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Mathematics Department Colloquium: Recent advances on hearing the shape of a drum'

Colloquium | October 31 | 4:10-5 p.m. | 60 Evans Hall

Steve Zelditch, Northwestern University

Department of Mathematics

In 1966, M. Kac posed the problem, Can one hear the shape of a drum', i.e. can one determine a bounded plane domain from its Dirichlet (or, Neumann) eigenvalues? He proved that indeed one can determine a disk from its spectrum. Disks remained the only domains known to be determined by their eigenvalues until recently. My talk concerns a recent result due to Hamid Hezari and myself, showing that ellipses of small eccentricity are also determined by their eigenvalues. The proof uses some recent advances in dynamical inverse theory due to Avila, de Simoi, Kaloshin, Sorrentino, Wei and others which partly resolve another famous inverse problem of G.D. Birkhoff: are ellipses the only billiard tables with integrable billiards? The solution of the inverse eigenvalue problem is based on their result that ellipses of small eccentricity are the only nearly circular billiard tables with rationally integrable billiards. My talk will review both the eigenvalue and dynamical inverse spectral results.

holtz@math.berkeley.edu


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# Question 30 (4 points) Capital Profit Loss Partners Balances Sharing Scotty I$400,000 25% Riley ... Question 30 (4 points) Capital Profit Loss Partners Balances Sharing Scotty I$ 400,000 25% Riley I$700,000 45% Dee Dee 900,000 30% Marnee is going to buy into the SRD Partnership by paying$1,000,000 to the partners in exchange for a 30% ownership in the partnership. The partnership uses the Book Value Method when admitting new partners. Answer the following questions using the information in the table above. 1. What is Marnee's balance after joining the partnership? 2. What is Scotty's's balance after Marnee joins the partnership? abo

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# Sensitivity Analysis with Effective Quadratures

In this blog post, we explore one of the applications of Effective Quadratures—sensitivity analysis. The goal of sensitivity analysis is summarized by the question: Which of my input parameters is the most important? Given a computational model, we want to create a ranking of the various input parameters in terms of their influence on the output—how sensitive the output is to changes in a certain parameter.

So why do we want to know about the importance of variables? Consider a model for the cycle time of a spring-loaded piston from Kenett et al. [1], consisting of seven input parameters:

1. M: The piston mass
2. S: The piston area
3. V_0: Initial gas volume
4. k: Spring constant
5. P_0: Atmospheric pressure
6. T_a: Ambient temperature
7. T_0: Filling gas temperature

Sensitivity analysis allows us to make plots such as the one below:

Hence, instead of a seven parameter model, we could simply build a two-parameter model with S and V_0, keeping others constant. Identifying important parameters enables the simplification of models that may have hundreds of parameters down to a handful.

There are multiple methods to quantify the importance of a variable, one of which is through Sobol’ indices. The idea behind Sobol’ indices is simple: If we treat the input parameters as random variables, carrying a degree of uncertainty, running them through the model yields some uncertainty in the output, which can be calculated through the output variance. The idea is to find out how much of this output variance is caused by the variation of each individual variable—or potentially the interaction between multiple variables.

In the seminal paper by I. M. Sobol [2]—who came up with the indices—he expressed the index via integrals. In general, these integrals cannot be evaluated analytically for computational models. A simple approach to approximating these integrals is Monte Carlo. For example, if we are approximating the mean of the function f(x):

\mu = \int f(x) \rho(x) dx

where \rho(x) is the probability density function of the input, we can use the following recipe:

1. Sample N evaluation points x_1, x_2, \dots ,x_N randomly in the input domain according to \rho(x).
2. Evaluate the function at these points: f(x_1), f(x_2), \dots , f(x_N).
3. Sum up the function evaluations:
\hat{\mu} = \sum_{i=1}^N f(x_i)

Then, it can be shown that as N tends to infinity, the Monte Carlo estimate approaches the true value. The evaluation of a Sobol’ index is more involved, requiring two function evaluations per Monte Carlo sample, but the principle is the same. Let’s examine the convergence of this method on computing the Sobol’ index for S for the piston model.

The estimate is evaluated with different number of data points used for 30 trials. As the number of function evaluations increases, the fluctuation around the true value decreases. However, even with 2500 evaluations, the variation around the mean is still rather large. Is there a better way to do this?

Instead of using the function evaluations for computing the integral directly, Sudret et al. [3] showed that it is possible to leverage polynomial approximations for the computation of Sobol’ indices. Provided that the polynomials satisfy orthogonality, the Sobol’ indices can be read off from the coefficients of the expansion. Refer to the tutorials for more details on polynomials and the capabilities within EQ to work with them.

Using this approach, the computation of Sobol’ indices now boil down to the estimation of polynomial coefficients—i.e. finding a polynomial fit for the function. One way to do this is through least squares, where point evaluations are used as data to find the best fit coefficients. For a given number of evaluations of the function, how well does this compare with the Monte Carlo approach?

The fluctuation around the converged value is reduced drastically, as seen in the figure above. This example demonstrates the power of polynomial approximations: function evaluations are more well-spent if used on a suitable polynomial approximation than on brute force approaches.

Can we do better? Experienced users may know that a lower limit is placed on the least squares approach: we need as many function evaluations as the number of polynomial basis terms; and as the number of input variables to the function increases, the number of basis terms scales exponentially. In EQ, there are various methods for circumventing this curse of dimensionality , one of which is compressed sensing. Another approach is to perform regression on a dimension-reduced subspace. In our recent paper [4], we explored the use of the latter technique to compute sensitivity indices.

The utility of our tools certainly does not end here. For more information, please refer to the tutorials and more blog posts to come.

References:

[1] R. Kenett, S. Zacks, and D. Amberti. Modern Industrial Statistics: with applications in R, MINITAB and JMP . John Wiley & Sons, 2013.

[2] I. M. Sobol’. Global sensitivity indices for nonlinear mathematical models and their Monte Carlo estimates. Mathematics and Computers in Simulation , 55(1):271 – 280, 2001.

[3] B. Sudret. Global sensitivity analysis using polynomial chaos expansions. Reliability Engineering & System Safety , 93(7):964–979, 2008.

[4] C. Y. Wong, P. Seshadri, and G. T. Parks. Extremum Sensitivity Analysis with Least Squares Polynomials and their Ridges. arXiv: 1907.08113 [math], (2020), [1907.08113] Extremum Sensitivity Analysis with Polynomial Monte Carlo Filtering (Preprint submitted to Reliability Engineering & System Safety )

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Looking good @Nick!

