividing net income by total assets. The higher the ratio, the better the company is at using their assets to generate income. ROA was developed by DuPont
to show how effectively assets are being used. It is also a measure of how much the company relies on assets to generate profit.
Return on Assets
The return on assets ratio is net income divided by total assets. That can then be broken down into the product of profit margins and asset turnover.
Components of ROA
ROA can be broken down into multiple parts. The ROA is the product of two other common ratios - profit margin and asset turnover. When profit margin and asset turnover are multiplied together, the
denominator of profit margin and the numerator of asset turnover cancel each other out, returning us to the original ratio of net income to total assets.
Profit margin is net income divided by sales, measuring the percent of each dollar in sales that is profit for the company. Asset turnover is sales divided by total assets. This ratio measures how
much each dollar in asset generates in sales. A higher ratio means that each dollar in assets produces more for the company.
Limits of ROA
ROA does have some drawbacks. First, it gives no indication of how the assets were financed. A company could have a high ROA, but still be in financial straits because all the assets were paid for
through leveraging. Second, the total assets are based on the carrying value of the assets, not the market value. If there is a large discrepancy between the carrying and market value of the assets,
the ratio could provide misleading numbers. Finally, there is no metric to find a good or bad ROA. Companies that operate in capital intensive industries will tend to have lower ROAs than those who
do not. The ROA is entirely contextual to the company, the industry and the economic environment. | {"url":"https://ftp.worldpossible.org/endless/eos-rachel/RACHEL/RACHEL/modules/en-boundless-static/www.boundless.com/finance/textbooks/boundless-finance-textbook/analyzing-financial-statements-3/profitability-ratios-39/return-on-total-assets-207-601/index.html","timestamp":"2024-11-12T17:33:48Z","content_type":"text/html","content_length":"20226","record_id":"<urn:uuid:3c84f4cb-844b-449f-beaf-d944afee3a73>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00010.warc.gz"} |
Chamfered Icosahedron -- from Wolfram MathWorld
A chamfered icosahedron, also called a tritruncated rhombic triacontahedron, is a polyhedron obtained by chamfering a regular icosahedron. The illustration above shows increasing amounts of
chamfering applied to the regular icosahedron.
An equilateral chamfered icosahedron may be constructed by appropriate choice of the edge length ratio for chamfering. The unit equilateral chamfered icosahedron has surface area given by the root on
an 8th-order polynomial, volume
and is implemented in the Wolfram Language as PolyhedronData["EquilateralChamferedIcosahedron"]. | {"url":"https://mathworld.wolfram.com/ChamferedIcosahedron.html","timestamp":"2024-11-12T15:09:58Z","content_type":"text/html","content_length":"53146","record_id":"<urn:uuid:b742c36a-21d5-46e7-981f-f03d406a1887>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00817.warc.gz"} |
DIMACS Series in
Discrete Mathematics and Theoretical Computer Science
VOLUME Thirty Six
TITLE: "Discrete Mathematics in the Schools"
EDITORS: Joseph G. Rosenstein, Deborah S. Franzblau and Fred S. Roberts. Published by the American Mathematical Society and the National Council of Teachers of Mathematics
A PostScript version of this document
Discrete Mathematics in the Schools:
An Opportunity to Revitalize School Mathematics
Joseph G. Rosenstein
This article serves as an introduction in four different but overlapping ways:
• As an introduction to a volume advocating discrete mathematics in the schools, it outlines the case for this position.
• As an introduction to a collection of thirty-four diverse articles, it provides some context for those articles.
• As an introduction to the 1992 conference which led to this volume, it provides information about the conference and its themes.
• As an introduction to my perspective as conference organizer, author, and editor, it summarizes the main reasons for my involvement in this enterprise.
The author's perspective
Starting at the end, which is of course the beginning, there are two major reasons for my ongoing efforts to promote discrete mathematics in the schools --- that in two major ways, discrete
mathematics offers an opportunity to revitalize school mathematics.
• Discrete mathematics offers a new start for students. For the student who has been unsuccessful with mathematics, it offers the possibility for success. For the talented student who has lost
interest in mathematics, it offers the possibility of challenge.
• Discrete mathematics provides an opportunity to focus on how mathematics is taught, on giving teachers new ways of looking at mathematics and new ways of making it accessible to their students.
From this perspective, teaching discrete mathematics in the schools is not an end in itself, but a tool for reforming mathematics education.
These two themes first appeared in a concept document that I developed in January 1991 and that grew out of the first two years of my experience directing the Leadership Program in Discrete
Mathematics, an NSF-funded teacher enhancement program for high school teachers, at Rutgers University.^1 Participants reported changes in their classrooms, in their students, and in themselves.
Their successes taught us that discrete mathematics was not just another piece of the curriculum. Many participants reported success with a variety of students at a variety of levels, demonstrated a
new enthusiasm for teaching in new ways, and proselytized among their colleagues and administrators. These two themes are discussed further in this article in sections entitled Discrete mathematics:
A new start for students and Discrete mathematics: A vehicle for improving mathematics education.
The October 1992 Conference
These two views of discrete mathematics --- as a new start for students and as a vehicle for improving mathematics education --- seemed to me to establish an agenda for those interested in both
discrete mathematics and mathematics education. If discrete mathematics could have a significant impact on mathematics education, how can that impact be actualized? This question led to a conference
entitled ``Discrete Mathematics in the Schools: How Do We Make an Impact?"
The Conference took place on October 2-4, 1992 at Rutgers University and was sponsored by the Center for Discrete Mathematics and Theoretical Computer Science (DIMACS), an NSF-funded Science and
Technology Center. It brought together thirty-three educators who had been involved in a variety of ways in introducing discrete mathematics in the schools; see Appendix A for a list of conference
participants. The concept document containing the two themes described above was distributed in advance of the conference and was reflected in the opening presentation at which I welcomed and
challenged the conference participants.
The conference program was designed to inform the participants about various perspectives of discrete mathematics and its role in K--12 education, and about all of the various activities taking place
that promoted discrete mathematics in the schools. An abbreviated version of the program, showing presentations and session titles, appears in Appendix B. Presentations were followed by extended
One outcome of the discussions at the conference was the Vision Statement which appears at the beginning of this volume. Two major points of the Vision Statement were that ``discrete mathematics is
an exciting and appropriate vehicle for working toward and achieving these goals" (referring to the goals of those striving to improve mathematics education), and that ``discrete mathematics needs to
be introduced into the curriculum for its own sake'' because of the increasing importance and prevalence of its applications.
What is discrete mathematics?
It is, of course, natural for K--12 teachers and administrators, as well as parents and the press, to ask this question. Unfortunately, it is not an easy question to answer. The problem is that the
phrase ``discrete mathematics'' does not refer to a well-defined branch of mathematics --- like algebra, geometry, trigonometry, or calculus --- but rather encompasses a variety of loosely-connected
concepts and techniques. Moreover, it is not a branch of mathematics which is generally familiar to the public. At the dedication ceremony of DIMACS as a Center in 1989, then-Governor Thomas Kean
(NJ) quipped that, before participating in this ceremony, his impression was that discrete mathematics was what accountants did behind closed doors. That may be a common initial impression of
discrete mathematics.
I have found that one effective way of answering the question is by giving lots of examples of the kinds of situations where the mathematics that is used is ``discrete''. Though not actually defining
discrete mathematics, the examples give a flavor of what comprises discrete mathematics, and also helps to demystify the phrase. Here is the list that we are currently using in one of the brochures
of the Leadership Program in Discrete Mathematics; this list contains examples that we anticipate will make sense to the teachers that we hope to attract to the program.
• What is the quickest way to sort a list of names alphabetically?
• Which way of connecting a number of sites into a telephone network requires the least amount of cable?
• Which version of a lottery gives the best odds?
• If each voter ranks the candidates for President in order of preference, how can a consensus ranking of the candidates be obtained?
• What is the best way for a robot to pick up items stored in an automated warehouse?
• How does a CD player interpret the codes on a CD correctly even if the CD is scratched?
• How can an estate be divided fairly?
• How can ice cream stands be placed at various street corners in a town so that at any corner there is a stand which is at most one block away?
• How can representatives be apportioned fairly among the states using current census information?
These problems --- and many others from different areas within discrete mathematics --- share several important characteristics. They are easily understood and discussed, readily seen as dealing with
real-world situations, and can be explored without extensive background in school mathematics. This is discussed in more detail in the following section. Although I have used this
``definition-by-examples'' of discrete mathematics for a number of years, in the spring of 1996, as the New Jersey Department of Education was preparing to present its recommendations for mathematics
standards to the State Board of Education, I was told that I had to provide a ``real definition'' for the document. So here is discrete mathematics as it appears in New Jersey's Core Curriculum
Content Standards:
Discrete mathematics is the branch of mathematics that deals with arrangements of discrete objects. It includes a wide variety of topics and techniques that arise in everyday life, such as how to
find the best route from one city to another, where the objects are cities arranged on a map. It also includes how to count the number of different combinations of toppings for pizzas, how best to
schedule a list of tasks to be done, and how computers store and retrieve arrangements of information on a screen. Discrete mathematics is the mathematics used by decision-makers in our society,
from workers in government to those in health care, transportation, and telecommunications. Its various applications help students see the relevance of mathematics in the real world.
In This Volume. Two articles in Section 3 of this volume address directly the question, ``What is discrete mathematics?" Stephen Maurer's article explores a number of possible charactizations of
discrete mathematics, none of which proves to be fully satisfactory. Joseph Rosenstein's article provides an extended elaboration of the description above, as it appears in the New Jersey Mathematics
Curriculum Framework.
Why introduce discrete mathematics into the curriculum?
A number of different arguments have been presented for including discrete mathematics in the school curriculum; these arguments can each be viewed against the backdrop of the problems posed above.
Discrete mathematics is:
○ Applicable In recent years, topics in discrete mathematics have become valuable tools and provide powerful models in a number of different areas.
○ Accessible In order to understand many of these applications, arithmetic is often sufficient, and many others are accessible with only elementary algebra.
○ Attractive Though easily stated, many problems are challenging, can interest and attract students, and lend themselves to exploration and discovery.
○ Appropriate Both for students who are accustomed to success and are already contemplating scientific careers, and for students who are accustomed to failure and perhaps need a fresh start in
In This Volume. A number of articles in this volume illustrate and elaborate on these reasons for incorporating discrete mathematics into the curriculum. Several articles that particularly address
each of the above themes are provided below.
○ Applicable The articles by Henry Pollak, Fred Roberts, John Dossey, and Eric Hart address the applications of discrete mathematics and how it provides models for real-world situations.
○ Accessible The articles by Janice Kowalczyk, Susan Picker, Nancy Casey and Michael Fellows, Joseph Rosenstein, Valerie DeBellis, Rob\-ert Jamison, and Evan Maletsky show, for example, how
discrete mathematics can be used in elementary and middle school grades.
○ Attractive The articles by Patrick Carney, Nancy Casey, Reuben Settergren, and Margaret Cozzens discuss how discrete mathematics excites student interest.
○ Appropriate The articles by Nancy Casey, Susan Picker, Bret Hoyer, and L. Charles Biehl discuss how discrete mathematics is appropriate for students who need a fresh start in mathematics. Other
articles in this volume discuss how discrete mathematics can be combined with and enhance existing topics like algebra (Bret Hoyer, Philip Lewis), precalculus (John Dossey, Joan Reinthaler, James
Sandefur), calculus (Robert Devaney), and computer science (Peter Henderson, Vera Proulx).
Discrete mathematics: A new start for students
The traditional topics of school mathematics --- arithmetic, algebra, geometry, etc. --- are of course important; without a good grounding in these topics, students will be seriously disadvantaged in
career options. And the nation will continue to have a serious shortfall in technically skilled personnel.
However, many students find school mathematics to be a serious stumbling block, and ultimately give up. The most frequently prescribed remedy for students who have failed in school mathematics
appears, unfortunately, to be more of the same. And ``more of the same'' usually means not only repetition of content, but also repetition of method. Thus, many students come to see school
mathematics only as a set of unintelligible procedures, which is not surprising since they were never given an opportunity to explore concepts meaningfully and apply them in new situations.
At the other end of the spectrum, many talented students also find school mathematics to be uninteresting and irrelevant, and thus opt for other careers. For these students, who are looking for a
spark of life and challenge in mathematics, a frequent response is ``wait until you get to calculus''; but many have lost interest by the time they get to calculus.
Discrete mathematics offers a new start. For the student who has been unsuccessful in mathematics, discrete mathematics offers the possibility of success. Students who have encountered mathematics
which they can do successfully are encouraged to take another look at the mathematics at which they have failed. Students who have found that they can solve meaningful problems gain a sense of
empowerment. Teachers in the Leadership Program have reported that, for students who have a history of failure in mathematics, being able to use terminology and solve problems in areas with which
other school personnel --- teachers and guidance counselors, as well as students --- are unfamiliar is a very heady experience.
The ranks of students who have been unsuccessful in mathematics contain a disproportionate number of minorities and women. Such students, who have given up hope of ever learning school mathematics,
can become interested in and can learn discrete mathematics since they do not associate it at the outset with routine school mathematics. Teachers in the Leadership Program in Discrete Mathematics
have used discrete mathematics successfully with these students in all types of schools, including those in urban areas.
For the talented student who has lost interest in mathematics, discrete mathematics offers the possibility of challenge. Discrete mathematics serves as a natural context for many of the puzzle-like
questions that intrigue the talented student, offers open-ended problems which quickly lead to the frontiers of knowledge, and provides easy access to applications which mathematicians are now making
in a variety of real-life situations. One can imagine students engaged in discrete mathematics saying ``This is how I would like to spend my professional life'', as well as ``This is fun''.
In This Volume. See the articles cited under ``accessible'', ``attractive'', and ``appropriate'' in the previous section.
Discrete mathematics: A vehicle for improving mathematics education.
The introduction of new material into the curriculum affords a particular opportunity to infuse new instructional techniques at the same time. When there is no specific body of material that
districts and teachers feel obligated to ``cover'', there is clearly ``time'' for experimentation --- with computers, with group learning, with problem solving. When the problems are new to the
teachers, and close to the cutting edge of knowledge, there is greater acceptance of a classroom open to discussion, to reasoning together, and to the excitement of discovering new solutions which
are not ``in the book''.
Moreover, as teachers become familiar with these techniques and see that they work with their students in their own classrooms, they will adapt them for use in their other classes. Those teachers who
have taken the time from traditional teacher-oriented instruction to try these learner-oriented techniques know that the time is well spent. The difficulty is in getting them to try.
Discrete mathematics offers a wealth of new material and, more important in this context, consists of many topics which lend themselves readily to approaches to learning that are recommended in the
national reports: discovery learning, experimentation, problem solving, cooperative learning, use of technology. With discrete mathematics, students can easily become involved in the doing of
mathematics, can see themselves as ``mathematicians'' rather than as followers of routine instructions.
In This Volume. Nancy Casey and Michael Fellows argue in their article that only if they use discrete mathematics will K--4 teachers have sufficiently rich mathematical content to properly address
the process standards of ``reasoning, problem-solving, communications, and connections'' stressed in the NCTM Standards.^2 Other articles focus on how discrete mathematics can help teachers achieve
educational objectives such as teaching students mathematical communication (Rochelle Leibowitz), reasoning (Susanna Epp), and problem-solving (Margaret Cozzens, Peter Henderson), and change public
perceptions of mathematics (Joseph Malkevitch). The article by Joseph Rosenstein and Valerie DeBellis discusses the impact of the Leadership Program in Discrete Mathematics on the activities of its
Resources for introducing discrete mathematics in the schools
At the time of the conference, there were relatively few resources available to teachers interested in including discrete mathematics in their classrooms and curricula. Increasingly in recent years,
in part because discrete mathematics is addressed in the NCTM Standards, more effort has been placed both on developing materials related to discrete mathematics and to incorporating discrete
mathematics activities in textbooks. As a result of the efforts of the Leadership Program in Discrete Mathematics and the ``Implementation of the NCTM Standard in Discrete Mathematics Project''
program directed by Margaret Kenney at Boston College and other sites across the country, there are now nearly 2000 teachers who have had extensive exposure to discrete mathematics; many of them have
been taking leadership roles, developing curriculum materials and making presentations at conferences.
In This Volume. The article by Deborah Franzblau and Janice Kowalczyk, based on recommendations of teachers in the Leadership Program in Discrete Mathematics, provides an extensive review of
available print and video resources. Two articles, one by Eric Hart and the other by Nancy Crisler, Patience Fisher, and Gary Froelich, discuss texts for high school students which include discrete
mathematics. Two articles, one by Nate Dean and Yanxi Liu, and the other by Mario Vassallo and Anthony Ralston, discuss discrete mathematics software. Two articles, by Harold Bailey and L. Charles
Biehl, discuss high school courses in discrete mathematics. And the article by Joseph Rosenstein and Valerie DeBellis discusses the Leadership Program in Discrete Mathematics.
Speaking for the editors, the conference participants, and the authors, we hope that this volume will be a major contribution both to facilitating the use of discrete mathematics in K--12 schools and
to demonstrating the potential of discrete mathematics as a vehicle to improve mathematics education and revitalize school mathematics.
Department of Mathematics, Rutgers University
E-mail address: joer@dimacs.rutgers.edu
1. The NSF-funded Leadership Program in Discrete Mathematics is co-sponsored by the Center for Discrete Mathematics and Theoretical Computer Science (DIMACS) and the Rutgers Center for Mathematics,
Science, and Computer Science Education (CMSCE). Although originally (in 1989-1991) for high school teachers, the Leadership Program subsequently (beginning in 1992) also enrolled middle school
teachers, and now (since 1995) focuses on K--8 teachers. See the article by Rosenstein and DeBellis in this volume for further information about the Leadership Program.
Appendix A
Discrete Mathematics in the Schools:
How Do We Make an Impact?
October 2--4, 1992
Conference Participants
NAME STATE AFFILIATION (at time of conference)
Bailey, Harold F. NY College of Mount Saint Vincent
Biehl, L. Charles DE McKean HS, Wilmington
Carrs, Marjorie University of Queensland, Brisbane, Australia
Crisler, Nancy MO Pattonville School Dist., St. Louis County
Dance, Rosalie MD Ballou Science/Math HS, Takoma Park
DeBellis, Valerie NJ Rutgers University
Dean, Nathaniel NJ Bellcore
Epp, Susanna IL DePaul University
Fellows, Michael University of Victoria, British Columbia, Canada
Froelich, Gary ND Bismarck HS
Hart, Eric IA Maharishi International University
Henderson, Peter NY SUNY Stony Brook
Hoover, Mark NJ Educational Testing Service
Hoyer, Bret IA John F. Kennedy HS, Cedar Rapids
Kenney, Margaret MA Boston College
Kowalczyk, Janice RI Teacher Education and Computer Center
Lacampagne, Carol B. DC U.S. Department of Education
Leibowitz, Rochelle MA Wheaton College
Lewis, Philip G. MA Lincoln Sudbury Regional HS
Malkevitch, Joseph NY York College (CUNY)
Maltas, James IA Malcolm Price Laboratory School, University of Northern Iowa
Maurer, Stephen PA Swarthmore College
McGraw, Sue Ann OR Lake Oswego HS
Piccolino, Anthony NJ Montclair State College
Picker, Susan NY Office of the Superintendent, Manhattan Public Schools
Pollak, Henry NY Columbia University
Proulx, Viera MA Northeastern University
Reinthaler, Joan DC The Sidwell Friends School
Roberts, Fred NJ Rutgers University
Rosenstein, Joseph G. NJ Rutgers University
Saks, Michael NJ Rutgers University
Vassallo, Mario NY SUNY Fredonia
Yunker, Lee IL Community HS Dist. 94, West Chicago
Appendix B
Discrete Mathematics in the Schools:
How Do We Make an Impact?
October 2--4, 1992
Conference Program (Abbreviated)
Friday October 2
Presentation: Joseph G. Rosenstein
``Discrete mathematics as a new start for students and teachers''
Classroom Perspectives, Experiences, and Models --- Session 1
L. Charles Biehl --- ``Discrete mathematics for students of average ability''
Susan Picker --- ``Discrete mathematics: Giving remedial students a second chance''
Presentation: Stephen Maurer
``What is discrete mathematics: The many answers''
Classroom Perspectives, Experiences, and Models --- Session 2
Gary Froelich --- ``A semester discrete mathematics course at the high school level''
James Maltas --- ``Implementing a discrete mathematics course for non-math students''
Nancy Crisler --- ``My experiences as a teacher and math coordinator''
Philip Lewis --- ``Using a computer lab: Algorithms, algebra, and axioms''
Presentation: Joseph Malkevitch
``Discrete mathematics and the public's perception of mathematics''
Classroom Perspectives, Experiences, and Models --- Session 3
Rosalie Dance --- ``Integrating discrete and continuous approaches in secondary math''
Lee Yunker --- ``Current and future trends on discrete mathematics in the curriculum''
Presentation: Eric Hart
``Curriculum materials for discrete mathematics in the schools''
An overview and a taste of ...
For All Practical Purposes --- Joe Malkevitch and Tony Piccolino
COMAP Project --- Nancy Crisler and Gary Froelich
UCSMP materials --- Susanna Epp
CORE-PLUS --- Eric Hart
Several textbooks --- Lee Yunker
Saturday October 3
Programs for teachers
Georgetown project --- Rosalie Dance and Joan Reinthaler
NCTM project --- Peg Kenney and others
Iowa Project --- Eric Hart and others
Rutgers Project --- Joe Rosenstein and others
Classroom Perspectives, Experiences, and Models --- Session 4
Joan Reinthaler --- ``Teaching modeling to weak math students''
Sue Ann McGraw --- ``Integrating discrete mathematics into traditional math courses''
Bret Hoyer --- ``A discrete mathematics course using For All Practical Purposes''
Susanna Epp --- ``Strengthening thinking skills using discrete mathematics''
Rochelle Leibowitz --- ``Strengthening writing skills using discrete mathematics''
Anthony Piccolino --- ``Discrete mathematics: Making math accessible to all''
Presentation: Henry Pollak
`` The role of modeling in teaching discrete mathematics''
Presentation: Fred Roberts
``The role of applications in teaching discrete mathematics''
Presentation: Mario Vassallo
``Computer software for teaching discrete mathematics in the schools''
Presentation: Nate Dean
``What computer software is currently being developed?''
