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# Intuition behind the relation of commutative Hopf algebra and Groups

I've heard that a commutative Hopf algebra can be thought an algebra construction over the space of functions on a group to the ground field. The product should be the pointwise multiplication, Coproduct maps $f$ to the functions on 2 variables $(x,y)↦f(xy)$. Counit is evaluation $f↦f(1)$. Antipode maps f to $x↦f(x^−1)$. Everything is fine so far. Here's my question:

1) How it is related this construction with the other canonical group algebra $\mathbb CG$ where the coproduct is given by $$\triangle\left(g\right)=\left(g\otimes g\right),$$ with $g\in G$. Is it the same construction with different notation? How can I show it's the same?

2) I also don't get if there's an equivalence between commutative Hopf algebras and Groups and to which extent (does the group have to be finite or it can be a Lie Group)?

3) how this is related if it is with the definition of Quantum Groups as non-commutative nor co-commutative Hopf Algebras?

Please if someone can do a systematic and intuitive argument for 1 and for 2 which I suppose to be strictly related. Thank you

• Remember that the group algebra is a cocommutative but not in general a commutative Hopf algebra. So the commutative one is the dual of the group algebra, except you need the group to be finite for this to give something of the same size. – Tobias Kildetoft Mar 5 '18 at 9:55
• In general, a commutative Hopf algebra is the same as an affine group scheme. – Tobias Kildetoft Mar 5 '18 at 9:59
• In addition, the study of quantum groups is not the study of group schemes. Typical quantum groups arising in physics are the so-called Drinfeld-Jimbo algebras. They are not commutative nor co-commutative. There is no definition of a quantum group in general though. Though algebraists and functional analysts have a strong opinion on what a quantum group should be. – Mathematician 42 Mar 5 '18 at 10:02
• if possible could you clarify the problems express in point 1? I don't get the relation if any with this definition of Group Hopf Algebra and the other one – Dac0 Mar 5 '18 at 10:05
• For finite groups, the two are duals of each other. – Tobias Kildetoft Mar 5 '18 at 10:19

Suppose that $G$ is a finite group. We can build the usual hopf algebra associated to the vector space $\mathbb{C}[G]$, with the usual group algebra structure $$\eta: 1 \to e, \quad m: g \otimes h\mapsto gh$$ the coalgebra structure $$\varepsilon: g \mapsto 1, \quad \Delta: g \mapsto g \otimes g$$ where all of the maps above are defined on the basis of group elements.
Now, we can take the dual Hopf algebra $\mathbb{C}[G]^\vee$, which as a vector space are the functions on $G$: $\mathbb{C}[G]^\vee = \{f: G \to \mathbb{C}\}$, and has new operations being the transpose of all the old operations. Now check what happens to all of the Hopf algebra structure. The unit $\eta$ dualises to the new counit $\eta^\vee: \mathbb{C}[G]^\vee \to \mathbb{C}^\vee \cong \mathbb{C}$, defined by $\eta^\vee(f) = (f \circ \eta)(1)$, which on a function $f$ will be $\eta^\vee(f) = f(\eta(1)) = f(e)$. So the new counit maps a function to its value over the identity.
Similarly, you can dualise the other maps, and see what happens. Should should find that the new coproduct $m^\vee$ is a kind of convolution, taking an indicator function $f_g$ to the element $\sum_{xy = g} f_x \otimes f_y$, the new unit $\varepsilon^\vee$ takes $1$ to the constant function on $G$ with value $1$, and the new multiplication $\Delta^\vee$ is pointwise multiplication of functions.


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# Reply To: I have symptoms, but I was a high achiever

Home Welcome to the ADDitude Forums For Adults Symptoms, Diagnosis & Beyond I have symptoms, but I was a high achiever Reply To: I have symptoms, but I was a high achiever

#197088
lanusp
Participant

Thank you so much for your answers! Whatever the root of my difficulties is, you made me feel less alone.
I really needed some clarification about the high achieving thing. I will not try to force an undue diagnosis in myself, if that is the case; but I’ll make sure that, in case ADHD gets excluded, it is for a solid reason.

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# Browse Abstracts & Presentations - 2021 International Symposium on Molecular Spectroscopy by Author "Oka, Takeshi"

• (International Symposium on Molecular Spectroscopy, 2021-06-23)
Based on the the Coriolis interaction conceived by Teller and Tisza (1932), the fundamental Wilson-Howard Hamiltonian (WHH), formulated in 1936\footnote{Wilson, Jr., E. B. \& Howard, J. B. 1936 J. Chem. Phys. 4, 260}, ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-22)
The giant molecular cloud Sgr B2 in the $\sim{150}$ pc radius Central Molecular Zone (CMZ) of the Galactic center has played a central role in molecular radio astronomical observations. In spite of its large distance, over ...

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# 3 Predict the product of the following compound:Cl2

###### Question:

3 Predict the product of the following compound: Cl2

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# What is the answer of 2 1/3 + 3 1/3=

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What is the answer of 2 1/3 + 3 1/3=

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Finite-part integrals (Hadamard) integrals with Mathematica

1. Dec 6, 2006

traianus

Hello guys,
I have to calculate an integral defined in the Hadamard sense with MATHEMATICA, but I only found the option for the integrals defined in the CAUCHY sense.
Just to be clear, suppose to have the integral with limits -1 and +1 of (1/x^2). This integral does not exist in the usual sense, but in the Hadamard sense its value is -2. Now, how do you calculate (numerically) the Hadamard integrals if the functions are complicated using MATHEMATICA?

Last edited: Dec 6, 2006
2. Dec 14, 2006

traianus

So??? Nobody knows Hadamard finite-part integrals? Is it so difficult for everybody?

3. Dec 17, 2006

The concept of "Hadamard fnite integral" seems very curious to me could oyo provide a link where to find some info?.. perhaps you should use zeta regularization when calculating your integrals so:

$$\sum_{n=1}^{\infty}n^{s}=\zeta (-s)$$

4. Dec 19, 2006

traianus

Hello Karlisbad, I know a very good article: "Numerical evaluation of ypersingular integrals" by Giovanni Monegato, Journal of Computational and Applied Mathematics 50 (1994) 9-31.

I do not know how the zeta regularization works. Can you please explain? Thank you!

5. Dec 25, 2006

A good intro for undergraduates can be found at ..

http://math.ucr.edu/home/baez/twf_ascii/week126

Based on an Euler converntion about divergent series..

resumming let be the divergent series..

$$1+2^{s}+3^{s}+...................... \rightarrow \zeta (-s)$$ (1)

of course if s<0 then the series above makes sense (no pole at s=1) and is just the Riemann zeta function.. the algorithm of "Zeta regularization" makes some kind of analytic continuation (i'm not mathematician so i can't give a rigorous formulation by the way really hate math rigour.. ) and then the sum of the series (divergent) can be obtained (¿¿cheating??...if you believe this is cheating you should study renormalization in physics) as the negative values of the Riemann zeta function, for example if s=1 then..

$$1+2+3+......................=-1/12$$

if s is an even number then the "sum" is taken to be 0.

further info: "H.G Hardy divergent series" (you have it on E-mule but you need to download an .djvu viewer) :tongue2:

6. Dec 26, 2006

Gib Z

$$1+2+3+......................=-1/12$$

That is not true, you need the Ramanujan Summation Operator. It is complete nonsense without that.

7. Dec 27, 2006

I had a curious idea..let be a function f(x) with a singularity at x=1 then we define the function:

$$f(x)=f*(x)+\delta(x-1)$$ and by definition $$f*(1)=0$$

$$f(x)=f*(x)$$ for all x except x=1

as you can see both function diverge at x=1 however using this definiton:

$$\int_{0}^{2}dxf(x)=\int_{0}^{2}dx(f*(x)+\delta(x-1))=1+\int_{0}^{2}dxf*(x)$$ which is now finite...

8. Dec 28, 2006

PerennialII

9. Dec 28, 2006

Hurkyl

Staff Emeritus
It's only finite if f and $\delta$ have a singularity of the same order, and whose coefficients are additive inverses.

It's more typical to use analytically defined functions (like 1/x) to subtract off singularities of functions.

10. Dec 28, 2006

matt grime

Sigh. Jose might hate the rigour of mathematics, but without it you just get apparent nonsense like this thread. (it should be s<-1, for instance, in post 5, not s<0), not to mention setting out in a confusing manner material which makes it appear that the sum 1+2+3+... actually does exist in the normal sense of summation. Still, never mind, eh? Is * supposed to be convolution? Those aren't functions, by the way. Why is that integral of that 'function' 'now finite'? What if I let f be some truly horrendous function with an essential singularity (infinitely many negative powers in the Laurent expansion)?

If it helps, we don't object to 'non-rigorous' stuff at all; it just isn't maths in the rigorous formal proof sense. We do object to ill defined terms, confusing notation, indecipherable descriptions, and things that are so wrong it is impossible to know where to start to correct them. There is a difference between being rigorous and doing something clearly. No physicist would accept what you've written above as acceptable presentation either. It's nothing to do with mathematicians.

Last edited: Dec 28, 2006
11. Dec 29, 2006

- The problem with rigour is as perhaps user "eljose" has pointed before in his post is that you can put 'obstacles' to science, the same happened to infinitesimals $$dx$$ , the Dirac delta function $$\delta (x)$$ or "Feynmann (??) Path integrals" in QM or QFT (Field theory) to quantizy everthing..which can't even be defined (rigorously) in terms of Riemann sums¡¡ (sigh), i read a paper by A.Connes about "renormalization and Hopf Algebras" of course the paper were so much rigorous and math-defined that you coud hardly "extract" some info about it regarding renormalization, whereas there're lots and lots of math introduction without rigour in physics, conjectures or hypothesis that can be understood by almos everybody..that's why i critizy rigour..i don't hate mathematician i only say that sometimes they are a bit of "pedantry".

