text:
# Challenge: can you take the sqrt (n) using only one operation

1. Apr 22, 2012

### logics

The best known algorithm ('Babylonian') to take the square root of a number requires 3 operations, less than Newton's. I suppose it's also the fastest. Is it so?
Do you know a simpler or faster one?

Can you find a method that requires only 1 operation?

Last edited: Apr 22, 2012
2. Apr 22, 2012

### Hurkyl

Staff Emeritus
Yes: apply the square root operation.

3. Apr 23, 2012

### logics

square root is the problem to be solved, can it also be the solution?

P.S.: I apologize if I have been ambiguous: *(one of the 4 operations. +, -, x, : ). The algorithm should contain 1 op, but it can be iterated a few times, of course!
to simplify discussion and use a pocket calculator, let's choose (n = x²) n with less than 21 digits:
210.3456789², 987654.3012²... 597482²....etc

Last edited: Apr 23, 2012
4. Apr 23, 2012

### dodo

Yes; if you already know the solution to be, say, r, then you can find said solution by just one division, n/r.

Last edited: Apr 23, 2012
5. Apr 23, 2012

### Hurkyl

Staff Emeritus

So, we get to use "limit" as an operation too? What about "<" as well?

I'm not just being snarky here -- it is provably impossible to obtain $\sqrt{2}$ using just those four operations along with any integers you want, so "limit" or some other related concept is rather important.

Many algorithms for extracting roots rely on being able to tell whether one number is bigger than another, and so "<" is important.

6. Apr 24, 2012

### logics

We are only trying to solve a practical problem: we are trying to apply our contemporary wisdom to see we if can do any better than the Babylonian method, if that is really the best algorithm. (I'll tackle this in a separate post). Our solution should be more convenient even using pencil and paper. Do you think it is possible?

If n is not a square number ($\sqrt{2}$...$\sqrt{125348}$... etc), we have to stop sometime, somewhere, maybe after 1010 digits, but we must stop. Hope you agree. If a simple, convenient method requires '>','<' ,... or other, it is interesting to see it.
I was only responding to your (jocular?) post #2, excluding the obvious sqrt, log, ln, etc. I hope that is clear.
here: they decided to stop after 6 digits: 354.045, and it took them 5 rounds (iterations) = 15 (5*3 [+, :, :]) ops. to reach that result. Can you do any better?
I suggested to stop at 10 places (9digits plus the point) so that non specialists (without a computer program), like myself, can follow or take part in the discussion with just a pocket calculator.

If this is not clear I'll exemplify: my calculator says $\sqrt{125348}$ = 354.0451949 , the number is obviously truncated and I cannot know if it is rounded up or down and if it is correct to stop at 354.0451948.
That is why I suggested to start with a number with finite number of decimals ( like $\sqrt{125347.862}$) to know that exactly 354.045 is our goal.

P.S. Moreover, if you start with a finite n you can find its root even when many digits are missing: $\sqrt{623,470.122 xxx xxx x49}$, we can find the root: x = 789.601243

Last edited: Apr 24, 2012
7. Apr 24, 2012

### DonAntonio

Your whole post is pretty confusing, making your nick paradoxical. You talked of calculating the square root of a number

using "only one operation"...Did you mean using one operation several times, did you mean to use one unique

operation once? It should be obvious that either possibility has a negative answer: no, it is not possible IN GENERAL to do that

with only one operation, whatever "one operation" means to you.

Now, it is possible to evaluate the square root of any number to any degree of wanted precision by elementary operations,

namely +, -, *, /, and I don't know if this is the babylonian method.

DonAntonio

8. Apr 24, 2012

### logics

If you do not know it, probably that is the reason why you could not understand my post. I gave you a link, scroll up a little:
In the OP (post #1), I noted that Babylonian algorithm requires 3 operations for each iteration, as surely by now you have found out. In the title, if you're referring to that, I could not possibly explain everything.

I'm asking if you can find a simpler one (algorithm), one which requires less then 3 ops at each iteration. One op., of course, is the limit.
EDIT: Such an algorithm is likely to be less powerful, to converge slower, but it would require less then 15 ops. (in the #Example quoted), I suppose anyone can do better than wiki . I'm sure you are able at least to make a first guess more accurate than x0= 600

If you run the algorithm only once then you take directly sqrt(n), and you reach your goal with one operation.( 'ne plus ultra' !).

Besides the OP, you probably missed also post #3:
I hope it is clear, if it was not.

P.S. A nick is a nick, what's in a name? that which we call a rose....

Last edited: Apr 24, 2012
9. Apr 24, 2012

### DonAntonio

I read only what you wrote in your first post, which is exactly the following:

"The best known algorithm ('Babylonian') to take the square root of a number requires 3 operations, less than Newton's. I suppose it's also

the fastest. Is it so? Do you know a simpler or faster one?

Can you find a method that requires only 1 operation?"

No iterations, no links, nothing. That's what confused me and, judging by the first responses at least, apparently also others.

DonAntonio

10. Apr 24, 2012

### jgens

No such algorithm exists. Fix the number 2 for example and try to approximate the √2 using only one operation from {+,-,×,/}. If we choose either + or -, then we can either add or subtract 2 (or multiples of 2) from itself repeatedly, so the result will be 2n where n an integer. If we choose either × or /, then we can either multiply or divide 2 (or powers of two) from itself repeatedly, so the result will be 2n where n an integer. But neither 2n nor 2n can approximate √2 to more than say one or two decimal places.

11. Apr 24, 2012

### logics

Thanks jgens, that gives me a chance to compare methods and check if it is my terminology that generates misunderstanding:

here is: B method, is this procedure an algorithm? (If it cannod described as an algorithm, I stand corrected and I apologize).
This method requires 3 operations:

(x + a : x) : 2
...1....2....3
As wiki shows, the formula must be iterated 5 times and that adds up to 15 ops. (to find: sqrt{125348} = 354.045, 6-digit accuracy)

There is another method : N[ewton]. It is equivalent to B, but (because of the way it is usually formulated), it requires 4 operations:
(x * x + a) : 2 * x
.. 1...2.....3....4
(You can find out how many ops. are necessary to solve the problem, if N it is better than B)

Now, why should all the objections apply only to a hypothetical method A that requires 2 operations?.
If I am missing something and your objections are valid only for my hypothesis, please change n to 125347.862025 or any other n which you deem suitable, and let's change the title:
Can you find the square root of 125347.862025 (x= 354.045) with less than 15 operations?

Thanks for your help, jgens, I hope you'll give us the first concrete contribution: if you were in Alexandria or Miletus, or in the Middle Ages with quill and parchment , how would you find 354.045?

Last edited: Apr 24, 2012
12. Apr 24, 2012

### DonAntonio

Using the system my dad taught me and which he learnt when he was a kid, either from my grandpa or at school:

subdive the given number in groups of 2, from right to left, before and after the decimal point:

12´53´47.86´20´25

The above 2-groups will be the ones "going down" after each step is finished.

Begin with 12 at left: take the number whose square is closeset to 12 but less than or equal to it. In this case

it is 3 and we write down aside.

Square it and substract it from 12: we get 3 below 12 and we draw down the next pair, 53, thus getting 353.

Double 3, getting 6 (of course, this is just the double method in some disguise which makes, perhaps, easier to carry on by hand)

Find a digit x such that $6x\cdot x$ is as close as possible to 353 but no more than it: this digit is 5 and we

write down aside at the right of the 3, so 65*5 = 325, and substract this from 352, getting 28, and now draw down the next pair 47...etc.

As you can see, we need two operations for each digit in the answer: doubling the already found number and then multiplying, except

for the first digit which only required squaring it.

So in this case we need 12 - 1 = 11 operations.

DonAntonio

Pd Hmmm...perhaps the substractions must be counted also and thus I have more operations. A pity...

13. Apr 24, 2012

### Hurkyl

Staff Emeritus
Division by 2 is barely an operation. General division is more expensive than general multiplication, and much more than plain addition.

I checked GMP source code -- for small square roots ($N \in [0, 2^{64})$), they use start off with a table lookup (for very small numbers), and then use Newton's algorithm to approximate y ~ 1/√N, which has the advantage of requiring no general divisions:
$$y \leftarrow (1/2) (N y^3 + y)$$
with the very last iteration mixing in an extra multiplication by N so the result is an approximation x ~ √N
$$x \leftarrow (1/2) (y (Ny)^2 + (Ny))$$

GMP is actually computing an integer square root with remainder. Once it has the approximation, it computes x^2, and then increments it along until it gets the exact value of $\lfloor \sqrt{N} \rfloor$ (this doesn't require general multiplication: $(x+1)^2 = x^2 + (2x+1)$), and then obtains the remainder.

I haven't worked out the exact details of what it's doing for larger square roots.

14. Apr 25, 2012

### logics

That is true, but may lead to subjective evaluation: plain addition (n+n) is equivalent to division by 2 (n:2) but, arguably, easier. If we want to be more accurate we may count the micro-operations

Doesn't that apply to Newton's algorithm?

15. Apr 25, 2012

### coolul007

The sum of the first "n" odd numbers is n^2, conversely, when the sum matches the desired number it's the square root. To find the square root of 2 simply add pairs of zeros for the desired accuracy. I guess this is 2 addition operations and one compare.

16. Apr 26, 2012

### logics

Sorry, jgens, I did not reply directly to your post because I thought I was using the term 'algorithm' in a wrong way. You used Hurkyl's example, so I thought you were referring to his post, (which I still do not understand). Probably, as I said, I am missing something, could you tell me what?
The fact that Newton's algorithm (which uses only 3 different ops. +, x, :) is actually used (post #13) proves, I suppose a fortiori, that one can find a method using those four operations.

