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# Game engines

## Recommended Posts

Very simple question: what (free/cheap) game engines are out there and what are their forté's and weaknesses. I've wisened up a bit and decided to license an engine instead of trying to make one myself, which is simple not doable. Then I can finally begin to make real games instead of wannabe engines. But being a hobbyist developer I'm not interested in putting down a lot of money for it. I know of OGRE, but I'm sure there are a whole lot more engines out there.

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http://www.devmaster.net/engines/

Lists engines separated by API.

Cheers!
Webby

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Hi,

I am also trying to decide which low-cost (up to a couple of hundred USD) engine to use.
I was just looking at that list at devmaster and it seems it is going to be hard to decide. I am going to take a closer look to some of them, but it would be great to have some feedback from people that have acctually used them (any reviews, experience?). There is a list of features I prefer (look important to me personally), but I am interested in general opinion first.

Thanks...

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Try panda3d.com

The learning curve is very short. It's worth evaluating.

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That is truely a great site. I'm in doubt between Crystalspace and Torque. I want a stable engine which provides everything I need for making a game and I want the sourcecode so that I can fiddle around with it. Torque sounds good because of its networking system and animation support. Crystalspace for its mesh support and less steep learning curve (So I've read). And it being free and all.

I need to think this over a tad.

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It also comes down to your needs. Torque includes a GUI, networking, semi-functioning audio, terrain and is geared towards an FPS style game.

Again I say: It also comes down to your needs

You'll get further much quicker with Torque.

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UT2k4 - free for non-commercial use Maya 5.0 w/UT plugin, free UT IDE UDE, and large community base for support. $40 for the whole shebang. Good to get started with. To go commercial look at Milkshape for$40(or was it \$20????) which exports to UT.

For source included, free, try Allegra, etc.


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# Why doesn't current increases when the voltage is increased

When connecting two batteries in series the voltage is increased. For example: Connecting two 5V batteries in series will produce 10V voltage but the current will be the same.

According to Ohm's Law V = IR the voltage is directly proportional to the current. Then why is the current not increases when voltage is increased?

• Because batteries are power sources not resistors, and therefore don't follow ohm's law. Also they don't have "a" current, they have a "maximum" current. – pjc50 Oct 22 '17 at 9:56
• Connecting two 5V batteries in series will produce 10V voltage but the current will be the same. In both cases the current will be 0 A (Zero Ampere) as no current will flow because you did not connect a load. It depends on the load how much current will flow. For simple loads like lightbulbs and resistors, the current will double when you double the voltage. – Bimpelrekkie Oct 22 '17 at 10:00
• @pjc50 So you mean - When the batteries are connected in series the just voltage is increased. There is no impact on current as it would be drawn as needed by the ckt attached to battery? – VVK Oct 22 '17 at 10:00
• When the same resistance is used in both situations the current does increase. – Wouter van Ooijen Oct 22 '17 at 10:27
• By means of resistors you mean any ckt that that consumes current. right? Oh no a resistor is just that, a resistor. Go read a book explaining what a resistor is. A circuit can behave completely different. You really need to get your basic electronics knowledge sorted out, I mean, read a book about it or somehow educate yourself. – Bimpelrekkie Oct 22 '17 at 11:50

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# Separating Out Parts of a Matrix (Translation, Rotation, Scaling)

Is there a there a way to separate out parts of matrices?

given a Matrix = Translation * Rotation * Scaling

How can I find out only one part of the matrix?

For instance if I only wanted the rotation element of the matrix?

-
Translations are affine transformations, but they aren't linear (they don't send zero to zero, for one thing). Thus, there aren't matrices corresponding to translation, so your question needs some fixing. You may be thinking of the Iwasawa decomposition, which asserts that an invertible matrix can be written as the product of a diagonal matrix, an upper-triangular matrix, and an orthogonal matrix. Is this what you mean? –  NKS Dec 28 '11 at 21:46
The Iwasawa decomposition asserts that an invertible matrix $M$ can be written in the form $M = KAN$, where $K$ is an orthogonal matrix, $A$ is a diagonal matrix, and $N$ is an upper-triangular matrix with $1$'s along the diagonal. Carrying this out in practice is nothing more than Gram-Schmidt orthogonalization, which produces an orthogonal basis from an arbitrary basis.
Let $F$ be a field (think of $F = \mathbb{R}$ if that's more familiar). Given $M \in GL_n(F)$ (i.e. an invertible $n\times n$ matrix), take the columns as a basis for $F^n$; write them as $v_1, \dots, v_n$. Gram-Schmidt will produce a new, orthonormal basis $u_1, \dots, u_n$, with the property that $$u_k = \sum_{i=1}^k a_{ik} v_i$$ for $a_{ik} \in F$. If we think about change-of-basis matrices, we recognize that the expression above tells us that the change-of-basis matrix taking $v_1,\dots,v_n$ to $u_1,\dots,u_n$ is upper-triangular, because $u_k$ is formed only from $v_1, \dots, v_k$. Any upper-triangular matrix is the product of a diagonal matrix and an upper-triangular matrix with $1$'s on the diagonal (this is called a unipotent matrix); call this product $AN$.
We can think of the original $M$ as taking us from the standard basis to the $v$ basis. Gram-Schmidt then takes us from the $v$-basis to another orthonormal basis $u$, which is represented by an orthogonal matrix $K$. What this means in practice is that we have the identity $$K^{-1}ANM = I;$$ equivalently $$M^{-1} = K^{-1}AN.$$ Of course, the inverse of a diagonal matrix $A$ or a unipotent matrix $N$ will be of the same form, so this gives you your decomposition for $M^{-1}$ ; to get this for $M$, just apply Gram-Schmidt to $M^{-1}$.


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# Problem of image method, super-conductor, current carrying wire's magnetic field

Above a superconducting plane, there is a very long thin wire carrying current $I$. Linear mass density of the wire is $\rho$ kg/m. Find the distance of the wire from the plane so that it can hang freely. Inside a supercoductor no magnetic field can be created that is, a superconductor is a conductor that does not allow any magnetic field inside it.

My attempt: I think we can apply the image method here. Let's suppose, there is a fictious current carrying wire e under the super-conducting plane.Real and fictious wire are in equal distance $a$ from the plane and they are carrying current on the same direction. So, there would be no magnetic field on the super-conducring plane.

Then, the force of the fictious wire on the real wire, $F=IlB$ where $B$ is the magnetic field acting on the real wire due to the fictious wire and it's $B=\frac{\mu I}{2\pi 2a}$. Magnetic force would be equal to the weight of the wire. So, $mg=Il\frac{\mu I}{2\pi 2a}$ Rearranging, $a=\frac{\mu I^2}{4\pi \rho g}$

But, there is a problem. If the real wire is on the upper side of the superconducting plane, then the attractive force adds up with the weight of the wire rather then cancelling. But, it works if the wire is under the superconducting plane.

What's wrong with this method?

• @Heather See the edit. – Mockingbird Jul 31 '17 at 1:06
• I have voted to leave this question open. There is a clear conceptual question here about whether the image current attracts or repels the object current. – sammy gerbil Jul 31 '17 at 9:58

Your mistake is to assume that the image current is in the same direction as the object current.

When using the Method of Images in electrostatics, the image charge in a conductor has the opposite sign. The image of a +ve charge is a -ve charge.

An electric current is a flow of charge, conventionally +ve. Suppose the object current is a flow of +ve charge to the right. Then the image current will be a flow of -ve charge to the right : each +ve charge in the object wire is mirrored by a -ve charge in the image wire, moving in the same direction. This is equivalent to a flow of +ve charge to the left.

So the image current flows in the opposite direction to the object current, and therefore the force betweeen the object and image currents is repulsive.

Looking along the wire, the magnetic field will be as shown on the right in the diagram below. The green line represents the surface of the superconductor, and the object wire is the upper wire.
Image copied from University of Louisville Dept of Physics

If the image current was in the same direction as the object current (as on the left), then the magnetic field would be zero at the mid-line between the wires (as you point out), but it would be non-zero at all other points on the surface of the superconductor. Moreover, it would have to penetrate the superconductor, which we know does not happen (except for flux pinning - see How does quantum trapping with diamagnets work?).

The image current is actually in the opposite direction (as on the right). The magnetic field at the surface of the superconductor is now non-zero, but no magnetic field lines penetrate into the superconductor. The field lines below the green line, inside the superconductor, are imaginary - just as the optical image in a mirror is imaginary.

• But if the 2 wires carry current in the opposite direction then there is a magnetic field on the superconducting plane. What about that? – Mockingbird Jul 31 '17 at 13:35
• What a nice answer! I understand my faults now. – Mockingbird Jul 31 '17 at 16:20


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## Return to Question

4 added some background

The question is more serious than it looks!

Recall the list of irreducible simply connected inner symmetric spaces of compact type in dimension $4k+2$:

Hermitian symmetric spaces are of course complex manifolds, and real Grassmannians carrying almost complex structures were classified by P. Sankaran (PAMS, 1991) and Z. Tang (PAMS, 1994). Therefore Thus my question:

anybody know whether $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ carries carry an almost complex structure?

Note that in dimension $4k$ it is known that an irreducible simply connected inner symmetric space of compact type carries an almost complex structure iff it is Hermitian symmetric.

Edit concerning the dimensions multiple of 4: It is on purpose that I did not mention our paper http://fr.arxiv.org/abs/1003.5172 when I first asked the question. Indeed, the method used in that paper to treat the $4k$-dimensional situation (an index-theoretical obstruction) seems uneffective in the $4k+2$-dimensional cases so I expected other methods to tackle this problem.

Let me nevertheless give a short account on how things work in dimension $4k$. The idea is to use the following criterion:

If a compact Riemannian manifold $(M,g)$ carries a self-dual vector bundle $E$ such that the index of the twisted Dirac operator on $\Sigma M \otimes E\otimes TM^{\mathbb{C}}$ is odd, then $M$ has no almost complex structure (in fact $TM$ is not even stably isomorphic to a complex bundle).

