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# 4.5: Ampère's Law

## A Magnetic Analog to Gauss's Law for Electricity

We have already stated that Gauss's law for magnetic field is trivial, as there are no magnetic monopoles, but it turns out that there is a separate mathematical law that works for magnetic sources in the same way that Gauss's law works for electric sources. It incorporates "enclosed" sources, and allows us to use symmetrical situations to solve for fields using methods simpler than integrating Biot-Savart's law.

This magnetic version of the electrical Gauss's law is called Ampère's law, and since it can't involve enclosed point sources, it instead deals with lines of current, which either circle back on themselves to form a closed circuit, or are infinitely-long (and circle-back on themselves at infinity). But how do we "enclose" a line of current? In the case of charge, it was enclosed if there was no way to remove the charge from the Gaussian surface without breaking through the surface (i.e. the surface has no holes in it). In the case of Ampere's law, we consider a current to be enclosed by an imaginary closed path – called an Ampèrian circuitrather than a surface. Such a current is enclosed when there is no way to move the line of current out without it breaking through the Ampèrian circuit.

Figure 4.5.1 – Defining an Enclosed Current

There are two ways that a current won't be enclosed: If there is a break in the path around the wire, so that the wire can slide through it, or is the segment of wire is finite in length and does not form a closed loop. If the loop of wire is closed, then the Ampèrian circuit is "linked" with it, and the current is enclosed. If the wire is infinitely long, the wire similarly cannot escape the Ampèrian circuit without breaking through it, so its current is also enclosed. [Mathematically, we generally define a current to be "enclosed" by a closed path if it pierces every possible surface that is bounded by the closed path. Clearly there are stretched surfaces we can construct with the closed path as it border that do not allow a finite-length segment to pierce it.]

With this definition of enclosed current in place, we have Ampère's law:

$\oint \limits_{\rm {path}}\overrightarrow B \cdot \overrightarrow {dl} = \mu_oI_{enclosed}$

The integral is performed along any Ampèrian circuit that goes completely around the enclosed current, in the same way that the integral for Gauss's law works for any closed surface that encloses the charge. All of the properties of Gauss's law have analogous properties for Ampère's law:

• Enclosed is defined in terms of ability to remove charge/current from surface/closed path without breaking through the enclosure.
• The sign of the charge inside a Gaussian surface is related to the positive direction of the area vector. If the total flux is in the negative direction (opposite to the area orientation), then the enclosed charge is negative. Similarly, we define a positive direction of circulation for an Ampèrian circuit, and if the direction of the magnetic field of the current has the same circulation orientation as that of the Ampèrian circuit, then that current is "positive," otherwise it is "negative."
• There can be both positive and negative charges/currents enclosed in a surface/closed path, and these are combined to give a net charge/current.
• We can use symmetry to solve for a field. This usually means that the field is either parallel or perpendicular to the surface/closed path, and that it has a constant magnitude on that surface/closed path.
• The shape of the surface/closed path is not relevant, as long as it is closed.

Figure 4.5.2 – Comparing Gauss's Law to Ampère's Law

## Applications of Ampère's Law

The applications of Ampère's law are much like those of Gauss's law – symmetry is used to compute fields. Before we do any of these, let's confirm a result we already have – the field on the axis of a circular loop of radius $$r$$ (Equation 4.4.10). We need to choose a useful circuit – it needs to enclose the current, and it needs to have a direction that matches well with the direction of the magnetic field that we know from symmetry. To this end, we will choose an Ampèrian circuit that follows the $$z$$-axis from $$-\infty$$ to $$+\infty$$, and from there circulates back around on itself at infinity. The integral of this infinitely-distant segment of the closed path will vanish, because the field weakens to zero this far away, so all that is left is the integral along the $$z$$-axis.

Figure 4.5.3 – Ampèrian Circuit for Loop

The magnetic field on the axis is in the same direction as the circuit along the $$z$$ axis, and while we don't know the angle between the field and the direction of the path elsewhere, it doesn't matter, because the field vanishes. So the closed-path line integral gives:

$\oint \overrightarrow B \cdot \overrightarrow {dl} = \int\limits_{z-axis} Bdz + \int\limits_{R=\infty}\overrightarrow B \cdot \overrightarrow {dl} = \dfrac{\mu_oIr^2}{2}\int\limits_{-\infty}^{+\infty} \dfrac{dz}{\left(r^2+z^2\right)^{\frac{3}{2}}} + 0 = \dfrac{\mu_oIr^2}{2}\left[\dfrac{z}{r^2\sqrt{r^2+z^2}}\right]_{-\infty}^{+\infty} = \mu_oI$

Since the enclosed current within this Ampèrian circuit is in fact the current in the circular loop, and since this current has a positive orientation for this closed path (it is coming out of the page and the path is counterclockwise, satisfying the right-hand-rule), Ampère's law is confirmed for this case.

Next let's use Ampère's law in a more traditional fashion – to compute the magnetic field in a symmetric physical situation. We'll look at the field of a solenoid. The solenoid has a high degree of symmetry, mainly thanks to our assumption that it is essentially infinite in length. Consider the direction of the field inside the solenoid. There is no reason to expect that it would diverge from being parallel with the axis anywhere, because then what would make the point where this divergence occurs so special? It looks like every other point in the infinite solenoid, so the field direction should be the same as every other point in the solenoid. Furthermore, it should not get any stronger along the axis – why would the field be a greater magnitude at one point along the axis than any other?

Of course, knowing that the field inside the solenoid is parallel to the axis everywhere and equal strength as a function of its position along the axis inside the solenoid doesn't tell us that the field strength is also uniform as a function of distance from the axis – perhaps it gets stronger as the position gets closer to the wire coils? For example, the field may look like this:

Figure 4.5.4 – Speculation About the Magnetic Field Inside a Solenoid

We can check this speculation using Ampère's law. We do this by constructing a rectangular Ampèrian circuit inside the solenoid that has two sides parallel to the field, and two sides perpendicular to the field.

Figure 4.5.5 – An Ampèrian Circuit for the Interior of the Solenoid

Now we apply Ampère's law. First of all, there is no electric current flowing through this closed path, so the integral around the path will be zero:

$\oint \overrightarrow B \cdot \overrightarrow {dl} = \mu_oI_{enclosed} = 0$

Second, the magnetic field is perpendicular to two of the sides of the rectangle, so the dot product of the field with the displacements during that portion of the circuit is zero, giving zero integral along the vertical segments. And the field strength doesn't change along the direction of the axis, while the field is parallel to these parts of the path, so the dot products and integrals are easy to perform:

$\begin{array}{l} \int\limits_{vertical} \overrightarrow B \cdot \overrightarrow {dl} && = && 0 \\ \int\limits_{top} \overrightarrow B_{top} \cdot \overrightarrow {dl} && = && \int \left(-B_{top}\;dl\right) && = && -B_{top}L \\ \int\limits_{bottom} \overrightarrow B_{bottom} \cdot \overrightarrow {dl} && = && \int \left(+B_{bottom}\;dl\right) && = && +B_{bottom}L \end{array}$

Adding all four contributions to the integral and setting it equal to zero (no enclosed current) gives us the result that $$B_{top} = B_{bottom}$$, which means that the field strength at different distances from the axis inside the solenoid is actually the same. The magnetic field in the interior of a solenoid is completely uniform.

Notice that a rectangular Ampèrian circuit parallel to the axis drawn outside the solenoid must give us the same result: The magnetic field must be uniform out there as well. But that means that the field is the same strength infinitely-far from the solenoid as it is just outside of it. Naturally the field is zero infinitely-far away, which means that we can conclude that the magnetic field is zero everywhere outside the solenoid.

Finally, we can also use Ampère's law to determine the strength of the uniform magnetic field inside the solenoid, without going through the integration of an infinite number of loops, as we did in the previous section. We accomplish this by constructing a rectangular Ampèrian circuit parallel to the axis with one side inside the solenoid, and one outside of it.

Figure 4.5.6 – An Ampèrian Circuit to Compute Field Inside a Solenoid

As before, we see that the contributions to the integral by the vertical sides of the rectangular path are zero, because inside the solenoid the field is perpendicular to the path (and outside the solenoid the field is zero). This time, however, the contribution by one of the horizontal sides (the one outside the solenoid) is also zero, since there is no field. The side of the rectangle inside the solenoid is easy to integrate, so putting them all together into Ampère's law gives:

$\mu_oI_{enclosed}=\oint \overrightarrow B \cdot \overrightarrow {dl} = BL\left(inside\right)+ 0 \left(vertical\right)+ 0 \left(outside\right)$

Now we have to determine the enclosed current. From the diagram, we can see using the right-hand-rule for the Ampèrian circuit that the current going into the page is positive. The diagram shows 6 wires enclosed, but we will call this a more generic $$N$$. If the current flowing through the wire is $$I$$, then the enclosed current is then $$+NI$$. Plugging this in above gives:

$\mu_oNI = BL \;\;\;\Rightarrow \;\;\; B = \mu_o\dfrac{N}{L}I = \mu_o n I$

This matches the result we found previously (Equation 4.4.13). And as was the case with Gauss's law for electric fields, the mathematical gymnastics required is greatly reduced using this method.

Example $$\PageIndex{1}$$

Consider a toroidal solenoid. This is a solenoid that is finite in length, and has had its axis bent such that the two open ends are closed upon each other, forming a classic doughnut (or bagel) shape. When we look at this device, from our perspective, the current is passing through the wires coming out of the page from the center of the toroid, and going into the page on the outside of the toroid (see the diagram below). use Ampère's law to find the field inside and outside the solenoid.

