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+7 (3412) 91 60 92

## Archive of Issues

Russia Nizhni Novgorod
Year
2021
Volume
31
Issue
2
Pages
331-349
 Section Mathematics Title On totally global solvability of evolutionary equation with unbounded operator Author(-s) Chernov A.V.ab Affiliations Nizhni Novgorod State Technical Universitya, Nizhni Novgorod State Universityb Abstract Let $X$ be a Hilbert space, $U$ be a Banach space, $G\colon X\to X$ be a linear operator such that the operator $B_\lambda=\lambda I-G$ is maximal monotone with some (arbitrary given) $\lambda\in\mathbb{R}$. For the Cauchy problem associated with controlled semilinear evolutionary equation as follows $x^\prime(t)=Gx(t)+f\bigl( t,x(t),u(t)\bigr),\quad t\in[0;T];\quad x(0)=x_0\in X,$ where $u=u(t)\colon[0;T]\to U$ is a control, $x(t)$ is unknown function with values in $X$, we prove the totally (with respect to a set of admissible controls) global solvability subject to global solvability of the Cauchy problem associated with some ordinary differential equation in the space $\mathbb{R}$. Solution $x$ is treated in weak sense and is sought in the space $\mathbb{C}_w\bigl([0;T];X\bigr)$ of weakly continuous functions. In fact, we generalize a similar result having been proved by the author formerly for the case of bounded operator $G$. The essence of this generalization consists in that postulated properties of the operator $B_\lambda$ give us the possibility to construct Yosida approximations for it by bounded linear operators and thus to extend required estimates from “bounded” to “unbounded” case. As examples, we consider initial boundary value problems associated with the heat equation and the wave equation. Keywords semilinear evolutionary equation in a Hilbert space, maximal monotone operator, totally global solvability UDC 517.957, 517.988, 517.977.56 MSC 47J05, 47J35, 47N10 DOI 10.35634/vm210212 Received 28 August 2020 Language Russian Citation Chernov A.V. On totally global solvability of evolutionary equation with unbounded operator, Vestnik Udmurtskogo Universiteta. Matematika. Mekhanika. Komp'yuternye Nauki, 2021, vol. 31, issue 2, pp. 331-349. References Chernov A.V. A majorant criterion for the total preservation of global solvability of controlled functional operator equation, Russian Mathematics, 2011, vol. 55, no. 3, pp. 85-95. https://doi.org/10.3103/S1066369X11030108 Kalantarov V.K., Ladyzhenskaya O.A. The occurrence of collapse for quasilinear equations of parabolic and hyperbolic types, Journal of Soviet Mathematics, 1978, vol. 10, issue 1, pp. 53-70. https://doi.org/10.1007/BF01109723 Sumin V.I. The features of gradient methods for distributed optimal control problems, USSR Computational Mathematics and Mathematical Physics, 1990, vol. 30, no. 1, pp. 1-15. https://doi.org/10.1016/0041-5553(90)90002-A Sumin V.I. Funktsional'nye vol'terrovy uravneniya v teorii optimal'nogo upravleniya raspredelennymi sistemami. Chast' I. Vol'terrovy uravneniya i upravlyaemye nachal'no-kraevye zadachi (Functional Volterra equations in the theory of optimal control of distributed systems. Part I. Volterra equations and controlled initial boundary value problems), Nizhny Novgorod: Nizhny Novgorod State University, 1992. Lions J.-L. Contrôle des systèmes distribués singuliers, Paris: Bordas, 1983. https://zbmath.org/?q=an:0514.93001 Fursikov A.V. Optimal control of distributed systems. Theory and applications, Providence, RI: AMS, 2000. https://zbmath.org/?q=an:1027.93500 Plotnikov V.I., Sumin V.I. Optimization of distributed systems in Lebesgue space, Siberian Mathematical Journal, 1981, vol. 22, no. 6, pp. 913-929. https://doi.org/10.1007/BF00968060 Chernov A.V. On the convergence of the conditional gradient method in distributed optimization problems, Computational Mathematics and Mathematical Physics, 2011, vol. 51, no. 9, pp. 1510-1523. https://doi.org/10.1134/S0965542511090077 Chernov A.V. On total preservation of solvability of controlled Hammerstein-type equation with non-isotone and non-majorizable operator, Russian Mathematics, 2017, vol. 61, no. 6, pp. 72-81. https://doi.org/10.3103/S1066369X1706010X Kobayashi T., Pecher H., Shibata Y. On a global in time existence theorem of smooth solutions to a nonlinear wave equation with viscosity, Mathematische Annalen, 1993, vol. 296, no. 2, pp. 215-234. https://doi.org/10.1007/BF01445103 Lu G. Global existence and blow-up for a class of semilinear parabolic systems: a Cauchy problem, Nonlinear Analysis: Theory, Methods and Applications, 1995, vol. 24, no. 8, pp. 1193-1206. https://doi.org/10.1016/0362-546X(94)00190-S Cavalcanti M.M., Domingos Cavalcanti V.N., Soriano J.A. On existence and asymptotic stability of solutions of the degenerate wave equation with nonlinear boundary conditions, Journal of Mathematical Analysis and Applications, 2003, vol. 281, no. 1, pp. 108-124. https://doi.org/10.1016/S0022-247X(02)00558-9 Saito H. Global solvability of the Navier-Stokes equations with a free surface in the maximal $L_p$-$L_q$ regularity class, Journal of Differential Equations, 2018, vol. 264, no. 3, pp. 1475-1520. https://doi.org/10.1016/j.jde.2017.09.045 Sumin V.I. Optimization of controlled generalized Volterra systems, Cand. Sci. (Phys. and Math.) Dissertation, Gorkii, 1975, 158 p. (In Russian). Sumin V.I. On functional Volterra equations, Russian Mathematics, 1995, vol. 39, no. 9, pp. 65-75. https://zbmath.org/?q=an:0861.45004 Sumin V.I. Functional Volterra equations in the mathematical theory of optimal control of distributed systems, Dr. Sci. (Phys. and Math.) Dissertation, Nizhny Novgorod, 1998, 346 p. (In Russian). Sumin V.I. Controlled functional Volterra equations in Lebesgue spaces, Vestnik Nizhegorodskogo Universiteta. Seriya Matematicheskoe Modelirovanie i Optimal'noe Upravlenie, 1998, no. 2 (19), pp. 138-151 (in Russian). Sumin V.I., Chernov A.V. Volterra operator equations in Banach spaces: the stability of existence of global solutions, NNSU, Nizhny Novgorod, 2000, 75 p. Deposited in VINITI 25.04.00, no. 1198-V00 (in Russian). Chernov A.V. Volterra operator equations and their application in the optimization theory of hyperbolic systems, Cand. Sci. (Phys. and Math.) Dissertation, Nizhny Novgorod, 2000, 177 p. (In Russian). Chernov A.V. A majorant-minorant criterion for the total preservation of global solvability of a functional operator equation, Russian Mathematics, 2012, vol. 56, no. 3, pp. 55-65. https://doi.org/10.3103/S1066369X12030085 Chernov A.V. On total preservation of global solvability for a Goursat problem associated with a controlled semilinear pseudoparabolic equation, Vladikavkazskii Matematicheskii Zhurnal, 2014, vol. 16, no. 3, pp. 55-63 (in Russian). http://mi.mathnet.ru/eng/vmj513 Chernov A.V. On the totally global solvability of a controlled Hammerstein type equation with a varied linear operator, Vestnik Udmurtskogo Universiteta. Matematika. Mekhanika. Komp'yuternye Nauki, 2015, vol. 25, issue 2, pp. 230-243 (in Russian). https://doi.org/10.20537/vm150207 Chernov A.V. On a majorant-minorant criterion for the total preservation of global solvability of distributed controlled systems, Differential Equations, 2016, vol. 52, no. 1, pp. 111-121. https://doi.org/10.1134/S0012266116010092 Sumin V.I. The problem of sustainability of existence global solutions of controlled boundary value problems and Volterra functional equations, Vestnik Nizhegorodskogo Universiteta. Matematika, 2003, no. 1, pp. 91-107 (in Russian). https://zbmath.org/?q=an:1067.49002 Sumin V.I., Chernov A.V. Volterra functional-operator equations in the theory of optimization of distributed systems, Systems Dinamics and Control Processes (SDCP-2014): Proceedings of Int. Conf. Dedicated to the 90th Anniversary of the birth of Acad. N.N. Krasovskii, Yekaterinburg: UMTS UPI, 2015, pp. 293-300 (in Russian). Sumin V.I. Controlled Volterra functional equations and the contraction mapping principle, Trudy Instituta Matematiki i Mekhaniki UrO RAN, 2019, vol. 25, no. 1, pp. 262-278 (in Russian). https://doi.org/10.21538/0134-4889-2019-25-1-262-278 Chernov A.V. Preservation of the solvability of a semilinear global electric circuit equation, Computational Mathematics and Mathematical Physics, 2018, vol. 58, no. 12, pp. 2018-2030. https://doi.org/10.1134/S0965542518120096 Sumin V.I. Equiquasinilpotency: definitions, conditions, examples of application, Vestnik Tambovskogo Universiteta. Seriya Estestvennye i Tekhnicheskie Nauki, 2010, vol. 15, issue 1, pp. 453-466 (in Russian). Chernov A.V. On total preservation of global solvability of a differential operator equation, Prikladnaya Matematika i Voprosy Upravleniya, 2017, no. 2, pp. 94-111 (in Russian). https://elibrary.ru/item.asp?id=29675515 Chernov A.V. On totally global solvability of controlled second kind operator equation, Vestnik Udmurtskogo Universiteta. Matematika. Mekhanika. Komp'yuternye Nauki, 2020, vol. 30, issue 1, pp. 92-111 (in Russian). https://doi.org/10.35634/vm200107 Sumin V.I. Volterra functional-operator equations in the theory of optimal control of distributed systems, IFAC-PapersOnLine, 2018, vol. 51, issue 32, pp. 759-764. https://doi.org/10.1016/j.ifacol.2018.11.454 Balakrishnan A.V. Applied functional analysis, New York-Heidelberg-Berlin: Springer-Verlag, 1976. https://zbmath.org/?q=an:0333.93051 Hille E., Phillips R.S. Functional analysis and semi-groups, Providence, R.I.: American Mathematical Society (AMS), 1957. https://zbmath.org/?q=an:0078.10004 *Funktsional'nyi analiz (Functional analysis) / S.G. Krein (Ed.), Moscow: Nauka, 1979. Gajewski H., Gröger K., Zacharias K. Nichtlineare Operatorgleichungen und Operatordifferentialgleichungen (Nonlinear operator equations and operator differential equations), Berlin: Akademie, 1974. https://zbmath.org/?q=an:0289.47029 Krein S.G. Lineinye differentsial'nye uravneniya v banakhovom prostranstve (Linear differential equations in a Banach space), Moscow: Nauka, 1967. Ramaswamy M., Shaiju A.J. Construction of approximate saddle-point strategies for differential games in a Hilbert space, Journal of Optimization Theory and Applications, 2009, vol. 141, issue 2, pp. 349-370. https://doi.org/10.1007/s10957-008-9478-z Dautray R., Lions J.-L. Mathematical analysis and numerical methods for science and technology. Vol. 5. Evolution problems, Berlin: Springer-Verlag, 1992. https://zbmath.org/?q=an:0755.35001 Brezis H. Functional analysis, Sobolev spaces and partial differential equations, New York: Springer, 2011. https://doi.org/10.1007/978-0-387-70914-7 Full text


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# What are the distinctions in between significant Linux distributions? Will I see?

