text:
# chaptertitle name and chapter number in one line but diffrent edge of the page

This is what I want:

Introduction <---all the way on the right edge of the page---> Chapter 1

But, the catch is that I have a 5cm width margin note space in a two sided document. I am working with the book class if that helps...

• Welcome to TeX.SX! Please help us help you and add a minimal working example (MWE) that illustrates your problem. Reproducing the problem and finding out what the issue is will be much easier when we see compilable code, starting with \documentclass{...} and ending with \end{document}. – user31729 Jul 28 '15 at 15:03
• Do you want the string "Chapter #" to be flushed to the right edge of the page? – Gonzalo Medina Jul 28 '15 at 15:56

Another solution with titlesec, which works also with multiline titles:

\documentclass[twoside]{book}

\usepackage[right=5cm, showframe]{geometry}
\usepackage[x11names]{xcolor}
\usepackage[explicit]{titlesec}

\titleformat{\chapter}{\Huge\bfseries}{\rlap{\hspace*{\dimexpr\textwidth + \marginparsep}\color{IndianRed3}\chaptername{} \thechapter}}{0pt}{#1}

\usepackage{lipsum}

\begin{document}
\mainmatter
\chapter{Introduction}

\lipsum[1-3]

\setcounter{chapter}{8}
\chapter{Another chapter. \\Yet another chapter.}

\lipsum[4-6]

\end{document}

• Have you tried \setcounter{chapter}{10}\chapter{Introduction}? – Ruben Jul 28 '15 at 17:24
• I only answered the O.P.'s requirements. In such a case he/she may modify the layout: change the values of marginparwidth, or decide the word ‘Chapter’ and chapter number will be in a tabular environment, in different rows. Inevitably there can be problem putting the label in the margin. It's up to the user to adapt the different parameters. – Bernard Jul 28 '15 at 17:39
• Where did the OP mention that the solution will be used with a document having less then ten chapters? – Ruben Jul 28 '15 at 17:53
• As I said, he can modify the parameters so as to fit marginparwidth: setting it to 5.5cm will do. Unless he uses Roman numbering. The code is pretty easy to adapt. Or make the label on two lines if he wants a Chapter MDCCCXVIII. I'm afraid there's no universal solution that works out of the box. – Bernard Jul 28 '15 at 18:03

I think you are most flexible if you redefine the \@makechapterhead by hand. (I'll add a remark with the reasons right before the complete code.)

Two parboxes, the first as wide as the \textwidth and the second as your right margin (5cm), side by side wrapped into a \hbox should do the job

\hbox{%
\parbox[t]{\the\textwidth}{\Huge\bfseries\raggedright#1}
\parbox[t]{5cm}{\huge\bfseries\raggedleft\@chapapp\space \thechapter}
}

As you see 'Chapter 1' gets pushed exactly to the edge. However, I would recommend to set the second \parbox to a little less then your right margin. I used 4.5cm in the final solution. Additionally I would add a little offset to the width of the first \parbox, so the final redefinition of \@makechapterhead is:

\def\@makechapterhead#1{%
\vspace*{50\p@}%
{\parindent \z@ \normalfont
\ifnum \c@secnumdepth >\m@ne
\if@mainmatter
\dimen0=\dimexpr\textwidth-2em\relax
\hbox{%
\parbox[t]{\the\textwidth}{\Huge\bfseries\raggedright#1}
\parbox[t]{4.5cm}{\huge\bfseries\raggedleft\@chapapp\space \thechapter}
}
\par\nobreak
\fi
\fi
\interlinepenalty\@M
\vskip 40\p@
}
}

Now a few remarks:

• You need the e-TeX extension to compile the code. It should be in your distro by default. I just mention it for the sake of completeness.
• With this solution you can set the marginparwidth without interfering with the chapter number (see the last pic).
• You are able to deal with very long chapter headings (see the last pic).

Here is the complete code:

\documentclass[twoside]{book}
\usepackage{lipsum}
\usepackage[right=5cm,marginparwidth=2.5cm,marginparsep=2em]{geometry}
\usepackage{marginnote}

\makeatletter
\vspace*{50\p@}%
{\parindent \z@ \normalfont
\ifnum \c@secnumdepth >\m@ne
\if@mainmatter
\dimen0=\dimexpr\textwidth-2em\relax
\hbox{%
\parbox[t]{\the\dimen0}{\Huge\bfseries\raggedright#1}
\parbox[t]{4.5cm}{\huge\bfseries\raggedleft\@chapapp\space \thechapter}
}
\par\nobreak
\fi
\fi
\interlinepenalty\@M
\vskip 40\p@
}
}
\makeatother

\begin{document}
\chapter{Long long long long long long long Introduction}

\marginnote{Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam erat, sed diam voluptua.}
\lipsum
\end{document}


text:
# pandas.Series.dt.nanoseconds¶

Series.dt.nanoseconds

Number of nanoseconds (>= 0 and less than 1 microsecond) for each element.

format: plain


text:
# The General Relativity Metric and Flat Spacetime

Gold Member
Ok. Then let me put it in this way:

$${dp}{=}{R}{d}{\theta}$$

$${dq}{=}{R}{Sin}{(}{\theta}{)}{d}{\phi}$$

That is not a coordinate transform.

It's actually worse than that! This statement:

$$dq = R \sin \theta \; d\phi$$

is a lie! The right-hand side is not 'd' of anything, so it is incorrect to call this one-form 'dq'.

Remember that for any p-form $\omega$, we must have $dd\omega = 0$. Taking 'd' of both sides of Anamitra's equation yields

\begin{align*} ddq &= R \; d ( \sin \theta \; d\phi) \\ 0 &= R \cos \theta \; d\theta \wedge d\phi \end{align*}
which is clearly false.

Edit: I have deleted the rest of my post. It was a careful explanation of the issues Anamitra would face if he tried to follow the procedure he has outlined. I am offended, however, at the idea of doing his work for him.

Last edited:
Gold Member
$${dq}{=}{R}{Sin}{(}{\theta}{)}{d}{\phi}$$
Because we must have$$dq = \frac{\partial q}{\partial \theta} d\theta + \frac{\partial q}{\partial \phi} d\phi$$your equation implies not only$$\frac{\partial q}{\partial \phi} = R \, \sin \, \theta$$but also$$\frac{\partial q}{\partial \theta} = 0$$As others have pointed out in different ways, there is no simultaneous solution to both those equations.

Mentor
The right-hand side is not 'd' of anything, so it is incorrect to call this one-form 'dq'.
there is no simultaneous solution to both those equations.
Thanks, I liked both of these explanations better than mine. I didn't realize the problem was that serious, but I guess I should have.

Gold Member
OK, I think at this point you will not be convinced unless you actually do a computation. So let's simplify this as much as possible and do just 2 dimensions. Take the following metric for the sphere,

$$ds^2 = d\theta^2 + \sin^2 \theta \; d\phi^2,$$
and show us how to obtain a globally flat coordinate system, using your method.

$${ds}^{2}{=}{dx}^{2}{+}{dy}^{2}{+}{dz}^{2}$$

With $${x}^{2}{+}{y}^{2}{+}{z}^{2}{=}{R}^{2}{=}{const}$$

$${x}{=}{R}{Sin}{\theta}{cos}{\phi}$$
$${y}{=}{R}{Sin}{\theta}{sin}{\phi}$$
$${z}{=}{R}{Cos}{\phi}$$

Putting on a constraint that r=1:

\begin{align*} ds^2 &= dr^2 +r^2 d\theta^2+r^2 sin^2(\theta) d\phi^2\\ (ds|_{r=1})^2 &= 0 + d \theta^2 + sin^2(\theta)d\phi^2 \end{align*}

Nobody disagrees that you can embed a 2D curved surface like a sphere into a flat 3D space. That doesn't make the embedded 2D space flat.

Use your procedure to "flatten" the sphere. Or if a sphere is too easy then use your procedure to flatten the Schwarzschild spacetime.

Non-local velocities[in cosmology or elsewhere] should not be a problem since we have a flat spacetime in the physical context. Parallel-Transport is not so serious an issue in flat spacetime.

We have this mapping of the surface of a sphere, r=1, and we're able to make this relationship of

$$ds^2 = d\theta^2 + \sin^2 \theta \; d\phi^2,$$

However, there are no distinguishing features of that sphere that would tell anyone the value of θ or Φ. As far as they are concerned, the physical distances at the pole are identical to the physical distances at the equator. If you told a person at the pole that he could only travel in the theta direction he would say you were crazy. Yet, that's exactly what this coordinate system in (θ,Φ) would say. If you told someone walking around the pole that he was traveling at the same speed (in dΦ/dt) as someone flying around the equator once every three seconds, he would think you are crazy. Yet, that's exactly what this coordinate system in (θ,Φ) would say.

But that is certainly not the units that people living in that coordinate system would use. They would use the units from the higher dimensional flat coordinate system.

(Edit: or perhaps, they would all imagine themselves to be traveling along an equator, so that sin(θ)=1 and ds2 =dθ2+dΦ2 = dx2 + dy2)

Last edited:
Gold Member
I just came across page 64 of the http://www.blatword.co.uk/space-time/Carrol_GR_lectures.pdf" [Broken] where some of the ambiguity is introduced.

Within this, it is made clear that via parallel transport, there is no "well defined notion of relative velocity" namely because there is no well-defined notion of relative direction. However, I think in a static model (such as the surface of a sphere, and (possibly?) the Schwarzschild metric) there should still be a well-defined notion of "speed." i.e. the non-directional relative motion.

When Carroll gets to the bottom of the page, though, he invokes a non-static model; "the metric of spacetime between us and the galaxies has changed", and "the universe has expanded," suggesting a scale factor changing with respect to time, a(t), Under these circumstances, even the speed would be ambiguous, I guess.

(Although I may be reading too much into it... Carroll does not specifically say the metric has changed "over time." He may just mean the change in the metric through space.)