Solution

It's best to start by looking at a cross-section of the solenoid, slicing through the "bagel" parallel to the plane of the page . This shows wires on the inside with current coming out of the page, and wires on the outside with current going into the page.

From symmetry, we can conclude that the magnetic field at any fixed distance from the center is the same magnitude, though the field strength may vary when that distance is changed. Also, symmetry demands that the field direction will have to be tangent to a circle in the plane of the page and concentric to the center of the toroid.

Constructing an Ampèrian circuit that is a circle in the plane of the page will then be parallel to whatever field is present anywhere, making the integral easy to do. If the radius of the circular path is $$r$$, then:

$\oint \overrightarrow B \cdot \overrightarrow {dl} = \oint B\;dl = B\oint dl = B\left(2\pi r\right) \nonumber$

If we choose the circuit to be outside the toroid (or inside the doughnut hole), then the enclosed current is zero, which tells us that the magnetic field outside the toroid is zero. Like a straight solenoid, a toroidal solenoid confines the magnetic field to its interior. To get the field inside the solenoid, choose the radius of the circular path such that it is inside the solenoid:

This time the enclosed current is the total number of turns in the solenoid $$N$$ multiplied by the current through the wire $$I$$, giving us the field inside:

$B\left(2\pi r\right) = \mu_o N I \;\;\;\Rightarrow\;\;\; B=\dfrac{\mu_o N I}{2\pi r} \nonumber$

Notice that this differs from the field in a straight solenoid in two ways: First, the field is not uniform within the solenoid – it is stronger closer to the center than it is near the outside of the solenoid. And second, it is the total number of turns (which is now finite) rather than the turn density that determines the field strength. Interestingly, the turn density for the toroid can be adjusted to "turns per radian" rather than turns per meter. This "angular turn density" is simply $$\frac{N}{2\pi}$$.

## Local (Differential) Form of Ampère's Law

Just as with the case of Gauss’s law, we can use a theorem from vector calculus to convert Ampère’s law into local form. The name of the theorem in question is Stokes' theorem, and it states that for a vector field $$\overrightarrow B\left(\overrightarrow r\right)$$:

$\oint \overrightarrow B \cdot \overrightarrow {dl} = \int \left(\overrightarrow \nabla \times \overrightarrow B\right)\cdot d\overrightarrow A\;,$

where the closed path borders the area over which the area integral is performed. If we now apply Ampère's law to this and note that the current that passes through the area equals the flux of the current density, we get:

$\int \left(\overrightarrow \nabla \times \overrightarrow B\right)\cdot d\overrightarrow A = \mu_o I_{encl} = \mu_o \int \overrightarrow J \cdot d\overrightarrow A \;\;\;\Rightarrow\;\;\; \overrightarrow \nabla \times \overrightarrow B = \mu_o \overrightarrow J$

As with the local form of Gauss's law, this provides us with a means for doing calculations that would otherwise be more difficult using the integral form.

This page titled 4.5: Ampère's Law is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Tom Weideman directly on the LibreTexts platform.


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stepresp

Step-signal response for rational object and `rationalfit` function object

Syntax

``[outputsignal,tout] = stepresp(h,ts,n,trise)``

Description

example

````[outputsignal,tout] = stepresp(h,ts,n,trise)` computes the time domain response of a rational function object, `h`, to a step signal based on the number of samples, `n` and the rise time, `trise`.```

Examples

collapse all

Calculate the step response of a rational function object from the file `passive.s2p`. Read `passive.s2p`.

```S = sparameters('passive.s2p'); freq = S.Frequencies;```

Get S11 and convert to a TDR transfer function.

```s11 = rfparam(S,1,1); Vin = 1; tdrfreqdata = Vin*(s11+1)/2;```

Fit to a rational function object.

`tdrfit = rationalfit(freq,tdrfreqdata);`

Define parameters for a step signal. Define parameters for a step signal