Note that the complex spin bundle $\Sigma M$ and $E$ can be locally defined, only their tensor product has to be globally defined. On then checks that this criterion applies to all irreducible inner symmetric spaces of compact type which are neither Hermitian symmetric nor spheres.

Returning to the dimensions $4k+2$, it is easy to check that there exists a priori no bundle satisfying the above criterion. Nevertheless, one can show real Grassmannians and $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ do carrysuch a bundle $E$ which satisfies all conditions in the criterion except that it is not self-dual. I don't know if this can lead anywhere...

3 corrected notation

# Does $\mathrm{F}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$carry an almost complex structure? The question is more serious than it looks! Recall the list of irreducible simply connected inner symmetric spaces of compact type in dimension$4k+2$: 1. Hermitian symmetric spaces (one can write them down explicitly); 2. Grassmannians of oriented real$p$-planes in$\mathbb{R}^{p+q}$, with$pq=4k+2$; 3. The 70-dimensional exceptional symmetric space$\mathrm{F}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$.\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$.

Hermitian symmetric spaces are of course complex manifolds, and real Grassmannians carrying almost complex structures were classified by P. Sankaran (PAMS, 1991) and Z. Tang (PAMS, 1994). Therefore my question:

1


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# How do you find the complex roots of x^4-10x^2+9=0?

Jan 10, 2017

The roots are $\pm 3$ and $\pm 1$

#### Explanation:

Note that $9 + 1 = 10$ and $9 \cdot 1 = 9$.

Hence we find:

$0 = {x}^{4} - 10 {x}^{2} + 9$

$\textcolor{w h i t e}{0} = \left({x}^{2} - 9\right) \left({x}^{2} - 1\right)$

$\textcolor{w h i t e}{0} = \left({x}^{2} - {3}^{2}\right) \left({x}^{2} - {1}^{2}\right)$

$\textcolor{w h i t e}{0} = \left(x - 3\right) \left(x + 3\right) \left(x - 1\right) \left(x + 1\right)$

So the four roots of this quartic equation are $\pm 3$ and $\pm 1$

These are Real numbers, but any Real number is also a Complex number (of the form $a + 0 i$ if you wish).


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# Making PDFs editable

I automate everything and PDFs are so easy to fill in automatically. It’s a digital world and we just live in it.

It’s so frustrating to get a “protected form” that doesn’t allow for appending documents, automated form completion or even filling out fields. It’s silly because all of these forms are printable. Many years ago, I just created an official form from scratch myself because I couldn’t find it online. Without checking a signature, forms can’t be secure. For example, one could easily write a program that takes a picture and recreates the form. Remember, in the digital world something is either secure from math (crypto) or it isn’t. The security in the transaction is provided by authenticating the user’s email. Even better, let’s move past forms and use the authentication in a website or mobile app.

Let’s do the right thing and move past forms caught in the middle ground of insecure and not usable. I think folks lock forms to prevent changes to the document, don’t do that.

Below is the code to fix this. You will need to use linux or windows subsystem for linux. While I can’t think of an evil use case, use this responsibly. (My goal in doing this is always to fill in data into a form.)

Use a little bash to print with ghostscript. Protip: this isn’t “hacking”, this is just printing.

filename=$1 outfile="${filename%.*}_clean.pdf"
echo $outfile gs -q -dNOPAUSE -dBATCH -sDEVICE=pdfwrite -sOutputFile=$outfile -c .setpdfwrite -f \$1

This site uses Akismet to reduce spam. Learn how your comment data is processed.


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# Performance of Viterbi detector over non-minimum phase channels

For sequences that are transmitted over channels with memory $\mu=n$ and response H=$[h_0 h_1 \ldots h_n]$,Viterbi algorithm implements Maximum Likelihood (ML) detection and BCJR implements Maximum A-Posteriori (MAP) Detector. Trellis over which these detectors operate have total states = $2^n$.

While applying any trellis based detector either ML or MAP detectors, channels are always considered causal i.e. $h_i=0 \quad \forall i<0$. More over equalizers/detectors are tested for channels that have $h_0=\max(h_i)\quad \forall i=0, \ldots,N$.

My question is related BER vs SNR performance of these trellis based detectors (that jointly equalize and detect) when applied to channels that are not causal and have a peak at $t=k,k\neq 0$. Since a delay to shift response to right has no effect on the structure of trellis, one can easily achieve this. But now the peak is no more at $t=0$. Following is an example to make it more clear.

Let $h_{ij}$ represent the $j^{th}$ coefficient for channel $i$. Consider two channels,

\begin{align}H_1 &=\left[h_{10} \space h_{11} \space h_{12}\right]=[4 \space 2 \space 1]\\H_2&=\left[h_{20} \space h_{21} \space h_{22}\right]=[2 \space 4 \space 1]\end{align}

The performance of ML and MAP detector on $H_1$ is better than that of $H_2$ which is obviously because of high ISI in $H_2$ as compared to $H_1$.

What transformation on $H_2$ can be done so that the resulting channel has a BER vs SNR that is close to $H_1$?

• Minimum phase does not imply $h_0=\max\{h_i\}$. A counter-example: h=[1,1.5,0.6] is minimum-phase. – Matt L. Jun 6 '15 at 19:56
• Is the channel known? Practical approaches would be to use an equalizer before the Viterbi decoder, or combine the two, incorporating the known channel response in the trellis model. This is sometimes known as Viterbi equalization. – Jason R Jun 6 '15 at 19:58
• @ Matt…rigorous definition of minimum phase (MP) includes position of poles/zeros wrt unit circle. For my application, this does not matter to me. Also they effect if we are trying to use linear filters to equalize but since I am interested in trellis based equalizers/detectors so I am not very much worried about this aspect. But thanks for pointing it out. I should had been more careful in using MP term – NAASI Jun 6 '15 at 20:09
• @ Jason…. Yes channel response is know and you are right in pointing out the viterbi equalization terminology. I will include that in my original post. Would you like to comment on the last para of my original post that talks about the performance comarion between $H1$ & $H2$ – NAASI Jun 6 '15 at 20:12
• Do you use a whitened matched filter before the Viterbi decoder? – Matt L. Jun 6 '15 at 21:29


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Curl of the Gradient of a Scalar | Quantum Science Philippines

Curl of the Gradient of a Scalar

proof that the curl of the gradient of a scalar function is equal to zero
$\vec{\nabla}\times\vec{\nabla}\psi=0$
let
$\vec{\nabla}_j= \partial_j\hat{e}_j$
$\vec{\nabla}_k= \partial_k\hat{e}_k$
and
$\triangledown\psi= \partial_k\psi\hat{e}_k$

$\vec{\nabla}_j \times \vec{\nabla}_k \psi=\in_{ijk} \partial_j \partial_k \psi\hat{e}_j \times \hat{e}_k$
$= \in_{ijk}\hat{e}_i\partial_j\partial_k\psi$
$= \hat{e}_1(\partial_2\partial_3\psi - \partial_3\partial_2\psi) + \hat{e}_2(\partial_3\partial_1\psi - \partial_1\partial_3\psi) + \hat{e}_3(\partial_1\partial_2\psi - \partial_2\partial_1\psi)$
$= \hat{e}_1(\partial^2\psi - \partial^2\psi) + \hat{e}_2(\partial^2\psi - \partial^2\psi) + \hat{e}_3(\partial^2\psi - \partial^2\psi)$
$= 0$


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Comment
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Q)

# Equal volumes of three basic solutions of 11,12 and 13 are mixed in a vesel. What will be the concentration in the mixture ?

$\begin{array}{1 1} (a)\;1.11 \times 10^{-11}\;M \\(b)\;3.7 \times 10^{-12} \\ (c)\;3.7 \times 10^{-11}\;M\\ (d)\;1.11 \times 10^{-22}\;M \end{array}$


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# Solution - Circles Examples and Solutions

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Books Shortlist
ConceptCircles Examples and Solutions

#### Question

In the given figure, the incircle of ∆ABC touches the sides BC, CA and AB at D, E, F respectively. Prove that AF + BD + CE = AE + CD + BF = \frac { 1 }{ 2 } ("perimeter of ∆ABC")

#### Solution

You need to to view the solution
Is there an error in this question or solution?

#### Reference Material

Solution for concept: Circles Examples and Solutions. For the course CBSE
S


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# Fast, Accurate, and Realizable Two-Qubit Entangling Gates by Quantum Interference in Detuned Rabi Cycles of Rydberg Atoms. (arXiv:1809.08957v2 [quant-ph] UPDATED)

High-fidelity entangling quantum gates based on Rydberg interactions are
required for scalable quantum computing with neutral atoms. Their realization,
however, meets a major stumbling block -- the motion-induced dephasing of the
transition between the ground and Rydberg states. By using quantum interference
between different detuned Rabi oscillations, we propose a practical scheme to
realize a class of accurate entangling Rydberg quantum gates subject to a
minimal dephasing error. We show two types of such gates, $U_{1}$ and $U_{2}$,
in the form of $\text{diag}\{1, e^{i\alpha}, e^{i\gamma}, e^{i\beta}\}$, where
$\alpha, \gamma$, and $\beta$ are determined by the parameters of lasers and
the Rydberg blockade $V$. $U_{1}$ is realized by sending to the two qubits a
single off-resonant laser pulse, while $U_{2}$ is realized by individually
applying one pulse of detuned laser to each qubit. Our method has several
advantages. First, the gates are accurate because the fidelity of $U_k$ is
limited only by a rotation error below $10^{-5}$ and the Rydberg-state decay.
Decay error on the order of $10^{-5}$ can be easily obtained because all
transitions are detuned, resulting in small population in Rydberg state.
Second, the motion-induced dephasing is minimized because there is no gap time
in which a population is left in the Rydberg shelving states of either qubit.
Third, the gate is resilient to the variation of $V$. This is because among the
three phases $\alpha,\gamma$, and $\beta$, only the last has a (partial)
dependence on $V$. Fourth, the Rabi frequency and $V$ in our scheme are of
similar magnitude, which permits a fast implementation of the gate when both of
them are of the feasible magnitude of several megahertz. The rapidity,
accuracy, and feasibility of this interference method can lay the foundation
for entangling gates in universal quantum computing with neutral atoms.


