text:
# Footnotes without numbering and without horizontal rule

This is similar to Footnotes without numbering question here Footnotes without numbering but it won't let me add a comment and this isn't an answer..

When I add \let\footnoterule\relax as in order to suppress the horizontal rule as suggested in the comment to the answer, such as this MWE:

\documentclass{article}
\let\svthefootnote\thefootnote
\textheight 1in
\newcommand\blankfootnote[1]%
{%
\let\thefootnote\relax\footnotetext{#1}%
\let\footnoterule\relax%
\let\thefootnote\svthefootnote%
}%
\let\svfootnote\footnote
\renewcommand\footnote[2][?]%
{%
\if\relax#1\relax%
\blankfootnote{#2}%
\else%
\if?#1\svfootnote{#2}\else\svfootnote[#1]{#2}\fi%
\fi
}
\begin{document}
First Footnote here\footnote[]{footnote 1} next footnote\footnote{footnote 2}
Force footnote number 5 here\footnote[5]{footnote 3 numbered 5}.

Another\footnote[]{footnote 4} footnote here.

And foot note\footnote{footnote 5} 5 here.
\end{document}

It suppresses the horizontal rule for all footnotes. How can I suppress the horizontal rule for some footnotes but not others?

• Welcome to TeX.se. Please don't post code fragments. Instead put the fragment into a compilable minimal document that people can play with. Also, you can format code by selecting it and clicking on the {} icon. May 30, 2020 at 17:06
• You might look at footmisc and manyfoot packages. Manyfoot creates different types of footnotes, usually with different formats or counters. It can also add extra rules between types of footnotes. So it the dafault rule is turned off and the unnumbered is first, you might get a rule only before the second (numbered) type of footnote. Or you could just use a bottom float. May 30, 2020 at 21:15

Just for fun:

\documentclass{article}
\usepackage{newfloat}
\DeclareFloatingEnvironment[fileext=cmt,placement=b]{Comment}
\setcounter{bottomnumber}{3}% max bottom floats per page
\newcommand{\comment}[1]{\begin{Comment}\normalfont\footnotesize\parindent=\bibindent #1\end{Comment}}

\usepackage{lipsum}

\begin{document}

Comment here\comment{First comment} plus some more text.

\lipsum[1-5]

Second comment.\comment{Second comment}

First footnote.\footnote{First footnote.}

\lipsum[6-10]

\end{document}


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# zbMATH — the first resource for mathematics

A multiplicity result for periodic solutions of forced nonlinear second order ordinary differential equations. (English) Zbl 0586.34038
The main result of this paper is a multiplicity result concerning $$2\pi$$- periodic solutions of the Liénard differential equation (1) $$x''+f(x)x'+g(t,x)=s$$ where s is a real parameter, f and g are continuous functions, and g is $$2\pi$$-periodic in t. Assuming that g satisfies the condition $$\lim_{| x| \to +\infty}g(t,x)=+\infty$$ (uniformly in t), it is shown that a number $$s_ 0$$ exists with the following property: i) for $$s<s_ 0$$, equation (1) has no periodic solution; ii) for $$s=s_ 0$$, equation (1) has at least one $$2\pi$$-periodic solution; iii) $$s>s_ 0$$, equation (1) has at least two $$2\pi$$-periodic solutions. That result is prepared by partial results along the same lines for more general second order equations. The method of proof is based on the use of upper and lower solutions and on the additivity property of the topological degree.

##### MSC:
 34C25 Periodic solutions to ordinary differential equations 34C15 Nonlinear oscillations and coupled oscillators for ordinary differential equations
Full Text:

format: plain


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# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Proposition 7.2.3.13. Suppose we are given a pullback diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}' \ar [d] \ar [r]^-{F} & \operatorname{\mathcal{C}}\ar [d]^-{\pi } \\ \operatorname{\mathcal{D}}' \ar [r]^-{\overline{F}} & \operatorname{\mathcal{D}}. }$

If $\pi$ is a cocartesian fibration and $\overline{F}$ is right cofinal, then $F$ is right cofinal.

Proof. By virtue of Corollary 7.2.1.14, it will suffice to prove Proposition 7.2.3.13 in the special case where $\overline{F}$ is right anodyne. Let $S$ be the collection of all morphisms of simplicial sets $\overline{F}: \operatorname{\mathcal{D}}' \rightarrow \operatorname{\mathcal{D}}$ having the property that, for every cocartesian fibration $\pi : \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$, the induced map $F: \operatorname{\mathcal{D}}' \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}$ is right anodyne. We wish to show show that every right anodyne morphism belongs to $S$. It follows immediately from the definitions that $S$ is weakly saturated, in the sense of Definition 1.4.4.15. It will therefore suffice to show that $S$ contains every horn inclusion $\Lambda ^{n}_{i} \hookrightarrow \Delta ^ n$ for $0 < i \leq n$. In other words, we are reduced to proving Proposition 7.2.3.13 in the special case where $\operatorname{\mathcal{D}}= \Delta ^ n$ is a standard simplex and $\overline{F}$ is the inclusion of the horn $\Lambda ^{n}_{i} \subseteq \Delta ^ n$.

Applying Corollary 5.3.4.9, we deduce that there exists a diagram of $\infty$-categories $\mathscr {G}: [n] \rightarrow \operatorname{QCat}$ and a scaffold $\lambda : \underset { \longrightarrow }{\mathrm{holim}}( \mathscr {G} ) \rightarrow \operatorname{\mathcal{C}}$ for the cocartesian fibration $\pi$. We then have a commutative diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_{i} \times _{\Delta ^ n} \underset { \longrightarrow }{\mathrm{holim}}(\mathscr {G}) \ar [r] \ar [d]^{F'} & \Lambda ^{n}_{i} \times _{\Delta ^ n} \operatorname{\mathcal{C}}\ar [d]^{F} \\ \underset { \longrightarrow }{\mathrm{holim}}(\mathscr {G}) \ar [r]^-{\lambda } & \operatorname{\mathcal{C}}, }$

where $F'$ is right anodyne (Lemma 7.2.3.11) and therefore right cofinal (Proposition 7.2.1.3). Lemma 5.3.6.4 guarantees that horizontal maps are categorical equivalences, so that $F$ is also right cofinal (Corollary 7.2.1.21). $\square$


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# The Earth Is To A Soccer Ball, As The Moon Is To What?

Algebra Level 1

Beckham wants to build a model of the Solar system. If he decided that the earth would be represented by a regulation sized soccer ball (roughly spherical and approximately 11 cm radius), which of the following would be the best representation of the moon?

Note: the actual radius of the earth and moon are approximately 6368 km and 1738 km, respectively.

×


text:
# Math Help - Actual formulas for trigonometric functions? (no calculator)

1. ## Actual formulas for trigonometric functions? (no calculator)

Hi,

I need to know the actual formulas for the trigonometric functions: sine, cosine, tangent, arcsine, arccosine, arctangent, etc.

What I need is a function, like that of a calculator, for example cosine, which only takes the radian value of angle and returns a value, just like the calculator, but I need to manually calculate it. Those are formulas/procedures I need but I have never been able to find.

From what I can see, for example Cosine, would take the angle in Radians, and would asume a Radius of 1. Since Cosine of an angle is equals to Rx/R, and since Radius is assumed to be 1 and the angle is known, only Rx would need to be determined (presumably would be the direct value Cosine would return), maybe with Pythagoras, but I can't see how if Pythagoras formulas are supposed to use Sine, Cosine, etc., as well. Then I need that you tell me the actual procedure for manually finding Sine, Cosine, and the rest of functions.

If you can also, provide a rationale as to what the different functions are doing. If you can't provide a very extensive explanation, the formulas alone would suffice and I would try to figure them out myself.

If you can, please tell me also where I can find a full, in-depth analysis of the unit circle, like Pythagoras would relate and develop the different components, with the formulas I asked for above, and why they are being developed a particular way, so I can study further by myself, and any other web resource and/or book I should refer to make sense of this normally canned-only functions.

Any information will be welcome, as I see the need of being able to calculate these trigonometric functions manually, without using the calculator or tools, to greatly and really improve math understanding.

2. Originally Posted by x87
Hi,

I need to know the actual formulas for the trigonometric functions: sine, cosine, tangent, arcsine, arccosine, arctangent, etc.

What I need is a function, like that of a calculator, for example cosine, which only takes the radian value of angle and returns a value, just like the calculator, but I need to manually calculate it. Those are formulas/procedures I need but I have never been able to find.

From what I can see, for example Cosine, would take the angle in Radians, and would asume a Radius of 1. Since Cosine of an angle is equals to Rx/R, and since Radius is assumed to be 1 and the angle is known, only Rx would need to be determined (presumably would be the direct value Cosine would return), maybe with Pythagoras, but I can't see how if Pythagoras formulas are supposed to use Sine, Cosine, etc., as well. Then I need that you tell me the actual procedure for manually finding Sine, Cosine, and the rest of functions.

If you can also, provide a rationale as to what the different functions are doing. If you can't provide a very extensive explanation, the formulas alone would suffice and I would try to figure them out myself.

If you can, please tell me also where I can find a full, in-depth analysis of the unit circle, like Pythagoras would relate and develop the different components, with the formulas I asked for above, and why they are being developed a particular way, so I can study further by myself, and any other web resource and/or book I should refer to make sense of this normally canned-only functions.

Any information will be welcome, as I see the need of being able to calculate these trigonometric functions manually, without using the calculator or tools, to greatly and really improve [your] math understanding.
Unless you have studied power series I don't see how the functions you're asking for can possbly be of any use to "really improve math understanding."

Use Google: power series cosine.

3. For most angles, there are no simple formulas. A few can be done simply. For example, it is easy to see that cos(0)= 1, sin(0)= 0 and then tan(0)= 0/1= 0, cot(0)= 1/0 is undefined, sec(0)= 1/0 is undefined, csc(0)= 1/1= 1.