```Ts = 1.0e-11; N = 10000; Trise = 1.0e-10;```

Calculate the step response for TDR and plot it

```[tdr,t1] = stepresp(tdrfit,Ts,N,Trise); figure plot(t1*1e9,tdr) ylabel('TDR') xlabel('Time (ns)')```

Input Arguments

collapse all

Rational function object, specified as a `rationalfit` or `rational` object.

Data Types: `double`
Complex Number Support: Yes

Sample time of the input signal, specified as a positive scalar integer in seconds.

Data Types: `double`

Number of samples, specified as a positive scalar integer.

Data Types: `double`

Time taken for step signal to reach maximum value, specified as a positive scalar integer in seconds.

Data Types: `double`

Output Arguments

collapse all

Computed step response, returned as a real finite vector.

Data Types: `double`

Sample time of the output signal, returned as a nonnegative vector of time values with discrete step size `ts` corresponding to the data in `outputsignal` in seconds.

Data Types: `double`

collapse all

Step Signal Equation

RF Toolbox™ uses the following equation to for the step signal:

`$\left\{\begin{array}{l}U\left(k{t}_{s}\right)=k{t}_{s}/{t}_{rise},\\ U\left(k{t}_{s}\right)=1,\end{array}\text{ }\text{ }\begin{array}{c}0\le k<\left({t}_{rise}/{t}_{s}\right)\\ \left({t}_{rise}/{t}_{s}\right)\le k\le N\end{array}$`

The following figure illustrates the construction of this signal.

Version History

Introduced in R2010a

format: plain


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# Evidence Action — Impact Evaluation of Iron and Folic Acid Supplementation ("Phase 2")

Published: April 2019

Note: This page summarizes the rationale behind a grant to Evidence Action that GiveWell has recommended to Good Ventures. Evidence Action staff reviewed this page prior to publication.

In March 2019, GiveWell recommended that Good Ventures makes a grant of $800,000 to Evidence Action for the monitoring and evaluation of the Phase 2 pilot stage of its technical assistance to the Government of India's national school-based iron and folic acid (IFA) supplementation program. ## Background In December 2018, Evidence Action's Beta Incubator received a GiveWell Incubation Grant to pilot a project to provide technical assistance to the Indian government's large-scale school-based IFA supplementation targeting children and adolescents. For more details on that grant, see this page. ## About the grant In March 2019, GiveWell recommended that Good Ventures make a grant of$800,000 to Evidence Action for the monitoring and evaluation of the Phase 2 pilot stage of its technical assistance to the Government of India's IFA supplementation program.

Evidence Action told us that it will undertake the following key monitoring activities with this grant. This is a tentative plan that is subject to change, but we believe that it is indicative of the type of activities that it will undertake:

• Monitor the change in coverage caused by Evidence Action's technical assistance to the IFA program. Evidence Action will try to do this in a few main ways:
• Difference-in-differences evaluation at the state and district level. Evidence Action will measure and compare changes in IFA supplementation coverage between baseline and follow up in a sample of schools between treatment and control districts in Rajasthan and Madhya Pradesh, the two states in which Evidence Action will pilot its technical assistance. It will randomly choose (within implementation feasibility constraints) which districts are treatment and control.1
• High frequency process monitoring. Evidence Action will also conduct weekly surveys in a random sample of schools to measure several other outcomes that may drive changes in coverage. For example, this includes drug availability, and teachers' and children's awareness and knowledge of the program.2
• Qualitative information from key stakeholders. Evidence Action and/or GiveWell will conduct interviews with key stakeholders, such as block or district government officials, to collect additional information on what level of coverage might have otherwise been achieved in the counterfactual, for example what other technical assistance partners might have worked with the Government of India in the treatment districts had Evidence Action not provided technical assistance.3
• Monitor the change in rates of anemia. Increasing coverage of IFA is intended to decrease rates of anemia. Evidence Action will directly measure anemia rates through low-cost haemoglobin tests at baseline and follow-up in both treatment and control districts using HemoCue machines.4
• Track government costs. Evidence Action will use a "government cost tracker" tool to collect information on the costs of the program.5

Evidence Action is still finalising many of the details of its monitoring and evaluation plans, particularly the approach to the difference-in-differences (including the sample size in terms of both numbers of districts and schools). To do so, it is working with Kartini Shastry, Associate Professor of Economics at Wellesley College, and IDinsight. GiveWell expects to receive further details about these plans in April 2019. Evidence Action intends to conduct the baseline monitoring in July or August 2019.

## Risks and reservations

The main risk of this grant is that we don't learn much about whether Evidence Action's technical assistance increased coverage of the Government of India's program. This might occur for a few reasons:

• Small sample size for the difference-in-differences. It is our understanding that only a small number of districts will be included in the difference-in-differences analysis. Randomising across a small number of districts, it is unlikely that systematic differences between the treatment and control groups wash out. We may have lasting doubts about whether any effect estimated in the difference-in-differences analysis is a true causal effect of the program as opposed to just picking up different pre-program trends in coverage between the treatment and control districts caused by other factors.
• Selection of control districts for the difference-in-differences. For the difference-in-differences estimates to tell us the effect of the program, coverage in the control group should be representative of what would have happened in the counterfactual in the districts in which Evidence Action intervened. Evidence Action currently plans to drop some districts from selection into the study (and therefore the control group), e.g. districts in which there are many other technical assistance partners.6 We hope to work out a plan with Evidence Action to mitigate any selection effects and help us best understand the counterfactual of Evidence Action's activities, but this plan has not been finalized.
• Measure of coverage. We don't yet have details on the methodology that will be used to measure coverage. For example, we do not know whether enumerators will directly observe children taking supplements, or whether Evidence Action will rely on answers to survey questions.
Document Source
Evidence Action, IFA Phase 2 Monitoring Considerations slide deck Source
• 1.

Evidence Action told us this in a conversation on 4th February 2019. We do not yet have details on the implementation feasibility constraints.

• 2.

"Monitoring survey conducted in random and targeted of sample schools and AWCs [anganwadi centers] every week on drug availability and sufficiency, regularity of administration and consumption of drugs, knowledge of teachers and children, IEC [information, education and communication] materials/activities etc", slide 13 in Evidence Action, IFA Phase 2 Monitoring Considerations slide deck.

• 3.

"Qualitative insights would include:

• Informal interviews with district/block officials conducted by EA staff at various levels (frequency TBD)
• Structured/Formal exploratory discussions and interviews
• Interviews with key officials
• Structured group discussion with stakeholders (teachers, AWWs, block officials etc.)
• Observational and informal insights"
• 4.

"Additional costs... for two rounds in two states for basic Hb level testing using HemoCue machines", slide 7 in Evidence Action, IFA Phase 2 Monitoring Considerations slide deck.

• 5.

"Cost data collected quarterly by State Project Co-od. (M&E) from districts and state using a “government cost tracker” tool using similar channels to DtW cost data collection", slide 14 in Evidence Action, IFA Phase 2 Monitoring Considerations slide deck.

• 6.

"Drop districts due to security, terrain, density, districts with extensive partner presence - to check data on which point in the process it makes most sense to drop, districts where there may be significant changes in boundaries planned", slide 15 in Evidence Action, IFA Phase 2 Monitoring Considerations slide deck.


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## View Blog | Read Bio

While visiting friends and family over the holidays, I was surprised by the number of people asking how I am paying for graduate school.

Lots of loans?

Not a lot of people know, so I thought I’d share: attending graduate school in a hard science is typically free.

You see, generally, when someone attends graduate school in a field such as Astronomy, Biology, Chemistry, Math, Physics…, their tuition is paid for in full and they get a salary on top of that.

This is the case as long as you do work for the department by being a Teaching Assistant, Research Assistant, Grader…or some job. Typical grad student salary is roughly between $15,000 and$25,000 a year, plus health benefits. (Pay, of course, depends on the institution, department, job, and the % of time committed.)

One difference between graduate students in physics doing research in “theory” vs “experiment” is that, generally, there are more positions and more money available for students to do experimental research. Students who work on theory research typically teach some or most of their years in graduate school, while students who work on experimental research typically do not teach past their 2nd year of graduate school (and are paid entirely as a research assistant). That’s only in general.

So, different strokes for different folks (see others’ posts about doing research in theory: 1, 2, 3), but in the end, either way, grad school tuition is free, and students get paid on top of that.


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# 9.1.1 Connect to Web Data by Copy Link Address

## Summary

Data Connectors are an ideal way to create a "live link" to data files that are periodically updated. Traditional file import requires the user to re-import data each time source data are modified. However, once a Connector is added to a workbook, it can automatically query the data source each time the project is opened, thus simplifying your data management tasks.

In this tutorial, we'll show you how to connect to "real-time" earthquake data on the United States Geological Survey (USGS) website, how to configure the connected project to automatically import data on open, and how to manage data caching so that you are assured of getting the most up-to-date data each time you open your project.

Minimum Origin Version Required: Origin 2021

## What you will learn

• How to connect to a web-stored CSV file by copying the URL.
• How to use Origin's free Maps Online App to create a background image for geographic data.
• How to use the Plot Setup dialog with Data Connector-imported data, saving us the need to rearrange columns.
• How to save and reopen a project that uses a Web Connector (CSV).