Presentation: Michael Fellows
``Discrete mathematics and computer science in the elementary schools''
``How Do We Make an Impact?''
Organizing our suggestions
Structuring Sunday's discussions
Sunday October 4
Viera Proulx --- ``Computer science in high school''
Peter Henderson --- ``Computer science, discrete mathematics, and problem solving''
Mark Hoover --- ``Assessment and discrete mathematics''
Harold Bailey --- ``Assessing current practice in discrete mathematics''
``How Do We Make an Impact?''
Work sessions in smaller groups
Reports from groups
The next steps | {"url":"http://dimacs.rutgers.edu/archive/Volumes/schools/intro.html","timestamp":"2024-11-02T12:29:10Z","content_type":"text/html","content_length":"30852","record_id":"<urn:uuid:92a8c997-a87a-42e6-b7cf-6472507c90ac>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00899.warc.gz"} |
Apparent Coefficient of Thermal Expansion of Mercury
Task number: 1807
A glass sphere with a capillary tube was filled to the brim with mercury of the volume 400 cm^3 at temperature of 0 °C. In a heat bath with temperature of 100 °C, 6.228 cm^3 of mercury overflowed.
Determine the apparent coefficient of volume expansion of mercury.
• Hint – what is the apparent coefficient of expansion
The apparent coefficient of volume expansion is dependent on the container in which the liquid is placed. It states a change in the volume of the liquid in relation to the volume of the
container, which, of course, also changes with temperature.
• Analysis
First, we must realize the difference between a real and apparent coefficient of volume expansion of liquid. While the real coefficient indicates the change of volume of liquid with increasing or
decreasing temperature, the apparent coefficient refers to the container in which the liquid is placed and whose volume also changes with temperature. We can visualize this problem in such a way
that there is a scale on the container which indicates the volume of the liquid inside the container, and when the container “stretches”, the scale of the container also “stretches”; therefore
the scale does not show the real, but rather the apparent volume of the liquid.
From the fact that we know how much mercury spilled from the container when it was heated to the temperature of 100 °C, we can determine the apparent coefficient of expansion for both the
container and liquid. We start from a known relationship for the thermal volume expansion of liquids where we know all the variables except the unknown coefficient. We can evaluate and calculate
this coefficient from this equation.
• Given values
t[0] = 0 °C initial temperature
V[0] = 400 cm^3 initial volume of mercury
t[1] = 100 °C temperature of bath
ΔV = 6.228 cm^3 amount of spilled mercury
β[a] = ? apparent coefficient of expansion of mercury
• Solution
When solving this task, we need to realize the difference between the real and apparent coefficient of expansion of a liquid, which is described in the section Analysis. The apparent coefficient
β[a] of a liquid and a container is therefore given by the relation
\[ \mathrm{\Delta}V_{\mathrm{dif}}=\beta_{\mathrm{a}} V_0(t_1-t_0), \]
where V[0]is the initial volume of liquid, t[1] − t[0] is the temperature difference and ΔV[dif] indicates the difference between the volume change of the liquid and the volume change of the
container in which the liquid is placed. In this task, the difference of volume changes ΔV[dif] určen právě objemem rtuti ΔV, is determined by the volume of mercury ΔV, which spilled out of the
container. Therefore, we can write
\[ \mathrm{\Delta}V=\beta_{\mathrm{a}} V_0\left(t_1-t_0\right) \]
and from here we can evaluate β[a]
Numerical solution
Volumes ΔV and V[0] can be substituted in cm^3, because β is determined by the proportion of the two volumes and therefore we can then reduce the fraction.
\[\beta_{\mathrm{a}}=\frac{\mathrm{\Delta}V}{V_0\left(t_1-t_0\right)}=\frac{6.228}{400\left(100-0\right)}\,{\ ^{\circ}\mathrm{C}^{-1}}=156\cdot{10^{-6}}{\ ^{\circ}\mathrm{C}^{-1}}\]
• Answer
The apparent coefficient of volume expansion of mercury for our container is 156·10^−6 °C^−1.
• Apparent coefficient of expansion
The apparent coefficient of mercury in an iron container can also be calculated from the volume expansion of the two materials.
The change of the volume of mercury is:
\[ \mathrm{\Delta}V_{\mathrm{m}}=\beta_{\mathrm{m}} V_0\left(t_1-t_0\right), \]
where β[m] is the coefficient of volume expansion of mercury, V[0] is the volume of mercury at initial temperature and t[1] − t[0] is the temperature difference.
The internal volume of the container changes with the same coefficient of expansion as the volume of the material from which the container is made. We can visualize this in such a way that we
tightly place an iron object into the iron container so that it completely fits. If we warm up the container with the object inside, the container and the object must expand in the same manner
(they are made of the same material); the object therefore still completely fits. This means that the inner volume of the container changes with the same coefficient of expansion as the material
from which the container is made.
The change in the internal volume of the container V[c] is therefore given by:
\[ \mathrm{\Delta}V_{\mathrm{c}}=\beta_{\mathrm{c}} V_0\left(t_1-t_0\right), \]
where β[c] is the coefficient of volumetric expansion of the material of the container, V[0] is the internal volume of the container at an initial temperature and t[1] − t[0] is the temperature
At the beginning, mercury has the same volume as the container, i.e. V[0]. When we subtract the equations for the volume changes, we obtain a relationship for the apparent coefficient:
\[ \mathrm{\Delta}V_{\mathrm{m}}-\mathrm{\Delta}V_{\mathrm{c}}=\beta_{\mathrm{m}} V_0\left(t_1-t_0\right)-\beta_{\mathrm{c}} V_0(t_1-t_0), \] \[ \mathrm{\Delta}V_{\mathrm{m}}-\mathrm{\Delta}V_{\
mathrm{c}}=\left(\beta_{\mathrm{m}}-\beta_{\mathrm{c}}\right) V_0\left(t_1-t_0\right). \]
The left side of the equation tells us how much mercury overflowed the container and on the right side there is a difference of the two expansion coefficients instead of one coefficient. This
difference is the unknown apparent coefficient of mercury β[a].
We deduced that:
\[ \beta_{\mathrm{a}}=\beta_{\mathrm{m}}-\beta_{\mathrm{c}}. \]
If we substitute table values:
β[m] = 2·10^−4 °C^−1
β[c] = 36·10^−6 °C^−1
we obtain:
\[ \beta_{\mathrm{a}}=\beta_{\mathrm{m}}-\beta_{\mathrm{c}}=\left(2\cdot{10^{-4}}-36\cdot{10^{-6}}\right)\,\mathrm{^{\circ}C^{-1}}=1.64\cdot{10^{-4}}\,\mathrm{^{\circ}C^{-1}}. \]
The table value therefore differs from the measured value by about 5 %. | {"url":"https://physicstasks.eu/1807/apparent-coefficient-of-thermal-expansion-of-mercury","timestamp":"2024-11-11T16:37:39Z","content_type":"text/html","content_length":"32454","record_id":"<urn:uuid:c6b91ae6-3779-4f89-95ae-d55b9abd83b9>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00612.warc.gz"} |
Calculus Center's blog
As a new semester begins it is important to start collecting the Calculus resources you will need. In this blog post I want to give you the top 10 Calculus websites. They are not ranked in any
particular order. Instead I have categorized them by videos, calculators, textbooks and solutions. This categorization is more art than science as many of these sites have content in more than one
category. And what criteria did I judge these sites on? Free, big, useful, and authoritative. Let's get started. | {"url":"https://www.calculussolution.com/blogs/calculus-center","timestamp":"2024-11-03T04:19:59Z","content_type":"application/xhtml+xml","content_length":"60165","record_id":"<urn:uuid:ba722d4f-7733-4727-b7a6-6c67967adfce>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00592.warc.gz"} |
A Gentle Introduction
Fourth Edition
January 2020 | 536 pages | SAGE Publications, Inc
The Fourth Edition of Statistics: A Gentle Introduction shows students that an introductory statistics class doesn’t need to be difficult or dull. This text minimizes students’ anxieties about math
by explaining the concepts of statistics in plain language first, before addressing the math. Each formula within the text has a step-by-step example to demonstrate the calculation so students can
follow along. Only those formulas that are important for final calculations are included in the text so students can focus on the concepts, not the numbers. A wealth of real-world examples and
applications gives a context for statistics in the real world and how it helps us solve problems and make informed choices.
New to the Fourth Edition are sections on working with big data, new coverage of alternative non-parametric tests, beta coefficients, and the "nocebo effect," discussions of p values in the context
of research, an expanded discussion of confidence intervals, and more exercises and homework options under the new feature "Test Yourself."
Included with this title:
The password-protected Instructor Resource Site (formally known as Sage Edge)
offers access to all text-specific resources, including a test bank and editable, chapter-specific PowerPoint® slides.
Learn more
About the Author
Chapter 1: A Gentle Introduction
How Much Math Do I Need to Do Statistics?
The General Purpose of Statistics: Understanding the World
Liberal and Conservative Statisticians
Descriptive and Inferential Statistics
Experiments Are Designed to Test Theories and Hypotheses
Eight Essential Questions of Any Survey or Study
On Making Samples Representative of the Population
Experimental Design and Statistical Analysis as Controls
The Language of Statistics
On Conducting Scientific Experiments
The Dependent Variable and Measurement
Measurement Scales: The Difference Between Continuous and Discrete Variables
Types of Measurement Scales
Rounding Numbers and Rounding Error
History Trivia: Achenwall to Nightingale
Chapter 1 Practice Problems
Chapter 1 Test Yourself Questions
Chapter 2: Descriptive Statistics: Understanding Distributions of Numbers
The Purpose of Graphs and Tables: Making Arguments and Decisions
A Summary of the Purpose of Graphs and Tables
Shapes of Frequency Distributions
Grouping Data Into Intervals
Advice on Grouping Data Into Intervals
The Cumulative Frequency Distribution
Cumulative Percentages, Percentiles, and Quartiles
Non-normal Frequency Distributions
On the Importance of the Shapes of Distributions
Additional Thoughts About Good Graphs Versus Bad Graphs
History Trivia: De Moivre to Tukey
Chapter 2 Practice Problems
Chapter 2 Test Yourself Questions
Chapter 3: Statistical Parameters: Measures of Central Tendency and Variation
Measures of Central Tendency
Choosing Among Measures of Central Tendency
Uncertain or Equivocal Results
Correcting for Bias in the Sample Standard Deviation
How the Square Root of x2 Is Almost Equivalent to Taking the Absolute Value of x
The Computational Formula for Standard Deviation
The Sampling Distribution of Means, the Central Limit Theorem, and the Standard Error of the Mean
The Use of the Standard Deviation for Prediction
Practical Uses of the Empirical Rule: As a Definition of an Outlier
Practical Uses of the Empirical Rule: Prediction and IQ Tests
History Trivia: Fisher to Eels
Chapter 3 Practice Problems
Chapter 3 Test Yourself Questions
Chapter 4: Standard Scores, the z Distribution, and Hypothesis Testing
The Classic Standard Score: The z Score and the z Distribution
More Practice on Converting Raw Data Into z Scores
Converting z Scores to Other Types of Standard Scores
Interpreting Negative z Scores
Testing the Predictions of the Empirical Rule With the z Distribution
Why Is the z Distribution So Important?
How We Use the z Distribution to Test Experimental Hypotheses
More Practice With the z Distribution and T Scores
Summarizing Scores Through Percentiles
History Trivia: Karl Pearson to Egon Pearson
Chapter 4 Practice Problems
Chapter 4 Test Yourself Questions
Chapter 5: Inferential Statistics: The Controlled Experiment, Hypothesis Testing, and the z Distribution
Hypothesis Testing in the Controlled Experiment
Hypothesis Testing: The Big Decision
How the Big Decision Is Made: Back to the z Distribution
The Parameter of Major Interest in Hypothesis Testing: The Mean
Nondirectional and Directional Alternative Hypotheses
A Debate: Retain the Null Hypothesis or Fail to Reject the Null Hypothesis
The Null Hypothesis as a Nonconservative Beginning
The Four Possible Outcomes in Hypothesis Testing
Significant and Nonsignificant Findings
Trends, and Does God Really Love the .05 Level of Significance More Than the .06 Level?
Directional or Nondirectional Alternative Hypotheses: Advantages and Disadvantages
Did Nuclear Fusion Occur?
Conclusions About Science and Pseudoscience
The Most Critical Elements in the Detection of Baloney in Suspicious Studies and Fraudulent Claims
Can Statistics Solve Every Problem?
History Trivia: Egon Pearson to Karl Pearson
Chapter 5 Practice Problems
Chapter 5 Test Yourself Questions
Chapter 6: An Introduction to Correlation and Regression
Correlation: Use and Abuse
A Warning: Correlation Does Not Imply Causation
Another Warning: Chance Is Lumpy
Correlation and Prediction
The Four Common Types of Correlation
The Pearson Product–Moment Correlation Coefficient
Testing for the Significance of a Correlation Coefficient
Obtaining the Critical Values of the t Distribution
If the Null Hypothesis Is Rejected
Representing the Pearson Correlation Graphically: The Scatterplot
Fitting the Points With a Straight Line: The Assumption of a Linear Relationship
Interpretation of the Slope of the Best-Fitting Line
The Assumption of Homoscedasticity
The Coefficient of Determination: How Much One Variable Accounts for Variation in Another Variable—The Interpretation of r2
Quirks in the Interpretation of Significant and Nonsignificant Correlation Coefficients
Reading the Regression Line
Final Thoughts About Multiple Regression Analyses: A Warning About the Interpretation of the Significant Beta Coefficients
Significance Test for Spearman’s r
Point-Biserial Correlation
Testing for the Significance of the Point-Biserial Correlation Coefficient
Testing for the Significance of Phi
History Trivia: Galton to Fisher
Chapter 6 Practice Problems
Chapter 6 Test Yourself Questions
Chapter 7: The t Test for Independent Groups
The Statistical Analysis of the Controlled Experiment
One t Test but Two Designs
Assumptions of the Independent t Test
The Formula for the Independent t Test
You Must Remember This! An Overview of Hypothesis Testing With the t Test
What Does the t Test Do? Components of the t Test Formula
What If the Two Variances Are Radically Different From One Another?
The Power of a Statistical Test
The Correlation Coefficient of Effect Size
Another Measure of Effect Size: Cohen’s d
Estimating the Standard Error
History Trivia: Gosset and Guinness Brewery
Chapter 7 Practice Problems
Chapter 7 Test Yourself Questions
Chapter 8: The t Test for Dependent Groups
Variations on the Controlled Experiment
Assumptions of the Dependent t Test
Why the Dependent t Test May Be More Powerful Than the Independent t Test
How to Increase the Power of a t Test
Drawbacks of the Dependent t Test Designs
One-Tailed or Two-Tailed Tests of Significance
Hypothesis Testing and the Dependent t Test: Design 1
Design 1 (Same Participants or Repeated Measures): A Computational Example
Design 2 (Matched Pairs): A Computational Example
Design 3 (Same Participants and Balanced Presentation): A Computational Example
History Trivia: Fisher to Pearson
Chapter 8 Practice Problems
Chapter 8 Test Yourself Questions
Chapter 9: Analysis of Variance (ANOVA): One-Factor Completely Randomized Design
A Limitation of Multiple t Tests and a Solution
The Equally Unacceptable Bonferroni Solution
The Acceptable Solution: An Analysis of Variance
The Null and Alternative Hypotheses in ANOVA
The Beauty and Elegance of the F Test Statistic
How Can There Be Two Different Estimates of Within-Groups Variance?
What a Significant ANOVA Indicates
Degrees of Freedom for the Numerator
Degrees of Freedom for the Denominator
Determining Effect Size in ANOVA: Omega Squared (w2)
Another Measure of Effect Size: Eta (h)
History Trivia: Gosset to Fisher
Chapter 9 Practice Problems
Chapter 9 Test Yourself Questions
Chapter 10: After a Significant ANOVA: Multiple Comparison Tests
Conceptual Overview of Tukey’s Test
Computation of Tukey’s HSD Test
What to Do If the Number of Error Degrees of Freedom Is Not Listed in the Table of Tukey’s q Values
Determining What It All Means
On the Importance of Nonsignificant Mean Differences
Chapter 10 Practice Problems
Chapter 10 Test Yourself Questions
Chapter 11: Analysis of Variance (ANOVA): One-Factor Repeated-Measures Design
The Repeated-Measures ANOVA
Assumptions of the One-Factor Repeated-Measures ANOVA
Determining Effect Size in ANOVA
Chapter 11 Practice Problems
Chapter 11 Test Yourself Questions
Chapter 12: Factorial ANOVA: Two-Factor Completely Randomized Design
The Most Important Feature of a Factorial Design: The Interaction
Fixed and Random Effects and In Situ Designs
The Null Hypotheses in a Two-Factor ANOVA
Assumptions and Unequal Numbers of Participants
Chapter 12 Practice Problems
Chapter 12 Test Yourself Problems
Chapter 13: Post Hoc Analysis of Factorial ANOVA
Main Effect Interpretation: Gender
Why a Multiple Comparison Test Is Unnecessary for a Two-Level Main Effect, and When Is a Multiple Comparison Test Necessary?
Multiple Comparison Test for the Main Effect for Age
Warning: Limit Your Main Effect Conclusions When the Interaction Is Significant
Multiple Comparison Tests
Interpretation of the Interaction Effect
Writing Up the Results Journal Style
Exploring the Possible Outcomes in a Two-Factor ANOVA
Determining Effect Size in a Two-Factor ANOVA
History Trivia: Fisher and Smoking
Chapter 13 Practice Problems
Chapter 13 Test Yourself Questions
Chapter 14: Factorial ANOVA: Additional Designs
Overview of the Split-Plot ANOVA
Two-Factor ANOVA: Repeated Measures on Both Factors Design
Overview of the Repeated-Measures ANOVA
Key Terms and Definitions
Chapter 14 Practice Problems
Chapter 14 Test Yourself Questions
Chapter 15: Nonparametric Statistics: The Chi-Square Test and Other Nonparametric Tests
Overview of the Purpose of Chi-Square
Overview of Chi-Square Designs
Chi-Square Test: Two-Cell Design (Equal Probabilities Type)
The Chi-Square Distribution
Assumptions of the Chi-Square Test
Chi-Square Test: Two-Cell Design (Different Probabilities Type)
Interpreting a Significant Chi-Square Test for a Newspaper
Chi-Square Test: Three-Cell Experiment (Equal Probabilities Type)
Chi-Square Test: Two-by-Two Design
What to Do After a Chi-Square Test Is Significant
When Cell Frequencies Are Less Than 5 Revisited
Other Nonparametric Tests
History Trivia: Pearson and Biometrika
Chapter 15 Practice Problems
Chapter 15 Test Yourself Questions
Chapter 16: Other Statistical Topics, Parameters, and Tests
Health Science Statistics
Additional Statistical Analyses and Multivariate Statistics
A Summary of Multivariate Statistics
Chapter 16 Practice Problems
Chapter 16 Test Yourself Questions
Appendix A: z Distribution
Appendix B: t Distribution
Appendix C: Spearman’s Correlation
Appendix D: Chi-Square ?2 Distribution
Appendix E: F Distribution
Appendix F: Tukey’s Table
Appendix G: Mann–Whitney U Critical Values
Appendix H: Wilcoxon Signed-Rank Test Critical Values
Appendix I: Answers to Odd-Numbered Test Yourself Questions
Student Study Siteedge.sagepub.com/coolidge4e
The open-access Student Study Site makes it easy for students to maximize their study time, anywhere, anytime. It offers flashcards that strengthen understanding of key terms and concepts, as well as
learning objectives that reinforce the most important material.
For additional information, custom options, or to request a personalized walkthrough of these resources, please contact your sales representative.
Instructor Teaching Siteedge.sagepub.com/coolidge4e
The open-access Student Study Site makes it easy for students to maximize their study time, anywhere, anytime. It offers flashcards that strengthen understanding of key terms and concepts, as well as
learning objectives that reinforce the most important material.
For additional information, custom options, or to request a personalized walkthrough of these resources, please contact your sales representative.
Statistics is generally not a dynamic topic. But Coolidge is able to break it down in a way that is manageable. His discussion of each type of analyses is easily accessed by the table of contents and
accurately depicted in the index. This is especially important for this generation of learners who want easy access to the specific information that is necessary without waiting through extraneous
concepts. Coolidge also describes contemporary and specific examples of how miss use of data can have an impact in real world circumstances. This is beneficial because it makes a true connection with
the power that a statistical researcher holds.
It is the only book on the market that covers important advanced techniques such as repeated measures ANOVA and multiple regressions, using SPSS.
Westminster College, Fulton, Missouri
The book is written to address a broad range of student ability. It is helpful to students without a strong background in mathematics.
Department of Psychology and Sociology, Tuskegee University
Good introductory book on statistics. Perfect for first-time statistics students, since concepts are presented simply but clearly.
As an instructor, I would want a hard-copy of this book.
Education, Carolina University
September 10, 2021
I don't think I ever received this book
Criminology/Criminal Just Dept, University Of Memphis
September 28, 2021
Adopted as a recommended text for students interested in diving more deeply into some of the concepts we cover (all too briefly) in a refresher course for incoming Masters of Public Policy students.
Political Science Dept, University Of Utah
February 10, 2020 | {"url":"https://uk.sagepub.com/en-gb/asi/statistics/book255370","timestamp":"2024-11-05T03:53:48Z","content_type":"text/html","content_length":"160695","record_id":"<urn:uuid:ca281fe1-70ac-4ed8-9905-bc0c253c8c38>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00432.warc.gz"} |
Count Frequency of Category Values in Pandas - Data Science Parichay
In this tutorial, we will look at how to count the frequency of values in a Pandas category type column or series with the help of some examples.
How to get a count of category values in a Pandas series?
You can apply the Pandas series value_counts() function on category type Pandas series as well to get the count of each value in the series. The following is the syntax –
# count of each category value
It returns the frequency for each category value in the series. It also shows categories (with count as 0) even if they are not present in the series.