Yours is sound reasoning, but it is founded on the assumption that any method must use B's (or N) logics (the weighted mean).
You can't prove, by that, that no such algorithm exists.
I do not believe anyone can prove that such method is impossible. As you know, proving that something is impossible is a tough job.
(Which operation is suitable, seems obvious.)
As I said in a previous edit, any schoolboy can do better than wiki and find the right x0, and that would mean only 4 iterations (that is 12 ops.)
What is the best 'first-guess' you can make?

Last edited: Apr 26, 2012
17. Apr 26, 2012

### phylotree

I don't know if this is flawed but a pretty good read to go on.

Given x, now you are asked to find √x. You then will know x's integral sup and inf values, supposing they are a, b then you will have $a ≤ \sqrt{x} ≤ b$. You can then use $c=\frac{a+b}{2}$ for each iteration to approximate the result
How many operations are needed depending upon how many decimal values you want to find .

18. Apr 26, 2012

### coolul007

Here is how one takes the square root of 2 by using addition, the square root of 2 00 is 14 sums of the odd numbers 1+3+5+7+9+11+13+15+17+19+21+23+25+ 27=196
square root of 2 00 00 1s 141 sums, 2 00 00 00 is 1414 sums, etc. Not to say it's efficient, but it will work with only addition.

19. Apr 26, 2012

### logics

Conversely, one may start with n (=x²) and use only subtraction, if 215² = 46225:

46225 -1, -3, -5.....etc. It would take (x =) 215 operations, using this elementary logics.

But subtraction can be more efficient, as it allows to improve the basic, simple algorithm. Remember you have the right, following B (or N) scheme, ( see wiki-link above), to find a suitable x0, performing micro-operations in your mind. If you guess the first digit and the number of digits (4'62'25 $\rightarrow$ 2 | 0 | 0|), you may start subtracting 4 | 00 | 00 | : 46225 - 40000 = 6225, and then continue subtracting (200 x 2 +1) 401,...etc:

(n² - 40000 = ) 6225 - 401, - 403, - 405....etc. That would take only 15 + 1 operations

But, of course, it takes a more ingenious method, a more sophisticated logic to outwit Newton and Babylonian logics, weighted mean is simple but powerful.

Last edited: Apr 27, 2012
20. Apr 27, 2012

### Altrepair

I am no expert here at all, but is there a reason why no one posted this algorithm: http://www.homeschoolmath.net/teaching/sqr-algorithm-why-works.php on square roots?? It appears to use the least amount of operations that the OP was asking for. It was used back in the days of elementary school of the old days before the calculator was around.


text:
Galois extension

In mathematics, a Galois extension is an algebraic field extension E/F that is normal and separable; or equivalently, E/F is algebraic, and the field fixed by the automorphism group Aut(E/F) is precisely the base field F. One says that such an extension is Galois. The significance of being a Galois extension is that the extension has a Galois group and obeys the fundamental theorem of Galois theory. [1]

A result of Emil Artin allows one to construct Galois extensions as follows: If E is a given field, and G is a finite group of automorphisms of E with fixed field F, then E/F is a Galois extension.

Characterization of Galois extensions

An important theorem of Emil Artin states that for a finite extension E/F, each of the following statements is equivalent to the statement that E/F is Galois:

Examples

Adjoining to the rational number field the square root of 2 gives a Galois extension, while adjoining the cube root of 2 gives a non-Galois extension. Both these extensions are separable, because they have characteristic zero. The first of them is the splitting field of X2 − 2; the second has normal closure that includes the complex cube roots of unity, and so is not a splitting field. In fact, it has no automorphism other than the identity, because it is contained in the real numbers and X3 − 2 has just one real root.

An algebraic closure $\bar K$ of an arbitrary field $K$ is Galois over $K$ if and only if $K$ is a perfect field.

References

1. ^ See the article Galois group for definitions of some of these terms and some examples.

format: plain


text:
# Reducing degeneracy in constrained (convex) optimization problem

DISCLAIMER: I've edited the question repeatedly for clarity and to target the most relevant answer.

I have the following general problem $$\min \|h_1\cdot h_2\|^2$$ such that $$\|g_1\wedge g_2-h_1\wedge h_2\|^2 = 0,$$ where $h_i,g_i\in \mathbb{R}^n\wedge \mathbb{R}^n$ are skew-symmetric matrices, $A\cdot B$ denotes matrix multiplication of matrices $A$ and $B$, and $\lambda\in \mathbb{R}$. $A\wedge B$ is the geometric product of $A$ and $B$.

In order to make the problem solve nicer, i.e., no infeasibility trap at $h_1=0$, for instance, I've reformulated the problem with the following constraints. $$\|g_1\wedge g_2-\lambda h_1\wedge h_2\|^2 = 0,\\ \lambda\ge 0,\\ \|h_1\|^2=1,\\ \|h_2\|^2=1$$

My test case is $n=6$ and $g_1=g_2=e_{12}+e_{34}+e_{56}$ (a worst case scenario). Obviously, the problem is degenerate due to symmetry as well as unitarily invariant on $\mathbb{R}^6$. I've introduced $\lambda$ and constraints 2 and 3 in order to exclude some slack in the solution. My expected outcome is $h_1 =\frac{1}{\sqrt{12}}\left(2e_{12}+e_{34}+e_{56}\right)$ and $h_2=\frac{1}{\sqrt{2}}\left(e_{34}+e_{56}\right)$ with a minimum of 1/12 and $\lambda=\sqrt{24}$.

I've used IPOPT to implement the minimzation, but it fails to find this solution. Instead a solution is found that is slightly worse. I'm a little perplexed as all functions are convex in each variable, but I'm no expert in NLP.

EDIT: As per @Geoff Oxberry, the problem is not convex. As a matter of fact, the initial guess of $h_1=g_1$ and $h_2=g_2$ is a maximum of the objective function under the constraints.

So how do I improve my initial guess or change constraints to improve solutions?

• I don't quite understand your notation. Specifically, I'm confused about what the wedge ($\wedge$) means, because it's not a logical "and" in this case. Almost everything else sort of makes sense. Also, the degeneracy of the problem isn't apparent to me (probably because of the notation issues). Could you explain the degeneracy in more detail? – Geoff Oxberry Mar 1 '12 at 18:13
• $\wedge$ is the geometric product. – Deathbreath Mar 1 '12 at 18:25
• Geometric product? – Geoff Oxberry Mar 1 '12 at 18:27
• Let $V\wedge V = V\otimes V/\{v\otimes v|v\in V\}$. See here for annother take on it. – Deathbreath Mar 1 '12 at 18:34
• I've amended the title to reduce the emphasis on convexity. The involved functions are convex in each single variable, but maybe not overall. – Deathbreath Mar 1 '12 at 19:06

First, $\|h_{1}\|^{2} = 1$ and $\|h_{2}\|^{2} = 1$ are not convex constraints, even though the functions on the left-hand sides are convex. The easiest way to picture this in a simpler case is to note that the unit circle (or sphere) is not a convex set because the line between any two points on the circle (or sphere) is not contained in the set.
Second, the function $\|h_{1} \cdot h_{2}\|^{2}$ is probably not convex because it contains bilinear terms. Bilinear terms are known not to be convex, and I doubt that composing them with a 2-norm changes the non-convexity.
• I've thought about it, and have come to the conclusion that indeed the problem is not generally convex, but really the weirdness is that $(h_1,h_2)=(g_1,g_1)$ is actually a maximum. IPOPT fails in the restoration phase at a nearly feasible point. – Deathbreath Mar 1 '12 at 20:43
• I'm considering changing the constraints. 3 and 4 turn into $\le 1$ and I'll bound $\lambda$ as soon as I know a good upper bound. I'm thinking about adding $\mbox{tr } h_1 -\mbox{tr } h_2 \le 0$ and $\mbox{tr } (h_1\cdot h_2) \le 0$. – Deathbreath Mar 2 '12 at 1:09
• The trace of $h_{1} \cdot h_{2}$ is almost certainly a nonconvex constraint. However, the difference of the two traces is linear and should be fine. – Geoff Oxberry Mar 3 '12 at 5:23
• I've actually dropped both those constraints as they did nothing to improve the solution, but the bounds on $\lambda$ and 3/4 are good. I'm looking into ways to characterize the degeneracies and adding perturbations to lift them. – Deathbreath Mar 3 '12 at 13:16


text:
# Conformal covariance of massless free nets

Fernando Lledó
June 16, 2000
In the present paper we review in a fibre bundle context the covariant and massless canonical representations of the Poincare' group as well as certain unitary representations of the conformal group (in 4 dimensions). We give a simplified proof of the well-known fact that massless canonical representations with discrete helicity extend to unitary and irreducible representations of the conformal group mentioned before. Further we give a simple new proof that massless free nets for any helicity value are covariant under the conformal group. Free nets are the result of a direct (i.e. independent of any explicit use of quantum fields) and natural way of constructing nets of abstract C*-algebras indexed by open and bounded regions in Minkowski space that satisfy standard axioms of local quantum physics. We also give a group theoretical interpretation of the embedding ${\mathfrak I}$ that completely characterizes the free net: it reduces the (algebraically) reducible covariant representation in terms of the unitary canonical ones. Finally, as a consequence of the conformal covariance we also mention for these models some of the expected algebraic properties that are a direct consequence of the conformal covariance (essential duality, PCT--symmetry etc.).

Keywords:
none


text:
# How to draw parts of a line dotted with TikZ?

I want to create a straight line between two nodes in a tikzpicture interrupted in the middle by a dotted portion. This should work whatever the orientation of the line.