12. Dec 29, 2006

matt grime

Who's putting obstacles there? Physicisist don't require the same level rigour that mathematicians do, so what? Point out one example where rigour held back the development of Feynmann path integrals, please.

You aren't trained to read mathematics research papers; they aren't written with people with undergrad physics degrees in mind. They are written by professionals for professionals.

Unrigorous work often helps, and mathematicians use it all the time. Your problem is not that mathematicians are pedantic about rigor. Your problems with getting your work read have been catalogued at length many times; you have had far more peer review than anyone else I can think of; you have yet to address more than one comment (to include proper references, occasionally).

Pointing out that saying 'let f be a function with singularity at 1' is useless is not pedantry (that would be pointing out the grammatical errors, which is not helpful). There are infinitely many such functions. Which one? It is important because what kind of singularity you have affects what follows in your post, making it just plain wrong. And no, you can't get away with saying 'oh, choose one that makes it work'. You don't appear to make any effort to avoid mistakes, so we have no idea what is 'wrong but just a typo' and what is 'just plain wrong'.

Last edited: Dec 29, 2006


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# Universal reference frame based on the speed of light/special relativity

Everything I've read has said that there is no universal static reference frame, but based on my understanding of special relativity I don't understand why it can't be determined.

Here is an experiment I can think of that would seem to determine "universal static", what am I missing? Assuming 1 dimensional space for a second, our reference frame (a spaceship) is moving and an indeterminate speed $V_0$. We take two probes with atomic clocks and send one in each direction at some fraction of the speed of light ($V_0+V_p$ and $V_0-V_p$). After a period of time, the probes return at a slower speed than they left (taking advantage of the fact that time dilation effects increase exponentially with velocity). Upon returning, the clock in the probe sent in the positive direction ($V_0+V_p$) should show less time past than the clock on our spaceship, and the probe sent in the opposite direction ($V_0-V_p$) should show that more time has past than the clock on our spaceship, this tells us which direction we are moving with respect to "universal static", and with repeated experiments we can determine our exact velocity relative to it.

• How can you send probes out at $V_0+V_p$ and $V_0-V_p$ if you don't know what $V_0$ is? – Brian Moths Nov 30 '17 at 20:21
• You don't know what V0 is, that's what you are determining experimentally, but if you fire a probe in both directions, their speeds will be V0+Vp and V0−Vp – Steven Pick Nov 30 '17 at 20:38
• No, their speeds will only be $V_0+V_p$ and $V_0-V_p$ if you fire them with the appropriate velocities, which can't be determined unless you know $V_0$, so this experiment is impossible to conduct in the first place. – Brian Moths Nov 30 '17 at 20:41
• Vp is known, V0 is unknown, so the resulting velocities with respect to the to be determined universal reference frame is V0+Vp and V0−Vp. What is impossible about that? – Steven Pick Nov 30 '17 at 20:51
• That's not how velocities add. Also, time dilation is not exponential in velocity. You seem to assume that an absolute reference frame exist, and time dilation is wrt to this frame. This is not how relativity works. – fqq Nov 30 '17 at 21:10

It is hardly possible to detect “universal” frame this way. Non-inertial clock will dilate at the same magnitude when it meets inertial clock again despite of direction of its motion.

Your question looks like the twin paradox in the preferred frame theory (Lorentz Ether theory). It is often said that this theory is empirically equivalent to the SR. That means, this theory predicts the same outcome of your thought experiment (or any other experiment) as Special Relativity. In Lorentz theory, moving observer lives by his own “local time”, which is different from “absolute, universal” or “ether” time.

https://en.wikipedia.org/wiki/Lorentz_ether_theory

It should be noted, that there is no "paradox" related to the ether theory. In the ether theory, the paradox that has arisen in the depths of special relativity is resolved by means of elementary algebraic methods.

We can simulate the both cases that you propose in the framework of the ether theory, so as to see whether it is possible to detect absolute frame.

1) Let’s consider what would happen if one of two twins who are at rest in the ether at one point, flies at speed $v$ to a distant point and then after a while returns to twin $A$ remaining at rest.

If for the twin flying in the ether his “local time” characterizing the rate of physical processes in his body and the pace of the movement of his clock on both segments of his flight (there and back) slows down due to interaction with the ether, then the lapse of his “local time” will be $1/\sqrt {1-v^2/c^2}$ times less than for the twin at rest in the ether, and the “travelling” twin will get less “old”. The turn of the travelling twin, provided it is virtually instantaneous, has no practical effect on the ratio of times of both twins.

2) Now let's calculate, what will happen if the two twins are flying side by side in the ether at speed $v$ – with their “local time” passing slower – then one of them stops, staying at rest in the ether for some time, then catching up with the travelling twin. I think it is equivalent to your probe, that was sent in “opposite direction”. The twin who continued his flight in the ether with no information about the fact of his motion in the ether perceives this maneuver of his brother as a round trip to a distant point.

An obvious answer is that, since according to the ether theory after the twin’s stop in the ether his time will pass faster than the “local time” of his twin brother who continues his flight, and then when the twin stopping in the ether after some time catches up with the missing brother, he will age more than the latter. The “local” time of the twin catching up with his flying brother will actually flow slower than for the flying brother. This is due to the faster speed of the twin catching up with his brother. As a result, the brother making a stop in the ether will age not more, but less than his twin brother who has not interrupted his flight.

Let us demonstrate that if the proper times of the motion there and back of the non-inertial twin relative to the inertial twin are equal, then for the non-inertial twin it will take $1/\sqrt {1-v^2/c^2}$ times less time than for the moving inertial twin, and the non-inertial twin will age less.

Let at the time of stop of one of the twins in the ether the clocks of the parting twins show zeros. Suppose that after making a stop for some time, the twin who has lagged behind, at the moment $t_1$ of the ether time when his clock (because of the stop) was showing this time, left at speed $u$, such that $v<u<c$, following his brother flying away from him. The distance between the twins at the start of the twin who has left behind is equal to $vt_1$. Setting out, the twin left behind will catch up with the twin flying at a constant speed $v$ at the point in time $t_2$, having spent the time equal to $vt_1/(u-v)$. During this period, by the clock of the twin following the flying away brother at speed u, there will be a lapse of proper time, which is $1/\sqrt {1-v^2/c^2}$ times less than the ether time and equals $vt_1\sqrt{1-(u/c)^2}/(u-v)$. Let us assume the velocity $u$ such that the proper time $t’_2-t’_1$ of the catching up twin is numerically equal to the time $t_1$ of his stay at rest relative to the ether, i.e. $t’_2-t’_1=t_1$ or

$$t_1=vt_1\sqrt {1-(u/c)^2/(u-v)} (1)$$

This equation meets the condition under which the twin spends the same proper time on a trip to a distant point and back. By elementary transformations of the equation (1) we can obtain the value of velocity $u$, which is equal to $\frac {2v}{1+(v/c)^2}$ . Substituting this value in the expression for the time $vt_1/(u-v)$ required for the return of the twin, and summing the time $vt_1/(u-v)$ and the time $t_1$, we obtain the ether time spent by the lagging behind twin on the stop and return to the flying twin. This time is equal to $2t_1/(1-v^2/c^2)$. Since the clock of the inertial twin flying at a speed $v$ go $1/\sqrt {1-v^2/c^2}$ times slower than the clock at rest in the ether, the flying twin will determine the time spent by the lagging behind twin on the stop and return to the flying twin as a quantity meeting the equality:

$$t’_2=2t_1/\sqrt {1-(v/c)^2}$$

Since the time elapsed for the non-inertial twin by the moment of his return is numerically equal to $2t_1$, and the time of the inertial twin is numerically equal to $2t_1/\sqrt{1-(v/c)^2}$, then the lapse of time for the non-inertial twin is $1/\sqrt {1-v^2/c^2}$ times shorter, and he has aged less than the inertial twin has.

This way we can see, that non–inertial twin (or clock) will show $1/\sqrt {1-v^2/c^2}$ less time than inertial one, despite of direction of its motion in ether. We get absolutely the same result as in the Special Relativity, but resolution of the paradox is very simple.

https://arxiv.org/abs/1201.1828


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# Deriving the Lorentz force from velocity dependent potential

We can achieve a simplified version of the Lorentz force by $$F=q\bigg[-\nabla(\phi-\mathbf{A}\cdot\mathbf{v})-\frac{d\mathbf{A}}{dt}\bigg],$$ where $\mathbf{A}$ is the magnetic vector potential and the scalar $\phi$ the electrostatic potential.

How is this derivable from a velocity-dependent potential $$U=q\phi-q\mathbf{A}\cdot\mathbf{v}?$$

I fail to see how the total derivative of $\mathbf{A}$ can be disposed of and the signs partially reversed. I'm obviously missing something.

Hints: Use

$$\frac{\partial U}{\partial {\bf v}}= -q{\bf A},$$

and the defining property of a velocity-dependent potential:

$${\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}.$$

See e.g. Herbert Goldstein, Classical Mechanics and Wikipedia for more details.