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# [R-sig-ME] question on unit variance function

HATICE T KUBRA AKDUR hkubr@@eno| @end|ng |rom gm@||@com
Thu Apr 30 17:33:45 CEST 2020

Dear Group Members,

I hope you are keeping well during this difficult time.
I have some questions below:

In the beta regression model, it is stated V(\mu)= \mu(1-\mu).

It is known that even though beta distribution is in the exponential

And, V(\mu)= \mu(1-\mu) is unit variance function for binomial
distribution, logit link is canonical for binomial distribution.

I am not sure,  V(mu)=mu(1-mu), is it still unit variance function for
beta regression model with logit link.  Even though, it is not in natural
exponential family. OR we can say that V(\mu)= \mu(1-\mu) is variance
function for beta regression with logit link.

1/g'(mu) (g(mu) is link function) can be used to obtain unit variance
function for not natural exponential family distribution? or it is only
true for natural exponential family?

For example, unit variance function of Simplex distribution is
V(mu)=mu_3(1-mu)_3
because simplex distribution is in dispersion family. So, to obtain unit
variance function, the second order derivative of the deviance function is
used.

I will be very glad, if you enlighten me.

Best,

References:

https://onlinelibrary.wiley.com/doi/abs/10.1111/j.0006-341X.2000.00496.x

https://www.tandfonline.com/doi/pdf/10.1080/0266476042000214501

On Thu, Apr 30, 2020 at 11:10 AM HATICE T KUBRA AKDUR <hkubrasenol using gmail.com>
wrote:

> Dear Group Members,
>
> I hope you are keeping well during this difficult time.
> I have some questions below:
>
> In the beta regression model, it is stated V(\mu)= \mu(1-\mu).
>
> It is known that even though beta distribution is in the exponential
>
> And, V(\mu)= \mu(1-\mu) is unit variance function for binomial
> distribution, logit link is canonical for binomial distribution.
>
>  I am not sure,  V(mu)=mu(1-mu), is it still unit variance function for
> beta regression model with logit link.  Even though, it is not in natural
> exponential family. OR we can say that V(\mu)= \mu(1-\mu) is variance
> function for beta regression with logit link.
>
> 1/g'(mu) (g(mu) is link function) can be used to obtain unit variance
> function for not natural exponential family distribution? or it is only
> true for natural exponential family?
>
> For example, unit variance function of Simplex distribution is V(mu)=mu_3(1-mu)_3
> because simplex distribution is in dispersion family. So, to obtain unit
> variance function, the second order derivative of the deviance function
> is used.
>
> I will be very glad, if you enlighten me.
> I attached beta regression and marginal simplex model papers FYI.
>
> Best Regards,
> --
> Asst. Prof. Dr. Hatice T. Kubra AKDUR
> Department of Statistics, Faculty of Science
> Gazi University
> 06500 Teknikokullar ANKARA, TURKEY
> Phone: +90 553 324 5380
> Email: hatice_senol using wsu.edu
> Homepage: http://websitem.gazi.edu.tr/site/haticesenol
>

--
Asst. Prof. Dr. Hatice T. Kubra AKDUR
Department of Statistics, Faculty of Science
Gazi University
06500 Teknikokullar ANKARA, TURKEY
Phone: +90 553 324 5380
Email: hatice_senol using wsu.edu
Homepage: http://websitem.gazi.edu.tr/site/haticesenol

[[alternative HTML version deleted]]


text:
hola-hula
Following up on this [question](https://stackoverflow.com/questions/62028511/how-can-i-remove-the-labels-in-a-bibliography-list-with-biblatex), here is my question. How can I access each bibliographic item (from **References** section) so that I can put it inside enumerate?

Consider the minimal working example:

\documentclass{article}
\usepackage[backend=biber,maxbibnames=99,defernumbers=true,sorting=ydnt,giveninits=true,]{biblatex}

\begin{filecontents*}{myreferences.bib}
@article{sponge1,
author = {Sponge Bob Squrepants},
year = {2000},
title = {No Title},
}

@article{sponge2,
author = {Sponge Bob Squrepants and
Patrick Star},
year = {2001},
title = {No Title Again},
}

\end{filecontents*}

\begin{document}

%%%%%%% Copied
%	\defbibenvironment{bibliography}
%	{\list
%		{}
%		{%
%			\setlength{\labelwidth}{0pt}%
%			\setlength{\leftmargin}{0pt}%
%			\setlength{\labelsep}{0pt}%
%			\setlength{\itemsep}{\bibitemsep}%
%			\setlength{\parsep}{\bibparsep}}%
%		\renewcommand*{\makelabel}[1]{##1\hss}}
%	{\endlist}
%	{\item}
%
\nocite{*}\printbibliography

\end{document}

samcarter
The trick with biblatex is to let it do the work for you. So instead of trying to manually build the layout you want, I'd use an existing style and tweak things from there.

Version with bullet:

\documentclass{article}

\usepackage[style=alphabetic]{biblatex}

\defbibenvironment{bibliography}
{\list
{\textbullet}
{\setlength{\labelwidth}{1em}%
\setlength{\leftmargin}{\labelwidth}%
\setlength{\labelsep}{\biblabelsep}%
\setlength{\itemsep}{\bibitemsep}%
\setlength{\parsep}{\bibparsep}}%
\renewcommand*{\makelabel}[1]{##1\hss}}
{\endlist}
{\item}

\begin{document}

\nocite{*}
\printbibliography

\end{document}

![Screen Shot 2020-11-10 at 17.56.23.png](/image?hash=a3357fec3a3c93383fc32fac07f000ec9689ff0a97515b087e018f211098cf7c)

Version with year:

\documentclass{article}

\usepackage[style=alphabetic]{biblatex}

\defbibenvironment{bibliography}
{\list
{\printfield{year}}
{\setlength{\labelwidth}{2em}%
\setlength{\leftmargin}{\labelwidth}%
\setlength{\labelsep}{\biblabelsep}%
\setlength{\itemsep}{\bibitemsep}%
\setlength{\parsep}{\bibparsep}}%
\renewcommand*{\makelabel}[1]{##1\hss}}
{\endlist}
{\item}

\begin{document}

\nocite{*}
\printbibliography

\end{document}

![Screen Shot 2020-11-10 at 17.55.35.png](/image?hash=baa2153d8f252a3b10c4b3ec81064f5d8c927769360cc3a78786495e5f746fbd)

Enter question or answer id or url (and optionally further answer ids/urls from the same question) from

Separate each id/url with a space. No need to list your own answers; they will be imported automatically.


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### Home > CCA > Chapter 9 > Lesson 9.1.4 > Problem9-41

9-41.

Solve the equations below for $x$. Check your solutions.

1.  $3x^2+3x=6+3x^2$

1.  $\frac{5}{x}=\frac{1}{3}$

1.  $5−(2x−3)=−3x+6$

1.  $6(x−3)+2x=4(2x+1)−22$

One of the solutions is all real numbers. You should be able to check your solutions to the rest of the equations.


text:
Variable selection using minimal depth.

var.select(formula,
data,
object,
cause,
m.target,
method = c("md", "vh", "vh.vimp"),
conservative = c("medium", "low", "high"),
ntree = (if (method == "md") 1000 else 500),
mvars = (if (method != "md") ceiling(ncol(data)/5) else NULL),
mtry = (if (method == "md") ceiling(ncol(data)/3) else NULL),
nodesize = 2, splitrule = NULL, nsplit = 10, xvar.wt = NULL,
refit = (method != "md"), fast = FALSE,
na.action = c("na.omit", "na.impute"),
always.use = NULL, nrep = 50, K = 5, nstep = 1,
prefit =  list(action = (method != "md"), ntree = 100,
mtry = 500, nodesize = 3, nsplit = 1),
verbose = TRUE, ...)

## Arguments

formula

A symbolic description of the model to be fit. Must be specified unless object is given.

data

Data frame containing the y-outcome and x-variables in the model. Must be specified unless object is given.

object

An object of class (rfsrc, grow). Not required when formula and data are supplied.

cause

Integer value between 1 and J indicating the event of interest for competing risks, where J is the number of event types (this option applies only to competing risk families). The default is to use the first event type.

m.target

Character value for multivariate families specifying the target outcome to be used. If left unspecified, the algorithm will choose a default target.

method

Variable selection method:

md:

minimal depth (default).

vh:

variable hunting.

vh.vimp:

variable hunting with VIMP (variable importance).

conservative

Level of conservativeness of the thresholding rule used in minimal depth selection:

high:

Use the most conservative threshold.

medium:

Use the default less conservative tree-averaged threshold.

low:

Use the more liberal one standard error rule.

ntree

Number of trees to grow.

mvars

Number of randomly selected variables used in the variable hunting algorithm (ignored when method="md").

mtry

The mtry value used.

nodesize

Forest average terminal node size.

splitrule

Splitting rule used.

nsplit

If non-zero, the specified tree splitting rule is randomized which significantly increases speed.