Let’s look at some examples of getting a count of each value in a categorical column in Pandas. First, let’s create a dataframe that we will be using throughout this tutorial –
import pandas as pd
# create pandas dataframe
df = pd.DataFrame({
"Year": [2015, 2016, 2017, 2018, 2019],
"Winner": ["A", "B", "B", "A", "A"],
"Runners-up": ["C", "C", "A", "B", "C"]
# convert to category type
df["Winner"] = df["Winner"].astype("category")
df["Runners-up"] = df["Runners-up"].astype("category")
# display the dataframe
Year Winner Runners-up
0 2015 A C
1 2016 B C
2 2017 B A
3 2018 A B
4 2019 A C
We now have a dataframe containing the information on the winners and the runners-up of a tri-university sports competition. You can see that the column, “Winner” is of category dtype and contains
the winning university’s name for the given year.
Let’s now see how many times each university won from the above dataframe. For this, we apply the Pandas value_counts() function on the “Winner” column.
# count of each category in Winner column
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A 3
B 2
Name: Winner, dtype: int64
You can see that university “A” won three times and university “B” won two times.
Notice that we do not get values for the university “C”. This is because “C” does not occur in the “Winners” column. The categories are inferred by the values present in the column which are just “A”
and “B”.
# display the Winner column
0 A
1 B
2 B
3 A
4 A
Name: Winner, dtype: category
Categories (2, object): ['A', 'B']
Now, you can explicitly specify the categories for a categorical column (or series) in Pandas. Let’s also add “C” as one of the valid categories for the “Winner” column using the add_categories()
# add "C" to the categories for the Winner column
df["Winner"] = df["Winner"].cat.add_categories("C")
# display the Winner column
0 A
1 B
2 B
3 A
4 A
Name: Winner, dtype: category
Categories (3, object): ['A', 'B', 'C']
Note that we’re not changing any of the records as such, we’re just adding an additional possible value for this categorical column.
Now, if you apply the Pandas value_counts() function, you get the count of occurrence of each category value irrespective of whether it occurs in the series or not.
# count of each category in Winner column
A 3
B 2
C 0
Name: Winner, dtype: int64
Now we get the number of times each university won the tournament. University “A” won three times, “B” won two times and “C” won 0 times.
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Ever get the feeling you've been cheated - wrong in fact, wrong in theory?
So how does Chess.com's system work anyway?
I don't really know, and I don't particularly want to speculate, not more than I'm obliged to. It ought to be up to them to explain themselves, not up to me.
But I also don't know
• whether that system has been assessed independently, and even if so, how thoroughly and how expertly
• how much it risks (and is understood to risk) catching the wrong people as well as the right ones
• how much its reliability may vary (and is understood to vary) according to the sample size of games
• how much it may depend (and is understood to depend) on fallible human inputs, human judgments and so on.
I don't know. But I do know that Chess.com aren't in possession of a foolproof system. Of course they aren't, because there's no such thing as a foolproof system. And I do know that they are wrong in
this particular instance.
What I
, however, is that their method to some degree involves looking at the moves you have played, and seeing how many match with the preferred choice of a computer program. Whether they do anything else,
or what precisely their criteria are, who knows. (But how reliable those criteria are - on that, I
have a well-informed opinion.)
One question this raises is - since there is such a thing as theory in chess, when in the game do they start scrutinising? Presumably not on move one. But if not, at what point
the matching begin? If they start too early, when in fact you're still in book (because book use is permitted in these games) isn't that a point where errors can be committed? Because moves which
you're finding from a printed source are being marked down as moves you're finding with a program?
Let me give you an example. Let me give you several examples.
When you finish a game on Chess.com, you get a little game report, which includes some basic computer analysis, and a chart that looks like this.
What it means precisely, I couldn't say, but I can guess what
Best Move
means, and what
means. And I can guess that 99.3 is a high figure, whatever it means precisely and however they're calculating it. The game it refers to is this one.
[Site "Chess.com"] [Date "2019.07.01"] [White "passy234"] [Black "Justinpatzer"] [Result "0-1"] [WhiteElo "2055"] [BlackElo "2149"] [EndDate "2019.07.05"] [Termination "Justinpatzer won by
resignation"] 1. Nf3 d5 2. g3 Bg4 3. Bg2 c6 4. c4 e6 5. O-O Nf6 6. Qb3 Qb6 7. Qc2 Nbd7 8. cxd5 exd5 9. d3 Qc5 10. Qb3 Qb6 11. Qc2 Qc5 12. Qd1 Bd6 13. Nc3 O-O 14. Be3 Qa5 15. a3 Rfe8 16. b4 Qd8 17.
Rc1 a5 18. Qb3 Qe7 19. Rb1 axb4 20. axb4 Ne5 21. Nxe5 Bxe5 22. Rfe1 d4 0-1
So we've got a twenty-two move minature, in which Black plays five moves of theory, and then turns over White in short order with an extremely high Accuracy rate. Which is pretty suspicious, isn't
Except it isn't. Because this, which suggests that theory ends after five moves on each side
and which would mean that the players were playing their own moves from this position
is wrong. Very wrong.
In fact Black was playing published theory until
move sixteen
Specifically, he was following Petrosian v Vovhannisyan,
Lake Sevan 2015
, which you can see below (to move 14, but as there was a repetition, we had played two more moves apiece) as it appears on page 202
of Delchev and Semkov,
Attacking the English/Reti
, Chess Stars, 2016
which I have on my bookshelves.
Which is how I came to be in the position below, after Black's 16...Qd8, before I had to play any moves of my own.
White then varied with 17 Rc1. So my original contribution consisted of five moves - five very ordinary moves - and then, after a simple blunder by White
a very obvious pawn fork to win the game.
Suddenly the game looks very different, doesn't it? Suddenly it's perfectly normal, unexceptional. Suddenly there's nothing odd about it
at all
Here's a few more examples. This game, for instance.
[Site "Chess.com"] [Date "2018.12.25"] [White "Dave_1969"] [Black "Justinpatzer"] [Result "1/2-1/2"] [WhiteElo "2015"] [BlackElo "2096"] [EndDate "2019.01.08"] [Termination "Game drawn by agreement"]
1. d4 Nf6 2. c4 e6 3. Nc3 Bb4 4. e3 O-O 5. Ne2 d5 6. a3 Bd6 7. c5 Be7 8. b4 b6 9. Nf4 c6 10. Be2 Nbd7 11. Nd3 a5 12. Bd2 Ba6 13. O-O Qc7 14. f4 axb4 15. axb4 Bc4 16. Qc2 Rfb8 17. Rfb1 Qb7 18. Bf3 Ne8
19. Ne5 Nxe5 20. fxe5 Ra6 21. Rxa6 Qxa6 22. Qd1 Nc7 23. Ra1 Qb7 24. e4 b5 25. exd5 cxd5 1/2-1/2
According to Chess.com, we have seven moves of theory for White, and six for Black
which is a funny thing to say anyway, since Black's seventh is forced. But leaving that aside, it would get us this position, where Black is apparently starting from scratch.
But he's not starting from scratch. he's not even halfway there. Because in my theory here goes as far as
move twenty
, as per Guliev-Arkhipov,
Dubai 1999
. You may have to hunt about a bit there to locate the score, or you can find it, as I did, on page 174
of Emms,
The Nimzo-Indian move by move
, Everyman, 2011
which I have on my bookshelves and which brought me to this position without yet having had to play any of my own moves.
After 21 Rxa6 I once again played five moves - in my view even less distinguished moves than in the previous example - and I then asked for a draw.
(Why would Black, by the way, be using computer assistance in order to play a dull draw in twenty-five moves? Nobody knows. But none of this makes sense.)
Then there's this game, another draw as it happens.
[Site "Chess.com"] [Date "2019.01.08"] [White "rembooooo"] [Black "Justinpatzer"] [Result "1/2-1/2"] [WhiteElo "2076"] [BlackElo "2111"] [EndDate "2019.01.22"] [Termination "Game drawn by agreement"]
1. e4 e5 2. Nf3 Nc6 3. Bc4 Bc5 4. O-O Nf6 5. c3 Nxe4 6. d4 exd4 7. cxd4 d5 8. dxc5 dxc4 9. Qe2 Qd3 10. Re1 f5 11. Nc3 O-O 12. Nxe4 fxe4 13. Qxe4 Bf5 14. Qf4 Be6 15. Ne5 Qd5 16. Qg3 Nxe5 17. Qxe5 Qxe5
18. Rxe5 Rfe8 19. f4 Rad8 20. Be3 Bd5 21. Bd4 Bc6 22. Rxe8+ Bxe8 23. Bc3 Bc6 24. Re1 Kf7 25. Re5 Rd1+ 26. Kf2 Rc1 27. g4 Rh1 28. Kg3 Rf1 29. Rf5+ Kg8 30. Rg5 g6 31. f5 Rf3+ 32. Kh4 Kf7 33. fxg6+ hxg6
34. Re5 Rf2 35. h3 Rf3 36. Re1 Bd5 37. a4 c6 38. a5 a6 39. Re2 Rd3 40. Rf2+ Rf3 41. Re2 Rd3 42. Rf2+ Rf3 1/2-1/2
in which according to Chess.com
we played five moves of theory for White, and four for Black, reaching this position.
In fact Black at least was following all
moves of a recommended line on page 379
of Bologan,
Bologan's Black Weapons
(etc), New In Chess, 2014
so original play only started here.
Again, why Black would user computer assistance to play forty-two dull moves for a draw is a question for Chess.com. It's not a question they seem keen on answering.
One more, for now.
[Site "Chess.com"] [Date "2019.07.07"] [White "MRValero"] [Black "Justinpatzer"] [Result "1-0"] [WhiteElo "2182"] [BlackElo "2167"] [EndDate "2019.07.26"] [Termination "MRValero won by resignation"]
1. e4 e5 2. Nf3 Nc6 3. Bb5 Nf6 4. Qe2 Bc5 5. c3 O-O 6. O-O Re8 7. d3 a6 8. Ba4 h6 9. Be3 Bxe3 10. fxe3 d6 11. Nbd2 b5 12. Bb3 Be6 13. Bc2 Qe7 14. h3 d5 15. exd5 Nxd5 16. Qf2 g6 17. e4 Nf4 18. Nxe5
Nxh3+ 19. gxh3 Nxe5 20. Qg3 Qg5 21. Qxg5 hxg5 22. d4 Nc4 23. Nxc4 Bxc4 24. Rf2 Be6 25. Kh2 c6 26. Rg2 Kg7 27. Rxg5 Rh8 28. Rg3 Rh4 29. Rf1 a5 30. a3 b4 31. axb4 axb4 32. Rf2 b3 33. Bd3 Ra1 34. Rd2
Rc1 35. Re3 g5 36. Rg2 Kh6 37. Rg1 Rxg1 38. Kxg1 g4 39. hxg4 Kg5 40. Kg2 Rxg4+ 41. Rg3 Kf4 42. Rxg4+ Bxg4 43. Kf2 f6 44. c4 c5 45. dxc5 Bd7 46. Be2 Kxe4 47. c6 Bc8 48. Bd1 Kd4 49. Bxb3 Kc5 50. c7 Kd6
51. Bc2 Kxc7 52. Ke3 Bh3 53. b4 Kc6 54. Bd3 Kd6 55. Kf4 Kc6 56. Ke3 Kd6 57. Kd4 1-0
You may remember some fragments from this game in the previous posting (and as I asked then, why would Black be using a computer to lose a long game with lots of errors in it?) but in this one,
theory supposedly takes us to move five
which is this position.
But in fact, I was following, by transposition, Meijers v Crouan, Sautron 2009. No link, sorry, but here it is on page 99
of Lysyj and Ovetchkin,
The Berlin Defence
, Chess Stars, 2012
so I was still in theory up to here.
I've not been through all my games for this exercise, by the way, wishing to bore neither myself nor the reader. But even so, I'm aware that there are more examples than just these four.
I don't, of course, know for sure - or really at all - how Chess.com come to their conclusions, nor what role may be played by the Accuracy or Book figures.
I don't know, because they do not say. But I know that some of what they
say is a manifest nonsense.
And what
if you try to draw conclusions from nonsensical data?
[For the record, I referred the first two of these queries to Chess.com, and they ignored them. I've added the other two to this piece, because why not.]
3 comments:
Avital Pilpel said...
A friend of mine, a master, just won a blitz game on chess.com with "98% accuracy". As he showed me, the only "blunder" chess.com thinks he made was the horrible move 1.Nf3 ... everything else
was "pefect". The chess.com accuracy algorithm is nonsense.
Avital Pilpel said...
*Perfect*, not "pefect", of course.
I suppose this is an urban legend.
1. Nf3 surely was not a "blunder" but a "book move", also at chess.com. (Btw, also the category "perfect" does not exist there, but I would not stick to that wording ..).
Maybe you can ask your "friend" for a screenshot of the analysis, then I would believe it ... | {"url":"https://lostontime.blogspot.com/2019/10/ever-get-feeling-youve-been-cheated_89.html","timestamp":"2024-11-11T21:29:16Z","content_type":"text/html","content_length":"82793","record_id":"<urn:uuid:51117391-6522-449f-bad1-88658be5aaea>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00451.warc.gz"} |
Stacking in a Higher Interest Rate Environment
This article explores the relevance of leverage in investment strategies amid higher interest rates. The decision to use leverage should be based on the expected returns of assets rather than current
interest rates. By emphasizing the importance of understanding risk premiums, the article highlights that leverage can remain beneficial even in environments with higher interest rates.
Return Stacking, Capital Efficiency, Diversified Alternatives
Coming out of a decade-long low interest rate environment, many investors are questioning whether utilizing leverage remains a prudent idea now that interest rates are meaningfully above zero.
Our answer is that the level of interest rates should not impact the decision to utilize leverage. If leverage makes sense in a low-interest rate environment, then it should make sense in a high
interest rate environment.
Risk Premiums and the Risk-Free Rate
At the foundation of any discussion about return is the risk-free (RF) rate, which is the rate of return earned for investing in a theoretical, riskless asset. Although the risk-free rate doesn’t
actually exist, a typical proxy is the yield of U.S. Treasuries, since those are as close to “risk-free” as an investor can get.
Intuitively, for any investor to make an investment in anything that is not riskless, they would require the expected return on that investment to be higher than the risk-free rate (this is not
strictly true, as insurance instruments offer an obvious counter-point, but is sufficiently true for most investments). Put simply: the investor expects to get paid to be willing to bear the risk. As
we have been known to say, “No Pain, No Premium”. Generally speaking, if a risky asset has an expected return lower than the risk-free rate, the price of that asset should come down such that the
expected return goes up and properly compensates the investor.
For any asset, the expected return above the risk-free rate is called a “risk premium.” For example, in equities we have the “Equity Risk Premium” (ERP) and in bonds, we have the “Term Risk Premium”
(TRP) (there are many other risk premiums that we could list, but to keep this article succinct, we will focus on the ERP and TRP).
Expected Return = Risk-Free Return + Expected Risk Premium
In Figure 1, we provide a stylized decomposition of expected returns of equities and bonds into their individual components.
Figure 1: Expected Returns on Cash, Bonds, and Equities
Source: Newfound Research. For illustrative purposes only. Consult your financial advisor before making any financial decisions.
From Figure 1, we can see that the expected returns for bonds and equities are composed of both the risk-free rate, as well as their respective risk premiums. These risk premia are defined as the
expected excess return of the asset above the risk-free rate.
Applying Leverage to an Asset
If an investor wants to gain exposure to the S&P 500 index, most investors will operate in the cash market, using cash to purchase a share of the SPDR S&P 500 ETF Trust (ticker: SPY) or the
underlying stocks themselves.
Another method, however, could be to borrow money (using leverage) to purchase the exposure. If we borrow the money, we will have to pay back the borrowed amount plus the interest accrued, but we
would keep whatever we made in excess of that amount. So, if the S&P 500 returns more than the interest we must pay, we came out ahead by borrowing. In this scenario, the excess return of the S&P 500
is what we expect to make by borrowing, so long as the investor can borrow close to the risk-free rate.
This latter method is effectively how a S&P 500 futures contract works.
One minor difference that we would be remiss to mention is the interest rate. If an investor goes to a bank to borrow money, the interest rate charged would likely be significantly higher than the
yield on U.S. Treasury Bills, while futures contracts embed an interest rate remarkably close to the yield on 1-3 Month U.S. Treasuries.
Taken together, an investor who purchases an S&P 500 futures contract expects to earn the equity risk premium of equities and their realized return will be equal to the return of the S&P 500 minus
the risk-free rate (assuming the embedded borrowing cost in S&P 500 futures equals the risk-free rate).
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Since return stacking inherently uses leverage (typically via futures contracts) to stack returns into a portfolio, the risk premium of the assets in the portfolio is of great interest to investors.
However, what the section above hopefully illuminates is that the actual level of interest rates does not.
In fact, since the return of every asset can be decomposed into the risk-free rate and its excess return, and futures contracts effectively isolate the excess return, the choice whether to use
leverage based upon the level of the risk-free rate implies that there is a relationship between the level of the risk-free rate and the size of the expected risk premium. In other words, it’s a
market timing decision!
To further illustrate this, in Figure 2, we build a 100/100 stock/bond portfolio to analyze how the expected returns change as we introduce leverage to the portfolio.
Hypothetical Expected Return Decomposition of a 100/100 Portfolio
Source: Newfound Research. For illustrative purposes only. All expected returns are hypothetical and should not be relied on for investment decisions. This depiction assumes that the interest charged
on leverage is equal to the risk-free rate which may not be realistic. Consult your financial advisor before making any financial decisions.
From Figure 2, we can surmise, that stacking bonds on an equity portfolio leaves the portfolio with three things: the risk-free rate, the ERP, and the TRP. Even though we are implicitly incurring
borrow costs by applying leverage, we clear that hurdle so long as the ERP and/or the TRP are positive. Since we would expect risky assets to have a positive excess return, a logical conclusion is
that the level of interest rates is inconsequential to applying leverage.
The Return Stacking landscape is ever evolving, go deeper by connecting with a team member.
Do We Even Need to Talk about Leverage Here?
While we tend to apply the lens of leverage due to our affinity for return stacking, does this conversation even require the application of leverage?
The short answer is “no,” but the long answer utilizes a concept that we already applied earlier in this article.
If we assume that the risk-free rate is 5%, for us to hold any allocation to a risky asset then we need to have the expectation that that risky asset will return greater than 5%, or at least properly
compensate the investor for the risk taken.
To further develop this intuition, let’s detail two potential investments:
1. An investment that will return 5% in one year, guaranteed.
2. An investment that will return a maximum of 5% in one year, with a 50% probability of losing 5%.
No rational investor would select the second option, as the first is guaranteed and provides the same maximum gain.
Without even applying leverage, the investor is better off allocating to the risk-free asset, since it is irrational to allocate to the second choice.
If an investor has the view that equities or bonds will underperform cash (i.e. the ERP and TRP are negative), then that investor should allocate to cash. This decision is true regardless of whether
the position is levered or not.
If, however, the risk-free rate goes up and we believe that it is no longer prudent to use leverage, what we’re actually saying is, “because the risk-free rate went up, we believe the risk premia of
our assets has gone down.” Again, this is inherently a market timing decision based upon the level of the risk-free rate. (That is not to say that there isn’t a potential relationship between the
risk-free rate and levels of risk premia due to aggregate investor risk preferences, but they should be contemplated explicitly.)
To concretely summarize this post, we posit the following:
• Levered exposures should earn (approximately) the excess return of the asset class.
• The choice of whether to invest in an asset should be driven by its expected risk premium and potential benefit to the portfolio as a whole.
• For higher interest rates to imply that leverage is no longer attractive, it must mean there exists a relationship between the level of interest rates and expected risk premiums. In other words,
we must believe we can use the level of interest rates as a market timing signal! | {"url":"https://www.returnstacked.com/stacking-in-high-interest-rate-environments/","timestamp":"2024-11-07T13:11:32Z","content_type":"text/html","content_length":"246074","record_id":"<urn:uuid:a3098012-1a90-4406-a542-5ba0837fb5ac>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00703.warc.gz"} |
Monads, Explained Without Bullshit
Monads, Explained Without Bullshit (18 Jul 2020)
there's a CS theory term, "monad," that has a reputation among extremely online programmers as being a really difficult to understand concept that nobody knows how to actually explain. for a while, i
thought that reputation was accurate; i tried like six times to understand what a monad is, and couldn't get anywhere. but then a friend sent me Philip Wadler's 1992 paper "Monads for functional
programming" which turns out to be a really good explanation of what a monad is (in addition to a bunch of other stuff i don't care about). so i'm gonna be repackaging the parts i like of that
explanation here.
math jargon is pretty information-dense for me, though, and my eyes tend to glaze over pretty quickly, so i'll be using Rust (or an idealized version thereof) throughout this post instead of math.
so, a monad is a specific kind of type, so we can think of it like a trait:
trait Monad {
// TODO
Rust has two types that will be helpful here, because (spoilers) it turns out they're both monads: Vec and Option. now, if you've worked with Rust before, you might be thinking "wait, don't you mean
Vec<T> and Option<T>?" and that's a reasonable question to ask, since Rust doesn't really let you just say Vec or Option by themselves. but as it happens, the monad-ness applies not to a specific Vec
<T> but to Vec itself, and the same goes for Option. which means what we'd like to do is say
impl Monad for Vec {}
impl Monad for Option {}
but Rust won't let us do that because we can't talk about Vec, only Vec<T>. this is (part of) why Rust doesn't have monads. so let's just kinda pretend that's legal Rust and move on. what operations
make a monad a monad?
Wadler calls this operation unit, and Haskell calls it return, but i think it is easier to think of it as new.
trait Monad {
fn new<T>(item: T) -> Self<T>;
new takes a T and returns an instance of whatever monad that contains that T. it's pretty straightforward to implement for both Option and Vec:
impl Monad for Option {
fn new<T>(item: T) -> Self<T> { Some(item) }
impl Monad for Vec {
fn new<T>(item: T) -> Self<T> { vec![item] }
we started out with a stuff, and we made an instance of whatever monad that contains that stuff.