I've been making use of Linux in some kind periodically for the previous 10 years approximately, yet have actually just actually remained in to it for the previous couple of months.

What are the significant distinctions in between circulations? Inevitably I recognize that plan monitoring and also branding are the large ones, yet would certainly it be less complicated to do job x with circulation y? Why so?

0
2019-05-04 21:30:20
Source Share

As others have actually recommended the plan monitoring, default software program selections (specifically desktop computer setting), and also arrangement documents formats are one of the most noticeable distinctions in between circulations.

I nonetheless assume more vital after that this is various ideologies that the circulations adhere to. Among others:

• Ubuntu/Fedora - User experience, regular user interface, great deals of constructed in visual devices.
• Arc - Simple and also tidy, reducing side software program
• Gentoo - Extremely personalized, resource based
• Debian - Lightweight, unfailing, security over new software program
0
2019-05-16 21:57:17
Source

Even though the plan supervisors are various, my understanding (though I do not make use of Linux consistently) is that the majority of plans are readily available some means or an additional with any kind of distro. Additionally, the window systems/ desktop computer settings are various, yet you can install alternates anyhow. The only various other point I can consider now would certainly be the straightforward distinction in areas bordering distros.

I'm rather certain that you'll have the ability to locate a means to do whatever you intend to do.

0
2019-05-08 06:15:38
Source

I assume 2 major distinction is

• Package monitoring
• Default Window Manager (Gnome, KDE, XFCE)

Just indicating these 2 large distinction will certainly rather transform your practices in doing job. As an example you can conveniently install a plan on Ubuntu making use of apt yet not so straightforward when making use of rpm. Having a distinction in Window Manager can additionally transform just how you connect with the system.

0
2019-05-08 06:06:03
Source

There are 2 significant tastes of Linux distros around. Debian and also Red Hat based circulations make up a huge percent of the circulations around.

The significant distinction in between them is the plan monitoring. If they are Debian based, they more than likely usage the dpkg & apt/deb system ; if it's a Red Hat system, it more than likely usages yum/rpm. A great deal of distros turn up due to the fact that a person was disappointed with the plan monitoring, so most will certainly have some kind of visual user interface that is various, yet the underlying system coincides.

If you find out just how to make use of apt-get and also yum, you'll cover 80% of the distros around, and also 99% of the systems you'll more than likely experience.

Each distro will certainly do something a little various with the window supervisor. The majority of significant distros pick in between KDE and also Gnome, with Gnome appearing to be the preferred one presently. The wonderful feature of Linux though, is that you can transform the window supervisor if you like (go Window Maker!).

If you find out the command line, there is not "a lot" of a distinction, yet the GUI adjustments in between distros will most definitely guide your selection. Additionally, some distros ship with vehicle drivers that will certainly make it very easy for sure equipment like video clip cards.

I very advise Linux Mint for residence systems, and also vanilla Debian, Ubuntu (web server version), or CentOS for web servers.

0
2019-05-08 03:58:25
Source

format: plain


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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$a^{28}$
Using the laws of exponents, the given expression, $(a^{-4})^{-7}$, is equivalent to \begin{array}{l} a^{-4(-7)} \\\\= a^{28} .\end{array}


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# Equivalent capacitance in infinite grid

1 vote
588 views

There is an infinite grid in which there is a capacitance C between any two point . Calculate the equivalent capacitance between A and B .

My try :

I am not able to find the relation bewteen $q_1$ and Q .

Can you explain your try? It is not obvious what you are doing.

The 4 capacitors connected to A do not have the same charge, nor the same voltage. By symmetry the 2 perpendicular to AB have the same charge and voltage, but the 2 in the line with AB do not.

The infinite grid of unit resistors has an equivalent resistance of $\frac12$ between adjacent nodes. The rules for addition of capacitors in series and parallel are the same for addition of resistors in parallel and series respectively. So we expect the equivalent capacitance of an infinite grid of unit capacitors to be the inverse of the result for resistors, ie 2.

The same argument works for equivalent capacitance between any two nodes $(0,0)$ and $(m,n)$. If the equivalent resistor network has resistance $R_{m,n}$ units then the capacitor network has capacitance $C_{m,n}=1/R_{m,n}$ units between the same nodes.

The trick for the infinite resistor network is to use symmetry and superposition.

Imagine case A : that current $I_{in}=1A$ enters the network at node A, spreads out symmetrically in every outward direction and comes out at a circular conducting ring at infinity. Separately there is case B : current $I_{out}=1A$ enters at the conducting ring at infinity, spreads inward symmetrically through the network and comes out at node B. Case B is the reverse of case A, shifted along by one node, so the currents in corresponding branches are the same.

In case A, $1A$ is injected at A so by symmetry the current in branch AB will be $\frac14A$. In case B we have the same result that a current of $\frac14A$ flows along branch AB in the same direction AB.

The problem we have is the superposition of cases A and B : a current of $1A$ flows in at A and exits at B. The currents in and out at infinity ($\frac34 A$) cancel out so that all current which goes in at A comes out at B.

In branch AB the two currents are in the same direction, making a total of $\frac12 A$. If branch AB has unit resistance then the voltage across the network between AB is $V_{out}=\frac12V$. The equivalent resistance of the network between A and B is $$R_{AB}=\frac{V_{out}}{I_{in}}=\frac{\frac12 V}{1A}=\frac12 \Omega$$

The same trick works for capacitance, except that this time we are injecting charge instead of current.

We now inject charge of $Q_{in}=+1\mu C$ at A (case A). Without any charge removed at B the charges on each capacitor adjacent to A, especially that in branch AB, are $\pm \frac14 \mu C$.

In case B we remove charge $Q_{out}=+1\mu C$ at B, which is equivalent to injecting charge $-1\mu C$. The charges on each capacitor adjacent to B, including that along AB, are $\pm \frac14 \mu C$, again with the +ve side connected to A and the -ve side to B.

Superposing the two cases (injection at A, removal at B) the total charge on capacitor AB is $Q_{AB}=\frac12 \mu C$ and the voltage across this unit capacitor is $V_{out}=\frac12 V$. The equivalent capacitance of the network is $$C_{eq}=\frac{Q_{in}}{V_{out}}=\frac{1\mu C}{\frac12 V}=2 \mu F$$

answered Jun 17, 2018 by (28,448 points)
selected Jun 19, 2018 by koolman


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# Better Lambda logging with Python Powertools

When writing Lambdas in Python there are some things I usually have to do in order to get them working. It usually goes something like this:

1. Create python lambda
2. Dump the event so I can inspect it and see what I need
3. Make changes
5. Rinse and Repeat

So the initial Lambda looks something like this:

But lately, I’ve discovered some solutions to make this better.

## Generate test events

First off, rather than creating a Lambda, dumping the event to view it and continuing from there, use SAM CLI to create example events:

That outputs this:

This results in example events you can save for testing and customize as needed. Many events (but not all) are supported. Now you can use something like sam local invoke and point to that event for testing. Coupled with either moto or localstack you can use it for testing purposes locally.

## Better Lambda Logging

For a long time I’ve just used print() statements in Lambdas to get what I wanted out of them. This was a big mistake. First off, AWS extends the logging format used by the default Python logging library, so just by using that you get additional context of log level, and invocation in the log output. But there is even a better solution, Lambda Powertools Python.

By using this logging functionality you can get better structured logs in CloudWatch Logs and even automatically include the event. The example Lambda to invalidate a CloudFront cache below uses it to output the event to the logs, and also logs some additional info in a structured format.

Resulting in logs that look like this:

Now the resulting logs are much easier to review visually, easier to parse if needed and easier to query with CloudWatch Log Insights. All with just an additional dependency and adding a few lines of code to the Lambda.


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CGAL 4.7 - 2D and 3D Linear Geometry Kernel
Kernel::CoplanarOrientation_3 Concept Reference

## Definition

Refines:
AdaptableFunctor (with four arguments)
CGAL::coplanar_orientation()

## Operations

A model of this concept must provide:

Orientation operator() (const Kernel::Point_3 &p, const Kernel::Point_3 &q, const Kernel::Point_3 &r, const Kernel::Point_3 &s)
Let p be the plane defined by the points p, q, and r. More...

Orientation operator() (const Kernel::Point_3 &p, const Kernel::Point_3 &q, const Kernel::Point_3 &r)
If p,q,r are collinear, then CGAL::COLLINEAR is returned. More...

## Member Function Documentation

 Orientation Kernel::CoplanarOrientation_3::operator() ( const Kernel::Point_3 & p, const Kernel::Point_3 & q, const Kernel::Point_3 & r, const Kernel::Point_3 & s )

Let p be the plane defined by the points p, q, and r.

Note that the order defines the orientation of p. The function computes the orientation of points p, q, and s in p: Iff p, q, s are collinear, CGAL::COLLINEAR is returned. Iff p and the plane defined by p, q, and s have the same orientation, CGAL::POSITIVE is returned; otherwise CGAL::NEGATIVE is returned.

Precondition
p, q, r, and s are coplanar and p, q, and r are not collinear.
 Orientation Kernel::CoplanarOrientation_3::operator() ( const Kernel::Point_3 & p, const Kernel::Point_3 & q, const Kernel::Point_3 & r )

If p,q,r are collinear, then CGAL::COLLINEAR is returned.