Similarly, cos(pi/2)= 0, sin(pi/2)= 1 and then tan(pi/2)= 1/0 is undefined, cot(pi/2)= 0/1= 0, sec(pi/2)= 1/1= 1, cosc(pi/2)= 1/0 is undefined.

For pi/4, consider a right triangle with one acute angle pi/4. Since pi/4+ pi/4= pi/2, the other angle is also pi/4. This is an isosceles triangle- the two legs are of the same length. If we take the length of those legs to be 1, the length of the hypotenuse is \sqrt{1^2+ 1^2}= \sqrt{2}. Then sin(pi/4)= 1/\sqrt{2}= \sqrt{2}/2, cos(pi/4)= 1/\sqrt{2}= \sqrt{2}/2, tan(pi/4)= 1/1=1, tan(pi/4)= 1/1= 1, sec(pi/4)= \sqrt{2}/1= \sqrt{2}, csc(pi/4)= \sqrt{2}/1= \sqrt{2}.

For pi/3, consider an equilateral triangle. It has all three angles equal and there sum must be pi. Each is pi/3. If you drop a line from one vertex perpendicular to the opposite side, it is easy to show, geometrically, that it bisects that side. If we take the side of the equilateral triangle to be 1, we have two right triangles, with hypotenuse of length 1, side adjacent to the pi/3 angle of length 1/2 and so the other leg is of length \sqrt{1- (1/2)^2}= \sqrt{1- 1/4}=\sqrt{3/4}= \sqrt{3}/2. Now we have sin(pi/3)= (\sqrt{3}/2)/1)= \sqrt{3}/2, cos(pi/3)= (1/2)/1= 1/2,tan(pi/3)= (\sqrt{3}/2)/(1/2)= \sqrt{3}, cot(pi/3)= (1/2)/(\sqrt{3}/2)= 1/\sqrt{3}= \sqrt{3}/3, sec(pi/3)= 1/(1/2)= 2, csc(pi/3)= 1/(\sqrt{3}/2}= 2/\sqrt{3}= 2\sqrt{3}/3.

pi/6+ pi/3= pi/2 so pi/6 is the complement to pi/2 and we can just swap the functions and their "co" functions above. sin(pi/6)= cos(pi/3)= 1/2, cos(pi/6)= sin(pi/3)= \sqrt{3}/2, tan(pi/6)= cot(pi/3)=\sqrt{3}/3,sec(pi/6)= csc(pi/3)= 2\sqrt{3}/3, csc(pi/6)= sec(pi/3)= 2.


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Last edited by Dizilkree
Monday, April 20, 2020 | History

5 edition of Differential Equations With Derive found in the catalog.

# Differential Equations With Derive

• 262 Want to read
• 2 Currently reading

Written in English

Subjects:
• Mathematics / Differential Equations,
• Computers,
• Derive,
• Differential Equations,
• Educational Software - General,
• Mathematics,
• Software,
• Computers / Educational Software / General

• The Physical Object
FormatPaperback
Number of Pages284
ID Numbers
Open LibraryOL9431138M
ISBN 10096236293X
ISBN 109780962362934

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### Differential Equations With Derive by David S. Arney Download PDF EPUB FB2

Used textbook “Elementary differential equations and boundary value problems” by Boyce & DiPrima (John Wiley & Sons, Inc., Seventh Edition, c ). Many of the examples presented in these notes may be found in this book.

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Included are most of the standard topics in 1st and 2nd order differential equations, Laplace transforms, systems of differential eqauations, series solutions as well as a brief introduction to boundary value problems, Fourier series and partial differntial. If you are learning differential equations on your own this derivation can be found Differential Equations With Derive book Schaum's Solved Problems in Differntial Equations Chapter 12 Problem #8.

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So let me write that down. So a traditional equation, maybe I shouldn't say traditional equation, differential equations have been around for a while.

Elementary Differential Equations with Boundary Value Problems is written for students in science, en-gineering,and mathematics whohave completed calculus throughpartialdifferentiation.

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Solution Methods for Separable First Order ODEs (p) () g x dx du x h u Typical form of the first order differential equations: () in which h(u) and g(x) are given functions. By re‐arranging the terms in Equation () the following form with the left‐hand‐side (LHS). Differential equations relate a function with one or more of its derivatives.

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text:
# Prove/Disprove that the set of rational numbers is uncountable (Real Analysis)

Cantor digonalization method is used to prove that an open interval for real numbers $(0,1)$ is not countable. Since rational numbers also have decimal expansions, we could try to use this same method to prove that the set of rational numbers in $(0,1)$ is not countable. Explain why we could not conclude that the set of rational numbers in $(0,1)$ is uncountable using this method.

What i tried

Cantor diagonalizaton works by writing all the real numbers in the interval $(0,1)$ as an array of matrix and then finding a real number that does not belong to this set of array of matrix by making the diagonal entries of the real number different from that of the set of array of numbers in the matrix thus a contradiction. While for the set of rational numbers it can be expressed in the form of $\frac{a}{b}$ where gcd(a,b)=1, We do the same thing as for the rational numbers but i think this set of rational numbers somehow always tend to appear in the array of matrix no matter how we try thus this method of proof dosent apply for the case of rational number. But im unsure how. Could someone explain this to me. Thanks

• When you construct the "new" real out of your lists of real, you use the fact that any decimal expansion corresponds to a real. It isn't the case for rationals. – Max Jan 26 '17 at 18:08
• Max, did explain it. You use the method to create a number not on the list. But you have no reason to assume the new number is rational. So the method concludes that if the rationals are countable than there exists at least one number that is not rational. Which is ... valid. But not relevant. – fleablood Jan 26 '17 at 18:19

Its not the process but the conclusion of the process that fails. When you write real numbers in the interval (0,1) as an array of matrix and then find a number by making the diagonal entries of the real number different from that of the set of array of numbers in the matrix, the resulting number is again a real number.

When you apply the same trick to the rational numbers the resulting number may not be rational. Remember being rational means the decimal expression either terminates or repeating. How do you ensure these properties in the resulting number?

• To be pedantic, I would say this method proves: Either the rationals are uncountable or if they are countable there exists at least one irrational number. This is a valid conclusion and valid proof. Unfortunately its a weak and wimpy result. – fleablood Jan 26 '17 at 18:24
• I can see that the Cantor number (in the case of the list of rationals) is not terminating. But I don't see why it is impossible to create a new repeating number. – zoli Jan 26 '17 at 18:27
• @zoli of course it is possible to get a number which is repeating, i.e.rational but not all of them are of this form. – tessellation Jan 26 '17 at 18:34
• @zoli hypothetically, without knowing that the rationals actually are countable, we don't know that it wouldn't be. But we we can't conclude that it would, so we can't use this as an argument that the number would be rational. And if the number could be irrational... the argument is useless. (But as we know, for different reasons, that the rationals are countable, we know the new number can't be rational because all the rationals are on the list.) – fleablood Jan 26 '17 at 18:50

Proof that the set of decimal expansions from $(0,1=.99999.....]$ are uncountable:

If they were countable we could list them all. Then we could manipulate them all to get a number not on the list. That new number is a decimal expansion. But that contradicts that all decimal expansions were on the list.

Proof that the set of rationals in decimal representation from $(0,1=.99999.....]$ are uncountable:

If they were countable we could list them all. Then we could manipulate them all to get a number not on the list. $\color {red} {\text{ That new number is a rational in decimal representation.}}$ But that contradicts that all rationals in decimal representation were on the list.

!!!BUT!!! There is no reason to believe that bit in red is true!

So instead we'd have to write:

If they were countable we could list them all. Then we could manipulate them all to get a number not on the list. If $\color {red} {\text{ that new number is a rational in decimal representation.}}$ $\color {blue} {\text{then that would}}$ contradict that all rational in decimal representation were on the list.

And from that we can conclude the true (but wimpy) result:

Either the rationals in decimal representation for $(0,1=.99999....]$ are uncountable OR if they are countable then there exist decimal representations in $(0,1]$ that are not rational.

The rationals are countable.

Here is a intuitive sketch.

Make a table. Put natural numbers across the top and down the left column. the body of the table is the ratio of the number at the top of the table over the number in the right column.

This is a representation of the full set of $\mathbb Q+$

Now you can run a serpentine line that runs through every entry in the table. this maps the rationals to $\mathbb N$

More formally:

There exists an injection from $\mathbb Q\to \mathbb Z\times \mathbb N$ i.e. $\frac pq = (p,q)$

the Cartesian product of countable sets is a countable set.

• yes, but the question is not whether the rationals are uncountable but why that "proof" that they are uncountable fails. – fleablood Jan 26 '17 at 18:24


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# Math Help - Need help fast

1. ## Need help fast

Two cyclists, one averaging 10 miles per hour and the other 12 miles per hour, start from the same town at the same time. If they travel in opposite directions, after how long will they be 66 miles apart?

2. Originally Posted by j_dryden
Two cyclists, one averaging 10 miles per hour and the other 12 miles per hour, start from the same town at the same time. If they travel in opposite directions, after how long will they be 66 miles apart?
$Distance=10x+22x=22x$

$66=22x\Rightarrow{x=3}$

Verify 12*3+10*3=36+30=66


text:
# tsls: Two stage least squares estimation In gmm: Generalized Method of Moments and Generalized Empirical Likelihood

## Description

Function to estimate a linear model by the two stage least squares method.

## Usage

 1 tsls(g,x,data)

## Arguments

 g A formula describing the linear regression model (see details below). x The matrix of instruments (see details below). data A data.frame or a matrix with column names (Optionnal).

## Details

The function just calls gmm with the option vcov="iid". It just simplifies the the implementation of 2SLS. The users don't have to worry about all the options offered in gmm. The model is

Y_i = X_iβ + u_i

In the first step, lm is used to regress X_i on the set of instruments Z_i. The second step also uses lm to regress Y_i on the fitted values of the first step.