Wadler calls it *, Haskell calls it "bind" and spells it >>=, but i think flat_map is the best name for it.
trait Monad {
fn flat_map<T, U, F: Fn(T) -> Self<U>>(data: Self<T>, operation: F) -> Self<U>;
we have an instance of our monad containing data of some type T, and we have an operation that takes in a T and returns the same kind of monad containing a different type U. we get back our monad
containing a U.
as you may have guessed by how i named it, flat_map is basically just Iterator::flat_map, so implementing it for Vec is fairly straightforward. for Option it's literally just and_then.
impl Monad for Option {
fn flat_map<T, U, F: Fn(T) -> Self<U>>(data: Self<T>, operation: F) -> Self<U> {
impl Monad for Vec {
fn flat_map<T, U, F: Fn(T) -> Self<U>>(data: Self<T>, operation: F) -> Self<U> {
so in theory, we're done. we've shown the operations that make a monad a monad, and we've given their implementations for a couple of trivial monads. but not every type implementing this trait is
really a monad: there are some guarantees we need to make about the behavior of these operations.
monad laws
(written with reference to the relevant Haskell wiki page)
like how there's nothing in Rust itself to ensure that your implementation of Add doesn't instead multiply, print a dozen lines of nonsense, or delete System32, the type system is not enough to
guarantee that any given implementation of Monad is well-behaved. we need to define what a well-behaved implementation of Monad does, and we'll do that by writing functions that assert our Monad
implementation is reasonable. we're going to have to also cheat a bit here and deviate from actual Rust by using assert_eq! to mean "assert equivalent" and not "assert equal"; that is, the two
expressions should be interchangeable in every context.
first off, we have the "left identity," which says that passing a value into a function through new and flat_map should be the same as passing that value in directly:
fn assert_left_identity_holds<M: Monad>() {
let x = 7u8; // this should hold for any value
let f = |n: u8| M::new((n as i16) + 3); // this should hold for any function
assert_eq!(M::flat_map(M::new(x), f), f(x));
next, we have the "right identity," which says that "and then make a new monad instance" should do nothing to a monad instance:
fn assert_right_identity_holds<M: Monad>() {
let m = M::new('z'); // this should hold for any instance of M
assert_eq!(M::flat_map(m, M::new), m);
and last but by no means least we have associativity, which says it shouldn't matter the sequence in which we apply flat_map as long as the arguments stay in the same order:
fn assert_associativity_holds<M: Monad>() {
let m = M::new(false); // this should hold for any instance of M
let f = |data: bool| if data { M::new(3usize) } else { M::new(7usize) }; // this should hold for any function
let g = |data: usize| M::new(vec!["hello"; data]); // this should hold for any function
M::flat_map(M::flat_map(m, |x: bool| f(x)), g),
M::flat_map(m, |x: bool| M::flat_map(f(x), g))
so now we can glue all those together and write a single function that ensures any given monad actually behaves as it should:
fn assert_well_behaved_monad<M: Monad>() {
but. why
well. monads exist in functional programming to encapsulate state in a way that doesn't explode functional programming (among other things, please do not @ me). Rust isn't a functional programming
language, so we have things like mut to handle state.
there's a bit of discussion in Rust abt how monads would be actually implemented - the hypothetical extended Rust that i use here is not actually what anyone advocates for, you can look around for
yourself if you care - but even the people in that discussion seem to not really explain why Rust needs monads. so all of this doesn't really build up to anything. but hey, now (with luck) you
understand what monads are! i hope you find that rewarding for its own sake. i hope i do, too. | {"url":"http://gemifedi.boringcactus.com/2020/07/18/monads-without-the-bullshit.html","timestamp":"2024-11-12T09:28:44Z","content_type":"text/html","content_length":"29950","record_id":"<urn:uuid:927edc67-1814-40ba-b592-e224db664c15>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00700.warc.gz"} |
Electric play dough - a bit more explanation
A friend and her daughter tried our
electric play dough
activity and enjoyed it. However, she gave me some helpful feedback - they would have found it easier had there been a bit more explanation of the science behind it so my friend (who isn't a
scientist) would have been able to better predict what would and wouldn't work. If you read this, thanks for the helpful comment!
Here's an attempt at a bit of further explanation for electric play dough circuits, but I've not studied anything to do with electricity for a long time, so if anyone can improve it then please do
leave a comment!
It's not at all designed to be the explanation to give a small child, but to assist grown ups with a hazy memory to be more confident in trying the activity with children.
What is voltage?
You can think of this as the 'push' of electrons around a circuit, such that a greater voltage means a battery is more able to 'push' the electricity around. More properly, its the difference in
electrical potential energy between two points. You measure voltage in units called Volts (abbreviated to V).
What is current?
You can think of this as the 'flow' of electrons around a circuit, such that a higher current means more electricity is flowing. The flow of electrons needs to be around a complete circuit, i.e.
there needs to be a route from one terminal of the battery to the other. Current is the amount of charge flowing per unit time and is measured in a unit called Amperes, commonly referred to as Amps
(abbreviated to A).
What is resistance?
This how difficult it is for the charge to flow around a circuit, i.e. how much something reduces the current flowing. Electrical resistance is measured in something called Ohms (abbreviated to Ω).
An analogy
If you think of a container of water with a pipe coming out of the bottom, the charge is analogous to the amount of water in the container, and the voltage analogous to the water pressure in the
pipe. So the more water in your container, the greater the pressure in the pipe at the bottom. Taking this analogy, the current is the amount of water flowing through the pipe. If you have a higher
voltage (pressure of water), then the current (amount of water flowing) is higher.
If you had two identical water containers as above, but one had a wider pipe than the other, the water would flow faster out of that one i.e. the resistance is lower and the current (flow rate) is
higher. If you want the water in the narrower pipe to flow at the same rate, you'd need more water in the container to increase the water pressure; similarly if you want to increase the current in an
electrical circuit with more resistance then you need to increase the voltage.
Ohm's law
You might remember this one from school, Ohm's law can be expressed as the formula V=I.R (where I is current). In other words, voltage = current x resistance. If you rearrange this, you can see that
current = voltage / resistance, or resistance = voltage / current, and this lets you work out what will happen e.g. if you increase the voltage or the resistance in your circuit.
So what's this got to do with electric play dough?
LEDs need sufficient current (in the right direction) to make them light. To get enough current in the
electric play dough
I blogged about, you need a 9 Volt battery. This is because play dough is a poor conductor compared to a copper wire, so it's got a lot of resistance. To get enough current, you need more voltage
than a 1.5 Volt battery (e.g. an AA) can provide; when I tried one LED with two lumps of play dough at each end of the AA battery, there wasn't even a faint glow from the LED.
You shouldn't put the LEDs straight across the 9V battery terminals because this causes too much current flows through the LED and it will burn out after lighting brightly for a while. Without the
resistance provided by the play dough, the current is too high.
You need a complete circuit for the current to flow and the LEDs to light. This is why when we made
traffic lights
, the spoon acted as a switch in completing a circuit that contained one (or two) of the LEDs. You also need to force the current to flow through the LED by leaving a gap between the play dough - if
you just stick both ends of the LED into one blob of play dough and connect it to a battery, the current doesn't pass through the LED and there's no light.
circuit with 5 LEDs in series
(in a line one after the other in the circuit) didn't work i.e. none of the LEDs lit. This can be explained by the lack of current. There's too much resistance in the circuit in total, so the current
flowing around the circuit is just too low to make the LEDs emit light. If you take out some of the lumps of play dough (which add resistance) and some of the LEDs, the current increases. With more
current, the LEDs glow more brightly.
In the simple parallel circuits we made, the current flowing through each LED is higher than if the same number of LEDs were in series. The LEDs therefore light more brightly in parallel than in
series. You can think about this from the perspective of voltage - the difference in charge across each LED (the voltage) is higher than for the LEDs in the series circuit so the current is greater.
If you want more LEDs in your electric play dough, the trick is therefore to make at least some of them in parallel. We managed 5 in parallel when we were
making circuits with cutlery
. We've built a few circuits where the LEDs are a mixture of series and parallel - in the one below, the LEDs lit nicely!
Our most recent play dough circuit - an elephant holding parallel LEDs with its trunk!
[Edited on 1/9/19 to take account of comment below re: definition of voltage]
1. Hello from the electric play dough friend's husband! :-) I've been following your blog with interest, and not least am very pleased to know what an ossicone is!
Being pedantic, I'm afraid voltage isn't the difference in *charge* between 2 points but the difference in electrical potential energy (hence 1 Volt is 1 Joule per Coulomb). Sorry...
Also, as well as the resistance reason for LEDs not lighting with AA batteries (or lots in series on a 9V battery), there's the fact that they each have a minimum voltage they need to see across
them before they will light up - which is related to the wavelength of the colour they emit by E=hf, hence red LEDs will light at a lower voltage than blue ones. If there is enough voltage but
high resistance then you'll often see them lighting up really dimly. But please don't ask me as a lowly engineer to explain anything quantum!
We hope to catch up with you all again soon.
1. Thanks for the clarification, you are of course correct on voltage. I blame too little sleep in recent months! I will edit above when I get a chance!
2. Thanks for a very clear explanation for a non-scientist :). It's always good to learn something new! | {"url":"https://www.ossiconesoxygen.com/2019/09/electric-play-dough-bit-more-explanation.html","timestamp":"2024-11-08T01:58:26Z","content_type":"application/xhtml+xml","content_length":"120686","record_id":"<urn:uuid:b3bfcbc5-fa90-414b-a092-bd2dcf1bbb9e>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00699.warc.gz"} |
Which Way Is East? A Solution
Wisconsin Geospatial News
Which Way Is East? A Solution
In an earlier post, I challenged readers to find a solution to a cartographic problem: What is the latitudinal deviation (at a chosen meridian) between a given parallel (a line of latitude) and a
great circle arc tangent to that parallel at a given point, where the great circle is the projection of a line heading east on a compass.
Recall (from the earlier post) that all parallels bear the same angular relationship to every meridian, along the entire length of the parallel. In other words, at every point on the parallel, the
curve bears due east and west.
The green dashed line in Figure 1 is an example for the parallel at 45 degrees North. In other words, the green line describes a true east-west line. If you started at Point A and followed the green
line, you would maintain the same angular relationship to north (i.e., 90 degrees).
On the other hand, if you placed a compass at Point A with the arrow pointing north, the projection of the imaginary line pointing east on the compass would follow a much different path. This line
would describe a great circle of the earth gradually departing southerly from the true parallel. This is shown in Figure 1 by the pink dashed line.
Fig 1. The earth’s graticule, with an illustration of the problem
Your challenge was to find the distance D along the meridian at 30 degrees W. The value of 30 has no special significance. It is just an example.
Unfortunately, I did not receive any responses to the challenge! Nevertheless, I have developed a solution, which I share below.
A Solution
There are several ways to solve this problem. My approach is based on solving spherical triangles using trigonometric rules that have analogs in more conventional Euclidean geometry. (This turned out
to be easier than I expected, given that I have no background in spherical trigonometry.) The key to solving the problem is to define a set of spherical triangles (triangles whose sides are formed by
great circle arcs) for which we know at least three of the six angles and sides. With those three pieces of information, the unknown angles and sides can be computed. One other thing to note about
spherical triangles is that their sides can be expressed as angles, since they are arcs of great circles that subtend an angle at the center of the sphere.
Figure 2 shows the calculations involved to find the latitude of Point C and the length of arc D. Points A, B and C are the same as on Figure 1. Point E is a new point located on the Equator 90
degrees east of Point A. Point E is on the great circle arc shared with Points A and C. For any point at any latitude, the great circle arc heading east from that point will intersect the Equator at
a point 90 degrees to the east.
Fig 2. Details on the solution
Since Point E has known latitude and longitude (0 degrees N and 0 degrees W in our example), we can use it to construct a spherical triangle with vertices at Point A, the North Pole, and Point E. For
this triangle, we know the length (angle) of Arc P connecting Point A to the North Pole, the length (angle) of Arc R connecting Point E to the North Pole, and Angle v between Arcs P and R. (The angle
is 90 degrees by definition.)
We can use these values to compute Angle z, the angle formed between Arc R and the great circle arc. The equation to compute Angle z is shown in Figure 2.
Given Angle z, we now have the three pieces of information (Angle z, Angle x, and Arc R) we need to compute Angle y, the angle formed between the great circle arc and Arc Q (connecting Point C to the
North Pole). The equation for Angle y is shown in Figure 2.
Given Angle y, we can now compute the length (angle) of Arc Q based on Angles x, y and z. This gives us the latitude of Point C, which leads to the latitudinal difference between Points B and C,
which is the angle represented by Arc D. We can now compute the length of Arc D.
The length is approximately 2052.17 km. So, if you followed the projected direction from your compass, thinking you were going east, you would be over 2000 km off-course after travelling 60 degrees
Notes on Precision and Calculations
Given that we are already generalizing angles and distances by relying on a spherical earth model, we should not expect too much precision in the results. To keep the precision of distances (in km)
somewhere between 1 km and 1 m, I use 2 decimal places for distances. This implies about 4 decimal places for coordinates in decimal degrees, and about 6 decimal places for angles expressed as
All calculations are done on a sphere of unit radius. For the final calculation of the length of Arc D, a radius of 6378 km is used, which translates into 111.32 km per degree. All calculations are
done in radians since most trigonometric functions in software expect inputs in radians, not degrees.
Note also that there may be rounding errors in the return values of trigonometric functions, depending on which software is used and the defined precision of floating point numbers.
Extension to Other Locations
Obviously, Point A is just one of an infinite number of points that could be selected as an example. The solution described here will generally work with any location, subject to some caveats.
First, the deviation between Points B and C only needs to be computed for a 90-degree wedge to the east of Point A, wherever Point A might be located. The pattern of deviations simply repeats itself
in a mirror image as you move around the earth. You can see this in the visualization below, which shows the trace of the great circle passing through Point A. In other words, it shows the location
of Point C as this point moves around the earth given different longitudinal values. (Incidentally, this visualization was created in Microsoft Excel. Did you know that Excel provides basic mapping
Fig. 3. Click to view visualization
The equations will not work without modification if Point A is in the southern hemisphere, but the values for the southern hemisphere are just mirror images of those in the northern hemisphere, so
the equations can still be used with slight modifications.
A problem will also occur if Point A and Point E are on opposite sides of 180 degrees East/West. In this case, a solution would be to transform the negative longitude values into positive longitudes
ranging from 0 degrees at the Prime Meridian to 360 degrees as you move east around the globe.
Confirmation of Results
Can we be sure the results are accurate? To confirm the results, I compared them to those obtained by another method, described here by Dr. Louis Strous. His example #4 (“A Great Circle in a Certain
Direction Through a Known Point”) involves converting latitude and longitude to 3-dimensional Cartesian coordinates (x, y, z) and then computing the location of a point at a known distance from a
starting point traveling in a known direction. I adapted this method to our problem by simplifying it to account for the fact that we are always traveling in an easterly direction. I used the Solver
in Excel to find the latitude of Point C by iteratively changing the distance value until a solution was obtained. Aside from rounding error, results are identical to the method described above.
Thanks to Mike Hasinoff for checking my math and being interested in the problem, Dr. Louis Strous for his inspiration, and this Wikipedia page. | {"url":"https://www.sco.wisc.edu/2024/06/12/which-way-is-east-a-solution/","timestamp":"2024-11-10T14:41:31Z","content_type":"text/html","content_length":"78088","record_id":"<urn:uuid:d52ee1d9-3893-4cb6-9663-54f1070d4df8>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00494.warc.gz"} |
Excel Formula Error - Too Many Arguments - Comma or Parenthesis Error? | Microsoft Community Hub
Excel Formula Error - Too Many Arguments - Comma or Parenthesis Error?
I am trying to create the following formula -
I keep returning too many arguments error - I've tried my best to adjust commas/parenthesis in a logical way to alleviate, but cannot figure it out!
If I remove the final portion of the formula, it returns the correct information and reads the formula correctly:
However, I need that final formula (IF(F2=P5,1%) added to make the formula usable for my workbook.
Please let me know if anyone knows what I'm missing or sees a resolution.
Thank you
• HansVogelaar This works! Thank you so much! And now makes much more sense. Have a wonderful day. | {"url":"https://techcommunity.microsoft.com/discussions/excelgeneral/excel-formula-error---too-many-arguments---comma-or-parenthesis-error/4001502","timestamp":"2024-11-10T16:08:53Z","content_type":"text/html","content_length":"235536","record_id":"<urn:uuid:e156c46f-d906-4f79-8cec-45c890931718>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00623.warc.gz"} |
Introduction | Kickflow Documentation
What is Kickflow?
Kickflow is a decentralized platform that enables community funding for projects on Tezos. Through the concept of Quadratic Funding, it gives the community the power to take the best projects
Besides being a general crowdfunding platform, Kickflow conducts periodic match funding rounds, where entries receive an optimal share from a sponsor-funded pool.
Kickflow is governed by a DAO that manages the funding rounds and operates on a token voting mechanism using the Kickflow Governance Token ($KFL).
The need for Quadratic Funding
Match funding schemes for public goods suffer from a fundamental weakness- the match amount for a project is proportional to the sum of contributions it receives during a stipulated period of time.
This has its benefits, but it entirely rules out individual preferences.
Think of a situation where two projects A and B have received a contribution of $1000 from the community. A has received it from 10 unique contributors, whereas B from just 2 unique contributors.
Even though it signifies A has a better reach and preference than B, both of them end up receiving an equal amount of match from the funding pool.
To solve this problem, we bring in CLR* matching. Here, instead of being linear, the match is calculated as the square of the sum of the square roots of the individual contributions.
Let's say A received $100 each from 10 contributors whereas B received $500 from 2 contributors. If we use CLR matching, A's match would be $10,000 whereas B's match would be $2000. Here clearly the
individual preferences have been factored in.
For a more in-depth study of Quadratic Funding, view the original paper by Vitalik Buterin.
*CLR refers to Capital-constrained Liberal Radicalism. It brings about a democratic essence to capital distribution amongst public goods. | {"url":"https://docs.kickflow.io/","timestamp":"2024-11-13T07:45:19Z","content_type":"text/html","content_length":"160508","record_id":"<urn:uuid:05b02dd5-c446-475d-8955-7ede038f8730>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00862.warc.gz"} |
Re: [Inkscape-devel] Recent change to ellipse to path conversion method?
30 Nov 2013 30 Nov '13
2:50 p.m.
The behaviour of the tail_error(B, n) function in build_from_sbasis() in the file sbasis-to-bezier.cpp has been reevaluated. The term tail_error(B, 2) is too large since it is of order 4 in theta,
for an arc, while the error currently produced is of order 6 in theta. For this reason the term tail_error(B, 3) was investigated numerically to see how it behaves. By numerical printouts at
different theta, it was found that for a unit circle the tail_error(B, 3) term is roughly given by : 0.000022*(theta)^6. This agrees surprisingly well with the formula given above for the case where
the sbasis is fit at t = 0.5. The formula given above was: error = (2/27)*(theta/4)^6. The difference between these results is only 20%, which I find surprising given the very different histories of
the two calculations. In any event, I would like to propose that the term tail_error(B, 3) be used to determine the error estimate when bisecting the sbasis curves to meet the required tolerance. If
this is done, numerical tests indicate that the dividing point in switching from 4 Beziers to 8 Beziers for a full circle will occur at r = 407 instead of the current r = 8.18.
cheers, Alvin
-- View this message in context: http://inkscape.13.x6.nabble.com/Recent-change-to-ellipse-to-path-conversion... Sent from the Inkscape - Dev mailing list archive at Nabble.com. | {"url":"https://lists.inkscape.org/hyperkitty/list/inkscape-devel@lists.inkscape.org/message/EYEXWQH6K352HNCHZPYIOEDAVSAKPDOS/","timestamp":"2024-11-13T04:51:47Z","content_type":"text/html","content_length":"14155","record_id":"<urn:uuid:006ae26e-1a80-40cb-9c0e-d8f2eccaf82f>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00324.warc.gz"} |
Dynamic shapes
Dynamic shapes¶
Code: symbolic_shapes.py
See also: The dynamic shapes manual
Deep learning compilers commonly only work for static shapes, that is to say, they produced compiled programs which only work for a single specific configuration of input shapes, and must recompile
if any input shape changes. This assumption works great for the majority of commonly run deep learning models today, but there are a few situations where it is insufficient:
• Some dimensions, such as batch size or sequence length, may vary. For example, an inference service performing adaptive batching will execute inference requests with varying batch sizes depending
on how many requests it received within its batching window. We may also want to consider padding out variable size sequences only to the maximum sequence length within a batch, which may vary
from batch-to-batch.
• Some models exhibit data-dependent output shapes, that is to say, the size of their outputs and intermediates may depend on the actual input data which may vary across runs. For example,
detection models may first generate a variable number of potential bounding boxes before running a more expensive image recognition model to identify if the subject is in a bounding box. The
number of bounding boxes is data dependent.
• One particularly important case of data-dependent shapes occurs when dealing with sparse representations, such as sparse tensors, jagged tensors, and graph neural networks. In all of these cases,
the amount of data to be processed depends on the sparse structure of the problem, which will typically vary in a data-dependent way.
In supporting dynamic shapes, we chose not to support dynamic rank programs, e.g., programs whose inputs tensors change in dimensionality, as this pattern rarely occurs in real-world deep learning
programs, and it avoids the need to reason inductively over symbolic lists of shapes.
Abridged public API¶
The default dynamic behavior in PyTorch 2.1 is:
• PT2 assumes everything is static by default
• If we recompile because a size changed, we will instead attempt to recompile that size as being dynamic (sizes that have changed are likely to change in the future). This generalization may fail
(e.g., because user code does a conditional branch on the size in question or missing dynamic shapes support in PT2). If you are trying to understand why PT2 has overspecialized some code, run
with TORCH_LOGS=dynamic and look for “eval” entries that say when guards are added and why.
• If you know ahead of time something will be dynamic, you can skip the first recompile with torch._dynamo.mark_dynamic(tensor, dim).
• If you say torch.compile(dynamic=False), we will turn off automatic dynamic shapes on recompiles and always recompile for each distinct size. Conversely, if you say torch.compile(dynamic=True),
we will try to make everything as dynamic as possible. This is mostly useful for small operators; if you try it on a big model it will (1) probably crash PT2 and (2) run slow for no good reason.