If not, then p,q,r define a plane p. The return value in this case is either CGAL::POSITIVE or CGAL::NEGATIVE, but we don't specify it explicitly. However, we guarantee that all calls to this predicate over 3 points in p will return a coherent orientation if considered a 2D orientation in p.


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1. ## Quick question

If $\left = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right)$

What is $\left$?

If $\left = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right)$

What is $\left$?
so square both sides of your initial equation
$(x)^2 = \frac{1}{n^2}\left(x_{1}+x_{2}+...+x_{n}\right)^2$

and evaluate further if you know what the x1,x2,....etc are

so square both sides of your initial equation
$(x)^2 = \frac{1}{n^2}\left(x_{1}+x_{2}+...+x_{n}\right)^2$

and evaluate further if you know what the x1,x2,....etc are
It's not $\left^2$ I want to find. It's $\left
$
.

Basically I'm trying to solve for $x$ the equation:

$2.61 = \sqrt{\left-\left^2}$ to find the value of $x$, where $2.61$ is the $\text{standard deviation}$ of the data $8,7, 4, 10, 6, x$.

I know that $\left = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right) = \frac{1}{6}\left(8+7+4+10+6+x\right) = \frac{1}{6}\left(35+x\right)$.

$\Rightarrow \left = \left[\frac{1}{6}\left(35+x\right)\right]^2 = \frac{1}{36}\left(35+x\right)^2$.

So $2.61 = \sqrt{\left-\frac{1}{36}\left(35+x\right)^2}$

But I've no idea how to get $\left$.

It's not $\left^2$ I want to find. It's $\left
$
.

Basically I'm trying to solve for $x$ the equation:

$2.61 = \sqrt{\left-\left^2}$ to find the value of $x$, where $2.61$ is the $\text{standard deviation}$ of the data $8,7, 4, 10, 6, x$.

I know that $\left = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right) = \frac{1}{6}\left(8+7+4+10+6+x\right) = \frac{1}{6}\left(35+x\right)$.

$\Rightarrow \left = \left[\frac{1}{6}\left(35+x\right)\right]^2 = \frac{1}{36}\left(35+x\right)^2$.

So $2.61 = \sqrt{\left-\frac{1}{36}\left(35+x\right)^2}$

But I've no idea how to get $\left$.
$s = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \overline{X} }$.

5. Originally Posted by mr fantastic
$s = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \overline{X} }$.
$\left = \frac{\displaystyle\sum_{i=1}^n x_i^2}{n}?$

6. Just tell me how to calculate, or give me definition of, $\left$ and I will take care of the rest.

Just tell me how to calculate, or give me definition of, $\left$ and I will take care of the rest.
$ = \frac{x_1^2 + x_2^2 + x_3^2 + .... + x_n^2}{n}$.
$ = \frac{x_1^2 + x_2^2 + x_3^2 + .... + x_n^2}{n}$.


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# No limit

1. Oct 2, 2007

### azatkgz

f(x)=1,xeQ
f(x)=0,x in not eQ

It's easy to understand that this function has no limit as x tends to 0,but how we can prove it with mathematical formulas.

And what's the lim[e^n-(1+1/n)^n^2]

Last edited: Oct 2, 2007
2. Oct 2, 2007

### quasar987

Prove it with the definition of limit.

Can you find an epsilon such that no delta-nbhd of 0 is mapped entirely in an epsilon-nbhd of f(0)=1?

Or use the characterisation in terms of sequences. The limit is f(0) iff for every sequence converginf to 0, the image sequence converges to f(0). Can you find a sequence such that the image sequence does not converge to f(0)?

Last edited: Oct 2, 2007
3. Oct 3, 2007

### azatkgz

quasar987,
I could not solve with sequences.Can u plz post the solution.

4. Oct 3, 2007

### HallsofIvy

Staff Emeritus
So far you haven't shown that you have tried anything at all! What have you tried?

Have you thought about what f(x) is if x is close to 0 and rational?
What f(x) is if x is close to 0 and irrational?
COULD f(x) be "close" to some limit if x is any number close to 0?


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# In NMinimize, how to incorporate constraints on positive semidefiniteness of a matrix which is not the matrix variable being optimised?

## The problem

I want to solve the following problem for symmetric matrix $$X$$:

\begin{aligned} \min_{X\succ 0} \; & -\log(\det(X)) & \\ \text{subject to} \; & \begin{pmatrix} X & X\bar A^T\\ \bar AX & X \end{pmatrix} \succeq 0 \\& 1-g_i^TXg_i\geq 0 \\ & 1-h_l^TKXK^Th_l\geq 0\\ & \alpha \mathcal{T}^{-1}\succeq X \end{aligned}

where $$\bar A, K, \mathcal{T}=\mathcal{T}^T, G, H$$ are matrices and $$g_i, h_l$$ are the $$i^{th}, l^{th}$$ rows of $$G, H$$ respectively and the indices $$i, l$$ range from 1 upto the number of rows in $$G, H$$ respectively.

## My attempts

I have managed to solve this without the two semidefinite constraints (the working code is below) but I am not able to figure out how to incorporate the two semidefinite constraints.

### Code without the semidefinite constraints

This code successfully solves the problem by using Cholesky decomposition, which would automatically ensure $$X$$ is positive definite if the diagonal elements of $$L$$ are positive.

Abar={{0., 1.}, {-0.000165779, 4.97343*10^-7}};
T={{0.104851, 0.00645711}, {0.00645711, 0.106041}};
K={{-56.8, 0.000029995}};
alpha=0.2;

X = With[{L = {{a, 0}, {b, c}}}, L.Transpose[L]];
MAT = (K.X.Transpose[K])[[1]][[1]];

cons=True;

G = {{1/15, 0}, {-1/15, 0}, {0, 1/100}, {0, -1/100}};
H = {{1/10}, {-1/10}};
Do[cons = cons && 1 - G[[i]].X.G[[i]] >= 0, {i, Dimensions[G][[1]]}];
Do[cons = cons && 1 - H[[i]][[1]]*MAT*H[[i]][[1]] >= 0, {i, Dimensions[H][[1]]}];

NMinimize[{-Log[Det[X]], cons && a \[Element] Reals && a > 0 && b \[Element] Reals && c\[Element]Reals && c > 0}, {a, b, c}]

The output is {-5.73644,{a->0.176056,b->0.0299951,c->100.}}.

### Code with semidefinite constraints

Now this code doesn't work. I read this answer and used the Thread[Eigenvalues[M]>=0] for my semidefinite constraints since both matrices in those constraints are symmetric. But it doesn't work. The following is my code.

Abar={{0., 1.}, {-0.000165779, 4.97343*10^-7}};
T={{0.104851, 0.00645711}, {0.00645711, 0.106041}};
K={{-56.8, 0.000029995}};
alpha=0.2;

X = With[{L = {{a, 0}, {b, c}}}, L.Transpose[L]];
MAT = (K.X.Transpose[K])[[1]][[1]];

G = {{1/15, 0}, {-1/15, 0}, {0, 1/100}, {0, -1/100}};
H = {{1/10}, {-1/10}};
Do[cons = cons && 1 - G[[i]].X.G[[i]] >= 0, {i, Dimensions[G][[1]]}];
Do[cons = cons && 1 - H[[i]][[1]]*MAT*H[[i]][[1]] >= 0, {i, Dimensions[H][[1]]}];

NMinimize[{-Log[Det[X]], cons && a \[Element] Reals && a > 0 && b \[Element] Reals && c\[Element]Reals && c > 0}, {a, b, c}]

The errors it gives are like GreaterEqual::nord: Invalid comparison with 1.25754 -1.57974i attempted. So for some reason, it seems to be getting imaginary eigenvalues or something which should happen since all matrices are symmetric.

### Other thoughts

I tried thinking about using the $$LDL^T$$ decomposition of positive semidefinite matrices with the condition that diagonal terms of the unique $$D$$ are non-negative or the fact that all principal minors of a positive semidefinite matrix are non-negative but couldn't figure out a way to incorporate the first one and the second one doesn't seem promising because as a trial case, I put the constraint that all leading principal minors of the semidefinite matrix are non-negative but the output was just NMinimize[...], no errors nothing just the input copied as output.

I would be extremely grateful if someone could help me out here! Thank you very much.

I got it and am posting this answer for others who might have this question in future. Of course, I am a beginner and any comments and suggestions on this will be helpful and welcome.

The SemidefiniteOptimization function is the way to go. It allows one to specify semidefinite cone constraints, which is same as a positive semidefiniteness constraint. The objective needs to be linear though, which as stated in my problem, is not. But fortunately, this example (on minimization of ellipsoid area) in this function's documentation tackles exactly that and we can equivalently write the objective function negative of trace of X. One more caveat is that numerical solvers (many if not all) do not prefer positive definiteness constraints and it's no different for this function, in fact, there's no way to give a positive definiteness constraint directly (one may use Cholesky decomposition and stuff but it lead to quadratic objective function. There may be a way to make that work but I haven't put much thought into that). Instead we ensure positive definiteness with some margin. I did it by constraining $$X-0.001I_{2\times 2}$$ to be inside a second order semidefinite cone, i.e., positive semidefinite.