## Value

'tsls' returns an object of 'class' '"tsls"' which inherits from class '"gmm"'.

The functions 'summary' is used to obtain and print a summary of the results. It also compute the J-test of overidentying restriction

The object of class "gmm" is a list containing at least:

 coefficients k\times 1 vector of coefficients residuals the residuals, that is response minus fitted values if "g" is a formula. fitted.values the fitted mean values if "g" is a formula. vcov the covariance matrix of the coefficients objective the value of the objective function \| var(\bar{g})^{-1/2}\bar{g}\|^2 terms the terms object used when g is a formula. call the matched call. y if requested, the response used (if "g" is a formula). x if requested, the model matrix used if "g" is a formula or the data if "g" is a function. model if requested (the default), the model frame used if "g" is a formula. algoInfo Information produced by either optim or nlminb related to the convergence if "g" is a function. It is printed by the summary.gmm method.

## References

Hansen, L.P. (1982), Large Sample Properties of Generalized Method of Moments Estimators. Econometrica, 50, 1029-1054,

## Examples

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 n <- 1000 e <- arima.sim(n,model=list(ma=.9)) C <- runif(n,0,5) Y <- rep(0,n) Y[1] = 1 + 2*C[1] + e[1] for (i in 2:n){ Y[i] = 1 + 2*C[i] + 0.9*Y[i-1] + e[i] } Yt <- Y[5:n] X <- cbind(C[5:n],Y[4:(n-1)]) Z <- cbind(C[5:n],Y[3:(n-2)],Y[2:(n-3)],Y[1:(n-4)]) res <- tsls(Yt~X,~Z) res

gmm documentation built on June 20, 2017, 3:01 p.m.


text:
Assignment 4 - Control Theory 2

Problem 1

Let G(s) be the plant transfer function. The closed-loop transfer function is
 T(s) = G(s) 1+G(s) .
Let wp be such that |G(jwp)| = 1 and ÐG(jwp) = f, then prove that
|T(jwp)| =  1

 2sin fm 2
where phase margin fm = 180 + f,      -180° £ f £ 180°.

Problem 2

Let
 C(s) = 1+s/wz 1+s/wp ;       wp > wz
and wl be the frequency at which ÐC(jwl) is maximum over all w > 0. Show that
 wl = Ö wz wp ,

 fl D = ÐC(jwl) = sin-1 æè wp/wz - 1 wp/wz + 1 öø , and

 20log10|C(jwl)| = 10log10(wp/wz).
The ratio wp/wz is often expressed as m. Make a table of a few important values of m and fl.

Problem 3

Figure 1: Block Diagram: Problem 3
Let the true system transfer function in the above Figure 1 be modelled using a multiplicative uncertainty as [G\tilde](s) = G(s)(1 + L(s)). The model we have available for design is G(s) and let the uncertainty magnitude be, lm(jw) = |L(jw)|. The closed-loop system sensitivity function is such that
 êê 1 1+tildeG(jw) êê < 1 pm(j w) .
Prove that for a stable feedback system to satisfy the sensitivity function constraint the following inequality should be satisfied:
 êê G(jw) 1+G(jw) êê lm(jw) + êê 1 1+G(jw) êê pm(jw) < 1.

Problem 4

Figure 2: Block Diagram: Problem 4
In the above Figure 2, let
 G(s) = 100 (s/10+1)(s/20+1) .
Design controller C(s) such that,
1. The closed-loop system has zero steady-state error to unit step reference input;

2. The crossover frequency is above 500 rad/s;

3. There is a minimum 60° phase margin.

Home

File translated from TEX by TTH, version 3.40.
On 08 Aug 2003, 16:46.


text:
# Detecting a lazer beam that hits near but not on a sensor

Joined Feb 21, 2020
23
I'm making myself a lazer tag system and on the boards I have assembled photo-transistors that are spaced between 1 and 2.5 inches apart. I'm trying to make it so the areas hit between the sensors will be recognized.

So far, I have tried the following and got results as follows:

Result: Sensitivity was excellent but data rate is terrible. I'm aiming for 2400bps but In that test I might have been lucky to get 300bps

Tried: lower resistor (82K)
Result: Sensitivity a bit small but the data rate recognized by the sensor is acceptable.

Tried: Direct line-of-sight with test circuits from PC serial port to PC serial port, but lazer module and photo-transistor setup is the same and still got 5V. (serial port connects through MAX232)
Result: almost no problem with 2400bps, could get 9600bps if beam directly hits the top of photo-transistor itself.

I don't know if I can do anything electrically to solve my problem but when I looked at digikey, I saw they sell convex LED lens caps that fit on the light and can accept a beam from any angle which I think will help a bit, but what if the beam hits directly between the two sensors? Is there a specific product (that digikey could sell) I can use to take that beam and make it travel across the item so that (at least one of) the two sensors can sense the light?

Only thing that comes across the top of my head is coarse-sanded plexiglass but I'm not sure if that would be good enough?

Any ideas?

#### KeithWalker

Joined Jul 10, 2017
796
De-focus the laser a little so that illuminates a larger area.

#### djsfantasi

Joined Apr 11, 2010
6,335
I'm making myself a lazer tag system and on the boards I have assembled photo-transistors that are spaced between 1 and 2.5 inches apart. I'm trying to make it so the areas hit between the sensors will be recognized.

So far, I have tried the following and got results as follows:

Result: Sensitivity was excellent but data rate is terrible. I'm aiming for 2400bps but In that test I might have been lucky to get 300bps

Tried: lower resistor (82K)
Result: Sensitivity a bit small but the data rate recognized by the sensor is acceptable.

Tried: Direct line-of-sight with test circuits from PC serial port to PC serial port, but lazer module and photo-transistor setup is the same and still got 5V. (serial port connects through MAX232)
Result: almost no problem with 2400bps, could get 9600bps if beam directly hits the top of photo-transistor itself.

I don't know if I can do anything electrically to solve my problem but when I looked at digikey, I saw they sell convex LED lens caps that fit on the light and can accept a beam from any angle which I think will help a bit, but what if the beam hits directly between the two sensors? Is there a specific product (that digikey could sell) I can use to take that beam and make it travel across the item so that (at least one of) the two sensors can sense the light?

Only thing that comes across the top of my head is coarse-sanded plexiglass but I'm not sure if that would be good enough?

Any ideas?
Coarse sanding a plexiglass screen is inexpensive. Try it!

#### djsfantasi

Joined Apr 11, 2010
6,335
And if you’re making the sensor boards, can you try one that’s a little more dense? Photo transistors no more than 0.3” apart!

#### oz93666

Joined Sep 7, 2010
737
yes ...Keith is right ...De-focus ...Some cheap lasers have the lens secured to a threaded screw which you can adjust ....

This one has a knurled end so you can change the focus ($2 eBay) Thread Starter #### mike_canada Joined Feb 21, 2020 23 And if you’re making the sensor boards, can you try one that’s a little more dense? Photo transistors no more than 0.3” apart! I already made them. and they're placed close to the lights so if someone aims at the lights then the unit works but if someone aims in-between it might not register. I'm not in the process of redoing 40 complex boards just to order triple the number of parts only to make parts closer. yes ...Keith is right ...De-focus ...Some cheap lasers have the lens secured to a threaded screw which you can adjust .... This one has a knurled end so you can change the focus ($2 eBay)
The question is, is that particular lazer legal for use in Canada? From what I recall, the law is not to use lazers with a grade stronger than 3B I think.

One disadvantage with the focus idea is that it makes skilled players sad because they want to practice aiming not have a small beam become a large circle at the target.

Joined Feb 21, 2020
23
Let's just say my coarse idea works. question is how coarse must I go? and don't they already have such plastic-like pieces for sale at digikey? I don't know if diffuser is the right name. I mean I want to type in their search box to find the coarse pieces I need so I can spread the light because that seems to be my only decent option right now since I don't want cheaters in my game. As for focusing, I think I can get away with ordering convex lens caps and put them on the lazer beam light output.

#### djsfantasi

Joined Apr 11, 2010
6,335
I already made them. and they're placed close to the lights so if someone aims at the lights then the unit works but if someone aims in-between it might not register. I'm not in the process of redoing 40 complex boards just to order triple the num
Ok, I understand your position. Just be sure that you understand that you may not be able to get anything to work with that restriction.

Good luck!

#### DickCappels

Joined Aug 21, 2008
6,303
Yes, diffuser is the name of the filter you need to place in front of the phototransistors. The diffusor will need to be placed some distant out in front of the phototransistors so there is room for the light to scatter from where it strikes the diffuser to the phototransistors.

Joined Feb 21, 2020
23
ok, so I'm trying to think of how I can order such a diffuser. I do have a local shop near me that makes custom products from various different plastics, but I'm not sure which plastic material to start with.

This is their website: https://paplastics.com/

#### DickCappels

Joined Aug 21, 2008
6,303
I once changed a bunch of clear LEDs to diffused LEDs by painting them with clear fingernail powder and then dusting them with talcum powder.

Also got nice results by roughing up the epoxy lenses with medium grit sand paper. You just need to let light through your diffuser and have it scatter as i goes through.

Edit:
You may also want to check out arts and crafts stores for materials that are similar to these.

#### Bernard

Joined Aug 7, 2008
5,463
I believe that all lasers shown in post # 9 can be re focused, just reviewed refocusing of one; @ 7 ft. tight, spot went from about 4 in- 1"- .2"- 1"- 6". A small spring holds lens against end piece.

Joined Feb 21, 2020
23
Ok so my searches for diffuser (and from another source, opal) material has resulted me into looking into translucent ceiling plates which the hardware salesman at Home Depot also identified as "diffusers" as well.