The Guard Model¶
When considering how to add support for dynamic shapes to TorchDynamo and TorchInductor, we made a major design decision: in order to reuse decompositions and other preexisting code written in Python
/C++ targeting the PyTorch API, we must be able to trace through dynamic shapes. Unlike a fully symbolic system which might capture both branches of a conditional, we always pick one branch and
specialize our trace under the assumption that we only use this trace when we would have made the same choice for that branch in the future. To do this, we maintain a “hint” for every symbolic size
saying what its concrete value is at compile time (as TorchDynamo is a just-in-time compiler, it always knows what the actual input sizes are.) When we perform a condition on a tensor, we simply
consult the hint to find out which branch to take.
This greatly simplifies the symbolic shape formulas we produce, but means we have a much more involved system for managing guards. Consider, for example, the following program:
def f(x, y):
z = torch.cat([x, y])
if z.size(0) > 2:
return z.mul(2)
return z.add(2)
The final IR we will compile with TorchInductor will either be torch.cat([x, y]).add(2) or torch.cat([x, y]).mul(2) (with the condition flattened away), but to determine which branch we are in, we
would need to know the size of z, an intermediate. Because TorchDynamo must know upfront if a compiled trace is valid (we do not support bailouts, like some JIT compilers), we must be able to reduce
z.size(0) as an expression in terms of the inputs, x.size(0) + y.size(0). This is done by writing meta functions for all operators in PyTorch which can propagate size information to the output of a
tensor without actually performing computation on the node.
Overall architecture¶
Symbolic shapes workflow:
1. When we start compiling a frame in Dynamo, we allocate a ShapeEnv (attached to FakeTensorMode) which keeps track of symbolic shapes state.
2. We allocate symbolic sizes for tensors on entry (what is static or dynamic is a policy decision, with some knobs).
3. We propagate the symbolic sizes through operators, maintaining both (1) FX IR so that we can faithfully export symbolic compute, and (2) Sympy expressions representing the size vars, so we can
reason about them.
4. When we condition on symbolic sizes, either in Dynamo tracing or in Inductor optimization, we add guards based on the conditional. These can be induced from both Python and C++.
5. These guards can induce further simplifications on symbolic variables. For example, if you assert s0 == 4, we can now replace all occurrences of s0 with 4.
6. When we’re done tracing and optimizing, we install all of these guards with the compiled code; the compiled code is only reusable if all the guards evaluate true.
Important files:
• C++ SymInt API: c10/core/SymInt.h, SymFloat.h, SymBool.h
• Python SymInt API: torch/__init__.py (look for SymInt/SymFloat/SymBool)
• C++ plumbing: c10/core/SymNodeImpl.h, torch/csrc/utils/python_symnode.h, torch/csrc/jit/python/init.cpp
• Python infrastructure: torch/fx/experimental/symbolic_shapes.py
• Other important files: torch/_subclasses/fake_tensor.py, torch/_meta_registrations.py, decomps, PrimTorch refs
Abridged internal API¶
Understanding the Python class hierarchy:
• SymInt/SymFloat/SymBool: these are user-visible classes that simulate their int/float/bool counterparts. If you add two SymInts, we give you a new SymInt that symbolically tracks that the integer
addition had occurred.
• SymNode: this is the internal structure (accessible via e.g., symint.node) which holds the actual symbolic tracking info. SymNode is type erased; this makes it more convenient to represent
mixed-type operations. Note that technically you don’t have to call into Python SymNode from SymInt; for example, XLA’s C++ SymNodeImpl would take the place of SymNode.
• ShapeEnv: per-compile context state which keeps track of all the free symbols and guards we have accumulated so far. Every SymNode records its ShapeEnv (but not vice versa; SymNodes only get used
if they participate in a guard).
C++ is fairly similar:
• c10::SymInt/SymFloat/SymBool: user-visible classes that simulate int/float/bool.
• c10::SymNode/SymNodeImpl: analogous to SymNode
• There is no ShapeEnv in C++; for ease of debugging, the entire symbolic reasoning apparatus is in Python.
When you write code that is traceable with make_fx, it must be able to deal with SymInt/SymFloat/SymBool flowing through it. The dynamic shapes manual gives some guidance for how to do this.
DimDynamic policy¶
Symbolic reasoning:
• Value ranges
• Sympy usage notes
• Constraints
• DimDynamic/Constraint
Unbacked SymInts¶
To resolve control flow, we check the hint, aka actual value, of a symbolic integer to determine which branch to go. However, in some cases, we may not have a hint: so-called unbacked symbolic
integers arise when a size variable emerges from a data-dependent operation like .nonzero() or .item(). It is illegal to perform control flow on these symbolic integers, so we must graph break on
these operations.
Naively implemented, this is too restrictive: most PyTorch programs will immediately fail if you try to do anything with unbacked symbolic integers. Here are the most important enhancements to make
this actually work:
• On tensor creation, PyTorch precomputes a lot of data about a tensor; for example, if you use empty_strided to create a tensor, we will eagerly sort the strides and determine if the tensor is
non-overlapping and dense. Sorts produce a lot of guards. However, it is more common to produce a tensor directly with a higher-level API like empty, which is guaranteed to produce a
non-overlapping and dense tensor. We modified PyTorch to avoid needlessly recomputing these properties.
• Even if nontrivial compute is needed, sometimes a property is never actually queried at all. Making these precomputed properties lazy allows us to avoid guarding on an unbacked symbolic integer
unless it is actually needed.
• The data in an integer tensor is generally not known to be non-negative. However, we provide an API constrain_range whereby a user can specify that a size is bounded above and below by known
In future versions of PT2 (beyond PT2.1), we will extend our reasoning system to infer that an unbacked symbolic integer is size-like based on usage. For example, if you pass the result of an .item()
call to a factory function like torch.empty, we will automatically infer that the result is a size (because if it was not, it would fail.) This assumption would get validated at runtime, raising an
error if it was not fulfilled. | {"url":"http://pytorch.org/docs/2.2/torch.compiler_dynamic_shapes.html","timestamp":"2024-11-10T01:44:42Z","content_type":"text/html","content_length":"55681","record_id":"<urn:uuid:8fd6baef-d17f-436e-80f5-6bff2ad253de>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00019.warc.gz"} |
All Categories
They always say you spend the beginning of the year setting up routines that will allow your students to thrive throughout the school year. While establishing routines can be daunting it does not
have to be! Read on to see how I use simple no prep activities to help my students fall in love with math while mastering our class math routines!
Click here to read all about (and get a freebie!) my "no prep" beginning of the year packets for Kindergarten, First Grade, and Second Grade Math!
Let them Explore!
What kid or adult can resist tinkering with new objects put in front of them? I sure can't! We know that if we want kids to use the manipulatives to solve math problems in a specific way, we first
need to give them opportunities for open ended play with the same materials.
Students are inherently interested when they see math manipulatives. While an adult might look at a pile of unifix cubes and think they are for counting or learning math concepts, kids see so much
more. I have had kids try to build towers taller than a friends, sort the cubes by color, build a "staircase" of rainbow cubes, and so much more! All of this "play" is actually the foundation of
many math concepts.
By investing a small amount of time into open ended manipulative play you are allowing students to explore so that when they need to complete a concrete task with the manipulatives they can, the
novelty has worn off somewhat.
Open ended play is also a great opportunity to practice classroom routines like cleaning up, answering to a call and response, moving around the room. It also allows for practice of social emotional
skills like sharing and problem solving!
Moving On to Paper and Pencil Tasks
While kids can learn so much from manipulatives the reality is they also need to be able to complete paper and pencil tasks. However, these don't have to be boring worksheets filled with endless
equations. I like to start the year with beginning of the year themed worksheets that are accessible to most students. I find that by giving them math worksheets that they can successfully complete
I am building up their math confidence! This plays dividends as harder concepts are introduced throughout the year.
I like to use math centers or rotations so I can pull small groups. As my students become more comfortable with math manipulatives I start dividing the class-- some completing paper and pencil
tasks and some exploring manipulatives. This allows kids to keep practicing their independence and the classroom routines and leaves me to float around helping anyone who needs it! I rotate
activities so every student gets to experience both sides.
Eventually, I will add a third group- one working with me! At this point the paper and pencil group and math manipulatives group are working independently. This allows me to start on the formal math
curriculum, knowing that my students have mastered the routine. Throughout the year the manipulative and paper and pencil tasks get more complex but the structure stays the same!
Check out my seasonal math worksheets here!
Read more of my tips for center organization here!
Check out how I use digital math centers for virtual and in person learning!
Click here for my Eureka Math centers!
3 Comments
I’ve said it many times and I will say it many more--
math centers
are my absolute favorite part of the day! I love watching students explore mathematical concepts by completing centers and games. They learn so much more through the hands-on interactions (and
conversations with their classmates) than they would simply completing math worksheets or in whole group lessons. When we abruptly switched to virtual learning, I mourned the loss of math centers.
But after some research I have found a way to give students intentional and fun activities to mimic the center experience they would have in the classroom.
I am using Boom Learning for my “digital math centers”
All of the virtual math activities mentioned here are available both on
Boom Learning
Teachers Pay Teachers
Why Boom Learning?
Boom cards are self-correcting! I love self-correcting activities both in person and virtually because students get in the moment feedback on how they are doing. In the classroom I use lots of
self-correcting puzzles. On Boom only the correct answer is accepted (this is different from what happens in Google resources or Seesaw). If students get the answer wrong they are able to try again
(we all know that making mistakes and trying again is a huge part of the learning process!)
Boom cards have a lot of flexibility in what students can be asked to do. In some activities I have them count the number of objects and type the total. In others they match shapes to real life
examples or numerals to quantities and drag them together. In some they use digital counters to build sets to match numbers or sets that have one more or one less. Some activities have students
click on the correct answer (a number, a shape, an object—the sky is the limit).
You can use a Boom account or you can use these activities without one! If you have the Boom account you can track student progress on individual decks which will allow you to see where
misconceptions are and plan for reteaching. If you don’t want or can’t have student accounts that is okay!!! You can still assign students the decks and they complete them from the link. You won’t
be able to see how they do but they will still get the practice. In the classroom I don’t usually see how students complete math centers because I am generally working with a small group!
The links can work with a variety of platforms- you can assign them in Google classroom, link them in Canvas, or include them in an email to parents!
Virtual Math Centers
I use the Eureka Math (EngageNY) curriculum so have based my activities off of their scope and sequence. However, they all meet the Common Core standards and are applicable to any pre-k or
kindergarten classroom! So far I have 3 groupings of digital math activities- counting and cardinality, geometry, and measurement and data.
Digital Counting Centers
I have built in different types of activities to give students practice counting sets up to 10. In some activities like Ice Cream Counting they count the sets and drag and drop them next to the
correct number. In others like Rocket Counting, Fish Counting, and Counting Dogs students count the pictures and type the number into the correct box. My favorite activities are the ones where
students build sets to match the number given (Ocean Counting, Night Sky, and Butterfly Counting). I like these because any combination of the number is marked as correct- if you are making a set of
5 you can do 5 fish straight across, 2 in the top and 3 in the bottom, every other box, etc. This helps students learn that numbers can be represented in many different ways!
I have also grouped some activities into my counting bundle that teach about one more and one less. There are virtual math activities where students build sets with one more or one less than a given
number. There are also two digital math activities where they count the objects and then click the number that represents one more or one less.
Shapes and geometry math activities for the distance learning classroom! Great virtual math activities for kindergarten.
Digital Geometry Centers
In the classroom I love teaching geometry by building shape museums of everyday objects. We talk all about the attributes and how we know what type of shape each object is. I tried to digitalize
this by having students identify real life objects that are certain shapes. Also, by giving them opportunities to drag and drop real life objects to the correct 2d or 3d shape.
Sorts are some of my favorite activities. I have included a 2d (flat shape) sort, a 3d (solid shape) sort, and a sort where students differentiate between flat and solid shapes! I think this
differentiation could be difficult without hands on practice but am confident that the real-life pictures will help students distinguish between 2D and 3D shapes.
Digital Measurement and Data Centers
Measurement is definitely an activity that is best done with hands on experiences. I plan to have students find objects around the house and compare them based on height, length, weight, and volume.
I created Boom cards to review these concepts. For height and length there are 3 activities- one where they order objects from shortest to tallest or longest to shortest, one where they sort objects
as taller/shorter or longer/shorter than a given object, and finally one where they count the cubes to measure the height or length of different objects.
For weight I have an activity where they order objects from heaviest to lightest or lightest to heaviest. There is also an activity where they see a scale with an object on one side of it. They then
choose between two given objects as to which could correctly complete the image. The students will have to think about the weight of the different objects as well as the positioning of the scale (is
the known object lighter and up in the air or heavier and low down?)
My measurement activities also include comparing quantities. There is a digital activity where they count the two types of animals and decide if the quantities are the same or different. This then
progresses to an activity where they count three sets of animals, type the correct numbers in and then click on the animal that has the most or fewest. There is a third more difficult activity as
well. Students compare two groups of animals and complete a comparison statement about what they see—for example 3 is less than 5 or 8 is more than 6.
Using Digital Math Activities
I am excited to use these virtual math centers with my students! I think they will be a good bridge between the in class experience and virtual learning. I also think they will have a place in the
brick and mortar classroom. I plan to use them as a computer center during my normal math rotation.
How are you adapting math centers to virtual learning? I would love to learn from your experiences
Check out my virtual math activities on
BOOM learning
and on
Teachers Pay Teachers!
2 Comments
1 Comment
My students are OBSESSED with all competitive games. Right now they are loving our "POW" games! These games are so simple to prep and perfect to play in small groups or during centers.
This number recognition game is a perfect math center activity!
POW is simple to play but very engaging!
Place all the cards face down. Students take turns picking a card. If they can identify the number on the card they get to keep the card, if not it goes back in the pile.
If a student picks the "POW!" card they have to put all of their cards back! The POW card is then taken out of circulation and the game continues.
The player with the most card at the end wins!
Fill out the form below to get your FREEBIE!
This freebie includes two versions of POW- recognizing numbers to 120 and recognizing numbers to 20.
Check out other versions of the "POW" game! There are addition, subtraction, and sight word fluency versions!
1 Comment
1 Comment
My students have been planning their Halloween costumes since approximately September 5th, the parents are busy planning our class Halloween party, and I have been making Halloween math worksheets
to try and capitalize on the excitement!
I made a no-prep Math worksheet for Kindergarten, First Grade, and Second Grade. Each grade level has 20 Halloween themed worksheets that can be used individually or as a packet. All of the
worksheets are Common Core aligned but work great in any classroom.
Fill out the form below to get your HALLOWEEN MATH FREEBIE!
1 Comment
I can't believe that it is almost time to go back to school! I have put together some Back to School math worksheets to help with the transition back into the school year.
I use these worksheets as independent work as students are getting used to the routines of the classroom and as I gradually introduce math centers. They could also be used as morning work or
There are packets for Kindergarten math, First Grade Math, and Second Grade Math. Each grade level includes 20 Common Core aligned, no prep worksheets. Click on the pictures below to see the full
Enter your email address below to get my Back to School Math Freebie!
0 Comments
7 Comments
I love teaching math! I especially love teaching math in small groups while the rest of the students are exploring and learning through math centers.
For the past few years I have worked at schools that use Eureka Math (formerly known as Engage NY). Eureka Math emphasizes conceptual understanding of math concepts. As a result I think it is a
great curriculum to pair with math centers and exploration. I think math centers are a perfect time to incorporate play into the classroom.
Math Centers and Play:
Play has so many great benefits for students. They learn:
• Problem solving
• Social skills including sharing, taking turns, compromising, resolving conflicts
• Fine motor skills
• Vocabulary
• Independence
• Creativity
While math centers in my classroom are not quite as free as other play times, I believe that they still get many of the benefits of play while also strengthening their math knowledge.
I give students free choice to choose any of the math centers I put out. They are able to pick the one they most want to work on and as soon as they feel they have completed it they may move on to
another one. Some students gravitate to the same centers over and over while others rotate more quickly, either way they are learning! Sometimes during small group learning we talk about how certain
centers are practicing specific math concepts and that can sometimes guide their choices.
I give students choice in what center they do so it is important that the centers are aligned to the curriculum. I have created a set of math centers that are easy to prep and align to Eureka Math
Module 1 for Kindergarten.
Module 1 is broken down into 8 topics. I made 3-4 centers per topic for a total of 25 centers. Each center is different but there are some common themes.
• Write the Room- students find cards around the room and count the objects on it. They write the number on the recording sheet. There is also a version where students count and record the objects
and then write what number is one more and a version where students find the set and the number that is one less.
• Count and Sort- Students count the sets and sort them by total. There are three versions- numbers to 3, numbers to 8, and numbers to 10.
• Number Writing and Building Sets-
□ Playdough Numbers- Students practice making the numeral with playdough and building sets with that many. There are two versions- numbers 0-5 and numbers 0-10.
□ Dry Erase Numbers- Students trace the number, fill in ten frames with that many, draw that many in a set, and circle the numeral. There are versions for 0-5 and 0-10 included.
• Puzzles- I love puzzles! I have included puzzles for number recognition, number sentences, and one more/one less.
Check out the full collection of centers below. Scroll down to get a freebie of the two centers that align with Topic A!
7 Comments
It's that time of year- the count down for summer break is on! It's always one of the craziest times of year for me between squeezing in instructional content, end of year testing, packing up the
classroom, organizing the end of year celebration, and enjoying the final days with my students!
I always send students home for the summer with books and a packet of summer themed activities and worksheets to keep their mind sharp over the summer. I hope that the summer activities will prevent
the summer slump!
I have created summer activities and worksheets for Kindergarten, First Grade, and Second Grade. Each grade level has 20 summer worksheets- 10 summer math worksheets and 10 summer literacy
worksheets. Worksheets are Common Core aligned and designed to review standards from the grade students just finished so they are ready to be all-stars in the upcoming school year!
Click on each picture to see the grade level summer activity packet! Scroll down and enter your email address to get a free sample of my summer activities and worksheets!
Check out some of my easy, no-prep End of School Year activities!
0 Comments
I love using literacy activities and centers in the pre-k, kindergarten, first grade, and second grade classroom. I think they increase engagement and support social-emotional learning. I use them
to compliment direct instruction and give students more hands-on practice with content.
I structure my centers to match the scope and sequence of our ELA curriculum. This year we are using the Wit and Wisdom curriculum for our literacy block. Wit and Wisdom is Common Core aligned and
tells teachers which standards will be targeted during each Module.
I take this scope and sequence and create literacy centers that practice the same Common Core standards but in a different way. The centers are not linked to the Wit and Wisdom curriculum but do
match the standards progression.
I am very excited about the new centers I made to go with First Grade Wit and Wisdom Module 1! Each center is Common Core aligned and can be completed by students independently.
Read about each center below. Click on each picture to see it in my
TPT store
Narrative Writing Prompts:
The different task cards challenge students to think of a time they read a book that meets certain criteria- a book you read outside, a non-fiction book, etc. (Common Core Standard:
Adding Details
: Students read sentences that match the picture on the card. They then improve the sentence by adding adjectives, verbs, or other details. (Common Core Standards:
Punctuation Sort
: I love sort centers! They are very engaging! Students read sentences and decide if they need a period, question mark, or exclamation point. Students then write their own sentences for the
punctuation marks (Common Core Standards:
, and
Parts of Speech Task Cards
: Students mark up the sentences on the task cards using the given code- circle the nouns, underline verbs, and box adjectives. Students then write the words they found under the correct part of
speech on the recording sheet. (Common Core Standards:
L.K.1, L.1.1, L.2.1).
Vocabulary Clip Card Activity: Students use the context to determine what the underlined vocabulary word means. Students clip or mark the definition for the underlined word. They then pick a
vocabulary word and fill out the recording sheet- writing and illustrating a sentence with the vocabulary word. You can have this center for FREE!!! Just fill in the form down below!
Center organization can be challenging.
Read about how I try to keep pieces from getting mixed up or lost
Read about other ways
I supplement Wit and Wisdom in my classroom
2 Comments
It is almost Earth Day! A great day to teach students about caring for the planet, recycling, etc. Also, a great reminder for us as teachers to save paper and recycle that huge pile of graded papers
that has a way of piling up!
I created Earth Day Math Activities for kindergarten, first grade, and second grade! The worksheets are all Common Core aligned but can be used in any classroom. Each packet covers a variety of
skills including word problems, comparing numbers, graphing, etc.
Click on each grade level to see the Earth Day activities. Scroll down and enter your email address to get the Freebie!!!
1 Comment
I have used Eureka Math for the past few years and I like to teach math in small groups. I have students rotate between 3 activities: Eureka lesson with teacher, math centers,
ST Math
(click to read about how I use ST Math).
I love this system- I think you can really target what each group needs and push students to learn more. But, the transitions can be a little tricky or time consuming.
It seems like the second half of Eureka Math Kindergarten is all about number bonds, addition, and subtraction. So I created a warm up routine that was no-prep and practiced these skills!
Every time a group transitioned to my rotation they immediately made a number bond about the students in their group. This gave me a few minutes to help the other groups get settled or to reset my
This was a great prompt because the groupings changed every day and the students could be creative!!!
Some common groupings in my class were:
• Boys and Girls
• T-shirt and Sweater
• Sitting up and Laying down
• Hair bows and No hair bows
• Shorts and Pants
Once students finished the number bond they would write an equation to match it. This was an awesome time to differentiate. Some students wrote the part-part-whole equation but others were able to
write the whole fact family!! This helped develop conceptual understanding of how parts and wholes go together in addition and subtraction equations.
This also led to many great teaching moments! As you can see in the above picture one of the equations is wrong: she wrote 3-7=10. We talked about how it was awesome that she went above and beyond
to write multiple equations (yay for risk taking!!) for her number bond and then solved each equation to find the one that was incorrect. She was able to explain that she should have started with
the whole number when subtracting and correct it herself! I love helping students to think through the concepts behind the mathematical principles and encouraging them to take risks and try tricky
Some days (depending on timing) we had each student share their number bond with the groups, sometimes they shared with a partner, and some days I just picked one or two students to share.
This was a very engaging warm-up, the students got very creative in their groupings (rides the bus home and is a pick-up)! It was also easy to differentiate based on the abilities of each math
group. Scroll down to get your freebie!
I have made a printable version of this math warm up! It can also be used as a challenge problem for early finishers!
You can use it as a paper and pencil activity or laminate the papers and use with dry erase markers so it can be used many times.
If you don't have math groups you could have students make number bonds of the people at their table, the members of their family, etc. Get creative!