Here is an expected output:

The best would be to have a line style defined for this instead of each time using clip to remove the straight line and then tweak a dotted one to fit the hole. A minimal code for the solution would look like

\documentclass[a4paper, 12pt]{report}

\usepackage{tikz}

\tikzset{
middle dotted line/.style={
thick,
}
}% end of tikzset

\begin{document}

\begin{tikzpicture}
\node (A) at (0, 3) {};
\node (B) at (5, 3) {};
\node (C) at (0, 2) {};
\node (D) at (5, 0) {};
\draw[middle dotted line] (A) -- (B);
\draw[middle dotted line] (C) -- (D);
\end{tikzpicture}

\end{document}

• Welcome to TeX.SE! Apr 5 at 13:13
• I take it \draw (A) -- (B) node[midpoint, fill=white, sloped]{$\cdots$}; won't do? Apr 5 at 15:26

Inspired by Mark Wibrow

\documentclass[tikz, border=5]{standalone}
\usetikzlibrary{decorations.pathreplacing,calc}
\tikzset{%
middle dotted line/.style={
decoration={show path construction,
lineto code={
\draw[#1] (\tikzinputsegmentfirst) --($(\tikzinputsegmentfirst)!.3333!(\tikzinputsegmentlast)$);,
\draw[dotted,#1] ($(\tikzinputsegmentfirst)!.3333!(\tikzinputsegmentlast)$)--($(\tikzinputsegmentfirst)!.6666!(\tikzinputsegmentlast)$);,
\draw[#1] ($(\tikzinputsegmentfirst)!.6666!(\tikzinputsegmentlast)$)--(\tikzinputsegmentlast);,
}
},
decorate
},
}
\begin{document}

\begin{tikzpicture}
\draw[middle dotted line] (0,0)--(4,1);
\draw[middle dotted line={line width=1.5pt}] (1,1)--(3,4);
\end{tikzpicture}

\end{document}

Using the spath3 library (at time of writing, needs the development version from github).

\documentclass{article}
%\url{https://tex.stackexchange.com/q/639726/86}
\usepackage{tikz}
\usetikzlibrary{spath3}

\begin{document}

\begin{tikzpicture}
\draw[spath/save=path,ultra thick, red] (0,0) -- +(3,0);
\draw[spath/split at keep start={path}{1/3},spath/use=path];
\draw[spath/split at keep end={path}{2/3},spath/use=path];
\draw[spath/split at keep middle={path}{1/3}{2/3},spath/use=path,dotted];

\draw[spath/save=path,ultra thick, red] (0,-1) .. controls +(6,-1) and +(-6,-1) .. +(3,0);
\draw[spath/split at keep start={path}{1/3},spath/use=path];
\draw[spath/split at keep end={path}{2/3},spath/use=path];
\draw[spath/split at keep middle={path}{1/3}{2/3},spath/use=path,dotted];

\end{tikzpicture}
\end{document}

It works by splitting the path into three components and then rendering each component with a different style. As Henri mentions in the comments, to have a genuine solid and dotted line requires (at least) two draw commands - one for each style. I use one for each segment since combining the first and last segments would actually lead to slightly more complicated code.

As it uses the spath3 library, it works on arbitrary paths.

Here's the result:

• This should be the accepted answer, because it works on arbitrarily curved paths. The currently accepted answer only works on straight paths. Apr 6 at 13:01
• It's a forlorn hope, I know, but it would be useful if the downvoter explained what they felt was bad about this answer. I try to be helpful on this site and when I miss the mark then the only way I'll improve is if I know what went wrong. Apr 7 at 5:52

Without guru's tricks try this:

\documentclass[tikz, border=1cm]{standalone}
\begin{document}
\begin{tikzpicture}
\path (0,0) coordinate (A) (3,5) coordinate (B);
\path (A)--(B) node [inner sep=0pt,pos=.333] (C) {};
\path (A)--(B) node [inner sep=0pt,pos=.666] (D) {};
\draw (A)--(C);
\draw (D)--(B);
\draw[dotted] (C)--(D);
\end{tikzpicture}
\end{document}

Output:

• If you use the calc library then you can do your coordinate calculations all in one go without creating extra nodes. (It feels sufficiently close to this answer that I don't want to post a separate answer just for that, but you could add an example using calc) Apr 5 at 17:40
• @AndrewStacey Thanks for your suggestion. But I post my answer just for its simplicity, without the help of others packages or libraries. I know that I can calculate the coordinates of my intermediates points C(1,5/3) and D(2,10/3): the parameter pos of Tikz make the job! Apr 5 at 21:55
• Fair enough. I've added a separate calc answer then as I think it's a useful example to show how the calc library can shorten the code. Apr 5 at 22:57
• an code-cleaning suggestion: using coordinate is more suitable than node. \documentclass[tikz, border=1cm]{standalone} \begin{document} \begin{tikzpicture} \path (0,0) coordinate (A) --(3,5) coordinate (B) coordinate[pos=1/3] (C) coordinate[pos=2/3] (D); \draw (A)--(C) (D)--(B); \draw[dotted] (C)--(D); \end{tikzpicture} \end{document} Apr 6 at 9:44

With the subpath function and time along a path, Asymptote handles this task very easy, for any curve.

unitsize(1cm);

// for curvy curves
path p=(0,-1) .. controls (0,-1)+(6,-1) and (3,-1)+(-6,-1) .. (3,-1);
real t1=.4,t2=.6;   // times on path
path q1=subpath(p,0,t1);
path q2=subpath(p,t1,t2);
path q3=subpath(p,t2,1);
draw(q1^^q3,magenta);
draw(q2,magenta+dotted);

// of course it works for straight lines
path pline=(0,0)--(3,1);
real s1=.3,s2=.7;   // times on path

path q1line=subpath(pline,0,s1);
path q2line=subpath(pline,s1,s2);
path q3line=subpath(pline,s2,1);
draw(q1line^^q3line,blue);
draw(q2line,blue+dotted);

shipout(bbox(5mm,invisible));

• It's great seeing how other technologies approach problems. Now all we need is a pstricks solution to complete the set! Apr 6 at 13:39

A minor variation of Raffaele Santoro's answer using the calc library to shorten the code. It produces the same output.

\documentclass[tikz, border=1cm]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\path (0,0) coordinate (A) (3,5) coordinate (B);
\draw (A)--($(A)!1/3!(B)$) ($(A)!2/3!(B)$) -- (B);
\draw[dotted] ($(A)!1/3!(B)$) -- ($(A)!2/3!(B)$);
\end{tikzpicture}
\end{document}
`


text:
GAP

## 20 publications using GAP published in 2019

[AEJ19] Aldwaik, S., Edjvet, M., and Juh\'asz, A., Asphericity of positive free product length 4 relative group presentations, Forum Math., 31 (1) (2019), 49–68.

[AH19] Araya, M. and Harada, M., On the classification of linear complementary dual codes, Discrete Math., 342 (1) (2019), 270–278.

[ABC19] Ara\'ujo, J., Bentz, W., and Cameron, P. J., Orbits of primitive $k$-homogenous groups on $(n-k)$-partitions with applications to semigroups, Trans. Amer. Math. Soc., 371 (1) (2019), 105–136.

[BBL19] Badr, E., Bars, F., and Lorenzo Garc\'ia, E., On twists of smooth plane curves, Math. Comp., 88 (315) (2019), 421–438.

[BCJ19] Ballester-Bolinches, A., Cosme-Ll\'opez, E., and Jiménez-Seral, P., Some contributions to the theory of transformation monoids, J. Algebra, 522 (2019), 31–60.

[B19] Bors, A., Finite groups with a large automorphism orbit, J. Algebra, 521 (2019), 331–364.

[BR19] Bouc, S. and Romero, N., The Whitehead group of (almost) extra-special $p$-groups with $p$ odd, J. Pure Appl. Algebra, 223 (1) (2019), 86–107.

[CH19] Creedon, L. and Hughes, K., Derivations on group algebras with coding theory applications, Finite Fields Appl., 56 (2019), 247–265.

[DS19] De Bruyn, B. and Sahoo, B. K., On minimum size blocking sets of the outer tangents to a hyperbolic quadric in $\rm PG(3,q)$, Finite Fields Appl., 56 (2019), 31–57.

[FM+19] Fawcett, J. B., Müller, J., O'Brien, E. A., and Wilson, R. A., Regular orbits of sporadic simple groups, J. Algebra, 522 (2019), 61–79.

[G19] Greer, M., Simple right conjugacy closed loops, in Nonassociative mathematics and its applications, Amer. Math. Soc., Providence, RI, Contemp. Math., 721 (2019), 151–163.

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# Prove there is no maximum perimeter trapezoid inscribed in a circle

I want to prove that there is no maximum perimeter trapezoid inscribed in a circle.

Edit: the way my textbook defined a trapezoid does not allow for a square to fit in the trapezoid definition. If your definition is different than that, this problem is the equivalent of proving that the maximum perimeter trapezoid inscribed in a circle is a square.

I know (and know how to prove) that an inscribed trapezoid must be an isosceles trapezoid.

I also know that the maximum perimeter quadrilateral inscribed in a circle is a square (but don't know how to prove it).

I know that the inscribed trapezoid can get as close to being a square as we like, but it can never be a square. That is why there is no maximum perimeter inscribed trapezoid. But I do not know how to prove that the square is the maximum perimeter quadrilateral inscribed in a circle.

I think that I should approximate the inscribed square by an inscribed trapezoid, but when I tried that I couldn't prove that the perimeter of the square was greater than that of the trapezoid.