He showed me about 5 different varieties of diffusers they had in stock and each one had different ripples to it, but I can't figure out which one is best for my purpose. Sadly, each one costs $15 because they only sell them as 2x4 (2 feet by 4 feet) sheets, and I'm not willing to spend$100+ just to take a tiny corner off each sheet for testing.

The question is which diffuser is the best from the varieties home depot or any other hardware store would have in stock?

I tried craft and office stores and got no luck.

#### jpanhalt

Joined Jan 18, 2008
9,057
Dispersion of the incident light is one way. Have you considered a fluorescent compound, such as rhodamine. It's incredibly cheap for the amount you would need. Suspend/dissolve it in some clear finish and paint the off-target areas.

Joined Feb 21, 2020
23
I don't know what "incident light" is. Maybe you mean incadescent?

But the source of light in my case is a red lazer beam. And where would I be able to get such a compound already made?

#### jpanhalt

Joined Jan 18, 2008
9,057
"Incident" light means light that hits it. Definitely not"incandescent;" although, incandescent light can produce incident light.

If you shine a light with the proper wavelength on something with fluorescein (https://en.wikipedia.org/wiki/Fluorescein) in it, the whole thing lights up. There are many substance that fluoresce with red light. Rhodamine B (https://en.wikipedia.org/wiki/Rhodamine_B) is one such substance and is really cheap.

#### djsfantasi

Joined Apr 11, 2010
6,335
I’ve used clear polystyrene sheet from Home Depot. It comes in approximately 3’ x 4’ sheets. Then, I use coarse sandpaper in a hand rotary / orbital sander to roughen up one side. If needed, I hand sand to even out the scratches. While I’ve never used it in your application, I have used it to construct large (1’) 7-segment displays.

Coating the non-sanded side with Rhodamine suspended in a clear finish may optimize the effect

Last edited:

#### MisterBill2

Joined Jan 23, 2018
5,749
I'm making myself a lazer tag system and on the boards I have assembled photo-transistors that are spaced between 1 and 2.5 inches apart. I'm trying to make it so the areas hit between the sensors will be recognized.

So far, I have tried the following and got results as follows:

Result: Sensitivity was excellent but data rate is terrible. I'm aiming for 2400bps but In that test I might have been lucky to get 300bps

Tried: lower resistor (82K)
Result: Sensitivity a bit small but the data rate recognized by the sensor is acceptable.

Tried: Direct line-of-sight with test circuits from PC serial port to PC serial port, but lazer module and photo-transistor setup is the same and still got 5V. (serial port connects through MAX232)
Result: almost no problem with 2400bps, could get 9600bps if beam directly hits the top of photo-transistor itself.

I don't know if I can do anything electrically to solve my problem but when I looked at digikey, I saw they sell convex LED lens caps that fit on the light and can accept a beam from any angle which I think will help a bit, but what if the beam hits directly between the two sensors? Is there a specific product (that digikey could sell) I can use to take that beam and make it travel across the item so that (at least one of) the two sensors can sense the light?

Only thing that comes across the top of my head is coarse-sanded plexiglass but I'm not sure if that would be good enough?

Any ideas?
Sanding a suitable plastic piece is a very excellent idea, I have used similar. You will need a few sensors, but not a lot of them, and you will need quite a bit of gain . But it can work very well.


text:
# Is nature symmetric between particles and antiparticles?

Is nature symmetric with respect to presence of particles? Do we have an antiparticle for every particle thought of? Are there any proven examples where we don't have an antiparticle? And what about antiparticle of a photon (we know it can also behave as a particle)?

-
do we discuss symmetry and anti-symmetry only when we have charged particles? –  bubble Jun 25 '11 at 4:05
@Dan you mean to say that we also discuss symmetry or anti symmetry based on spin? –  bubble Jun 25 '11 at 4:16
Are you just asking whether every known particle has an antiparticle (and vice-versa)? –  David Z Jun 25 '11 at 4:32
@David yes thats what I am asking –  bubble Jun 25 '11 at 4:33

The anti-particle for any particle is obtained by charge C and parity P conjugation. C is the operation that interchanges positive and negative charges and P is the operation that reflects in a mirror. The combined operation of CP must produce a particle of the same mass. This is a theorem of relativistic quantum field theory due to CPT symmetry. This other particle is either the same particle or an antiparticle with opposite charge and/or chirality.

Some particles such a photons, gluons, Z bosons, pions, Higgs, graviton etc, do not have anti-particles because they are invariant under the CP transformation. You can say that they are their own anti-particle. This can only happen for particles without electric charge and with no chirality.

In principle the QCD color charge is also reversed for an anti-particle. This suggests that a gluon should not be regarded as its own anti-particle, but since colourless states are never seen the distinction cannot really be made in any operational sense.

All known particles which are their own anti-particle are bosons, but it is also possible for a fermion to be its own anti-particle if it is a Majorana spinor rather than a Dirac spinor. No known fermions are of this type (unless neutrinos are Majorana) but they exist in SUSY models.

Observed elementary particles that do have anti-particles include all the quarks and leptons (except possibly the neutrinos) and the charged W bosons. Any composite particle also has an antiparticle made of the anti-particles of its constituents. This can only be its own anti-particle if all its constituents are (e.g. a glueball), or if it is made of particle/anti-particle combinations as is the case for pions.

-
Nice reminder on the Majorana spinors. –  Carl Brannen Jun 26 '11 at 4:07
if boson is an anti-particle of boson can two bosons stay together or they annihilate ? –  bubble Jun 26 '11 at 12:32
If you put a charged particle and its anti-particle together they can form a bound state. After a short instant they will annihilate, if they don't decay individually first. For example you can combine an electron and a positron to form positronium. Same thing could be done with bosons in principle but there are no good examples. Does that answer the question? –  Philip Gibbs - inactive Jun 26 '11 at 17:17

Theoretically, for every known particle there is an anti-particle. But bosons are their own antiparticle, for example, the $\pi^+$ which is (mostly) composed of an up-quark and a down-antiquark $u\bar{d}$, has as an anti particle the $\pi^- = d\bar{u}$. The antiparticle of the photon is just the photon, while the antiparticle of the $W^+$ is the $W^-$.

Composite particles, such as the hydrogen atom, also have corresponding antiparticles, but not all of them have been observed. So far the heaviest antimatter atom observed is anti-helium, but theoretically, it seems clear that anti-uranium could be produced.

-
Not all bosons are their own antiparticle. Only the photon, Z and gluon are. –  Kasper Meerts Jun 25 '11 at 11:46
To be somewhat pedantic, the heaviest antimatter atom that has been observed is anti-hydrogen. While antimatter atomic nuclei of deuterium, tritium, Helium-3, and Helium-4 (aka alpha particle) have been produced, to my knowledge the actual atoms (bound states of the anti-nuclei and positrons) have not yet been observed. –  Willie Wong Jun 25 '11 at 12:09
@Willie: "Anti-Helium Discovered in Relativistic Heavy Ion Collider Experiment" sciencedaily.com/releases/2011/04/110424152441.htm –  Hans de Vries Jun 25 '11 at 21:27
@Hans: at the bottom of the news article you cite, you have the link to the actual journal article, whose title is: "Observation of the antimatter helium-4 nucleus". :-) I have that particular issue of Nature sitting right next to me on the couch right now (I just read the issue three days ago), so I was pretty sure about what I wrote. –  Willie Wong Jun 26 '11 at 0:05
@Willie; A "helium nuclei" is an atom. If this is news to you, look up the wikipedia definitions of "ion" and "atom". –  Carl Brannen Jun 26 '11 at 4:00


text:
# What is the combining form meaning calcium?

Calcium is an ELEMENT; $Z = 20$. And it is the 5th most metal in the Earth's crust. As an alkaline earth, i.e. Group 2 of the Periodic Table, it commonly forms $C {a}^{2 +}$ ions. And salts such as $C a C {O}_{3}$, $C a C S {O}_{4}$, and $C a {\left(O H\right)}_{2}$ are very common.....


text:
# Is there a way to detect math formatting in Stack Overflow posts, a la Math Overflow? [duplicate]

See this question on Math Overflow for reference. It has several math equations that are "detected" by either surrounding the equations in dollar signs ($) or starting newlines and beginning with a backslash (\). Trying$AA + AB = AC\$ doesn't work here, unfortunately.

Is there any way to add that kind of mathematical detection to Stack Overflow posts? If not, can it be added?

This answer brought the missing feature to my attention.

## marked as duplicate by Deduplicator, Community♦Oct 22 '15 at 14:29

• I know. I also know that it's never going to be added to the site, hence the workaround. One other workaround is to use HTML entities like Δ which renders correctly in an actual post, if not in comments. – theB Oct 22 '15 at 16:13


text:
# Hamel basis and Banach spaces

Suppose $X$ is a linear space and $X$ has a Hamel basis with uncountable number of elements. Does there exist a norm on $X$ such that $X$ is a Banach space with respect to this norm?

I can answer this question when the cardinality $\lambda$ of Hamel basis of $X$ is not less than $\mathfrak{c}$.
Consider Banach space $\ell_\infty(\Lambda)$. Clearly $\operatorname{Card}(\ell_\infty(\lambda))=\lambda\times \operatorname{Card}(\mathbb{C})=\lambda$, so cardinalities of Hamel bases of $\ell_\infty(\lambda)$ and $X$ coincide. Therefore, there is some linear isomorphism between $X$ and $\ell_\infty(\Lambda)$. This isomorphism induces complete norm on $X$.
The cardinality $\lambda$ of a Hamel basis of an infinite-dimensional Banach space always satisfies $\lambda^\omega =\lambda$. Choose your favourite uncountable cardinal without this property.


text:
# AES-256 for Windows XP

Scripts and instructions listed below include AES-256 support for Windows XP, which, among other things, will allow Internet Explorer 8 browser to open sites that use HTTPS protocol with 256-bit encryption.