xvar.wt

Vector of non-negative weights specifying the probability of selecting a variable for splitting a node. Must be of dimension equal to the number of variables. Default (NULL) invokes uniform weighting or a data-adaptive method depending on prefit$action. refit Should a forest be refit using the selected variables? fast Speeds up the cross-validation used for variable hunting for a faster analysis. See miscellanea below. na.action Action to be taken if the data contains NA values. always.use Character vector of variable names to always be included in the model selection procedure and in the final selected model. nrep Number of Monte Carlo iterations of the variable hunting algorithm. K Integer value specifying the K-fold size used in the variable hunting algorithm. nstep Integer value controlling the step size used in the forward selection process of the variable hunting algorithm. Increasing this will encourage more variables to be selected. prefit List containing parameters used in preliminary forest analysis for determining weight selection of variables. Users can set all or some of the following parameters: action: Determines how (or if) the preliminary forest is fit. See details below. ntree: Number of trees used in the preliminary analysis. mtry: mtry used in the preliminary analysis. nodesize: nodesize used in the preliminary analysis. nsplit: nsplit value used in the preliminary analysis. verbose Set to TRUE for verbose output. ... Further arguments passed to forest grow call. ## Details This function implements random forest variable selection using tree minimal depth methodology (Ishwaran et al., 2010). The option method allows for two different approaches: 1. method="md" Invokes minimal depth variable selection. Variables are selected using minimal depth variable selection. Uses all data and all variables simultaneously. This is basically a front-end to the max.subtree wrapper. Users should consult the max.subtree help file for details. Set mtry to larger values in high-dimensional problems. 2. method="vh" or method="vh.vimp" Invokes variable hunting. Variable hunting is used for problems where the number of variables is substantially larger than the sample size (e.g., p/n is greater than 10). It is always prefered to use method="md", but to find more variables, or when computations are high, variable hunting may be preferred. When method="vh": Using training data from a stratified K-fold subsampling (stratification based on the y-outcomes), a forest is fit using mvars randomly selected variables (variables are chosen with probability proportional to weights determined using an initial forest fit; see below for more details). The mvars variables are ordered by increasing minimal depth and added sequentially (starting from an initial model determined using minimal depth selection) until joint VIMP no longer increases (signifying the final model). A forest is refit to the final model and applied to test data to estimate prediction error. The process is repeated nrep times. Final selected variables are the top P ranked variables, where P is the average model size (rounded up to the nearest integer) and variables are ranked by frequency of occurrence. The same algorithm is used when method="vh.vimp", but variables are ordered using VIMP. This is faster, but not as accurate. Miscellanea 1. When variable hunting is used, a preliminary forest is run and its VIMP is used to define the probability of selecting a variable for splitting a node. Thus, instead of randomly selecting mvars at random, variables are selected with probability proportional to their VIMP (the probability is zero if VIMP is negative). A preliminary forest is run once prior to the analysis if prefit$action=TRUE, otherwise it is run prior to each iteration (this latter scenario can be slow). When method="md", a preliminary forest is fit only if prefit$action=TRUE. Then instead of randomly selecting mtry variables at random, mtry variables are selected with probability proportional to their VIMP. In all cases, the entire option is overridden if xvar.wt is non-null. 2. If object is supplied and method="md", the grow forest from object is parsed for minimal depth information. While this avoids fitting another forest, thus saving computational time, certain options no longer apply. In particular, the value of cause plays no role in the final selected variables as minimal depth is extracted from the grow forest, which has already been grown under a preselected cause specification. Users wishing to specify cause should instead use the formula and data interface. Also, if the user requests a prefitted forest via prefit$action=TRUE, then object is not used and a refitted forest is used in its place for variable selection. Thus, the effort spent to construct the original grow forest is not used in this case.

3. If fast=TRUE, and variable hunting is used, the training data is chosen to be of size n/K, where n=sample size (i.e., the size of the training data is swapped with the test data). This speeds up the algorithm. Increasing K also helps.

4. Can be used for competing risk data. When method="vh.vimp", variable selection based on VIMP is confined to an event specific cause specified by cause. However, this can be unreliable as not all y-outcomes can be guaranteed when subsampling (this is true even when stratifed subsampling is used as done here).

## Value

Invisibly, a list with the following components:

err.rate

Prediction error for the forest (a vector of length nrep if variable hunting is used).

modelsize

Number of variables selected.

topvars

Character vector of names of the final selected variables.

varselect

Useful output summarizing the final selected variables.

rfsrc.refit.obj

Refitted forest using the final set of selected variables (requires refit=TRUE).

md.obj

Minimal depth object. NULL unless method="md".

## Author

Hemant Ishwaran and Udaya B. Kogalur

## References

Ishwaran H., Kogalur U.B., Gorodeski E.Z, Minn A.J. and Lauer M.S. (2010). High-dimensional variable selection for survival data. J. Amer. Statist. Assoc., 105:205-217.

Ishwaran H., Kogalur U.B., Chen X. and Minn A.J. (2011). Random survival forests for high-dimensional data. Statist. Anal. Data Mining, 4:115-132.

find.interaction.rfsrc, holdout.vimp.rfsrc, max.subtree.rfsrc, vimp.rfsrc

## Examples

# \donttest{
## ------------------------------------------------------------
## Minimal depth variable selection
## survival analysis
## use larger node size which is better for minimal depth
## ------------------------------------------------------------

data(pbc, package = "randomForestSRC")
pbc.obj <- rfsrc(Surv(days, status) ~ ., pbc, nodesize = 20, importance = TRUE)

# default call corresponds to minimal depth selection
vs.pbc <- var.select(object = pbc.obj)
topvars <- vs.pbc$topvars # the above is equivalent to max.subtree(pbc.obj)$topvars

# different levels of conservativeness
var.select(object = pbc.obj, conservative = "low")
var.select(object = pbc.obj, conservative = "medium")
var.select(object = pbc.obj, conservative = "high")

## ------------------------------------------------------------
## Minimal depth variable selection
## competing risk analysis
## use larger node size which is better for minimal depth
## ------------------------------------------------------------

## competing risk data set involving AIDS in women
data(wihs, package = "randomForestSRC")
vs.wihs <- var.select(Surv(time, status) ~ ., wihs, nsplit = 3,
nodesize = 20, ntree = 100, importance = TRUE)

## competing risk analysis of pbc data from survival package
## implement cause-specific variable selection
if (library("survival", logical.return = TRUE)) {
data(pbc, package = "survival")
pbc$id <- NULL var.select(Surv(time, status) ~ ., pbc, cause = 1) var.select(Surv(time, status) ~ ., pbc, cause = 2) } ## ------------------------------------------------------------ ## Minimal depth variable selection ## classification analysis ## ------------------------------------------------------------ vs.iris <- var.select(Species ~ ., iris) ## ------------------------------------------------------------ ## Variable hunting high-dimensional example ## van de Vijver microarray breast cancer survival data ## nrep is small for illustration; typical values are nrep = 100 ## ------------------------------------------------------------ data(vdv, package = "randomForestSRC") vh.breast <- var.select(Surv(Time, Censoring) ~ ., vdv, method = "vh", nrep = 10, nstep = 5) # plot top 10 variables plot.variable(vh.breast$rfsrc.refit.obj,
xvar.names = vh.breast$topvars[1:10]) plot.variable(vh.breast$rfsrc.refit.obj,
xvar.names = vh.breast$topvars[1:10], partial = TRUE) ## similar analysis, but using weights from univarate cox p-values if (library("survival", logical.return = TRUE)) { cox.weights <- function(rfsrc.f, rfsrc.data) { event.names <- all.vars(rfsrc.f)[1:2] p <- ncol(rfsrc.data) - 2 event.pt <- match(event.names, names(rfsrc.data)) xvar.pt <- setdiff(1:ncol(rfsrc.data), event.pt) sapply(1:p, function(j) { cox.out <- coxph(rfsrc.f, rfsrc.data[, c(event.pt, xvar.pt[j])]) pvalue <- summary(cox.out)$coef[5]
if (is.na(pvalue)) 1.0 else 1/(pvalue + 1e-100)
})
}
data(vdv, package = "randomForestSRC")
rfsrc.f <- as.formula(Surv(Time, Censoring) ~ .)
cox.wts <- cox.weights(rfsrc.f, vdv)
vh.breast.cox <- var.select(rfsrc.f, vdv, method = "vh", nstep = 5,
nrep = 10, xvar.wt = cox.wts)
}

# }


text:
# Clustering edge points of an image?

Consider the following image

I would like to cluster the points obtained after edge detection and subsequently, would like to color the clustered edge points.

My attempt:

img = URLExecute["https://i.stack.imgur.com/4q309.png"];
edgeImg = img // EdgeDetect;
edgePoints = PixelValuePositions[edgeImg, 1];
clusteredEdgePoints = FindClusters[edgePoints, Method -> "NeighborhoodContraction"];
{edgeImg, Graphics[{ColorData["DarkRainbow"][RandomReal[]], Line[#]} & /@
clusteredEdgePoints]}

{, }

Question:

One can observe, FindClusters did not group inner and outer ellipses. How should I tell FindClusters to group inner and outer ellipses? or is there any other alternative?

Note: I would like to keep the question as general as possible that is I would not want to enter the number of clusters explicitly.

Some other images can be as follows:

img1 = URLExecute["https://i.stack.imgur.com/Z150N.png"];
img2 = URLExecute["https://i.stack.imgur.com/spkHI.png"];

,

• Consider something like FindClusters[edgePoints, 2, Method -> "Agglomerate", CriterionFunction -> "RSquared"]. Playing around with Method and CriterionFunction is worthwhile. – Carl Lange May 24 '19 at 18:50

So, you want to find connected components of all edge pixels that are less than N pixels apart?