• A square is a trapezoid. A very special case of a trapezoid, maybe, but still definitely a trapezoid. The same way as, for example, a square is a quadrilateral. – dxiv Oct 30 '16 at 4:07
• That depends on your definition. My text book doesn't consider a square to be a trapezoid. I will edit the question to add this information. Thanks for pointing that out. – Wheepy Oct 30 '16 at 4:10
• That must be one odd book. A trapezoid is defined everywhere as a quadrilateral with (at least) two parallel sides. Following the same (broken) logic your book would argue that the trapezoid is not a quadrilateral. – dxiv Oct 30 '16 at 4:26
• It's certainly an odd way of defining it for someone who has never seen it, but this way is chosen by some authors for teaching preferences. That way, when they say "trapezoid", they mean "a trapezoid that is not a square", when they say "rectangle" they mean "a rectangle that is not a square", etc. Both these ways of defining are possible and lead to the same geometry, apart from names. There is, however, no way of defining a trapezoid without fitting the definition of a quadrilateral. Saying "a trapezoid that is not a quadrilateral" doesn't make any sense in any possible definition. – Wheepy Oct 30 '16 at 4:39
• @dxiv I hang out in elementary school classrooms and see this definition problem a lot. Is a square a rectangle, or does the definition of rectangle exclude all sides equal? The latter is sometimes better "for teaching preferences" but it's a disaster kids will have to unlearn later when they do more mathematics. And you could define quadrilateral to be "a four sided polygon with no symmetries". That would be consistent with "a square is not a rectangle". But ugly! – Ethan Bolker Oct 30 '16 at 14:00

I shall prove the statement in the title, given the book's definition of a trapezoid. We may assume the trapezoid $T$ inscribed in the unit circle, with horizontal parallel sides $y=a$ and $y=b$, whereby $-1<a<b<1$, and $b>0$. If $a>0$ we can enlarge the perimeter of $T$ by replacing $a$ by $-a$. The four vertices of $T$ can then be written as $$(\pm \cos\psi,-\sin\psi), \quad(\pm\cos\phi,\sin\phi)$$ with $0\leq\psi<{\pi\over2}$ and $0<\phi<{\pi\over2}$. The perimeter $L$ of $T$ computes to $$L(\phi,\psi)=4\sin{\phi+\psi\over2}+2\cos\phi+2\cos\psi\ ,$$ such that $$L_\phi=2\cos{\phi+\psi\over2}-2\sin\phi,\qquad L_\psi=2\cos{\phi+\psi\over2}-2\sin\psi\ .$$ If $\psi=0$ one has $L_\psi=2\cos{\phi\over2}>0$, hence increasing $\psi$ will increase $L$. If both $\phi$ and $\psi$ are $>0$ we can locally increase $L$ unless $L_\phi=L_\psi=0$. The latter would lead to $\psi=\phi$ and then to $\cos\phi=\sin\phi$, hence $\psi=\phi=45^\circ$, which is not feasible here.

A trapezoid is a quadrilateral with at least one pair of opposite sides being parallel. A square fits this definition. Any theorem applying to a trapezoid also applies to a square.

• That depends on your definition. My text book doesn't consider a square to be a trapezoid. I will edit the question to add this information. Thanks for pointing that out. – Wheepy Oct 30 '16 at 4:11
• @Robert. A trapezoid is a quadrilateral with exactly one pair of opposite sides being parallel. A square fits the definition of a rectangle, not of a trapezoid. – imranfat Oct 30 '16 at 4:34
• @imranfat Would you please provide a reference. That would be a very unusual definition in my books. – dxiv Oct 30 '16 at 4:38
• The first line on that page: "a convex quadrilateral with at least one pair of parallel sides is referred to as a trapezoid". I'll leave it at that, and if your book defines it otherwise then I'd seriously recommend a different book. – dxiv Oct 30 '16 at 4:45
• So if you prove some arbitrary property for trapezoids, then you'd have to prove it again for parallelograms, rectangles, squares etc. It's technically fine to have your own non-standard definitions as long as you spell them out (and +1 you did that in the edit to your question), but that will invite miscommunications down the road. That's (just) my opinion, and this is my last comment here. – dxiv Oct 30 '16 at 4:50

Hint: assume the convex cyclic quadrilateral $ABCD$ includes the center $O$ of its circumscribed circle or radius $r$ (otherwise find a suitable one with larger perimeter). Then each angle $\widehat{AOB}\le \pi$ and $\sum_{cyc}\widehat{AOB} = 2 \pi$. Given that the perimeter is $P = \sum_{cyc} 2 r \sin \widehat{AOB}/2$ and $\sin x$ is concave on $(0,\pi)$ it follows by Jensen's inequality that:

$$P/2r = \sum_{cyc} \sin \frac{\widehat{AOB}}{2} \le 4 \sin\left(\frac{1}{4} \sum_{cyc} \frac{\widehat{AOB}}{2}\right) = 4 \sin \frac{\pi}{4} = 2 \sqrt{2}$$

with equality iff all angles $\widehat{AOB}/2$ are equal i.e. the quadrilateral is a square.

Note the symmetries of the circle, you should be able to convince yourself that the maximum peremiter quadrilateral is one that maximizes symmetries (if there was a way to make the perimeter larger on one side of a line of symmetry the same action can always be taken by the other side) this leaves the square and the the crossed square as possible shapes, the crossed square has greater perimeter then the square but it is not the limit of expanding the symmetries of a trapezoid, therefore a square has the maximum perimeter of not self intersecting quadrilaterals.

• That's good intuition, but not a proof. At face value, the exact same argument would work to prove that the square is the inscribed quadrilateral of minimum permieter. – dxiv Oct 30 '16 at 5:53
• yeah but i didn't want to give a full proof just a starting point. one can show that the square is fixed at either max or min then simply check against a non square rectangle. @dxiv – shai horowitz Oct 30 '16 at 5:55
• you can also prove the shortest perimeter of an inscribedshape is a rectangle with the 2 lengths approaching 0 yielding a min perimeter of 2d. – shai horowitz Oct 30 '16 at 5:58
• this minimization fixed point becomes important in proving that the square is the quadrilateral of most area with min perimeter. – shai horowitz Oct 30 '16 at 6:09