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Collected Papers of John Milnor: VII. Dynamical Systems (1984–2012)search
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Collected Papers of John Milnor: VII. Dynamical Systems (1984–2012)
Hardcover ISBN: 978-1-4704-0937-1
Product Code: CWORKS/19.7
List Price: $175.00
MAA Member Price: $157.50
AMS Member Price: $140.00
eBook ISBN: 978-1-4704-1715-4
Product Code: CWORKS/19.7.E
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Hardcover ISBN: 978-1-4704-0937-1
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Collected Papers of John Milnor: VII. Dynamical Systems (1984–2012)
Hardcover ISBN: 978-1-4704-0937-1
Product Code: CWORKS/19.7
List Price: $175.00
MAA Member Price: $157.50
AMS Member Price: $140.00
eBook ISBN: 978-1-4704-1715-4
Product Code: CWORKS/19.7.E
List Price: $159.00
MAA Member Price: $143.10
AMS Member Price: $127.20
Hardcover ISBN: 978-1-4704-0937-1
eBook ISBN: 978-1-4704-1715-4
Product Code: CWORKS/19.7.B
List Price: $334.00 $254.50
MAA Member Price: $300.60 $229.05
AMS Member Price: $267.20 $203.60
• Collected Works
Volume: 19; 2014; 592 pp
MSC: Primary 37; 32; 30; 14
This volume is the seventh in the series “Collected Papers of John Milnor.” Together with the preceding Volume VI, it contains all of Milnor's papers in dynamics, through the year 2012. Most of
the papers are in holomorphic dynamics; however, there are two in real dynamics and one on cellular automata. Two of the papers are published here for the first time.
The papers in this volume provide important and fundamental material in real and complex dynamical systems. Many have become classics, and have inspired further research in the field. Some of the
questions addressed here continue to be important in current research. In some cases, there have been minor corrections or clarifications, as well as references to more recent work which answers
questions raised by the author. The volume also includes an index to facilitate searching the book for specific topics.
Graduate students and research mathematicians interested in complex dynamical systems.
This item is also available as part of a set:
□ Cover
□ Frontispiece
□ Title page
□ Copyright page
□ Contents
□ Preface
□ Dedicated to Adrien Douady (1935-2006)
□ Acknowledgments
□ Addendum: Updated References
□ Introduction
□ Notes on surjective cellular automaton-maps (Unpublished manuscript of 1984)
□ Tsujii’s monotonicity proof for real quadratic maps (Unpublished manuscript of 2000)
□ Local connectivity of julia sets: Expository lectures
□ On rational maps with two critical points
□ Periodic orbits, external rays and the Mandelbrot set: An expository account
□ Pasting together Julia sets–A worked out example of mating
□ On Lattés maps
□ Elliptic curves as attractors in ℙ², Part 1: Dynamics
□ Schwarzian derivatives and cylinder maps
□ Cubic polynomial maps with periodic critical orbit, part I
□ Cubic polynomial maps with periodic critical orbit, part II: Escape regions
□ Errata to “Cubic polynomial maps with periodic critical orbit, Part II: Escape regions”
□ Hyperbolic components with an appendix by A. Poirier
□ Index
□ Back Cover
□ This is a remarkable collection of contributions, which has been carefully and thoroughly edited and updated, with much mathematics of great importance. I hope it finds its home not in the
dustier parts of libraries with bound volumes of collected works that simply comprise collated papers but in the shelves and on the desks of active researchers interested in dynamics, where
it belongs.
Thomas B. Ward, Zentralblatt MATH
• Book Details
• Table of Contents
• Additional Material
• Reviews
• Requests
Volume: 19; 2014; 592 pp
MSC: Primary 37; 32; 30; 14
This volume is the seventh in the series “Collected Papers of John Milnor.” Together with the preceding Volume VI, it contains all of Milnor's papers in dynamics, through the year 2012. Most of the
papers are in holomorphic dynamics; however, there are two in real dynamics and one on cellular automata. Two of the papers are published here for the first time.
The papers in this volume provide important and fundamental material in real and complex dynamical systems. Many have become classics, and have inspired further research in the field. Some of the
questions addressed here continue to be important in current research. In some cases, there have been minor corrections or clarifications, as well as references to more recent work which answers
questions raised by the author. The volume also includes an index to facilitate searching the book for specific topics.
Graduate students and research mathematicians interested in complex dynamical systems.
This item is also available as part of a set:
• Cover
• Frontispiece
• Title page
• Copyright page
• Contents
• Preface
• Dedicated to Adrien Douady (1935-2006)
• Acknowledgments
• Addendum: Updated References
• Introduction
• Notes on surjective cellular automaton-maps (Unpublished manuscript of 1984)
• Tsujii’s monotonicity proof for real quadratic maps (Unpublished manuscript of 2000)
• Local connectivity of julia sets: Expository lectures
• On rational maps with two critical points
• Periodic orbits, external rays and the Mandelbrot set: An expository account
• Pasting together Julia sets–A worked out example of mating
• On Lattés maps
• Elliptic curves as attractors in ℙ², Part 1: Dynamics
• Schwarzian derivatives and cylinder maps
• Cubic polynomial maps with periodic critical orbit, part I
• Cubic polynomial maps with periodic critical orbit, part II: Escape regions
• Errata to “Cubic polynomial maps with periodic critical orbit, Part II: Escape regions”
• Hyperbolic components with an appendix by A. Poirier
• Index
• Back Cover
• This is a remarkable collection of contributions, which has been carefully and thoroughly edited and updated, with much mathematics of great importance. I hope it finds its home not in the
dustier parts of libraries with bound volumes of collected works that simply comprise collated papers but in the shelves and on the desks of active researchers interested in dynamics, where it
Thomas B. Ward, Zentralblatt MATH
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Lesson 11
Percentages and Double Number Lines
11.1: Fundraising Goal (5 minutes)
This warm-up is the first time students are asked to find \(A\%\) of \(B\) when \(B\) is not 100 or 1.
Students may approach the problem in a few different ways, with or without filling in values on the provided double number line diagram. For example, they may understand from the fundraising context
of the problem that \$40 is 100% because it is the amount of the goal. From there, they may simply find half of \$40 for the 50% value and add that value to \$40 to find the 150% value. Other
students may use equivalent ratio reasoning to calculate the value at 50% and 150%. As students work, notice the different strategies used and any misconceptions so they can be addressed during
Remind students that in the previous lesson, we found percentages of 100 and of 1 using double number lines. Explain that in this lesson we will find percentages of other numbers. Give students 2
minutes of quiet work time, and follow with a whole-class discussion. Encourage students to create a double number line to help them answer the questions if needed.
Student Facing
Each of three friends—Lin, Jada, and Andre—had the goal of raising $40. How much money did each person raise? Be prepared to explain your reasoning.
1. Lin raised 100% of her goal.
2. Jada raised 50% of her goal.
3. Andre raised 150% of his goal.
Anticipated Misconceptions
Students may be surprised by a percentage greater than 100. If they are puzzled by this, explain that Andre raised more money than the goal.
Activity Synthesis
Consider displaying this double number line for all to see.
Invite a few students to share their solving strategies. One way to highlight the different techniques students use is to invite several students to explain how they calculated the amount of money
raised by Andre. We can find 150% of 40 in several ways. For example, we can add the values of 50% of 40 and 100% of 40. We can also reason that since \(100 \boldcdot (1.5) = 150\) then \(40 \
boldcdot (1.5) = 60\), which means \$60 is 150% of \$40.
If not uncovered in students’ explanations, explain that we are finding percentages of \$40, since this number—not 100 cents or 1 dollar—is the fundraising goal for the three friends. Since 100% of a
goal of \$40 is \$40, the 100% and \$40 are lined up on the double number line.
Students who relied on the visual similarity between, for example, \$0.25 is 25% of 1 dollar in the previous lesson find this strategy unworkable here (as \$50 is not 50% of \$40). To encourage
students to use their understanding of equivalent ratios to reason about percent problems, ask the class to explain—either when the above misconception arises or as a closing question—why \$50 is not
50% of \$40, but 50% of 100 cents is 50 cents.
11.2: Three-Day Biking Trip (15 minutes)
In this activity, students find percentages of a value in a non-monetary context. They begin by assigning a value to 100% and reasoning about other percentages.
The double number line is provided to communicate that we can use all our skills for reasoning about equivalent ratios to reason about percentages. Providing this representation makes it more likely
that students will use it, but it would be perfectly acceptable for them to use other strategies.
Monitor for students using these strategies:
• Use a double number line and reason that since 25% is \(\frac14\) of 100%, 25% of 8 is \(8\boldcdot \frac14 = 2\). They may then skip count by the value for the first tick mark to find the values
for other tick marks.
• Reason about 125% of the distance as 100% of the distance plus 25% of the distance, and add 8 and \(\frac14\) of 8.
• Reason about 75% of 8 directly by multiplying 8 by \(\frac34\) and 125% of 8 by multiply 8 by \(\frac54\).
Arrange students in groups of 3–4. Provide tools for making a large visual display. Give students 2–3 minutes of quiet think time. Encourage students to use the double number line to help them answer
the questions if needed. Afterwards, ask them to discuss their responses to the last two questions with their group and to create a visual display of one chosen strategy to be shared with the class.
Representation: Internalize Comprehension. Activate or supply background knowledge. Allow students to use calculators to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing
Student Facing
Elena biked 8 miles on Saturday. Use the double number line to answer the questions. Be prepared to explain your reasoning.
1. What is 100% of her Saturday distance?
2. On Sunday, she biked 75% of her Saturday distance. How far was that?
3. On Monday, she biked 125% of her Saturday distance. How far was that?
Activity Synthesis
For each unique strategy, select one group to share their displays and explain their thinking. Sequence the presentations in the order they are presented in the Activity Narrative. If no students
mention one or more of these strategies, bring them up. For example, if no one thought of 125% of the distance hiked as 100% plus 25%, present that approach and ask students to explain why it works.
Speaking, Listening: MLR7 Compare and Connect. As students prepare a visual display of how they made sense of the last question, look for groups with different methods for finding 125% of 8 miles.
Some groups may reason that 25% of the distance is 2 miles and 100% of the distance is 8 miles, so 125% of the distance is the sum of 2 and 8. Others may reason that the product of 100 and 1.25 is
125. Since 125 is 125% of 100, then 125% of 8 miles is the product of 8 and 1.25. As students investigate each other’s work, encourage students to compare other methods for finding 125% of 8 to their
own. Which approach is easier to understand with the double number line? This will promote students’ use of mathematical language as they make sense of the connections between the various methods for
finding 125% of a quantity.
Design Principle(s): Cultivate conversation; Maximize meta-awareness
11.3: Puppies Grow Up (15 minutes)
Previously students were asked to find various percentages given 100% of a quantity. Here they are asked to find 100% of quantities given other percentages. The context does not explicitly state that
the values being sought (the adult weights of two puppies) are the values for 100%, so students will first need to make that connection.
Double number lines continue to be provided as a reasoning tool, but students may use a table of equivalent ratios or other methods. Those who use double number lines are likely to find them
effective for the first question (find 100% of a quantity given 20%) but less straightforward for the second question (find 100% of a quantity given 30%). Since 100 is not a multiple of 30, students
may use strategies such as subdividing the double number line into intervals of 10% and scaling up from there to find the value of 100%.
Arrange students in groups of 2. Give students 3–4 minutes of quiet think time and then time to share their responses with their partner. Encourage students to refer to diagrams in previous
activities if they are not sure how to get started. Students may need help interpreting the question to understand that 100% corresponds to the puppy’s adult weight.
Writing, Speaking, Listening: MLR1 Stronger and Clearer Each Time. After students have had the opportunity to determine the adult weight of Jada’s puppy, ask students to write a brief explanation of
their process. Ask each student to meet with 2–3 other partners in a row for feedback. Provide students with prompts for feedback that will help them strengthen their ideas and clarify their language
(e.g., “Can you explain how…?”, “You should expand on...”, etc.). Students can borrow ideas and language from each partner to refine and clarify their original explanation. This will help students
refine their explanation and learn about other methods for finding the adult weight of a puppy.
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness
Student Facing
1. Jada has a new puppy that weighs 9 pounds. The vet says that the puppy is now at about 20% of its adult weight. What will be the adult weight of the puppy?
2. Andre also has a puppy that weighs 9 pounds. The vet says that this puppy is now at about 30% of its adult weight. What will be the adult weight of Andre’s puppy?
3. What is the same about Jada and Andre’s puppies? What is different?
Student Facing
Are you ready for more?
A loaf of bread costs $2.50 today. The same size loaf cost 20 cents in 1955.
1. What percentage of today’s price did someone in 1955 pay for bread?
2. A job pays $10.00 an hour today. If the same percentage applies to income as well, how much would that job have paid in 1955?
Anticipated Misconceptions
Students may stop before they reach 100% or go further than 100%. If this happens, explain that in this situation, the adult weight is at exactly 100%. Students may not use equal-sized increments
between the tick marks they draw and label.
Activity Synthesis
Invite previously identified students to share their work. Start with someone who solved the first question using a double number line as follows, and follow with increasingly efficient strategies.
Keep the number line displayed for all to see and to refer to throughout discussion.
If no students reasoned with a table, display this abbreviated table, or illustrate one student’s approach and organize the steps in a table.
Follow a similar flow when discussing strategies for solving the second problem: start with a double number line and, if not mentioned by students, discuss how a table such as this one can be an
efficient tool.
Representation: Internalize Comprehension. Use color and annotations to illustrate connections between representations. For example, on a display, illustrate one student’s approach on both a double
number line and a table. Support connections by highlighting how each step appears in each representation.
Supports accessibility for: Visual-spatial processing; Conceptual processing
Lesson Synthesis
If you know the percentage and 100%, then you can find the percentage with a double number line by putting the value assigned to 100% opposite the tick mark labeled 100%. For example, if we want to
find some percentage of 50 pounds, we can label 100% and 50 like this:
Display the table for all to see. Questions for discussion:
• What situation might this double number line represent?
• This says that 100% of 50 is 50. Where can we place some other percentages of 50?
• (If no one mentions a percentage greater than 100%) What about 110% of 50? Where would we place it? How would it be labeled?
11.4: Cool-down - A Medium Bottle of Juice (5 minutes)
Student Facing
We can use a double number line to solve problems about percentages. For example, what is 30% of 50 pounds? We can draw a double number line like this:
We divide the distance between 0% and 100% and that between 0 and 50 pounds into ten equal parts. We label the tick marks on the top line by counting by 5s (\(50 \div 10 = 5\)) and on the bottom line
counting by 10% (\(100 \div 10 =10\)). We can then see that 30% of 50 pounds is 15 pounds.
We can also use a table to solve this problem.
Suppose we know that 140% of an amount is \$28. What is 100% of that amount? Let’s use a double number line to find out.
We divide the distance between 0% and 140% and that between \$0 and \$28 into fourteen equal intervals. We label the tick marks on the top line by counting by 2s and on the bottom line counting by
10%. We would then see that 100% is \$20.
Or we can use a table as shown. | {"url":"https://im.kendallhunt.com/MS/teachers/1/3/11/index.html","timestamp":"2024-11-03T22:25:11Z","content_type":"text/html","content_length":"123364","record_id":"<urn:uuid:dc34540b-fc98-4f92-b910-517d92ba55eb>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00877.warc.gz"} |
Mastering the Art of Finding Acceleration with Two Velocities: A Comprehensive Guide
Acceleration is a fundamental concept in physics that describes the rate of change in an object’s velocity over time. To determine the acceleration of an object, you can use the relationship between
the initial and final velocities, as well as the time interval during which the change in velocity occurred. In this comprehensive guide, we will delve into the intricacies of calculating
acceleration using two velocities, providing you with a thorough understanding of the underlying principles and practical applications.
Understanding the Acceleration Formula
The primary formula used to calculate acceleration with two velocities is:
a = Δv / Δt
– a is the acceleration (in units of m/s^2)
– Δv is the change in velocity (in units of m/s)
– Δt is the change in time (in units of s)
This formula represents the average acceleration over a given time interval, calculated by dividing the change in velocity by the change in time. It’s important to ensure that the units of velocity
and time are consistent when using this formula.
Calculating Acceleration Using the Δv/Δt Formula
Let’s consider a practical example to illustrate the application of the Δv/Δt formula:
Suppose an object has an initial velocity of 5 m/s and a final velocity of 15 m/s, and the time interval between these two velocities is 10 seconds. To calculate the acceleration, we can use the
following steps:
1. Determine the change in velocity:
Δv = v_final - v_initial
Δv = 15 m/s - 5 m/s = 10 m/s
2. Determine the change in time:
Δt = t_final - t_initial
Δt = 10 s - 0 s = 10 s
3. Calculate the acceleration using the Δv/Δt formula:
a = Δv / Δt
a = 10 m/s / 10 s = 1 m/s^2
Therefore, the object is accelerating at a rate of 1 m/s^2.
Alternative Acceleration Formula: [v(f) – v(i)] / [t(f) – t(i)]
Another way to calculate acceleration with two velocities is by using the formula:
a = [v(f) - v(i)] / [t(f) - t(i)]
– v(f) is the final velocity
– v(i) is the initial velocity
– t(f) is the final time
– t(i) is the initial time
This formula calculates the acceleration by finding the difference between the final and initial velocities and dividing it by the difference between the final and initial times.
Let’s apply this formula to the same example:
1. Determine the final and initial velocities:
v(f) = 15 m/s
v(i) = 5 m/s
2. Determine the final and initial times:
t(f) = 10 s
t(i) = 0 s
3. Calculate the acceleration using the formula:
a = [v(f) - v(i)] / [t(f) - t(i)]
a = [15 m/s - 5 m/s] / [10 s - 0 s]
a = 10 m/s / 10 s = 1 m/s^2
The result is the same as the previous example, confirming that the object is accelerating at a rate of 1 m/s^2.
Considering the Direction of Acceleration
It’s important to note that the acceleration formulas discussed so far do not specify the direction of the acceleration. Acceleration is a vector quantity, meaning it has both magnitude and
direction. If the direction of acceleration is relevant to your analysis, you must consider it separately.
For example, if an object is moving in the positive x-direction and its velocity increases, the acceleration would be in the positive x-direction. Conversely, if the velocity decreases, the
acceleration would be in the negative x-direction.
To incorporate the direction of acceleration, you can use the appropriate sign convention (positive or negative) when calculating the change in velocity (Δv) or the difference between the final and
initial velocities [v(f) – v(i)].
Practical Applications and Examples
Calculating acceleration with two velocities has numerous practical applications in various fields, including:
1. Kinematics: Analyzing the motion of objects, such as the acceleration of a car during a race or the acceleration of a falling object due to gravity.
2. Dynamics: Studying the forces acting on an object and their relationship to the object’s acceleration, as in the case of Newton’s second law of motion.
3. Engineering: Designing and optimizing the performance of mechanical systems, such as the acceleration of a rocket or the braking system of a vehicle.
4. Sports Science: Evaluating the performance of athletes, such as the acceleration of a sprinter or the deceleration of a basketball player during a jump shot.
Here’s an example problem to illustrate the practical application of the acceleration formulas:
Problem: A car accelerates from a stop (0 m/s) to a speed of 20 m/s in 5 seconds. Calculate the acceleration of the car.
1. Determine the initial and final velocities:
v_initial = 0 m/s
v_final = 20 m/s
1. Determine the change in time:
Δt = t_final - t_initial
Δt = 5 s - 0 s = 5 s
2. Calculate the acceleration using the Δv/Δt formula:
a = Δv / Δt
a = (20 m/s - 0 m/s) / 5 s
a = 4 m/s^2
Therefore, the car is accelerating at a rate of 4 m/s^2.
Mastering the art of finding acceleration with two velocities is a crucial skill in physics and various engineering disciplines. By understanding the underlying principles and applying the
appropriate formulas, you can accurately determine the acceleration of an object and gain valuable insights into its motion and the forces acting upon it.
Remember, the key to success in this topic is to practice solving a variety of problems, familiarize yourself with the different formulas, and develop a strong conceptual understanding of the
relationship between velocity, time, and acceleration.
The techiescience.com Core SME Team is a group of experienced subject matter experts from diverse scientific and technical fields including Physics, Chemistry, Technology,Electronics & Electrical
Engineering, Automotive, Mechanical Engineering. Our team collaborates to create high-quality, well-researched articles on a wide range of science and technology topics for the techiescience.com
All Our Senior SME are having more than 7 Years of experience in the respective fields . They are either Working Industry Professionals or assocaited With different Universities. Refer Our Authors
Page to get to know About our Core SMEs. | {"url":"https://techiescience.com/how-to-find-acceleration-with-two-velocities/","timestamp":"2024-11-04T11:10:52Z","content_type":"text/html","content_length":"103811","record_id":"<urn:uuid:7c510d79-673e-46ea-b7e7-8ed5b6489016>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00500.warc.gz"} |
Spring Surprise: Projectile Motion made Fun, Mathematical and Real!
Roberta Tevlin, Editor, OAPT Newsletter
Edited by Tim Langford
Projectile motion often involves a lot of mathematical problem-solving that is overly simplified and highly contrived. Football players do not stop to calculate the range before making a pass.
Invading armies might want to make calculations for siege weapons, but these tend to be too complicated (trebuchets) or involve too much energy loss (catapults). Guess and check, was probably the
preferred technique. Fortunately there is a cheap and reliable projectile launcher that you can use to show that physics works. Your students will be able to use it to hit a target on their first
shot by using calculations for conservation of energy and projectile motion.
How to Make a Projectile Launcher of Minimum Destruction
Find one of those long springs for demonstrating wave motion, preferably one that has been abused and is no longer uniform. Cut this into short springs ranging from 6 cm to 14 cm. Use pliers to bend
a loop at one end by 90 o to form a hook that a finger can grab. Then bend a few loops at the other end by 180o so it can be hooked onto a nail at the end of a board.
Figure 1: A short, medium and long projectile.
Hammer a nail onto the end of a piece of two-by-four that is about a foot long. The nail should be a finishing nail or else you have to remove the head from a regular nail. (This prevents the spring
from catching on the nail when launched.) 
Figure 2: The fully constructed spring and launcher.
Aim the board away from people, stretch the spring, let go, and watch it fly! The spring is a great projectile for two reasons:
• It is very dense, so that air resistance has little effect on it and
• It is very elastic, so that most of the stored elastic energy is converted into kinetic energy.