Last edited by a moderator:
Staff Emeritus
There's a well defined notion of a static observer in a static space-time like the Scwarzschild space-time and there is an unambiguous notion of the speed of any object relative to the static observer that's co-located with him.

Cosmological space-times are for the most part not static, so this trick won't work for them.

Gold Member
Flatlanders living on a sphere:

"That way." (pointng in two opposite directions)

"And what direction did you just come from?"

"Sorry, that direction does not exist here. I would require a third degree of freedom to tell you, explicitly."

Last edited:
Gold Member
There's a well defined notion of a static observer in a static space-time like the Scwarzschild space-time and there is an unambiguous notion of the speed of any object relative to the static observer that's co-located with him.

Cosmological space-times are for the most part not static, so this trick won't work for them.

I still have to read more. Maybe when you put enough static schwarzschild metrics together, and rub them against each other, it becomes a nonstatic space-time.

Anamitra
By holding the other coordinates constant, you are able to do the integral, sure. But when you take the exterior derivative of both sides, you obtain

$$dT = \frac{\partial}{\partial x^\alpha} \Big( \int^{x^0} \sqrt{g_{00}(t',x^1,x^2,x^3)} \; dt' \Big) dx^\alpha$$

which is a closed 1-form, but it is not a 1-form that fits nicely into your metric! It has a bunch of $dx^1,\ dx^2,\ dx^3$ terms in addition to $dx^0$. You will run into a similar problem with the rest of your integrals.

You should see now why the above claim is false. The point is that you have a system of differential equations (i.e., a bunch of 1-forms which you assume can be written as $dx^\alpha$ for some coordinate functions $x^\alpha$), and this system of differential equations is not integrable in any finite-sized open region, in general.

If the GR metric is valid only in a path dependent situation. I would request Ben Niehoff to go through the following posting:
https://www.physicsforums.com/showpost.php?p=3436925&postcount=21

Right here you've lied. By calling it $dT$, you are claiming that the one-form $\sqrt{g_{00}} dt$ is closed (i.e., locally exact). But as others have pointed out, this is not true unless $g_{00}$ happens to be a function of $t$ alone.

The Science Advisor considers it essential to use impolite language to express himself---possibly he feels that it should provide a special parameter to the forum


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# Additions To Esther

\$3.99

The Additions to Esther are variances between the original Hebrew and the Greek Septuagent where approximately six chapters of additional text are interspersed throughout the ten chapter book. A large portion of this additional narrative are the prayers of Mordecai and Queen Esther before she risks her life before the King.

Category:

## Description

Esther is a Biblical text beloved by Jews and Christians alike. The Additions to Esther are variances between the original Hebrew and the Greek Septuagent where approximately six chapters of additional text are interspersed throughout the ten chapter book. A large portion of this additional narrative are the prayers of Mordecai and Queen Esther before she risks her life before the King.

The Apocryphal books are considered to be Scripture in the Roman Catholic and Orthodox Christian traditions while being considered very important books of the church that should be read for deeper understanding for most Protestant traditions.


text:
Long Wang, "Robust strong stabilizability of interval plants: it suffices to check two vertices", Systems and Control Letters, 1995, Vol. 26, pp. 133--136.

In this paper, the author considers controlled objects (called plants in control theory) that can be described by differential equations of the following type: $${d^nx\over dt^n}+b_{n-1}{d^{n-1}x\over dt^{n-1}}+...+b_0= {d^mu\over dt^m}+a_{m-1}{d^{m-1}u\over dt^{m-1}}+...+a_0,$$ where $x(t)$ is the state of the plant at the moment $t$, and $u(t)$ is the value of the control applied at the moment $t$. The plant is called proper if $m\le n$. A plant is called stabilizable if there exists a controller, i.e., another differential equation that describe $u$ in terms of $x$ so that the system of these two equations is stable (crudely speaking, small deviations from the initial conditions tend to 0 as $t\to\infty$). A plant is called strongly stabilizable if it can be stabilized by a proper controller that is not only stable with the plant attached, but also stable by itself. There exists a simple method for checking whether a given plant is strongly stabilizable.

In real life, we often only know the {\it intervals} $[a^-_i,a^+_i]$ and $[b^-_j,b^+_j]$ of possible values of $a_i$ and $b_j$. Given these intervals, we want to check whether we can conclude that the actual plant (with some unknown values $a_i\in [a^-_i,a^+_i]$ and $b_j\in [b^-_j,b^+_j]$) is strongly stabilizable, or there are some values $a_i$ and $b_j$ from the given intervals for which the plant is not strongly stabilizable.

One necessary condition for strong stabilizability is known: that the polynomials $a(s)=s^m+\sum a_i s^i$ and $b(s)=s^n+\sum b_js^j$ (where $s=i\omega$ is related to the parameter $\omega$ of the Fourier transform) have no common complex roots in the unstable zone ${\rm Re} (s)\ge 0$.

The author shows that under this condition, all plants from a given interval plant are strongly stabilizable iff the following two of them are: we take any vector $b_0\in [b^-_0,b^+_0], \ldots,b_{n-1}\in [b^-_{n-1},b^+_{n-1}]$, and take $a_i=a^-_i$ ($0\le i\le m-1$) for the first plant and $a_i=a^+_i$ ($0\le i\le m-1$) for the second plant.

Thus, to check strong stabilizability of an interval plant $I$, it suffices to check two vertices of the box that describes all possible plants from $I$.


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Trapezoids and Kites | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Geometry - Second Edition Go to the latest version.

# 6.5: Trapezoids and Kites

Created by: CK-12

## Learning Objectives

• Define and find the properties of trapezoids, isosceles trapezoids, and kites.
• Discover the properties of midsegments of trapezoids.
• Plot trapezoids, isosceles trapezoids, and kites in the coordinate plane.

## Review Queue

1. Draw a quadrilateral with one set of parallel lines.
2. Draw a quadrilateral with one set of parallel lines and two right angles.
3. Draw a quadrilateral with one set of parallel lines and two congruent sides.
4. Draw a quadrilateral with one set of parallel lines and three congruent sides.
5. Draw a quadrilateral with two sets of congruent sides and no parallel sides.

Know What? A traditional kite, seen at the right, is made by placing two pieces of wood perpendicular to each other and one piece of wood is bisected by the other. The typical dimensions are included in the picture. If you have two pieces of wood, 36 inches and 54 inches, determine the values of $x$ and $2x$. Then, determine how large a piece of canvas you would need to make the kite (find the perimeter of the kite).

## Trapezoids

Unlike parallelograms, trapezoids have only one set of parallel lines. The other two sides have no restrictions.

Trapezoid: A quadrilateral with exactly one pair of parallel sides.

Examples look like:

Isosceles Trapezoid: A trapezoid where the non-parallel sides are congruent.

The third trapezoid above is an example of an isosceles trapezoid. Think of it as an isosceles triangle with the top cut off. Isosceles trapezoids also have parts that are labeled much like an isosceles triangle. Both parallel sides are called bases.

## Isosceles Trapezoids

Previously, we introduced the Base Angles Theorem with isosceles triangles. The theorem states that in an isosceles triangle, the two base angles are congruent. This property holds true for isosceles trapezoids. The two angles along the same base in an isosceles trapezoid will also be congruent. This creates two pairs of congruent angles.

Theorem 6-17: The base angles of an isosceles trapezoid are congruent.

Example 1: Look at trapezoid $TRAP$ below. What is $m \angle A$?

Solution: $TRAP$ is an isosceles trapezoid. So, $m \angle R = 115^\circ$, by Theorem 6-17. To find $m \angle A$, set up an equation.

$115^\circ + 115^\circ + m \angle A + m \angle P & = 360^\circ\\230^\circ + 2m \angle A & = 360^\circ \rightarrow m \angle A = m \angle P\\2m \angle A & = 130^\circ\\m \angle A & = 65^\circ$

Notice that $m \angle R + m \angle A = 115^\circ + 65^\circ = 180^\circ$. These angles will always be supplementary because of the Consecutive Interior Angles Theorem from Chapter 3. Therefore, the two angles along the same leg (or non-parallel side) are always going to be supplementary. Only in isosceles trapezoids will opposite angles also be supplementary.

Example 2: Write a two-column proof.

Given: Trapezoid $ZOID$ and parallelogram $ZOIM$

$\angle D \cong \angle I$

Prove: $\overline{ZD} \cong \overline{OI}$

Solution:

Statement Reason
1. Trapezoid $ZOID$ and parallelogram $ZOIM, \ \angle D \cong \angle I$ Given
2. $\overline{ZM} \cong \overline{OI}$ Opposite Sides Theorem
3. $\angle I \cong \angle ZMD$ Corresponding Angles Postulate
4. $\angle D \cong \angle ZMD$ Transitive PoC
5. $\overline{ZM} \cong \overline{ZD}$ Base Angles Converse
6. $\overline{ZD} \cong \overline{OI}$ Transitive PoC

In this example we proved the converse of Theorem 6-17.

Theorem 6-17 Converse: If a trapezoid has congruent base angles, then it is an isosceles trapezoid.

Next, we will investigate the diagonals of an isosceles triangle. Recall, that the diagonals of a rectangle are congruent AND they bisect each other. The diagonals of an isosceles trapezoid are also congruent, but they do NOT bisect each other.

Isosceles Trapezoid Diagonals Theorem: The diagonals of an isosceles trapezoid are congruent.

Example 3: Show $TA = RP$.

Solution: This is an example of a coordinate proof. Here, we will use the distance formula to show that $TA = RP$, but with letters instead of numbers for the coordinates.

$TA & = \sqrt{(x-d)^2+(0-y)^2} && RP = \sqrt{(x-d-0)^2+(y-0)^2}\\& = \sqrt{(x-d)^2+(-y)^2} && \quad \ \ = \sqrt{(x-d)^2+y^2}\\& = \sqrt{(x-d)^2+y^2}$

Notice that we end up with the same thing for both diagonals. This means that the diagonals are equal and we have proved the theorem.

## Midsegment of a Trapezoid

Midsegment (of a trapezoid): A line segment that connects the midpoints of the non-parallel sides.