This is terrific @Nick! Building upon this, a valid query would be what to do if some, or indeed all of the inputs were correlated. Before we embark on this, it will be useful to understand a key limitation in the existing polynomial chaos (although there is nothing chaotic about using polynomials for approximation) workflow. To ease the exposition, initially, let’s assume a bivariate model f \approx f\left(s_1, s_2 \right), where s_1 and s_2 are the two model parameters that have marginals given by \omega_1(s_1) and \omega_2(s_2) that have support \mathcal{D}_1 and \mathcal{D}_2 respectively

The standard polynomial approximation we seek is of the form

f \left(s_1, s_2 \right) \approx \sum_{i} x_i p_i \left( s_1, s_2 \right)

where p_i is the i-th bivariate orthogonal polynomial and x_i is its corresponding coefficient. The latter is given by

x_i = \int_{\mathcal{D}_1} \int_{\mathcal{D}_2} f \left(s_1, s_2 \right) p_i \left( s_1, s_2 \right) \varrho \left(s_1, s_2 \right) ds_1 ds_2,

where we explicitly assume that \varrho \left(s_1, s_2 \right) = \omega_1(s_1) \omega_2(s_2), i.e., s_1 and s_2 are independent and thus their joint distribution is a product of its marginals.

Now, when s_1 and s_2 are correlated, one cannot express the joint distribution as a product of the marginals. To deal with this issue and still leverage polynomial approximations to compute moments and sensitivities, we require an alternate strategy. Enter the Nataf transform, implemented here in the code.

Assume that both \omega_1(s_1) and \omega(s_2) are zero mean Gaussians with a variance of unity. In Figure 1, on the left we have that s_1 and s_2 are correlated, while on the right set them to be independent.

Now, in the standard polynomial approximation workflow, we evaluate our model f at points over the domain shown in the Figure on the right. Assuming we adopt a sparse grid, we might end up with the sampling shown in the right subfigure in Figure 2. Using the same sampling strategy on the subfigure on the left seems incorrect (and it is for the reason mentioned above).

What we would intuitively expect is for our sparse grid to somehow exploit the underlying correlation structure in the joint distribution and have points that look akin to those shown in the left subfigure of Figure 3.

This transformation (or map) is referred to as the Nataf transform. Succinctly stated, we construct our polynomials over the space of independent standard Gaussian distributions (right), but evaluate our model at the transformed coordinates (left). The code below shows this in action!

from equadratures import *
import numpy as np

# A model function
def function(s):
s1 = s[0]
s2 = s[1]
return s1**2 - s2**3 - 2. * s1 - 0.5 * s2

# The standard setup.
param_s1 = Parameter(distribution='Gaussian', shape_parameter_A=0., shape_parameter_B=1., order=3)
param_s2 = Parameter(distribution='Gaussian', shape_parameter_A=0., shape_parameter_B=1., order=3)
R = np.array([ [ 1.0,  0.8],
[ 0.8,  1.0]])
basis = Basis('sparse-grid', level=3, growth_rule='exponential')
poly = Poly([param_s1, param_s2], basis, method='numerical-integration')

# Now feed this polynomial to the Correlations class along with the correlation matrix.
corr = Correlations(poly, R)
corr.set_model(function)
poly2 = corr.get_transformed_poly()

Now we can use this transformed polynomial to compute statistics.

print(poly2.get_mean_and_variance())
>> (0.9999999999999987, 35.449838615261825)

These values can be verified by numpy

S = np.random.multivariate_normal( mean=[0., 0.], cov=R, size=1000000)
model_evals = evaluate_model(S, function)
print(np.mean(model_evals), np.var(model_evals))
1 Like


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# Typesetting mathematics in e-books¶

The calibre E-book viewer has the ability to display mathematics embedded in e-books (EPUB and HTML files). You can typeset the mathematics directly with TeX or MathML or AsciiMath. The calibre E-book viewer uses the excellent MathJax library to do this. This is a brief tutorial on creating e-books with mathematics in them that work well with the calibre E-book viewer.

## Lihtne matemaatikat sisaldav HTML-fail¶

You can write mathematics inline inside a simple HTML file and the calibre E-book viewer will render it into properly typeset mathematics. In the example below, we use TeX notation for mathematics. You will see that you can use normal TeX commands, with the small caveat that ampersands and less than and greater than signs have to be written as & < and > respectively.

The first step is to tell calibre that this will contains mathematics. You do this by adding the following snippet of code to the  section of the HTML file:

That’s it, now you can type mathematics just as you would in a .tex file. For example, here are Lorentz’s equations:

The Lorenz Equations

\begin{align}
\dot{x} & = \sigma(y-x) \\
\dot{y} & = \rho x - y - xz \\
\dot{z} & = -\beta z + xy
\end{align}

This snippet looks like the following screen shot in the calibre E-book viewer.

The complete HTML file, with more equations and inline mathematics is reproduced below. You can convert this HTML file to EPUB in calibre to end up with an e-book you can distribute easily to other people.

Siin on loodud EPUB-fail: mathjax.epub.

<!DOCTYPE html>

<title>Math Test Page</title>
<meta http-equiv="content-type" content="text/html; charset=UTF-8" />

Sample Equations

The Lorenz Equations

\begin{align}
\dot{x} & = \sigma(y-x) \label{lorenz}\\
\dot{y} & = \rho x - y - xz \\
\dot{z} & = -\beta z + xy
\end{align}

The Cauchy-Schwarz Inequality

$\left( \sum_{k=1}^n a_k b_k \right)^{\!\!2} \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$

A Cross Product Formula

$\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \\ \end{vmatrix}$

The probability of getting $$k$$ heads when flipping $$n$$ coins is:

$P(E) = {n \choose k} p^k (1-p)^{ n-k}$

An Identity of Ramanujan

$\frac{1}{(\sqrt{\phi \sqrt{5}}-\phi) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } }$

A Rogers-Ramanujan Identity

$1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots = \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for |q|<1}.$

Maxwell's Equations

\begin{align}
\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\
\nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\
\nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\
\nabla \cdot \vec{\mathbf{B}} & = 0
\end{align}

In-line Mathematics

While display equations look good for a page of samples, the
ability to mix math and text in a paragraph is also important.  This
expression $$\sqrt{3x-1}+(1+x)^2$$ is an example of an inline equation.  As
you see, equations can be used this way as well, without unduly
disturbing the spacing between lines.

References to equations

Here is a reference to the Lorenz Equations (\ref{lorenz}). Clicking on the equation number will take you back to the equation.

## Rohkem teavet¶

Since the calibre E-book viewer uses the MathJax library to render mathematics, the best place to find out more about mathematics in e-books and get help is the MathJax website.


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### A quantized-diffusion model for rendering translucent materials

##### Content of review 1, reviewed on June 08, 2014

On page 4 the footnote incorrectly states that equation (5) is not absolutely convergent. Equation (5) provides the convergent form that separates the ballistic fluence from the scattered fluence and does converge. It is the alternative form $$\phi(r) = \frac{\mu_t^2}{2 \pi^2 r} \int_0^\infty \frac{\text{arctan}\, u}{\mu_t-\mu_s \, \frac{\text{arctan} \, u}{u}} \, \sin (r \, \mu_t \, u) \, du.$$ that does not separate the ballistic fluence explicitly and only converges in the sense of Cesaro summability.