However, nothing is perfect, and part of the challenge for the students is to adjust their calculated pullback distance to account for the energy losses. The students find that multiplying their
calculated values by 1.1 seems to work well.
I have some inflatable minions which make a great target. The group is sufficiently large — about 40 cm in diameter — that a good number of the shots are successful. The minions also make a
satisfying thwack! when they are hit. A cardboard box will also work as a target, but it won’t be as amusing.
Figure 3: The target.
Building the Challenge Slowly
I have developed a set of six work sheets (with answers!) that can be used for six different one-hour lessons. You can do all six or choose just one or two. (The links to these files are provided at
the end of this article.) The first worksheet establishes Hooke’s Law. The second one turns the spring’s potential energy into gravitational energy. I challenge the students to calculate the pullback
distance that will shoot the spring to get as close to the ceiling as possible without hitting. The third class has the students calculate how to fire the springs horizontally to hit the minions.
This is the simplest version of projectile motion. I stack tables and stools to get the different launch heights.
Figure 4: A simple way to vary the height.
The fourth sheet has students fire the spring from the ground at 45°. The minions are about the same height as the launcher, so you can assume that the vertical displacement is zero. I set the launch
angle by mounting the board on an adjustable inclined plane. Arbor Scientific has
something similar
but it costs $69 US. You could easily put a hinge between two boards and attach a protractor.
Figure 5: Set-up for an angled launch.
The fifth and a sixth sheets involve more complicated calculations. The spring is shot at angles that are not at 45° and with launch heights that are well-above the ground.
Building the Understanding from Concrete to Abstract
These contests involve lots of math. There are kinematic formulas for constant velocity and constant acceleration and trigonometric equations for components of velocity. Conservation of energy
involves equations for kinetic, gravitational and spring potential energy. The students can easily get lost in the abstract math. Therefore the lessons have the students start by examining and
discussing the problem without equations. They are asked to sketch the path of the spring, to draw the components of the velocity and to discuss their answers.
After they have a good feel for the concepts, they do a practice calculation where everyone uses the same numbers. This provides a quick check on their math skills. Next, each group does a practice
contest where each group makes the necessary measurements and calculations for their particular spring. Then they do a test shot using the pullback distance that they calculated. If they miss the
target by a large amount, they have an opportunity to check what they did wrong in their calculations. If they are short by a small amount, they have a good idea of how they must adjust their
pullback distance to compensate for the variables that have not been included in the calculation, i.e. air resistance, friction, rotation of the spring etc.
The groups that measure, calculate, communicate and launch carefully are able to hit the target on their first shot almost every time. This provides strong motivation to the rest of the class to
improve their work.
Where does Spring Surprise fit into the Curriculum?
These worksheets will fit most easily in 12U physics after Unit B (Dynamics) and C (Energy and Momentum) have been covered or in 12C physics after Unit B (Motion), C (Mechanical Systems) and E
(Energy Transformations). The activities can function as a hands-on review and summary of projectile motion and conservation of energy.
The springs could also be used earlier in the grade 12 courses or even in a grade 11 course, if you don’t use spring potential energy. Here are a couple of ideas:
• Projectile Motion: Launch the spring horizontally. Measure the height and horizontal distance travelled and calculate the initial speed. Fire the spring at an angle and measure the range.
Calculate the initial speed. These two speeds should be the same if you use the same spring and pullback distance.
• Conservation of Kinetic and Potential Energy: Fire the spring straight up and measure the maximum height reached. Calculate the initial speed. Once again, this speed should match the speeds
calculated using projectile motion calculations.
In each example above, you can also check the calculated speed by filming the motion from the side and using a slow-motion camera to measure the distance travelled between two frames.
Credit: I would like to thank Jonathan Orling, who showed me and the other teachers at Einstein Plus*, how you can use these springs as projectiles. He says that he learnt about them from Mr. Seppela
in his high school physics class in 1985!
*Einstein Plus is a fantastic one-week summer camp at the Perimeter institute of Theoretical Physics. You can find out more about it here:
https://www.perimeterinstitute.ca/outreach/teachers/programs-and-opportunities/einsteinplusClick here
to download worksheets for the six lessons, with answers. | {"url":"http://newsletter.oapt.ca/files/spring-surprise.html","timestamp":"2024-11-14T00:02:28Z","content_type":"text/html","content_length":"29616","record_id":"<urn:uuid:82f30aaa-0a19-4959-b4b5-3d1a337b13eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00301.warc.gz"} |
Axes Objects -- Defining Coordinate Systems for Graphs :: Axes Properties (Graphics)
Axes Objects -- Defining Coordinate Systems for Graphs
MATLAB uses graphics objects to create visual representations of data. For example, a two-dimensional array of numbers can be represented as lines connecting the data points defined by each column,
as a surface constructed from a grid of rectangles whose vertices are defined by each element of the array, as a contour graph where equal values in the array are connected by lines, and so on.
In all these cases, there must be a frame of reference that defines where to place each data point on the graph. This frame of reference is the coordinate system defined by the axes. Axes orient and
scale graphs to produce the view of the data that you see on screen.
MATLAB creates axes to define the coordinate system of each graph. Axes are always contained within a figure object and themselves contain the graphics objects that make up the graph.
Axes properties control many aspects of how MATLAB displays graphical information. This section discusses some of the features that are implemented through axes properties and provides examples of
how to uses these features.
The table in the axes reference page listing all axes properties provides an overview of the characteristics affected by these properties.
Axes Properties Labeling and Appearance Properties
© 1994-2005 The MathWorks, Inc. | {"url":"http://matlab.izmiran.ru/help/techdoc/creating_plots/axes_pr2.html","timestamp":"2024-11-11T09:50:47Z","content_type":"text/html","content_length":"3330","record_id":"<urn:uuid:940465f3-5240-4ead-a004-b502e56b32b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00675.warc.gz"} |
If the force acting on a body is 40 n and the acceleration is 4-Turito
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If the force acting on a body is 40 N and the acceleration is 4 m/ s^2, what can be the mass of the body?
A. 160 Kg
B. 10 Kg
C. 0.1 Kg
D. 1600 Kg
F= ma
The correct answer is: 10 Kg
• F = ma
• Given that, F= 40N, a=4m/s^2
• So, m = F/a = 40/4
• m= 10 Kg
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Direct Fourier Transform
Direct Fourier Transform#
Functions used to compute the discretised direct Fourier transform (DFT) for an ideal interferometer. The DFT for an ideal interferometer is defined as
\[V(u,v,w) = \int B(l,m) e^{-2\pi i \left( ul + vm + w(n-1)\right)} \frac{dl dm}{n}\]
where \(u,v,w\) are data space coordinates and where visibilities \(V\) have been obtained. The \(l,m,n\) are signal space coordinates at which we wish to reconstruct the signal \(B\). Note that the
signal correspondes to the brightness matrix and not the Stokes parameters. We adopt the convention where we absorb the fixed coordinate \(n\) in the denominator into the image. Note that the data
space coordinates have an implicit dependence on frequency and time and that the image has an implicit dependence on frequency. The discretised form of the DFT can be written as
\[V(u,v,w) = \sum_s e^{-2 \pi i (u l_s + v m_s + w (n_s - 1))} \cdot B_s\]
where \(s\) labels the source (or pixel) location. If only a single correlation is present \(B = I\), this can be cast into a matrix equation as follows
\[V = R I\]
where \(R\) is the operator that maps an image to visibility space. This mapping is implemented by the im_to_vis() function. If multiple correlations are present then each one is mapped to its
corresponding visibility. An imaging algorithm also requires the adjoint denoted \(R^\dagger\) which is simply the complex conjugate transpose of \(R\). The dirty image is obtained by applying the
adjoint operator to the visibilities
\[I^D = R^\dagger V\]
This is implemented by the vis_to_im() function. Note that an imaging algorithm using these operators will actually reconstruct \(\frac{I}{n}\) but that it is trivial to obtain \(I\) since \(n\) is
known at each location in the image.
im_to_vis(image, uvw, lm, frequency[, ...]) Computes the discrete image to visibility mapping of an ideal interferometer:
vis_to_im(vis, uvw, lm, frequency, flags[, ...]) Computes visibility to image mapping of an ideal interferometer:
africanus.dft.im_to_vis(image, uvw, lm, frequency, convention='fourier', dtype=None)[source]#
Computes the discrete image to visibility mapping of an ideal interferometer:
\[{\Large \sum_s e^{-2 \pi i (u l_s + v m_s + w (n_s - 1))} \cdot I_s }\]
image of shape (source, chan, corr) The brighness matrix in each pixel (flatten 2D array per channel and corr). Note not Stokes terms
uvw coordinates of shape (row, 3) with u, v and w components in the last dimension.
lm coordinates of shape (source, 2) with l and m components in the last dimension.
frequencies of shape (chan,)
convention{‘fourier’, ‘casa’}
Uses the \(e^{-2 \pi \mathit{i}}\) sign convention if fourier and \(e^{2 \pi \mathit{i}}\) if casa.
dtypenp.dtype, optional
Datatype of result. Should be either np.complex64 or np.complex128. If None, numpy.result_type() is used to infer the data type from the inputs.
complex of shape (row, chan, corr)
africanus.dft.vis_to_im(vis, uvw, lm, frequency, flags, convention='fourier', dtype=None)[source]#
Computes visibility to image mapping of an ideal interferometer:
\[{\Large \sum_k e^{ 2 \pi i (u_k l + v_k m + w_k (n - 1))} \cdot V_k}\]
visibilities of shape (row, chan, corr) Visibilities corresponding to brightness terms. Note the dirty images produced do not necessarily correspond to Stokes terms and need to be
uvw coordinates of shape (row, 3) with u, v and w components in the last dimension.
lm coordinates of shape (source, 2) with l and m components in the last dimension.
frequencies of shape (chan,)
Boolean array of shape (row, chan, corr) Note that if one correlation is flagged we discard all of them otherwise we end up irretrievably mixing Stokes terms.
convention{‘fourier’, ‘casa’}
Uses the \(e^{-2 \pi \mathit{i}}\) sign convention if fourier and \(e^{2 \pi \mathit{i}}\) if casa.
dtypenp.dtype, optional
Datatype of result. Should be either np.float32 or np.float64. If None, numpy.result_type() is used to infer the data type from the inputs.
float of shape (source, chan, corr)
im_to_vis(image, uvw, lm, frequency[, ...]) Computes the discrete image to visibility mapping of an ideal interferometer:
vis_to_im(vis, uvw, lm, frequency, flags[, ...]) Computes visibility to image mapping of an ideal interferometer:
africanus.dft.dask.im_to_vis(image, uvw, lm, frequency, convention='fourier', dtype=numpy.complex128)[source]#
Computes the discrete image to visibility mapping of an ideal interferometer:
\[{\Large \sum_s e^{-2 \pi i (u l_s + v m_s + w (n_s - 1))} \cdot I_s }\]
image of shape (source, chan, corr) The brighness matrix in each pixel (flatten 2D array per channel and corr). Note not Stokes terms
uvw coordinates of shape (row, 3) with u, v and w components in the last dimension.
lm coordinates of shape (source, 2) with l and m components in the last dimension.
frequencies of shape (chan,)
convention{‘fourier’, ‘casa’}
Uses the \(e^{-2 \pi \mathit{i}}\) sign convention if fourier and \(e^{2 \pi \mathit{i}}\) if casa.
dtypenp.dtype, optional
Datatype of result. Should be either np.complex64 or np.complex128. If None, numpy.result_type() is used to infer the data type from the inputs.
complex of shape (row, chan, corr)
africanus.dft.dask.vis_to_im(vis, uvw, lm, frequency, flags, convention='fourier', dtype=numpy.float64)[source]#
Computes visibility to image mapping of an ideal interferometer:
\[{\Large \sum_k e^{ 2 \pi i (u_k l + v_k m + w_k (n - 1))} \cdot V_k}\]
visibilities of shape (row, chan, corr) Visibilities corresponding to brightness terms. Note the dirty images produced do not necessarily correspond to Stokes terms and need to be
uvw coordinates of shape (row, 3) with u, v and w components in the last dimension.
lm coordinates of shape (source, 2) with l and m components in the last dimension.
frequencies of shape (chan,)
Boolean array of shape (row, chan, corr) Note that if one correlation is flagged we discard all of them otherwise we end up irretrievably mixing Stokes terms.
convention{‘fourier’, ‘casa’}
Uses the \(e^{-2 \pi \mathit{i}}\) sign convention if fourier and \(e^{2 \pi \mathit{i}}\) if casa.
dtypenp.dtype, optional
Datatype of result. Should be either np.float32 or np.float64. If None, numpy.result_type() is used to infer the data type from the inputs.
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How do you find the integral of int sin^5(x)cos^8(x) dx? | HIX Tutor
How do you find the integral of #int sin^5(x)cos^8(x) dx#?
Answer 1
$- {\cos}^{9} \frac{x}{9} + \frac{2 {\cos}^{11} x}{11} - {\cos}^{13} \frac{x}{13} + C$
#I=int sin^5x cos^8xdx=intsinx sin^4x cos^8x dx#
#I=int sinx (sin^2x)^2 cos^8xdx=int (1-cos^2x)^2 cos^8x sinx dx#
#cosx=t => -sinxdx=dt => sinxdx=-dt#
#I=int (1-t^2)^2 t^8 (-dt) = -int (1-2t^2+t^4)t^8 dt#
#I=-int (t^8-2t^10+t^12) dt = -t^9/9+(2t^11)/11-t^13/13+C#
#I=-cos^9x/9 + (2cos^11x)/11 - cos^13x/13 + C#
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Answer 2
To find the integral of (\int \sin^5(x)\cos^8(x) , dx), you can use trigonometric identities and integration by parts. Here's how:
1. Start by using the identity (\sin^2(x) = 1 - \cos^2(x)) to rewrite (\sin^5(x)) as ((1 - \cos^2(x))^2\sin(x)).
2. Expand ((1 - \cos^2(x))^2) using the binomial theorem.
3. Now you have an integral involving powers of (\sin(x)) and (\cos(x)), which you can integrate using integration by parts.
4. Let (u = \cos(x)) and (dv = \sin^4(x)\cos^4(x) , dx), then differentiate (u) to find (du) and integrate (dv) to find (v).
5. Apply integration by parts formula: (\int u , dv = uv - \int v , du).
6. Substitute the values of (u), (du), (dv), and (v) into the integration by parts formula and evaluate the integral.
7. Repeat the process if necessary until you obtain an expression that can be easily integrated.
Following these steps will lead you to the solution of the integral (\int \sin^5(x)\cos^8(x) , dx).
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Huawei Certification
Guaranteed To Pass the Huawei Certifications Exam With Study Material
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Huawei Braindumps – Exact Questions and Answers Form of Syllabus
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Math Is Fun Forum
Re: Forum Features
Re: Forum Features
Registered: 2010-06-20
Posts: 10,610
Re: Forum Features
hi Billy,
Welcome to the forum!
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Registered: 2019-11-26
Posts: 1
Re: Forum Features
Hey! Its me Billy - Mr New Guy
From: Planet Mars
Registered: 2016-11-15
Posts: 821
Re: Forum Features
Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam
From: Planet Mars
Registered: 2016-11-15
Posts: 821
Re: Forum Features
Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam
Registered: 2014-05-21
Posts: 2,436
Re: Forum Features
MathsIsFun wrote:
Other Tags
It appears that these no longer work...
Registered: 2010-06-20
Posts: 10,610
Re: Forum Features
mathaholic wrote:
Does it work?
Use square brackets.
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Forum Features
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
I can do Latex commands for x^2 and x^3 by using x² and x³, but when I try :xsup4 or any exponent higher than 4, it comes out wrong. What is the proper LaTex code for say, x^5? I've searched here and
by Googling.
Last edited by Eric Vai (2020-07-09 08:09:53)
Registered: 2005-06-28
Posts: 48,283
Re: Forum Features
Please see ' LaTeX - A Crash Course' in 'Help me' topic.
x², x³ etc. are useful symbols in the forum (a shortcut).
In LaTeX, it is
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
Re: Forum Features
Thank you Ganesh.
Registered: 2020-09-20
Posts: 1
Re: Forum Features
Excellent share
Re: Forum Features
$/Huge $ Hi
I'm just testing it out.
Quote Of The Month:
'Whether it's the best of times or the worst of times, it's the only time we've got.' - Art Buchwald.
Re: Forum Features
Nope, does not work
Quote Of The Month:
'Whether it's the best of times or the worst of times, it's the only time we've got.' - Art Buchwald.
Re: Forum Features
Just testing this one out now.
Quote Of The Month:
'Whether it's the best of times or the worst of times, it's the only time we've got.' - Art Buchwald.
Re: Forum Features
Nope, again
Quote Of The Month:
'Whether it's the best of times or the worst of times, it's the only time we've got.' - Art Buchwald.
Re: Forum Features
Quote Of The Month:
'Whether it's the best of times or the worst of times, it's the only time we've got.' - Art Buchwald.
Re: Forum Features
YES YES YES Yes I can use this!
Quote Of The Month:
'Whether it's the best of times or the worst of times, it's the only time we've got.' - Art Buchwald.
Re: Forum Features
What is 1000000000 ÷ 11
Quote Of The Month:
'Whether it's the best of times or the worst of times, it's the only time we've got.' - Art Buchwald.
Re: Forum Features
My life=∞
Quote Of The Month:
'Whether it's the best of times or the worst of times, it's the only time we've got.' - Art Buchwald.
Re: Forum Features
Quote Of The Month:
'Whether it's the best of times or the worst of times, it's the only time we've got.' - Art Buchwald.
Registered: 2022-10-11
Posts: 22
Re: Forum Features
This post has been eaten by a biscuit
xp8_4b23021eaz57840c4d2 My jokes | {"url":"http://www.mathisfunforum.com/viewtopic.php?pid=428109","timestamp":"2024-11-04T11:02:20Z","content_type":"application/xhtml+xml","content_length":"33598","record_id":"<urn:uuid:3e9be7fa-735f-4db1-a5bb-fcb886121ced>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00081.warc.gz"} |
How to Use IF ELSE in Excel: 2024 Guide for Beginners - Solve Your Tech
How to Use IF ELSE in Excel: 2024 Guide for Beginners
Excel is a powerful tool that can handle complex calculations and data analysis. One of the essential functions in Excel is the IF ELSE function, which allows you to set conditions and return
specific values depending on whether the condition is met or not. In this article, we’ll walk you through how to use IF ELSE in Excel.
Step by Step Tutorial: How to Use IF ELSE in Excel
The IF ELSE function in Excel is a logical function that checks whether a condition is met, and if so, it returns one value; if not, it returns another. Here’s a step-by-step guide on how to use it.
Step 1: Open Excel and Select a Cell
Open your Excel spreadsheet and click on the cell where you want the IF ELSE function to appear.
Selecting a cell is the first step in any Excel function. Make sure you choose the correct cell where you want your result to be displayed.
Step 2: Type the IF Formula
Type the IF formula into the selected cell. The basic structure of the IF formula is =IF(logical_test, value_if_true, value_if_not_true).
Typing the IF formula correctly is crucial. Ensure you start with an equal sign, followed by IF, and then your conditions and values inside the parentheses.
Step 3: Define the Condition
Define the condition you want to test. This goes into the logical_test part of the formula.
Your condition can be any logical expression that Excel can evaluate as true or false. For example, you could use “=A1>10” to check if the value in cell A1 is greater than 10.
Step 4: Set the Value if True
Set the value that you want to be returned if the condition is met. This is the value_if_true part of the formula.
Deciding on the value to return if the condition is true is important. This can be a number, text, or even another Excel function.
Step 5: Set the Value if False
Set the value that you want to be returned if the condition is not met. This is the value_if_not_true part of the formula.
Just like the value if true, the value if false can be a number, text, or another function. It’s what Excel will return if the condition does not hold true.
After completing these steps, Excel will display the result based on the condition you set. If the condition is true, it will show the value_if_true; if it’s false, it will show the
Tips for Using IF ELSE in Excel
• Make sure your logical_test is set up correctly; otherwise, Excel won’t be able to evaluate it.
• Use quotation marks around text values in your IF formula.
• Nesting IF functions can help you evaluate multiple conditions, but try not to nest too many, as it can get complicated quickly.
• Remember that Excel functions are not case-sensitive.
• Debugging IF formulas can be tricky. Use the F9 key to evaluate parts of your formula to find errors.
Frequently Asked Questions
What is an IF ELSE statement?
An IF ELSE statement in Excel is a conditional statement that returns one value if a condition is true and a different value if it’s false.
Can IF ELSE statements be nested in Excel?
Yes, IF ELSE statements can be nested within each other to evaluate multiple conditions.
How many conditions can an IF ELSE statement have in Excel?
Excel supports up to 64 nested IF conditions.
Can IF ELSE statements in Excel handle text?
Yes, IF ELSE statements can return text values as well as numbers.
Can I use the IF ELSE function to perform calculations?
Absolutely, the IF ELSE function can return the results of calculations as well as static values.
1. Open Excel and select a cell.
2. Type the IF formula.
3. Define the condition.
4. Set the value if true.
5. Set the value if false.
Learning how to use IF ELSE in Excel can significantly enhance your data analysis and decision-making skills within your spreadsheets. The function might seem daunting at first, but with a little
practice, you’ll find it incredibly useful for handling multiple scenarios and conditions in your data. Always remember to double-check your formulas and make sure your conditions are logical and
correctly entered. Also, don’t shy away from using Excel’s help resources or online communities if you get stuck. Excel is a powerful tool, and mastering the IF ELSE function is a step towards
becoming an Excel wizard. Now go ahead, give it a try and watch your Excel skills grow!
Matthew Burleigh has been writing tech tutorials since 2008. His writing has appeared on dozens of different websites and been read over 50 million times.
After receiving his Bachelor’s and Master’s degrees in Computer Science he spent several years working in IT management for small businesses. However, he now works full time writing content online
and creating websites.
His main writing topics include iPhones, Microsoft Office, Google Apps, Android, and Photoshop, but he has also written about many other tech topics as well. | {"url":"https://www.solveyourtech.com/how-to-use-if-else-in-excel-2024-guide-for-beginners/","timestamp":"2024-11-02T17:52:00Z","content_type":"text/html","content_length":"245846","record_id":"<urn:uuid:3e29e002-aab0-4db2-bc03-8bbdc4636459>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00038.warc.gz"} |
Slider Interface - Dyoptr
Slider Interface
Learn how to use the slider interface to change a parameter over a range of values and visualize the impact to analyses and 3D view.