There is only one midsegment in a trapezoid. It will be parallel to the bases because it is located halfway between them. Similar to the midsegment in a triangle, where it is half the length of the side it is parallel to, the midsegment of a trapezoid also has a link to the bases.

Investigation 6-5: Midsegment Property

Tools Needed: graph paper, pencil, ruler

1. Draw a trapezoid on your graph paper with vertices $A(-1, 5), \ B( 2, 5), \ C(6, 1)$ and $D(-3, 1)$. Notice this is NOT an isosceles trapezoid.
2. Find the midpoint of the non-parallel sides either by using slopes or the midpoint formula. Label them $E$ and $F$. Connect the midpoints to create the midsegment.
3. Find the lengths of $AB, \ EF$, and $CD$. Can you write a formula to find the midsegment?

Midsegment Theorem: The length of the midsegment of a trapezoid is the average of the lengths of the bases, or $EF=\frac{AB+CD}{2}$.

Example 4: Algebra Connection Find $x$. All figures are trapezoids with the midsegment.

Solution:

a) $x$ is the average of 12 and 26. $\frac{12+26}{2}=\frac{38}{2}=19$

b) 24 is the average of $x$ and 35.

$\frac{x+35}{2}&=24\\x+35&=48\\x&=13$

c) 20 is the average of $5x-15$ and $2x-8$.

$\frac{5x-15+2x-8}{2}&=20\\7x-23& =40\\7x&=63\\x&=9$

## Kites

The last quadrilateral we will study is a kite. Like you might think, it looks like a traditional kite that is flown in the air.

A few examples:

From the definition, a kite is the only quadrilateral that we have discussed that could be concave, as with the case of the last kite. If a kite is concave, it is called a dart.

The angles between the congruent sides are called vertex angles. The otherangles are called non-vertex angles. If we draw the diagonal through the vertex angles, we would have two congruent triangles.

Given: $KITE$ with $\overline{KE} \cong \overline{TE}$ and $\overline{KI} \cong \overline{TI}$

Prove: $\angle K \cong \angle T$

Statement Reason
1. $\overline{KE} \cong \overline{TE}$ and $\overline{KI} \cong \overline{TI}$ Given
2. $\overline{EI} \cong \overline{EI}$ Reflexive PoC
3. $\triangle EKI \cong \triangle ETI$ SSS
4. $\angle K \cong \angle T$ CPCTC

Theorem 6-21: The non-vertex angles of a kite are congruent.

Theorem 6-22: The diagonal through the vertex angles is the angle bisector for both angles.

The proof of Theorem 6-22 is very similar to the proof above for Theorem 6-21. If we draw in the other diagonal in $KITE$ we find that the two diagonals are perpendicular.

Kite Diagonals Theorem: The diagonals of a kite are perpendicular.

To prove that the diagonals are perpendicular, look at $\triangle KET$ and $\triangle KIT$. Both of these triangles are isosceles triangles, which means $\overline{EI}$ is the perpendicular bisector of $\overline{KT}$ (the Isosceles Triangle Theorem, Chapter 4). Use this information to help you prove the diagonals are perpendicular in the review questions.

Example 5: Find the other two angle measures in the kites below.

a)

b)

Solution:

a) The two angles left are the non-vertex angles, which are congruent.

$130^\circ + 60^\circ + x + x & = 360^\circ\\2x & = 170^\circ\\x & = 85^\circ \qquad \quad \text{Both angles are}\ 85^\circ.$

b) The other non-vertex angle is also $94^\circ$. To find the fourth angle, subtract the other three angles from $360^\circ$.

$90^\circ + 94^\circ + 94^\circ + x & = 360^\circ\\x & = 82^\circ$

Be careful with the definition of a kite. The congruent pairs are distinct. This means that a rhombus and square cannot be a kite.

Example 6: Use the Pythagorean Theorem to find the length of the sides of the kite.

Solution: Recall that the Pythagorean Theorem is $a^2+b^2=c^2$, where $c$ is the hypotenuse. In this kite, the sides are all hypotenuses.

$6^2+5^2& =h^2 && \ 12^2+5^2=j^2\\36+25& =h^2 && 144+25=j^2\\61& =h^2 && \qquad 169=j^2\\\sqrt{61} & =h && \qquad \ 13=j$

## Kites and Trapezoids in the Coordinate Plane

Example 7: Determine what type of quadrilateral $RSTV$ is. Simplify all radicals.

Solution: There are two directions you could take here. First, you could determine if the diagonals bisect each other. If they do, then it is a parallelogram and you could proceed like the previous section. Or, you could find the lengths of all the sides. Let’s do this option.

$RS & = \sqrt{(-5-2)^2+(7-6)^2} && ST=\sqrt{(2-5)^2+(6-(-3))^2}\\& = \sqrt{(-7)^2+1^2} && \quad \ =\sqrt{(-3)^2+9^2}\\& = \sqrt{50}=5\sqrt{2} && \quad \ = \sqrt{90}=3\sqrt{10}$

$RV& =\sqrt{(-5-(-4))^2+(7-0)^2} && VT=\sqrt{(-4-5)^2+(0-(-3))^2}\\& = \sqrt{(-1)^2+7^2} && \quad \ =\sqrt{(-9)^2+3^2}\\& = \sqrt{50}=5\sqrt{2} && \quad \ =\sqrt{90}=3\sqrt{10}$

From this we see that the adjacent sides are congruent. Therefore, $RSTV$ is a kite.

Algebra Review: From now on, this text will ask you to “simplify the radical.” From Algebra, this means that you pull all square numbers (1, 4, 9, 16, 25, ...) out of the radical. Above $\sqrt{50}=\sqrt{25 \cdot 2}$. We know $\sqrt{25}=5$, so $\sqrt{50}=\sqrt{25 \cdot 2}=5\sqrt{2}$.

Hint: If you are only given a set of points when determining what type of quadrilateral a figure is, always plot the points and graph. The visual will help you decide which direction to go.

Example 8: Determine what type of quadrilateral $ABCD$ is. $A(-3, 3), \ B(1, 5), \ C(4, -1), \ D(1, -5)$. Simplify all radicals.

Solution: First, graph $ABCD$. This will make it easier to figure out what type of quadrilateral it is. From the graph, we can tell this is not a parallelogram. Find the slopes of $\overline{BC}$ and $\overline{AD}$ to see if they are parallel.

Slope of $\overline{BC}=\frac{5-(-1)}{1-4}=\frac{6}{-3}=-2$

Slope of $\overline{AD}=\frac{3-(-5)}{-3-1}=\frac{8}{-4}=-2$

We now know $\overline{BC} \ || \ \overline{AD}$. This is a trapezoid. To determine if it is an isosceles trapezoid, find $AB$ and $CD$.

$AB & =\sqrt{(-3-1)^2+(3-5)^2} && ST = \sqrt{(4-1)^2+(-1-(-5))^2}\\& = \sqrt{(-4)^2+(-2)^2} && \quad \ \ = \sqrt{3^2+4^2}\\& = \sqrt{20}=2\sqrt{5} && \quad \ \ = \sqrt{25}=5$

$AB \neq CD$, therefore this is only a trapezoid.

Example 9: Determine what type of quadrilateral $EFGH$ is.

$E(5, -1), \ F(11, -3), \ G(5, -5), \ H(-1, -3)$

Solution: To contrast with Example 8, we will not graph this example. Let’s find the length of all four sides.

$EF & = \sqrt{(5-11)^2+(-1-(-3))^2} && FG = \sqrt{(11-5)^2+(-3-(-5))^2}\\& = \sqrt{(-6)^2+2^2} && \quad \ = \sqrt{6^2+2^2}\\& = \sqrt{40}=2\sqrt{10} && \quad \ =\sqrt{40}=2\sqrt{10}$

$GH & = \sqrt{(5-(-1))^2+(-5-(-3))^2} && HE = \sqrt{(-1-5)^2+(-3-(-1))^2}\\& = \sqrt{6^2+(-2)^2} && \quad \ = \sqrt{(-6)^2+(-2)^2}\\& = \sqrt{40}=2\sqrt{10} && \quad \ =\sqrt{40}=2\sqrt{10}$

All four sides are equal. That means, this quadrilateral is either a rhombus or a square. The difference between the two is that a square has four $90^\circ$ angles and congruent diagonals. Let’s find the length of the diagonals.

$EG & = \sqrt{(5-5)^2+(-1-(-5))^2} && FH = \sqrt{(11-(-1))^2+(-3-(-3))^2}\\& = \sqrt{0^2+4^2} && \quad \ \ = \sqrt{12^2+0^2}\\& = \sqrt{16}=4 && \quad \ \ = \sqrt{144}=12$

The diagonals are not congruent, so $EFGH$ is a rhombus.

Know What? Revisited If the diagonals (pieces of wood) are 36 inches and 54 inches, $x$ is half of 36, or 18 inches. Then, $2x$ is 36. To determine how large a piece of canvas to get, find the length of each side of the kite using the Pythagorean Theorem.

$18^2+18^2& =s^2 && 18^2+36^2=t^2\\324& =s^2 && \qquad 1620=t^2\\18\sqrt{2} & \approx 25.5 \approx s && \quad \ \ 18\sqrt{5} \approx 40.25 \approx t$

The perimeter of the kite would be $25.5 + 25.5 + 40.25 + 40.25 = 131.5$ inches or 11 ft, 10.5 in.

## Review Questions

1. $TRAP$ an isosceles trapezoid. Find:
1. $m \angle TPA$
2. $m \angle PTR$
3. $m \angle ZRA$
4. $m \angle PZA$
2. $KITE$ is a kite. Find:
1. $m \angle ETS$
2. $m \angle KIT$
3. $m \angle IST$
4. $m \angle SIT$
5. $m \angle ETI$
3. Writing Can the parallel sides of a trapezoid be congruent? Why or why not?
4. Writing Besides a kite and a rhombus, can you find another quadrilateral with perpendicular diagonals? Explain and draw a picture.
5. Writing Describe how you would draw or construct a kite.