3.2 KB

#### File name:

AES256_for_WindowsXP.vbs

133

374.0 KB

#### File name:

AES256_for_WindowsXP_ENG.EXE

136

426 Bytes

#### File name:

AES256_for_WindowsXP.bat

42

### Attention!

Windows XP system libraries do not officially support AES-256, and this script is just a workaround. Use the script only if you understand what the AES-256 algorithm is needed for, and you know how to restore the system in case of unforeseen critical errors.

### Why do I need AES-256 support?

Due to the fact that Windows XP does not support the AES-256 encryption algorithm, many Internet sites running HTTPS do not open in Internet Explorer 8 under this operating system. For the same reason, many programs that depend on Internet Explorer or use system encryption libraries do not work properly (for example, read about an error due to which Skype cannot connect to the Internet).

### How to enable AES-256 support manually?

Officially, Windows XP does not support AES-256, and it is unlikely that the algorithm will ever be supported. However, Windows XP has an “older brother”, Windows Embedded POSReady 2009 (the operating system is designed for POS-terminals, ATM, self-service checkouts and others). For the English version of this OS there is an update KB3081320, which adds support for AES-256, but it cannot be installed on Windows XP.

Fortunately, this update is very simple, and most importantly, I managed to find a simple way to add AES-256 support for Windows XP. The plan is as follows:
I saved it as «KB3081320.exe», so that it would be more convenient to write commands.

2. Extract all files with the command:
C:\path\to\installer\KB3081320.exe /x:C:\KB3081320

3. Replace system libraries with copies from the folder C:\KB3081320\SP3QFE
For your convenience, I uploaded them to the server: dssenh.dll, rsaenh.dll and schannel.dll

In case you do not know how to replace system libraries, you can do it in this way:
1. Find the required library in each of these folders:
• %windir%\ServicePackFiles\i386\
• %windir%\system32\dllcache\
• %windir%\system32\
2. Rename the library from these folders to something else
3. Copy the new library to each of these folders
4. Restart the computer

### How to enable AES-256 by installing KB3081320?

Since KB3081320 can be installed only on Windows Embedded, we should use a hack to turn our computer into an “ATM”. Of course, it will not give money (although, who knows), but at least it will deceive the update installer. Therefore, let’s start:
2. Run the installer WindowsXP-KB3081320-x86-Embedded-ENU.exe
3. Remove from the system the presence of “ATM” using PosReady_Disable.reg
4. Restart the computer (do this only after you removed “ATM”)

### What algorithm does my computer support?

To check which encryption algorithm is used on your computer or to find out if you enabled AES-256 support: start Internet Explorer → click “Help” → choose “About” → and check string “Cipher Strength” (for example, if your computer supports AES-256, here is specified “Cipher Strength: 256-bit”).

I will be grateful for your feedback and additions. Please do not hesitate to leave comments – this is very important for me and, especially, for blog visitors.

1. 0 0 0
Dave,
Correct me if I'm wrong, but I think it might be a good idea to suggest that users change the files in \dllcache\ first. I followed your instructions and rebooted, and nothing changed. I think that Windows File Protection immediately repaired \system32\ with the originals from \dllcache\ (before I replaced them).

As a side note: I did not have a %windir%\ServicePackFiles\i386\ directory. I think this is because I never installed a service pack -- SP3 was slipstreamed onto my installation disc. This was fine, and did not affect the process.

I should mention that after my second reboot this did in fact work for me, beautifully. I've been looking for a solution to Windows XP HTTPS issues for almost a year now. Skype immediately began showing link previews again, which it had not done in a couple of weeks, and gave me a banner ad (which I don't even remember the last time I saw). I'm looking forward to seeing if all of the HTTPS errors I've been getting in Opera and Chrome will go away now, but unfortunately I cannot remember any of the sites it had been happening on.

Last but not least, I think your spam filter is filtering Opera 12. :'D
1. 0 0 0
Hi Dave! Thank you for your notices. I am very pleased that I was able to help you. In truth, most people believe that IE8 on Windows XP cannot support AES256. Until recently, I also thought so. However, some days ago, thanks to one visitor, I found out that it can be enabled if you “convert” Windows XP to Windows Embedded POSReady 2009, install some updates and change some registry parameters. Investigating this information thoroughly, I found and developed this simple solution.

I myself tested this solution manually and never had problems. Nevertheless, I will update the instruction, as this is indeed a correct remark. And you are right about ServicePackFiles folder — it exists only if Windows was upgraded to SP3.

The only question: what do you mean by “your spam filter is filtering Opera 12”? Can you explain please?
1. 0 0 0
Dave,
Sorry, I was mistaken! I've encountered comment form spam filters in the past that seemed to filter Opera 12 for some reason. That's not actually the case here. The error was actually a result of me running NoScript (disabling JavaScript, essentially) -- your form submits via AJAX, and if scripting is turned off, it submits incorrectly and fails with an error message of "An Error Was Encountered. The action you have requested is not allowed.".

By the way, do you know if there is any method that tricks TLS 1.1 and 1.2 into working in IE8 on XP? Even with this change, the checkboxes for them do not appear. (I swear I've seen them in the past, though...)
1. 0 0 0
I apologize, but, unfortunately, I did not investigate this question on how to enable TLS 1.1 and 1.2. If you can find the answer, I will be grateful for any hints.

In the meantime, I updated the article, added new scripts, and thanks to kb80 upgraded DLLs to a more secure patch.
2. 2 +2 0
kb80,
Btw, the security update KB3081320 supersedes KB3055973 and contains the most current version of the dlls, so if you apply them from KB3081320 instead, the system also will be protected from the MiTM vulnerability, according to the article.

1. 1 +1 0
Yeah! That’s cool! Thank you very much! I updated the article and added some new things.
1. 0 0 0
Dave,
Have you considered posting a new article linking back to this one, mentioning that you fixed a security flaw? Some folks (such as myself) may have installed it already, and are unaware that it's been updated, and will only find out if they happen to visit again for some reason.
1. 0 0 0
As far as I know, all people interested in such articles always subscribe to notifications or regularly visit it. And you are a living example :)
3. 1 +1 0
Rumo,
I have discovered that once the PosReady key is introduced in the system registry it will never more be deleted or changed (so that PosReady_Disable.reg has no effect). All subkeys under the WPA key (like that one) are protected by DRM — or something like that. This can be most of time harmless, but one or another rogue program can refuse to install in what they identify as a PosReady system.

Other than that, thanks for the tip. I have replaced the files in \Windows\system32 and \Windows\system32\dllcache from outside Windows XP (in a dual-boot machine), and it solved the problem with Skype. No ill effects so far.

[Updated ]
I've submitted a previous comment about a problem with the hack you recommend for installing KB3081320, but apparently it has been removed. I think people should be warned that the change made in their systems by PosReady_Enable.reg is irreversible.

[Updated ]
Now that my first comment is back, my subsequent comment has become superfluous — and so this one...

Thanks.
1. 0 0 0
Hi Rumo! Thank you for your comments and sorry for being late. One of these days I’m going to test it and I will update the article.

[Updated ]
Tested it! You are absolutely right. I checked, and found that the value is not deleted if you restart the computer. Therefore, it is important to remove it before restarting the computer.
1. 0 0 0
Rumo,
Hi! I think that it's not possible to remove it even before restarting the computer. Please correct me if I'm wrong.
1. 0 0 0


text:
# For a fixed dominant weight $\lambda$, are almost all dominant weights in the same coset above it?

First some notation as in e.g. the book by Humphreys on Lie Algebras.

Let $$E$$ be an Euclidean space with inner product $$(-,-)$$, and denote $$\langle v,w \rangle = \frac{2(v,w)}{(w,w)}$$. Let $$\Phi$$ be an irreducible root system on $$E$$, so $$\langle \beta, \alpha \rangle \in \mathbb{Z}$$ and $$\alpha - \langle \beta, \alpha \rangle \beta \in \Phi$$ for all $$\alpha, \beta \in \Phi$$. Fix a set of simple roots $$\alpha_1$$, $$\ldots$$, $$\alpha_l$$.

Let $$\Lambda$$ be the set of weights, i.e. the set of $$\lambda \in E$$ such that $$\langle \lambda, \alpha \rangle \in \mathbb{Z}$$ for all $$\alpha \in \Phi$$.

Let $$\Lambda^+$$ be the set of dominant weights, that is, the set of $$\lambda \in \Lambda$$ such that $$\langle \lambda, \alpha_i \rangle \geq 0$$ for all $$i$$.

We have the usual partial order on $$\Lambda$$, by defining $$\mu \preceq \lambda$$ iff $$\lambda - \mu = \sum_{i = 1}^k k_i \alpha_i$$ for some integers $$k_i \in \mathbb{Z}_{\geq 0}$$.

It is well known that for a fixed $$\lambda \in \Lambda^+$$, there are only finitely many $$\mu \in \Lambda^+$$ such that $$\mu \preceq \lambda$$ (See for example 13.2 in Humphreys). But is the following true?

Fix $$\lambda \in \Lambda^+$$. Then for all but finitely many $$\mu \in \Lambda^+$$ with $$\lambda - \mu \in \mathbb{Z}\Phi$$, we have $$\lambda \preceq \mu$$.

If the answer is yes, this could be used to give a different solution to a previous question asked here: link.