I further cross-checked the answer obtained here with that from the well-tested YALMIP package in MATLAB and they match upto 2-3 decimal places. Here is the code:

Abar={{0., 1.}, {-0.000165779, 4.97343*10^-7}};
T={{0.104851, 0.00645711}, {0.00645711, 0.106041}};
K={{-56.8, 0.000029995}};
alpha=0.2;

X = {{a, b}, {b, c}};
MAT = (K.X.Transpose[K])[[1]][[1]];

cons = VectorGreaterEqual[{ArrayFlatten[{{X, X.Transpose@Abar}, {Abar.X, X}}],0},{SemidefiniteCone,4}] && VectorGreaterEqual[{alpha*Inverse@T - X,0},{SemidefiniteCone,2}]
&& VectorGreaterEqual[{X - 1/1000*IdentityMatrix[n],0},{SemidefiniteCone,2}];

G = {{1/20, 0}, {-1/20, 0}, {0, 1/5}, {0, -1/5}};
H = {{1/30}, {-1/30}};
Do[cons = cons && 1 - G[[i]].X.G[[i]] >= 0, {i, Dimensions[G][[1]]}];
Do[cons = cons && 1 - H[[i]][[1]]*MAT*H[[i]][[1]] >= 0, {i, Dimensions[H][[1]]}];

X = X /. SemidefiniteOptimization[-Tr[X], cons && a \[Element] Reals && b \[Element]Reals && c \[Element] Reals, {a, b, c}];


text:
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The completion of the reaction is usually shown by a change of color caused by a substance called an indicator, and to calculate conc of unknown(acid or base) from the known (acid or base) conc, we use this formela formula M1*V1 =M2 * V2, You need to start with a reaction. Did you see bubbles? $$\ce{Na_2CO_3 +HCl ->NaHCO_3 + NaCl}$$ Aqueous sodium carbonate titration with HCl solution. Na2CO3 + 2HCl ===> 2NaCL + H20 + CO2. As: $0.0358 >> K_\mathrm{a1}$,justifies the further approximation of dropping the $[\ce{H+}]$ term in the denominator. in titration, one solution (solution 1) is added to another solution (solution 2) until a chemical reaction between the components in the solutions has run to completion. How did you do the titration? Shouldn't some stars behave as black holes? So if … // What would you average with 6.3665 to get second eq. This reaction is a kind of acid neutralisation reaction, whereby an acid reacts with a carbonate to form carbon dioxide gas, salt and water. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Na2CO3 (aq) + 2HCl (aq) --------> CO2 (g) + 2NaCl (aq… $$\pu{V_\ce{HCl}}=31.25\times{10^{-3}}$$ Still have questions? Should I call the police on then? $$\pu{V_\ce{HCl}}=62.25\times{10^{-3}}$$ This is bicarbonate. Okay, so I have had to do an experiment at school for chemistry. pt. Can anyone please help? The reaction between sodium carbonate and hydrochloric acid takes place in two stages: Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq) (1) NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l) (2) Titration of 50 mL of Na2CO3 (0.1M) with HCl (0.2M) Calculation of pH values at key points: Solution pH Na2CO3(aq) + 2HCl(aq) --> 2NaCl(aq) + H2CO3(aq) I have to write the Complete Ionic equation, spectator ions and Net Ionic equation. Hydrochloric acid solution may be titrated against sodium carbonate solution using methyl orange indicator. It only takes a minute to sign up. $$K_\mathrm{a1} =4.3\times{10^{-7}} =\frac{[\ce{H^+}]^2}{0.0358}$$ $$[\ce{H_3O^+}]=\sqrt\frac{K_\mathrm{a1}K_\mathrm{a2}[\ce{NaHCO_3}]+K_\mathrm{a1}K_\mathrm{w}}{K_\mathrm{a1}+[\ce{NaHCO_3}]}$$, $$[\ce{H_3O^+}]=\sqrt\frac{(4\cdot{3}\times{10^{-7}}\times{4\cdot{8}\times{10^{-11}}}\times{0.056})+(4\cdot{3}\times{10^{-7}}\times{10^{-14}})}{(4\cdot{3}\times{10^{-7}}+0.056)}\approx{5.35\times{10^{-9}}}\pu~{M}$$ I know that the carbonic acid quickly decomposes into: H2CO3 (aq) --> H2O(l) + CO2(g) but im unsure of which chemical equation to use for the complete and net ionic equations. At the first equivalence point: The titration works because twice as much HCl in moles was used as was the Na2CO3. Should recorded lectures be provided for students when teaching a math course online? $$\text{moles of}~ \ce{HCl} =2\times{3.125\times{10^{-3}}}=\pu{V_\ce{HCl}\times{0.1}}$$ Has anyone seriously considered a space-based time capsule? Does not work because you need 2 moles of HCl for every mole of Na2CO3. when this happens. Figuring out from a map which direction is downstream for a river? The titration reaction is CO3 (2-) + 2H + - --> H2O + CO2 What is the concentration of HCl if 35.7 mL of HCl was needed to to react with 0.521 g of Na2CO3 to reach the end point? Solution 1 is called the titrant, and we say that it is used to titrate solution 2. Use MathJax to format equations. The bubbles were the CO2 escaping from the solution. When weak base is titrated with a strong acid solution is slightly acidic at end point. $$\pu{pH}=\frac{1}{2}\times{(pK_\mathrm{a1}+pK_\mathrm{a2} )}$$ THis foumula suitable when $K_\mathrm{a1}<<[\ce{HCO_3^-}]\text{with concentrated solution,but neglegt water autoionization and assume} [\ce{H2CO3}] = [\ce{CO3-} ]$, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…, “Question closed” notifications experiment results and graduation, Calculating pH for titration of weak base with strong acid, Calculating molecular weight of an acid based on titration data, Predicting pH of an acidic salt of an weak acid and weak base, Hydrolysis of salts containing amphiprotic anion, Acid-base titration: Calculate pKa with only three values given, How to calculate pH of the Na2CO3 solution given ambiguous Ka values, Finding mass percentage of components of a solid mixture. Write the formula for the compound that has the atoms and, or groups in the order given: 3 Fe, and two groups made up of 1 As and 4 O.? site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Today, you will use it to find the concentration of dilute hydrochloric acid by titration. which makes you more jittery coffee or tea? The bubbles were the CO2 escaping from the solution. This process is known as standardising the hydrochloric acid. Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. The chief was seen coughing and not wearing a mask. View Titration of Na2CO3 - 2020-1.pdf from CHEM 3611 at University of the West Indies at Mona. also this cant be extended to calculate the second equivalence point. $$K_\mathrm{a1} = 4.3\times{10^{–7}} =\frac{[\ce{H^+}]^2}{0.0358-[\ce{H+}] }$$ View Titration of Na2CO3 - 2020-1.pdf from CHEM 3611 at University of the West Indies at Mona. La phénolphtaléine convient au premier stade et le méthylorange est le meilleur pour la seconde. At eq. In Star Trek TNG Episode 11 "The Big Goodbye", why would the people inside of the holodeck "vanish" if the program aborts?

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text:
# Using a determinant to find the area of the triangle (deriving the formula)

Sunwoo Bae
Homework Statement:
Shown in the text
Relevant Equations:
Det(A) = det(A^T) determinant of transpose equals determinant of the original matrix

This is the question. The following is the solutions I found:

I understand that the first line was derived by setting one vertex on origin and taking the transpose of the matrix. However, I cannot understand where the extra row and column came from in the second line. Can anyone explain how the formula is derived?

Thank you!

$$\det \begin{pmatrix}a & b & 1 \\ c & d & 0 \\ e & f & 0\end{pmatrix}= \det \begin{pmatrix} c & d \\ e & f\end{pmatrix}$$


text:
43

# ANGIOTENSIN CONVERTING ENZYME INHIBITORY ACTIVITY AND ANTIOXIDANT ACTIVITIES OF AQUEOUS EXTRACT OF COMBRETUM MICRANTHUM LEAVES

CHAPTER ONE

1.0 INTRODUCTION

1.1 Background of the Study

The word hypertension is defined as a persistence increase in systemic arterial blood pressure (Sembulingam and Sembuligam, 2006). Clinically, when the systolic pressure remains elevated above 140mmHg and diastolic pressure remains elevated above 90mmHg, it is considered as hypertension. The prevalence varies with age, race, education, occupation and many other variables (Benowitz, 2009). In Nigeria for example, the true incidence of hypertension remains unknown but its prevalence among male and female is estimated to be 11.2% with age adjusted figure of 9.3% (Nurudeenet al., 2013). This translates into approximately 13.4 million Nigerians becoming hypertensive at the age of 15years and above, using the projected national population census figure of 120million (Akinkungbe, 1998). In fact, hypertension is reported to be next to malaria as most serious health problems in developing tropical countries (Agunwa, 1988).

The global dimension of hypertension is immense, as it ranks the most common cardiovascular ailment afflicting about one billion people in the world and causing roughly 7.1 million deaths annually (Brundtland, 2002). Hypertension is said to be the most common cardiovascular disease among Africans and congestive cardiac failure its commonest complication (Akinkungbe 1972, 1985). Earlier studies suggested that hypertension was rare in African population (Sharper et al., 1969, Pobeeet al., 1977), however, epidemiological transition, urbanization, adoption of urban and foreign lifestyles and improved case findings, among others, have made hypertension more prevalent as shown in some studies (Cooper et al., 1998). The last Nigerian National Non-communicable Disease Survey (NNCDS) conducted in 1997 reported 11.4% prevalence of adult hypertension, varying from 14.8% in urban to 9.8% in rural residences respectively.

Contents


text:
## Theory of Combinatorial Algorithms

Prof. Emo Welzl and Prof. Bernd Gärtner

# Mittagsseminar (in cooperation with M. Ghaffari, A. Steger and B. Sudakov)

 Mittagsseminar Talk Information

Date and Time: Tuesday, March 19, 2013, 12:15 pm

Duration: 30 minutes

Location: CAB G51

Speaker: Benjamin Doerr (MPI Saarbruecken)

## Mastermind With Many Colors

We analyze the general version of the classic guessing game Mastermind with $n$~positions and $k$~colors. Since the case $k \le n^{1-\eps}$, $\eps>0$ constant, is well understood, we concentrate on larger numbers of colors. For the most prominent case $k = n$, our results imply that Codebreaker can find the secret code with $O(n \log \log n)$ guesses. This bound is valid also when only black answer-pegs are used. It improves the $O(n \log n)$ bound first proven by Chv\'atal (Combinatorica 3 (1983), 325--329). We also show that if both black and white answer-pegs are used, then the $O(n \log\log n)$ bound holds for up to $n^2 \log\log n$ colors. These bounds are almost tight as the known lower bound of $\Omega(n)$ shows. Unlike for $k \le n^{1-\eps}$, simply guessing at random until the secret code is determined is not sufficient. In fact, we show that an optimal non-adaptive strategy (deterministic or randomized) needs $\Theta(n \log n)$ guesses.

Previous talks by year:   2018  2017  2016  2015  2014  2013  2012  2011  2010  2009  2008  2007  2006  2005  2004  2003  2002  2001  2000  1999  1998  1997  1996

Information for students and suggested topics for student talks


text:
# How do you solve the system 3x + y = 6 and 6x + 2y = 8 ?

##### 1 Answer
May 17, 2015

Divide both sides of the second equation by 2 and we get $3 x + y = 4$

There's a disagreement with the first equation, so this system has no solution.


text:
Introduction to Hamiltonian Monte Carlo

One thing that has been occupying my head in the past couple of weeks has been HMC and how it can be used in large data/large model context. HMC stands for Hamiltonian Monte Carlo and it’s the de facto Bayesian method for sampling due to it’s speed. Before getting into big datasets and big models, let me motivate this problem a little bit.