For questions 6-11, find the length of the midsegment or missing side.

Algebra Connection For questions 12-17, find the value of the missing variable(s). Simplify all radicals.

For questions 18-21, determine what type of quadrilateral $ABCD$ is. $ABCD$ could be any quadrilateral that we have learned in this chapter. If it is none of these, write none.

1. $A(1, -2), \ B(7, -5), \ C(4, -8), \ D(-2, -5)$
2. $A(6, 6), \ B(10, 8), \ C(12, 4), \ D(8, 2)$
3. $A(-1, 8), \ B(1, 4), \ C(-5, -4), \ D(-5, 6)$
4. $A(5, -1), \ B(9, -4), \ C(6, -10), \ D(3, -5)$
5. $A(-2, 2), \ B(0, 1), \ C(2, 2), \ D(1, 5)$
6. $A(-7, 4), \ B(-4, 4), \ C(0, 0), \ D(0, -3)$
7. $A(3, 3), \ B(5, -1), \ C(7, 0), \ D(5, 4)$
8. $A(-4, 4), \ B(-1, 2), \ C(2, 4), \ D(-1, 6)$
9. Write a two-column proof of Theorem 6-22. Given: $\overline{KE} \cong \overline{TE}$ and $\overline{KI} \cong \overline{TI}$ Prove: $\overline{EI}$ is the angle bisector of $\angle KET$ and $\angle KIT$
10. Write a two-column proof of the Kite Diagonal Theorem. Given: $\overline{EK} \cong \overline{ET}, \ \overline{KI} \cong \overline{IT}$ Prove: $\overline{KT} \perp \overline{EI}$ $^\star$ Use the hint given earlier in this section.
11. Write a two-column proof of the Isosceles Trapezoid Diagonals Theorem using congruent triangles. Given: $TRAP$ is an isosceles trapezoid with $\overline{TR} \ || \ \overline{AP}$. Prove: $\overline{TA} \cong \overline{RP}$
12. Explain why the segments connecting the midpoints of the consecutive sides in a kite will always form a rectangle.
13. Explain why the segments connecting the midpoints of the consecutive sides in an isosceles trapezoid will always form a rhombus.

Feb 23, 2012

Dec 23, 2014


text:
# embedding text into vector spaces

Tags: ml, articles

• Word embedding is an nlp technique to map text or words to vectors

• Allows vector space operations to happen, such as summing or computing the distances of vectors

• Once words are generated into individual vectors, combine them into text vectors (aka document or sentence vecotrs)

• Easy way typically sums or averages the vectors together
• Two snippets of text are discovered by mapping both of them into vector space and finding the distances between the vectors

• Typically uses the angular distance
• nearest neighbor can be used

• high dimensional vectors of word embeddings typically break down
• approximate nearest neighbor must be used (ANN)
• assuming 4 billion 200-dimensional query vectors


text:
# Sparsest similar matrix

Given a square matrix A (say with complex entries), which is the sparsest matrix which is similar to A?

I guess it has to be its Jordan normal form but I am not sure.

Remarks:

• A matrix is sparser than other if it has less nonzero entries.

• Two square $$n \times n$$ matrices $$A,C$$ are similar if there exists and invertible matrix $$P$$ such that $$A = P^{-1}CP$$

• If the matrix is diagonalisable, clearly its Jordan form is the sparsest one in its similarity class because you cannot get fewer than $\operatorname{rank}(A)$ entries. – user1551 Oct 19 at 11:32
• My intuition says the same, but apart from trivial cases ($A$ is diagonaliseable) I do not see an immediate proof of that. – TZakrevskiy Oct 19 at 11:32
• Simulposted to MO, mathoverflow.net/questions/313224/sparsest-similar-matrix DON'T DO THAT! – Gerry Myerson Oct 19 at 12:00

## 1 Answer

The companion matrix to the polynomial $$(x^2-1)^2=1-2x^2+x^4$$ is $$\pmatrix{0&0&0&-1\cr1&0&0&0\cr0&1&0&2\cr0&0&1&0\cr}$$ which has Jordan form $$\pmatrix{1&1&0&0\cr0&1&0&0\cr0&0&-1&1\cr0&0&0&-1\cr}$$ which has more nonzero entries.

• The following paper claims that the companion matrix is the sparsest in a certain sense: doi.org/10.1016/j.laa.2012.08.017 – Julian Kuelshammer Oct 19 at 13:24
• That is an interesting example. But if we call the first matrix $A$ and the second one $C$, to me it seems (motivated by Jordan normal form definition) among all matrices that are NOT equal to $A$, $C$ is the sparsest similar matrix to $A$. Can your example be generalized to prove my statement is incorrect? – abolfazl Oct 19 at 16:41
• You can always get a different but equally sparse matrix by conjugating by some diagonal and/or permutation matrix . . . – Noam D. Elkies Oct 19 at 18:34


text:
5

# Hepatitis â‚¬No Hepatitis CROW TOTALTattoo(s)25113No tattoo491513Column Total579626...

## Question

###### Hepatitis â‚¬No Hepatitis CROW TOTALTattoo(s)25113No tattoo491513Column Total579626

Hepatitis â‚¬ No Hepatitis C ROW TOTAL Tattoo(s) 25 113 No tattoo 491 513 Column Total 579 626

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text:
## Find points when tangent to ellipse is horizontal

I have this question...

[A] The cone with equation $z^2=x^2+y^2$ and the plane with equation $2x+3y+4z+2=0$ intersect in an ellipse. Write an equation of the plane normal to this ellipse at the point P(3, 4, -5).

[B] It is apparent from geometry that the highest and lowest points of the ellipse from [A] are those points where its tangent line is horizontal. Find those points.

For part [A], I took the gradient vectors of the two functions and used their cross product to find the ellipse $x-2y+z+10=0$. I know this answer is correct, but I don't have an answer to part B to check against. Any insight for part B would be greatly appreciated!

Thanks!


text:
# What is the relationship between ketone and enol?

Nov 28, 2016

An enol derivative, typically, $R C \left(- O H\right) = C {H}_{2}$, derives from a carbonyl species. It may be deprotonated to give the $\text{enolate}$, $R C \left(- {O}^{-}\right) = C {H}_{2}$.

#### Explanation:

Carbonyl compounds typically undergo enol/ketone tautomerism, i.e. for acetone we can write:

${H}_{3} C - C \left(= O\right) C {H}_{3} r i g h t \le f t h a r p \infty n s {H}_{2} C = C \left(- O H\right) C {H}_{3}$

This equilibrium lies strongly to the left, but it may be manipulated.

The alcohol, $\text{the enol}$, contains a protic hydrogen, that can be potentially deprotonated to give an enolate, the anion:

${H}_{2} C = C \left(- O H\right) C {H}_{3} + \text{Base"rightleftharpoonsH_2C=C(-O^(-))CH_3+"Base"""^(-)H""^(+)}$

The enolate anion itself is conceived to be in equilibrium with the carbanion:

${H}_{2} C = C \left(- {O}^{-}\right) C {H}_{3} r i g h t \le f t h a r p \infty n s {H}_{2}^{-} C - C \left(= O\right) C {H}_{3}$.

This $\text{carbanion/enolate}$ equilibrium is the classic strategy employed in $C - C$ bond formation. Carbanions may be formed from malonic acid derivatives, and ethylacetoacetate. For more details consult your organic text.


text:
# When was it found that a function could be decomposed as a sum of even and odd functions?

I'm not sure if it's a good question but I was reading about the decomposition of any function, f(x), as a sum of even and odd functions; f(x) is not an even or odd function.

Is it possible to know when this fact (or, theorem?) was discovered and who found it?

• I imagine most any of the "greats" (Darboux, Baire, Lebesgue, Borel, Hilbert, etc.) who was faced with situations in which this was useful (e.g. maybe early Fourier series work in which decompositions into an even part and and odd part was used) were able to quickly recognize this (if not express it the way we would today), but how far back one can find a reasonably explicit record of such a decomposition does seem somewhat interesting. I can imagine this being "widely known" over a number of years, and maybe occasionally appearing in 1800s literature such as (continued) – Dave L Renfro Aug 20 at 10:37
• The Educational Times (had a widely participated in section on math problems and such), Nouvelles Annales de Mathématiques, Zeitschrift für Mathematik und Physik, Cambridge Mathematical Journal, L'Intermédiaire des Mathématiciens. – Dave L Renfro Aug 20 at 10:54
• Thank you, @DaveLRenfro. You referenced Fourier series which appeared for the first time in 1807. Wouldn't Fourier be familiar with such a decomposition in one way or another? Most of the 'greats' you mentioned were born after 1850. So the question still remains: How far back one can find a reasonably explicit record of such a decomposition? – PG1995 Aug 20 at 21:15

It depends partly on what "function" means. According to a definition prevalent before Dirichlet it meant something expressed by a formula and expandable into a power series. The decomposition claim then is obvious, as long as one introduced the notions of even and odd and related them to even and odd powers. Euler does both in Traiectoriarum Reciprocarum Solutio (1727), XIX-XXI (he uses Latin words pares and impares). Luckily, this is one of the papers translated by Ian Bruce:

"§17. In the first place functions are to be noted that I call even, of which this is the property: that they remain unchanged if $$- x$$ is put in place of $$x$$. All the powers of $$x$$ with even exponents are of this kind, and of fractions with even numerators and odd denominators. Then, any such functions of powers of this kind composed either by addition or subtraction, or by multiplication or division, or hence any such construction raised to sum are likewise even powers, such as $$x^{\frac45}$$, $$(ax^2+bx^{\frac23})^n$$.

§18. In the second place, I take heed of odd functions, which in short produce the negative of these, if $$x$$ is changed to $$-x$$. All the powers of x with odd exponents, such as $$x$$ itself, $$x^3$$, $$x^5$$, etc. are of this kind; or fractions of which the numerators and denominators are odd; also functions composed of these powers either added or subtracted, and which are also dignified to be raised to odd exponents, such as $$x^{\frac45}$$, $$(ax^3+bx^{\frac57})^3$$."