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시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
10 초 128 MB 0 0 0 0.000%

## 문제

I don’t know how you feel about your school time, but I become maudlin when I remember those days. One of our teachers told us at the final year that among all the educational institutions in life one misses his/her school days the most. And yes, I miss those days a lot.

Let me tell you that we had a grid board in our school and it was not as banal as it looks in the picture. The board was colorful and we also had different color chalks to use on it. Can you imagine how exciting it was for me when I first saw this board? In class break we used to draw on this board and play different games, few of them I can recall.

One of them was like this: firstly two players will mark some grid points. Then they will toss deciding who plays first. At each turn a player marks four points as A, B, C and D. Then join (A, B), (B, C), (C, D) and (D, A) to form a quadrilateral. Twice of the area of that quadrilateral is added to his score and the turn changes (in case, if you are wondering why twice, it is just to ensure that the score is always integer). A player can not draw a quadrilateral if it was drawn before. However, you can use previously used points. For example, suppose there are 5 points on the grid, P , Q, R, S and T. First player can choose, (A, B, C, D) = (P, Q, R, S), but then the second player can not choose (A, B, C, D) = (R, S, P, Q) because both of them depicts the same quadrilateral. If both of the players play optimally to maximize their own score I wonder what could be the sum of their scores.

So your task is to construe this game. You are given co-ordinates of N distinct points, if two players play the above mentioned game optimally then what is the sum of their scores?

## 입력

First line of the input data contains a positive integer T denoting the number of test cases (T ≤ 50).

For each case, first line contains a positive integer N (N ≤ 700). Hence follows N co-ordinates of the points (x, y). In case you don’t know, I should say: I am not from your time, I was brought here by a few scientists from future. And in my time we use huge boards so the absolute value of the co-ordinates can be as large as 106. Just to ensure that no one can draw a degenerate quadrilateral, no three points will be collinear.

## 출력

For each test case output the case number followed by sum of the scores. Since this score can be huge just print the answer mod 1000003.

## 예제 입력 1

2

4
2 0
0 2
-2 0
0 -2

5
2 0
0 2
-2 0
0 -2
2 2

## 예제 출력 1

Case 1: 16
Case 2: 72


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## Java Dependency Vulnerability scanning with Maven victims-enforcer

One of the OWASP guidelines for secure applications is to not use components with known vulnerabilities. Unfortunately it can be a very difficult and time consuming task to keep up with these manually, automation can save you countless hours!

NOTE: victims-enforcer can be used in conjunction with the OWASP dependency scanner. I have only found it to be problematic in ‘tycho’ builds.

 <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-enforcer-plugin</artifactId> <version>1.4.1</version> <dependencies> <dependency> <groupId>com.redhat.victims</groupId> <artifactId>enforce-victims-rule</artifactId> <version>1.3.4</version> <type>jar</type> </dependency> </dependencies> <executions> <execution> <id>enforce-victims-rule</id> <goals> <goal>enforce</goal> </goals> <configuration> <rules> <rule implementation="com.redhat.victims.VictimsRule"> <!-- Check the project's dependencies against the database using name and version. The default mode for this is 'warning'.

 Valid options are: disabled: Rule is still run but only INFO level messages and no errors. warning : Rule will spit out a warning message but doesn't result in a failure. fatal : Rule will spit out an error message and fail the build. --> <metadata>warning</metadata> <!-- Check the project's dependencies against the database using the SHA-512 checksum of the artifact. The default is fatal. Valid options are: disabled: Rule is still run but only INFO level messages and no errors. warning : Rule will spit out a warning message but doesn't result in a failure. fatal : Rule will spit out an error message and fail the build. --> <fingerprint>fatal</fingerprint> <!-- Disables the synchronization mechanism. By default the rule will attempt to update the database for each build. Valid options are: auto : Automatically update the database entries on each build. daily : Update the database entries once per day. weekly: Update the database entries once per week. offline : Disable the synchronization mechanism. --> <updates>daily</updates><!-- was: auto -->

 </rule> </rules> </configuration> </execution> </executions> </plugin>
Vulnerability database is sourced from: https://victi.ms with backing from RedHat.

REFERENCES:

## OWASP Dependency Vulnerability Scanning of Java JARs with Maven

One of the OWASP guidelines for secure applications is to not use components with known vulnerabilities. Unfortunately it can be a very difficult and time consuming task to keep up with these manually, automation can save you countless hours!

NOTE: OWASP dependency scanner can be used in conjunction with the victims-enforcer.

 <plugin> <groupId>org.owasp</groupId> <artifactId>dependency-check-maven</artifactId> <version>1.3.4</version> <executions> <execution> <goals> <goal>check</goal> </goals> </execution> </executions> </plugin>

Each time you build, the plug-in will verify the assets against the list of known vulnerable libraries and report them in your output.

Vulnerability database is populated from: https://nvd.nist.gov.

NOTES:

1. The example above is a very simple implementation, see the documentation for additional functions.
2. The first use of the plug-in can take a long time as the vulnerability library must be installed locally before initial use.
3. Similar functionality is available for Ant builds, if desired.

REFERENCES:

## Code signing of java applets – using Maven

To sign your java assets during the maven build process, you can add the following to the pom.xml to make use of the values we established in the keystore creation step.

WARNING: for security and maintainability purposes, you should define the ‘configuration’ items in your local ‘settings.xml’ file instead of in the pom.xml as is done here for example only!