If you are new to Bayesian modelling, I have a little primer on the topic so I will assume for the most part you are familiar with basic Bayesianism. Let’s say we build our cute model by constructing a likelihood $p(y|x, \theta)$ and prior $p(\theta)$, then to obtain the posterior, we can use Bayes rule:

$$p(\theta| x, y) = \frac{p(y | x, \theta)p(\theta)}{\int_\theta p(y | x, \theta)p(\theta)}$$

Now when we can work out the integral at the bottom analytically, all is well, we just plug it in and BAM, out comes the posterior. Unfortunately, this is a rather special case and more often than not, we don’t know what that integral is. We usually know the posterior up to a constant:

$$p(\theta| x, y) \propto f(\theta) = p(y | x, \theta)p(\theta)$$

If you are from the machine learning crowd, you can think of this $f$ as taking in as the input the weights of the neural network and outputting the loss of the neural network with those weights on your whole data.

Sampling is hard

First thing almost everyone learns in undergraduate maths is the trapezoid rule for estimating the integral. Basically, you take equally spaced points in your space, evaluate the function at those points and multiply by some volume factor. To achieve an error rate of $\epsilon$, you are going to need $O(\epsilon^{-2p})$ number of points. Let’s put some real numbers here, suppose we want an error rate of 0.01 for say ResNet-152 which has 60m parameters. This would require roughly $10^{200,000,000}$ points to evaluate. Just to put this into perspective, there is an estimate number of $10^{80}$ atoms in the observable universe!

Enter in Markov chain Monte Carlo

Markov chain Monte Carlo is a simple idea: you build a Markov chain, which has its invariant measure $p(\theta| x, y)$, then with some regularity assumptions you get that the chain converges to this distribution.

For convenience, let’s just call our measure of interest $\pi(\theta) = p(\theta| x, y)$ and remember that we can only compute this up to some unknown constant with $f(\theta)$. We can specify a Markov chain by its transition probabilities $K(\theta' | \theta)$ which notes the probability of moving to state $\theta'$ when we are in state $\theta$. The detailed balance equations state the Markov chain with transition probabilities $K$ has a unique stationary distribution $\pi$ if

$$\pi(\theta) K(\theta' | \theta) = \pi(\theta') K(\theta | \theta').$$

Under certain regularity assumptions we can also show that the Markov chain converges to this stationary distribution as time tends to infinity.

So how can we construct a Markov chain with $\pi$ as the stationary measure when we only know it up to a constant?

There is an easy way of doing this using the Metrapolis Hastings algorithm. The idea is given we are at $x$, we are going to propose a new location $x'$ using the pdf $\zeta(x'| x)$. Then you build a Markov chain like this:

1. Given $X_t$ we move as follows.
2. Sample $\xi_t$ from $\zeta(\cdot | X_t)$.
3. Set the acceptance probability to be $a_t := \min\left\{1, \frac{f(\xi_{t})\zeta(X_t| \xi_t)}{f(X_t)\zeta(\xi_t| X_t)}\right\}$

then the transition to the next step is given by

$X_{t+1} = \begin{cases} \xi_t & \text{with probability } a_t \\ X_{t} & \text{with probability }1 - a_t \end{cases}$

In words, we have a proposal $\xi_t$ for each step. If the proposal $\xi_{t}$ has a high enough probability compared to $X_{t}$, we accept it, otherwise we reject the proposal.

Writing out the transition probabilities, you can see easily that they satisfy the detailed balance equation because $\frac{f(\xi_{t})}{f(X_t)}=\frac{\pi(\xi_{t})}{\pi(X_t)}$.

The most obvious proposal distribution is given by a normal distribution: $$\zeta(x'| x) = \frac{1}{(2\pi)^{d/2}} e^{-\frac{1}{2}(x' - x)^T (x' - x)}$$ and for this distribution we have a key properly which is $\zeta(x'| x) = \zeta(x| x')$ so that the acceptance probability is just the ratio of the proposed points.

In the implementation below, I have worked with the negative log densities, e.g. $-\log \pi$ which is called surface. Generally it’s easier to do the computations above in log land and it will become even more so when we look at Hamiltonian Monte Carlo.

import numpy as np
from torch.distributions import MultivariateNormal
import matplotlib.pyplot as plt
import torch
plt.style.use('ggplot')

def generate_cov(state):
corr = state.uniform()
cov = state.uniform(0, 0.5) * np.array([[1, corr], [corr, 1]])
return cov

class TestSurface(object):
def __init__(self, n=10):
self.n = n
state = np.random.RandomState(2021)
self.rvs = [
MultivariateNormal(
torch.tensor(state.uniform(-1, 1, size=(2))),
torch.tensor(generate_cov(state))
)
for _ in range(n)
]

def __call__(self, x):
output = 0.0
for rv in self.rvs:
output += torch.exp(rv.log_prob(x)) / self.n
return -torch.log(output)

class GaussianMH(object):
def __init__(self, surface, scale=0.5):
self.surface = surface
self.scale = scale
self.state = np.random.RandomState(1234)

def step(self, x=None):
if x is None:
x = np.array([0.0, 0.0])
sample = self.state.normal(size=2, scale=self.scale)
proposal = x + sample
if self.is_accepted(x, proposal):
return proposal
return x

def is_accepted(self, x_old, x_new):
new_log_pba = - self.surface(torch.tensor(x_new))
old_log_pba = - self.surface(torch.tensor(x_old))
accept_pba = (new_log_pba - old_log_pba).exp().detach().numpy()
if accept_pba > 1:
return True
if self.state.rand(1) < accept_pba:
return True
return False

surface = TestSurface(15)

mh = GaussianMH(surface)
walk = [mh.step()]
for _ in range(100):
walk.append(mh.step(walk[-1]))
walk = np.stack(walk, axis=0)

fig = plt.figure(figsize=(12, 12))
ax = plt.axes(projection='3d')
x = np.linspace(-1, 1, 100)
y = np.linspace(-1, 1, 100)
X, Y = np.meshgrid(x, y)
input = torch.tensor(np.stack([X.reshape(-1), Y.reshape(-1)], axis=1))
Z = surface(input).numpy().reshape(*X.shape)

ax.scatter(walk[:, 0], walk[:, 1],
surface(torch.tensor(walk)).numpy(),
marker='x', c='blue', alpha=1.0, label='Accepted')
ax.contour3D(X, Y, Z, 100)
ax.view_init(35, 35)
ax.set_xlabel('')
ax.set_ylabel('')
ax.set_zlabel('Negative log probability')
fig.show()

Hamiltonian Monte Carlo

Metropolis-Hastings algorithm with Gaussian updates do not scale well to high dimensions. The reason for this is because in high dimensions, the probability that selecting a random point has a high enough density is pretty low and just gets lower as we increase the dimension. So in high dimensions, it’s essential to make sure we propose points in a way that respects the density we are trying to sample from.

The idea of the Hamiltonian Monte Carlo is to introduce an auxiliary variable and sample jointly from this bigger space. For this, we look at the space $(\theta, r) \in \mathbb{R}^{d + d}$ where $r \in \mathbb{R}^d$ is called the momentum (we will see why in a bit). Next the joint density on this space is given by:

$$q(\theta, r) = \pi(\theta) \mathcal N(r| 0, \Sigma) \propto f(\theta) \mathcal N(r| 0, \Sigma)$$ where $\Sigma$ is a parameter which we can pick. In words, the measure $q$ is the product of two independent random variables, $\pi$ on the $\theta$ part and a normal random variable on the momentum part.

For convenience, let: $$U(\theta) = -\log f(\theta) \qquad K(r) = -\log \mathcal N(r| 0, \Sigma)$$ so that $$q(\theta, r) = \frac{1}{Z}e^{-U(\theta)} e^{-K(r)}.$$

Now imagine that we have a ball at position $\theta$ with momentum $r$, then we make this ball move on the log density $U$ by using Hamiltonian dynamics: \begin{aligned} \frac{d\theta}{dt} & = \frac{dK}{dr} \\ \frac{dr}{dt} & = \frac{dU}{d\theta} \end{aligned}

Let’s refer to the solution at time $t$ by $(\theta_t, r_t)$ from some initial point. The really cool thing about the Hamiltonian dynamics is that it retains the same density throughout: $$q(\theta_t, r_t) = q(\theta_{t'}, r_{t'}) \qquad \forall , t, t' >0$$

OK, so now we are set to describe how we can use this. First, we are going to sample from the joint density $q$, after which, we can just ignore $r$ samples which will result in samples from $\pi$. We can use the deterministic proposal like this: $$\zeta(\theta', r' | \theta, r) = \delta_{\text{H}_t(\theta, r)}(\theta', -r')$$ where $\text{H}_t$ is the Hamiltonian dynamics applied for time $t>0$ (which is a parameter we can adjust). So in this case, our proposal is deterministic.

Why do we have $-r'$ in the equation above? Well if you look closely at the dynamics, you will realise that if $(\theta_t, r_t) = \text{H}_t(\theta, r)$, then $(\theta, r) = \text{H}_t(\theta_t, -r_t)$, so adding the minus sign gives us reversibility. Because the Hamiltonian dynamics gives us a point with the same probability as the starting point, this means the Metropolis-Hastings acceptance ratio is always 1, i.e. we always accept the proposal generated in this form. This means that this transition leaves the measure $q$ invariant.

Instead of applying Metropolis-Hastings updates repeatedly, we can utilise an other trick. The reason for this trick is to reduce the autocorrelation of the samples, because remember Metropolis-Hastings guarantees that samples are from the target distribution, but they might be very correlated, which is obviously not ideal. To do this, we can just sample the momentum variable $r$ at each step. To conclude, the HMC algorithm is:

1. Sample $r \sim N(0, \Sigma)$ where $\Sigma$ is a parameter that the user selects.
2. Run Hamiltonian dynamics for time t and record the resulting $\theta_t$.

Since both of those steps leave the measure $q$ invariant we are guaranteed under regularity assumptions to sample from the measure $\pi$ as we run more steps.