text:
# [tex-eplain] Eplain 3.1

Wed Dec 6 19:53:26 CET 2006

```On Tue, Dec 05, 2006 at 06:36:07PM -0500, John R. Culleton wrote:
> Same old bug appears:
> ...
> (./eplain.tex
> ! No room for a new \count .

I've tried it with the new and the old eplain, and it works with both,
so it must be something in contex which has changed.  My context is
very old:

ConTeXt  ver: 2005.01.31  fmt: 2006.4.21  int: english  mes: english

so I guess, to debug this, I'll need to install a new context, which I
heard is not simple (which version do you have, btw?).  I'll try to do
this, but it might not be soon, maybe someone who has a recent context
can debug this?

> Since new versions are tested with LaTeX would it hurt to test them with
> cntext also?

I'll do that, although as you can see it's not guaranteed to work with
a newer (or older?) context than mine...


text:
inside dollar symbols: $eq$. As your desire to program complex automation and process control functions increases then higher level ladder logic symbols can be used. This section is a consideration of such logic functions. In additions to the four logical functions outlined above, Microsoft Excel provides 3 "conditional" functions - IF, IFERROR and IFNA. There are two series of symbols for logic gates. Hot Network Questions How to sketch a right-angled triangles? First-Order and Second-Order Logic. The English expression "It is not the case that . ¥First-order logic (FOL) models the world in terms of ÐObjects, which are things with individual identities ÐProperties of objects that distinguish themtfrom other objects ... variable symbol, or an n-place function of n terms. ." Returns TRUE if any argument is TRUE. In logic, a set of symbols is commonly used to express logical representation. The function of these contacts is to check for a positive or negative change in the input signal. Functions of Flowchart Symbols. TRUE function. We compared “AND” and “OR” logic function blocks with their ladder logic counterpart. How is first-order logic a strong enough logic for the foundations of mathematics? We fix an enumerable set Fun of function symbols. Symbol L a T e X Symbol L a T e X Symbol L a T e X Symbol … In arguments of the logical functions, you can use cell references, numeric and text values, Boolean values, comparison operators, and other Excel functions. Combined called combinatorial logic. NOT function. Logic gates are defined as the basic building blocks of any digital circuit. S/N FUNCTION CATEGORY DESCRIPTION USAGE; 01 : AND : Logical : Checks multiple conditions and returns true if they all the conditions evaluate to true. . The traditional symbols have distinctive shapes making them easy to recognise, they are widely used in industry and education. Logic gates are the basic building blocks used typically in the field of Digital Electronics. As the name of the ladder logic symbols also tells, these functions are made to register either positive or negative changes. In addition to the link above What is the difference between a predicate and a function, you might want to read the distinctions provided by Wikipedia.See, e.g. Logic NOR Gates are available using digital circuits to produce the desired logical function and is given a symbol whose shape is that of a standard OR gate with a circle, sometimes called an “inversion bubble” at its output to represent the NOT gate symbol with the logical operation of the NOR gate given as. Logic Operators Quantifiers Deduction symbols See also References External links The following information is provided for each mathematical symbol: Symbol ... Function grows faster than \omega ω U+03C9 Symbol Usage Interpretation Article LaTeX HTML Unicode First or second derivative of function Reverses the logic of its argument. Logic gate that functions as an AND and a NAND : Logic gate exclusive NOR XNOR gate + info: Logic gate that functions as an OR and a NOR : Equivalent to logic gate XNOR + info: Logic buffer : Logic inverter NOT + info: Logic buffer tri-state + info : Negated logic buffer: Differential : Logic driver: Logic Gate Symbols, British System (BS 3939) SWITCH function. Essentially, a single function block can take the place of an entire line of ladder logic. Each gate performs a specific logic function behind which there is an equivalent transistor based electronic circuit. Each function and predicate symbol has an arity k>0. Additionally, the third column contains an informal definition, the fourth column gives a short example, the fifth and sixth give the Unicode location and name for use in HTML documents. •If there are n people and m locations, representing the fact that some person moved from one location to another requires nm2 separate symbols. Logical Symbols. Excel logical functions - facts and figures. First-Order Logic (First-Order Predicate Calculus) 2 Propositional vs. Predicate Logic •In propositional logic, each possible atomic fact requires a separate unique propositional symbol. 9. Basically, there are two different bit logic functions or operations in FBD. ; In short, a predicate is a (strictly Boolean-valued) function, but a function … In LaTeX, there are several ways to create equations: start with $$and end with$$. Schmitt inverter: The Schmitt inverter is a logic gate designed with the hysteresis phenomenon due to the positive feedback inside the gate. Viewed 64 times 0 $\begingroup$ Consider the following definition. Logic signs and symbols. Boolean Function. Learning the basic ladder logic symbols will give you a solid foundation. There are 3 basic logic gates- AND Gate, OR Gate, NOT Gate. Completeness of first-order logic with interpreted function symbols. probability function: probability of event A: P(A) = 0.5: P(A ⋂ B) probability of events intersection: ... Logic symbols.
2020 function symbols logic


text:
KU Physics & Astronomy LOcally Organized Assembly (PALOOZA) meeting 2020-2021

Feb 23 – 26, 2021
Zoom Passcode: 990417
America/Chicago timezone

Performance of CMS Endcap Precision Timing Sensors

Feb 23, 2021, 6:05 PM
10m
Zoom Link: https://kansas.zoom.us/j/92389769214 (Zoom Passcode: 990417)

Zoom Passcode: 990417

Zoom Link: https://kansas.zoom.us/j/92389769214 Zoom Passcode: 990417 Zoom Meeting ID: 923 8976 9214

Speaker

Margaret Lazarovits

Description

The Large Hadron Collider, a particle accelerator in Geneva, Switzerland, will enter its High Luminosity (HL) phase, whose upgrade has already started. This phase will include an increase in particle collisions, achieving instantaneous luminosities of 5 to 7 times the nominal luminosity, $1 \times 10^{34}$ cm$^{-2}$ s$^{-1}$. This increase gives more opportunities for rare processes to come to light, however it also means that there is more overlapping particle interactions, called pileup, in the detectors. To address this added pileup, the Compact Muon Solenoid (CMS) detector at the LHC will install a new precision timing detector, the MIP Timing Detector (MTD), as part of the upgrades for the HL-LHC. The sensors comprising the endcap section of the MTD are silicon low gain avalanche detectors (LGADs). From extensive tests at the Fermilab test beam facility, it has been shown that prototypes of these sensors perform in accordance to our expectations and requirements for the MTD. Specifically, these sensors have demonstrated a timing resolution within 30-40 picoseconds.

Type of contribution Oral contribution (20 minutes)


text:
# One-sided measurement error: $\widetilde{X} = X - \eta$, $\eta\geq0$. Is $E[X|\widetilde{X}=\widetilde{x}]$ increasing in $\widetilde{x}$?

Let $X\geq0$, $\eta\geq0$ and $X,\eta$ independent. We measure $X$ with a one-sided error: $\widetilde{X} = X - \eta$.

Is $E[X|\widetilde{X}=\widetilde{x}]$ increasing in $\widetilde{x}$?

• I was reading $\eta$ as a parameter (in accordance with statistical convention, and for which the answer is trivial), but then it makes no sense to declare it as independent of $X$ (since it's a fixed quantity). ... so just to confirm, $\eta$ is another random variate? [If so, why not stick to the convention of uppercase Roman letters?] – Glen_b Sep 6 '18 at 7:19
• Yes, $\eta$ is a random variable. $\eta$ looks like an n as in noise and is typically chosen in the context of measurement error, e.g. $\eta$ on Wikipedia. – robust Sep 6 '18 at 22:59
• Ah, thanks, in the context of an error term it makes sense. Which variable are you taking expectation over in $E[X|\widetilde{X}=\widetilde{x}]$? I guess this is $E_X[X|\widetilde{X}=\widetilde{x}]$ but did you intend $E_\eta[X|\widetilde{X}=\widetilde{x}]$ (or perhaps something else) instead? – Glen_b Sep 6 '18 at 23:26
• In this case it doesn't matter what variable the expectation is taken over, because conditioning on $\widetilde{X}$ implies that $X$ and $\eta$ determine each other. – robust Sep 7 '18 at 4:55

Assuming the expectation is over $X$:

$E_X[X|\widetilde{X}=\widetilde{x}]=E_X[X|X-\eta=\widetilde{x}]=E_X[X|X=\widetilde{x}+\eta]=\widetilde{x}+\eta$,

which is not increasing in $\widetilde{x}$ because of the $\eta$.

However, $E_\eta[E_X(X|\widetilde{x})]=\widetilde{x}+E(\eta)$ would (naturally) be increasing in $\widetilde{x}$.

• Can you clarify what you mean by "because of the $\eta$"? Also, why is $E_\eta[E_X(X|\widetilde{x})] = \widetilde{x}$? – robust Sep 7 '18 at 4:47
• The second thing is an error on my part (now corrected): I took something you'd refer to as a noise term to be mean $0$ but it's not in your case. With the respect to the first thing, because $\eta$ is a random variable; you could increase $\tilde{x}$ but $\tilde{x}+\eta$ might not increase with it. – Glen_b Sep 7 '18 at 8:44
• I don't follow $E_\eta[E_X(X|\widetilde{x})]=\widetilde{x}+E(\eta)$: Taking $E_\eta$ on the outside expectation implies that the conditioning variable $\widetilde{X}$ in the inside expectation also varies, and hence the right-hand side cannot contain a simple $\widetilde{x}$. On the other hand, $E_{\eta|\widetilde{X}=\widetilde{x}}[E_X(X|\widetilde{x})]=\widetilde{x} + E(\eta|\widetilde{X}=\widetilde{x})$, which by itself doesn't provide a definite answer to the question. – robust Sep 7 '18 at 17:30
• "Taking Eη on the outside expectation implies that the conditioning variable X˜ in the inside expectation also varies," ... -- I don't think so – Glen_b Sep 8 '18 at 2:13
• Take the setup from my answer, and let $\widetilde{x}=4$. In that setup $E[\eta]=1$, and hence $\widetilde{x} + E[\eta]=5$, which is larger than any value in the support of $X$, $\eta$ and $\widetilde{X}$. – robust Sep 8 '18 at 3:19

No, not in general.

Let $P(X=3)=P(X=4)=P(\eta=0)=P(\eta=2)=0.5$.

Then,

• $E[X|\widetilde{X}=1] = 3$
• $E[X|\widetilde{X}=2] = 4$
• $E[X|\widetilde{X}=3] = 3$
• $E[X|\widetilde{X}=4] = 4$

which is neither increasing nor decreasing in $\widetilde{x}$.

If we modify the original question, and require $X$ and $\eta$ to have continuous or common support, the counterexample remains valid if sufficent probability mass is put on above values.


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# Space groups: Herman-Maugin notation, diagrams and Wyckoff positions

I am currently taking an Introduction to Crystallography course. After studying and understanding the punctual symmetry groups (2mm, 4/m 2/m 2/m, 32,...), the flat groups and the diverse elements of symmetry that exist: helical axes, sliding planes, symmetry planes, among others.

I have doubts when building the diagram of a certain spatial group. For example, in an exercise, I am asked to construct the diagram of the spatial group Pmab.

Pmab implies the existence of: an ordinary plane of symmetry perpendicular to the crystallographic axis a, a sliding plane of type a perpendicular to the crystallographic axis b, a sliding plane of type b perpendicular to the crystallographic axis c.

Also, Pmab is an abbreviated notation, because, in reality, there are also two helical binary axes parallel to the axes "a" and "b", and a binary axis parallel to the axis "c".

On the other hand, Pmab belongs to the orthorhombic (rhombic) system and derives from the point symmetry group 2/m 2/m 2/m 2/m.

This is all the information I get from the Herman-Maugin notation. If there is something wrong that I have understood, I hope you will tell me.

Once here I don't know how to start building the diagram, which would look like this:

As far as Wyckoff positions are concerned, I am often asked for them, but I am not quite sure what they are.

• How do you know that the two binary axes along "a" and "b" are helical? – Ivan Neretin Jul 3 '19 at 12:53
• I know, because in the orthorhombic system, simplified notation tends to obviate them. But I don't understand why – aprendiendo-a-programar Jul 3 '19 at 13:13
• That would explain how do you know there are axes in the first place. That's right, but that's not what I asked. How do you know they are helical? – Ivan Neretin Jul 3 '19 at 13:45
• I don't know, I said they're helical, because that's how it appears on the diagram. – aprendiendo-a-programar Jul 3 '19 at 14:42
• Well, they are helical all right, but you need to know how to deduce that from "Pmab" alone. – Ivan Neretin Jul 3 '19 at 14:49

Well, knowing crystallography implies being able to deduce the picture above solely from the group's name $$\rm Pmab$$, and that's how it is done.

First you write down the symmetry elements mentioned explicitly in the group's name (that is, $$\rm m,\;a$$, and $$\rm b$$ in our case; in other examples there could have been axes among them, or even worse) and orient those accordingly, which you already know how to do. Then you combine these elements in all possible ways to find out what else is hidden beneath.