Euler did not have much use for them, or decomposition into them, although, as Barnett notes:

"In fact, the expressions $$(e^x + e^{-x})/2$$ and $$(e^x - e^{-x})/2$$ do make an appearance in Volume I of Euler's Introductio in analysin infinitorum (1745, 1748). Euler's interest in these expressions seems natural in view of the equations $$\cos x=(e^{\sqrt{-1}x} + e^{-\sqrt{-1}x})/2$$ and $$\sqrt{-1}\sin x=(e^{\sqrt{-1}x} + e^{-\sqrt{-1}x})/2$$ that he derived in this text. However, Euler's interest in what we call hyperbolic functions appears to have been limited to their role in deriving infinite product representations for the sine and cosine functions. Euler did not use the word hyperbolic in reference to the expressions... nor did he provide any special notation or name for them.

In a Hacker News discussion Decomposing a function into its even and odd parts David Goldblatt suggested that the idea might go back to earlier times:

"I don't have any references for you right now, but I would be surprised if the usage of the terms doesn't pre-date your citation of Euler by at least several decades. The basic technique for deriving Taylor series is known to date to at least 1671, to the correspondence between James Gregory and John Collins; it also shows up in letters from de Moivre to Johann Bernoulli in 1694 and from Leibniz to Bernoulli in 1708. Newton had a geometric means of deriving the coefficients of power series when he was writing the Principa (published in 1687), as demonstrated by the proof of his tenth proposition, and he included a description of the algebraic technique in an early draft of his Quadrature of Curves (but removed it before publication in 1706)... Perhaps nobody bothered to give the concept a name until Euler in 1727, and perhaps nobody even considered the concept with regard to a more general class of functions until even later".

• Thank you very much! So, it wouldn't be wrong to conclude that the idea of decomposition was around during the time of Euler (1707-1783), and possibly it could pre-date it. It'd also suggest that Fourier should be aware of it before he came up with his Fourier series in 1807. – PG1995 Aug 21 at 2:14
• @PG1995 Miller's Earliest Known Uses of Some of the Words of Mathematics mentions that "even function is found in English in 1809 in a translation of Legendre’s 1792 paper “Mémoire Sur Les Transcendantes Elliptiques”", and Lagrange's Théorie des fonctions analytiques (1797) promoted the power series notion of a function. So it is likely that Fourier was aware of it, although one would have to search his works to be sure, these things do not always happen logically. – Conifold Aug 21 at 3:00
• Thanks a lot! I really appreciate your help. – PG1995 Aug 21 at 3:12


text:
# Change the orientation of Arrow of xtwocell in the 2cell option for xy

I'm trying to make a diagram similar to this (from wikipedia)

Except that I want to the double arrow $\eta$ two be pointing from node $B$ toward the middle of arrow $X$ (it is, after all, a natural transformation from $RF$ to $X$).

I'm trying to use the \xtwocell command of the 2cell package for xy. In subsection 3 (Excursions) of section 25.4 (extra options) of chapter 25 (two-cell) on page 47 of the xyrefer manual, it is stated

\xtwocell[⟨hop⟩]{⟨displace⟩}... connects to the ⟨pos⟩ displaced by ⟨displace⟩ from the relative cell location specified by ⟨hop⟩. The displacement can be any string of valid XY-pic commands, but they must be enclosed within a group {...}. When the cell location is the target, a null grouping {} must be given. When used with the <\omit> nudge, such excursions allow a labelled Arrow to be placed anywhere within an XY-pic diagram; furthermore the Arrow can be oriented to point in any direction.

Of course the manual gives no examples and makes no further comment about how to change the orientation of the Arrow (in this manual, capitalized "Arrow" refers to the double arrow between arrows, i.e. the 2-cell).

What does he mean by "any string of valid XY-pic commands"? What may the displacement argument for \xtwocell be? Any experiments I tried (eg <1>,dr,[0,1]) typing in the curly braces just ended up in the $A$ node text, so not being interpreted as a displacement or position or orientation. How do I change the orientation of the Arrow? How do I achieve the "orientation in any direction" that was promised me?

Here is my simple markup

\xymatrix{
A\ar[d]\ar[r]\xtwocell[0,1]{}\omit{<2>} & B\\
C\ar@{.>}[ur] &
}

which gives the following image

I want to change the direction that double arrow is pointing, point it to the corner object $C$, or reverse it. I can shift it by changing the number in \omit{<2>}, but how to change the orientation?

(Google showed me that the same question is asked here on the xy mailing list, but never answered)

See also my question at Label the source/target object in the middle of compositemaps from 2cell package of xy? for another attempted method at this diagram.

-

If you would consider using tikz-cd, this can easily be achieved with

• a named node (name=X) and
• the to node styles (see below).

The argument of \arrow that usually denotes the direction/target node will be ignored.

For a shorter arrow I’d use the short style from my answer to your other question Label the source/target object in the middle of compositemaps from 2cell package of xy?

You can use the keys short shift >, short shift < and short center to adjust the arrow.

## Code A

\documentclass[tikz,convert=false]{standalone}
\usepackage{tikz-cd}
\pgfkeys{
/tikz/commutative diagrams/to node/.style={%
/tikz/execute at begin to=\edef\tikztotarget{#1}},
/tikz/commutative diagrams/to node*/.style={
/tikz/commutative diagrams/to node={#1},
/tikz/shorten >=\pgfkeysvalueof{/tikz/commutative diagrams/to node shorten >}},
/tikz/commutative diagrams/to node shorten >/.initial=+.5ex}
\begin{document}
\begin{tikzcd}[column sep=large]
A \rar[][name=X]{X}
\dar[swap]{F}
& B \\
C \urar[dashed][swap]{R}
\arrow[Rightarrow, to node*=X]{}{\eta}
\end{tikzcd}
\end{document}

## Code B

\documentclass[tikz,convert]{standalone}
\usetikzlibrary{decorations.pathreplacing}
\usepackage{tikz-cd}
\newcommand*{\tikzcdset}{\pgfqkeys{/tikz/commutative diagrams}}
\tikzcdset{
to node/.style={%
/tikz/execute at begin to=\edef\tikztotarget{#1}}}
\makeatletter
\tikzcdset{
short shift >/.initial=+.6em,
short shift </.initial=+.6em,
short center/.initial=.5,
short/.style={
to path={
\pgfextra
\pgfinterruptpath
\path (\tikztostart) coordinate (@aux1) -- (\tikztotarget) coordinate (@aux2);
\endpgfinterruptpath
\pgfpointdiff{\pgfpointanchor{@aux1}{center}}{\pgfpointanchor{@aux2}{center}}%
\pgfmathveclen@{\pgfmath@tonumber{\pgf@x}}{\pgfmath@tonumber{\pgf@y}}%
\edef\pgfpathlength{\pgfmathresult pt}%
\pgfcoordinate
{@@aux1}
{\pgfpointlineatdistance
{(\pgfkeysvalueof{/tikz/commutative diagrams/short center})*\pgfpathlength-\pgfkeysvalueof{/tikz/commutative diagrams/short shift <}}
{\pgfpointanchor{@aux1}{center}}
{\pgfpointanchor{@aux2}{center}}}%
\pgfcoordinate
{@@aux2}
{\pgfpointlineatdistance
{(\pgfkeysvalueof{/tikz/commutative diagrams/short center})*\pgfpathlength+\pgfkeysvalueof{/tikz/commutative diagrams/short shift >}}
{\pgfpointanchor{@aux1}{center}}
{\pgfpointanchor{@aux2}{center}}}%
\endpgfextra
(@@aux1) -- (@@aux2) \tikztonodes}}
}
\makeatother
\begin{document}
\begin{tikzcd}[column sep=large]
A \rar[][name=X]{X}
\dar[swap]{F}
& B \\
C \urar[dashed][swap]{R}
\arrow[Rightarrow, to node=X, short]{}{\eta}
\end{tikzcd}
\end{document}

## Output B

-
Yes, I would consider using this package. Is it possible to make the double arrow shorter? Seeing a 2cell span the diagram like that is a little jarring. –  Joe Hannon Aug 22 '13 at 5:48
@JoeHannon Could you elaborate on that? You can shorten the arrow at both ends with the keys shorten > and shorten < as shown in the example. There are also ways to shorten the line by a certain percentage of the path length in various different ways. It is also possible to place a normalsized \Rightarrow between C and X similar to the small arrow in your linked question. Feel free to edit your question with a mock-up of the image you want to achieve. –  Qrrbrbirlbel Aug 22 '13 at 6:32
In your diagram, it seems like the doublearrow spans from the lower-left corner to your invisible node. But typically we want 2cells to be shorter, hovering somewhere in the center of our commutative diagram. Sorry I can't be more specific, but I can say that the 2-cells displayed by the 2cell package are the right size (not surprising since 2cell was invented by a cat theorist for his book). If I can get a 2-cell of the size given by 2cell, but with the direction given by your tikz-cd example, then I will be happy. I did not understand your instructions about the shorten > keys, though. –  Joe Hannon Aug 22 '13 at 6:59
@JoeHannon I am not familiar with commutative diagrams or their vocabulary. Do you want something like this? –  Qrrbrbirlbel Aug 22 '13 at 8:46
Yes, that arrow does look shorter, and does not span the diagram. –  Joe Hannon Aug 22 '13 at 19:54


text:
# Uniform bounds for preperiodic points in families of twists

@inproceedings{Levy2012UniformBF,
title={Uniform bounds for preperiodic points in families of twists},
author={Alon Y. Levy and Michelle Manes and Bianca Thompson},
year={2012}
}
• Published 19 April 2012
• Mathematics
Letbe a morphism of PN defined over a number field K. We prove that there is a bound B depending only onsuch that every twist of � has no more than B K-rational preperiodic points. (This result is analagous to a result of Silverman for abelian varieties (10).) For two specific families of quadratic rational maps over Q, we find the bound B explicitly.
9 Citations