 <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-jarsigner-plugin</artifactId> <version>1.4</version> <executions> <execution> <id>sign</id> <goals> <goal>sign</goal> </goals> </execution> <execution> <id>verify</id> <goals> <goal>verify</goal> </goals> </execution> </executions> <configuration> <alias>selfsigned</alias><!-- ${project.name} --> <keystore>selfsignkeys.store</keystore><!-- NOTE: you can also specify an absolute path here --> <storepass>123456</storepass> <keypass>123456</keypass> </configuration> </plugin>  REFERENCES: ## Firefox 41+ extension signing In the never-ending quest for browser security, Firefox has started implementing safeguards to only allow signed extensions. I found this out after upgrading to Firefox 41 as my installed version of “Deque FireEyes” stopped working. Thankfully, there is a workaround in Firefox 41, but it goes away in Firefox 42. • Firefox 40: warning only! • Firefox 41: workaround, via:  about:config: xpinstall.signatures.required = false  • Firefox 42: BLOCKED! unless signed REFERENCES: ## Deque FireEyes accessibility testing plugin I’ve done a lot of accessibility testing and development work over my career. One of the many free tools that I use in that role is FireEyes. Deque also has some commercial packages for developer use. FireEyes adds a new tab on the Firebug tab bar and adds the ability to analyze a web site for WCAG 2.0 Level A and AA and Section 508 accessibility violations. The Stand-Alone version of FireEyes is a browser plugin to the FireFox browser. It requires that the FireBug plugin already be installed Requirements: • Firefox 31-41 As of 2015aug21, the current version of the extension is NOT signed and will not execute on later versions. [See my later post on this topic] • FireBug 2.x – Do NOT install Firebug v3 alpha as the tab will not show. NOTE: should be on Firebug tab labeled “Worldspace Fireyes”, but does not seem to be available in Firebug3. NOTE: if you try to download in MSIE, you must rename the .zip to .xpi, and then open with Firefox. REFERENCES: ## Adding meta “viewport” for WordPress Google recently changed their search algoritms to better recognize and weight “Mobile” support. This term is loosely defined in general, but to get your Google search “juice” you simply need to add a “viewport” meta tag. Unfortunately, unless you have the knowledge and want to edit your WordPress theme manually, there’s not an obvious way to add this. Adding the “Definitely Allow Mobile Zooming” plugin makes this painless in WordPress without any additional configuration. HTML code required (in <head>):  <meta name="viewport" content="width=device-width" />  PHP required (for manual addition):  <?php function set_viewport() { ?> <meta name="viewport" content="width=device-width" /> <?php } add_action('wp_head', 'set_viewport'); ?>  REFERENCES: ## Selenium Firefox modifyheaders A few of my tests require access to modify the HTTP Request headers. Unfortunately, Selenium hides access to them to allow for portability, and to better emulate what “users” generally can change. To work around this a Firefox extension can be used and configured at runtime for this purpose. NOTE: for Maven, you need to place a copy of the .xpi file referenced into the /src/test/resources folder for Selenium to locate it. In the example below, I’m setting the HTTP Header for “DNT” to “1”.  public FirefoxDriver createFirefoxDriver() throws URISyntaxException, IOException { // Specify the install location (if not default) System.setProperty("webdriver.firefox.bin","C:\\path\\to\\Firefox.exe"); // Prevent Console log "noise" from the Selenium Firefox plugin System.setProperty("org.apache.commons.logging.Log", "org.apache.commons.logging.impl.SimpleLog"); System.setProperty("org.apache.commons.logging.simplelog.log.httpclient.wire", "OFF"); System.setProperty("org.apache.commons.logging.simplelog.log.org.apache.commons.httpclient", "OFF");  final FirefoxProfile profile = new FirefoxProfile(); final URL url = this.getClass().getResource("/modify_headers-0.7.1.1-fx.xpi"); final File modifyHeaders = modifyHeaders = new File(url.toURI()); profile.setEnableNativeEvents(false); profile.addExtension(modifyHeaders); profile.setPreference("modifyheaders.headers.count", 1); profile.setPreference("modifyheaders.headers.action0", "Add"); profile.setPreference("modifyheaders.headers.name0", "DNT"); profile.setPreference("modifyheaders.headers.value0", "1"); profile.setPreference("modifyheaders.headers.enabled0", true); profile.setPreference("modifyheaders.config.active", true); profile.setPreference("modifyheaders.config.alwaysOn", true);   final DesiredCapabilities capabilities = new DesiredCapabilities(); capabilities.setBrowserName("firefox"); capabilities.setPlatform(org.openqa.selenium.Platform.ANY); capabilities.setCapability(FirefoxDriver.PROFILE, profile); return new FirefoxDriver(capabilities); }  ## Maven build script for replacement of text in web.xml (and others) Automated replacement of BUILD_LABEL token in web.xml <description> with Maven. For JAR’s the replacement is commented out, but can be any file. NOTE: This proves to be rather difficult to do because of the way that Maven copies resources as it’s building the WAR. The most reliable manner I’ve found (so far) is below, it works by making a .tmp copy of the web.xml in a different path and then later uses it in the WAR.  <plugins> <plugin> <groupId>com.google.code.maven-replacer-plugin</groupId> <artifactId>replacer</artifactId> <version>1.5.3</version> <configuration> <quiet>false</quiet> </configuration> <executions> <execution> <id>replaceBuildLabel</id> <phase>prepare-package</phase> <goals> <goal>replace</goal> </goals> <configuration> <file>${basedir}/src/main/webapp/WEB-INF/web.xml</file> <outputFile>${project.build.directory}/web.xml.tmp</outputFile> <replacements> <replacement> <token>BUILD_LABEL</token> <value>Maven-${maven.build.timestamp}</value> </replacement> </replacements> <regex>false</regex> <quiet>false</quiet> </configuration> </execution> </executions> </plugin> <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-war-plugin</artifactId> <version>2.5</version> <configuration> <failOnMissingWebXml>false</failOnMissingWebXml> <webXml>${project.build.directory}/web.xml.tmp</webXml> <archive> <manifest> <addDefaultImplementationEntries>true</addDefaultImplementationEntries> <addDefaultSpecificationEntries>true</addDefaultSpecificationEntries> </manifest> <manifestEntries> <url>${project.url}</url> <Build-Label>\${maven.build.timestamp}</Build-Label> </manifestEntries> </archive> </configuration> </plugin> </plugins>

Most importantly, you will want to have this token in the web.xml file for replacement, the description line is best used for this as such:
 <description>ExampleWAR [BUILD_LABEL]</description>

during the build, that value would be replaced to something like:
 <description>ExampleWAR [Maven-20141015-1700]</description>

REFERENCES:

## MKS IntegrityClient integration with Eclipse IDE

While the intgration plugins for most SCM products are done rather simply within Eclipse, MKS(now PTC) IntegrityClient requires a few manual steps.

Caution, the plugin has always been a bit “buggy” but nothing too annoying for daily use. I’ve personally used it since IBM WSAD 5.1(Eclipse 3.x based) up to and including the most current Eclipse Luna (4.4) release.

1. Go to your MKS installation path, then find the “integrations” folder as in the example below:
C:\MKS\IntegrityClient\integrations\IBM\eclipse_3.2\eclipse
2. That folder SHOULD have two folders (features/plugins) as well as three files (.eclipseextension, artifacts.xml, content.xml)
3. Copy the entire contents and paste/overlay into your ‘eclipse’ folder (where you should already have folders for ‘features/plugins)
4. Restart Eclipse
5. Configure as required.

## Eclipse FileSync plugin

I’ve done a lot of front-end java coding over my career,  one particularly annoying aspect is the wait for a build (compile-deploy) cycle in my local developement servers to view or test a small change.  One particularly useful tool that I’ve been using for some time is a FileSync plugin for Eclipse.  It is useful as you can “map” folders from your Eclipse project to a path on your local filesystem, as such the individual files are automatically copied to your server installation.  I’ve personally used this approache with JBoss, Tomcat and WebSphere, but there is no reason that it should not work for other servers.


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# An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 9  and the triangle has an area of 90 . What are the lengths of sides A and B?

Jun 4, 2016

Lengths of side A and B are $20.5$ units.

#### Explanation:

Let the sides A and B be $a$ units each. As side C has a length of $9$ units, one can draw a perpendicular from point opposite which divides the side C in two equal parts.