Finally there is a final step. In practice we can’t simulate the Hamiltonian dynamics exactly. In the example below, I have used the Leapfrog method for this. This means that our proposed state will have some degree of error and thus will not quite have the same density under $q$. This means in practice we have to run a Metropolis-Hastings proposal step with the proposal at the end:

1. Work out the acceptance probability $a_t = \min\left\{1, \frac{f(\theta_t)}{f(\theta)}\right\}$ and accept the sample with probability $a_t$.
from tqdm.autonotebook import tqdm
class HMC(object):
def __init__(self, surface, step_size=1e-4, path_length=100):
self.surface = surface
self.step_size = step_size
self.path_length = path_length
self.state = np.random.RandomState(1234)

def step(self, x=None, momentum=None):
if momentum is None:
momentum = torch.tensor(self.state.normal(size=2))
if x is None:
x = torch.tensor([0.0, 0.0])
# Leap-frog method on numerically solving the Hamiltonian eq
point = self.surface(x)
point.backward()
momentum = momentum - 0.5 * self.step_size * x.grad
x += self.step_size * momentum
point = self.surface(x)
point.backward()
momentum = momentum - 0.5 * self.step_size * x.grad
return x.detach(), momentum.detach()

def _calculate_hamiltonian(self, x, momentum):
# Negative log density of normal up to constant
momentum_term = (momentum ** 2).sum()
x_term = self.surface(x)
return x_term + momentum_term

def is_accepted(self, x_old, x_new, momentum_old, momentum_new):
accept_pba = (
self._calculate_hamiltonian(x_old, momentum_old) -
self._calculate_hamiltonian(x_new, momentum_new)
).exp()
if accept_pba > 1:
return True
if torch.rand(1) < accept_pba:
return True
return False

def walk(self, x=None, momentum=None):
x, momentum = self.step(x=x, momentum=momentum)
for _ in range(self.path_length - 1):
x, momentum = self.step(x=x, momentum=momentum)
return x, momentum

hmc = HMC(surface, step_size=0.1, path_length=10)
walks = []
is_accepted = []
x = torch.tensor([0.0, 0.0])
momentum = torch.tensor(hmc.state.normal(size=2))
for _ in tqdm(range(100)):
x_new, momentum_new = hmc.walk(x=x, momentum=momentum)
is_accepted.append(hmc.is_accepted(x, x_new, momentum, momentum_new))
walks.append(np.copy(x_new.detach().numpy()))
x = x_new
momentum = torch.tensor(hmc.state.normal(size=2))
walks = np.stack(walks, axis=0)
is_accepted = np.array(is_accepted)

fig = plt.figure(figsize=(12, 12))
ax = plt.axes(projection='3d')
x = np.linspace(-2, 2, 100)
y = np.linspace(-2, 2, 100)
X, Y = np.meshgrid(x, y)
input = torch.tensor(np.stack([X.reshape(-1), Y.reshape(-1)], axis=1))
Z = surface(input).numpy().reshape(*X.shape)
accepted_walk = walks[is_accepted, :]
rejected_walk = walks[~is_accepted, :]
ax.scatter(accepted_walk[:, 0], accepted_walk[:, 1],
surface(torch.tensor(accepted_walk)).numpy(),
marker='x', c='blue', alpha=1.0, label='Accepted')
ax.scatter(rejected_walk[:, 0], rejected_walk[:, 1],
surface(torch.tensor(rejected_walk)).numpy(),
marker='x', c='red', alpha=1.0, label='Rejected')
ax.contour3D(X, Y, Z, 100)
ax.view_init(45, 20)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
ax.legend()
ax.set_title('Hamiltonian Monte Carlo samples')
fig.show()

Where to from here?

HMC is a great way to sample in high dimensions because the acceptance probability is going to be very close $1$ (it’s the error between the discrete ODE vs continuous ODE). This makes it a really nice tool. I have below some further materials if you would like to get a deeper dive into HMC.


text:
# How do you find the intercepts, asymptotes and graph y=5^x?

Jan 25, 2016

y-intercept = ( 0 , 1 ) This function has no x-intercept
and no asymptotes.

#### Explanation:

To find y-intercept let x = 0 : $y = {5}^{0} = 1$

To find x-intercept let y = 0 : ${5}^{x} = 0 \textcolor{b l u e}{\text{ has no solution}}$

This function is continually increasing and so no asymptote.

graph{5^x [-10, 10, -5, 5]}


text:
## anonymous 3 years ago how far can a car travel in 40 minutes if it is traveling at an average speed of 50 m\ph

• This Question is Open
1. karatechopper

|dw:1357186275081:dw|

2. karatechopper

Can ya work from there?

3. anonymous

i really need help! canu help??

4. karatechopper

I did...

5. karatechopper

I gave you a hint!

6. anonymous

how did u get 60??

7. karatechopper

60 minutes in an hour. :)

8. karatechopper

Why the rush brother?

9. anonymous

oooooooooohhhh!!! thanks

10. karatechopper

;)

11. karatechopper

No problem!

12. anonymous

a car travels 440 miles in 8 hour. write a proportion to find the average speed of the car

13. karatechopper

This time you try!

14. anonymous

I got 33.33 miles. I'll try to draw out my steps next.

15. anonymous

|dw:1357180828464:dw|

16. karatechopper

I got something totally dif..

17. anonymous

r t = d$50*\frac{40}{60}=\frac{100}{3}=33\text{ and}\frac{1}{3} \text{ miles}$


text:
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# In the xy-plane, at what two points does the graph of y=(x+a)(x+b)

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In the xy-plane, at what two points does the graph of y = (x + a)(x + b) intersect the x-axis?

(1) a + b = -1
(2) The graph intersects the y-axis at (0, -6)

Originally posted by misterJJ2u on 21 Jan 2007, 00:36.
Last edited by Bunuel on 05 Jun 2019, 07:29, edited 3 times in total.
Renamed the topic and edited the question.
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In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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Updated on: 05 Jun 2019, 07:31
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1
93
In the xy-plane, at what two points does the graph of y = (x + a)(x + b) intersect the x-axis?

(1) a + b = -1
(2) The graph intersects the y-axis at (0, -6)

X-intercepts of the function $$f(x)$$ or in our case the function (graph) $$y=(x+a)(x+b)$$ is the value(s) of $$x$$ for $$y=0$$. So basically the question asks to find the roots of quadratic equation $$(x+a)(x+b)=0$$.

$$(x+a)(x+b)=0$$ --> $$x^2+bx+ax+ab=0$$ --> $$x^2+(a+b)x+ab=0$$.

Statement (1) gives the value of $$a+b$$, but we don't know the value of $$ab$$ to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of $$y$$ when $$x=0$$ --> $$y=(x+a)(x+b)=(0+a)(0+b)=ab=-6$$. We know the value of $$ab$$ but we don't know the value of $$a+b$$ to solve the equation.

Together we know the values of both $$a+b$$ and $$ab$$, hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.

Hope it's clear.
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Originally posted by Bunuel on 22 Oct 2009, 13:28.
Last edited by Bunuel on 05 Jun 2019, 07:31, edited 2 times in total.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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21 Jan 2007, 02:49
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For me (C)

y= (x+a)(x+b) intersect the x-axis is equivalent to search the solutions of :
(x+a)(x+b) =0
<=> x^2 + (a+b)*x + a*b = 0 (1)

From 1
a+b = -1
implies:

(1) <=> x^2 -x + a*(-a-1) = 0

we need to know a.

INSUFF.

From 2
(0, -6) is the Y interceptor. That implies:

a*b = -6.

we need another equation with a & b, in other words, the values of a and b.

INSUFF.

Both (1) and (2)
a+b = -1
a*b = -6

These equations give us (a,b), what we need to find the roots.

SUFF.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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22 Oct 2009, 13:45
1
Bunuel wrote:
In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis?
(1) a+b=-1
(2) The graph intersects y-axis at (0,-6)

Q is in what two points graph intersect the x-axis. Which means we should be able to find two points: (x1,0) and (x2,0), so basiacally the roots of equation: (x+a)(x+b)=0.

y=(x+a)(x+b)=x^2+x(a+b)+ab. --> We must now the values of (a+b) and ab.

(1) a+b=-1 --> not sufficient.
(2) x=0, y=-6 --> -6=ab not sufficient

(1)+(2) a+b=-1, ab=-6 --> y=x^2+x(a+b)+ab=x^2-x-6 --> x intersection means y=0 --> 0=x^2-x-6 --> (x+2)(x-3)=0 -
-> x1=-2, x2=3. Sufficient

OMG!!

I didnt realize that I need to find two points!!
thats why when i ended up with -2 and 3, i got confused because i was left with two values!!!

how did you find ab=-6?
(2) x=0, y=-6 --> -6=ab not sufficient

did you use the formula
y=mx+b?

thanks,
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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22 Oct 2009, 13:51
2
(2) The graph intersects y-axis at (0,-6) --> x=0, y=-6 --> y=x^2+x(a+b)+ab --> -6=0^2+0(a+b)+ab --> ab=-6.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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24 Dec 2009, 10:37
2
y=(x+a)(x+b) when y=0
To solve this one, what do we need to know?
Obviously a or b, which are not stated in the information (1) & (2)
So one rule advise by MGMAT Book, always expand when the information given is factorized or Factorized when the information given is expended.
We know from a quadratic expression, the x axis intersect when y=0
So let's expand, and we have x^2 + (a + b)x + ab = 0
So now, we need to know ab and (a+b) to solve this equation
Therefore the correct answer is C, since only both information taken together permit to answer the question.
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In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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15 Oct 2010, 01:05
1
Right, so when the graph intersects the x-axis, y is 0. Those points are clearly x=-a and x=-b

(1) a+b=1. Not sufficient to know a or b
(2) We have point (0,-6). Plus it into the equation to get -6=ab. Again not sufficient to know either number

(1+2). Now we have 2 equations, sufficient. a-6/a=1; a^2-a-6=0; a=3 or a=-2 Hence b=-2 or b=3 .. Either way the points are 3 and -2

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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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25 Dec 2010, 04:49
Bunuel wrote:
In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis?
(1) a+b=-1
(2) The graph intersects y-axis at (0,-6)

X-intercepts of the function $$f(x)$$ or in our case the function (graph) $$y=(x+a)(x+b)$$ is the value(s) of $$x$$ for $$y=0$$. So basically the question asks to find the roots of quadratic equation $$(x+a)(x+b)=0$$.

$$(x+a)(x+b)=0$$ --> $$x^2+bx+ax+ab=0$$ --> $$x^2+(a+b)x+ab=0$$.

Statement (1) gives the value of $$a+b$$, but we don't know the value of $$ab$$ to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of $$y$$ when $$x=0$$ --> $$y=(x+a)(x+b)=(0+a)(0+b)=ab=-6$$. We know the value of $$ab$$ but we don't know the value of $$a+b$$ to solve the equation.