Essentially, any symmetry element is but a linear transformation that takes any vector $$\bf\vec x$$ to another vector $$\bf A\vec x+\vec b$$, where $$\bf A$$ is the $$3\times3$$ matrix of our transformation (rotation or reflection or otherwise), and $$\bf\vec b$$ is the shift, that is, a half-translation along the axis for a screw axis, or a half-translation in whatever direction for a glide plane, or $$\bf\vec0$$ in all other cases. The typical matrices are: $$\begin{array}{cc} \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} & \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix}& \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix}\\ \text{Reflection \perp X} & \text{Rotation about X}& \text{Center} \end{array}$$

Given two symmetry elements, you simply combine the two transformations by doing some matrix multiplication, which I sincerely recommend, if only to know how it feels. Alternatively, you may rely upon the wisdom of the predecessors, which says:

1. Two perpendicular planes give a twofold axis parallel to both.
2. A plane and a parallel twofold axis give another plane, perpendicular to the plane and parallel to the axis.
3. A plane and a perpendicular twofold axis give a center.
4. Two perpendicular twofold axes give a third axis, perpendicular to both.
5. All of the above remains true if you change
• every "plane" to "plane or a glide plane" and
• every "axis" to "axis or a screw axis".
6. When some of the combined elements have shift in them, the following happens:
• If the shift is parallel to the resulting element, it becomes a part of the said element (that is, the result is going to be a screw axis or a glide plane).
• If the shift is perpendicular to the resulting element, it moves the said element by half the value of the shift (that is, by $$1\over4$$ of a translation).
7. If you end up with a center of symmetry someplace other than at the origin, you relocate your origin there.

Now try to apply the rules 1-7 and see where this gets you.

• Comments are not for extended discussion; this conversation has been moved to chat. – andselisk Jul 4 '19 at 16:07


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# Rothe-Hagen identity

In mathematics, the Rothe-Hagen identity is a mathematical identity valid for all complex numbers ($x, y, z$) except where the denominators vanish:

$sum_\left\{k=0\right\}^nfrac\left\{x\right\}\left\{x+kz\right\}\left\{x+kz choose k\right\}frac\left\{y\right\}\left\{y+\left(n-k\right)z\right\}\left\{y+\left(n-k\right)z choose n-k\right\}=frac\left\{x+y\right\}\left\{x+y+nz\right\}\left\{x+y+nz choose n\right\}.$

Search another word or see Rothe-Hagen identityon Dictionary | Thesaurus |Spanish


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Title: Kindergarten counting practice 1-10 Author: K5 Learning Subject: Navigation. Chemistry: Counting Atoms Question Preview (ID: 2474) Solve Each Problem And Check The Right Answer. Conversions within the American system (such as pounds to ounces, the number of feet in a mile, the number of inches in a foot, etc). Also, this practice will also help students to count the atoms in a chemical formula whether it is a single molecule of a compound or multiple molecules. Wikibooks Ideally, every possible organic compound should have a name from which an unambiguous structural … ACT Test Tips. If you counted 10 molecules very second for 50 years without stopping you would have counted only 1.6 × 10 10 molecules. 10. For example, a dozen is defined as 12 objects, and a pound is defined as 16 ounces. Counting numbers 1–10 is a fundamental building block of math and number sense skills. DJC Kids Media kindergarten, pre-school and ESL Easy reader phonics based picture books and videos. Is your kid crazy about counting? Watch all of our videos ad free with our app (desktop, apple, or android):https://www.thesingingwalrus.tv/Only $4.99 USD per month and$44.99 USD for a year!\"Counting Numbers 1-10\" is a part of a fun series of educational videos for kids. 3 counting atoms PowerPoint chemistry worksheets covering multiatomic ions, formulas with parenthesis for the paperless digital classroom, Microsoft Teams, distance learninganswer key included⭐paper version⭐Google Slides version⭐Google Forms version⭐Boom Cards version⭐⭐⭐BIG Bundle of Chemistry⭐⭐⭐Cre. While in its simplest representation, the 18-electron rule is explained in that a closed, stable noble gas configuration of ns 2(n-1)d 10 np 6 … Feb 22, 2016 - This worksheet will serve as a practice to help students distinguish one element from another in a variety of chemical formulas. Counting Numbers | Numbers 1-10 lesson for children - YouTube Start studying Organic Chemistry Alkane Names 1-10 carbons. Don't forget to subscribe to our Youtube channel! Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop. An alkyl group is a radical of with a certain number of carbons and is of the general formula C n H 2 n + 1 {\displaystyle {\text{C}}_{n}{\text{H}}_{2n+1}} that has had one of its hydrogens removed, freeing up one of its bonds. Chemists needed a name that can stand for a very large number of items. Counted numbers are exact: there are two chairs in the photograph. Another way to determine the power of 10 in scientific notation is to count the number of places you need to move the decimal point to get a numerical value between 1 and 10. handwriting worksheets) for parents and teachers. A mass reported as 0.5 grams is implied to be known to the nearest tenth of a gram and not to the hundredth of a gram. Here's a fun and unique way for students to practice counting atoms! Measured numbers have a limited number of significant figures. in a compound, and form more comple… Number writing, sequencing and word/number matching skills help your child get ready for kindergarten. This particular resource used the following sources: http://www.boundless.com//chemistry/definition/exact-numbers a) 6 x 10… There are over 20 chemistry problems in this sweet and delicious picture. Metric or SI (Le Système International d'Unités) prefix are based on powers of ten.They are modifiers on the root word that tell us the unit of measure. Students will count the number of atoms in each formula and color based on the key! In contrast, measured numbers always have a limited number of significant digits. HAVE FUNNNNN! generates 1 strongly bonding and n–1 weakly bonding, nonbonding or antibonding molecular orbitals. You count the colonies on this plate and find that it has 179 colonies. Play games to get them! As he counts along, he'll get stronger at writing numbers, too. : ... Scientific notation takes the form of M x 1 On where 1 I M < 10 and 'n" represents the number of decimal places to be moved. The 1:10 dilution plate has many colonies, maybe more than 300. This allows it to bond to a carbon (or oxygen, etc.) Flickr An exact number can only be expressed in one way and cannot be simplified any further. CC BY-SA 3.0. http://en.wikibooks.org/wiki/AP_Chemistry/The_Basics Eighty-seven people attended the lecture. This number chart gives him a real counting challenge in naming the numbers up to 100. (adsbygoogle = window.adsbygoogle || []).push({}); Exact numbers are either defined numbers or the result of a count. Search this site. In short, 1 mol contains 6.022 140 76 × 10 23 of the specified particles.. To determine the number of molecules in a value of moles for a substance, Avogadro’s number must be used. This means that 0.5 mL of the 1:100 dilution contains 179 CFU. Thus we can say that a total of n+1 bonding molecular orbitals are generated in the process of cluster formation, and the number of electrons required to fill these bonding molecular orbitals is 2n+2. CC BY-SA 3.0. http://commons.wikimedia.org/wiki/File:Waage.Filter.jpg 92 Chemistry: Matter and Change ... confusion when counting zeros! • 1 mole is the amount of atoms in 12 g of pure carbon-12, or 6.02 1023 representative particles, which is any kind of particle – an atom, a molecule, a formula unit, an electron, an ion, etc. Attendance Policy. Counting Practice from 1 to 10 Kindergarten Counting Worksheet 1) 1 4 9 Count by 1 from 1 to 10 2) 7 9 10 Count by 1 from 1 to 10 3) 1 8 9 Count by 1 from 1 to 10 4) 3 5 7 Count by 1 from 1 to 10. In chemical nomenclature, the IUPAC nomenclature of organic chemistry is a method of naming organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry (IUPAC). You May Use A Periodic Table, The Mole Conversions Handout, And A Calculator. There are fifteen books on the shelf. Learn 1 10 chemistry prefixes with free interactive flashcards. Visit BYJUS to learn more about it. Exact numbersThese numbers are either defined numbers or result of a count. Exact numbers have an infinite number of significant figures, but they often appear as integers. Boundless Learning We have learned to express these numbers as powers 10. How many grams are in 1.00 mole of Li? Students love to color and this … One Time Payment $10.99 USD for 2 months: Weekly Subscription$1.99 USD per week until cancelled: Monthly Subscription \$4.99 USD per month until cancelled: Annual … 15. In order to correctly convert one metric unit to another, you will need to determine which of two prefixes represents a bigger amount and then determine the exponential "distance" between them. CHEMISTRY: COUNTING ATOMS IN COMPOUNDS WORKSHEET INSTRUCTIONS: Write the quantity of atoms of each element opposite the formula of the compound for the quantity of formula units and molecules shown: 24 For example: 5P203 For example: (5x3=ì15 4. Come and join our community on Facebook, or subscribe to our Youtube Channel! SECTION 10.1 Measuring Matter The number of places equals the power of 10. They have an infinite number of significant figures. Basically, that's why this particular unit was invented. An exact number has absolutely no uncertainty in it. Chemistry Expectations and Information (Home) Syllabus. This Counting 1-10 workbook from the Get Ready! 16. Because 12.000g of carbon is an easily manageable amount of substance to use, the mole became our standard unit for counting the numbers of atoms or molecules. [print questions] Answers? This number is positive if you move the decimal point to the right and negative if you move the decimal point to the left. Exact numbers have an infinite number of significant figures, but they often appear as integers. CC BY-SA. https://www.flickr.com/photos/docnic/2870499877/ For example, a dozen is defined as 12 objects, and a pound is defined as 16 ounces. Use this diverse collection of basic math worksheets to guide your children on their exploration of numbers. It can be used as an introductory video before showing the numbers song. Think Dr. Seuss meets Sandra Boynton! 18. series delivers big, how-to-count learning fun in a compact size. In that case you can go onto the 1:100 plate. It's a lot easier to write the word 'mole' than to write '6.02x10 23 ' anytime you want to refer to a large number of things. Mole: unit for counting and one mole of any substance is equal to 6.022 x 10 23 particles of that substance. Metric Prefixes. 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X 10 23 eggs high-quality, openly licensed content from around the Internet Use! Along, he 'll get stronger at writing numbers, too number is called Avogadro ’ s number standard is! Math worksheets to guide your children on their exploration of numbers the standard form is large. Needed a name that can stand for a substance 6.022 x 10 of! In a value of moles for a substance, Avogadro ’ s number 140 ×. To reach 100 it has 179 colonies our Youtube channel chemistry problems in this sweet and delicious picture find it... Video before showing the numbers song the photograph chemistry teachers is 6.02x10 23 teachers. Counting challenge in naming the numbers up to 100 be simplified and have an infinite numbers of significant figures but... 1.6 × 10 10 molecules the specified particles or oxygen, etc. children their... Useful in cases where the counting of substances is difficult because of their small size and higher! 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# Mathematical Statistics and Data Analysis - Solutions

### Chapter 7, Survey Sampling

#### Solution 19

For the first part, we need to find the constant $\, k \,$ such that the interval $\, (-\infty, \bar X + ks_{\bar X}) \,$ is a 90% confidence interval for $\, \mu \,$.

We shall use the same function here $\, z(\alpha) \,$ as used in the book. Also let $\, Z \,$ denotes the standard normal function.

Thus we have:

Thus for 90% confidence interval we need $\, \alpha = 0.9 \,$. And $\, k = z(1-\alpha) \,$. By the definition of $\, z \,$, we have $\, z(\alpha) = \Phi^{-1}(1-\alpha) \,$. Thus $\, k= \Phi^{-1}(1-(1-\alpha)) = \Phi^{-1}(0.9) \,$

Now we just need to lookup in the standard normal table to find $\, k = 1.28 \,$.

Second part is also similar where we need to find the value of $\, k \,$ for the one sided confidence interval $\, (\bar X - ks_{\bar X}) \,$.

We can proceed from the last result:

Thus for 95% confidence interval we need $\, \alpha = 1-0.95=0.05 \,$. And $\, k = z(\alpha) \,$. By the definition of $\, z \,$, we have $\, z(\alpha) = \Phi^{-1}(1-\alpha) \,$. Thus $\, k= \Phi^{-1}(1-\alpha) = \Phi^{-1}(0.95) = 1.645 \,$.

$$\tag*{\blacksquare}$$


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# Programmatically setting Emacs frame size

My emacs (on Windows) always launches with a set size, which is rather small, and if I resize it, it's not "remembered" at next start-up.

I've been playing with the following:

``````(set-frame-position (selected-frame) 200 2) ; pixels x y from upper left
(set-frame-size (selected-frame) 110 58) ; rows and columns w h
``````

which totally works when I execute it in the scratch buffer. I put it in my .emacs, and although now when I start the program, I can see the frame temporarily set to that size, by the time `*scratch*` loads, it resets back to the small default again.

Can anyone help me fix up the above code so that it "sticks" on start-up?

-
See this previous question as well: How do I set the size of emacs’ window? –  Chris Conway Dec 2 '08 at 22:59

Here's what I use in my ~/.emacs:

(add-to-list 'default-frame-alist '(left . 0))

(add-to-list 'default-frame-alist '(top . 0))

(add-to-list 'default-frame-alist '(height . 50))

(add-to-list 'default-frame-alist '(width . 155))

-
I had given up on this problem for a while ago, using your code fixed it finally :) –  Rene Saarsoo Dec 6 '08 at 19:42

Did you try this : emacs -geometry 110x58+200+2 &

Found at :

-

For emacs on windows, I generally put it in the registry.

``````HKCU\Software\GNU\Emacs\
Emacs.Geometry REG_SZ "245x74"
``````

(This keeps machine-local settings out of my .emacs file, which I share with many other machines...)

-

Between Emacs 19 and 21, I used to have a perfectly working .emacs that did exactly what you are looking for, by distinguishing between default-frame and initial-frame:

``````  (setq default-frame-alist '((foreground-color . "LightGray")
(background-color . "Black")
(cursor-color . "Medium Sea Green")
(width . 80)
(height . 36)
(vertical-scroll-bars . right)))
``````

and

``````  (setq initial-frame-alist
(cons '(width . 96)
(cons '(height . 72)
(cons '(menu-bar-lines . 1)
initial-frame-alist))))
``````

Alas, when I "upgraded" to Emacs 23.2 the above doesn't fully anymore. Emacs is not my career. I am using it as a tool, so I can't devote too much time delving into understanding why. So I simply worked around the problem by adding to the Emacs Windows XP shortcut "-geometry 96x72". So the shortcut target now looks:

``````C:\emacs-23.2\bin\runemacs.exe -geometry 96x72
``````
-

None of the solutions posted before could solve my resizing problem, so I thought I'd share an additional solution: upon start-up, Emacs 23 uses a certain font, before switching to the user-defined one; as a consequence, if this font is large, Emacs also reduces the window size so that it fits in the screen. Then, the user font is applied, and the window might appear smaller than what desired by the user. Thus, adding the following in my ~/.Xresource solved the problem:

``````Emacs.font: -*-fixed-*-*-*-*-8-*-*-*-*-*-*-*
Emacs.dialog*.fontList:        -*-fixed-*-*-*-*-8-*-*-*-*-*-*-*
``````

This makes Emacs use a small font during start-up, so that its window is not too large and is not truncated. Of course, you might want to customize the font you want here, or in the Emacs init file.

-
``````(setq initial-frame-alist '(


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# The effect of scale of predictor variables in regression techniques

In polynomial regression, it is recommended to center predictor input variables to break multi colinear relationships of x to x^2.

From Wikipedia: The underlying monomials can be highly correlated "For example, x and x2 have correlation around 0.97 when x is uniformly distributed on the interval (0, 1). "

When a variable x is between -1 and 1, x^2 makes the magnitude smaller while when x is outside of that range, x^2 makes x's magnitude larger.

Making the variable into an integer variable could change the behavior.

E.g.

df$x=round((df$x - mean(df\$))*100)

Any opinions on the scale especially in regards to interval [-1,1] vs [-100,100]

It is common to normalize predictors subtracting the mean and dividing by the standard deviation when doing inference analysis but this question pertains to regression prediction.

Asking a similar question in regards to natural log, a variable that has a range (0,1] has a dramatically different transformed value than [1,100].

log(seq(0.1,1,.1)) #mostly negative
log(seq(0.1,1,.1)*100) #rather positive

If the predictor variable in the case of log happened to be sometimes less than 1 and others greater than 1, that could make the transformation act a little "wild". Would it be best to transform the variable to be within (0,1] or [1,] but not both?


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We report the discovery of a Jupiter-mass planet orbiting an M-dwarf star that gave rise to the microlensing event OGLE-2011-BLG-0265. Such a system is very rare among known planetary systems and thus the discovery is important for theoretical studies of planetary formation and evolution. High-cadence temporal coverage of the planetary signal, combined with extended observations throughout the event, allows us to accurately model the observed light curve. However, the final microlensing solution remains degenerate, yielding two possible configurations of the planet and the host star. In the case of the preferred solution, the mass of the planet is ${{M}_{{\rm p}}}=0.9\pm 0.3\ {{M}_{{\rm J}}}$, and the planet is orbiting a star with a mass $M=0.22\pm 0.06\ {{M}_{\odot }}$. The second possible configuration (2σ away) consists of a planet with ${{M}_{{\rm p}}}=0.6\pm 0.3\ {{M}_{{\rm J}}}$ and host star with $M=0.14\pm 0.06\ {{M}_{\odot }}$. The system is located in the Galactic disk 3–4 kpc toward the Galactic bulge. In both cases, with an orbit size of 1.5–2.0 AU, the planet is a "cold Jupiter"—located well beyond the "snow line" of the host star. Currently available data make the secure selection of the correct solution difficult, but there are prospects for lifting the degeneracy with additional follow-up observations in the future, when the lens and source star separate.

### OGLE-2011-BLG-0265Lb: A JOVIAN MICROLENSING PLANET ORBITING AN M DWARF

#### Abstract

We report the discovery of a Jupiter-mass planet orbiting an M-dwarf star that gave rise to the microlensing event OGLE-2011-BLG-0265. Such a system is very rare among known planetary systems and thus the discovery is important for theoretical studies of planetary formation and evolution. High-cadence temporal coverage of the planetary signal, combined with extended observations throughout the event, allows us to accurately model the observed light curve. However, the final microlensing solution remains degenerate, yielding two possible configurations of the planet and the host star. In the case of the preferred solution, the mass of the planet is ${{M}_{{\rm p}}}=0.9\pm 0.3\ {{M}_{{\rm J}}}$, and the planet is orbiting a star with a mass $M=0.22\pm 0.06\ {{M}_{\odot }}$. The second possible configuration (2σ away) consists of a planet with ${{M}_{{\rm p}}}=0.6\pm 0.3\ {{M}_{{\rm J}}}$ and host star with $M=0.14\pm 0.06\ {{M}_{\odot }}$. The system is located in the Galactic disk 3–4 kpc toward the Galactic bulge. In both cases, with an orbit size of 1.5–2.0 AU, the planet is a "cold Jupiter"—located well beyond the "snow line" of the host star. Currently available data make the secure selection of the correct solution difficult, but there are prospects for lifting the degeneracy with additional follow-up observations in the future, when the lens and source star separate.
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# Science:Infinite Series Module/Appendices/Proof of The Ratio Test/Absolute Convergence

The ratio test requires the idea of absolute convergence. Given any infinite series Σak, we can introduce the corresponding series

${\displaystyle \sum _{k=1}^{\infty }|a_{k}|=|a_{1}|+|a_{2}|+|a+3|+\ldots }$

whose terms are the absolute values of the original series. We can explore whether this corresponding series converges, leading us to the following definition.

Definition: Absolute Convergence
The infinite series

${\displaystyle \sum _{k=1}^{\infty }a_{k}}$

is absolutely convergent if the series of absolute values

${\displaystyle \sum _{k=1}^{\infty }|a_{k}|}$

is convergent.

A few simple examples demonstrate the concept of absolute convergence.

### Example: Convergent p-Series

The infinite series

${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k^{2}}}}$

is absolutely convergent because

${\displaystyle \sum _{k=1}^{\infty }{\Bigg |}{\frac {(-1)^{k}}{k^{2}}}{\Bigg |}=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}}$

is a convergent p-series (p =2).