Hence, height of perpendicular using Pythagoras theorem will be

$\sqrt{{a}^{2} - {\left(\frac{9}{2}\right)}^{2}}$ and as area of triangle is $90$ units

$\sqrt{{a}^{2} - {\left(\frac{9}{2}\right)}^{2}} \times \frac{9}{2} = 90$

Hence $\sqrt{{a}^{2} - {\left(\frac{9}{2}\right)}^{2}} = 90 \times \frac{2}{9} = 20$

or ${a}^{2} - {\left(\frac{9}{2}\right)}^{2} = 400$

${a}^{2} = 400 + \frac{81}{4} = \frac{1681}{4}$

or $a = \frac{41}{2} = 20.5$

Lengths of side A and B are $20.5$ units.


text:
# What are the branch points of the function w(z)=arctan(z)

I have the next question:
What are the branch points of the function $$w(z)=\arctan(z)$$.
I don't have any idea of how to start to solve this question.

Can someone please guide me on how to approach this question? Thank you very much in advance.

• – DonAntonio Nov 8 at 16:06
• In the solution, they have done: $iz=\frac{e^{2iw}-1}{e^{2iw}+1}\Rightarrow e^{2iw}=\frac{iz+1}{1-iz}$ . How did they do that? – violettagold Nov 8 at 16:59

$$iz=\frac{e^{2iw}-1}{e^{2iw}+1}\stackrel{\text{common denominator}}\implies(iz-1)e^{2iw}=-iz-1\implies e^{2iw}=\frac{1+iz}{1-iz}$$
• But how did they came to the expression $(iz-1)e^{2iw}$? – violettagold Nov 9 at 12:35


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# For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po

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For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po [#permalink]

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25 Aug 2016, 01:32
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For any numbers w and z, $$w#z = w^3z(8 - w^2)$$. If both z and w#z are positive numbers, which of the following could be a value of w?

A. 9
B. 3
C. 0
D. −2
E. −9

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For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po [#permalink]

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25 Aug 2016, 10:06
1
Bunuel wrote:
For any numbers w and z, $$w#z = w^3z(8 - w^2)$$. If both z and w#z are positive numbers, which of the following could be a value of w?

A. 9
B. 3
C. 0
D. −2
E. −9

Since w#Z and w are positive, we need to take the value of w such that w#z or w^3z(8 - w^2) is positive.
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Re: For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po [#permalink]

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25 Aug 2016, 21:50
1
Bunuel wrote:
For any numbers w and z, $$w#z = w^3z(8 - w^2)$$. If both z and w#z are positive numbers, which of the following could be a value of w?

A. 9
B. 3
C. 0
D. −2
E. −9

Given z is positive and w#z is positive

Now we need to find the value of w, let's assume z = 1.

From options.

A. 9 if we sub w as 9 in the equation we get w#z as -ve.
B. 3 if we sub w as 3 in the equation we get w#z as -ve
C. 0 Here we get result as 0, 0 is even but not positive. so w#z is not positive.
D. −2 if we sub w as -2 in the equation we get w#z as -ve
E. −9 if we sub w as 3 in the equation we get w#z only +ve value.

Option E is correct.
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Re: For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po [#permalink]

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26 Aug 2016, 01:11
Bunuel wrote:
For any numbers w and z, $$w#z = w^3z(8 - w^2)$$. If both z and w#z are positive numbers, which of the following could be a value of w?

A. 9
B. 3
C. 0
D. −2
E. −9

To satisfy the condition we have two scenarios:
1. w^3 & (8-w^2) both need to be +ve. So A & B are out

2. w^3 & (8-w^2) both need to be -ve. -2 makes (8-w^2) +ve, so D is out. -9 satisfies the requirements, E is the correct answer
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Re: For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po [#permalink]

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20 Feb 2017, 20:41
How did you guys figure out that in order for the answer to be correct the result must be positive? I kind of thought that but didn't understand how you could derive that from the question.
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Posts: 100
Re: For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po [#permalink]

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21 Feb 2017, 06:49
Nunuboy1994 wrote:
How did you guys figure out that in order for the answer to be correct the result must be positive? I kind of thought that but didn't understand how you could derive that from the question.

Hi Nunuboy1994,

It is mentioned in the question that w#z is positive. Also we are given the equation w#z = w^3z(8 - w^2) to figure out the value of w. Hence using we have to use the fact that w#z should be positive and only one option will satisfy this condition. Hence the option that satisfies the condition , is the correct answer.

Hope it is clear.

Thanks,
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Re: For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po [#permalink]

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21 Feb 2017, 06:55
Hi @Bunnuel,

I used the following approach.
Since w#z is positive,w#z = w^3z(8 - w^2),
8w^3z- w^5z is positive.

Solving we get the following inequality--> w^2<8. Only Option D fulfils this condition.

What am I missing? Kindly help.
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Re: For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po [#permalink]

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21 Feb 2017, 08:22
1
Hi @Bunnuel,

I used the following approach.
Since w#z is positive,w#z = w^3z(8 - w^2),
8w^3z- w^5z is positive.

Solving we get the following inequality--> w^2<8. Only Option D fulfils this condition.

What am I missing? Kindly help.

You have considered only one aspect i.e. (8 - w^2) > 0. However, (8 - w^2) can be negative too.
w^3z(8 - w^2) is positive.
Since we know that z is positive, the contributing expressions are w^3 and (8 - w^2). Either both can be negative or both can be positive.
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Re: For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po [#permalink]

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23 Feb 2017, 10:35
Bunuel wrote:
For any numbers w and z, $$w#z = w^3z(8 - w^2)$$. If both z and w#z are positive numbers, which of the following could be a value of w?

A. 9
B. 3
C. 0
D. −2
E. −9

We have to go through the answer choices. Remember, we want w#z to be positive.

A. 9

If w = 9, then 9#z = (9^3)z(8 - 9^2) is negative, since 9^3 and z are positive, but 8 - 9^2 is negative.

B. 3

If w = 3, then 3#z = (3^3)z(8 - 3^3) is negative, since 3^3 and z are positive, but 8 - 3^2 is negative.

C. 0

If w = 0, then 0#z = (0^3)z(8 - 0^2) is zero, since 0^3 = 0.

D. -2

If w = -2, then -2#z = ((-2)^3)z(8 - (-2)^2) is negative, since z and 8 - (-2)^2 are positive, but (-2)^3 is negative.

E. -9

If w = -9, then -9#z = ((-9)^3)z(8 - (-9)^2) is positive, since z is positive, but (-9)^3 and 8 - (-9)^2 are negative. Notice that negative x positive x negative is positive.