• Thank you Qmechanic, for your gracious hint, better than an answer. :) Actually the first relation is the one I missed. – Valentina Sep 14 '13 at 16:15

Velocity-dependent potential is not strictly a potential. Lagrange equations say that

$$\frac{d}{dt}\frac{\partial L}{\partial \bf{v}} = \frac{\partial L}{\partial \bf{r}}$$

You have $L = L_0 - U$ where $L_0$ corresponds to free motion (e.g. $L_0 = mv^2/2$ or $L_0 = -mc^2\sqrt{1-(v/c)^2}$).

If $U$ does not contain $\bf{v}$ you have ${\partial L}/{\partial \bf{v}} = \bf{p}$ and so $\dot{\bf{p}} = -\nabla U$.

In this case, however $U$ contains $\bf{v}$ so on the you have

$$\frac{d}{dt}\left({\bf{p}} + q\bf{A}\right) = -\nabla U$$


text:
Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# Why is the Lewis structure of ozone important?

Why? Because it's a simple predictor of molecular shape.

The ozone molecule is O_3, and each O centre contributes 6 electrons to the valence shell.

A reasonable Lewis structure would be: O=O^(+)-O^(-). Because, around the central oxygen, there are 5 electrons (2 from the double bond, 1 from the single bond, and 2 from the lone pair), we assign this centre a positive charge, and of course we can assign each terminal oxygen a negative charge alternately by resonance.

Given the Lewis structure we predict by VSEPR a bent molecule with /_O-O-O <= 120""^@. What do we find experimentally? A bent molecule with intermediate O-O bonds; /_O-O-O = 116.8""^@.

Thus, by simply knowing how to draw a Lewis structure, counting the electrons, and using VSEPR, we have predicted the structure of a gaseous molecule, which we can't see, but we can smell. I think that is pretty clever given the (short!) time we spent on the problem.


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Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp015m60qv67x
 Title: Refuge: Syrian Refugee Incorporation and the Production of New Minorities Authors: Gowayed, Hebatalla Advisors: Zelizer, VivianaDuneier, Mitchell Contributors: Sociology Department Keywords: genderimmigrationInternationalpolicyracerefugees Subjects: Sociology Issue Date: 2018 Publisher: Princeton, NJ : Princeton University Abstract: Drawing on two years of ethnographic observations and in-depth interviews, this dissertation examines the experiences of Syrian refugees. I focus on the lives of 18 Syrian families resettled in the United States. In the final chapter, I compare their experiences to Syrians resettled in Canada, who sought asylum in Germany, and who were resettled or sought asylum in Italy. I show that the Syrian refugees in this study, in all four countries, experienced three sets of crucial losses in their displacement: (1) human capital, (2) social relations—in terms of patterns of family life as well as relations outside of the household—and (3) a sense of security—or sense of legal and interpersonal safety. In light of these losses, I examine what I call refugees’ process of recuperation—or how these men and women rebuilt these resources within the opportunities and constraints of a resettlement or asylum system. Through this focus on process, and through the cross-national comparison, I contribute to existing literature on refugees by centering the role of the state in facilitating or hindering refugees’ production of these resources. I argue that refugees’ ability to produce human capital, their management of social relations and their sense of security are the product of the opportunities and constraints imposed by immigration laws and social service systems that comprise resettlement and asylum policy. These systems reflect policy-makers’ understanding of their humanitarian obligation to people outside of their borders, and those within them who are in need of social services. Because the resources I examine are crucial determinants for immigration outcomes, I argue that the state shapes the socioeconomic mobility of these new newcomer refugees by facilitating or blocking the recuperation of these resources, and thus shapes the kinds of racial and economic minorities they can become. URI: http://arks.princeton.edu/ark:/88435/dsp015m60qv67x Alternate format: The Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog: catalog.princeton.edu Type of Material: Academic dissertations (Ph.D.) Language: en Appears in Collections: Sociology

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# Math Help - Terminal arm

1. ## Terminal arm

The point (-2,3) lies on the terminal arm of position angle

I dont know how to solve this question, I had one where it was simular but gave an equation like y=cotx and I couldnt solve that either.

I know sin=3 and cos=-2 but that is it.

2. Originally Posted by brentwoodbc
The point (-2,3) lies on the terminal arm of position angle

I dont know how to solve this question, I had one where it was simular but gave an equation like y=cotx and I couldnt solve that either.
I still don't know your question. If you're trying to find $\cot \theta$, then the answer would be $-\frac{2}{3}$, because in the cartesian plane, with point (x, y) on the terminal ray, $\cot \theta = \frac{x}{y}$.

I know sin=3 and cos=-2 but that is it.
First, watch your notation. You can't write a trig function without an angle. You must write something like
$\sin \theta = 3$ or $\cos \theta = -2$.
Second, neither value is possible. Sine or cosine of any angle will be between -1 and 1 inclusive. If you're using the point (x, y) to figure out sine and cosine, you need these definitions:
$\sin \theta = \frac{y}{r}$ and
$\cos \theta = \frac{x}{r}$.

In other words, you forgot to divide each by r. (r is easy to find, it's just $\sqrt{x^2 + y^2}$).

01

3. Originally Posted by brentwoodbc
The point (-2,3) lies on the terminal arm of position angle


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# Is there any rule for nonexistent limits?

I have 2 questions when dealing with nonexistent limits:

Q1: If $\lim\limits_{x\to a} f(x)=L\neq 0$ and $\lim\limits_{x\to a}g(x)$ does not exist, then is it correct to conclude that $\lim\limits_{x\to a} f(x)g(x)$ does not exist ? I feel the statement true but fail to prove. I tried using Cauchy test for divergence but it did not work. I appreciate your help.

Q2: Suppose $\lim\limits_{x\to a} f(x)=b$ and $\lim\limits_{y\to b}g(y)$ does not exist. I could find examples in which $\lim\limits_{x\to a} g(f(x))$ exists or does not exist. My question is when does it diverge ? What are conditions needed to imply that $\lim\limits_{x\to a} g(f(x))$ does not exist ?

Q1: Yes, this is valid. To see this, write $h(x)=f(x)g(x)$. If $\lim h(x)$ existed, then since $\lim f(x)$ exists and is not $0$, this would imply $\lim h(x)/f(x)$ exists. But $h(x)/f(x)=g(x)$, and we know $\lim g(x)$ does not exist! Thus $\lim h(x)$ cannot exist.
Q2: You can conclude that $\lim g(f(x))$ does not exist if $f$ is open at $a$. This means that for any open interval $(d,e)$ containing $a$, the image $f((d,e))$ contains an open interval around $b$. Roughly speaking, this says that $f(x)$ takes every possible value near $b$ as $x$ approaches $a$. One more concrete condition that implies $f$ is open at $a$ is that $f$ is continuous on an open interval containing $a$ and has neither a local maximum nor a local minimum at $a$.
• The assumption $f$ is open at $a$ is needed to ensure that there exists a sequence $\left\{y_n=f(x_n)\right\} \in A \ \left\{b\right\}$ approaching $b$ when $x\to a$. Is that right ? – Bungbu Nov 19 '15 at 4:39
• Well, not just that there exists a sequence, but that for any sequence $y_n$ approaching $b$, you can choose a sequence $x_n$ approaching $a$ such that $y_n=f(x_n)$ (at least, for all sufficiently large $n$). – Eric Wofsey Nov 19 '15 at 4:40


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# Loop-cut won't go all the way around

Alright so I've seen some postings about loop cuts and done some research. I am very new to blender and modeling overall. I am trying to loop cut this concrete barrier. Now I'm following a tutorial and kind of did my own thing in some areas. I cleaned it up following some instructions in another answer here on stack exchange.

I can't for the life of me figure out why I can loop cut around the whole bottom:

...but not the whole middle:

I see this marked as answered by another question. That question refers to ngons and inside faces. Now once again I'm totally new to this. I looked on the inside. I don't believe I have inside faces or any ngons by the terms I've read online in my research.

Here is a wireframe picture and the inside view:

Maybe I'm missing something the answers on other posts are saying. If so and you know what it is I'd be very appreciative of what it is.

• See the first answer on the other post. You have a ngon on the bottom of your mesh. – iKlsR Oct 17 '14 at 4:21
• Just tried removing that ngon because I realized it. It has no effect on the loop cut for the area I'm trying to work on. Any ideas? – CrazyGrunt Oct 17 '14 at 4:38
• I initially marked as a duplicate of this since the answer is on the other question but since you are new, I think this could get an answer. – iKlsR Oct 17 '14 at 4:40
• – gandalf3 Oct 17 '14 at 6:19
• – gandalf3 Aug 27 '15 at 8:57

• It depends, in this case, it was hidden as it was right on top of the other, I noticed because the edge cut stopped there instead of the large bottom face. Additionally, you can sometimes use W > Remove Doubles to check for and clear these. – iKlsR Oct 17 '14 at 5:09


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The electron configuration for magnesium is 1s^2 2s^2 2p^6 3s^2. What is the distribution of electrons in the electron shells of a magnesium atom?

$2 : 8 : 2$


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# Difference between revisions of "2007 AMC 12A Problems"

## Problem 1

One ticket to a show costs $20$ at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?

$\mathrm{(A)}\ 2\qquad \mathrm{(B)}\ 5\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 15\qquad \mathrm{(E)}\ 20$

## Problem 2

An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. It is filled with water to a height of 40 cm. A brick with a rectangular base that measures 40 cm by 20 cm and a height of 10 cm is placed in the aquarium. By how many centimeters does the water rise?

$\mathrm{(A)}\ 0.5\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 1.5\qquad \mathrm{(D)}\ 2\qquad \mathrm{(E)}\ 2.5$

## Problem 3

The larger of two consecutive odd integers is three times the smaller. What is their sum?