• Looking back at my choice of notation in 1972 and earlier, I'd now want to make clearer use of coroots and such rather than relying so much on pointed brackets. – Jim Humphreys May 10 '19 at 20:10
• Since the root lattice has finite index in the weight lattice, the restriction that $\lambda-\mu$ be in the root lattice does not affect the answer. – Victor Protsak May 18 '19 at 1:20
• @VictorProtsak: If $\lambda - \mu \not\in \mathbb{Z}\Phi$, then $\lambda \not\preceq \mu'$ for every dominant weight $\mu'$ in the coset $\mu + \mathbb{Z}\Phi$. So it is needed if $\Lambda \neq \mathbb{Z}\Phi$. – Mikko Korhonen May 18 '19 at 8:03
• Let $P_{\geq 0}$ denote the (real) cone generated by the fundamental weights and $Q_{\geq 0}$ the (real) cone generated by the positive roots (note that these are polar dual cones, with $P_{\geq 0}$ strictly contained in $Q_{\geq 0}$). It should be that for any $v \in P_{\geq 0}$ we have $P_{\geq 0}\setminus (v+Q_{\geq 0})$ is a bounded, nonconvex polytope. – Sam Hopkins May 19 '19 at 1:36
• In fact, I think it's not even important that $v\in P_{\geq 0}$ for this to be true. In other words, the answer to your question is yes, even for not necessarily dominant $\lambda \in \Lambda$. – Sam Hopkins May 19 '19 at 1:43

To me the answer seems to be yes.

Let $$\varpi_i$$ be the $$i$$th fundamental dominant weight. Recall first that since $$\Phi$$ is irreducible, for all $$i$$ we have $$\varpi_i = \sum_{j = 1}^l q_{ji} \alpha_j$$ for $$q_{ji} \in \mathbb{Q}$$ with $$q_{ji} > 0$$ for all $$j$$. (This can be seen either case-by-case by inverting the Cartan matrix, or with a general proof as in Exercise 13.8 of Humphreys' book).

Let $$\lambda \in \Lambda^+$$ and write $$\lambda = c_1\alpha_1 + \cdots + c_l\alpha_l$$ for $$c_i \in \mathbb{Q}$$.

Now consider $$\mu \in \Lambda^+$$ such that $$\lambda - \mu \in \mathbb{Z}\Phi$$. Write $$\mu = a_1 \varpi_1 + \cdots + a_l\varpi_l$$ for $$a_i \in \mathbb{Z}_{\geq 0}$$. I claim that there is an $$N > 0$$ depending only on $$\Phi$$ and the $$c_i$$ such that if $$\mu \not\succeq \lambda$$, then $$a_i \leq N$$ for all $$i$$. Consequently the number of such $$\mu$$ is finite.

For this note that $$\mu \not\succeq \lambda$$ if and only if $$\mu = c_1'\alpha_1 + \cdots + c_l'\alpha_l$$ with $$c_t' < c_t$$ for some $$t$$. Now $$c_t' = \sum_{i = 1}^l a_i q_{ti}$$, so $$c_t' < c_t$$ implies that $$a_iq_{ti} < c_t$$ for all $$i$$. Hence we can take $$N = \max \{c_j/q_{ji}\}_{i,j}$$.

• Can you specify what $c_i'$ is in your argument? I am trhying to understand the details. – Jim Humphreys May 19 '19 at 1:10
• @JimHumphreys: In this case $c_i' = \sum_{j = 1}^l a_j q_{ij}$. (I added this detail to the answer). – Mikko Korhonen May 19 '19 at 7:29

Here's basically the same answer as Mikko Korhonen, but written in a way that's slightly easier for me to understand.

Let me use $$\leq$$ to denote the partial order on all of the vector space $$E$$ with $$u \leq v$$ for $$u,v \in E$$ if and only if $$v-u = \sum_{i=1}^{n}a_i \alpha_i$$ with all $$a_i \geq 0$$ (but not necessarily integral). Considering $$\leq$$ instead of $$\preceq$$ means I don't have to deal with the different classes of the root lattice mod the weight lattice.

Fix any $$v \in E$$. We claim there are only finitely many dominant weights $$\mu \in \Lambda^{+}$$ for which we don't have $$v \leq \mu$$, from which the desired claim obviously follows. Indeed, let $$\omega$$ be a fundamental weight. As Mikko explains (and as is also fundamental to the answer I gave to the linked question), we have $$\omega = \sum_{i=1}^{n} a_i \alpha_i$$ with all $$a_i > 0$$. Hence clearly there is some $$m \geq 0$$ so that $$v \leq m\omega$$. Now let $$M$$ be the maximum over all fundamental weights of such $$m$$. Then writing $$\mu = \sum_{i=1}^{n} c_i \omega_i$$, the only way we could fail to have $$v \leq \mu$$ is if $$c_i < M$$ for all $$i$$. (Here we are using the fact that if $$v \leq u$$ and $$\nu \in \Lambda^{+}$$, then $$v \leq u+\nu$$, which can be seen again from the fact that writing $$\omega = \sum_{i=1}^{n} a_i \alpha_i$$ for any fundamental weight $$\omega$$, we have $$a_i > 0$$.) There are clearly only finitely many $$\mu=\sum_{i=1}^{n} c_i \omega_i \in \Lambda^{+}$$ with $$c_i < M$$ for all $$i$$.

EDIT:

Here is an even more general statement/context. Let $$V$$ be an n-dimensional Euclidean vector space with inner product $$\langle \cdot, \cdot \rangle$$ and let $$v_1,\ldots,v_n \in V$$ be a collection of vectors such that:

• $$v_1,\ldots,v_n$$ form a basis of $$V$$;
• $$\langle v_i , v_j \rangle \leq 0$$ for all $$i \neq j$$ (in other words, the vectors are pairwise non-acute);
• there is no nontrivial decomposition of $$V = V_1 \oplus V_2$$ into orthogonal subspaces $$V_1$$ and $$V_2$$ such that $$v_i \in V_1 \cup V_2$$ for all $$i$$ (this is an irreducibility condition- equivalently it says that if we draw the graph on the $$v_i$$ with $$v_i$$ adjacent to $$v_j$$ if $$\langle v_i, v_j\rangle < 0$$ then that graph will be connected).

Then let $$Q_{\geq 0} := \{ v=\sum_{i=1}^{n}a_iv_i, a_i \geq 0\}$$ be the cone generated by the $$v_i$$. And let $$P_{\geq 0} := \{v \in V\colon \langle v,w\rangle \geq 0 \textrm{ for all w\in Q_{\geq 0}}\}$$ be the dual cone to $$Q_{\geq 0}$$.

Then the claim is that for any $$v\in V$$ we have that $$P_{\geq 0} \setminus (v+Q_{\geq 0})$$ is a bounded subset of $$V$$.

To prove this, observe that any nonzero $$w \in P_{\geq 0}$$ (in particular, any generator of this cone) has all $$a_i > 0$$ when we write $$w=\sum_{i=1}^{n} a_i v_i$$. The reason for this is that the matrix $$M=(\langle v_i, v_j \rangle)$$ is a nonsingular, irreducible $$M$$-matrix, and hence $$M^{-1}$$ (which expresses the coordinates of the generators of $$P_{\geq 0}$$) is a matrix with all entries strictly positive (see e.g. Theorem A of https://core.ac.uk/download/pdf/82640451.pdf). Then we can apply the same argument as above to the generators of the cone $$P_{\geq 0}$$.

(The particular situation above corresponds to the $$v_i$$ being the simple roots and the cone $$P_{\geq 0}$$ being the dominant cone, i.e., cone spanned by the fundamental weights.)


text:
Nicholas T.

# A grocer mixed gumdrops worth $0.60 per pound with chocolate worth$0.52 per pound.

If she made 60 pounds of a mixture worth \$0.54 per pound, how many pounds of chocolates are in the mixture?

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# Understand onViewableItemsChanged in FlatList

post ime : 2020-01-21 category : React Native read :

If you want to get viewable items in the [FlatList], you had better take a look at the onViewableItemsChanged prop. For example, suppose you have a video list, and you want automatically play the video when the video is appearing on the screen for a few seconds. In iOS, there is visibleCells in UITableView to achieve this. In React Native, I am glad to tell you that FlatList has a more powerful property, onViewableItemsChanged. This article would help you better understand how to use theonViewableItemsChanged prop in the [FlatList], and how it works under the hood.

## What is onViewableItemsChanged

onViewableItemsChanged is a prop in VirtualizedList and FlatList. When you scroll a FlatList, the items showing on the FlatList change. Then, this function is called, telling you what current viewableItems are and what changed items are.

This function should be used together with viewabilityConfig. A specific onViewableItemsChanged will be called when its corresponding ViewabilityConfig‘s conditions are met.

Here is ViewabilityConfig

Here is the type of onViewableItemsChanged function:

## How to use it

Let’s look at two simple example.

Besides, supposed you have to implement different logic for items with 60% viewable region and those with 75% viewable region. You can use viewabilityConfigCallbackPairs, which contains an list of key/value objects, which define different viewability configurations and onViewableItemsChanged callbacks.

## How does onViewableItemsChanged works

### Viewable Region

The layout and viewable region information for VirtualizedList is stored in _scrollMetrics object. Through the nativeEvent in onScroll callback, VirtualizedList gets these layout information.

If it is a vertical VirtualizedList, the layout.layoutMeasurement.height in the nativeEvent is assigned to visibleLength; which is the height of viewable region here. Also, in a vertical VirtualizedList, the layout.layoutMeasurement.height is equal to viewportHeight.

### Overview

#### Different viewabilityConfig in one VirtualizedList

_viewabilityTuples is an array inside VirtualizedList to store ViewabilityHelper/onViewableItemsChanged pairs. This array is initialized in the constructor function.

If you define viewabilityConfigCallbackPairs, each viewabilityConfig will be used to initialize a different ViewabilityHelper object.

ViewabilityHelper is a utility class for calculating viewable items based on the viewabilityConfig and metrics, like the scroll position and layout.

As I mentioned before, in a VirtualizedList could has several ViewabilityHelper objects in _viewabilityTuples, containing different viewabilityConfig to handle different viewability conditions. Let’s take a look at some important props in ViewabilityHelper.

#### Items’ layout

In the overview graph, you can see a func _updateViewableItems called in many scenarios. For example, it is called in onScroll callback. Then, It calls viewabilityHelper.onUpdate to find out the viewable items, which appear in the viewport for VirtualizedList.