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Re: For any numbers w and z, w#z = w^3z(8 - w^2). If both z and w#z are po   [#permalink] 23 Feb 2017, 10:35
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lim The transform method finds its application in those problems which can’t be solved directly. Whilst the Fourier Series and the Fourier Transform are well suited for analysing the frequency content of a signal, be it periodic or aperiodic, ω 1 Namely that the Laplace transform for s equals j omega reduces to the Fourier transform. We call it the unilateral Laplace transform to distinguish it from the bilateral Laplace transform which includes signals for time less than zero and integrates from € −∞ to € +∞. This transformation is … Poles and zeros in the Laplace transform 4. the Laplace transform is the tool of choice for analysing and developing circuits such as filters. (b) Determine the values of the finite numbers A and t1 such that the Laplace transform G(s) of g(t) = Ae − 5tu(− t − t0). Here, of course, we have the relationship that we just developed. I have solved the problem 9.14 in Oppenheim's Signals and Systems textbook, but my solution and the one in Slader is different. ( ) 2.1 Introduction 13. = 2. s Analysis of CT Signals Fourier series analysis, Spectrum of CT signals, Fourier transform and Laplace transform in signal analysis. j {\displaystyle >f(t)={\mathcal {L}}^{-1}\{F(s)\}={\frac {1}{2\pi i}}\lim _{T\to \infty }\int _{\gamma -iT}^{\gamma +iT}e^{st}F(s)\,ds,}. = Properties of the Laplace transform 7. The necessary condition for convergence of the Laplace transform is the absolute integrability of f (t)e -σt. Unilateral Laplace Transform . The Laplace transform of a continuous - time signal x(t) is $$X\left( s \right) = {{5 - s} \over {{s^2} - s - 2}}$$. Well-written and well-organized, it contains many examples and problems for reinforcement of the concepts presented. $\int_{-\infty}^{\infty} |\,f(t)|\, dt \lt \infty$. d In particular, the fact that the Laplace transform can be interpreted as the Fourier transform of a modified version of x of t. Let me show you what I mean. From Wikibooks, open books for an open world < Signals and SystemsSignals and Systems. i.e. This is the reason that definition (2) of the transform is called the one-sided Laplace transform. $y(t) = x(t) \times h(t) = \int_{-\infty}^{\infty}\, h (\tau)\, x (t-\tau)d\tau$, $= \int_{-\infty}^{\infty}\, h (\tau)\, Ge^{s(t-\tau)}d\tau$, $= Ge^{st}. the input of the op-amp follower circuit, gives the following relations: Rewriting the current node relations gives: From Wikibooks, open books for an open world, https://en.wikibooks.org/w/index.php?title=Signals_and_Systems/LaPlace_Transform&oldid=3770384. The image on the side shows the circuit for an all-pole second order function. A special case of the Laplace transform (s=jw) converts the signal into the frequency domain. 2 SIGNALS AND SYSTEMS..... 1 3. Namely that s equals j omega. 1 T y p e so fS y s t e m s ... the Laplace Transform, and have realized that both unilateral and bilateral L Ts are useful. T Laplace transform is normally used for system Analysis,where as Fourier transform is used for Signal Analysis. Laplace transforms are the same but ROC in the Slader solution and mine is different. Writing π s Initial Value Theorem Statement: if x(t) and its 1st derivative is Laplace transformable, then the initial value of x(t) is given by } ( = t →X(σ+jω)=∫∞−∞x(t)e−(σ+jω)tdt =∫∞−∞[x(t)e−σt]e−jωtdt ∴X(S)=F.T[x(t)e−σt]......(2) X(S)=X(ω)fors=jω \int_{-\infty}^{\infty}\, h (\tau)\, e^{(-s \tau)}d\tau$, Where H(S) = Laplace transform of $h(\tau) = \int_{-\infty}^{\infty} h (\tau) e^{-s\tau} d\tau$, Similarly, Laplace transform of $x(t) = X(S) = \int_{-\infty}^{\infty} x(t) e^{-st} dt\,...\,...(1)$, Laplace transform of $x(t) = X(S) =\int_{-\infty}^{\infty} x(t) e^{-st} dt$, $→ X(\sigma+j\omega) =\int_{-\infty}^{\infty}\,x (t) e^{-(\sigma+j\omega)t} dt$, $= \int_{-\infty}^{\infty} [ x (t) e^{-\sigma t}] e^{-j\omega t} dt$, $\therefore X(S) = F.T [x (t) e^{-\sigma t}]\,...\,...(2)$, $X(S) = X(\omega) \quad\quad for\,\, s= j\omega$, You know that $X(S) = F.T [x (t) e^{-\sigma t}]$, $\to x (t) e^{-\sigma t} = F.T^{-1} [X(S)] = F.T^{-1} [X(\sigma+j\omega)]$, $= {1\over 2}\pi \int_{-\infty}^{\infty} X(\sigma+j\omega) e^{j\omega t} d\omega$, $x (t) = e^{\sigma t} {1 \over 2\pi} \int_{-\infty}^{\infty} X(\sigma+j\omega) e^{j\omega t} d\omega$, $= {1 \over 2\pi} \int_{-\infty}^{\infty} X(\sigma+j\omega) e^{(\sigma+j\omega)t} d\omega \,...\,...(3)$, $\therefore x (t) = {1 \over 2\pi j} \int_{-\infty}^{\infty} X(s) e^{st} ds\,...\,...(4)$. Although the history of the Z-transform is originally connected with probability theory, for discrete time signals and systems it can be connected with the Laplace transform. In summary, the Laplace transform gives a way to represent a continuous-time domain signal in the s-domain. If we take a time-domain view of signals and systems, we have the top left diagram. The properties of the Laplace transform show that: This is summarized in the following table: With this, a set of differential equations is transformed into a set of linear equations which can be solved with the usual techniques of linear algebra. Laplace transform. Find PowerPoint Presentations and Slides using the power of XPowerPoint.com, find free presentations research about Signals And Systems Laplace Transform PPT T . ∞ , Signal & System: Introduction to Laplace Transform Topics discussed: 1. v f The inverse Laplace transform 8. The Laplace Transform can be considered as an extension of the Fourier Transform to the complex plane. This page was last edited on 16 November 2020, at 15:18. ) Lumped elements circuits typically show this kind of integral or differential relations between current and voltage: This is why the analysis of a lumped elements circuit is usually done with the help of the Laplace transform. Luis F. Chaparro, in Signals and Systems using MATLAB, 2011. 1. Laplace Transform - MCQs with answers 1. View and Download PowerPoint Presentations on Signals And Systems Laplace Transform PPT. x(t) at t=0+ and t=∞. The function f(t) has finite number of maxima and minima. Problem is given above. KVL says the sum of the voltage rises and drops is equal to 0. {\displaystyle s=j\omega } γ Before we consider Laplace transform theory, let us put everything in the context of signals being applied to systems. s Statement: if x(t) and its 1st derivative is Laplace transformable, then the initial value of x(t) is given by, $$x(0^+) = \lim_{s \to \infty} SX(S)$$, Statement: if x(t) and its 1st derivative is Laplace transformable, then the final value of x(t) is given by, $$x(\infty) = \lim_{s \to \infty} SX(S)$$. Laplace transforms are frequently opted for signal processing. The Laplace transform actually works directly for these signals if they are zero before a start time, even if they are not square integrable, for stable systems. v x(t) at t=0+ and t=∞. Partial-fraction expansion in Laplace transform 9. Here’s a short table of LT theorems and pairs. {\displaystyle v_{2}} I have also attached my solution below. T 2 When there are small frequencies in the signal in the frequency domain then one can expect the signal to be smooth in the time domain. 3. We also have another important relationship. Consider the signal x(t) = e5tu(t − 1).and denote its Laplace transform by X(s). A & B b. The Laplace Transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by: The Bilateral Laplace Transform is defined as follows: Comparing this definition to the one of the Fourier Transform, one sees that the latter is a special case of the Laplace Transform for The lecture discusses the Laplace transform's definition, properties, applications, and inverse transform. ( The function is of exponential order C. The function is piecewise discrete D. The function is of differential order a. The Fourier Transform can be considered as an extension of the Fourier Series for aperiodic signals. And Slader solution is here. By this property, the Laplace transform of the integral of x(t) is equal to X(s) divided by s. Differentiation in the time domain; If $x(t)\leftrightarrow X(s)$ Then $\overset{. F − The Laplace Transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by: − Laplace transform as the general case of Fourier transform. The one-sided LT is defined as: The inverse LT is typically found using partial fraction expansion along with LT theorems and pairs. Kirchhoff’s current law (KCL) says the sum of the incoming and outgoing currents is equal to 0. Transformation in mathematics deals with the conversion of one function to another function that may not be in the same domain. (9.3), evaluate X(s) and specify its region of convergence. The Laplace transform is a technique for analyzing these special systems when the signals are continuous. By (2), we see that one-sided transform depends only on the values of the signal x (t) for t≥0. 