$\mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \mathrm{(E)}\ 20$

## Problem 4

Kate rode her bicycle for 30 minutes at a speed of 16 mph, then walked for 90 minutes at a speed of 4 mph. What was her overall average speed in miles per hour?

$\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$

## Problem 5

Last year Mr. Jon Q. Public received an inheritance. He paid $20\%$ in federal taxes on the inheritance, and paid $10\%$ of what he had left in state taxes. He paid a total of 10500 for both taxes. How many dollars was his inheritance?

$(\mathrm {A})30000$ $(\mathrm {B})32500$ $(\mathrm {C})35000$ $(\mathrm {D})37500$ $(\mathrm {E})40000$

## Problem 6

Triangles $ABC$ and $ADC$ are isosceles with $AB=BC$ and $AD=DC$. Point $D$ is inside triangle $ABC$, angle $ABC$ measures 40 degrees, and angle $ADC$ measures 140 degrees. What is the degree measure of angle $BAD$?

$\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60$

## Problem 7

Let $a, b, c, d$, and $e$ be five consecutive terms in an arithmetic sequence, and suppose that $a+b+c+d+e=30$. Which of $a, b, c, d,$ or $e$ can be found?

$\textrm{(A)} \ a\qquad \textrm{(B)}\ b\qquad \textrm{(C)}\ c\qquad \textrm{(D)}\ d\qquad \textrm{(E)}\ e$

## Problem 8

A star-polygon is drawn on a clock face by drawing a chord from each number to the fifth number counted clockwise from that number. That is, chords are drawn from 12 to 5, from 5 to 10, from 10 to 3, and so on, ending back at 12. What is the degree measure of the angle at each vertex in the star polygon?

$\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 24\qquad \mathrm{(C)}\ 30\qquad \mathrm{(D)}\ 36\qquad \mathrm{(E)}\ 60$

## Problem 9

Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?

$\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78$

## Problem 10

A triangle with side lengths in the ratio $3 : 4 : 5$ is inscribed in a circle with radius 3. What s the area of the triangle?

$\mathrm{(A)}\ 8.64\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 5\pi\qquad \mathrm{(D)}\ 17.28\qquad \mathrm{(E)}\ 18$


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# derivative of inverse matrix

###### Theorem 1.

Suppose $A$ is a square matrix depending on a real parameter $t$ taking values in an open set $I\subseteq\mathbbmss{R}$. Further, suppose all component functions in $A$ are differentiable, and $A(t)$ is invertible for all $t$. Then, in $I$, we have

 $\frac{dA^{-1}}{dt}=-A^{-1}\frac{dA}{dt}A^{-1},$

where $\frac{d}{dt}$ is the derivative.

###### Proof.

Suppose $a_{ij}(t)$ are the component functions for $A$, and $a^{jk}(t)$ are component functions for $A^{-1}(t)$. Then for each $t$ we have

 $\sum_{j=1}^{n}a_{ij}(t)a^{jk}(t)=\delta_{i}^{k}$

where $n$ is the order of $A$, and $\delta_{i}^{k}$ is the Kronecker delta symbol. Hence

 $\sum_{j=1}^{n}\frac{da_{ij}}{dt}a^{jk}+a_{ij}\frac{da^{jk}}{dt}=0,$

that is,

 $\frac{dA}{dt}A^{-1}=-A\frac{dA^{-1}}{dt}$

from which the claim follows. ∎

Title derivative of inverse matrix DerivativeOfInverseMatrix 2013-03-22 14:43:52 2013-03-22 14:43:52 matte (1858) matte (1858) 7 matte (1858) Theorem msc 15-01


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• # Abusing the Linker to Minimize Compilation Time

General and Gameplay Programming

## Abstract

In this article, I will describe a technique that reduces coupling between an object that provides access to many types of objects ("the provider") and its users by moving compile-time dependencies to link-time. In doing so, we can reduce the amount of unnecessary compiling of the provider's dependencies whenever the provider is changed. I will then explore the benefits and costs to such a design.

## The Problem

Typically in a game, a provider object is needed to expose a variety of services and objects to many parts of the game, much like a heart pumps blood throughout the body. This provider object is a sort of "context object", which is setup with the current state of the game and exposes other useful objects. Such a class could look something like listing 1, and an example of use could look something like listing 2.  // // Listing 1: // ServiceContext.h #pragma once // Dependencies class Game; class World; class RenderService; class ResourceService; class PathfindingService; class PhysicsService; class LogService; // ServiceContext // Provides access to a variety of objects class ServiceContext { public: Game* const game; World* const world; RenderService* const render; PathfindingService* const path; PhysicsService* const physics; AudioService* const audio; LogService* const log; };  Listing 1: The definition of a sample context object  // // Listing 2: // Foobar.cpp #include "Foobar.h" #include "ServiceContext.h" #include "PathService.h" #include "LogService.h" void Foobar::frobnicate( ServiceContext& ctx ) { if ( !condition() ) return; current_path = ctx.path->evaluate(position, target->position); if ( !current_path ) ctx->log("Warning","No path found!"); }  Listing 2: An example usage of the sample context object The ServiceContext is the blood of the program, and many objects depend on it. If a new service is added to ServiceContext or ServiceContext is changed in any way, then all of its dependents will be recompiled, regardless if the dependent uses the new service. See figure 1.
Figure 1: Recompilations needed when adding a service to the provider object
To reduce these unnecessary recompilations, we can use (abuse) the linker to hide the dependencies.

## The Solution

We can hide the dependencies by moving compile-time dependencies to link-time dependencies. With templates, we can write a generic get function and supply specialized definitions in its translation unit.  // // Listing 3: // ServiceContext.h #pragma once // Dependencies struct ServiceContextImpl; // ServiceContext // Provides access to a variety of objects class ServiceContext { public: // Constructors ServiceContext( ServiceContextImpl& p ); public: // Methods template T* get() const; private: // Members ServiceContextImpl& impl; }; // // ServiceContextImpl.h #pragma once // Dependencies class Game; class World; class RenderService; class ResourceService; class PathfindingService; class PhysicsService; class LogService; // ServiceContextImpl // Exposes the objects to ServiceContext // Be sure to update ServiceContext.cpp whenever this definition changes! struct ServiceContextImpl { Game* const game; World* const world; RenderService* const render; PathfindingService* const path; PhysicsService* const physics; AudioService* const audio; LogService* const log; };  Listing 3: The declarations of the two new classes  // // Listing 4: // ServiceContext.cpp #include "ServiceContext.h" #include "ServiceContextImpl.h" ServiceContext::ServiceContext( ServiceContextImpl& p ) : impl(p) { } // Expose impl by providing the specializations for ServiceContext::get template<> Game* ServiceContext::get() { return impl.game; } // ... or use a macro #define SERVICECONTEXT_GET( type, name ) \ template<> \ type* ServiceContext::get() const { \ return impl.name; \ } SERVICECONTEXT_GET( World, world ); SERVICECONTEXT_GET( RenderService, render ); SERVICECONTEXT_GET( PathfindingService, path ); SERVICECONTEXT_GET( PhysicsService, physics ); SERVICECONTEXT_GET( AudioService, audio ); SERVICECONTEXT_GET( LogService, log ); Listing 4: The new ServiceContext definition In listing 3, we have delegated the volatile definition of ServiceContext to a new class, ServiceContextImpl. In addition, we now have a generic get member function which can generate member function declarations for every type of service we wish to provide. In listing 4, we provide the get definitions for every member of ServiceContextImpl. The definitions are provided to the linker at link-time, which are then linked to the modules that use the ServiceContext. // // Listing 5: // Foobar.cpp #pragma once #include "Foobar.h" #include "ServiceContext.h" #include "PathService.h" #include "LogService.h" void Foobar::frobnicate( ServiceContext& ctx ) { if ( !condition() ) return; current_path = ctx.get()->evaluate(position, target->position); if ( !current_path ) ctx.get()->log("Warning","No path found!"); } Listing 5: The Foobar implementation using the new ServiceContext With this design, ServiceContext can remain unchanged and all changes to its implementation are only known to those objects that setup the ServiceContext object. See figure 2.
Figure 2: Adding new services to ServiceContextImpl now has minimal impact on ServiceContext's dependants
When a new service is added, Game and ServiceContextImpl are recompiled into new modules, and the linker relinks dependencies with the new definitions. If all went well, this relinking should cost less than recompiling each dependency.

## The Caveats

There are a few considerations to make before using this solution:
1. The solution hinges on the linker's support for "whole program optimization" and can inline the ServiceContext's get definitions at link-time. If this optimization is supported, then there is no additional cost to using this solution over the traditional approach. MSVC and GCC have support for this optimization.
2. It is assumed that ServiceContext is changed often during development, though usually during early development. It can be argued that such a complicated system is not needed after a few iterations of ServiceContext.
3. It is assumed that the compiling time greatly outweighs linking time. This solution may not be appropriate for larger projects.
4. The solution favors cleverness over readability. There is a increase in complexity with such a solution, and it could be argued that the complexity is not worth the marginal savings in compiling time. This solution may not be appropriate if the project has multiple developers.
5. The solution does not offer any advantage in Unity builds.

## Conclusion

While this solution does reduce unnecessary recompilations, it does add complexity to the project. Depending on the size of the project, the solution should grant a small to medium sized project with decreased compilation time.

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text:
896 views
Is the following grammar LL(1) ?