• How to manipulate Origin's caching mechanism so that data are always refreshed on opening the project.

## Connect to Web Data File by Copy Link Address

1. Start a new project. There should be one new workbook visible in the workspace.
2. Browse to this page on the USGS website: https://earthquake.usgs.gov/earthquakes/feed/v1.0/csv.php
3. In the right-hand column, you'll see under "Feeds", a list of links to CSV data files that are updated every minute.
4. Scroll down and under "Past 30 Days", right-click on "All Earthquakes" and choose Copy link address.
5. Return to Origin and with your workbook active, choose Data: Connect to Web.
6. Paste the link address you just copied into the URL box using CTRL + V and click OK.
7. Accept default settings in the CSV Import Options dialog box and click OK. The data from the USGS site are imported into the workbook. Note the green color of the Connector icon , indicating that the data connection is valid.

## Plotting Connector-imported Data with Plot Setup

In this portion of the tutorial, we will use OriginLab's free Maps Online App to create a base-map for a scatter plot of latitude, longitude, magnitude and depth data.

Because of the "locked" nature of Data Connector-imported data, we cannot rearrange columns. By default, Origin requires that X data columns be to the left of Y data columns. But in our locked worksheet, the X data (longitude) are to the right of our Y data (latitude). The Plot Setup dialog box works nicely in this situation, as we can designate columns "on the fly" without having to move things around (and lose the "live" connection to our data).

1. Press F10 to open the App Center. Click on the Search tab and with Category = All, enter "maps" into the search box. If the Maps Online App shows with a green checkmark beside it, you have already installed the App. If it displays the Download and Install icon , click the icon to install the App in your Origin software.
2. Click on the Apps Gallery (by default, docked to the right side of your workspace) and launch the Maps Online App. In the dialog box, click the Map drop-down and choose Nasa Earth Observations (NEO) WMS and set Layers to BlueMarbleNG, then click OK. We now have a base-map for plotting our earthquake data.
3. From the menu, choose Graph: Plot Setup (if not displayed, expand top and bottom panels of the dialog box). Set Available Data = Worksheets in Folder and you should see your all-month sheet in the top panel. Highlight the sheet in the top panel.
4. In the middle panel set Plot Type = Scatter and set longitude = X and latitude = Y, as shown. Now, click the Add button in the lower-right corner of the middle panel. Click OK to overlay the scatter plot of lat-lon data on your map.
5. Double-click on the plot to open Plot Details and recreate the settings show in the following screen-cap by setting these Symbol tab controls: Size = Col(E): "mag", Scaling Factor = 3, Size Formula = (s-s1)/(s2-s1), Symbol Color = By Points set to Map:Col(D):"depth". In summary, we are using latitude and longitude values to locate our scatter points on the map; we are using mag (magnitude) data to control symbol size and depth to control symbol color.
6. On the Tools toolbar, click the Add Color Scale button, then click on the color scale to open Color Scale Control dialog. On the Levels tab, Hide Head Level and Hide Tail Level. On the Layout tab, clear the Reverse Order box. On the Title tab, check the Show box. Click OK to close the dialog. You can also add a Bubble Scale to the map to show relative symbol sizes (since size is mapped to earthquake magnitude).
7. We can make a few cosmetic adjustments to our graph such as using Print/Dimensions settings to modify page size to more closely fit the aspect ratio of the map image. Additionally, we can add axis titles, and adjust fonts and font sizes using Mini Toolbar buttons.
 If you hover on a plotted point in your graph, you will see a customizable data point tooltip. Carefully right-click on the tooltip, choose Preferences and you can modify the tooltip to display all of your plot variables -- latitude, longitude, magnitude and depth, in any order and in any numeric format -- as you mouse over your plot.

## How to Manage Caching of Web Data

Data Connectors were conceived, in part, to facilitate import of very large data files. Connectors employ the strategy of not saving large file imports (> 500kB by default), thus saving on disk space. If you have a number of connected workbooks in your project file, you can selectively re-connect to files as needed, allowing you to import just the data that you need to access now.

In an effort to manage the number of times a connection is refreshed -- this takes time and can slow work, after all -- OriginLab developers added system variable @CFDT, which controls the length of time between Connector queries of source data. Data imported by Connect to Web is cached for a time period equal to the value of @CFDT seconds per 100kB of file size (e.g connected data in a 10MB file is cached for 900 * 100 = 90,000 sec or ~24 hr).

Additionally, there is a LabTalk command -- delete -web -- that will flush the cache on demand. For the purposes of our earthquake project, we would like to ensure that we are always getting the most up-to-date data, so we will make use of this.

In this portion of the exercise, we will configure our project so that:

• Data are automatically imported whenever the project is opened.
• The cache is flushed on project open so that we are certain of always having the most up-to-date information.
1. Choose File: Save Project As, give the file a name such as USGS_Earthquake_All_30d.opju and click Save.
2. Click on the worksheet's Connector icon and choose Auto Import > On Project Open.
3. From the Origin menu, choose View: Code Builder. Code Builder is Origin's integrated development environment (IDE) for developing LabTalk, Origin C and Python code. In the Workspace pane on the left side of the dialog, expand the Project branch and double-click on the ProjectEvents.OGS file. The file is loaded into Code Builder workspace. ProjectEvents.OGS is a special file that is saved with each Origin project and is used to store and execute LabTalk script at various event triggers such as upon opening of the project. We have already mentioned that we have a LabTalk command that can be used to flush the cached web data so we will customize ProjectEvents.OGS to run a simple line of script so that our web data are always refreshed on project open. Scroll up to the section [AfterOpenDoc], add the following two lines of script, then click the Save button on the Code Builder toolbar.
delete -web;
type -b "The web cache has been cleared. Project data have been refreshed."

ProjectEvents should look something like this:

4. Choose File: Save to save your project, along with the contents of ProjectEvents.OGS.
5. Now, close your project, then using File: Recent Projects, click on your recently-saved-and-closed project. An attention message should open stating that the "web cache has been cleared. Project data have been refreshed." Subsequently, the worksheet should fill with data and the plot of earthquakes on the world map, should be regenerated.


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Frequency vs. Wavelength

Frequency and wavelength are related and inversely proportional. Calculating one from the other is trivial.

The mathematical equation is: $λ = c ÷ f$

However, in radio, the equation is usually simplified by converting the frequency from hertz to megahertz, and replacing the speed of light with a proportionately smaller number: $λ ≈ 300 ÷ (f MHz)$

Try it!

Fill in either value.


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# Enumerative geometry via the A^1-degree

Speaker: Sabrina Pauli

Date: Sat, Sep 26, 2020

Location: Zoom

Subject:

Class: Scientific

### Abstract:

Morel's $A^1$ -degree in $A^1$-homotopy theory is the analog of the Brouwer degree in classical topology. It takes values in the Grothendieck-Witt ring $GW(k)$ of a field $k$, that is the group completion of isometry classes of non-degenerate symmetric bilinear forms. We can use the $A^1$ -degree to count algebro-geometric objects in $GW(k)$, giving an $A^1$-enumerative geometry over non-algebraically closed fields. Taking the rank and the signature recovers classical counts over the complex and the real numbers, respectively. For example, the count of lines on a smooth cubic surface enriched in $GW(k)$ has rank 27 and signature 3.


text:
# Formulas You’ll Need to Know for Mathematics Questions on the HESI Test

## Why do nurses have to know math?

That’s the question you’re probably asking yourself as you’re preparing for the HESI Mathematics test. Well, math is a skill you’ll need for keeping records of your patients, understanding trends, and organizing information. That’s why the admittance test for nurses includes math.

But don’t worry, we’re here to help you. In the following chart, you’ll find easy-to-remember formulas that will be useful to you during the HESI Mathematics test. You’ll have access to a basic calculator during the test, but not to any other resources, so make sure you learn these formulas. A good way to do that is to use them for our free practice questions and flashcards.

$\begin{array}{|c|c|c|c|} \hline \textbf{Category} & \textbf{Formula} & \textbf{Symbols} & \textbf{Comment} \\ \hline \text{Fractions} & \dfrac{a}{b} +\dfrac{c}{d} = \dfrac{(a\cdot d) + (c\cdot b)}{b\cdot d} & a,b,c,d = \text{any real number} & \text{simplify the fraction }\\ & & &\text{(if possible) } \\ \hline \text{Fractions} & \dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{a\cdot c}{b\cdot d} & a,b,c,d = \text{any real number} & \text{simplify the fraction }\\ & & &\text{(if possible) } \\ \hline \text{Fractions} & \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a\cdot d}{b\cdot c} & a,b,c,d = \text{any real number} & \text{simplify the fraction }\\ & & &\text{(if possible) } \\ \hline \text{Fractions} & a\frac{b}{c} = \dfrac{a\cdot c + b}{c} & a,b,c = \text{any real number} & \text{simplify the fraction }\\ & & &\text{(if possible) } \\ \hline \text{Percents} & a \cdot b \% = a \cdot \frac{b}{100} & a = \text{any real number} & \text{simplify} \\ & & b \% = \text{any percent} & \text{(if possible)} \\ \hline & x + a = b \rightarrow x = b-a & & \\ & x - a = b \rightarrow x = b+a & & \\ & x \cdot a = b \rightarrow x = b \div a & a, b = \text{constants} & \\ \text{Algebra} & x \div a = b \rightarrow x = b \cdot a & x = \text{variable} & \\ & x^a = b \rightarrow x = \sqrt[a]{b} & & \\ & \sqrt[a]{x} = b \rightarrow x = b^a & & \\ & a^x = b \rightarrow x = \dfrac{\log b}{\log a} & & \\ \hline \end{array}$


text:
## Thursday, December 14, 2017

### How well is India's land record digitisation programme doing: Findings from Rajasthan

by Anirudh Burman and Devendra Damle.

Good titles in land improve the security of land tenure, and also enable land holders to capitalise land, either by mortgaging it, or making productive use of it. Conversely, the lack of clear titles in land reduce the security of land tenure since titles are prone to challenge, and inhibit productive use of land since it is difficult to signal title over land (here). It is therefore essential for a well functioning land market to have a well developed land titling system.

The Government of India has been running a program for improving land titles called the Digital India Land Record Modernisation Program (DILRMP, titled the National Land Record Modernisation Program until 2015) since 2008. The DI-LRMP was initiated to improve land titles in India, with the ultimate objective of creating a system of conclusive titling. The title recorded with the state is considered conclusive proof of the title to land in a conclusive titling system.