Together we know the values of both $$a+b$$ and $$ab$$, hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.

Hope it's clear.

I get your explanation.But using a+b=-1 and ab=-6 and solving it I get (a-b)^2=25 .which means a-b=+5 or -5.So since I cant be sure of the value of a-b,I selected option e as the answer.Can you please help on this.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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25 Dec 2010, 08:49
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gautamsubrahmanyam wrote:
Bunuel wrote:
In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis?
(1) a+b=-1
(2) The graph intersects y-axis at (0,-6)

X-intercepts of the function $$f(x)$$ or in our case the function (graph) $$y=(x+a)(x+b)$$ is the value(s) of $$x$$ for $$y=0$$. So basically the question asks to find the roots of quadratic equation $$(x+a)(x+b)=0$$.

$$(x+a)(x+b)=0$$ --> $$x^2+bx+ax+ab=0$$ --> $$x^2+(a+b)x+ab=0$$.

Statement (1) gives the value of $$a+b$$, but we don't know the value of $$ab$$ to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of $$y$$ when $$x=0$$ --> $$y=(x+a)(x+b)=(0+a)(0+b)=ab=-6$$. We know the value of $$ab$$ but we don't know the value of $$a+b$$ to solve the equation.

Together we know the values of both $$a+b$$ and $$ab$$, hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.

Hope it's clear.

I get your explanation.But using a+b=-1 and ab=-6 and solving it I get (a-b)^2=25 .which means a-b=+5 or -5.So since I cant be sure of the value of a-b,I selected option e as the answer. Can you please help on this.

The question asks: at what two points the graph of $$y=(x+a)(x+b)$$ intersect the x-axis?

So we should find two points: (x1, 0) and (x2, 0), (points of intersection of the given graph with X-axis). Basically the question asks to find the roots of quadratic equation $$(x+a)(x+b)=0$$ --> $$x^2+(a+b)x+ab=0$$.

When we combine statement we have: $$a+b=-1$$ and $$ab=-6$$, so you should solve quadratic equation $$x^2-x-6=0$$ --> $$x_1=-2$$ and $$x_2=3$$ --> points of intersection are (-2, 0) and (3, 0). Check on the diagram:
Attachment:

MSP272919df1bed26600c1g0000673adafe5fce1fc2.gif [ 3.61 KiB | Viewed 132085 times ]
Hope it's clear.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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03 May 2011, 10:21
1
1
y=(x+a)(x+b) = x^2+(a+b)x+ab
this is parabola equation. intersects x-axis means y=0

st 1--> a+b = -1
x^2-x+ab=0 cannot solve the equation; not sufficient

st 2--> graph intersects y-axis at -6
i.e. ab = -6
x^2+(a+b)-6=0 cannot solve the eq, not sufficient.

Both together, x^2-x-6=0
Solving gives x = -2 or x=3

Hence C.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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03 May 2011, 18:33
5
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udaymathapati wrote:
Attachment:
M-Q19.JPG

At what two points does the graph of y = (x+a)(x+b) intersect the x axis?

You don't need to worry what the equation represents. Just think, what does 'intersection with x axis' imply? It means the y co-ordinate is 0.

0 = (x+a)(x+b)
or x = -a or -b
Hence the graph must intersect the x axis at points (-a, 0) and (-b, 0). We need the values of a and b now.

Statement 1: a + b = -1
Two variables, only one equation. Not sufficient.

Statement 2: Graph intersects the y axis at (0, -6).
At y axis, x = 0. This means when x = 0, y co-ordinate is -6.
Put these values in y = (x+a)(x+b) to get -6 = ab.
Again, two variables, one equation. Not sufficient alone.

Using both statements, we have two variables and two different equations so we will be able to find the values of a and b. It doesn't matter which is 'a' and which is 'b'. We find that the two of them are -3 and 2. Since we need the points (-a, 0) and (-b, 0), the required points are (3, 0) and (-2, 0). Sufficient.

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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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25 May 2011, 03:26
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1
To find x-intercepts, you set y=0 and solve. The question asks for the x-intercepts of y = (x+a)(x+b). Setting y=0, the question is asking when (x+a)(x+b) = 0. Since one of the factors on the left side must be 0, we find we have intercepts when x = -a and x = -b.

Statement 1 tells us the sum of a and b, which doesn't let us find their values alone.

Statement 2 gives us a point on the curve; if we plug that point into our equation, what we get must be true. So -6 = (0+a)(0+b), and ab = -6. So Statement 2 gives us the product of a and b, which isn't enough to find their individual values.

Together, we know that a+b = -1 and ab = -6. That is, we know the sum and product of two numbers. Notice this is what happens every time you factor a quadratic: you look for two numbers with a certain product and a certain sum. For example, if you were to factor the quadratic x^2 - x - 6, you'd look for two numbers which multiply to -6 and add to -1. So if a+b = -1 and ab = -6, then our solutions for a and b must be -3 and 2, in some order. Since our intercepts are at -a and -b, the intercepts are at x=3 and x=-2, so the information is sufficient.

Notice we can't figure out which value is equal to a and which is equal to b, but we don't need to do that to answer the question; we just need to know where the two points are, and not which point is a and which point is b.
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In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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24 Jul 2011, 21:40
blog wrote:
In the XY-plane, at what two points does the graph of Y = (X+a)(X+b) intersect the X-axis?

1. a+b = -1
2. The graph intersects the Y-axis at (0,-6).

The question stem asks us the points where the graph intersects the X axis. What does 'intersect the X axis' imply? It implies that Y = 0 at the points where the graph intersects the X axis because on the entire X axis, Y = 0.

Y = (X+a)(X+b) will intersect the X axis when Y = 0.
(X + a)(X + b) = 0
We will get two values for the X co-ordinate (since there are two points). The two values will be -a and -b. So basically, we are looking for the values of a and b. (If you remember your high school Math, this is a quadratic equation so its graph is an inverted U. It intersects the X axis at two points - Just to give you some perspective.)

Statement 1: a + b = -1
a and b can take infinite different values. Not sufficient alone.

Statement 2: The graph intersects the Y-axis at (0,-6)
'Intersect at Y axis' implies that X = 0.
When X = 0, we are given that Y co-ordinate is -6.
Let's put this in the given equation: Y = (X+a)(X+b)
-6 = ab
a and b can take infinite different values. Not sufficient alone.

Together, we know that a + b = -1 and ab = -6
Two different equations and two unknowns. We will be able to find the values of a and b. Hence together they are sufficient. (We don't actually need to find the values but you can see that a and b will be -3 and 2 giving us -a and -b as 3 and -2.)

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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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Updated on: 26 Oct 2011, 07:17
2
1
kotela wrote:
In the XY plane what two points does the graph of y=(x+a)(x+b) intersect the x -axis?

1. a+b=-1

2. The graph intersects the y-axis at (0,-6)

can anyone plz explain?

Intersect the x axis means y is 0.
the x coordinates will be -a and -b respectively.

1. a+b=-1
Insufficient

2.
-6=(0+a)(0+b)
ab=-6
Insufficient.

1+2
a+b=-1
ab=-6
a=-3, b=2
a=2, b=-3
Sufficient as we want two points we can say (-3,0) and (-2,0) without knowing the exact values of a & b.
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Originally posted by blink005 on 26 Oct 2011, 03:04.
Last edited by blink005 on 26 Oct 2011, 07:17, edited 2 times in total.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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26 Feb 2012, 23:50
I tried plugging in values in this one.

Started from option 2:
ab = -6
Possible values are (2,-3); (3,-2); (-6,1); (1,-6) --> INSUFFICIENT.

Option 1:
substitute for x and y in equation we get
a+b = -1
Several possible values such as (-3,2); (-8,7) and so on. --> INSUFFICIENT.

Combining both --> find from option 2 which possible value leads to a+b =-1, only one of the four choices does that (2,-3). Hence SUFFICIENT. Answer choice C.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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27 Feb 2012, 00:03
mourinhogmat1 wrote:
I tried plugging in values in this one.

Started from option 2:
ab = -6
Possible values are (2,-3); (3,-2); (-6,1); (1,-6) --> INSUFFICIENT.

Option 1:
substitute for x and y in equation we get
a+b = -1
Several possible values such as (-3,2); (-8,7) and so on. --> INSUFFICIENT.

Combining both --> find from option 2 which possible value leads to a+b =-1, only one of the four choices does that (2,-3). Hence SUFFICIENT. Answer choice C.

This is not a good idea to plug the numbers for this problem, it's better to understand the concept and you won't need any math at all (certainly you won't need to solve a+b=-1 and ab=-6).

Next, there are infinitely many values of a and b possible to satisfy ab=-6, not just four: notice that we are not told that a and b are integers only, so for example a=1/2 and b=-12 is also a solution.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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30 Apr 2012, 12:17
Stiv wrote:
In the xy-plane, at what two points does the graph of $$y= (x + a) (x + b)$$ intersect the x - axis?
1) $$a + b = -1$$
2) The graph intersects the y-axis at (0, -6)

So, it's important to know --- when the quadratic is given in factored form --- y= (x + a) (x + b) --- then we know the two roots, x = -a, and x = -b. Roots are the x-intercepts, the places where the graph intersects the x-axis. Basically, the prompt is asking us to find the values of a & b.

Statement #1: a + b = -1

One equation for two unknowns. Not enough to solve. Not sufficient.

Statement #2: The graph intersects the y-axis at (0, -6)

Plugging in x = 0 (the condition of the y-axis), we get y = (0+a)(0+b) = ab = -6

Again, one equation for two unknowns. Not enough to solve. Not sufficient.

Combined statements:
a + b = -1
ab = -6

Two equations with two unknowns ---> we can solve for the values of a & b, which will answer the question. Sufficient.

Here's another practice question on quadratics for practice.
http://gmat.magoosh.com/questions/120
When you submit your answer to that question, the next page will have a full video explanation.

Let me know if you have any further questions.

Mike
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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16 Jul 2016, 12:08
Top Contributor
JimmyWorld wrote:
In the xy plane, at what points does the graph of y=(x+a)(x+b) intersect the x-axis?

(1) a + b = -1
(2) The graph intersects the y axis at (0,-6)

Target question: At which two points of the graph does y=(x+a)(x+b) intersect the x-axis?