S→ aABbCD | ϵ

A→ ASd |  ϵ

B→ Sac| hC | ϵ

C→ Sf |Cg

D→ aBD | ϵ

Draw the LL(1) parsing table for the given grammar.

If our intention is to only check if this grammar is LL(1) or not then we dont need to do any of these first ,follow calculation then creating LL(1) parsing table and check if one cell contain multiple entries or not .

By looking at the grammar we can conclude that it is not LL(1) as it is left recursive .

The left recursive productions are

1)A→ ASd

2)C→ Cg.

now we can eliminate the left recursion and try to check for LL(1) or not but that is a topic for another question  we have to consider the grammar which we have given.

First we need to compute the first and follow set for the grammar.

 Non terminal First Follow $S$ $\left \{ a,\epsilon \right \}$ $\left \{ a,f,d,dollor \right \}$ $A$ $\left \{ a,d,\epsilon \right \}$ $\left \{ a,h,b,d \right \}$ $B$ $\left \{ a,h,\epsilon \right \}$ $\left \{ a,b,f,d,dollor \right \}$ $C$ $\left \{ a,f \right \}$ $\left \{ a,b,f,d,g,dollor \right \}$ $D$ $\left \{ a,\epsilon \right \}$ $\left \{ a,f,d,dollor \right \}$

Now we have to draw LL(1) parsing table:-

 $a$ $b$ $c$ $d$ $f$ $g$ $h$ $Dollor$ $S$ $S → aABbCD$ $S→ \epsilon$ $S->\epsilon$ $S->\epsilon$ $S->\epsilon$ $A$ $A-> ASd$  $A-> \epsilon$ $A-> \epsilon$ $A-> ASd$ $A-> \epsilon$ $A-> \epsilon$ $B$ $B-> Sac$ $B-> \epsilon$ $B-> \epsilon$ $B-> \epsilon$ $B-> \epsilon$ $B-> hC$ $B-> \epsilon$ $C$ $C-> Sf$ $C-> Sf$ $D$ $D-> aBD$ $D-> \epsilon$ $D-> \epsilon$ $D-> \epsilon$ $D-> \epsilon$

As we can see multiple entries in single cell then we can conclude it is not LL(1).

Left recursion present in grammar definitely not LL(1)

1
3,333 views


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# If KL-divergence's asymmetrical should I minimize KL(P||Q) or KL(Q||P)?

Suppose I am adjusting distribution $Q$ to make get a best fit for distribution $P$, should I minimize $KL(P||Q)$ or $KL(Q||P)$? What is the difference?

Related question:

Intuition on the Kullback-Leibler (KL) Divergence

• It is not a distance measure. It is not a metric. – Carl Sep 17 '17 at 4:06
• Can you clarify how your question differs from the linked thread? If it isn't what you want, please edit your question to state what you learned & what you still need to know. Then we can provide the information you need without just duplicating material elsewhere that already didn't help you. – gung Sep 18 '17 at 14:39

The KL divergence is not a distance this is why the alternative word divergence is used instead. If you want symmetry you can take the sum of $KL(Q||P)$ and $KL(P||Q)$ as mentioned in one of the answers in the linked post.
The intuition is that you do not know the true distribution $Q$ so you make an estimate or a guess of the true distribution $P$. The two may be in the same parametric family or they may not be similar at all. Therefore to understand some notion of how far your assigned probabilities (a view of the events) are from the true probabilities (how much the two perspectives diverge) you would take an expectation under your estimated probabilities. However, you would still like this divergence to be $0$ if you somehow specified the exact model, to ensure this you subtract that true value. Using the logarithm property $log(x/y) =log(x) - log(y)$ then gives you the KL-divergence equation.
You usually want $KL(Q||P)$. That's from $P$ to $Q$. I remember that it goes from right to left, just like the notation for conditional probabilities.
You want the expectation being taken with respect to the true distribution $P$. That way, sample averages can be assumed to converge to the true expectations, by the law of large numbers.


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# MathWiki

## WikiSandbox

Filed in: Main.WikiSandbox · Modified on : Fri, 01 Jun 12

Feel free to use this page to experiment with the Text Formatting Rules. Just click the "Edit Page" link at the bottom of the page.

For any real number $p > 0$, define $f_{n, p}:- \mathbb R^n \to \mathbb R$ by $$f_{n, p}(\bm{x})={|x_1^p+ \cdots+x_n^p|}^{\frac1p}=\|\bm{x}\|_p,$$ for $\bm{x }=(x_1,\dots,x_n)\in\mathbb R^n$.

$$\bm{x }=(x_1,x_2,\dots,x_n)\in\mathbb R^m\quad \hbox{all x_j\geq1}$$ $[n,k-2]$ (linear) code ${a^b}^c$

New test: $ab+bc+cd+d^e=1$.

Another test: $\lambda eadbe+bcd+cdx+dcd^euf=1-\int xyz\, dx + 2z$. abccdddd $n:= e/w/x/u.v$.


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# Preventing an output of a function from being assigned a value

Say I have a function

Example[n_Integer]:=n{x,y,z}

Example@4

{4 x, 4 y, 4 z}

Now, let us assign some values to x,y,z

x=7 y=7 z=8

And then we evaluate our function again

Example@4

{28, 28, 32}

How can I make it so that despite some values being assigned to the variables x,y,z the output of the function Example is only affected by its argument n? I have tried to use SetAttributes[Example,Protected] but that does not work. Could someone help?

• Can you provide more details of why you want to do this? I think you oversimplified your actual issue. – Carl Woll May 28 at 21:13
• @CarlWoll I produce a list of symbolic expression, for instance {-x,-y,x,y,x+y,y+z.} – amator2357 May 28 at 21:17
• So, something like: Example[n_Integer]:=(Clear[x,y,z]; n{x,y,z}) will work for you? – Carl Woll May 28 at 21:20
• @CarlWoll I produce a list of symbolic expressions, for instance l = {-x,-y,x,y,x+y,y+z}. I then write a function that assigns some expression to x and y, for example x={1,-1,0},y={0,1,-1} so then automatically the remaining symbols in l get assigned a value. Then I want to find a reflection of one of the expression in l in the hyperplane perpendicular to, say x. Once I do that, I'd get, for example, {-1,1,0}. Now, I want to know to which symbol in my initial l that value corresponds, if that makes sense. – amator2357 May 28 at 21:25
• This sounds very much like an XY problem. Can you maybe update your question to reflect what you are truly after? – Roman May 28 at 21:45

## 1 Answer

Just avoid setting x and y.

For example:

f2[l_, n_] := n l;

Clear[x, y, z]
masterl = {x, y, z};

Then you can substitute in values if you want:

xx = 3; yy = 4;
l2 = masterl /. {x -> xx, y -> yy}

(* {3, 4, z} *)

Or do some work with the original list:

f2[masterl, 5]

(* {5 x, 5 y, 5 z} *)


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# ideal gas — Engelska översättning - TechDico

kurs-assistent C. Haikarainen - PDF Gratis nedladdning

Various parts of it are also known as Charles' law, Boyle's law, Gay-Lussac's law, or Avogadro's law.. The formulation of this law played a crucial role in the development of the atomic theory (that is, the realization that matter really is made up of atoms), and in the development of thermodynamics. The Universal Gas Constant, Ru is independent of the particular gas and is the same for all "perfect" gases, and is included in of The Ideal Gas Law: p V = n Ru T   The Universal Gas Constant - Ru - in alternative Units · atm.cm3/(mol.K) : 82.057338 · atm.ft3/(lbmol.K) : 1.31443 · atm.ft3/(lbmol. R) : 0.73024 · atm.l/(mol.K ) :  Therefore, R can include energy units such as Joules or calories. Values for the gas constant R. Units. Value.

P - Pressure (atm). V- Volume (L). N - Moles (mol). R - Gas constant (0.08206) These molecules are non-interacting and obey the ideal gas law. a.

To reduce explosives are stored in a gas station, oil depot, or chemical plant, for example). which can be either C, U, J, E, B, A, I, R, Zor K, depending on your region).

## dict.cc gås English-Swedish Dictionary

N2O is P1V1 = n1R1T1. or  In Ideal Gas Law, there's gas constant R that we need to use in the formula PV= nRT. There are times when we are given different units and we need to do  7 Jul 2003 Answer. This is best answered by tracing the origins of the ideal gas law itself.

### Lag För Ideal Gas På Den Gröna Svart Tavlan Stock

Ordet "R" kan ha följande grammatiska funktioner: substantiv ideal gas law constant, Scientology. universal  Ordinary hydrogen gas consists of di-atomic molecules in which two H-atoms unite VALUE OF R In atmp.dm3/mole.k. Consider one mole of an ideal gas at S.T.P.

If further the Kraftverksteknik / JK. The first law x. W q At equal values of T: (dh / ds) p. = constant. For a perfect gas, h = C p. T,. (dh / ds) p. = constant. Solubility Equilibrium: Using a Solubility Constant (Ksp) in Calculations, Using Orbital Where r is the process strongest intermolecular force in h20=hydrogen bonding The ideal gas law is used to describe the behavior of ideal gases, but  RAPPORT CEA-R-6194 – B.R. Sehgal, P. Piluso, K. Trambauer, B. Adroguer, F. Fichot,.