To be conclusive, a titling system requires that (a) all land transactions be recorded by the government, (b) the applications for registering land titles be verified scrupulously before they are registered, (c) the status of a title once registered is final and not open to dispute (curtain principle - or drawing a curtain over past defects/ disputes), and (d) the registry be completely updated in relation to the status of titles as they are present on the ground (mirror principle). The DI-LRMP aims to create this system all over India through the following activities (see NLRMP Guidelines):

1. Digitisation of textual revenue records (contained in Record of Rights, revenue records are updated through "mutation", and are presumptive proof of the title recorded in the RoR) and registration records (records with the registration department - all agreements pertaining to transfers of land have to be registered, save specified exceptions), including maps.

2. Integration of the processes of mutation (RoR updation), registration and map generation/ creation.

3. Creation of modern record rooms, capacity building within the administrative machinery.

4. Allowing for processes of land registration to be initiated online in addition to physical processes.

5. Fresh land surveys and mapping in areas where required.

The completion of these activities is envisaged to take India close to a conclusive titling system. Our team was one among three coordinating research institutions to conduct the first assessment of the progress of the DI-LRMP. The NIPFP team chose the state of Rajasthan to conduct this study. Rajasthan is the largest state in India constituting 10.4 percent of the total area and contains 5.67 percent of the total population of India.

Our study was able to highlight:

1. Issues with the implementation of the DI-LRMP, at different stages of implementation; and

2. The issues faced by the local administration in the maintenance of land records and in implementing the DI-LRMP.

The removal of systemic and administrative issues highlighted in this report may make the implementation of the DI-LRMP more effective. The report based on the study is available here.

### Land records in Rajasthan

Work on computerisation of Records of Rights started in 1999-2000, well before the NLRMP was conceived. These activities got subsumed by the Land Records Computerisation project which was later subsumed by NLRMP.

Rajasthan's history presents some interesting challenges to this exercise. The state was originally composed of five main Riyasats (princely states), and several smaller ones. Each of these five Riyasats used different units to measure land. For example, a bigha in one Riyasat was the equivalent of 3000 sq.ft., while in another it was close to 1600 sq.ft. These differences still persist today. Not only are the units of measurement different, but some of the official terms used in relation to land records also vary across districts. The state government tried to introduce the metric system in 1976, but as per discussions with local officials and villagers, it was rarely used in practice by the local populace who were more comfortable with the old measurement systems. With the DI-LRMP there is a renewed effort to standardise units of measurement across the state.

1976 was also the year the state government last undertook re-survey operations across the state. The re-survey operations which will be undertaken under DI-LRMP will be the first since then. All maps currently in use were made using traditional techniques. They will be replaced by maps made using High Resolution Satellite Imagery.

In spite of early movements towards improvements in land records, the level of implementation of DILRMP in Rajasthan remains low. This is highlighted in the findings from our study.

### Scope of the study

Our study evaluated the implementation of DI-LRMP through four activities:

1. Collecting state-level data from government websites of the Department of Land Resources, Government of India, as well as the websites of the Departments in charge of Revenue and Registration, Government of Rajasthan;

2. Interviews with senior officials in the Departments of Revenue and Registration in the Government of Rajasthan;

3. Performing test-checks on the land record websites on the government of Rajasthan, and

4. Studying the implementation of DI-LRMP in two tehsils in Rajasthan.

The research team undertook the tehsil level study with the help of local retired revenue officials. This facilitated interactions with local officials in the revenue and registration offices in the two tehsils. In addition, the study involved a visit to five villages each in the two selected tehsils, interactions with villagers, and collection and verification of information relating to a sample of ten parcels in each village (a total of 50 parcels in each tehsil).

### Findings

#### State-level findings

We first collected state-level data from websites of the Rajasthan Government, and verified it in conversations with government officials. The state-level data reveals the following:

1. Most RoRs (revenue records) have been digitised, but maps have not been updated through modern survey methods and the maps available online at present are digitised copies of cadastral maps.

2. Our test-checks revealed that while most records of rights are available online, there was a significant volume of cases where the records were not available/accesible. There is a uniform problem of the absence of legacy records online.

3. The RoR is available only in paper form locally in a total of 1603 villages out of 47,918 villages.

4. Only a small proportion of the RoRs are available with digital signature of the designated official (3,632 villages out of 47,918), and RoRs of most villages are not available in a legally usable form (42,683 villages).

5. Most tehsils in Rajasthan have functional and usable maps, albeit in paper form. 88 tehsils out 242 have some proportion of damaged or mutilated maps. In all other tehsils, 90 percent or more of maps are in a usable condition.

6. While there has been some provisioning for online registration, most of the processes are still manual. Out of 527 Sub-Registrar Offices (SROs), 117 SROs have online systems for verifying documents and paying stamp fee duty.

7. The state of Rajasthan has made very little progress on integration of all three processes - mutation, registration and map generation. The process of registration alerts the revenue records database by noting the fact of registration in some form in 15 SROs, but this does not work the other way round, i.e. the process of changes in revenue records does not alert the registration database.

An analysis of the findings from the state highlights significant steps required to be taken for the implementation of the DILRMP. This was also confirmed by our study of DILRMP implementation in the two tehsils selected for the study.

#### Tehsil-level findings

Tehsil selection: The two tehsils selected for the study were Girwa and Uniara. Both tehsils are at different stages of implementing DI-LRMP:

1. Girwa: Girwa is part of the sub-humid southern plains agro-climatic zone. It is situated on and around the rocky hills of the Aravalli Range, at an average elevation of 540 meters. The climate is moderate year-round with moderate seasonal variation in temperature and humidity. Rainfall is scanty with little to moderate year-on-year variation. The main reason for selecting Girwa as one of the sample tehsils is that it is a good representative of the typical tehsil. It represents the typology of about 184 tehsils, out of 314 tehsils in Rajasthan in terms of the status of land records computerisation.

2. Uniara: Uniara is a tehsil in Tonk, Rajasthan and is situated approximately a three-hour drive away from Jaipur, the state capital. The tehsil is primarily agricultural, with little or no industrial activity. The tehsil is largely rural, with no large cities in its close vicinity. Uniara is part of the semi-arid eastern plains agro-climatic zone. The terrain is flat barring a few areas to the northwest. The climate is predominantly dry and there are large seasonal variations in temperature. Rainfall is scanty with large year-on-year variation. Uniara is cited as an example of a model tehsil in Rajasthan with respect to land records modernisation. It was also featured in the Success Stories of NLRMP report published by the Department of Land Records, Government of India.

#### Findings from Girwa and Uniara

Vacancies: We found significant vacancies in the local administrative units in both tehsils. Tables 1 and 2 provide details of the sanctioned and vacant positions in Girwa and Uniara respectively.

 Post Sanctioned Vacant Patwari 52 20 Land Revenue Inspector 13 0 Naib-Tehsildar 3 1 Tehsildar 1 0

 Post Sanctioned Vacant Patwari 54 35 Land Revenue Inspector 13 4 Naib-Tehsildar 3 1 Tehsildar 1 0

Accuracy of recorded land area: We measured a total of 99 parcels across the two tehsils (Figure 1). In 30 percent of the cases the area is within 5 percent of the area on record. In 25 percent of the parcels, the difference between the area on record as compared to the area as measured, is 10-20 percent. In 24 percent of the sample parcels, the difference between the area on record as compared to the measured area is more than 20 percent. It must be noted that digitised cadastral maps were not available for either tehsil for the purpose of our study. The deviation recorded here is the deviation from the area recorded in the RoR.

Figure 1: Difference Between Area on Record vs Measured

Causes for deviation from recorded area: Encroachment onto neighbouring land is the most frequently observed cause for deviation from the recorded area. We observed two types of encroachments:

1. Encroachment onto adjoining fields i.e. onto private property, and

2. Encroachment onto public property. This can be further subdivided into two types: (a) encroachment onto roads, nallahs and farm roads, and (b) encroachment onto pasture land, forest land and other government-owned land.

Figure 2 provides details on the causes for differences between recorded area and measured area.

Figure 2: Causes of Difference Between Area on Record vs Measured
Note: Only shows parcels where the difference is 10 percent or larger.

Information recorded in Record of Rights: We noted that a number of transactions and kinds of titles are not recorded in the RoRs at the tehsil level. For example, construction on agricultural land is not recorded if less than 500 square metres. Possession, independent of ownership is also not recorded. Encumbrances other than mortgages are also not recorded. This is an impediment in providing conclusive titles, as a number of rights over land are not recorded at all.

Administrative issues: We found local administration under-staffed, lacking basic infrastructure (at the patwar-mandal levels) including electricity in some places, and without network connectivity. Most of the administrative processes are manual. In addition, revenue administration is burdened with a number of tasks at best incidental to land revenue and record management. Some officials pointed out that land revenue collection, a core activity for the revenue department, does not seem like a feasible activity any longer as the costs of collecting revenue far exceed the revenue collection. Building administrative capacity is therefore an important challenge for the state.