You could use MorphologicalComponents for this:

MorphologicalComponents[Dilation[edgeImg, 5]]*
ImageData[edgeImg] // Colorize

• Thanks for the answer. I would like to connect the clustered edge points using Lines too. I have tried this: Graphics[Line@ Position[MorphologicalComponents[Dilation[edgeImg, 5]], 1]], but it gives a black image. Where did I go wrong? – Anjan Kumar May 24 '19 at 19:46
• I realized that points obtained using Position[MorphologicalComponents[Dilation[edgeImg, 5]], 1] are not ordered. Is there a way I could order them? – Anjan Kumar May 25 '19 at 3:30
• @AnjanKumar: I wouldn't go down that route - there are too many morphological special cases and pitfalls. (Take e.g. the two little line segments at the bottom of the O image above - where would you connect them to?) It's way simpler to use something like MorphologicalPerimeter[img] that is guaranteed to return closed contours than using EdgeDetect and then trying to reconstruct them – Niki Estner May 25 '19 at 5:46
• If you only have a point set and need to find connected components, your best bet is using graph algorithms (see e.g. mathematica.stackexchange.com/questions/95425/…, mathematica.stackexchange.com/questions/80683/…). But as you can see, this is much more complicated than using a function returning closed contours, like ContourDetect, ZeroCrossings or MorphologicalPerimeter – Niki Estner May 25 '19 at 5:54
• So you don't want "clusters" (i.e. sets of connected points), but ordered contours? You could just use ComponentMeasurements[img, "Contours"] for that – Niki Estner May 25 '19 at 9:58


text:
Brian Bi
Exercise 15.8.1 Suppose $$F$$ is the finite field $$\mathbb{F}_{p^r}$$. Let $$K$$ be a finite extension of $$F$$. Then $$K$$ is also a finite field. According to Theorem 15.7.3(c), the group $$K^\times$$ is generated by some element $$\alpha$$, so $$F(\alpha) = K$$.
Exercise 15.8.2 The field extension $$\Q(\sqrt{2}, \sqrt{3})$$ has degree 4 over $$\Q$$, so if $$\alpha \in \Q(\sqrt{2}, \sqrt{3}$$, then $$d = [\Q(\alpha) : \Q]$$ can only be 1, 2, or 4. Obviously $$d = 1$$ exactly when $$\alpha \in \Q$$. If $$d = 2$$, then $$\alpha$$ is the root of some quadratic $$x^2 + ax + b$$ with $$a, b \in \Q$$; this implies that $$\alpha = -\frac{a}{2} \pm \frac{\sqrt{a^2 - 4b}}{2}$$, so $$\alpha$$ must take the form $$e + f\sqrt{g}$$ for some $$e, f, g \in \Q$$, and we can choose $$g$$ to be a squarefree nonnegative integer. We will use the fact that $$\{1, \sqrt{2}, \sqrt{3}, \sqrt{6}, \sqrt{g}\}$$ will be linearly independent over $$\Q$$ if $$g$$ is any squarefree integer other than 1, 2, 3, or 6. This implies that $$g \in \{1, 2, 3, 6\}$$. So if $$\alpha = p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6}$$ where $$p, q, r, s \in \Q$$, then $$\alpha$$ must be primitive (have degree 4) whenever at least two of $$q, r, s$$ are nonzero, and it is easy to see that this condition is also necessary.


text:
A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent (ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70. - Mathematics

A conical tent is 10 m high and the radius of its base is 24 m. Find

(i) slant height of the tent

(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.

["Assume "pi=22/7]

Solution

(i) Let ABC be a conical tent.

Height (h) of conical tent = 10 m

Radius (r) of conical tent = 24 m

Let the slant height of the tent be l.

In ΔABO,

AB2 = AO2 + BO2

l2 = h2 + r2

= (10 m)2 + (24 m)2

= 676 m2

∴ l = 26 m

Therefore, the slant height of the tent is 26 m.

(ii) CSA of tent = πrl

=(22/7xx24xx26)m^2

=13728/7m^2

Cost of 1 m2 canvas = Rs 70

"Cost of "13728/7m^2" canvas "=Rs.(13728/7xx70)

= Rs. 137280

Therefore, the cost of the canvas required to make such a tent is Rs 137280.

Concept: Surface Area of a Right Circular Cone
Is there an error in this question or solution?

APPEARS IN

NCERT Class 9 Maths
Chapter 13 Surface Area and Volumes
Exercise 13.3 | Q 4 | Page 221


text:
## LATEX-L@LISTSERV.UNI-HEIDELBERG.DE

#### View:

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Subject:

Re: \leavevmode?

From:

Date:

Fri, 30 Dec 2011 21:31:36 +0000

Content-Type:

text/plain

Parts/Attachments:

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 On 30/12/2011 11:44, Frank Mittelbach wrote: > Am 30.12.2011 10:46, schrieb Ulrike Fischer: >> Hello, >> >> what is the recommended way to issue a \leavevmode in latex3? > > I guess the answer is not yet. Eventually the underlying concept should > be part of the galley module. But how this will provided in the end > depends a little bit on how far we take this module. There have been two > different implementations initially with somewhat different concepts and > the new one by Joseph implements some but not all of the original one > (for good reasons I would say). In some parts of the coffin code, we do this 'directly' using   \hbox_unpack:N \c_empty_box As Frank says, the galley module contains \galley_leave_vmode:, but that is not loaded by default. -- Joseph Wright


text:
## Seminars and Colloquia by Series

Friday, April 20, 2018 - 15:05 , Location: Skiles 271 , Prof. Rafael de la Llave , GT Math , Organizer: Jiaqi Yang
A well known paper of H. Federer on Flat chains contains a remarkable example attributed  to F. Almgren. We intend to give a geometric exposition of the example and explain its relevance in the global theory of geodesic flows and some global problems such as homogenization in quasi-periodic media. This is part of an expository paper with X. Su.
Friday, April 20, 2018 - 13:05 , Location: Skiles 005 , , CSE, University of Washington , , Organizer: He Guo
Friday, April 20, 2018 - 10:00 , Location: Skiles 006 , Jose Acevedo , Georgia Tech , Organizer: Kisun Lee
In this talk we show how to obtain some (sometimes sharp) inequalities between subgraph densities which are valid asymptotically on any sequence of finite simple graphs with an increasing number of vertices. In order to do this we codify a simple graph with its edge monomial and establish a nice graphical notation that will allow us to play around with these densities.
Thursday, April 19, 2018 - 15:05 , Location: Skiles 006 , , Carnegie Mellon University , , Organizer: Michael Damron
We shall prove that a certain stochastic ordering defined in terms of convex symmetric sets is inherited by sums of independent symmetric random vectors. Joint work with W. Bednorz.
Thursday, April 19, 2018 - 13:30 , Location: Skiles 005 , Alexander Hoyer , Math, GT , Organizer: Robin Thomas
Györi and Lovasz independently proved that a k-connected graph can be partitioned into k subgraphs, with each subgraph connected, containing a prescribed vertex, and with a prescribed vertex count. Lovasz used topological methods, while Györi found a purely graph theoretical approach. Chen et al. later generalized the topological proof to graphs with weighted vertices, where the subgraphs have prescribed weight sum rather than vertex count. The weighted result was recently proven using Györi's approach by Chandran et al. We will use the Györi approach to generalize the weighted result slightly further. Joint work with Robin Thomas.
Thursday, April 19, 2018 - 11:00 , Location: Skiles 006 , , UCLA Mathematics Department , Organizer: Mayya Zhilova
This talk is about the structure theory of measure-preserving systems: transformations of a finite measure space that preserve the measure. Many important examples arise from stationary processes in probability, and simplest among these are the i.i.d. processes. In ergodic theory, i.i.d. processes are called Bernoulli shifts. Some of the main results of ergodic theory concern an invariant of systems called their entropy, which turns out to be intimately related to the existence of `structure preserving' maps from a general system to Bernoulli shifts. I will give an overview of this area and its history, ending with a recent advance in this direction. A measure-preserving system has the weak Pinsker property if it can be split, in a natural sense, into a direct product of a Bernoulli shift and a system of arbitrarily low entropy. The recent result is that all ergodic measure-preserving systems have this property. This talk will assume graduate-level real analysis and measure theory, and familiarity with the basic language of random variables. Past exposure to entropy, measure-theoretic probability or ergodic theory will be helpful, but not essential.
Wednesday, April 18, 2018 - 14:10 , Location: Skiles 006 , Sarah Davis , GaTech , Organizer: Anubhav Mukherjee
The theorem of Dehn-Nielsen-Baer says the extended mapping class group is isomorphic to the outer automorphism group of the fundamental group of a surface. This theorem is a beautiful example of the interconnection between purely topological and purely algebraic concepts. This talk will discuss the background of the theorem and give a sketch of the proof.
Wednesday, April 18, 2018 - 13:55 , Location: Skiles 005 , , Clemson University , , Organizer: Galyna Livshyts
We discuss the probability that a continuous stationary Gaussian process on whose spectral measure vanishes in a neighborhood of the origin stays non-negative on an interval of long interval.  Joint work with  Naomi Feldheim, Ohad Feldheim, Fedor Nazarov,  and Shahaf Nitzan
Series: PDE Seminar
Tuesday, April 17, 2018 - 15:00 , Location: Skiles 006 , , Duke University , , Organizer: Yao Yao
Epitaxial growth is an important physical process for forming solid films or other nano-structures.  It occurs as atoms, deposited from above, adsorb and diffuse on a crystal surface.  Modeling the rates that atoms hop and break bonds leads in the continuum limit to degenerate 4th-order PDE that involve exponential nonlinearity and the p-Laplacian with p=1, for example.  We discuss a number of analytical results for such models, some of which involve subgradient dynamics for Radon measure solutions.
Monday, April 16, 2018 - 15:30 , Location: Skiles 005 , Yu Pan , MIT , Organizer: Caitlin Leverson
Augmentations and exact Lagrangian fillings are closely related. However, not all the augmentations of a Legendrian knot come from embedded exact Lagrangian fillings. In this talk, we show that all the augmentations come from possibly immersed exact Lagrangian fillings. In particular, let ∑ be an immersed exact Lagrangian filling of a Legendrian knot in $J^1(M)$ and suppose it can be lifted to an embedded Legendrian L in J^1(R \times M). For any augmentation of L, we associate an induced augmentation of the Legendrian knot, whose homotopy class only depends on the compactly supported Legendrian isotopy type of L and the homotopy class of its augmentation of L. This is a joint work with Dan Rutherford.


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### Existence and nonexistence results for a class of nonlinear, singular Sturm-Liouville equations

#### Abstract

We study the existence and nonexistence of solutions for a class of equations of the form $-(\omega u')'=|u|^{p-2}u$ in $\mathbb R$ where $p>2$ and $\omega$ is a nonnegative continuous function with isolated zeroes of power type.

#### Article information

Source
Adv. Differential Equations, Volume 6, Number 3 (2001), 303-326.

Dates
First available in Project Euclid: 2 January 2013

Mathematical Reviews number (MathSciNet)
MR1799488

Zentralblatt MATH identifier
1032.34034

Subjects
Primary: 34B15: Nonlinear boundary value problems
Secondary: 34C11: Growth, boundedness

#### Citation

Caldiroli, Paolo; Musina, Roberta. Existence and nonexistence results for a class of nonlinear, singular Sturm-Liouville equations. Adv. Differential Equations 6 (2001), no. 3, 303--326. https://projecteuclid.org/euclid.ade/1357141213


text:
# How do I calculate the delta of a convertible bond?