1. T1. ) Henderson–Hasselbalch equation: pH = pKa + log.

X  The ideal gas constant is also known as the molar gas constant, the gas constant This constant is written as R, and is a constant of proportionality (constant one side of a proportional relationship to make them equal) for the ide Definition of the universal (R) gas constant frequently used in ideal gas entities in the Ideal gas equation, such as pressure, temperature, and volume. In the ideal gas law, R is the ideal gas universal constant and has a value of 8.314 4621 Joules/(Mol K). This constant quantifies the relationship between the   22 May 2014 The constant factor in the equation of state for ideal gases. The universal gas constant, also known as the molar or ideal gas constant, is. R*  14 Oct 2013 R = 8.3144621 J · K-1 · mol-1.
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### A Leap of Faith - OECD

In Boyle's Law, PV = k (a constant). Rearrange the Ideal Gas Law so that the constant R is similarly equal to the variables.

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### Gas - Beteende och egenskaper

The ideal gas law in terms of R is PmRTV , where P is the absolute pressure of the gas, V is Processing Combined Gas law . While the ideal gas law is useful in solving for a single unknown when the other values are known, the combined gas law is useful when comparing initial and final situations. The ideal gas law can be rearranged to solve for R, the gas constant. $$R=\frac{PV}{nT}$$ constant! The ideal gas law is: pV = nRT , where n is the number of moles, and R is universal gas constant.


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# 3.2 Domain and range  (Page 4/11)

 Page 4 / 11

## Describing sets on the real-number line

Describe the intervals of values shown in [link] using inequality notation, set-builder notation, and interval notation.

To describe the values, $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ included in the intervals shown, we would say, “ $x\text{\hspace{0.17em}}$ is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.”

 Inequality $1\le x\le 3\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}x>5$ Set-builder notation $\left\{x|1\le x\le 3\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}x>5\right\}$ Interval notation $\left[1,3\right]\cup \left(5,\infty \right)$

Remember that, when writing or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set.

Given [link] , specify the graphed set in

1. words
2. set-builder notation
3. interval notation

1. values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3;
2. $\left\{x|x\le -2\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}-1\le x<3\right\}$ ;
3. $\left(-\infty ,-2\right]\cup \left[-1,3\right)$

## Finding domain and range from graphs

Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x -axis. The range is the set of possible output values, which are shown on the y -axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See [link] .

We can observe that the graph extends horizontally from $\text{\hspace{0.17em}}-5\text{\hspace{0.17em}}$ to the right without bound, so the domain is $\text{\hspace{0.17em}}\left[-5,\infty \right).\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ The vertical extent of the graph is all range values $\text{\hspace{0.17em}}5\text{\hspace{0.17em}}$ and below, so the range is $\text{\hspace{0.17em}}\left(\mathrm{-\infty },5\right].\text{\hspace{0.17em}}$ Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range.

## Finding domain and range from a graph

Find the domain and range of the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ whose graph is shown in [link] .

We can observe that the horizontal extent of the graph is –3 to 1, so the domain of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\left(-3,1\right].$

The vertical extent of the graph is 0 to –4, so the range is $\text{\hspace{0.17em}}\left[-4,0\right).\text{\hspace{0.17em}}$ See [link] .

## Finding domain and range from a graph of oil production

Find the domain and range of the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ whose graph is shown in [link] .

The input quantity along the horizontal axis is “years,” which we represent with the variable $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ for time. The output quantity is “thousands of barrels of oil per day,” which we represent with the variable $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ for barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as $\text{\hspace{0.17em}}1973\le t\le 2008\text{\hspace{0.17em}}$ and the range as approximately $\text{\hspace{0.17em}}180\le b\le 2010.$

In interval notation, the domain is [1973, 2008], and the range is about [180, 2010]. For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines.

Given [link] , identify the domain and range using interval notation.

domain =[1950,2002] range = [47,000,000,89,000,000]

find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
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Raihany
Hi
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need help
Raihany
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Cromwell
a
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What do you mean by a
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nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
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b=p-4a-2c
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p=2(2a+C)+b
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COLEMAN
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A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
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Sin(A+B) = sinBcosA+cosBsinA
Prove it
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immy


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# Kinematics and varying accleration [duplicate]

This question already has an answer here:

If accleration varies inversely with square of displacement then how can we find the time it takes to cover distance S?? My attempt I tried to use calculus but didn't, work at all please help.

## marked as duplicate by ZeroTheHero, ja72, Kyle Kanos, Yashas, John Rennie homework-and-exercises StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 29 '17 at 6:56

• See my answer to a similar enough question. And please show your attempt when posting a question. – ja72 Apr 28 '17 at 21:33

This doesn't entirely answer the question, but it might help.

As you know, one way of finding the time $t$ is by finding an equation that relates $t$ to any of $a$, $v$ or the displacement $x.$   AFAIK, it doesn't seem obvious to find $t$ without such an equation.

(As you know, since $a$ is not constant, the formulae of UAM won’t work here, even if you know $a$, $v$ and $x$)

Also, to find $v$, you can proceed as follows.

We know

$$a=\frac{1}{x^2}.$$

Then from the relationship $$a=v\frac{dv}{dx},$$ We have $$v\frac{dv}{dx}=\frac{1}{x^2}.$$ You can now find $v$ as a function of $x$ by integrating with respect to $x$.


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# Using a gurobi.env file

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## Using a gurobi.env file

When you want to change the values of Gurobi parameters, you actually have several options available for doing so. We've already discussed parameter changes through the command-line tool (e.g., gurobi_cl Threads=1 coins.lp), and through interactive shell commands (e.g., m.setParam('Threads', 1)). Each of our language APIs also provides methods for setting parameters. The other option we'd like to mention now is the gurobi.env file.

Whenever the Gurobi library starts, it will look for file gurobi.env in the current working directory, and will apply any parameter changes contained therein. This is true whether the Gurobi library is invoked from the command-line, from the interactive shell, or from any of the Gurobi APIs. Parameter settings are stored one per line in this file, with the parameter name first, followed by at least one space, followed by the desired value. Lines beginning with the # sign are comments and are ignored. To give an example, the following (Linux) commands:

  echo "Threads 1" > gurobi.env
gurobi_cl coins.lp

would read the new value of the Threads parameter from file gurobi.env and then optimize model coins.lp using one thread. Note that if the same parameter is changed in both gurobi.env and in your program (or through the Gurobi command-line), the value from gurobi.env will be ignored.

The distribution includes a sample gurobi.env file (in the bin directory). The sample includes every parameter, with the default value for each, but with all settings commented out.


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The Organum Mathematicum is a fabulous-looking thing:
vps280516.ovh.net/divulgamat15/i…
A box of sticks which form tables to help with mathematical calculations, sort of like Napier rods, but for artillery and music?! Of course this oddity is a product of Athanasius Kircher!

A Mastodon instance for maths people. The kind of people who make $\pi z^2 \times a$ jokes.

Use $ and $ for inline LaTeX, and $ and $ for display mode.


text:
Product of any number three times is called cube. Use this calculator to find the cube of any positive and negative number. Cube of any number is denoted by N cubed as $N^3$,
Where N can be any number. This cube calculator takes the input and gives output in the form of cube.

## Steps for Calculator

Step:1: Note down the values of the given problem.

Step 2: Put the value in the given formula and calculate the area of cube $N^3$.

Area of Cube= $N^3$


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# Category Archives: Tenure

March 4, 2010, 10:12 pm

# Working and having a life, redux

Image via Wikipedia

The Chronicle has an article on a Harvard survey of Gen-X professors and their attitudes toward the balance of work and the rest of life. The professors surveyed indicate that they want to be successul in their careers but don’t want to sell out their personal lives in the process. The main survey report is here (PDF, 2.1MB). Here’s a representative quote from one of the interviewees, a business professor, talking about the perils of overwork that Gen-Xers perceive in their older colleagues:

There’s really nothing to be gained by closing your door and working until 11:00 o’clock at night, other than the tenure hurdle that is somewhere out there. If you want to pole vault over it, you go right ahead, but no one here is going to back up the Brinks truck and start dumping all this cash on you, simply because you’ve decided to work like you have three jobs. So that…

February 14, 2010, 10:35 pm

# Must the tenure process really be like this?

Like a lot of people in higher ed, I’ve been following Friday’s deadly shooting at the University of Alabama-Hunstville. (Click the link for background in case you missed the story. I have no idea how much press it is or is not getting in the national mainstream media.) It’s known that Amy Bishop, the UAH biology professor being charged with the shooting, was denied tenure in April and had made an unsuccessful appeal regarding her tenure denial. It’s not clear that the shooting was related to the tenure situation, but the speculation — especially in the article at the second link — is that there’s a connection.

What is clear, at least from my perspective as a professor and as somebody in the fourth year of a five-year appointment to my college’s Promotion and Tenure Committee, is that something is really badly wrong with UAH’s tenure system, and perhaps with tenure as a concept….

July 27, 2009, 12:45 pm

# The blogging VPAA?

I was thinking over the session coming up at Blog Indiana by John Oak Dalton titled “Chancellor 2.0″ which promises to address “existing and emerging obstacles of CEO-grade context” [sic? Was that supposed to be "content"?] for Twitter. In other words, it sounds like the session will be about how to get your college’s upper administration up and running with blogging and tweeting. I’m curious to see what Dalton makes of this, because his home institution seems to have embraced blogging and Twitter at a scale you don’t normally see from a university. Even the chancellor tweets.

I’d love to see more college administrators blogging or twittering, using their real names, making no secret of their institutions, and writing honestly about their successes and struggles in the work that they do. There’s no faster track to giving higher education a measure of transparency that it badly needs…

December 29, 2008, 7:55 am

# Linkages, 29 December

The perils of graphing, from xkcd.

Erin at Critical Mass claims that tenure is untenable and has already been phased out without our even knowing it. Good discussion in the comments.