### Conclusion

Our report makes the case for legal and administrative changes based on the results of our study. It is important to realise that the process of digitisation must be accompanied by complementary legal changes and administrative capacity building. Rajasthan has an advantage in the fact that its digitisation program is not yet at an advanced stage and some of these issues can be fixed right now rather than later. Additionally, while digitisation may remove some dependency on human interaction with revenue officials, land record management will continue to require sufficient human capital.

Anirudh Burman and Devendra Damle are reasearchers at National Institute of Public Finance an Policy.

LaTeX mathematics works. This means that if you want to say $10 you have to say \$10.


text:
# Finding approximately rank-one submatrices with the nuclear norm and l1 norm

Authors
Type
Preprint
Publication Date
Nov 08, 2010
Submission Date
Nov 08, 2010
Identifiers
arXiv ID: 1011.1839
Source
arXiv
We propose a convex optimization formulation with the nuclear norm and $\ell_1$-norm to find a large approximately rank-one submatrix of a given nonnegative matrix. We develop optimality conditions for the formulation and characterize the properties of the optimal solutions. We establish conditions under which the optimal solution of the convex formulation has a specific sparse structure. Finally, we show that, under certain hypotheses, with high probability, the approach can recover the rank-one submatrix even when it is corrupted with random noise and inserted as a submatrix into a much larger random noise matrix.


text:
# Hamiltonian cycles in maximal planar graphs

Whitney proved long ago that every maximal planar graph with no separating triangles is Hamiltonian. However, the converse is not necessarily true, that is, there exist maximal planar graphs that are hamiltonian and have separating triangles. So what happens when a maximal planar graph has separating triangles? Is there still some condition under which we can say with certainty that it will be Hamiltonian as well?

One theorem of this kind was shown by Jackson and Yu.

Consider a maximal planar graph $G$ with a separating triangle $T$. $G$ can be decomposed into two graphs $G_1$ and $G_2$ such that $G_1 \cup G_2 = G$ and $G_1 \cap G_2 = T$. If $G_1$ or $G_2$ still has separating triangles, then decompose them similarly. Eventually, we will be left with a collection $S$ of maximal planar graphs without any separating triangles. Now, consider a graph whose nodes represent elements of the set $S$ and there is an edge between two nodes if they share a separating traingle in $G$. Let’s call this graph $D$. The theorem by Jackson and Yu states that if $D$ has maximum degree 3, then $G$ is Hamiltonian.

This way of decomposing maximal planar graphs has been popular. It was first used by William Cunningham and Jack Edmonds in 1980. They proved several interesting results about the decomposition graph $D$.

## 6 thoughts on “Hamiltonian cycles in maximal planar graphs”

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text:
## What are the uses of total internal reflection?

The phenomenon of total internal reflection of light is used in many optical instruments like telescopes, microscopes, binoculars, spectroscopes, periscopes etc. The brilliance of a diamond is due to total internal reflection. Optical fibre works on the principle of total internal reflection.

### How is total internal reflection used in Fibre optics?

Fiber optics involves the transmission of light down fibers of plastic or glass, applying the principle of total internal reflection. Cladding prevents light from being transmitted between fibers in a bundle. Diamonds sparkle due to total internal reflection coupled with a large index of refraction.

#### What are two uses of internal reflection?

Applications of total internal Reflection Many optical instruments use right-angled prisms to reflect a beam of light through 90° or 180° (By total internal reflection) such as cameras, binoculars, periscope, and telescope.

What is TIR write its application?

Answer: Total internal reflection is very useful. As a result, it has a number of applications that include: Use in right angled isosceles prism – These prisms can turn light through 90 and 180 degrees based on internal reflection. right angled isosceles prism are mostly used in various optical instruments.

What is total internal reflection and why is it important in optical fibers What is the critical angle?

If the angle of incidence is bigger than the critical angle, the refracted ray will not emerge from the medium, but will be reflected back into the medium. This is called total internal reflection. The critical angle occurs when the angle of incidence where the angle of refraction is $$\text{90}$$$$\text{°}$$.

## What are the applications of optical fibre?

Applications. Optical fiber is used by telecommunications companies to transmit telephone signals, Internet communication and cable television signals. It is also used in other industries, including medical, defense, government, industrial and commercial.

### What is an example of total internal reflection?

Mirage: It is an optical illusion that is responsible for the appearance of the water layer at short distances in a desert or on the road. Mirage is an example of total internal reflection which occurs due to atmospheric refraction.

#### What is optical fibre write two applications?

Optical fiber is used by telecommunications companies to transmit telephone signals, Internet communication and cable television signals. It is also used in other industries, including medical, defense, government, industrial and commercial.

What is optical Fibre write two applications?

How total internal reflection affects our daily life?

The examples of total internal reflection in daily life are; : For diamond, the value of critical angle is 23°. When the ray of light incident at each face of diamond make an angle greater than the critical angle and suffer total internal reflection at various faces in diamond. That’s why diamond shines brilliantly.

## How do diamonds use total internal reflection?

Light cannot easily escape a diamond, because its critical angle with air is so small. Most reflections are total, and the facets are placed so that light can exit only in particular ways—thus concentrating the light and making the diamond sparkle.


text:
2014-06-17T23:08:51-04:00

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
2014-06-18T00:44:41-04:00

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
We have to add  and
Here we can see that the denominators are not same, to make the denominators same, we will find the L.C.M. of both the denominators -

Since both the denominators don't have any factor in common, so the L.C.M. will be equal to their product - 30 * 29 = 870

Now we will multiply both the numerators and denominators with a number, so that the denominators may become equal to 870.

First let's make the denominator of  o

Let's think of a number, it should when multiplied by 30, gives 870, clearly such number is 29. So, we multiply both the denominator and numerator of
with 29-

Similarly, we multiply the numerator and denominator of  by 30 because 30 * 29 = 870

So,

Now we add both the terms -


text:
# Einstein Summation Convention / Lorentz Boost

1. Dec 22, 2007

### raintrek

Einstein Summation Convention / Lorentz "Boost"

1. The problem statement, all variables and given/known data

I'm struggling to understand the Einstein Summation Convention - it's the first time I've used it. Would someone be able to explain it in the following context?

Lorentz transformations and rotations can be expressed in matrix notation as

$$x^{\mu'} = \Lambda^{\mu'}\!_{\mu}\:x^{\mu}$$

Coordinates are defined by $$x^{\mu}$$ with $$\mu = 0,1,2,3$$, such that $$(x^{0}, x^{1}, x^{2}, x^{3}) = (ct, x, y, z)$$

I'm seeking clarification on the meanings of the various $$\mu, \mu'$$ indices in the matrix notation equation. Any help would be massively appreciated!

2. Dec 22, 2007

### ophase

Summation rule

$$\mu, \acute{\mu}$$ means that these are all different indices which can have 0,1,2,3 and cannot contract with each other. I mean for example, while $$\mu = 0$$, the other one can be $$\acute{\mu} = 3$$.
Einstein summation rule is that you have to sum the terms with same indices. So you should scan all the values of the indices.
So
$$x^{\acute{\mu}} = \Lambda^{\acute{\mu}}\!_{\mu}\:x^{\mu}$$ means that

$$x^0 = \Lambda^{0}\!_{0}\:x^{0}+\Lambda^{0}\!_{1}\:x^{1}+\Lambda^{0}\!_{2}\:x^{2}+\Lambda^{0}\!_{3}\:x^{3}$$
$$x^1 = \Lambda^{1}\!_{0}\:x^{0}+\Lambda^{1}\!_{1}\:x^{1}+\Lambda^{1}\!_{2}\:x^{2}+\Lambda^{1}\!_{3}\:x^{3}$$
$$x^2 = \Lambda^{2}\!_{0}\:x^{0}+\Lambda^{2}\!_{1}\:x^{1}+\Lambda^{2}\!_{2}\:x^{2}+\Lambda^{2}\!_{3}\:x^{3}$$
$$x^3 = \Lambda^{3}\!_{0}\:x^{0}+\Lambda^{3}\!_{1}\:x^{1}+\Lambda^{3}\!_{2}\:x^{2}+\Lambda^{3}\!_{3}\:x^{3}$$

Don't forget that each term is actually a matrix element. So write the values of the matrix elements above and obtain the Lorentz transformation equations.

3. Dec 22, 2007

### raintrek

Ah, so am I correct in thinking that whenever an expression contains one index as a superscript and the same one as a subscript, a summation over those values is implied?

Also, does the $$\Lambda^{\mu'}\!_{\mu}$$ just mean a matrix with $$\mu'$$ representing the row number and $$\mu$$ representing the column number?

so, in essence,

$$\Lambda^{0'}\!_{0} \Lambda^{0'}\!_{1} \Lambda^{0'}\!_{2} \Lambda^{0'}\!_{3}$$
$$\Lambda^{1'}\!_{0} \Lambda^{1'}\!_{1} \Lambda^{1'}\!_{2} \Lambda^{1'}\!_{3}$$
$$\Lambda^{2'}\!_{0} \Lambda^{2'}\!_{1} \Lambda^{2'}\!_{2} \Lambda^{2'}\!_{3}$$
$$\Lambda^{3'}\!_{0} \Lambda^{3'}\!_{1} \Lambda^{3'}\!_{2} \Lambda^{3'}\!_{3}$$

is the $$\Lambda^{\mu'}\!_{\mu}$$ matrix produced from your equations?

Last edited: Dec 22, 2007
4. Dec 22, 2007

### ophase

Yes you'r right.
But
$$\Lambda^{\musingle-quote}\!_{\mu}$$ matrix cannot produce from above equation. We'r talking about a new representation of Lorentz transforms.

You can find the $$\Lambda^{\musingle-quote}\!_{\mu}$$ matrix here --> http://en.wikipedia.org/wiki/Lorentz_transformation

5. Dec 22, 2007

### siddharth

Yes.

6. Dec 22, 2007

### HallsofIvy

Staff Emeritus
More correctly $\Lambda^{\mu'}_\mu$ means a tensor that can, in a given coordinates system, be represented by such a matrix.


text:
# Talk:Euler's rotation theorem

WikiProject Mathematics (Rated C-class, Mid-priority)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
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This article is within the scope of WikiProject Citizendium Porting, a collaborative effort to improve Wikipedia articles by working in any useful content from their Citizendium counterparts. If you would like to participate, you can visit the project page, where you can join the project and see a list of open tasks.
 Current with its Citizendium counterpart article as of: Jun 30, 2009

## quaternions

Why is this article more about quaternions than it is about Euler rotation? One reference to quaternion algebra is adequate. The topic has been obfuscated by a writer who wants to steer readers. 65.175.151.8 15:08, 24 October 2006 (UTC)Tim Golden BandTechnology.com

they are intimately connected, same thing in different disguises. IMHO, the section on quaternions could use some elaboration and clarification. Mct mht 01:42, 8 February 2007 (UTC)