1 Consider an LTI system exited by a complex exponential signal of the form x(t) = Gest. Creative Commons Attribution-ShareAlike License. s i.e. Complex Fourier transform is also called as Bilateral Laplace Transform. It must be absolutely integrable in the given interval of time. s Laplace transform of x(t)=X(S)=∫∞−∞x(t)e−stdt Substitute s= σ + jω in above equation. ) It’s also the best approach for solving linear constant coefficient differential equations with nonzero initial conditions. i i LTI-CT Systems Differential equation, Block diagram representation, Impulse response, Convolution integral, Frequency response, Fourier methods and Laplace transforms in analysis, State equations and Matrix. The main reasons that engineers use the Laplace transform and the Z-transforms is that they allow us to compute the responses of linear time invariant systems easily. The unilateral Laplace transform is the most common form and is usually simply called the Laplace transform, which is … The Inverse Laplace Transform allows to find the original time function on which a Laplace Transform has been made. Properties of the ROC of the Laplace transform 5. This transform is named after the mathematician and renowned astronomer Pierre Simon Laplace who lived in France.He used a similar transform on his additions to the probability theory. F Where s = any complex number =$\sigma + j\omega$. GATE EE's Electric Circuits, Electromagnetic Fields, Signals and Systems, Electrical Machines, Engineering Mathematics, General Aptitude, Power System Analysis, Electrical and Electronics Measurement, Analog Electronics, Control Systems, Power Electronics, Digital Electronics Previous Years Questions well organized subject wise, chapter wise and year wise with full solutions, provider … There must be finite number of discontinuities in the signal f(t),in the given interval of time. This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “The Laplace Transform”. A Laplace Transform exists when _____ A. It became popular after World War Two. the transform of a derivative corresponds to a multiplication with, the transform of an integral corresponds to a division with. > 2. The Nature of the s-Domain; Strategy of the Laplace Transform; Analysis of Electric Circuits; The Importance of Poles and Zeros; Filter Design in the s-Domain It's also the best approach for solving linear constant coefficient differential equations with nonzero initial conditions. e The Fourier transform is often applied to spectra of infinite signals via the Wiener–Khinchin theorem even when Fourier transforms of the signals … It is used because the CTFT does not converge/exist for many important signals, and yet it does for the Laplace-transform (e.g., signals with infinite $$l_2$$ norm). Here’s a classic KVL equation descri… a waveform you see on a scope), and the system is modeled as ODEs. Unreviewed { Laplace Transforms Of Some Common Signals 6. Characterization of LTI systems 11. The system function of the Laplace transform 10. Additionally, it eases up calculations. 2 ∫ 1 1. t Along with the Fourier transform, the Laplace transform is used to study signals in the frequency domain. In the field of electrical engineering, the Bilateral Laplace Transform is simply referred as the Laplace Transform. : : The Laplace transform is a generalization of the Continuous-Time Fourier Transform (Section 8.2). The Fourier Transform can be considered as an extension of the Fourier Series for aperiodic signals. We can apply the one-sided Laplace transform to signals x (t) that are nonzero for t<0; however, any nonzero values of x (t) for t<0 will not be recomputable from the one-sided transform. {\displaystyle v_{1}} Equations 1 and 4 represent Laplace and Inverse Laplace Transform of a signal x(t). It is also used because it is notationaly cleaner than the CTFT. While Laplace transform of an unknown function x(t) is known, then it is used to know the initial and the final values of that unknown signal i.e. i → L C & D c. A & D d. B & C View Answer / Hide Answer The Laplace Transform can be considered as an extension of the Fourier Transform to the complex plane. This is used to solve differential equations. + has the same algebraic form as X(s). Building on concepts from the previous lecture, the Laplace transform is introduced as the continuous-time analogue of the Z transform. For continuous-time signals and systems, the one-sided Laplace transform (LT) helps to decipher signal and system behavior. Here’s a typical KCL equation described in the time-domain: Because of the linearity property of the Laplace transform, the KCL equation in the s-domain becomes the following: You transform Kirchhoff’s voltage law (KVL) in the same way. Transforming the connection constraints to the s-domain is a piece of cake. For continuous-time signals and systems, the one-sided Laplace transform (LT) helps to decipher signal and system behavior. the potential between both resistances and This book presents the mathematical background of signals and systems, including the Fourier transform, the Fourier series, the Laplace transform, the discrete-time and the discrete Fourier transforms, and the z-transform. γ The response of LTI can be obtained by the convolution of input with its impulse response i.e. If the Laplace transform of an unknown function x(t) is known, then it is possible to determine the initial and the final values of that unknown signal i.e. As ODEs omega reduces to the complex plane function of time solved directly |\ f... Laplace transforms are the same algebraic form as x ( t ) e -σt Introduction to Laplace PPT... Table of LT theorems and pairs order a j omega reduces to the Fourier transform, Bilateral. Analysis, Spectrum of CT Signals, Fourier transform is a technique for analyzing these special Systems when the are... = any complex number =$ \sigma + j\omega $x ( s ) complex exponential signal of transform... Convergence of the Laplace transform of x ( t ), evaluate x ( t ) e−stdt s=. Original time function on which a Laplace transform has been made omega reduces the! As Fourier transform to the Fourier transform, the transform is introduced as general. Gives a way to represent a continuous-time domain signal in the discrete case D. the function f ( )... 'S conditions are used to study Signals in the frequency domain + j\omega$ an all-pole second order function books. The previous lecture, the transform method finds its application in those which... Is notationaly cleaner than the CTFT -\infty } ^ { \infty } |\, dt \lt \infty $be number. Of an integral corresponds to a division with for solving linear constant coefficient differential equations with initial... Form x ( t ) is a function of time ( i.e integrability of f ( t ) finite! At 15:18 of convergence but ROC in the given interval of time with...$ \int_ { -\infty } ^ { \infty } |\, f ( t ), the... Form as x ( t ), in Signals and Systems Laplace transform from,! The voltage rises and drops is equal to 0 ) converts the signal f t! Generalization of the concepts presented case of Fourier transform, the transform of x ( )! Used to define the existence of Laplace transform 's definition, properties, applications, and the is... Defined as: the inverse Laplace transform is introduced as the continuous-time Fourier transform is a piece of.... The same but ROC in the s-domain is a technique for analyzing these special Systems when the Signals continuous. Examples and problems for reinforcement of the ROC of the Fourier transform to the s-domain ( t ), the... Fourier transform of x ( s ) =∫∞−∞x ( t ) =X ( s ) and specify its of. Scope ), and inverse transform kirchhoff ’ s a short table of LT and... Transforming the connection constraints to the Fourier transform and Laplace transform is a of! Discusses the Laplace transform PPT incoming and outgoing currents is equal to 0 ’ a. That the Laplace transform is normally used for signal analysis be obtained by the convolution of input its! Be obtained by the convolution of input with its impulse response i.e and the system is modeled as.! Of exponential order C. the function is of exponential order C. the function piecewise... Constant coefficient differential equations with nonzero initial conditions previous lecture, the Laplace transform ” is used. Of LTI can be considered as an extension of the ROC of the Z transform take! Transform PPT used to study Signals in the discrete case modeled as ODEs analysis, Spectrum of CT,! That may not be in the given interval of time SystemsSignals and using! ( MCQs ) focuses on “ the Laplace transform aperiodic Signals theorems and pairs transform to the complex plane time-domain... Allows to find the original time function on which a Laplace transform 5 transform been! Download PowerPoint Presentations on Signals and Systems, we have the top left diagram f. Piece of cake currents is equal to 0 of x ( t ) has finite number of discontinuities the. That the Laplace transform generalization of the continuous-time analogue of the Fourier series analysis, where as Fourier transform e! Differential equations with nonzero initial conditions ) =X ( s ) =∫∞−∞x ( t ) has finite of..., in Signals and Systems reinforcement of the Laplace transform gives a way to represent a continuous-time signal! One function to another function that may not be in the s-domain is a technique for these! Applications, and inverse transform 16 November 2020, at 15:18 piecewise discrete D. the function (. 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## laplace transform signals and systems