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## References

SHOWING 1-10 OF 17 REFERENCES

We present a proof, which is conditional on the Birch and Swinnerton-Dyer Conjecture for a specific abelian variety, that there do not exist rational numbers x and c such that x has exact period N =
The theory of moduli of morphisms on P^n generalizes the study of rational maps on P^1. This paper proves three results about the space of morphisms on P^n of degree d > 1, and its quotient by the
Let ϕ:ℙ1 → ℙ1 be a rational map of degree d = 2 defined over ℚ and assume that f−1°ϕ° f = ϕ for exactly one nontrivial f ε PGL2 (ℚ−). We describe families of such maps that have ℚ‐rational periodic
• Mathematics
• 1995
It has been conjectured that for N sufficiently large, there are no quadratic polynomials in Q[z] with rational periodic points of period N. Morton proved there were none with N=4, by showing that
In this note we discuss some arithmetic and geometric questions concerning self maps of projective algebraic varieties.
where the Li(x) are homogeneous polynomials of degree 1 with coefficients which are algebraic numbers. One of the most interesting cases arises when the following two conditions are satisfied. (1)
Corollaire. — Soit d un entier ≥ 1 . Il existe un nombre réel B(d) tel que pour toute courbe elliptique E , définie sur un corps de nombres K de degré d sur Q , tout point de torsion de E(K) soit
* Provides an entry for graduate students into an active field of research * Each chapter includes exercises, examples, and figures * Will become a standard reference for researchers in the field


text:
# Quick question about gravity/time dilation

1. Dec 10, 2009

### Mzachman

I'm thinking about something, but I need to know this first...

Is the time dilation from gravity caused by a net gravitational force, or by the total magnitudes of the gravities added up?

For example, would a clock floating directly between two identical planets have time pass as if it were in open space, or as 2 x the difference in time that one planet at that distance from the clock would cause?

Thanks!

2. Dec 10, 2009

### Jolb

As far as I know, gravitational time dilation is only dependent on your relative gravitational potential. Therefore, if you're between two planets, the gravitational potential at you is more negative than some point at infinity. So if you were directly at the center of mass of two identical bodies of mass m in circular orbits, at a distance r from both bodies, then an observer at infinity would see your clock ticking slower than a clock orbiting a lone mass m at distance r, (ignoring angular velocities). However, I don't know enough GR to tell you whether it would be 2x slower--my intuitive guess is that it's between 1 and 2 times slower.

3. Dec 10, 2009

### drmichael

I believe you are correct Jolb. The observed time distortion is directly porportional to the position of the clock (assuming both gravitational bodies are identical in mass and distance from the clock). There is however a thoery being investigated that proposes that the rotational direction of the planets could play a factor in how they manifest their gravitational fields.

4. Dec 10, 2009

### bcrowell

Staff Emeritus
You're correct that the effect depends on the gravitational potential, and this is the point that's crucial in answering the OP's question. The potential is a scalar, not a vector, so there's no vector addition going on, and the effects don't cancel, they reinforce.

Basically the reason it can't depend on the field is that according to the equivalence principle, gravitational fields are observer-dependent.

In the limit of weak fields, the time dilation effect is simply proportional to the potential, and the potential adds linearly, so the result is that it's 2x slower. For stronger fields (e.g., if we put two black holes close together, rather than two planets), you would see deviations from this behavior. There are two reasons for this. One is that approximating time dilation as $\Delta\Phi$ is only an approximation to $e^{\Delta\Phi}$. The other is that general relativity is a nonlinear theory, so the potentials don't add linearly.

5. Dec 11, 2009

### Eric McClean

Gravitational time dilation is the effect of time passing at different rates in regions of different gravitational potential; the lower the gravitational potential (closer to the center of a massive object), the more slowly clocks run. Albert Einstein originally predicted this effect in his theory of relativity and it has since been confirmed by tests of general relativity.

This has been demonstrated by noting that atomic clocks at differing altitudes (and thus different gravitational potential) will eventually show different times. The effects detected in such experiments are extremely small, with differences being measured in nanoseconds.

Gravitational time dilation was first described by Albert Einstein in 1907[1] as a consequence of special relativity in accelerated frames of reference. In general relativity, it is considered to be a difference in the passage of proper time at different positions as described by a metric tensor of spacetime. The existence of gravitational time dilation was first confirmed directly by the Pound-Rebka experiment.

6. Dec 14, 2009

### Mzachman

Ok, that's what I was wondering. I was guessing that was the way it worked, but i wasn't sure.

Now, assuming that gravity "travels" at the speed of light, and that time dilation is based on gravitational potential, and not the net vector for gravity..... what are the implications of this for when the gravity from more and more of the tons of objects in the universe "reach" us? Will our clocks be constantly slowing down as time goes on? This seems to suggest some sort of weird thing where clocks will go infinitely slower and slower as time goes forwards and more and more gravitational potential "reaches" us, although we would not be able to sense this, correct? I'm not really sure what this would mean, but do you have any thoughts on it?

That's what I was originally thinking about, but I had to clarify that first point before I could discuss what I was actually thinking about.

7. Dec 15, 2009

### Al68

No. I think the key thing with gravitational time dilation is that it's not really a different phenomena than SR (velocity-based) time dilation, and it is manifested in accelerated reference frames whether a gravitational field is present or not. The role played by a gravitational field in this sense is that the only way for two clocks to be "stationary" in a gravitational field is if they are accelerated together in an accelerated reference frame, like earth's surface.

Consider this: Say there are two clocks, one at the top and bottom of a tall tower. From an inertial reference frame (freefall), at any specified time the clock at the bottom has a higher velocity than the clock at the top, and will therefore run slower relative to the top clock. From the accelerated frame in which the clocks are stationary, that's gravitational time dilation.

The same would be the case if the clocks were at the top and bottom of an accelerating rocket in deep space instead of a tower on earth.

8. Dec 15, 2009

### bcrowell

Staff Emeritus
You're correct that we would be unable to sense it. If all clocks in the universe were to slow down in lock-step, then there would be no way to detect it at all. Coordinates in general relativity, such as time, are completely arbitrary. You can do any smooth, one-to-one coordinate transformation at all, and the Einstein field equations are still valid. You get a description of all the same phenomena, with no empirically testable difference.

9. Dec 16, 2009

### Mzachman

That would suggest that it's based on the vector type net acceleration of gravity. Unless those can somehow add, so that in the two planet example we would be "accelerating" both directions and once, and therefore feel no "force." Hmm...

Ok, so technically it is possible then.... It just got me thinking about the "beginning of time" and all that... It's just a strange thought to me that gravity "slows down" time, but if everywhere has more gravity applied to it (and therefore should "slow down" everywhere), we just can't apply the same idea as if there was a difference in gravity between two spots causing the difference in time. So technically, to someone on the "outside" of our universe, gravity free, if they could somehow look in they would see us slowing down and slowing down infinitely slower and time progresses, but we're just going on with our daily lives happy as ever and unknowing. That's an interesting thought.

One last question. This may be dumb since I know very little about it. I just read QED by Feynman, and he talked about anti-matter being matter going backwards through time. I'm just curious, does this cause any problems with Einstein's relativity stuff?

Thanks for all the help guys, this is intriguing stuff (especially for a physicist to be.... :) ).


text:
# Show the group isomorphism $(\mathbb{Z}/n)^\times \cong (\mathbb{Z}/p_1^{k_1})^\times \times \cdots \times (\mathbb{Z}/p_n^{k_n})^\times$

When $r$ and $s$ are relatively prime we have the ring isomorphism $\mathbb{Z}/rs \cong \mathbb{Z}/r \times \mathbb{Z}/s$

Given a prime factorization of $n$ where $n = p_1^{k_1} \cdots p_n^{k_n}$

Show that for groups of units invertible under multiplication:

$(\mathbb{Z}/n)^\times \cong (\mathbb{Z}/p_1^{k_1})^\times \times \cdots \times (\mathbb{Z}/p_n^{k_n})^\times$

• Hint: Since you know $\mathbb{Z}/rs\simeq\mathbb{Z}/r\times\mathbb{Z}/s$ as rings, then you know that the units of one map to units to the other. This shows $\mathbb{Z}/rs^\times\simeq\mathbb{Z}/r^\times\times\mathbb{Z}/s^\times$ (there is something small to be checked here, namely that the units of $(\mathbb{Z}/r\times\mathbb{Z}/s)$ are $\mathbb{Z}/r^\times\mathbb{Z}/s^\times$). Then, use induction. – Michael Burr May 12 '15 at 20:52
• Thank you! Can you get me started on the induction proof of that last point. Thanks! – clay May 12 '15 at 20:55
• Induct on the number of distinct prime factors. The case where there is one distinct prime factor is easy ... – Michael Burr May 12 '15 at 21:07
• In the inductive step I need to show $\left(\mathbb{Z}/r \times \mathbb{Z}/s\right)^\times = \left(\mathbb{Z}/r\right)^\times \times \left(\mathbb{Z}/s\right)^\times$ for relatively prime $r$ and $s$. How can I do that? – clay May 12 '15 at 22:49

$\mathbb{Z}/n^{\times} = \left(\mathbb{Z}/\prod_{i=1}^{j} p_i ^{k_j}\right)^{\times}= \bigg(\mathbb{Z}/\left(\prod_{i=1}^{j-1} p_i^{k_i}\right)\cdot p_j^{k_j}\bigg)^{\times} \cong \bigoplus_{i=1}^j \left(\mathbb{Z}/p_i^{k_i}\right)^{\times}$
The product of the $p_i^{k_i}$ for $1 \leq i \leq j-1$ is relatively prime with $p_j^{k_j}$.
The induction hypothesis would be to suppose it is true up to $j-1$, then we show above that up to $j$ resolves to the base case.


text:
# A mixture of $Al(OH)_3$ and $Fe(OH)_3$ can be separated by

$\begin{array}{1 1}(a)\;HCL&(b)\;NH_4OH\\(c)\;NaOH&(d)\;HNO_3\end{array}$

Excess of NaOH with aluminium hydroxide dissolves white ppt.Where as ppt of ferric hydroxide does not dissolve with NaOH.
Hence (c) is the correct answer.


text:
# Hamiltonian path

A Hamiltonian cycle around a network of six vertices

In the mathematical field of graph theory, a Hamiltonian path (or traceable path) is a path in an undirected or directed graph that visits each vertex exactly once. A Hamiltonian cycle (or Hamiltonian circuit) is a Hamiltonian path that is a cycle. Determining whether such paths and cycles exist in graphs is the Hamiltonian path problem, which is NP-complete.