## unclear about number of parameters and quaternion significance

• Technically one needs only 3 numbers to represent an arbitrary rotation, since the rotation axis can be normalized, with its magnitude representing the angle (I suspect that this closer to what Euler's rotation theorem states)
yes, you're right. maybe that'll get corrected. Mct mht 01:42, 8 February 2007 (UTC)
• The article makes it sound as if any set of 4 numbers representing a rotation is called a quaternion. This is untrue, since a quaternion has a specific mathematical meaning, and is used to represent 3D rotation in a specific way.
right there also. current version can certainly be improved. rotations corresponds to unit quaternions, in a "1 to 2" way, meaning unit quaternions is a double cover of SO(3). Mct mht 01:42, 8 February 2007 (UTC)

## Show calculation of axis and angle of rotation

Should we show how to calculate the angle and axis of rotation? I have the formula here:

As a corollary of the theorem follows that the trace of a rotation matrix (R) is equal to:

$Trace[R] = 2 * cos(\theta) +1$

so s can be written as:

$s = \arccos{\left[ \frac{Trace(R) - 1}{2} \right]} = \arccos{\left[ \frac{r_{00} + r_{11} + r_{22} - 1}{2} \right]}$

Second, the unique axis of rotation is defined as:

$\vec{k} = \frac{1}{2 \sin{(\theta)}} * \left( \begin{matrix} r_{21}-r_{12} \\ r_{02}-r_{20} \\ r_{10}-r_{01} \end{matrix} \right)$

and the magnitude:

$|\vec{k}| = \frac{1}{2 \sin{(\theta)}} * \sqrt{(r_{21}-r_{12})^2 + (r_{02}-r_{20})^2 + (r_{10}-r_{01})^2}$

BUT I need a citation for these values...

## Excursion into matrix theory

The statement

An m×m matrix A has m orthogonal eigenvectors if and only if A is normal.

is false. The statement should read

An m×m matrix A can be diagonalised if and only if A is normal. An orthogonal set of m eigenvectors can be obtained if A is hermitian. The eigenvectors of a rotation matrix are not in general mutually orthogonal.

Really? "a matrix is normal if and only if its eigenspaces span Cn and are pairwise orthogonal with respect to the standard inner product of Cn" (Normal matrix#Consequences). Boris Tsirelson (talk) 10:33, 19 April 2010 (UTC)

## The matrix proof is wrong

The proof shows that a rotation matrix has an invariant vector. Period. It does not show that "in 3D space, any two Cartesian coordinate systems with a common origin are related by a rotation about some fixed axis". When saying "An algebraic proof starts from the fact that a rotation is a linear map in one-to-one correspondence withi a 3×3 rotation matrix R", you are already assuming what you want to show. That is, that the cartesian coordinates are related by a rotation about an axis. --190.188.3.11 (talk) 16:22, 23 June 2010 (UTC)

Two cartesian coordinates are related by a rotation matrix whose columns are the vectors of the second frame. The matrix has a real eigenvector v and is linear, therefore any vector k.v is also a eigenvector. This eigenspace is the rotation axis.--94.126.240.2 (talk) 07:36, 25 June 2010 (UTC)
You say that "Two cartesian coordinates are related by a rotation matrix whose columns are the vectors of the second frame." It seems this should be mentioned in the article, and proved too. Although it may seem obvious that an isometry S such that S (0) = 0 is linear, this has to be shown. So it seems that several things are being assumed because they are "obvious", making the proof not rigorous.--190.188.3.11 (talk) 04:12, 2 July 2010 (UTC)
I can't see any problems with it. The only thing it asserts is that the rotation can be represented by a matrix with the property
$\mathbf{R}^\mathrm{T}\mathbf{R} = \mathbf{R}\mathbf{R}^\mathrm{T} = \mathbf{E}$,
i.e. its inverse is its transpose. This is non-obvious but is easy enough to show by e.g. building the matrix up from rotations about three orthogonal axes. All rotations can be generated this way, otherwise gimbals would not work. Other than that the only thing established is that the determinant is +1 and not -1, but that follows from considering the volume-preserving nature of rotations. From these properties the unit root is found algebraically, and the existence of a fixed vector deduced from the theory of eigenvectors. --JohnBlackburnewordsdeeds 11:41, 2 July 2010 (UTC)
I do not have all these things very clear, but the article does not define what a rotation is. What is a rotation? The algebraic proof shows that a rotation matrix has an invariant vector; it does not show that two cartesian coordinates are related by a rotation. Reflextions don´t also preserve volume? As far as I know, that the determinant is 1 is shown using continuity arguments. The article as written does not seem to be correct.

--190.188.3.11 (talk) 20:24, 7 August 2010 (UTC)

The definition is in the first sentence: a rotation is "any displacement of a rigid body such that a point on the rigid body remains fixed". It also links to rotation group and rotation matrix for more algebraic treatments. Then in the first section it gives Euler's definition.
I think it gives it this general definition to make it clear that the definition does not assume there's a fixed axis: Euler's theorem is a statement that such an axis always exists, an a proof of it is a proof of the existence of such an axis, so it cannot be assumed. So a very general definition of rotation is needed. From this and from properties that you can assume (such as a rotation is always volume preserving, so the matrix has determinant 1) the theorem is proved.--JohnBlackburnewordsdeeds 20:55, 7 August 2010 (UTC)

## Incorrect rendering of alpha character in Safari browser

In the derivation that appears in the article, the Greek character alpha appears as one of the variables. Despite the fact that alpha is a common Greek character, it displays as an italicized "a" in the Safari browser, as I can confirm by copying the character and pasting it into Microsoft Word. The same process in Firefox gives the correct alpha character. Since both the letter a and alpha appear in the derivation, the derivation can be quite confusing when all alphas are replaced with the letter "a." This confusion can be minimized by replacing the letter alpha with the LaTex markup $\alpha$. It doesn't appear that this character is rendered as a PNG image, because the font size appears exactly the same as the other characters. In the Manual of Style (mathematics), it specifies that "if you enter a very simple formula using the mathematical notations like $L^p$ this will (in the default used by most users) not be displayed using a PNG image but using HTML, like this: Lp." Jdlawlis (talk) 04:49, 10 May 2011 (UTC)

It appears as a letter alpha to me. Here is a compact paragraph it appears in, copied in full:
Now Euler needs to construct point O in the surface of the sphere that is in the same position in reference to the arcs aA and αA. If such a point exists then:
The 'α' looks quite different from the 'a' in both the edit window and when rendered. It's worth remembering that many things control how things are rendered. The choice of browser (i.e. browser, OS/Platform and browser version), browser and OS settings, what fonts you have, the skin specified at Special:Preferences#preftab-1 and the Math formatting options on the same page. The latter is the one that varies the appearance of maths the most: it can mean that almost all [itex] tags are rendered as text or none are. It can even disable their interpretation altogether.
What this means is you can't make assumptions about how it will appear on other users machines from your own. In particular if you see a problem in an article, but the markup seems perfectly OK (and α, a Greek letter, is far from exotic or difficult for browsers to render) and is like many other articles, then it is more likely your setup needs adjusting. It is also why it says in the section of the MOS you highlight:
"Either form is acceptable, but do not change one form to the other in other people's writing."
--JohnBlackburnewordsdeeds 13:34, 10 May 2011 (UTC)

Fair enough. I'm only trying to highlight something that I find confusing. I don't think that I'm the only person who will run into this issue, but you have made your position clear. Jdlawlis (talk) 15:26, 10 May 2011 (UTC)

I feel this is an important theorem that can be presented in a somewhat simpler way. I will developed this in the near future and hope that you find it to be of value.Prof McCarthy (talk) 01:17, 23 June 2011 (UTC)

## Demonstration with geometric algebra.

I agree with the observations made by 190.188.3.11 in " The matrix proof is wrong ". It is at least incomplete, that is one should indeed prove that any two cartesian coordinate systems (with same orientation) are related by a rotation.

I will try to prove that with " geometric algebra " (I know, some people here will make a grimace ... ).

Let us consider a unit sphere which can slide on itself. After some displacement the extremities of an $(e_1,e_2,e_3)$ reference frame will be situated at $(f_1,f_2,f_3)$. We must be more precise : that means we define arbitrarily two new points $(f_1,f_2)$ with the only condition $\mathbf(f_1). \mathbf(f_2) =0$, which assures that we respect conservation of distances and angles. Of course $f_3$ is determined by ${f_1} \times {f_2} = {f_3}$.

Each point of the sphere can of course be defined by its (invariant) coordinates relative to both frames, by definition of the displacement of a rigid body.

We know that a rotation in $(R^3)$ can be characterized by a rotor like :

$(1) \qquad R=\cos(\theta /2)+ I a \sin(\theta /2) \qquad \qquad R^{\dagger}=\cos(\theta /2)- I a \sin(\theta /2)$

where $a$ is the unit vector of the rotation axis, and $\theta$ the rotation angle.

What we have to do now is to verify that the knowledge of the couples $(e_1,f_1)$ and $(e_2,f_2)$ is necessary and sufficient to determine uniquely the rotor and thus the axis and rotation angle.

We must have :

$(2) \qquad f_1=R^{\dagger} e_1 R \qquad \qquad f_2=R^{\dagger} e_2 R \qquad \qquad f_1.f_2=0$

We can write successively :

$(3) \qquad R f_1=e_1 R$

$(4) \qquad (\cos(\theta /2)+ I a \sin(\theta /2))f_1=e_1(\cos(\theta /2)+ I a \sin(\theta /2))$

$(5) \qquad \cos(\theta /2)f_1+\sin(\theta /2) I (a.f_1+a \wedge f_1)=\cos(\theta /2)e_1+\sin(\theta /2) I (a.e_1-a \wedge e_1)$

$(6) \qquad \cos(\theta /2)(f_1-e_1)+ \sin(\theta /2)I a \wedge (f_1+e_1)+\sin(\theta /2)I a . (f_1-e_1)=0$

and :

$(7) \qquad \cos(\theta /2)(f_2-e_2)+ \sin(\theta /2)I a \wedge (f_2+e_2)+\sin(\theta /2)I a . (f_2-e_2)=0$

By splitting in pseudoscalar and vectorial parts we get:

$(6 bis) \qquad a.(f_1-e_1)=0 \qquad \qquad \cos(\theta /2)(f_1-e_1)+\sin(\theta /2)I a \wedge (f_1+e_1)=0$

$(7 bis) \qquad a.(f_2-e_2)=0 \qquad \qquad \cos(\theta /2)(f_2-e_2)+\sin(\theta /2)I a \wedge (f_2+e_2)=0$

As a consequence of the scalar relations we can write :

$(8)\qquad a=\lambda I (f_1-e_1)\wedge(f_2-e_2)$

where $\lambda$ is the inverse of the norm of $(f_1-e_1)\wedge(f_2-e_2)$.

We may calculate : $\lambda=[(f_1-e_1)^2(f_2-e_2)^2-(f_1-e_1).(f_2-e_2)]^{(-1/2)}$

As for the vectoriel relations, they should be coherent. After replacing $a$ by its value and simplifying in (6bis) and (7bis) we get :

$(6ter) \qquad \cos(\theta /2)-\lambda \sin(\theta /2)(f_2-e_2).(f_1+e_1)=0$

$(7ter) \qquad \cos(\theta /2)+\lambda \sin(\theta /2)(f_1-e_1).(f_2+e_2)=0$

These equations coincide if :

$(9)\qquad -(f_2-e_2).(f_1+e_1)=(f_1-e_1).(f_2+e_2)$

That is true if and only if the condition $\mathbf (f_1) . \mathbf (f_2) =0$ is verified.