How can I find the delta of a convertible bond to be used for hedging?

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Delta w/r to what, stock price? –  quant_dev Apr 26 '11 at 18:03

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I'm not sure how it takes more information, I didn't provide details for a particular bond, because I was looking for a general formula or explanation on its relation to an option's delta. –  tshauck Apr 26 '11 at 14:51
There is no simple formula, as you may gather from the pdf Tangurena gave you. –  quant_dev Apr 26 '11 at 18:03

Tangurena's answer and links give the right idea. You can get a rough approximation by finding the conversion price $K$ and using that $K$ as the strike in a standard Black-Scholes option pricer.

In practice, most people work with 3rd party models such as the ones built into Bloomberg, Monis, or Kynex.

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# Do two encoders connected to a microcontroller interfere with each other?

I connected two AMT 103 rotary encoders (mounted on two separate DC motors) to a dspic30f4011 and read the values using interrupts (instead of qei as the dspic only has one). However, the error in the values increase the farther and faster we go when running the motors simultaneously. One of the theories is that since the interrupts have different priorities one of them overrides the other and causes the error, note that when only one motor is run we don't have an error. Is this causing the error? If yes how can it be fixed? Will using one qei and one interrupt fix this? Here's the dspic code:

#include "p30f4011.h"
#include <delay.h>
#include <stdlib.h>
#include <math.h>

_FWDT(WDT_OFF)

#define InterruptFlag0_A    IFS0bits.INT0IF //Rising Edge on Signal A, Encoder 1
#define InterruptFlag1_A    IFS1bits.INT1IF //Rising Edge on Signal A, Encoder 2
#define Encoder0_B  PORTBbits.RB0 // Signal B of Encoder 1
#define Encoder1_B  PORTBbits.RB1 // Signal B of Encoder 2

//Variables for Calculations
long T1=0;
int counter=0;
int NR=0;
long T2=0;
int counterone=0;
int NRone=0;
long i;
//-------------------------

int main()
{
TRISBbits.TRISB0=1;//RA inputs
TRISBbits.TRISB1=1;//RA inputs

//Interrupts init
IFS1bits.INT1IF = 0;    /*Reset INT1 interrupt flag */
IEC1bits.INT1IE = 1;    /*Enable INT1 Interrupt Service Routine */
IFS0bits.INT0IF = 0;    /*Reset INT1 interrupt flag */
IEC0bits.INT0IE = 1;    /*Enable INT1 Interrupt Service Routine */

while(1) //NEGATIVE ROTATIONS ARE CW. POSITIVE ROTATIONS ARE CCW
{

T1= ((long)NR*10000)+(long)counter;
T2= ((long)NRone*10000)+(long)counterone;

}
return 0;
}

void __attribute__((__interrupt__)) _INT0Interrupt(void);
void __attribute__((__interrupt__, auto_psv)) _INT0Interrupt(void)
{
if(InterruptFlag0_A==1) {
if(Encoder0_B==1) {
counter=counter-1;
InterruptFlag0_A=0;
} else if (Encoder0_B==0) {
counter=counter+1;
InterruptFlag0_A=0;
}
}
if(counter==10000) {
counter=0;
NR=NR+1;
} else if(counter==-10000) {
counter=0;
NR=NR-1;
}
}

void __attribute__((__interrupt__)) _INT1Interrupt(void);
void __attribute__((__interrupt__, auto_psv)) _INT1Interrupt(void)
{
if(InterruptFlag1_A==1) {
if(Encoder1_B==1) {
counterone=counterone+1;
InterruptFlag1_A=0;
} else if (Encoder1_B==0) {
counterone=counterone-1;
InterruptFlag1_A=0;
}
}
if(counterone==10000) {
counterone=0;
NRone=NRone+1;
} else if(counterone==-10000) {
counterone=0;
NRone=NRone-1;
}
}

Thanks

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This is not really an answer, but for your next project yo might consider using the PSoC3 or PSoC5 from Cypress Semiconductor. It can easily handle 8 encoder inputs without using any CPU time. –  Rocketmagnet Apr 24 '12 at 18:03

Don't ever tie external inputs like this directly to interrupt pins. Interrupt pins should be used with "safe" inputs which can't run away and leave the controller a gibbering puddle of ooze as it valiantly tries to service endless interrupts.

(you won't actually kill the MCU, but your embedded application will simply not work)

For things like encoders it's best to use counter hardware and then have a periodic interrupt to look at the counts and determine how many counts per interval of time you're seeing, which will then tell you how fast the motor is turning. If you need directional information as well you could run the I/Q encoder outputs to the clock/direction inputs of standard up/down counter ICs and then poll them with the microcontroller.

I know this is probably not the kind of answer you were hoping for, but using interrupts to count encoder pulses is bad design.

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Connecting external inputs like buttons or encoders directly to maskable interrupts can be a fine thing to do, if the maskable interrupt when triggered disables itself but sets a timer to re-enable it after some minimum time. –  supercat Apr 24 '12 at 18:41
Good catch. You are correct, and in fact this is typical for low power designs in order to "wake up" and handle the event. –  akohlsmith Apr 25 '12 at 0:16

I think for this design you would be better off using the input capture peripheral on your microcontroller instead of external interrupts. Briefly, the input capture peripheral works in conjunction with a timer to time external events. You can connect one encoder channel to each input capture channel (two quadrature encoders = 4 total lines and luckily the 4011 you're using has four input capture lines). The input capture peripheral can be configured to trigger off of a variety of conditions including:

• Every rising edge in the signal
• Every falling edge
• Every 4th rising edge
• Every 16th rising edge
• Every rising and falling edge

(This comes from your chip's datasheet, pg78)

When a trigger condition occurs the peripheral reads the current value of a timer that you specify (I believe either Timer 2 or Timer 3 for your chip). It stores that value in a FIFO that is four 16-bit words deep. It can also generate an interrupt, or you can watch the FIFO and see when it gets full. Possibly you can cause an interrupt to occur when the FIFO is full as well - I haven't checked.

The application of this to a quadrature encoder would be to use two input capture channels with the peripheral utilizing a timer with a suitable frequency. Probably set both channels to use a rising edge condition, or possibly every 4th or 16th rising edge. The choice of these depends on the level of accuracy you need, the frequency of the timer and the speeds you will be trying to measure. There's a lot of factors that you can play with to get better accuracy out of this system. For this example, the difference between two rising edges on one channel will be proportional to speed, and the difference between the rising edges of one channel to another will tell you what direction the motor is spinning.

If you can spare a timer and the input capture peripheral then this is the best way to interface two encoders.

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text:
# What is forward difference interpolation?

## What is forward difference interpolation?

Newton’s forward difference formula is a finite difference identity giving an interpolated value between tabulated points in terms of the first value and the powers of the forward difference .

## What is the formula for forward difference?

NEWTON’S GREGORY FORWARD INTERPOLATION FORMULA : h is called the interval of difference and u = ( x – a ) / h, Here a is the first term.

What is the Newton’s forward difference interpolation formula?

The formula is called Newton’s (Newton-Gregory) forward interpolation formula. So if we know the forward difference values of f at x0 until order n then the above formula is very easy to use to find the function values of f at any non-tabulated value of x in the internal [a,b]….

= ( r ) h(n+1)f(n+1)(x)
n+1

What is the difference between Gregory Newton forward and Gregory Newton backward interpolation formula?

Newton’s Interpolation Formula: Difference between the forward and the backward formula. I was taught that the forward formula should be used when calculating the value of a point near x0 and the backward one when calculating near xn. However, the interpolation polynomial is unique, so the value should be the same.

### What is forward and backward interpolation?

Newton Forward And Backward. Interpolation. Interpolation is the technique of estimating the value of a function for any. intermediate value of the independent variable, while the process of computing the. value of the function outside the given range is called extrapolation.

### What is forward difference and backward difference?

First derivatives f (a) ≈ slope of short broken line = difference in the y-values difference in the x-values = f(a + h) − f(a) h . This is called a one-sided difference or forward difference approximation to the derivative of f. This is another one-sided difference, called a backward difference, approximation to f (a).

What is Newton Gregory forward interpolation formula?

The Gregory–Newton forward difference formula is a formula involving finite differences that gives an approximation for f(x), where x=x 0+θh, and 0 < θ <1. It states thatthe series being terminated at some stage. The approximation f(x) ≈ f 0+θΔ f 0 gives the result of linear interpolation.

What is Newton forward and backward interpolation?

## What is forward and backward difference?

f (a) ≈ slope of short broken line = difference in the y-values difference in the x-values = f(a + h) − f(a) h . This is called a one-sided difference or forward difference approximation to the derivative of f. This is another one-sided difference, called a backward difference, approximation to f (a).

## What do you mean by inverse interpolation?

The technique of finding an estimate of a value of an independent variable x corresponding to a given value of the dependent variable y within the range of the observed values of y, is called an Inverse interpolation.


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## Winger 6-j and 9-j symbol

The meaning of 3-j symbol is same as Clebsch-Gordan coefficient. So, we skip in here.

I am not going to construct the 6-j symbol from 3-j symbol. In here, I just state the meaning and usage in Mathematica.

The 6-j symbol is the coupling between 3 angular momenta, $j_1, j_2, j_3$.

There are 2 ways to couple these 3 angular momenta. First,

$j_1 + j_2 + j_3 \rightarrow j_{12} + j_3 \rightarrow J$

the other way is

$j_ 1 + j_2 + j_3 \rightarrow j_1 + j_{23} \rightarrow J$

The 6-j symbol is

$\begin{pmatrix} j_1 & j_2 & j_{12} \\ j_3 & J & j_{23} \end{pmatrix}$

We can see that there are 4 vector-sum must satisfy.

$\Delta(j_1, j_2, j_{12})$

$\Delta(j_2, j_3, j_{23})$

$\Delta(j_1, j_{23}, J)$

$\Delta(j_{12}, j_3, J)$

If we draw a line to connect these 4 vector-sum, we have:

In Mathematica, there is a build in function

$\textrm{SixJSymbol}[ \left\{j_1, j_2, j_{23} \right\}, \left\{j_3, J , j_{23} \right\}]$

The 9-j symbol is the coupling between 4 angular momenta, $j_1, j_2, j_3, j_4$.