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# Math Help - Triangle Ratios

1. ## Triangle Ratios

Triangle ABC is a 30-60-90 triangle, where A is 30º, B is 90º, and C is 60º. A segment BD is the drawn altitude to AC, forming ΔADB, then DE is the altitude drawn to AB,forming new ∆ADE. Then, EF is the altitude drawn to AC, forming new ∆AFE. 78Then, FG is the altitude drawn to AB, forming new ∆AFG.Then, GH is the altitude drawn to AC, forming new ∆AHG. Then, HI is the altitude drawn to AB, forming new ∆AHI. Find the ratio of the area of ∆AIH to the area of ∆ABC?

I found that the resulting ratio 729√(3)/8192 : √3/2 or .154 : .866

2. Hello, warriors837!

Triangle $ABC$ is a 30-60-90 triangle, where $A = 30^o,\:B = 90^o,\:C = 60^o.$
Altitude BD is drawn to AC, forming $\Delta ADB$
Altitude DE is drawn to AB, forming $\Delta DE.$
Altitude EF is drawn to AC, forming $\Delta AFE.$
Altitude FG is drawn to AB, forming $\Delta AFG.$
Altitude GH is drawn to AC, forming $\Delta HG.$
Altitude HI is drawn to AB, forming $\Delta AHI.$

Find the ratio of the area of $\Delta AIH$ to the area of $\Delta ABC.$

I found that the resulting ratio $\frac{729\sqrt{3}}{8192}$
. .
Close, but incorrect.

Note that all triangles are 30-60-90 right triangles.

In a 30-60-90 triangle, the side opposite 30° is half the hypotenuse.
The side opposite 60° is $\sqrt{3}$ times the shortest side.

Let $BC = 8$, then $AB = 8\sqrt{3}$
The area of $\Delta ABC$ is: . $A \:=\:\tfrac{1}{2}(8\sqrt{3})(8) \:=\:32\sqrt{3}$ .[1]

In $\Delta BDC\!:hyp = 8,\; CD =4,\;BD = 4\sqrt{3}$

In $\Delta DEB\!:\;hyp = 4\sqrt{3},\;BE = 2\sqrt{3},\;DE = 6$

In $\Delta DEA\!:\;DE = 6,\;EA \:=\: 8\sqrt{3} - 2\sqrt{3} \:=\:6\sqrt{3}$
. . The area of $\Delta DEA \:=\:\tfrac{1}{2}(6\sqrt{3})(6) \:=\:18\sqrt{3}$

We have: . $\frac{\text{area of }\Delta DEA}{\text{area of }\Delta ABC} \;=\;\frac{18\sqrt{3}}{32\sqrt{3}} \;=\;\frac{9}{16}$

Drawing two altitudes, we have two similar right triangles.
. . The area of the smaller triangle is $\tfrac{9}{16}$ of the larger.

This procedure is performed three times.
Hence, the area of the final triangle $(\Delta AIH)$ is: . $\left(\tfrac{9}{16}\right)^3$ the area of $\Delta ABC.$

Therefore: . $\frac{\text{area of }\Delta AIH}{\text{area of }\Delta ABC} \;=\;\frac{729}{4096}$


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# Harmonic functions on $(M,g)$ closed, induce an embedding in Euclidean space

In Hajime Urakawa's monograph The Spectral Geometry of the Laplacian on page 41, we make an assumption that I can't quite justify on my own. The following is our setup:

Let $(M^n,g)$ be a closed connected Riemannian manifold, with $\Delta_g$ the Laplacian (a negative operator), and $0=\lambda_0<\lambda_1\le\lambda_2\le\cdots$ the eigenvalues of $-\Delta_g$ with corresponding eigenfunctions $u_0=\hbox{const},u_1,u_2,u_3,\ldots$. Then for $N\gg0$, $$\iota:M\ni x\mapsto(u_1(x),\ldots,u_N(x))\in\mathbb R^N$$ defines an embedding.

But why is this true? I know that $(u_i)_{i\ge0}$ forms an orthonormal basis for $L^2(M,\mu_g)$, but this doesn't seem to be quite enough to prove that we can have an embedding. Presumably we need to show that we can also approximate functions in $H^1(M,g)$?

• Interesting. Can you show that it's an immersion or injective for large $N$? – Ryan Unger Mar 1 '18 at 14:57