Hamiltonian paths and cycles are named after William Rowan Hamilton who invented the icosian game, now also known as Hamilton's puzzle, which involves finding a Hamiltonian cycle in the edge graph of the dodecahedron. Hamilton solved this problem using the icosian calculus, an algebraic structure based on roots of unity with many similarities to the quaternions (also invented by Hamilton). This solution does not generalize to arbitrary graphs.

Despite being named after Hamilton, Hamiltonian cycles in polyhedra had also been studied a year earlier by Thomas Kirkman, who, in particular, gave an example of a polyhedron without Hamiltonian cycles.[1] Even earlier, Hamiltonian cycles and paths in the knight's graph of the chessboard, the knight's tour, had been studied in the 9th century in Indian mathematics by Rudrata, and around the same time in Islamic mathematics by al-Adli ar-Rumi. In 18th century Europe, knight's tours were published by Abraham de Moivre and Leonhard Euler.[2]

### Definitions

A Hamiltonian path or traceable path is a path that visits each vertex of the graph exactly once. A graph that contains a Hamiltonian path is called a traceable graph. A graph is Hamiltonian-connected if for every pair of vertices there is a Hamiltonian path between the two vertices.

A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a cycle that visits each vertex exactly once. A graph that contains a Hamiltonian cycle is called a Hamiltonian graph.

Similar notions may be defined for directed graphs, where each edge (arc) of a path or cycle can only be traced in a single direction (i.e., the vertices are connected with arrows and the edges traced "tail-to-head").

A Hamiltonian decomposition is an edge decomposition of a graph into Hamiltonian circuits.

A Hamilton maze is a type of logic puzzle in which the goal is to find the unique Hamiltonian cycle in a given graph.[3][4]

### Examples

Orthographic projections and Schlegel diagrams with Hamiltonian cycles of the vertices of the five platonic solids – only the octahedron has an Eulerian path or cycle, by extending its path with the dotted one

### Properties

The Herschel graph is the smallest possible polyhedral graph that does not have a Hamiltonian cycle. A possible Hamiltonian path is shown.

Any Hamiltonian cycle can be converted to a Hamiltonian path by removing one of its edges, but a Hamiltonian path can be extended to Hamiltonian cycle only if its endpoints are adjacent.

All Hamiltonian graphs are biconnected, but a biconnected graph need not be Hamiltonian (see, for example, the Petersen graph).[9]

An Eulerian graph G (a connected graph in which every vertex has even degree) necessarily has an Euler tour, a closed walk passing through each edge of G exactly once. This tour corresponds to a Hamiltonian cycle in the line graph L(G), so the line graph of every Eulerian graph is Hamiltonian. Line graphs may have other Hamiltonian cycles that do not correspond to Euler tours, and in particular the line graph L(G) of every Hamiltonian graph G is itself Hamiltonian, regardless of whether the graph G is Eulerian.[10]

A tournament (with more than two vertices) is Hamiltonian if and only if it is strongly connected.

The number of different Hamiltonian cycles in a complete undirected graph on n vertices is ${\displaystyle (n-1)!/2}$ and in a complete directed graph on n vertices is ${\displaystyle (n-1)!}$. These counts assume that cycles that are the same apart from their starting point are not counted separately.

### Bondy–Chvátal theorem

The best vertex degree characterization of Hamiltonian graphs was provided in 1972 by the BondyChvátal theorem, which generalizes earlier results by G. A. Dirac (1952) and Øystein Ore. Both Dirac's and Ore's theorems can also be derived from Pósa's theorem (1962). Hamiltonicity has been widely studied with relation to various parameters such as graph density, toughness, forbidden subgraphs and distance among other parameters.[11] Dirac and Ore's theorems basically state that a graph is Hamiltonian if it has enough edges.

The Bondy–Chvátal theorem operates on the closure cl(G) of a graph G with n vertices, obtained by repeatedly adding a new edge uv connecting a nonadjacent pair of vertices u and v with deg(v) + deg(u) ≥ n until no more pairs with this property can be found.

Bondy–Chvátal Theorem (1976). A graph is Hamiltonian if and only if its closure is Hamiltonian.

As complete graphs are Hamiltonian, all graphs whose closure is complete are Hamiltonian, which is the content of the following earlier theorems by Dirac and Ore.

Dirac's Theorem (1952). A simple graph with n vertices ( ${\displaystyle n\geq 3}$) is Hamiltonian if every vertex has degree ${\displaystyle {\tfrac {n}{2}}}$ or greater.
Ore's Theorem (1960). A simple graph with n vertices ( ${\displaystyle n\geq 3}$) is Hamiltonian if, for every pair of non-adjacent vertices, the sum of their degrees is n or greater.

The following theorems can be regarded as directed versions:

Ghouila-Houiri (1960). A strongly connected simple directed graph with n vertices is Hamiltonian if every vertex has a full degree greater than or equal to n.
Meyniel (1973). A strongly connected simple directed graph with n vertices is Hamiltonian if the sum of full degrees of every pair of distinct non-adjacent vertices is greater than or equal to ${\displaystyle 2n-1}$.

The number of vertices must be doubled because each undirected edge corresponds to two directed arcs and thus the degree of a vertex in the directed graph is twice the degree in the undirected graph.

Rahman–Kaykobad (2005). A simple graph with n vertices has a Hamiltonian path if, for every non-adjacent vertex pairs the sum of their degrees and their shortest path length is greater than n.[12]

The above theorem can only recognize the existence of a Hamiltonian path in a graph and not a Hamiltonian Cycle.

Many of these results have analogues for balanced bipartite graphs, in which the vertex degrees are compared to the number of vertices on a single side of the bipartition rather than the number of vertices in the whole graph.[13]

### Existence of Hamiltonian cycles in planar graphs

Theorem (Whitney, 1931). A 4-connected planar triangulation has a Hamiltonian cycle.[14]
Theorem (Tutte, 1956). A 4-connected planar graph has a Hamiltonian cycle.[15]

### The Hamiltonian cycle polynomial

An algebraic representation of the Hamiltonian cycles of a given weighted digraph (whose arcs are assigned weights from a certain ground field) is the Hamiltonian cycle polynomial of its weighted adjacency matrix defined as the sum of the products of the arc weights of the digraph's Hamiltonian cycles. This polynomial is not identically zero as a function in the arc weights if and only if the digraph is Hamiltonian. The relationship between the computational complexities of computing it and computing the permanent was shown by Grigoriy Kogan.[16]


text:
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### Polynomial Arithmetics

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Multiplication of Polynomials

»  Numerical expressions can be multiplied
the multiplication distributes over addition or subtraction

→  [2^2+3(2)+1] xx[2^3] =2^5+3(2^4)+2^3

→  [3^2+3] xx[3^3-5] =3^5+3^3-5(3^2)-15

»  Polynomial multiplication follows the numerical expression multiplication laws
→  [x^2+3(x)+1] xx[x^3] =x^5+3(x^4)+x^3

→  [y^2+3] xx[y^3-5] =y^5+y^3-5(y^2)-15

### Multiplication of Polynomials

plain and simple summary

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Polynomials as a whole or each of the sub expression can be considered as numbers.

Multiplication is performed by distribution of multiplication into addition or subtraction.

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What is 4xx5?

• 45
• 20

The answer is '20'.

Given the numbers 4 and 5 are integers, What is 4xx5?

• an integer
• not an integer

If any two numbers are multiplied or added the result is another number.

In an algebraic expression, any sub-expression can be handled as if the sub-expression is a number.

which of the following equals 4xx(3+2)?

• 4 xx 3 + 4 xx 2
• 4xx3 + 2

The answer is '4 xx 3 + 4 xx 2'

Note the following

•  4 xx color(deepskyblue)((3+2)) = 4 xx color(deepskyblue)5 = 20

•  color(coral)(4 xx 3) + color(deepskyblue)(4 xx 2) = color(coral)(12) + color(deepskyblue)8 = 20

This can be generalized for any variables

color(coral)x xx color(deepskyblue)((y+z)) = color(coral)x xx color(deepskyblue)y + color(coral)x xx color(deepskyblue)z.

This is an important result to work with polynomials. The variables x, y, z in this can be any sub-expression of a polynomial.

For example: if x=a^3+b^2 then

color(coral)x xx color(deepskyblue)((y+z)) = color(coral)x xx color(deepskyblue)y + color(coral)x xx color(deepskyblue)z

is equivalently

color(coral)((a^3+b^2)) xx color(deepskyblue)((y+z)) = color(coral)((a^3+b^2)) xx color(deepskyblue)y + color(coral)((a^3+b^2)) xx color(deepskyblue)z

Multiply x and x^2+3.

• x^3+3
• x^3+3x

The answer is 'x^3+3x'. The multiplication is carried out as follows.

Similar to 4 xx(3+2) =4xx3 + 4 xx2, the multiplication of terms of polynomial are carried out,
color(coral)(x) xx (color(deepskyblue)(x^2+3)) = color(coral)(x) xx color(deepskyblue)(x^2) + color(coral)(x) xx color(deepskyblue)(3).

To understand multiplication of polynomial, one such multiplication is illustrated.