• this._scrollMetrics.visibleLength is used as viewportHeight
• this._createViewToken is used to construct a ViewToken object, which contains item data, index, key and isViewable flag of the item.
• this._getFrameMetrics is a function to get layout information of the item cell by index. The item layout is from getItemLayout prop of VirtualizedList or this._frames map. this._frames stores the itemKey/itemLayout pairs.
• By this.state, we know the range of the rendered items by first and last value. VirtualizedList updates these two values when the rendered items are changed.

### How to find out viewable items

In onUpdate method, it calls computeViewableItems to get viewableIndices. viewableIndices is an array of indexes of the viewable items. So, how does computeViewableItems work?

#### How to get the indexes of viewable items

In computeViewableItems in the ViewabilityHelper class, it iterates items from ${first} to ${last}. If an item is viewable, it will be stored in an array named viewableIndices.

From the code, we can see the top and bottom value is related to the screen coordinate. I drew a graph to show the relationship between metrics.offset, scrollOffset, metrics.length , top and bottom, to help you better understand the above code.

### What kind of item is viewable

An item is said to be viewable when it meets the following conditions
for longer than ${minimumViewTime} milliseconds (after an interaction if waitForInteraction is true): 1. the fraction of the item visible in the view area >= itemVisiblePercentThreshold. When it comes to the fraction of the item visible in the view area, we need to care about cases shown in the following graph. RN use Math.min(bottom, viewportHeight) - Math.max(top, 0) to calculate the viewable length. 2. Entirely visible on screen when the height of a item is bigger than the viewportHeight. ref code here ### Timer and Schedule In onUpdate func in ViewabilityHelper, if we define minimumViewTime value, the _onUpdateSync is scheduled to be called. It is the handler of the timeout. And, If after a few seconds,${minimumViewTime}, if some items aren’t longer viewable, the _onUpdateSync func, filter out these indices that have gone out of viewport.

In the above graph, at first, the _viewableIndices is from 1 to 9. Then the user scrolls the VirtualizedList and the _onUpdateSync is triggered after minimumViewTime. At this moment, the current _viewableIndices is from 2 to 10. So the item indexed 1 is filtered out.

### 6. How to get changed items

Comparing with the last time when onViewableItemsChanged is triggered, at this time to trigger onViewableItemsChanged. Some viewable items will be out of the screen, some hidden items will become viewable. In _onUpdateSync function, the preItems map stores the information about previous visible items, the previous means last time when VirtualizedList calls onViewableItemsChanged. Now it has a nextItems map, which stores the information about viewable items this time. Then it figures out the changed items by comparing these two maps. Then, it calls onViewableItemsChanged, passing viewableItems and changed` items.

Author : RY Zheng


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# 10.1: Spontaneous Processes

Entropy is a state function that is often erroneously referred to as the 'state of disorder' of a system. Qualitatively, entropy is simply a measure how much the energy of atoms and molecules become more spread out in a process and can be defined in terms of statistical probabilities of a system or in terms of the other thermodynamic quantities. Entropy is also the subject of the Second and Third laws of thermodynamics, which describe the changes in entropy of the universe with respect to the system and surroundings, and the entropy of substances, respectively.

### Statistical Definition of Entropy

Entropy is a thermodynamic quantity that is generally used to describe the course of a process, that is, whether it is a spontaneous process and has a probability of occurring in a defined direction, or a non-spontaneous process and will not proceed in the defined direction, but in the reverse direction. To define entropy in a statistical manner, it helps to consider a simple system such as in Figure 1. Two atoms of hydrogen gas are contained in a volume of $$V_1$$.

Figure 1. Two hydrogen atoms in a volume $$V_1$$

Since all the hydrogen atoms are contained within this volume, the probability of finding any one hydrogen atom in $$V_1$$ is 1. However, if we consider half the volume of this box and call it $$V_2$$,the probability of finding any one atom in this new volume is $$\frac{1}{2}$$, since it could either be in $$V_2$$ or outside. If we consider the two atoms, finding both in $$V_2$$, using the multiplication rule of probabilities, is $$\frac{1}{2} \times \frac{1}{2} =\frac{1}{4}$$. For finding 4 atoms in $$V_2$$ would be $$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}= \frac{1}{16}$$. Therefore, the probability of finding N number of atoms in this volume is $$\frac{1}{2}^N$$. Notice that the probability decreases as we increase the number of atoms.

If we started with volume $$V_2$$ and expanded the box to volume $$V_1$$, the atoms would eventually distribute themselves evenly because this is the most probable state. In this way, we can define our direction of spontaneous change from the lowest to the highest state of probability. Therefore, entropy S can be expressed as

$S=k_B \ln{\Omega} \tag{1}$

where $$\Omega$$ is the probability and $$k_B$$ is a proportionality constant. This makes sense because entropy is an extensive property and relies on the number of molecules, when $$\Omega$$ increases to $$W^2$$, S should increase to 2S. Doubling the number of molecules doubles the entropy.

So far, we have been considering one system for which to calculate the entropy. If we have a process, however, we wish to calculate the change in entropy of that process from an initial state to a final state. If our initial state 1 is $$S_1=K_B \ln{\Omega}_1$$ and the final state 2 is $$S_2=K_B\ln{\Omega}_2$$,

$\Delta S=S_2-S_1=k_B \ln \dfrac{\Omega_2}{\Omega_1} \tag{2}$

using the rule for subtracting logarithms. However, we wish to define $$\Omega$$ in terms of a measurable quantity. Considering the system of expanding a volume of gas molecules from above, we know that the probability is proportional to the volume raised to the number of atoms (or molecules), $$\alpha V^{N}$$ . Therefore,

$\Delta S=S_2-S_1=k_B \ln \left(\dfrac{V_2}{V_1} \right)^N=Nk_B \ln \dfrac{V_2}{V_1} \tag{3}$

We can define this in terms of moles of gas and not molecules by setting the $$k_{B}$$ or Boltzmann constant equal to $$\frac{R}{N_A}$$, where R is the gas constant and $$N_A$$ is Avogadro's number. So for a expansion of an ideal gas and holding the temperature constant,

$\Delta S=\dfrac{N}{N_A} R \ln \left(\dfrac{V_2}{V_1} \right)^N=nR\ln \dfrac{V_2}{V_1} \tag{4}$

because $$\frac{N}{N_A}=n$$, the number of moles. This is only defined for constant temperature because entropy can change with temperature. Furthermore, since S is a state function, we do not need to specify whether this process is reversible or irreversible.

### Thermodynamic Definition of Entropy

Using the statistical definition of entropy is very helpful to visualize how processes occur. However, calculating probabilities like $$\Omega$$ can be very difficult. Fortunately, entropy can also be derived from thermodynamic quantities that are easier to measure. Recalling the concept of work from the first law of thermodynamics, the heat (q) absorbed by an ideal gas in a reversible, isothermal expansion is

$q_{rev}=nRT\ln \dfrac{V_2}{V_1} \; . \tag{5}$

If we divide by T, we can obtain the same equation we derived above for $$\Delta S$$:

$\Delta S=\dfrac{q_{rev}}{T}=nR\ln \dfrac{V_2}{V_1} \;. \tag{6}$

We must restrict this to a reversible process because entropy is a state function, however the heat absorbed is path dependent. An irreversible expansion would result in less heat being absorbed, but the entropy change would stay the same. Then, we are left with $$\Delta S> q_{irrev}/T$$ for an irreversible process because $$\Delta S=\Delta S_{rev}=\Delta S_{irrev}$$. This apparent discrepancy in the entropy change between an irreversible and a reversible process becomes clear when considering the changes in entropy of the surrounding and system, as described in the second law of thermodynamics.

It is evident from our experience that ice melts, iron rusts, and gases mix together. However, the entropic quantity we have defined is very useful in defining whether a given reaction will occur. Remember that the rate of a reaction is independent of spontaneity. A reaction can be spontaneous but the rate so slow that we effectively will not see that reaction happen, such as diamond converting to graphite, which is a spontaneous process.

### The Second Law as Energy Dispersion

Energy of all types -- in chemistry, most frequently the kinetic energy of molecules (but also including the phase change/potential energy of molecules in fusion and vaporization, as well as radiation) changes from being localized to becoming more dispersed in space if that energy is not constrained from doing so. The simplest example stereotypical is the expansion illustrated in Figure 1.

The initial motional/kinetic energy (and potential energy) of the molecules in the first bulb is unchanged in such an isothermal process, but it becomes more widely distributed in the final larger volume. Further, this concept of energy dispersal equally applies to heating a system: a spreading of molecular energy from the volume of greater-motional energy (“warmer”) molecules in the surroundings to include the additional volume of a system that initially had “cooler” molecules. It is not obvious, but true, that this distribution of energy in greater space is implicit in the Gibbs free energy equation and thus in chemical reactions.

“Entropy change is the measure of how more widely a specific quantity of molecular energy is dispersed in a process, whether isothermal gas expansion, gas or liquid mixing, reversible heating and phase change, or chemical reactions.” There are two requisites for entropy change.

1. It is enabled by the above-described increased distribution of molecular energy.
2. It is actualized if the process makes available a larger number of arrangements for the system’s energy, i.e., a final state that involves the most probable distribution of that energy under the new constraints.

Thus, “information probability” is only one of the two requisites for entropy change. Some current approaches regarding “information entropy” are either misleading or truly fallacious, if they do not include explicit statements about the essential inclusion of molecular kinetic energy in their treatment of chemical reactions.

### References

1. This literal greater spreading of molecular energy in 3-D space in an isothermal process is accompanied by occupancy of more quantum states (“energy levels”) within each microstate and thus more microstates for the final macrostate (i.e., a larger $$\Omega$$ in R\ln{W}). Similarly, in any thermal process higher energy quantum states can be significantly occupied – thereby increasing the number of microstates in the product macrostate as measured by the Boltzmann relationship.
2. I. N. Levine, Physical Chemistry, 6th ed. 2009, p. 101, toward the end of “What Is Entropy?”
3. Chang, Raymond. Physical Chemistry for the Biosciences. Sausalito, California: University Science Books, 2005.