In this video, I will show you how we can use here the slider interface.
With the slider interface, we can define here different parameters and then we can insert these parameters here to the optical design editor instead of numerical values.
And here in the slider interface, we can define here for each parameter the minimum and the maximum value.
And here with the slider, we can then slide between this minimum and this maximum value.
So in this example, I’ve opened here this FISO interferometer, and we have inserted here the parameters here for this, measuring object, here for the position x, for the position y, and here for the
position z.
And when I’m sliding here, for example, the parameter x, so for the position x, we will see that here the interferogram at the bottom right corner will change by sliding because here the position x
of this measuring object here is changing. And for that also the interferogram is changing.
To open here the slider interface, we need to go here to this Windows icon. And here in the drop down menu, we just select here the slider interface.
And here inside the slider interface, we can add, parameters just by clicking here this button.
And here we can define the name of the parameter. So you can type an individual name, for example, parameter.
And here on the right side, we define the minimum and the maximum value. So for example, minus zero point zero two and plus zero point zero two.
And now we just insert here this parameter name into the optical design editor, for example, for the position x. So we just type in here the name of this, parameter. So in this case, parameter.
And if we will slide here now between this minimum and this maximum value, we will see that here this interferogram at the bottom right corner here is changing.
We can also add some math expressions here to the slider interface.
To do so, we just need to click here this button.
And here we can type in the name of our math expression. So for example, the name should be math.
And here on the right side, we can insert the math expression.
So we can calculate, for example, here with these parameters.
So we can, for example, insert here the square of x. So we just type in here the square of x, and now we can add here this name of the math expression in our optical design editor. So for example,
here the position x should be here this math expression with the name math.
And now we see that here also this interferogram has changed because here this value is now this expression here which we defined here in the slider interface.
Thanks for watching. | {"url":"https://dyoptr.com/slider-interface/","timestamp":"2024-11-14T17:40:17Z","content_type":"text/html","content_length":"126919","record_id":"<urn:uuid:905439cf-be4c-4b4f-a76d-9b1bc28d1245>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00442.warc.gz"} |
Roots of a Polynomial - MathsTips.com
A polynomial is defined as the sum of more than one or more algebraic terms where each term consists of several degrees of same variables and integer coefficient to that variables. x2−3×2−3,
5×4−3×2+x−45×4−3×2+x−4 are some examples of polynomials. The roots or also called as zeroes of a polynomial P(x) for the value of x for which polynomial P(x) is equal to 0. In other words, we can say
that polynomial P(x) will have the same value of x if x=r i.e. the value of the root of the polynomial that will satisfy the equation P(x) = 0. These are sometimes called solving the polynomial. The
degree of the polynomial is always equal to the number of roots of polynomial P(x).
In any polynomial, the root is that the value of the variable that satisfies the polynomial. Polynomial is an expression consisting of variables and coefficients of the form: $P_{n}(x)= a_{n}x^{n}+a_
{n-1}x^{n-1}+...+a_{0}$, where $a_{n}$ is not equal to zero and n refers to the degree of a polynomial and $a_{0}, a_{1},.... a_{n}$ are real coefficient. Thus, the degree of the polynomial gives
the idea of the number of roots of that polynomial. The roots may be different.
Example 1: Find the roots of the polynomial equation: $x^{2}+4x+4$
Solution: Given polynomial equation $x^{2}+4x+4$
By factoring the quadratic: $x^{2}+4x+4$ = $x^{2}+2x+2x+4 = 0$
x(x+2) + 2(x+2) = 0 therefore, (x+2)(x+2)=0
Set each factor equal to zero: x+2 =0 or x+2 = 0
So, x=-2 or x=-2 . Both the roots are same, i.e. -2.
Example 2: Find the roots of the polynomial equation: $2x^{3}+7x^{2}+3$
Solution: Given polynomial equation $2x^{3}+7x^{2}+3$
By factoring the quadratic: $2x^{3}+7x^{2}+3$ = $x(2x^{2}+6x+x+3) = 0$
x(2x(x + 3) + (x + 3)) = 0 therefore, x(2x + 1)(x + 3) = 0
Set each factor equal to 0: x = 0,2x+1 = 0,x+3 = 0
So, x = 0,x = $\frac{-1}{2}$,x = -3. Zeroes of polynomial are $\frac{-1}{2}$,-3,0.
Quadratic roots of Polynomial
Roots are the solution to the polynomial. The roots may be real or complex (imaginary), and they might not be distinct. A quadratic equation is $ax^{2}+bx+c=0$, where $aeq 0$ and $x = \frac{-b \pm \
If the coefficients a, b, c are real, it follows that: if $b^{2}-4ac > 0$ = the roots are real and unequal, if $b^{2}-4ac = 0$ = the roots are real and equal, if $b^{2}-4ac < 0$ the roots are
Example 1: Find the roots of the quadratic polynomial equation: $x^{2}-10x+26 = 0$
Solution: Given quadratic polynomial equation $x^{2}-10x+26 = 0$
So, a = 1,b = -10 and c = 26
By putting the formula as D = $b^{2}-4ac$ = 100 – 4 * 1 * 26 = 100 – 104 = -4 < 0
Therefore D < 0,so roots are complex or imaginary.
Now finding the value of x, using quadratic formula = $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ = $\frac{-(-10) \pm \sqrt{-4}}{2*1}$ = $\frac{10\pm 2\sqrt{-1}}{2}$ = $\frac{10\pm 2{i}}{2}$ = $5 \
pm i$
Therefore, the roots are 5 + i and 5 – i.
Example 2: Find the roots of the quadratic polynomial equation: $x^{4}-81$
Solution: Given polynomial equation: $x^{4}-81$ = $(x^{2})^{2}-9^{2}$ = $(x^{2}+9)(x^{2}-9)$ = $(x^{2}+9)(x^{2}-3^{2})$ = $(x^{2}+9)(x+3)(x-3)$.
So, roots are $x^{2}+9=0$ = $x^{2}=-9$
Therefore, x = $\sqrt{-9}$ = +3i,-3i (imaginary roots) and real roots are +3,-3.
Find the roots of polynomials by factoring:
1. $x^{2}+2x-15=0$
2. $x^{4}-13x^{2}=-36$
3. $x^{2}-14x+49$
4. $x^{2}-10x+25$
5. $x^{2}+2x-15$
Find the roots of the quadratic polynomial equation:
1. $x^{2}-5x+6$
2. $2x^{2}+7x-4$
3. $6y^{2}-13y+6$
4. $x^{2}+2x-8$
5. $x^{2}-4x+3$ | {"url":"https://www.mathstips.com/polynomial-roots/?utm_campaign=toc&utm_medium=prev&utm_source=polynomials","timestamp":"2024-11-10T21:53:02Z","content_type":"text/html","content_length":"71580","record_id":"<urn:uuid:03ada684-5b77-4491-b7ca-dd5161bd0d33>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00060.warc.gz"} |
Saturated Steam Enthalpy Calculator - GEGCalculators
Saturated Steam Enthalpy Calculator
The enthalpy of saturated steam depends on its temperature and pressure. At standard atmospheric pressure (1 bar or approximately 14.5 psi), saturated steam typically has an enthalpy of around 2,257
kJ/kg at 100°C. However, enthalpy values can vary with different pressures and temperatures.
Saturated Steam Enthalpy Calculator
Temperature (°C) Temperature (°F) Pressure (bar) Pressure (psi) Enthalpy (kJ/kg)
0°C 32°F 1.01325 14.7 2506
100°C 212°F 1.01325 14.7 2676
150°C 302°F 1.01325 14.7 2816
200°C 392°F 1.01325 14.7 2935
250°C 482°F 1.01325 14.7 3037
300°C 572°F 1.01325 14.7 3127
350°C 662°F 1.01325 14.7 3206
400°C 752°F 1.01325 14.7 3278
1. How do you calculate the Enthalpy of saturated steam? The enthalpy of saturated steam can be estimated using the formula: Enthalpy = Sensible Heat + Latent Heat
2. What is the Enthalpy of saturated steam table? The Enthalpy of saturated steam can be found in steam tables, which provide values at different temperatures and pressures.
3. What is the Enthalpy of saturated steam at 100°C? The enthalpy of saturated steam at 100°C is approximately 2,257 kJ/kg.
4. How do you calculate steam saturation temperature? The saturation temperature of steam can be calculated using the steam table or the equation: T_sat = T_boiling point at given pressure.
5. What is the enthalpy of saturated steam at 1 bar? The enthalpy of saturated steam at 1 bar (approximately 14.5 psi) is approximately 2,257 kJ/kg.
6. What is saturated steam in thermodynamics? Saturated steam in thermodynamics refers to steam that is in equilibrium with both liquid water and vapor at a specific temperature and pressure.
7. What is saturated enthalpy? Saturated enthalpy is the total heat content of saturated steam at a given temperature and pressure. It includes both sensible heat and latent heat.
8. What is the enthalpy of saturated steam at 30 Psig? The enthalpy of saturated steam at 30 Psig (pounds per square inch gauge) can be estimated at approximately 1,178 kJ/kg.
9. What is the formula for the quality of saturated steam? The formula for the quality (x) of saturated steam is: x = (Actual enthalpy – Liquid enthalpy) / (Vapor enthalpy – Liquid enthalpy)
10. How do you calculate the enthalpy of steam? Enthalpy of steam is calculated by considering both the sensible heat (due to temperature) and latent heat (due to phase change) using the formula:
Enthalpy = Sensible Heat + Latent Heat.
11. What is saturated steam at 300°C? Saturated steam at 300°C is steam that exists at a temperature of 300°C while being in equilibrium with liquid water at the same temperature and pressure.
12. What is the enthalpy of wet steam? The enthalpy of wet steam varies depending on the degree of wetness (quality). It can be calculated using the formula for enthalpy and the quality of the steam.
13. What is the critical temperature of saturated steam? The critical temperature of saturated steam is approximately 374.15°C (647.3 K) at a critical pressure of around 220.64 bar.
14. What temperature is dry saturated steam? Dry saturated steam exists at a single temperature and pressure and contains no liquid water. Its temperature depends on the pressure.
15. What is the Enthalpy of dry saturated steam at 10 bar? The enthalpy of dry saturated steam at 10 bar can be estimated at approximately 2,667 kJ/kg.
16. How hot is 70 psi steam? Steam at 70 psi (pounds per square inch) can have a temperature of approximately 154°C (309.2°F).
17. What is the temperature of 250 psi saturated steam? Saturated steam at 250 psi can have a temperature of approximately 202°C (395.6°F).
18. What is the specific enthalpy of steam at 60 bar? The specific enthalpy of steam at 60 bar can be estimated at approximately 2,795 kJ/kg.
19. What is saturated steam vs superheated steam? Saturated steam is in equilibrium with liquid water, while superheated steam has been further heated to a temperature above its saturation point,
making it dry and free of liquid water.
20. What is the difference between steam and saturated steam? Steam is a general term for the gaseous phase of water, while saturated steam specifically refers to steam that is in equilibrium with
liquid water at a given temperature and pressure.
21. Is saturated steam the same as superheated steam? No, saturated steam and superheated steam are different. Saturated steam contains both vapor and liquid water in equilibrium, while superheated
steam is dry and lacks liquid water.
22. How does saturation affect enthalpy? Saturation affects enthalpy by defining the state of steam. Enthalpy increases as steam is heated from a saturated state due to the addition of sensible heat.
23. How do you read a saturated steam table? To read a saturated steam table, you look up the pressure or temperature of interest and find the corresponding values for properties such as enthalpy,
entropy, and specific volume.
24. How do you calculate specific enthalpy? Specific enthalpy is calculated as the total enthalpy divided by the mass. It can be expressed as: Specific Enthalpy (h) = Total Enthalpy (H) / Mass (m).
25. What is the enthalpy of superheated steam? The enthalpy of superheated steam depends on the temperature and pressure. It can be found in steam tables for specific conditions.
26. What is saturated steam under pressure? Saturated steam under pressure refers to steam that is in equilibrium with liquid water and exists at a specific pressure.
27. How do you calculate saturated? To calculate saturated properties of steam, you need to know either the pressure or temperature, and then use a steam table or steam property equations.
28. How do you calculate the enthalpy of water vapor? The enthalpy of water vapor can be calculated using specific enthalpy equations and considering the temperature and pressure conditions.
29. At what temperature does saturated steam become superheated steam? Saturated steam becomes superheated steam when it is further heated to a temperature above its saturation point, typically by
passing it through a superheater.
30. Is wet steam saturated steam? No, wet steam is not saturated steam. Wet steam is a mixture of both vapor and liquid water, while saturated steam is purely vapor in equilibrium with liquid water.
31. How do you calculate water enthalpy? Water enthalpy is calculated using specific enthalpy equations for water, typically at different temperatures and pressures.
32. What are the three types of steam? The three types of steam are saturated steam, superheated steam, and wet steam.
33. Is saturation temperature the same as boiling point? Saturation temperature is often referred to as the boiling point, as it is the temperature at which a liquid boils and changes into vapor at a
given pressure.
34. What is enthalpy of dry steam? The enthalpy of dry steam is the total heat content of steam that contains no liquid water. It includes sensible heat and latent heat.
35. What temperature is saturation vapor? Saturation vapor temperature is the temperature at which a vapor becomes saturated, typically associated with its boiling point.
36. Can steam get hotter than 212 degrees? Yes, steam can get hotter than 212 degrees Fahrenheit (100 degrees Celsius) if it is superheated. Saturated steam at atmospheric pressure is 212°F (100°C),
but superheated steam can exceed this temperature.
37. How many psi is considered high-pressure steam? High-pressure steam typically starts at around 15 psi (pounds per square inch) and can go up to hundreds of psi in industrial applications.
38. What is the hottest steam can be? The temperature of steam can be extremely high in industrial processes, exceeding 1,000°C (1,832°F) in some cases, depending on the pressure and the degree of
39. How many BTU is a pound of saturated steam? A pound of saturated steam at atmospheric pressure carries approximately 970 BTUs (British Thermal Units) of energy.
40. Does saturation temperature of steam increase with pressure? Yes, as pressure increases, the saturation temperature of steam also increases. They have a proportional relationship.
41. Can steam be hotter than 100 degrees? Yes, steam can be hotter than 100 degrees Celsius (212 degrees Fahrenheit) if it is superheated. Saturated steam at atmospheric pressure is at 100°C, but
superheated steam can be heated to higher temperatures.
42. What is the enthalpy of 1 kg of steam at 70 bar? The enthalpy of 1 kg of steam at 70 bar depends on the temperature. It can be estimated to be around 3,500 kJ/kg or higher, depending on the
specific conditions.
43. How do you calculate the enthalpy of a steam turbine? The enthalpy change in a steam turbine can be calculated by determining the difference in enthalpy between the inlet and outlet steam, taking
into account any work done by or on the turbine.
44. What is the specific heat of a saturated vapor? The specific heat of a saturated vapor, such as saturated steam, varies with temperature and pressure. It can be found in steam tables.
45. Why do we use saturated steam? Saturated steam is used in various industrial processes for heating, sterilization, and power generation because of its consistent properties and predictable
46. Why is saturated steam better for heat transfer? Saturated steam is often preferred for heat transfer because it provides a constant temperature and releases latent heat during condensation,
making it efficient for heating applications.
47. What phase is saturated steam? Saturated steam is in the gas phase, specifically as water vapor, while it coexists with liquid water.
48. What is saturated steam in simple words? Saturated steam is steam that has reached its maximum moisture-carrying capacity at a given temperature and pressure, coexisting with liquid water.
49. What are the 5 types of steam? The five types of steam are saturated steam, superheated steam, wet steam, dry steam, and industrial steam (which may include different combinations).
50. What happens when you compress saturated steam? When you compress saturated steam, its pressure and temperature increase without changing its phase, which can be useful in various industrial
51. What is the theory of saturated steam? The theory of saturated steam is based on the thermodynamic principles that describe the behavior of water vapor at its saturation point, where it coexists
with liquid water.
52. Why is superheated steam used in the main propulsion turbines instead of saturated steam? Superheated steam is used in main propulsion turbines because it provides greater energy transfer
efficiency, allowing for better control and higher power output compared to saturated steam.
53. What phase is superheated steam? Superheated steam is in the gas phase and is entirely vapor with no liquid water content.
54. Does higher enthalpy mean more stable? Higher enthalpy does not necessarily mean more stable. Enthalpy is a measure of energy content, not stability. The stability of a substance depends on
various factors.
55. Is higher or lower enthalpy better? The choice of higher or lower enthalpy depends on the specific application. Higher enthalpy may provide more energy for certain processes, while lower enthalpy
may be desired for cooling applications.
56. What is the quality of saturated steam? The quality of saturated steam represents the proportion of dry vapor (steam) in a mixture of vapor and liquid water. It ranges from 0 (completely wet) to
1 (completely dry).
57. What is saturated steam at 300 used to heat? Saturated steam at 300°C can be used for various heating applications in industrial processes, such as in heat exchangers and for sterilization.
58. What is the relationship between steam pressure and Enthalpy? The relationship between steam pressure and enthalpy is that as pressure increases, the enthalpy of saturated steam also increases,
reflecting the additional energy stored in the steam.
59. What is the specific enthalpy of steam? Specific enthalpy of steam is the total enthalpy per unit mass of steam and is typically measured in units like kJ/kg or Btu/lb.
60. What is the easiest way to calculate enthalpy? The easiest way to calculate enthalpy is to use steam tables or software that provides property calculations for specific conditions of temperature
and pressure.
61. What are the two ways to calculate enthalpy? The two primary ways to calculate enthalpy are: a. Using steam tables or charts. b. Using the specific heat capacity and temperature change (for
sensible heat) along with latent heat of vaporization (for phase change).
62. What is the enthalpy of saturated steam and water? The enthalpy of saturated steam and water mixture depends on the specific conditions and the quality of the mixture.
63. What is the formula for enthalpy of dry saturated steam? The formula for the enthalpy of dry saturated steam is typically represented as: Enthalpy = C_p * (T – T_ref) + H_fg Where C_p is the
specific heat capacity, T is the temperature, T_ref is a reference temperature, and H_fg is the latent heat of vaporization.
64. What is the enthalpy of saturated steam at 1 bar? The enthalpy of saturated steam at 1 bar is approximately 2,257 kJ/kg.
65. What is saturated steam vs superheated steam? Saturated steam is in equilibrium with liquid water, while superheated steam has been heated beyond its saturation point and contains no liquid
66. What is the difference between saturated steam and superheated steam? The main difference is that saturated steam contains both vapor and liquid water, whereas superheated steam is dry and
contains only vapor. Superheated steam also has a higher temperature and enthalpy.
67. What is the difference between steam and saturated steam? Steam is a general term for water vapor, while saturated steam specifically refers to steam in equilibrium with liquid water at a given
temperature and pressure.
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Leave a Comment | {"url":"https://gegcalculators.com/saturated-steam-enthalpy-calculator/","timestamp":"2024-11-09T13:12:57Z","content_type":"text/html","content_length":"183166","record_id":"<urn:uuid:33a4b011-8c1f-4688-89b1-d42858c1389f>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00214.warc.gz"} |
Got Variance?
O.K., let's see if I can explain the potential issues here concisely.
What is the equity of a position? It is the value of the game, averaged over all possible ways of playing out the game (i.e., all possible subsequent sequences of dice rolls). We can't compute
this directly because we don't have infinite computing power. So we sample from the space of possible games by doing a rollout. Ignoring variance reduction for simplicity, what we get is a long
list of outcomes: some +1, some -1, some +2, some -2, some +3, some -3, some +4, some -4, etc. We can plot a histogram of these results, with the height of each bar being proportional to how
often we observed that result. The weighted average of the samples gives us an equity estimate.
Now, where does the confidence interval come from? This comes from the assumption that, in the long run, if we take enough samples, the shape of the histogram will approach a bell curve (a
Gaussian or normal distribution). The bell curve has certain properties, such as the fact that 95% of the area lies within 2 standard deviations of the mean. We can estimate the standard
deviation by taking the standard deviation of our samples. This is what the bots report as the 95% confidence interval: plus or minus 2 sample standard deviations from the sample mean (maybe they
do some slight correction to compensate for the fact that the sample standard deviation is a biased estimate of the true standard deviation even in the Gaussian case, but this is a technical
point that is not really relevant to the present discussion).
So what if anything is wrong with this procedure? The main issue is that the histogram may not, in fact, approach a bell curve in the limit. Perhaps it's slightly lopsided. Perhaps rare, large
values (i.e., very large cubes) occur with probability different from what the Gaussian assumption would predict. How can we test this?
The way to do it is to take extremely large samples and try to piece together the histogram directly, without making any assumptions about bell curves. For this we need something like the GNU
"View statistics" feature, which tells us how many trials ended with a single/double/triple win or loss with the cube on various values. This way we can see directly what the histogram looks like
and find out if indeed 95% of the area lies within 2 standard deviations of the mean.
In a money game, there's still some small chance that one could encounter difficulties because maybe the cube gets exponentially large an exponentially small percentage of the time, and no finite
amount of sampling will ever detect this. We can circumvent this problem by restricting ourselves to match play, where there's an upper limit on the size of the cube. Then, provided we take a
large enough sample, we will get an excellent approximation of the true (infinite) histogram.
What will we find if we carry out such an experiment? I don't know. However, it seems plausible to conjecture that different positions will give rise to different-looking histograms. Some
positions may lead to more lopsided graphs than others.
It's not so much that the bot is doing something wrong as that the Gaussian assumption is reasonable for some positions but no so reasonable for others. Without doing thorough analyses of the
type described above, we can't really hope to figure out which positions have funny-looking histograms and which ones don't. But conversely, if we do, we may get some insight into what sorts of
positions are likely to report misleading confidence intervals. | {"url":"https://www.bgonline.org/forums/webbbs_config.pl?noframes;read=68019","timestamp":"2024-11-10T07:29:40Z","content_type":"text/html","content_length":"15203","record_id":"<urn:uuid:d08b3672-2ad9-4f01-98c4-e1885666c752>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00807.warc.gz"} |