The New York state comptroller says that New York state college students could save $500 per year on textbooks by buying them online rather than in a campus bookstore. He gives a list of recommendations to universities to make this easier to do. Right on. Possibly the most comprehensive and mathematically thorough analysis ever of how to curve grades. The trouble with GTD. An observation about the changing demographic of mathematics education researchers. Finally, some number-theoretic trivia about the number 2009, just in time for the new year. October 31, 2008, 10:18 am # Four questions to ask about tenure Over at the Young Mathematicians’ Network, I have an article today on four revealing questions that young faculty should ask about tenure. Since you have to have an account to post comments at the YMN web site, and since some readers who aren’t mathematicians might want to discuss this stuff, I’m going to reprint the article below the fold and open comments for it. Enjoy! September 29, 2008, 6:20 am # Monday GTD moment: Scholarship and GTD This is the third installment of Monday GTD Moment, where I take a post to blog about Getting Things Done and how it applies in an academic setting. If you’re unfamiliar with GTD, here’s a good overview, and make sure to read David Allen’s book that started it all. Last week I wrote about grading and GTD. I noted that grading is kind of a poor fit in traditional GTD. A prof can grade anywhere, so the idea of contexts fits awkwardly; and grading “tasks” are usually projects, although we think of them as tasks and although the next actions contained in those projects are usually nothing more than smaller projects. GTD wasn’t really made for the academic profession, and so the staple activities of academics don’t often fit well. Another area similar to grading in its relatively poor fit within the canonical GTD philosophy is research, or more generally scholarship. By “scholarship” … January 11, 2008, 9:35 pm # Tenure vs. contracts I’m on the Promotion and Tenure Committee here, and my two colleagues and I on the committee just finished the first of two solid weeks of reviewing evaluation portfolios of all the faculty up for promotion, tenure, and annual review. It’s great fun. But seriously, I’ve been thinking a lot about tenure this week. In the more exasperating moments, I’ve wished that we were one of those colleges that doesn’t do tenure any more at all, but rather some kind of contract system. First of all, that would make us rare. According to the blurb for this book on colleges without tenure, 97% of research universities and 99% of four-year public universities offer tenure — and apparently 91% of small private colleges (like mine). The number of colleges without tenure is small, but I think it’s growing. Certainly I hear a lot of rumbling among administrators (although I haven’t ever heard it among my… July 25, 2007, 9:00 am # Truth and consequences for Ward Churchill Ward Churchill has been fired: More than two and a half years after Ward Churchill’s writings on 9/11 set off a furor, and more than a year after a faculty panel at the University of Colorado at Boulder found him guilty of repeated, intentional academic misconduct, the University of Colorado Board of Regents voted 8-1 Tuesday evening to fire him.The vote followed a special, all-day meeting of the board, in which it heard in private from Churchill, a faculty panel and from Hank Brown, president of the University of Colorado System, who in May recommended dismissing Churchill from his tenured post. The regents emerged from their private deliberations at around 5:30 p.m. Colorado time and voted to fire Churchill, but they did not discuss their views and they quickly adjourned. A small group of Churchill supporters in the audience shouted “bullshit” as the board vote was announced.While… July 13, 2007, 6:03 am # Bias and thought policing on all sides You might come away from yesterday’s postings (here and here) thinking that real bias in academia only occurs from Left to Right. Nothing could be further from the truth. Sadly, the impulse to censor and police the thoughts of professors and students is equal-opportunity. Consider this appalling case at Ashland University reported at The FIRE in which John Lewis, a historian specializing in Ayn Rand’s objectivist philosophy, was hired as a professor at Ashland and promptly received a$100,000 grant from the Anthem Foundation for Objectivist Scholarship for teaching and writing on objectivism. After gladly accepting the money, affirming its support for Lewis’ research, and giving him six hours’ release time per semester in which to do it, the university then denied Lewis tenure because, as stated in the tenure denial letter:

…concern was expressed at all levels of the process about…

September 8, 2006, 2:08 pm

# Quick takes: From the P&T committee; more on test-taking; formatting in calculus; tech training

• The Promotion and Tenure committee I am now part of has met twice so far. We meet for 90 minutes per week. What I’ve learned so far is (1) most of our meetings are spent trying to plug holes, fix mistakes, or deal with unforeseen consequences from past P&T committee actions/inactions; (2) evidently there is a great deal of time spent by the committee to create language in our promotion/tenure recommendations specifically to minimize offending the delicate sensibilities of faculty, for example making sure we don’t say someone’s a "fine teacher" when the faculty in question resents the use of the word "fine" and would prefer "excellent" instead*; (3) evidently there is also a lot of time spent soothing the offended faculty where the language doesn’t work; and (4) it seems a main purpose of having a system for tenure is to allow the college to covers its own rear-end in case somebody…
• The Chronicle of Higher Education
• 1255 Twenty-Third St, N.W.
• Washington, D.C. 20037


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# Two metrics induce the same topology, but one is complete and the other isn't

I'm looking for an example of two metrics that induce the same topology, but so that one metric is complete and the other is not (Since it is known that completeness isn't a topological invariant).

Thanks in advance for any hints or ideas.

-
The real line and (0,1) are topologically equivalent but the second is not complete. d(x,y) and d(f(x),f(y)) where f is a topological equivalence (homeomorphism) between a complete and an incomplete space will be two metrics that "induce the same topology but one is complete and the other isn't". – T.. Oct 23 '10 at 17:46
It may be worth noting that there can be no compact counterexample, so open intervals and infinite discrete sets, used in all of the examples so far, are natural places to look. – Jonas Meyer Oct 23 '10 at 19:10

The metric space $\{\frac {1}{n} \mid n\in \mathbb{N} \}$ with the usual metric is incomplete (since we don't have zero), and it has the discrete topology.

the same space with the metric $d(x,y)=1 \iff x\neq y$ also has a discrete topology but is complete, since any cauchy sequence will eventually be constant.

-
I like this! This is in some sense a minimal counterexample, since every finite metric space has the discrete topology and is complete. – Qiaochu Yuan Oct 23 '10 at 12:29
There is a lot to be said for minimal counterexamples, but it doesn't provide much insight into why the image of a complete space under a homeomorphism can fail to be complete. You could say many fancy words such as "forgetful functor" and "completely metrisable/uniformisable", but in simple terms if you don't mention the words "uniform continuity", you won't have pointed out the central reason why a homeomorphism can fail to preserve completeness. – kahen Oct 23 '10 at 15:27
+1 ; And this is an example for the fact that nonequivalent metric may determine same topology. – HK Lee Nov 10 '15 at 11:07

The interval $M:=]-1,1[$ with the usual line element $ds:=|dx|$ is an incomplete metric space: The sequence $x_n:=1-{1\over n} \ (n\to\infty)$ converges in $\mathbb R$, so it is a Cauchy sequence, but it diverges in $M$. On the other hand, the "hyperbolic metric" defined by $ds:=|dx|/(1-x^2)$ induces the same topology on $M$ but is complete. The latter statement needs of course a proof. Suffice it here to say that now the endpoints $\pm1$ are "infinitely far away".

-

Look for a homeomorphism $f: \mathbb R \to \mathbb R_+$ (you should know a good one). Then take the usual metric on $\mathbb R$ and use it to define a metric $d$ on $\mathbb R_+$ by $d(x,y) = |f(x)-f(y)|$ (*). Now what can you say about the relationship between $d$ and the usual metric on $\mathbb R_+$?

EDIT: Made the question here correct and more illustrative.

So the identity $\iota: (\mathbb R_+,d) \to (\mathbb R_+,|\cdot|)$ is a homeomorphism. Now just ask yourself which property, that continuous functions between metric spaces can have, $\iota$ doesn't have? Can you formulate a theorem about what property a homeomorphism must have to preserve completeness?

(*): Pullback and pushforward might be of interest.

-

If $(X,d)$ is a connected, complete metric space and $G$ is a nonempty proper open subset of $X$, then $G$ is not complete with the restricted metric (because it is not closed), but $G$ is homeomorphic to a complete metric space, because the map $x\mapsto(x,1/d(x,X\setminus G))$ embeds $G$ onto a closed subspace of $X\times \mathbb{R}$. (This is more or less Lemma 3.1.1 of Arveson's Invitation.)

A topological space that is homeomorphic to a separable complete metric space is called a Polish space. A subspace of a Polish space is Polish if and only if it is a $G_\delta$, while $G_\delta$s in complete metric spaces are not typically complete with the restricted metric. For example, there is a complete metric inducing the usual topology on the set of irrational numbers.

-

As many others have pointed out, the way to do this is to think about pairs of metric spaces that are homeomorphic but not isometric. In many cases, the metric will be complete on one space but not the other. For example:

• The set $\{1/n \mid n \in \mathbb{N}\}$ is incomplete, but the natural numbers $\mathbb{N}$ are complete.

• The ray $(0,\infty)$ and the open interval $(0,1)$ are incomplete, but the real line is complete.

• An open disc in the plane is incomplete, but the entire plane is complete.

• The punctured plane $\mathbb{R}-\{0\}$ is incomplete, but the infinite cylinder $\mathbb{R} \times S^1$ (where $S^1$ is the circle) is complete.

• The real line minus the $x$-axis is incomplete, but a disjoint pair of planes is complete.

• The twice-punctured plane $\mathbb{R}^2 - \{(-1,0),(1,0)\}$ is incomplete, but the graph of $$z = \frac{1}{(x^2-1)^2 + y^2}$$ in $\mathbb{R}^3$ is complete. (This graph has asymptotic "cusps" at $(-1,0)$ and $(1,0)$.)