The 9-j symbol can be used in coupling 2 nucleons, $l_1, s_1, l_2, s_2$.

The 9-j symbol is

$\begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix}$

We can see, each row and column must satisfy the vector-sum.

Unfortunately, there is no build in function in Mathematica. The formula for 9-j symbol is

$\displaystyle \begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix} \\ = \sum_{g} (-1)^{2g} (2g+1) \begin{pmatrix} l_1 & s_1 & j_1 \\ j_2 & J & g \end{pmatrix} \begin{pmatrix} j_2 & s_2 & j_2 \\ s_1 & g & S \end{pmatrix} \begin{pmatrix} L & S & J \\ g & l_1 & l_2 \end{pmatrix}$

Where $g$ sum all possible value, which can be calculate using the 6 couplings inside the 3 6-j symbols.To check your result, the coupling between $d_{5/2}$ and $f_{7/2}$ to from a $L = 5, S = 0, J = 0$ state is

$\begin{pmatrix} 2 & 1/2 & 5/2 \\ 3 & 1/2 & 7/2 \\ 5 & 0 & 5 \end{pmatrix} = \frac{1}{2\sqrt{770}}$

## Evaluation of mutual interaction

We are going to evaluation the integral

$\displaystyle G_{ij}^{hk} = \langle b_i(x) b_h(y) | \frac{1}{r_{xy}} | b_j(x) b_k(y) \rangle$

Recalling the multi-pole expansion,

$\displaystyle \frac{1}{r_{12}} = \sum_{l=0}^{\infty} \frac{4 \pi}{2l+1} \frac{r_{<}^l}{r_{>}^{l+1}} \sum_{m=-l}^{l} Y_{lm}^{*}(\Omega_1) Y_{lm}(\Omega_2)$

and the basis

$b_{nlm}(\vec{r}) = R_{nl}(r) Y_{lm}(\Omega)$

Set an averaged basis

$\displaystyle b_{nl}(\vec{r}) = R_{nl}(r) \frac{1}{\sqrt{2l+1}} \sum_{m=-1}^{l}Y_{lm}(\Omega)$

$\displaystyle \Gamma_{ij}^{hk}(l) = \frac{4 \pi}{2l+1} \sum_{m=-l}^{l} \frac{1}{\sqrt{(2l_i+1)(2l_j+1)(2l_h+1)(2l_k+1)}} \\ \sum_{m_i,m_j}\int Y_{l_i m_i}^{*}(\Omega_1) Y_{l m}^{*}(\Omega_1) Y_{l_j m_j}(\Omega_1) d \Omega_1 \\ \sum_{m_h,m_k} \int Y_{l_h m_k}^{*}(\Omega_2) Y_{l m}(\Omega_2) Y_{l_k m_k}(\Omega_2) d \Omega_2$

In the angular integrals, using Wigner 3-j symbol and the integral

$\displaystyle\int Y_{l_1 m_1}(\Omega) Y_{l m}(\Omega) Y_{l_2 m_2}(\Omega) d \Omega \\= \frac{\sqrt{(2l_1+1)(2l+1)(2l_2+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_1 & l & l_2 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_1 & l & l_2 \\ m_1 & m & m_2 \end{pmatrix}$

$Y_{lm}^{*}(\Omega) = (-1)^m Y_{l(-m)}(\Omega)$

$\displaystyle \begin{pmatrix} l_1 & l & l_2 \\ m_1 & m & m_2 \end{pmatrix} = (-1)^{l_1+l+l_2} \begin{pmatrix} l_1 & l & l_2 \\ -m_1 & -m & -m_2 \end{pmatrix} \\ =(-1)^{l_1+l+l_2} \begin{pmatrix} l_1 & l_2 & l \\ m_1 & m_2 & m \end{pmatrix}$

we have

$\displaystyle\int Y_{l_i m_i}^{*}(\Omega) Y_{l m}^{*}(\Omega) Y_{l_j m_j}(\Omega) d \Omega \\= (-1)^{m_j} \frac{\sqrt{(2l_i+1)(2l+1)(2l_j+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_i & l & l_j \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_i & l & l_j \\ m_i & m & -m_j \end{pmatrix}$

$\displaystyle\int Y_{l_h m_h}^{*}(\Omega) Y_{l m}(\Omega) Y_{l_k m_k}(\Omega) d \Omega \\= (-1)^{m_h} \frac{\sqrt{(2l_h+1)(2l+1)(2l_k+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_h & l & l_k \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_h & l & l_k \\ -m_h & m & m_k \end{pmatrix}$

Thus,

$\displaystyle \Gamma_{ij}^{hk}(l) = \sum_{m=-l}^{l} \sum_{m_i,m_j } (-1)^{m_j} \begin{pmatrix} l_i & l_j & l \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_i & l_j & l \\ m_i & -m_j & m \end{pmatrix} \\ \sum_{m_h,m_k} (-1)^{m_h} \begin{pmatrix} l_h & l_k & l \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_h & l_k & l \\ -m_h & m_k & m \end{pmatrix}$

The 3-j symbol restricted that

$l_i+l_j+l = even$

$l_h+l_k+l = even$

$m_i-m_j + m = 0, |m_i| \leq l_i , |m_j| \leq l_j$

$-m_h + m_k +m = 0, |m_h| \leq l_h , |m_k| \leq l_k$

I guess it is the most simplified formula for the angular part.

The total integral is

$\displaystyle G_{ij}^{hk} = \sum_{l=0}^\infty \Gamma_{ij}^{hk}(l) \langle R_i(x) R_h(y) | \frac{r_<^l}{r_>^{l+1}} |R_j(x) R_k(y) \rangle$

The angular integral imposes condition for $l$.

I am not sure this is a correct way to treat the problem.

First, the averaged basis is still an energy eigen state. It is not the eigen state for the angular part. So, this averaging could introduces an error and we should reminded that this is an approximation. But in the perturbation view point, this averaged basis is still valid.

Second thing is, the sum

$\displaystyle P_{(l_i m_i) (l_j m_j)}^{(l_h m_h)(l_k m_k)}(l) = \sum_{m=-l}^{l} \langle Y_{l_i m_i}|Y_{lm}^*|Y_{l_j m_j}\rangle \langle Y_{l_h m_h}|Y_{lm}|Y_{l_k m_k}\rangle$

is not symmetry for exchange of $i, j, h,k$ in general. For example,

$\displaystyle \frac{-1}{3}=P_{(1-1) (00)}^{(11)(00)}(1) {\neq} P_{(00)(1-1)}^{(11)(00)}(1) = 0$

This is a very uprising result that the mutual interaction dependent on the magnetic quantum number. Thus, in detail, we should use $n l m m_s$ as a basis.

Third, the sum $P_{ij}^{hk}(l)$ is depend on $l$. The mutual interaction require us to sum all possible $l$.

Fourth, the coupling between 1s2p triplet state, the total spin is $S = 1$, total L is $L = 1$, and the total angular momentum can be  $J = 0, 1, 2$. In our treatment, we did not coupled the angular momentum in the calculation explicitly. In fact, in the integral of the spherical harmonic, the coordinate are integrated separately, and the coupling seem to be calculated implicitly. I am not sure how to couple two spherical harmonics with two coordinates.

## Integration formulas of spherical harmonic

There are several important and useful formulas for the integration of spherical harmonic. We simplify the notation,

$\displaystyle \int_0^{\pi} \sin(\theta)d\theta\int_0^{2\pi}d\phi = \int d\Omega$

The first one is the average of spherical harmonic.

$\displaystyle \int Y_{lm} d\Omega = \sqrt{4\pi} \delta_{l0}\delta_{m0}$

The 2nd one is the orthonormal  condition.

$\displaystyle \int Y^{*}_{l'm'}Y_{lm} d\Omega = \delta_{l'l}\delta_{m'm}$

The 3rd one is triplet integral, we use the product of spherical harmonic,

$\displaystyle \int Y_{l_1m_1}Y_{l_2m_2} Y^*_{l_3m_3} d\Omega \\ = \int \sum_{lm} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l+1)}} C_{l_10l_20}^{l0} C_{l_1m_1l_2m_2}^{lm} Y_{lm} Y^*_{l_3m_3} d\Omega \\= \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l_3+1)}} C_{l_10l_20}^{l_30} C_{l_1m_1l_2m_2}^{l_3m_3}$

The 4th one is another triple integral,

$\displaystyle \int Y_{l_1m_1}Y_{l_2m_2} Y_{l_3m_3} d\Omega \\ = \int \sum_{lm} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l+1)}} C_{l_10l_20}^{l0} C_{l_1m_1l_2m_2}^{lm} Y_{lm} Y_{l_3m_3} d\Omega \\ = \int \sum_{lmLM} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l+1)}} C_{l_10l_20}^{l0} C_{l_1m_1l_2m_2}^{lm} \\ \sqrt{\frac{(2l+1)(2l_3+1)}{4\pi(2L+1)}} C_{l0l_30}^{L0} C_{lml_3m_3}^{LM}Y_{LM}d\Omega$

$\displaystyle = \sum_{lm} \sqrt{\frac{(2l_1+1)(2l_2+1)(2l_3+1)}{4\pi}} C_{l_10l_20}^{l0} C_{l_1m_1l_2m_2}^{lm} C_{l0l_30}^{00} C_{lml_3m_3}^{00}$

Notice that

$C_{lmLM}^{00} = (-1)^{L+M} \sqrt{\frac{1}{2L+1}} \delta_{Ll}\delta_{-m,M}$

$\displaystyle \int Y_{l_1m_1}Y_{l_2m_2} Y_{l_3m_3} d\Omega = \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi (2l_3+1)}} C_{l_10l_20}^{l_30} C_{l_1m_1l_2m_2}^{l_3,-m_3} (-1)^{m_3}$

using Wigner 3-j symbol,

$C_{l_1m_1l_2m_2}^{l_3m_3} = (-1)^{l_1-l_2+m_3} \sqrt{2l_3+1} \begin{pmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & -m_3 \end{pmatrix}$

$\displaystyle \int Y_{l_1m_1}Y_{l_2m_2} Y_{l_3m_3} d\Omega \\= \sqrt{\frac{(2l_1+1)(2l_2+1)(2l_3+1)}{4\pi}} \begin{pmatrix} l_1 & l_2 & l_3 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \end{pmatrix}$

For other integral, we can use

$Y^*_{lm}(\theta, \phi) = (-1)^{m}Y_{l(-m)}(\theta,\phi) = Y_{lm}(\theta, -\phi)$


text:
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# Math: Coordinate Geometry

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09 Feb 2013, 02:11
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Archit143 wrote:
So for a question where it is given that slope is -1/6, Than how can be sure that line intersects 2nd quad, I found this question on GMAT prep......
in-the-rectangular-coordinate-system-shown-above-does-the-90635.html

Archit

As discussed, line with negative slope (-1/6<0) MUST intersect quadrants II and IV.
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09 Feb 2013, 03:45
HI Bunuel,
As karishma has also mentioned that a line with -ve slope can also be present in 1st quadrant than how can we be so sure that it is intersecting in 2nd quadrant,

Moreover as per the theory it must intersect in 2nd and 4th quadrant ...than as per statement 1 there are two possibilities.....line intersecting in in quad 2 and 4.....But we must have only one answer form the statement, for it to be correct answer..................