(color(coral)(2x^2+3x+5))xx(color(deepskyblue)(x+7))

sub-expression (color(coral)(2x^2+3x+5)) is a number and holding it as a whole. Applying that number into x+7
quad =(color(coral)(2x^2+3x+5))xx color(deepskyblue)(x)
quad quad quad + (color(coral)(2x^2+3x+5)) xx color(deepskyblue)(7)

sub-expression (color(coral)(2x^2+3x)) as a number
quad =(color(coral)(2x^2+3x)) xx color(deepskyblue)(x) + color(coral)(5)xx color(deepskyblue)(x)
quad quad quad + (color(coral)(2x^2+3x)) xx color(deepskyblue)(7) +color(coral)(5) xx color(deepskyblue)(7)

quad =2x^3+3x^2+5x
quad quad quad + 14x^2+21x + 35

quad =2x^3+(3+14)x^2+(5+21)x+35

quad =2x^3+17x^2+26x+35

which of the following equals 4xx(3-2)?

• 4 xx 3 - 4 xx 2
• 4xx3 - 2

The answer is '4 xx 3 - 4 xx 2'

Note the following

•  4 xx color(deepskyblue)((3-2)) = 4 xx color(deepskyblue)1 = 4

•  color(coral)(4 xx 3) - color(deepskyblue)(4 xx 2) = color(coral)(12) - color(deepskyblue)8 = 4

This can be generalized for any variables

color(coral)x xx color(deepskyblue)((y-z)) = color(coral)x xx color(deepskyblue)y - color(coral)x xx color(deepskyblue)z.

Note the following

•  color(coral)(x_1) xx color(deepskyblue)((y_1+z_1)) = color(coral)(x_1) xx color(deepskyblue)(y_1) + color(coral)(x_1) xx color(deepskyblue)(z_1)

•  color(coral)(x_2) xx color(deepskyblue)((y_2-z_2)) = color(coral)(x_2) xx color(deepskyblue)(y_2) - color(coral)(x_2) xx color(deepskyblue)(z_2)

By substituting z_1 = -z_2 in the first result, the second result is derived. So these two are equivalent results. .

Multiply x and x^2-3.

• x^3-3
• x^3-3x

The answer is 'x^3-3x'. The multiplication is carried out as follows.

Similar to 4 xx(3-2) =4xx3 - 4 xx2, the multiplication of terms of polynomial are carried out,
color(coral)(x) xx (color(deepskyblue)(x^2-3)) = color(coral)(x) xx color(deepskyblue)(x^2) - color(coral)(x) xx color(deepskyblue)(3).

To understand multiplication of polynomial, one such multiplication is illustrated.

(color(coral)(2x^2-3x+5))xx(color(deepskyblue)(x-7))

sub-expression (color(coral)(2x^2-3x+5)) is a number and holding it as a whole. Applying that number into x-7
quad =(color(coral)(2x^2-3x+5))xx color(deepskyblue)(x)
quad quad quad + (color(coral)(2x^2-3x+5)) xx color(deepskyblue)((-7))

sub-expression (color(coral)(2x^2-3x)) as a number
quad =(color(coral)(2x^2-3x)) xx color(deepskyblue)(x) + color(coral)(5)xx color(deepskyblue)(x)
quad quad quad + (color(coral)(2x^2-3x)) xx color(deepskyblue)((-7)) +color(coral)(5) xx color(deepskyblue)((-7))

quad =2x^3-3x^2+5x
quad quad quad -14x^2+21x - 35

quad =2x^3+(-3-14)x^2 +(5+21)x-35

quad =2x^3-17x^2+26x-35

comprehensive information for quick review

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dummy

Multiplication of Polynomials: A polynomial is considered to be a number as a whole or the sub expressions as numbers.

Eg: (a+b+c^3)xx(p^2-q)
in this each of the following can be considered a number
a;
b;
c^3;
a+b;
b+c^3;
a+c^3;
a+b+c^3;
p^2;
-q;
(p^2-q).

Distributive property of multiplication over addition
x(y+z) = xy+xz
In this property, x, y, z can be any of the polynomial or the sub-expression as given above.

practice questions to master the knowledge

Exercise

practice questions to master the knowledge

Exercise

Which of the following equals (x^2-3)(y-4)

• x^2-3y-4
• x^2y-4x^2-3y+12

The answer is 'x^2y-4x^2-3y+12'

Progress

Progress

Note: This page is for students at 8th grade level who are not introduced to closure, commutative, associative, and distributive laws of numerical arithmetic.
What is 4 times 5?
45
45
20
20
The answer is ' 20 '.
Given the numbers 4 and 5 are integers, What is 4 times 5.
1
2
If any two numbers are multiplied or added the result is another number. ;; In an algebraic expression, any sub-expression is a number. The sub-expression can be handled as if it is a number.
Which of the following equals 4 times, 3 + 2.
1
2
The answer is 4 times 3, plus, 4 times 2
Note the following. 4 times 3 plus 2 = 4 times 5 = 20 ;; and 4 times 3 plus 4 times 2 = 12 + 8 =20 ;; This can be generalized for any variables ;; x multiplied y + z, = x multiplied y, + x multiplied z. This is an important result to work with polynomials. The variables x, y, z in this can be sub-expression of a polynomial. ;; An example is given with an algebraic expression a cube plus b squared.
multiply x and x squared + 3.
1
2
The answer is x cube plus 3x. The multiplication is carried out as follows.
To understand multiplication of polynomial, one such multiplication is illustrated.
Which of the following equals 4 times, 3 minus 2?
1
2
The answer is 4 times 3, minus, 4 times 2
Note the following. 4 times 3 minus 2 = 4 times 1 = 4 ;; and 4 times 3 minus 4 times 2 = 12 minus 8 =4 ;; This can be generalized for any variables ;; x multiplied y minus z, = x multiplied y, minus x multiplied z.
Note the given results. By substituting z 1 equals minus z 2 in the first result, the second result is derived. So these two are equivalent results.
multiply x and x squared minus 3.
1
2
The answer is x cube minus 3x. The multiplication is carried out as follows.
To understand multiplication of polynomial, one such multiplication is illustrated.
Polynomials as a whole or each of the sub expression can be considered as numbers. Multiplication is performed by distribution of multiplication into addition or subtraction.
Multiplication of Polynomials: A polynomial is considered to be a number as a whole or the sub expressions as numbers. An example and sub expressions of the given example are shown. ;; Distributive property of multiplication over addition, x multiplied y plus z = x y + x z. ;; In this property, x, y, z can be any of the polynomial or the sub-expression as given above.
Which of the following equals x squared minus 3, multiplied, y minus 4?
1
2
The answer is "x squared y minus 4 x squared minus 3 y + 12"

we are not perfect yet...


text:
Q

# Fabina borrows Rs 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum compounded annually. Who pays more interest and by how much?

Fabina borrows Rs 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Views

For Fabina,

Principal,P =Rs 12500

Simple interest Rate,R = 12% p.a.

Time period,n = 3 years.

$\therefore$ Simple Interest, SI at 12% for 3 years = $3\times\frac{12500\times12}{100}$ = Rs 4500

Principal,P =Rs 12500

Compound interest Rate,R = 10% p.a.

Time period,n = 3 years.

We know, Amount when interest is compounded annually,

$A =P(1+\frac{R}{100})^n$

$A=12500(1+\frac{10}{100})^3 = Rs 16637.50$

$\therefore Compound \:\:Interest, CI = A - P = Rs\: (16637.50 - 12500) = Rs \:4137.50$

$\therefore$ Fabina pays more interest and Rs (4500 - 4137.50) = Rs 362.50 more.

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text:
Database[SQLite] - Maple Programming Help

Home : Support : Online Help : Connectivity : Database Package : SQLite : Database/SQLite/ColumnCount

Database[SQLite]

 ColumnCount
 - get number of columns in prepared statement

 Calling Sequence ColumnCount( statement )

Parameters

 statement - prepared statement obtained using Prepare command

Description

 • The ColumnCount command returns number of columns in prepared statement.

Examples

 > $\mathrm{with}\left(\mathrm{Database}[\mathrm{SQLite}]\right)$
 $\left[{\mathrm{Attach}}{,}{\mathrm{Bind}}{,}{\mathrm{ClearBindings}}{,}{\mathrm{Close}}{,}{\mathrm{ColumnCount}}{,}{\mathrm{ColumnNames}}{,}{\mathrm{Execute}}{,}{\mathrm{Fetch}}{,}{\mathrm{FetchAll}}{,}{\mathrm{FetchRow}}{,}{\mathrm{Finalize}}{,}{\mathrm{Open}}{,}{\mathrm{Opened}}{,}{\mathrm{Prepare}}{,}{\mathrm{RESULT_BUSY}}{,}{\mathrm{RESULT_DONE}}{,}{\mathrm{RESULT_ROW}}{,}{\mathrm{Reset}}{,}{\mathrm{Step}}\right]$ (1)
 > $\mathrm{db}≔\mathrm{FileTools}[\mathrm{JoinPath}]\left(\left[\mathrm{kernelopts}\left('\mathrm{datadir}'\right),"SQLite","G20-Population.db"\right]\right):$
 > $\mathrm{connection}≔\mathrm{Open}\left(\mathrm{db}\right):$

Select all data from table - prepare statement

 > $\mathrm{stmt}≔\mathrm{Prepare}\left(\mathrm{connection},"SELECT * FROM population"\right)$
 ${\mathrm{stmt}}{≔}{"SQLite statement"}{,}{"SELECT * FROM population"}$ (2)

Number of columns

 > $\mathrm{ColumnCount}\left(\mathrm{stmt}\right)$
 ${21}$ (3)

Fetch all rows

 > $\mathrm{FetchAll}\left(\mathrm{stmt}\right)$
 $\left[\begin{array}{c}{\mathrm{53 x 21}}{\mathrm{Matrix}}\\ {\mathrm{Data Type:}}{\mathrm{anything}}\\ {\mathrm{Storage:}}{\mathrm{rectangular}}\\ {\mathrm{Order:}}{\mathrm{C_order}}\end{array}\right]$ (4)
 > $\mathrm{Finalize}\left(\mathrm{stmt}\right):$
 > $\mathrm{Close}\left(\mathrm{connection}\right):$

Compatibility

 • The Database[SQLite][ColumnCount] command was introduced in Maple 18.