Thus we get:

$(10) \qquad \cot(\theta /2)=\lambda (f_2-e_2).(f_1+e_1)=\lambda (f_2 .e_1-f_1 .e_2)$

and the rotation is unambiguously determined.

Obviously the same demonstration can be made starting with non-orthogonal points and the condition $f_1.f_2=e_1.e_2$ .

Chessfan (talk) 20:02, 20 January 2012 (UTC)

## Euler pole

As far as I know, the single rotation axis is typically called "Euler axis". I guess that the "Euler pole" is not the Euler axis, but just the fixed point around which the body rotates. See this image, for instance: http://www.earth.northwestern.edu/people/seth/202/new_2004/euler_pole.html Paolo.dL (talk) 16:30, 2 May 2013 (UTC)


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## Convex subsets of Hadamard manifolds

Given a subset $S$ of a Hadamard manifold $M$. Is there a curvature criterion (for $\partial S$) to decide whether $S$ is convex.

I am looking for a ganeralization of the following statement:

A connected subset of $\mathbb{R}^2$ bounded by a smooth curve $\gamma$ with $||\dot{\gamma}||=1$ is convex, if $\gamma$ "always turns in the same direction" - more formally if $t\mapsto det(\dot{\gamma}(t),\ddot{\gamma}(t))$ never changes sign.

-
I believe that the appropriate generalization is a positive semidefinite second fundamental form. – Deane Yang Jan 10 2011 at 14:33
@Henrik: Can you specify what you consider a convex subset of a Riemannian manifold? – Dror Atariah Jan 10 2011 at 14:48
convex for me means geodesically convex. is there another reasonable meaning? – Deane Yang Jan 10 2011 at 15:02
In Hadamard manifolds there is a unique geodesic between any two points so convexity can only mean geodesic convexity. – Igor Belegradek Jan 10 2011 at 15:08
Thank you. After some googling I found the paper "Locally convex hypersurfaces of negatively curved spaces" by S. Alexander (ams.org/journals/proc/1977-064-02/…), which adresses precisely this question. – HenrikRüping Jan 10 2011 at 15:48
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Problem of the Month
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Solution for April 2011

 The Problem: Does there exist a continuous real-valued function $f: \mathbb{ R}\rightarrow \mathbb{ R}\;$ that takes each of its values exactly twice? Does there exist a continuous real-valued function $f: \mathbb{ R}\rightarrow \mathbb{ R}\;$ that takes each of its values exactly three times? Answer. No for part (a); Yes for part (b).

Correct responses:
Correct solutions to the March problem were submitted by
 Lou Cairoli (USA) Mei-Hui Fang (Austria) Philippe Fondanaiche (France) Gruian Cornel (Romania) Tom Holens (Manitoba) Lars Huettenberger (Germany) Benoît Humbert (France) Ile Ilijevski (Macedonia) Kipp Johnson (USA) Omran Kouba (Syria) Antek Łączkowski (Poland) Matthew Lim (USA) Patrick J. LoPresti (USA) Fatemeh Naghipour (Regina) Paolo Perfetti (Italy) Shpetim Rexhepi (Macedonia) John T. Robinson (USA) Ignacio Somma Esteves Albert Stadler (Switzerland)

The solution:

Solution to part (a).

Preliminary comments. Our argument is based on two fundamental theorems about the real line:

The Intermediate Value Theorem. If $f(x)$ is a real-valued continuous function for $a \le x \le b$, and $r$ is any number between $f(a)$ and $f(b)$, then there is a number $c$ between $a$ and $b$ such that $f(c) = r$. (This theorem captures the intuitive notion that you can draw a continuous function without lifting your pencil off the paper.)

The Extreme Value Theorem. If $f(x)$ is a real-valued continuous function for $a \le x \le b$, then it must attain its maximum and minimum values on that interval. (In other words, there exist $c$ and $d$ in the in the closed interval from $a$ to $b$ such that $f(c) \ge f(x) \ge f(d)$ for all $x$ with $a \le x \le b$.)

For many centuries mathematicians used these theorems tacitly, thinking that they were obvious if they thought about them at all. It was only in the nineteenth century that they were carefully formulated and became part of the foundations of analysis. On the other hand, the properties do have intuitive appeal and should cause the reader no difficulty even if he has never seen them before. For further details see the relevant Wikipedia pages, or any calculus or analysis textbook. Perfetti spotted our problem as number $1.3.22$ on page 17 of Problems in Mathematical Analysis II, Continuity and erentiation, by W.J. Kaczor and M.T. Nowak, published by the American Mathematical Society (2001). All the submitted proofs were quite similar; our argument has been taken mostly from Patrick LoPresti, with additional ideas from the other submissions.

Nonexistence proof.
Suppose to the contrary that a function $f(x)$ were to exist that takes every value in its range exactly twice. Choose some number $r$ in its range; for convenience, let us take $r=0$. (The function $g(x) =f(x)-r$ would take each of its values exactly twice if and only if $f(x)$ does, so we could have started with $g$.) By hypothesis, then, $f(x)=0$ for exactly two values of $x$, say $a$ and $b$; that is, $f(a) = f(b) = 0$ and f is nonzero elsewhere.

Consider the behavior of $f(x)$ when $a<x<b$. Observe that $f(x)$ must be either entirely positive or entirely negative on this open interval, for if it were positive somewhere and negative somewhere else between $x=a$ and $x=b$, then by the Intermediate Value Theorem it would be zero somewhere in between, and that would be a third zero of $f(x)$, contrary to our hypothesis. Without loss of generality, suppose $f(x)$ is positive between $a$ and $b$. (Should it be negative, simply replace it by $-f(x)$.) Now, by the Extreme Value Theorem, f(x) attains its maximum value, call it $M$: $f(c) = M$ for some $c$ for which $a \le c \le b$. In fact, because $f(x)$ is zero at $a$ and $b$ and positive in between, $M$ must be positive, and thus $a < c < b$. By hypothesis, there is exactly one other value of $x$, say $x=d \ne c$, such that $f(x) = M$. Where can $d$ possibly be?

Suppose $d > b$. Then we have
$$a < c < b < d, \quad f(a) = 0, f(c) = M, f(b) =0, \mbox{ and } f(d) = M.$$
But by the Intermediate Value Theorem, $f(x)$ would necessarily assume every value from $0$ to $M$ as $x$ runs from $a$ to $c$, and again as $x$ runs from $c$ to $b$, and a third time for $x$ between $b$ and $d$. That is, $f(x)$ would necessarily assume each of its values at least three times, which is a contradiction. So $d$ cannot be greater than $b$. By similar reasoning, $d$ cannot be less than $a$. Certainly $d$ cannot equal $a$ or $b$ (because $f(d) = M$ is positive and $f(a) = f(b) = 0$). That leaves only the possibility that $d$ lies between $a$ and $b$. Suppose we have labeled them so that $c < d$. By definition, $M$ is the maximum value of $f(x)$ for $x$ between $a$ and $b$; in particular, between $x=c$ and $x=d$, $f(x) \le M$. But nowhere in that interval can $f(x)$ be equal to $M$ since $f(x)$ takes each of its values only twice. So $f(x)$ must be less than $M$ between $x=c$ and $x=d$. But that is likewise impossible: again by the Intermediate Value Theorem, $f(x)$ already takes on every positive value less than $M$ when $x$ is between $a$ and $c$, and a second time when $x$ is between $d$ and $b$, whence, by hypothesis, it cannot take any value less than $M$ for a third time. That eliminates all possible locations for $d$, and we conclude that it is not possible for $f(x)$ to take each of its values exactly twice.

Comment. Fang, Fondanaiche, Humbert, and LoPresti observed that, more generally, there can exist no continuous function that takes each of its values exactly $n$ times, if $n$ is an even number. A careful modification of our argument above will prove their claim.

Solution to part (b). Examples of continuous functions that take each of its values exactly times came in three varieties:

Type 1.

 For each integer $k$ define $$\small f(x) = \left\{ \begin{array}{lcc} 3(x-k) + k & 0 \le x-k \le \frac13 \\ -3(x-k)+2+k & \frac13 \le x-k \le \frac23 \\ 3(x-k) -2+k & \frac23 \le x-k \le 1 \end{array} \right ..$$

Figure 1: f(x)

Figure 1 is a slight modification of the GeoGebra file sent to us by Benoît Humbert. GeoGebra, as the name implies, combines graphics with algebra. The software can be downloaded for free, but it might take a while to figure out how to use it.

Type 2.

Figure 2: $g(x) = \frac2{3\pi}x + \sin x +\frac1{3\pi}\sin 2\pi$

Figure 2 is a copy of the WolframAlpha web page produced by Patrick LoPresti, which he obtained by typing

graph[2/(3pi)*x + sin(x) + 1/(3pi)*sin(2x)]

into the space provided on the page www.wolframalpha.com. You can explore the resulting figure there by moving your mouse around, and you can modify the equation at will.

Type 3.

Figure 3: For each integer $k$ define
$h(x) = (2k+1)- (-1)^k\cos x$, for $3k\pi \le x \le 3(k+1)\pi$.

Each of the three pictures clearly shows a function that takes every real value exactly three times: Note that the relative extreme points of each curve are evenly spaced; consider a portion of the curve (such as the N-shaped path shown in red in Figure 1 from $(0, 0)$ to $(1,1)$) that begins half-way between a relative minimum and a relative maximum, includes two extreme points, then ends half-way to the next relative maximum. We can think of it as a fundamental unit that is repeated throughout the graph. That portion of each graph takes on each $y$-value except the first and last exactly three times; the initial $y$-value equals the the relative minimum on the right and the relative maximum of the previous fundamental unit, while the final $y$-value equals the maximum on the left and the minimum of the next fundamental unit. This intuitive discussion is easily confirmed by the functions that accompany the first and third figures (labeled $f$ and $h$). The second formula takes a bit more work to verify: You must show that the derivative of $g(x)$ is zero at all odd multiples of $\frac\pi2$, and that
$$g\left(\frac{(4k+1)\pi}2\right) = g\left(\frac{(4k+7)\pi}2\right)=\frac43(k+1) \mbox{ for all integers } k.$$

Further comments. It should be clear, more generally, that there exists a continuous real-valued function that takes every value of its range exactly $n$ times if and only if $n$ is odd. Indeed, for any odd natural number $n$ you can easily modify any of the three figures to produce continuous functions that take every real value exactly $n$ times — just draw $n-1$ points in each fundamental unit. In terms of algebra, the equation for $h(x)$ becomes
$$h_n(x) = (2k+1)- (-1)^k\cos x, \; \mbox{ for }\; nk\pi \le x \le n(k+1)\pi.$$
The case where $n=5$ is shown in Figure 4. Compare that curve with the general version of the function $g_n(x) = \frac2{n\pi}x + \sin x +\frac1{n\pi}\sin 2x$ shown in Figure 5 for $n=5$.

Figure 4: $h_5(x)$ takes every real value exactly five times.

Figure 5: $g_5(x) = \frac2{5\pi}x + \sin x +\frac1{5\pi}\sin 2x$ takes every real value exactly five times.

It should likewise be clear that any open interval, finite or not, could serve as the range of a continuous function that takes each of its values exactly $n$ times when $n$ is odd: simply shrink the amplitude of each fundamental unit. When $n=3$ for example, to shrink the range of $h(x)$ to $-1< y < 1$, on each interval $3k\pi \le x \le 3(k+1)\pi$ for $k = 0, 1, 2, \dots$, define $h(x) = \frac{2^k-1}{2^k} + \frac1{2^{k+2}}(1- (-1)^k\cos x)$; extend the curve to the left by defining $h(-x) = -h(x)$. The right half of the function is shown in Figure 6.

Figure 6: The range of $y=h(x)$ for $x\ge 0$ is $0 \le y < 1$.

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