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## Linear Independence and Volumes of Parallelepipeds

If a parallelogram is defined by two vectorsandthen the area of the parallelogram is defined by( Notewhereis the angle betweenand

If the vectorsandare joined by a third vectorto form a solid shape, then the volume of the solid is the area of the base (which we may consider to be the area of the parallelepiped formed byand) multiplied by the height.is perpendicular to bothandso is in the direction of the vertical height. By taking the dot product ofwithand dividing bywe obtain the component ofperpendicular toThis is the height of the parallelepiped. Multiplying bygives the volume: This is illustrated below.

If the vectorsare in the same plane, then they are linearly dependent, since three vectors in a two dimensional space are linearly dependent. They all lie in the same plane and the height of the parallelepiped is zero. If a matrix is formed with the columns or rows consisting of the three vectors, the determinant of this matrix will be zero since the vectors are linearly dependent.


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2 added 983 characters in body

It could be Kronecker's determination of the sign of the Gauss sum by means of Cauchy's theorem. Already Gauss noted that the determination of the sign implies the law of quadratic reciprocity.

In response to the request for references:

Leopold Kronecker: Summirung der Gauss'schen Reihen ... J. Reine Angew. Math. 105 (1889), 267-268.

Also in volume 4 of his Werke, 297-300. (This was where I xeroxed it, so I can vouch for the page numbers, I have the pages in front of me right now).

Also in Landau's Elementare Zahlentheorie (together with two others, by Mertens and Schur), near the end of the book.

Also supposed to be in Ayoub: Introduction to the Analytic Theory of Numbers, but I am not familiar with his book, so I cannot vouch for this.

There is a later determination of the sign of the Gauss sum by contour integration, due to Mordell, which is quite accessible; it is in Chandrasekharan's Introduction to Analytic Number Theory, page 35--39. Chandrasekharan does a more general case.

Now, I have not claimed that Kronecker's proof was the one that Hilbert was thinking of. I cannot read the mind of a dead man (nor that of a living one).

1

It could be Kronecker's determination of the sign of the Gauss sum by means of Cauchy's theorem. Already Gauss noted that the determination of the sign implies the law of quadratic reciprocity.


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# Is it known whether there are ever infinitely many primes of the form $\prod_i p_i^{n_i} + 1$ where the $p_i$ are fixed primes but the $n_i$ can vary?

So if we fix finitely many primes $p_i$, where one $p_i$ is $2$, but let the powers $n_i$ on the $p_i$ vary, is it known whether it is ever possible to have infinitely many primes of the form $\prod_i p_i^{n_i} + 1$? Note this is sort of a soft question, if there are any choice of distinct primes $p_i$ that make it possible then I would be happy with that. The negation of the statement would be that for any choice of primes $p_i$, the set of primes so formed according to the question would be finite.

• – Michael Burr Jul 9 '15 at 23:19
• @MichaelBurr Yes thanks, that's exactly what motivated my question. But now I'm allowing any arbitrary products of powers of an a priori fixed set of primes. I was hoping my question would strike at the heart of the matter, or at least reveal our limitations in being able to answer the question. I.e., is it true that for ANY fixed finite set of primes, we cannot yet determine the answer? – user2566092 Jul 9 '15 at 23:22
• It's certainly not possible if the $p_i$ are all odd. – Robert Israel Jul 9 '15 at 23:25
• The next refinement would be what is the best $f$ we can use in the statement that there are an infinite number of primes $q$ of the form $q = 1 + \prod_i p_i ^ {n_i}$ with all $p_i \le f(q)$. This question asks if $f(q) = c$, and $f(q) = q + 1$ works. – Dan Brumleve Jul 9 '15 at 23:27
• @RobertIsrael Fair enough, I should have clarified because I implicitly knew that that was a trivial answer if all primes were odd. I'm modifying my question now. – user2566092 Jul 9 '15 at 23:28

What we can do (using Dirichlet's theorem on primes in arithmetic progressions) is say there are primes $q$ such that $q-1$ is divisible by each of a given finite set of primes, and not divisible by each of another given finite set of primes. You want $q-1$ to be divisible only by primes in a given finite set. Much, much harder.


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Revision history [back]

Diameter 'Answer In'/'Request In' fields not available with tshark/pyshark

Hi All,

I am using pyshark (a Python wrapper for tshark) to analyse traffic capture files, and I ran into a problem around finding Diameter Common Messages request/answer pairs.

The Wireshark GUI helps identifying the answer to a specific Diameter request packet with the field called Answer In. The Request In field helps the other way around. See the Wireshark Diameter Protocol Display Filter Reference, diameter.answer_in and diameter.answer_to field names.

Using the above information makes it easy to construct display filters in the Wireshark GUI for selecting specific request/answer pairs. Unfortunately, when using the tshark CLI on the same capture file, this field seems to be missing.

tshark -r d:/test.pcapng -Y diameter.answer_in  # Works in Wireshark GUI, not in tshark

What do I need to use these field names for filtering capture files with tshark?

My tshark version is: TShark (Wireshark) 2.9.0 (v2.9.0-0-gf4238cd2)

Many thanks, Bence


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# Tiling a Chessboard with Dominoes (Opposite Colors Removed)

This is a natural follow-up to our first gallery entry on the impossibility of tiling certain chessboards with dominoes.

Problem: Suppose we remove two squares from a chessboard which have opposite color. Is it possible to tile the remaining squares with 2-by-1 dominoes?

Solution:

From Honsberger's Mathematical Gems I.

Notice that if we remove two squares of opposite color, then there is only one way to place dominoes on the remaining squares according to this scheme (one cannot tile a domino across the “walls”). Removing two squares of opposite color ensures there is an even number of squares between the removed squares, and so the dominoes will fit as desired. In other words, the two “paths” from A to B, starting on any side of A, contain an even number of squares, and so placing a domino every two squares will fill the path with a good tiling.

Not only does this method provide a positive answer to the solution, it gives a constructive answer. If we had some odd desire to do so, we could write an algorithm which tiles any chessboard where two squares are removed, (moreover, it would run in linear time). It’s not obvious at first whether such an algorithm exists, as the average human attempt to tile a board with two random squares removed would likely require some backtracking. In other words, human solutions would generally be heuristic in nature, but this analysis gives a solution that any person could utilize.

# N Choose 2 is the Sum of the First N-1 Integers

Problem: Determine an arithmetic expression for $\binom{n}{2}$.

Solution: The following picture describes a bijection between the set of yellow dots and the set of pairs of purple dots:

In particular, selecting any yellow dots and travelling downward along diagonals gives a unique pair of blue dots. Conversely, picking any pair of blue dots gives a unique yellow dot which is the meeting point (the “peak”) of the inward diagonals. If we say the bottom row has $n$ elements, then the number of yellow dots is clearly $1 + 2 + \dots + (n-1)$, and the number of pairs in the last row is just $\binom{n}{2}$.

This proof is typical of the most elegant combinatorial proofs: come up with a bijection between one set of objects whose size is unknown and another set of objects whose size is easy to determine. Such a method of proof is ubiquitous in elementary combinatorics, and it extends even further to determine things about infinite sets. In higher mathematics, bijections are quite frequently used (along with other structure-preserving conditions) to determine when two things should be deemed equivalent. This concept is called isomorphism, and much of mathematics is devoted to classifying all objects in a category up to an appropriate notion of isomorphism. One particularly long instance of this is the classification of finite simple groups, but one which is easier to understand, and studied extensively by mathematicians since time immemorial, is the classification of Euclidean isometries (see here for the case of $\mathbb{R}^2$).

# Möbius Transformations are Isometries of a Sphere

We present a video on Möbius transformations and the geometry of the sphere. Anyone who has taken or will take complex analysis (that means you engineers!) should watch this. It shows not only the beautiful correspondence between the two, but it reveals the intuition behind a lot of complex analysis, when more often than not a student is left in the dust of rigorous formulas.

In short, this is a proof without words that the Möbius transformations are in correspondence with rigid motions of the unit sphere in $\mathbb{R}^3$. This 1-1 mapping is precisely Riemann’s stereographic projection. While we usually might see this in one context (showing $\mathbb{R}^2$ is the one-point compactification of the plane), it is not often connected to all of our interesting transformations in such a picturesque way.

Möbius transformations lay the foundation for much of hyperbolic geometry and other advanced topics in complex analysis. They’re quite fascinating.

# False Proof – 31.5 = 32.5

Problem: Show 31.5 = 32.5.

“Solution”:

Explanation: It appears that by shifting around the pieces of one triangle, we have constructed a second figure which covers less area! Since the first triangle has base length 13 and height 5, its area is 32.5. Clearly, the second figure has the same area minus a square of area 1, giving the second figure area 31.5.

In fact, by counting up the squares in each component, we do see that it is simply a shifting of pieces, so the areas of the two figures must be the same. Therefore, the fault lies in our first assumption: that the first figure is actually a triangle. Looking more closely, we see that the slope of the hypotenuse of the red component is $3/8$, while the blue triangle as slope $2/5$. If the first figure were a triangle, then the slope of the hypotenuse would be uniform. Since $2/5 \neq 3/8$, the first figure is not a triangle, and hence its area is not 32.5. In fact, counting up the area of each component, the areas of both figures are actually 32.

This false proof accentuates the need to question one’s intuition. In fact, nothing false was said by the picture alone. The two figures do truly have the same area. The falsity is entirely in the viewer’s interpretation of what the picture says (and the questionably moral intent of a trickster mathematician). So the lesson is: while intuition can be a useful and essential guide, it is often not rigorous enough to be proof, and it can create very dangerous false positives.