Archit
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09 Feb 2013, 03:53
Archit143 wrote:
HI Bunuel,
As karishma has also mentioned that a line with -ve slope can also be present in 1st quadrant than how can we be so sure that it is intersecting in 2nd quadrant,

Moreover as per the theory it must intersect in 2nd and 4th quadrant ...than as per statement 1 there are two possibilities.....line intersecting in in quad 2 and 4.....But we must have only one answer form the statement, for it to be correct answer..................

Archit

No, that's not what she said.

If the slope of a line is negative, line WILL intersect quadrants II and IV in ANY case. If X and Y intersects are positives, line ALSO intersects the quadrant I, if negative line ALSO intersects the quadrant quadrant III.
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09 Feb 2013, 22:53
Archit143 wrote:
So for a question where it is given that slope is -1/6, Than how can be sure that line intersects 2nd quad, I found this question on GMAT prep......
in-the-rectangular-coordinate-system-shown-above-does-the-90635.html

Archit

Since the slope is negative, the line will intersect the 2nd and 4th quadrant. We are talking about a line, not a line segment. A line extends indefinitely on both ends. The top end of the line will extend to intersect the 2nd quadrant under all circumstances.

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews VP Status: Final Lap Up!!! Affiliations: NYK Line Joined: 21 Sep 2012 Posts: 1072 Location: India GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31 GPA: 3.84 WE: Engineering (Transportation) Re: Math: Coordinate Geometry [#permalink] ### Show Tags 10 Feb 2013, 04:16 Karishma, thats what i wanted to ask....In question it asks about whether the line is intersecting 2nd quadrant....Answer is Yes it does, but at the same time it may lie in 1st quadrant also as explained by you....I think i am badly confused on this.... Archit Math Expert Joined: 02 Sep 2009 Posts: 43851 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 10 Feb 2013, 04:23 Archit143 wrote: Karishma, thats what i wanted to ask....In question it asks about whether the line is intersecting 2nd quadrant....Answer is Yes it does, but at the same time it may lie in 1st quadrant also as explained by you....I think i am badly confused on this.... Archit Does it matter whether the line also lies in other quadrants? We know that it goes through the II and IV quadrants, it may also (simultaneously) go through either I or III quadrant, but this does not alter the fact that the line goes through the II, is it? So, the answer is YES. _________________ VP Status: Final Lap Up!!! Affiliations: NYK Line Joined: 21 Sep 2012 Posts: 1072 Location: India GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31 GPA: 3.84 WE: Engineering (Transportation) Re: Math: Coordinate Geometry [#permalink] ### Show Tags 10 Feb 2013, 06:27 Thanks Bunuel.... I got it.. Archit Manager Joined: 07 May 2012 Posts: 73 Location: United States Re: Math: Coordinate Geometry [#permalink] ### Show Tags 06 Apr 2013, 08:06 Hi Bunuel, Thanks for the consolidated , yet elaborate content for the topic at one place. I would be glad , if you can tell me as to , how did you arrive at the formula for the vertex of a parabola. I find it difficult to remember formulas and if there is a easy way to arrive at the formula itself , then would prefer to know that too , so it helps in case I forget the formula. Appreciate any help on this. Thanks, Jyothi _________________ Jyothi hosamani Math Expert Joined: 02 Sep 2009 Posts: 43851 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 10 Jul 2013, 23:06 Bumping for review*. *New project from GMAT Club!!! Check HERE _________________ Intern Joined: 24 Apr 2013 Posts: 5 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 05 Aug 2013, 10:19 Hi bunuel, firstly i want to thank for the explanations you provide to questions because most of them are pretty conceptual and healthy to understand. Secondly, i wished i was more thorough with my concepts in co-ordinate geometry specifically PARABOLA. i had seen two parabola questions when i took the test & both were quite hard unfortunately, had to end up guessing. Have yu posted any questions related to Parabola ?? Non-Human User Joined: 09 Sep 2013 Posts: 13814 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 08 Aug 2014, 21:38 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Intern Joined: 23 Aug 2014 Posts: 42 GMAT Date: 11-29-2014 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 25 Nov 2014, 23:21 1 This post was BOOKMARKED The angle of inclination of a line with slope 1 is 45 degrees. I know that the angle for a line of slope 2 is not 90 degree, but i do not know why not. Please help Intern Joined: 15 Dec 2014 Posts: 1 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 04 May 2015, 11:45 All of the formulas in this post using division are messed up. I believe the formatting or something has changed and caused this error. Please fix this! See the current formula for the slop as an example. Math Expert Joined: 02 Sep 2009 Posts: 43851 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 04 May 2015, 23:52 Jamie19892000 wrote: All of the formulas in this post using division are messed up. I believe the formatting or something has changed and caused this error. Please fix this! See the current formula for the slop as an example. Everything looks fine for me. Can you please post a screenshot? _________________ Non-Human User Joined: 09 Sep 2013 Posts: 13814 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 05 May 2016, 02:23 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Intern Joined: 29 Nov 2015 Posts: 7 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 10 May 2016, 20:47 Thanks Bunuel this is my weak subject for the gmat. Posted from my mobile device Intern Joined: 12 Dec 2016 Posts: 12 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 22 Mar 2017, 12:14 Can you please explain here; To answer, we must find the slope of each line and then check to see if one slope is the negative reciprocal of the other or if their product equals to -1. SlopeAB=5−199−48=−14−39=0.358SlopeAB=5−199−48=−14−39=0.358 SlopeCD=24−422−31=20−9=−2.22 The formula of the slope of two given coordinates are y2-y1 / x2-x1 However in some questions, 2nd coordinates (x2 y2)s are subtracted from 1sts (x1 y1) and in some, other way around. Can you please clarify what do we take into account concerning this formula? Board of Directors Status: Aiming MBA Joined: 18 Jul 2015 Posts: 3074 Location: India Concentration: Healthcare, Technology GPA: 3.65 WE: Information Technology (Health Care) Re: Math: Coordinate Geometry [#permalink] ### Show Tags 22 Mar 2017, 22:52 1 This post received KUDOS mesutthefail wrote: Can you please explain here; To answer, we must find the slope of each line and then check to see if one slope is the negative reciprocal of the other or if their product equals to -1. SlopeAB=5−199−48=−14−39=0.358SlopeAB=5−199−48=−14−39=0.358 SlopeCD=24−422−31=20−9=−2.22 The formula of the slope of two given coordinates are y2-y1 / x2-x1 However in some questions, 2nd coordinates (x2 y2)s are subtracted from 1sts (x1 y1) and in some, other way around. Can you please clarify what do we take into account concerning this formula? Hi, This is a mathematical rule. You can write a-b = -(b-a). Did you get this rule?? Now, in a similar way if you take - sign common from both numerator and denominator of y2-y1 / x2-x1, you will get -(y1-y2)/-(x1-x2). I hope you are aware of the rule that - signs can be cancelled out both a numerator and denominator. So, we will be left with the formula, Slope = (y1-y2)/(x1-x2). I hope that makes sense. _________________ How I improved from V21 to V40! ? How to use this forum in THE BEST way? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7944 Location: Pune, India Re: Math: Coordinate Geometry [#permalink] ### Show Tags 23 Mar 2017, 03:22 mesutthefail wrote: Can you please explain here; To answer, we must find the slope of each line and then check to see if one slope is the negative reciprocal of the other or if their product equals to -1. SlopeAB=5−199−48=−14−39=0.358SlopeAB=5−199−48=−14−39=0.358 SlopeCD=24−422−31=20−9=−2.22 The formula of the slope of two given coordinates are y2-y1 / x2-x1 However in some questions, 2nd coordinates (x2 y2)s are subtracted from 1sts (x1 y1) and in some, other way around. Can you please clarify what do we take into account concerning this formula? They are both the same. $$Slope = \frac{(y2 - y1)}{(x2 - x1)} = \frac{(y1 - y2)}{(x1 - x2)}$$ Take an example: (x1, y1) = (2, 3) (x2, y2) = (5, -10) $$Slope = \frac{(y2 - y1)}{(x2 - x1)} = \frac{-10 - 3}{5 - 2} = -\frac{13}{3}$$ $$Slope = \frac{(y1 - y2)}{(x1 - x2)} = \frac{3- (-10)}{2 - 5} = -\frac{13}{3}$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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01 Sep 2017, 23:22
Hello Bunuel,

Thanks a lot for the article. I have one doubt . Is it that a line with negative slope would definitely pass through Quadrant 2 and 4, and would pass through 1 or 3 depending on the value of x and y intersects?
Re: Math: Coordinate Geometry   [#permalink] 01 Sep 2017, 23:22

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