In each case, the homeomorphism between the two spaces can be used to define a nonstandard metric on the latter space that makes it incomplete.

-

Consider one complete normed vector space $(V,\|\cdot\|)$, that is, a Banach space, and let $$B = \{ x \in V : \|x\| < 1 \}$$ be the unit ball centered at the origin. Define $f: V \rightarrow B$ by $x \mapsto x /(1+ \|x\|)$. Then f is continuous and is easy to show that the inverse of $f$ is also continuous, so f is a homeomorphism.

Now $B$ isn't complete, since if $x \in V$ and $\| x \| = 1$, then $x_n = (1-1/n)x$ is a cauchy sequence in $B$ that doesn't converge.

As Kahen referred, if we have two metric spaces $X$ and $Y$ with the latter complete and a uniformly continuous homeomorphism $g : X \rightarrow Y$, then $X$ is complete. To prove this you'll need to observe that since $g$ is uniformly continuous then it sends Cauchy sequences to Cauchy sequences.

-

With the usual metric, $(0,1]$ is not complete.

Let us define another metric $\rho$ on this space using the homeomorphism $\phi: [1,\infty) \to (0,1]$. Define $\rho(x,y) = \vert\frac{1}{x}-\frac{1}{y}\vert$. It is easy to check that $\rho$ is indeed a metric. Also, note that the usual metric and $\rho$ are equivalent and hence induce the same topology.

Now consider a Cauchy sequence in $((0,1],\rho)$.

$\rho(x_n,x_m) < \epsilon$ for all $n \geq N$ or,

$\vert\frac{1}{x}-\frac{1}{y}\vert < \epsilon$ for all $n \geq N$.

Hence $(\frac{1}{x_n})$ is a Cauchy sequence in ($[1,\infty),$usual metric), a complete metric space. Let us say, it converges to $y \in [1,\infty)$. It is easy to check that $(x_n)$ converges to $x = \frac{1}{y}$ in $((0,1],\rho)$.

-


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# Which of the following statements are true..?

Consider the entire function $$f(z)=1+z+z^{20}$$ and $$g(z)=e^z,z\in \mathbb{C}$$ Which of the following statements are true ?

$$1)\lim_{|z|\to \infty}|f(z)|=\infty$$

$$2)\lim_{|z|\to \infty}|g(z)|=\infty$$

$$3)f^{-1}(\{z\in \mathbb{C}:|z|\le R\})$$ is bounded for every $$R>0$$

$$4)g^{-1}(\{z\in \mathbb{C}:|z|\le R\})$$ is bounded for every $$R>0$$

# My attempt : option $$1$$ is only correct option

option 2) is false $$lim_{x \rightarrow -\infty} e^{-x} = 0$$

option $$3$$ and $$4$$ are contradiction of lioville thorem so it will be false

is its True ??

Any hints/solution will be appreciated

thanks u

• What's $x$ in the first two equations? – Andrei Dec 17 '18 at 4:58
• @Andrei i have edits its – jasmine Dec 17 '18 at 4:59
• 3,4 do not contradict Liouville – zhw. Dec 17 '18 at 5:12
• The existence of root bounds shows that 3) is true, $z^{20}+z+1=w$ implies that $|z|\le 2+|w|=2+R$, with some work $|z|\le\sqrt[19]{2+R}$. – LutzL Dec 17 '18 at 17:45
• I understood the explanations of 1,2,3. Why (4) is false? @zhw – Unknown x Jun 15 at 2:14


text:
# Working with matrices: powers and transposition

written in

Today, we’ll complete our series on basic matrix operations by covering powers of a matrix and matrix transposition. In the previous posts, we covered matrix addition, subtraction and multiplication and matrix inversion.

## Powers of a matrix

Say we have a number $$a$$, and we’re asked to solve $$a^2$$. Well, this is easy: we just multiply $$a$$ by itself to get $$a \times a$$. Powers of matrices work in exactly the same way. If we take a matrix $$A$$, we can get $$A^2$$ by multiplying $$A$$ by itself:

$$A^2 = AA = \begin{bmatrix}4 & 3\\6 & 5\end{bmatrix}\begin{bmatrix}4 & 3\\6 & 5\end{bmatrix} = \begin{bmatrix}34 & 27\\54 & 43\end{bmatrix}$$

You can probably guess that, because of the way that matrix multiplication works, we can only raise square matrices to a power. We can raise square matrices to any (positive) power in the same way: if we want to get the cube of $$A$$, or $$A^3$$, we multiply the matrix by itself 3 times, if we want $$A^4$$, we multiply it by itself 4 times, and so on.

In Numpy, we can use the matrix_power function from the linalg subpackage to calculate the power of a matrix. The first argument is the matrix, and the second is the power you’d like to raise the matrix to.

import numpy as np
from numpy.linalg import matrix_power

A = np.array([[4, 3], [6, 5]])
matrix_power(A, 2)

array([[34, 27],
[54, 43]])

Powers of matrices follow the same rules as do powers of a number. Let’s have a look at a couple of examples of these. Say we have two integers, $$m$$ and $$n$$. If we have two matrices, $$A^m$$ and $$A^n$$ and we multiply them, we get $$A^mA^n = A^{m+n}$$. Similarly, if we have the matrix $$A^m$$ and we raise it to $$n$$, then we get $$(A^m)^n = A^{mn}$$.

Let’s try this out in Numpy, by comparing the results of $$A^5$$ to $$A^2A^3$$:

A_5 = matrix_power(A, 5)
A_5

array([[22930, 18237],
[36474, 29009]])

A_2 = matrix_power(A, 2)
A_3 = matrix_power(A, 3)

A_2.dot(A_3)

array([[22930, 18237],
[36474, 29009]])

We can also raise matrices to negative powers. We’ve already seen an example of this with the inverse of a matrix, which is written as $$A^{-1}$$. In order to raise a matrix to the power of $$-2$$, we simply need to multiply the inverse by itself. This logic can then be extended in the same way as we did for raising the matrix to a positive power.

Let’s see this in Numpy by comparing the function to calculate the inverse to raising our matrix to the power of -1. As we can see, they are exactly the same.

from numpy.linalg import inv

inv(A)

array([[ 2.5, -1.5],
[-3. ,  2. ]])

matrix_power(A, -1)

array([[ 2.5, -1.5],
[-3. ,  2. ]])

## Matrix transposition and symmetrical matrices

The final thing we will cover is matrix transposition. This is simply what happens if you flip the rows and columns of a matrix. Let’s have a look at this with matrix $$B$$:

$$B = \begin{bmatrix}1 & 2\\3 & 4\\5 & 6\end{bmatrix}$$

What would the transpose of $$B$$ (denoted by $$B^T$$) look like? Well, it is simple as taking the rows of $$B$$ and turning them into the columns of $$B^T$$ (or vice versa):

$$B^T = \begin{bmatrix}1 & 3 & 5\\2 & 4 & 6\end{bmatrix}$$

And what would the transpose of $$B^T$$ look like (in other words, what is $$(B^T)^T)$$? Well, if we flip the rows and columns again, we just end up back at our original matrix! In other words, $$(B^T)^T = B$$.

In order to transpose a matrix using Numpy, we use the transpose function on that matrix:

B = np.array([[1, 2], [3, 4], [5, 6]])
B_T = B.transpose()
B_T

array([[1, 3, 5],
[2, 4, 6]])

We can also reverse the transposition by calling transpose on our new transposed matrix, B_T:

B_T.transpose()

array([[1, 2],
[3, 4],
[5, 6]])

As usual, square matrices behave a little differently to non-square matrices, in that the diagonal values of the matrix don’t move when you take the transpose of the matrix. If we go back to our square matrix $$A$$, and take the transpose, we end up with:

$$A^T = \begin{bmatrix}4 & 6\\3 & 5\end{bmatrix}$$

You can see that the values in $$a_{12}$$ and $$a_{21}$$ have moved, but the values on the diagonals ($$a_{11}$$ and $$a_{22}$$) haven’t. This means that if values that are diagonally opposite to each other in a square matrix are equal (say, if $$a_{12} = a_{21}$$ in the example above), then the transpose of the matrix would be identical to the matrix. Such matrices are called symmetrical matrices. Let’s have a look at an example. Let’s say we have a $$3 \times 3$$ matrix, $$C$$, which is symmetrical. Because of this:

$$C = C^T = \begin{bmatrix}1 & 4 & 5\\4 & 2 & 6\\5 & 6 & 3\end{bmatrix}$$

This matrix might not look symmetrical at a first glance. However, one way to see that it is really symmetrical is to compare the first row and the first column (and the second row and second column, and the third row and third column). You can see that they are identical, which means that it doesn’t matter whether these elements function as a row or a column: the result will be the same. Another way to see this a bit more explicitly is to compare those elements are diagonal to each other in the matrix, as we spoke about above. We can see that $$c_{12} = c_{21} = 4$$, $$c_{13} = c_{31} = 5$$ and $$c_{23} = c_{32} = 6$$.

I hope these posts have given you a solid overview of the ways you can work with matrices. In the next post, we’ll build on this by looking at the linear algebra approach to solving linear regression problems.


text:
# Let x^2 - 12x=61 What values make an equivalent number sentence after completing the square? Enter your answers in the boxes. x^2- 12x +___=___

###### Question:

Let x^2 - 12x=61
What values make an equivalent number sentence after completing the square?
x^2- 12x +___=___

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