### Example: Harmonic Series

The infinite series

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k}}\end{aligned}}}

is convergent (by the alternating series test), but is not absolutely convergent because

${\displaystyle \sum _{k=1}^{\infty }{\Bigg |}{\frac {(-1)^{k}}{k}}{\Bigg |}=\sum _{k=1}^{\infty }{\frac {1}{k}}}$

is the infamous harmonic series, which is not a convergent series.

It is possible for a series to be convergent, but not absolutely convergent (such series are termed conditionally convergent, but we do not need this definition for our purposes).


text:
# zbMATH — the first resource for mathematics

Haagerup approximation property for arbitrary von Neumann algebras. (English) Zbl 1335.46052
A finite von Neumann algebra $$M$$ is said to have the Haagerup approximation property if for some (equivalently, every) faithful normal tracial state $$\tau$$ on $$M$$ there exists a net of unital completely positive $$\tau$$-preserving maps $$\Phi_i$$ converging to the identity in the point weak$$^*$$-topology and such that their natural $$L^2(M, \tau)$$ versions are compact. This notion, motivated by the Haagerup property for discrete groups, was introduced by M. Choda [Proc. Japan Acad., Ser. A 59, 174–177 (1983; Zbl 0523.46038)] and further studied by P. Jolissaint [J. Oper. Theory 48, No. 3, 549–571 (2002; Zbl 1029.46091)].
In the present article, the authors extend this notion to arbitrary $$\sigma$$-finite von Neumann algebras, with the standard form framework as the starting point – so that the defining approximations are first viewed as operators $$T_i$$ on the canonical Hilbert space on which the von Neumann algebra acts. It is shown that in the finite case the notion coincides with that of M. Choda [Proc. Japan Acad., Ser. A 59, 174–177 (1983; Zbl 0523.46038)] and P. Jolissaint [J. Oper. Theory 48, No. 3, 549–571 (2002; Zbl 1029.46091)] and that given a faithful normal state $$\phi$$ one can always extract out of $$T_i$$ a corresponding family $$\Phi_i$$ of completely positive contractive $$\phi$$-preserving maps on the von Neumann algebra. The property is shown to be stable under taking crossed products by locally compact abelian groups, and under passing to expected subalgebras (even if the corresponding conditional expectation is not normal).
For further information on the recent developments on the Haagerup property for arbitrary von Neumann algebras we refer to [M. Caspers and A. Skalski, Commun. Math. Phys. 336, No. 3, 1637–1664 (2015; Zbl 1330.46057); Int. Math. Res. Not. 2015, No. 19, 9857–9887 (2015; Zbl 1344.46043)], another forthcoming preprint by the authors, entitled “Haagerup approximation property and positive cones associated with a von Neumann algebra”, arXiv:1403.3971, and their joint preprint with Narutaka Ozawa, “Haagerup approximation property via bimodules”, arXiv:1501.06293.

##### MSC:
 46L10 General theory of von Neumann algebras
##### Citations:
Zbl 0523.46038; Zbl 1029.46091; Zbl 1330.46057; Zbl 1344.46043
Full Text:
##### References:
 [1] C. Anantharaman-Delaroche, Amenable correspondences and approximation properties for von Neumann algebras, Pacific J. Math. 171 (1995), 309-341. · Zbl 0892.22004 [2] H. Ando and U. Haagerup, Ultraproducts of von Neumann algebras, J. Funct. Anal. 266 (2014), 6842-6913. · Zbl 1305.46049 [3] H. Ando, U. Haagerup and C. Winsløw, Ultraproducts, QWEP von Neumann algebras, and the Effros-Maréchal topology, J. Reine Angew. Math. (2014) (online). · Zbl 1382.46047 [4] H. Araki, Some properties of modular conjugation operator of von Neumann algebras and a non-commutative Radon-Nikodym theorem with a chain rule, Pacific J. Math. 50 (1974), 309-354. · Zbl 0287.46074 [5] F. Boca, On the method of constructing irreducible finite index subfactors of Popa, Pacific J. Math. 161 (1993), 201-231. · Zbl 0795.46044 [6] M. Brannan, Approximation properties for free orthogonal and free unitary quantum groups, J. Reine Angew. 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# [Answer] Which algebraic expression is a polynomial with a degree of 4? Geography S1 edg.

###### Answer: 9×4 – x3 – x/5 Mixtures and solutions
Which algebraic expression is a polynomial with a degree of 4? Geography S1 edg.

In mathematics a Möbius strip band or loop (US: / ˈ m oʊ b i ə s ˈ m eɪ-/ MOH-bee-əs MAY- UK: / ˈ m ɜː b i ə s /; German: [ˈmøːbi̯ʊs]) also spelled Mobius or Moebius is a surface with only one side (when embedded in three-dimensional Euclidean space) and only one boundary curve.The Möbius strip is the simplest non-orientable surface. It can be realized as a ruled surface.

Algebraic expression . The n th roots of unity are by definition the roots of the polynomial x n − 1 and are thus algebraic numbers. As this polynomial is not irreducible (except for n = 1) the primitive n th roots of unity are roots of an irreducible polynomial of lower degree called the cyclotomic polynomial and often denoted Φ n.

In mathematics the resultant of two polynomials is a polynomial expression of their coefficients which is equal to zero if and only if the polynomials have a common root (possibly in a field extension) or equivalently a common factor (over their field of coefficients).In some older texts the resultant is also called the eliminant.. The resultant is widely used in number theory eithe…


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This app uses the Newton forward difference method to generate the difference table, polynomial and approximate the value of this polynomial at the point say $x = x_{0}$. It also evaluates the $n^{th}$ derivative if specified.

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Array of $x$ values.
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Array of $y$ values.
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The variable to use as the unknown in the polynomial.
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Point at which to approximate the value of the generated polynomial.
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The Newton difference method to be used.
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Whether or not the point $x_{0}$ should be plotted on the graph.
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Whether or not to differentiate the polynomial. Select the order from the dropdown.
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Whether or not to add the derivative of the polynomial to the plot.
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Decimal points to be displayed.

Enter your valid inputs then click to display results.

Our applications use latest technologies to bring computational power to the web, and are the result of 10+ years of lecturing, training, research, programming and development.

Enter your valid inputs then click to display Python syntax.

Our applications use latest technologies to bring computational power to the web, and are the result of 10+ years of lecturing, training, research, programming and development.


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# Extensions of Chicken Nugget Theorem

Given that $$a,b,c\$$ are pairwise relatively prime, what is the largest number not expressible as a linear combination of $$a,b,c\$$?

What if $$a$$ and $$b$$ have common factor $$m$$?

What if $$a$$ and $$b$$ have common factor $$m$$, and $$b$$ and $$c$$ have common factor $$n$$?

• I don’t know how to do the notation correctly. Can anyone help me with that? – Lieutenant Zipp Apr 25 '19 at 21:22
• For starters, put dollar signs at each end of each attempted MathJax expression. – J.G. Apr 25 '19 at 21:23
• Out of curiousity...why do they call it the Chicken Nugget Theorem? – Mike Apr 25 '19 at 21:27
• – bounceback Apr 25 '19 at 21:29
• There was a Numberphile video on it a while back. I think it originally comes from people trying to order McDonald’s nuggets which came in boxes of 9 and 20. Or something. – Lieutenant Zipp Apr 25 '19 at 21:30

This is more of an extended comment than an answer to your specific questions.

This is called the Frobenius number of the semigroup generated by $$a,b$$ and $$c$$. The case for two generators, say $$a$$ and $$b$$ with $$a$$ and $$b$$ rel. prime, has a closed formula given by $$g(a,b)=ab-a-b.$$

However, in general for 3 generators it is known that there is no polynomial formula expressing the Frobenius number in terms of $$a,b$$ and $$c$$. The reference is F. Curtis (1990). "On formulas for the Frobenius number of a numerical semigroup". Mathematica Scandinavica. 67 (2): 190–192. You can find the paper here: https://www.mscand.dk/article/view/12330/10346

Your not all relatively prime cases, are restricted cases of the 2 number cases: $$a=md\land b=me \implies m(dx+ey)+cz=f$$


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# Gyula Lakos

Search this author in Google Scholar

Articles: 1

### On the $H$-ring structure of infinite Grassmannians

Gyula Lakos

Methods Funct. Anal. Topology 16 (2010), no. 3, 242-258

The $H$-ring structure of certain infinite dimensional Grassmannians is discussed using various algebraic and analytical methods but avoiding cellular arguments. These methods allow us to treat these Grassmannians in a greater generality.


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# Not getting proper height in unity when android device has notch or navigation bar on screen

Screen.height is not giving the proper resolution of an Android device when that device has a notch or navigation bar.

Due to this there is offset at bottom of screen:

Here I have set the Admob banner to display at the bottom. But it shows with some offset away from the bottom edge of the screen.

How can I get the proper height of the device in Unity and work with the notch on devices that have them?

• Your question is unclear. When you say Screen.height does not give proper resolution, do you mean it returns the space your game occupies and not the actual screen space? Because that's what supposed to happen, at least on android versions where the navigation bar is never minimised. Also you mention "Due to this, there is offset at the bottom of the screen". This does not make sense to me, are you saying that Unity produces a game that by default does not occupy the whole screen? Dec 6 '18 at 11:29
• I meant when i use Screen.height it give me 2156 but my screen resolution is 1080 X 2246. So i should get 2246 and I am new to unity Dec 6 '18 at 11:33

Screen.height returns the amount of pixels on the screen that are available to your app. In older versions of android the navigation bar would always remain static on the screen, eating a portion of it. This means that any app that runs does not have access on that portion of the screen. That is not an error, it's how the Android OS works, and in my opinion, it's the correct thing to do.
• @WrongTurn why do you need to find the notch area? The OS gives your game a specific space (Screen.width x Screen.height), you can only draw on that space. Android does not let developers mess with the notch so that it's not abused. Dec 6 '18 at 11:55