Let's examine the point where a line (or curve) crosses the x-axis. At the point of intersection, the point is on the x-axis, which means that the y-coordinate of that point is 0. So, for example, to find where the line y=2x+3 crosses the x-axis, we let y=0 and solve for x. We get: 0 = 2x+3
When we solve this for x, we get x= -3/2.
So, the line y=2x+3 crosses the x-axis at (-3/2, 0)

Likewise, to determine the point where y = (x + a)(x + b) crosses the x axis, let y=0 and solve for x.
We get: 0 = (x + a)(x + b), which means x=-a or x=-b
This means that y = (x + a)(x + b) crosses the x axis at (-a, 0) and (-b, 0)
So, to solve this question, we need the values of a and b

Aside: y = (x + a)(x + b) is actually a parabola. This explains why it crosses the x axis at TWO points.

Now let's rephrase the target question...
REPHRASED target question: What are the values of a and b?

Statement 1: a + b = -1
There's no way we can use this to determine the values of a and b.
Since we can answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The line intercepts the y axis at (0,-6)
This tells us that when x = 0, y = -6
When we plug x = 0 and y = -6 into the equation y = (x + a)(x + b), we get -6 = (0 + a)(0 + b), which tells us that ab=-6
In other words, statement 2 is a fancy way to tell us that ab = -6
Since there's no way we can use this information to determine the values of a and b, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined:
Statement 1 tells us that a+b = -1
Statement 2 tells us that ab = -6

Rewrite equation 1 as a = -1 - b
Then take equation 2 and replace a with (-1 - b) to get: (-1 - b)(b) = -6
Expand: -b - b^2 = -6
Set equal to zero: b^2 + b - 6 = 0
Factor: (b+3)(b-2) = 0
So, b= -3 or b= 2

When b = -3, a = 2 and when b = 2, a = -3
In both cases, the two points of intersection are (3, 0) and (-2, 0)
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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25 Feb 2018, 15:39
Hi All,

Like most questions on the GMAT, this question can be approached in a number of different ways. There's actually a great Algebra pattern/shortcut built into this question:

We given the equation Y = (X+A)(X+B) and we're asked at what 2 points the graph will intersect with the X-axis. This essentially comes down to the A and B. If we know their values, then we can answer the question. It's also worth noting that since we're multiplying, you can "flip-flop" the values of A and B and you'd have the same solution.

For example: (X+1)(X+2) is the same as (X+2)(X+1)…...

1) A + B = -1

There's no way to determine the exact values for A and B with this information.

IF...
A = 0, B = -1
A = 100, B = -101
Etc.
Different numbers for A and B would lead to different solutions.
Fact 1 is INSUFFICIENT

2) The graph intersects the Y-axis at (0, -6).

Now we have one of the points on the graph. Plugging it into the original equation gives us….

-6 = (0+A)(0+B)
-6 = AB

We have the same situation as in Fact 1: more than 1 possible solution.
A = 1, B = -6
A = 2, B = -3
Etc.
Fact 2 is INSUFFICIENT

Combined, we have…
A + B = -1
AB = -6

Here's where things get interesting. This is a "system" of equations, so we CAN solve it...the "catch" is that the answers would "flip flop":

If you did do the math, you'd have
A = -3, B = 2
OR
A = 2, B = -3

Since this is a graphing question, these two options provide the SAME solution.
Combined, SUFFICIENT

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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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13 Dec 2018, 09:18
If you prefer the explanation in video format:

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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)   [#permalink] 13 Dec 2018, 09:18

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## question

What would be the combined ratio for the following: An insurance company made $500,000 in premiums with losses of$20,000 and expenses of \$40,000. What is the combined ratio? SHOW ALL WORK OR DONT ANSWER. THANKS


text:
# How are forces related to decays?

How are decays related to forces, what is meant by particle X decays through the, say, strong force?

The way I understand forces is by how they change the acceleration of particles with the right charge (mass, electric etc), through F=ma, how does it cause one particle to turn into other?

How is it determined which force is responsible for which decays?

Can a particle decay through the gravitational interaction?

-

## 1 Answer

welcome to physics.SE

How are decays related to forces, what is meant by particle X decays through the, say, strong force?

It means that particle X was bound by the strong force to particle Y to form particle Z for a delta(t) and then it parts company.

The way I understand forces is by how they change the acceleration of particles with the right charge (mass, electric etc), through F=ma, how does it cause one particle to turn into other?

Your understanding is about the classical domain of forces.

Elementary particles necessarily belong to the quantum domain, one uses relativistic four vectors to describe them and forces are mediated by other elementary particles, graphically shown with Feynman diagrams, which are a short hand for the way one can calculate the probability of the decay. One can think of the mediating particle as a carrier of delta(p), where p is the four momentum, which transfers momentum and energy to the decay products from the binding energy of the original particle.

From a teacher site here is the Feynman diagram of the decay of a lambda baryon to a proton and a pion, mediated by the weak interaction:

Lambda contains a strange quark which is not stable under weak interactions. For a time the three quarks are bound together by the strong force, with gluon exchanges ( Not shown) but then the s quark decays into an up quark which is bound into a proton, and a W- weak boson off mass shell goes into an anti up and a down quark making a pi- .

How is it determined which force is responsible for which decays?

It has been determined experimentally and theoretically encoded into the Standard Model

Can a particle decay through the gravitational interaction?

We have not found elementary particles or resonances bound by the gravitational interaction so the answer is that experimentally it cannot. The reason is that the gravitational force is many orders of magnitude weaker than the other three forces that reign in the particle domain.

-


text:
Question

# Solve the system of equations using matrices.Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. begin{cases}x+y-z=-22x-y+z=5-x+2y+2z=1end{cases}

Matrices
Solve the system of equations using matrices.Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
$$\begin{cases}x+y-z=-2\\2x-y+z=5\\-x+2y+2z=1\end{cases}$$

2021-01-09
Step 1
Given:
The system of linear equations are,
$$x+y-z=-2$$
$$2x-y+z=5$$
$$-x+2y+2z=1$$
For applying Gauss-Jordan elimination method, the above system of equations can be represented in matrix form as,
$$\begin{bmatrix}1 & 1&-1&-2 \\2 & -1&1&5\\-1&2&2&1 \end{bmatrix}$$
Step 2
The above matrix can be converted into row echelon form as,
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & -3&3&9\\-1&2&2&1 \end{bmatrix} \left(\text{By applying: } R_2 \rightarrow R_2-2R_1\right)$$
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & -3&3&9\\0&3&1&-1 \end{bmatrix}\left(\text{By applying: } R_3 \rightarrow R_3+R_1\right)$$
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & 1&-1&-3\\0&3&1&-1 \end{bmatrix} \left(\text{By applying: } R_2 \rightarrow \frac{-1}{3}R_2\right)$$
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & 1&-1&-3\\0&0&4&8 \end{bmatrix} \left(\text{By applying: } R_3 \rightarrow R_3-3R_2\right)$$
Step 3
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & 1&-1&-3\\0&0&1&2 \end{bmatrix} \ \ \left(\text{By applying: } R_3 \rightarrow \frac{1}{4}R_3\right)$$
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & 1&0&-1\\0&0&1&2 \end{bmatrix} \ \ \left(\text{By applying: } R_2 \rightarrow R_2+R_3\right)$$
$$\begin{bmatrix}1 & 1&0&0 \\0 & 1&0&-1\\0&0&1&2 \end{bmatrix} \ \ \left(\text{By applying: } R_1 \rightarrow R_1+R_3\right)$$
Step 4
$$\begin{bmatrix}1 & 0&0&1 \\0 & 1&0&-1\\0&0&1&2 \end{bmatrix} \ \ \left(\text{By applying: } R_1 \rightarrow R_1-R_2\right)$$
Hence, the solutions of the given system of equations are x=1 , y=-1 and z=2


text:
## anonymous one year ago solve by using elimination.express answer as an ordered pair 2x-5y=-7 5x-3y=11

1. Michele_Laino

hint: we can multiply the first equation by 5, so the new first equation is: 10x-25y=-35 next we can multiply the second equation by 2, so we get: 10x-6y=22 then we have this equivalent system: $\Large \left\{ \begin{gathered} 10x - 25y = - 35 \hfill \\ 10x - 6y = 22 \hfill \\ \end{gathered} \right.$ now, please subtract the second equation, from the first one, what equation do you get?

2. anonymous

10x-19y=-13

3. Michele_Laino

hint: if we do that subtraction, we can write this: $\Large \left( {10x - 25y} \right) - \left( {10x - 6y} \right) = - 35 - 22$ please simplify

4. anonymous

im confused im i supposed to subtract -35 and 22

5. Michele_Laino

yes!: -35-22=-57

6. Michele_Laino

now, please simplify the left side

7. anonymous

25y and 6y = -31

8. Michele_Laino

hint: $\Large \begin{gathered} \left( {10x - 25y} \right) - \left( {10x - 6y} \right) = \hfill \\ \hfill \\ = 10x - 25y - 10x + 6y = ...? \hfill \\ \end{gathered}$

9. anonymous

-15-4=-19

10. Michele_Laino

what are -15 and -4 ?

11. anonymous

you said 10x-25y

12. Michele_Laino

here is the right step: |dw:1440873623698:dw|

13. anonymous

oooo

14. Michele_Laino

so we get this equation: $\Large - 19y = - 57$ please solve for y

15. anonymous

3

16. Michele_Laino

correct!

17. Michele_Laino

now, please replace y with 3 into the first equation of the original system, what do you get?

18. anonymous

15

19. Michele_Laino

yes! and Iwe gaet the subsequent equation: $\Large 2x - 15 = - 7$ please solve for x

20. Michele_Laino

we*

21. anonymous

4

22. Michele_Laino

correct! so the solution of your system is: x=4 y=3

23. anonymous

thank you for helping me i understand it now a little

24. Michele_Laino

:)


text:
# What is the derivative of sin(x^2 -2)^3?

$\frac{d}{\mathrm{dx}} \sin {\left({x}^{2} - 2\right)}^{3} = 6 x {\left({x}^{2} - 2\right)}^{2} \cos {\left({x}^{2} - 2\right)}^{3}$.
$\frac{d}{\mathrm{dx}} \sin {\left({x}^{2} - 2\right)}^{3} = \cos {\left({x}^{2} - 2\right)}^{3} \cdot 3 {\left({x}^{2} - 2\right)}^{2} \cdot 2 x$
$= 6 x {\left({x}^{2} - 2\right)}^{2} \cos {\left({x}^{2} - 2\right)}^{3}$.


text:
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