text:
# Anti-aliasing ASCII art

## Background

ASCII art is the practice of creating images by using ASCII text to form shapes.

Aliasing is the effect created by the large "pixels" of ASCII art, which are the size of characters. The image becomes blocky and hard to see. Anti-aliasing removes this effect by creating a gradient and by softening the hard edges of the ASCII art.

# The Challenge

Your challenge is to write the shortest program possible that will take a piece of ASCII art and will output a version that has been anti-aliased.

## What sort of anti-aliasing?

All of the ASCII art will consist of two types of symbols: Spaces and non-whitespace. For each non-whitespace character, your program must determine whether it is in a position where it need to be anti-aliased. If it is, then you need to replace it with the correct character. If it isn't, then the character stays the same.

How do you know if a character needs to be anti-aliased? The answer depends on the characters that are immediately above, below, left, and right of the character (not the diagonals). Here is a chart of when anti-aliasing is required, where ? and x can stand for any non-whitespace character.

 x? ->  d?
?      ?

?x  -> ?b
?      ?

?      ?
?x  -> ?F

?      ?
x? ->  Y?

x  ->  ;   Note: This character has been changed from _ to ;
?      ?

?      ?
x  ->  V

?x  -> ?>

x? ->  <?

x  ->  @

## Input (and example pre-anti-aliasing ASCII art)

First, there will be two lines of input (to STDIN), a number H followed by a number W. There will then be H lines of exactly W characters each (excluding the newline). These following lines will be the ASCII art that needs to be anti-aliased. Here is an example input (not beautiful, but a test):

7
9
888888
888888
999 98  7
666666
666666
6666
6

## Output (and example anti-aliased art)

Your program should output to STDOUT the ASCII art (of the same dimensions), which has been anti-aliased. Here is the output for the above input. Notice how the the border characters are treated as bordering whitespace.

  d8888>
d8888F
<99 98  @
Y6666b
Y6666>
Y66F
V

This might not look that good (due to the space between lines in the code block), it looks better with larger ASCII art, and the quality depends on the exact font used.

# Another Example

Input

12
18
xx  xxx  xxx
xxxx  xxx  xxx
xxxxxx  xxx  xxx
xxx  xxx  xxx  xxx
xxxx xxx  xxx  xxx
xxxxxx  xxx  xxx
xxxx  xxx  xxx
x  xx  xxx  xxx  x
xx    xxx  xxx  xx
xxx  xxx  xxx  xxx
xxxx  xxx  xxx  xx
xxxxx  xxx  xxx  x

Output

   db  <xb  <xb
dxxb  Yxb  Yxb
dxxxxb  Yxb  Yxb
dxx  xxb  xxb  xxb
Yxxb xxF  xxF  xxF
YxxxxF  dxF  dxF
YxxF  dxF  dxF
;  YF  dxF  dxF  ;
xb    dxF  dxF  dx
xxb  <xF  <xF  <xx
xxxb  Yxb  Yxb  Yx
Yxxx>  Yx>  Yx>  V

# Rules, Restrictions, and Notes

Your program should be written in printable ASCII characters only, so that we can make art out of the programs. Other than that, the standard code-golf rules apply.

• Since there are no answers yet, I have changed one character in the anti-aliasing chart. _ has become ; because it works better. – PhiNotPi Apr 11 '12 at 15:15
• This might be my favorite code golf question of all time. Working on like 4 different awesome solutions. – captncraig Apr 12 '12 at 1:46
• Although I am confused. You say diagonals don't count, but your diagrams all show question marks filling out the diagonals. From what I see in the examples, it may be safe to only look to the sides, but I get confused? Do diagonals ever matter? – captncraig Apr 12 '12 at 3:17
• No, the diagonals never matter. It will probably be more clear if I removed the diagonals from the chart. – PhiNotPi Apr 12 '12 at 3:23
• I think there might be a typo in your example; I believe the right-hand column should have Ys on the inside edge. Enjoyed coming up with the answer to this one though, good question :D – Ed James Apr 12 '12 at 19:16

### Ruby, 180 168 characters

gets
w=1+gets.to_i
f.zip(f[1..-1]+s=[" "],s+f,s*w+f,f[w..-1]+s*w){|a,*b|$><<"@;V#{a}>bF#{a}<dY#{a*5}"[a>" "?(b.map{|y|y>" "?1:0}*"").to_i(2):3]}  Another Ruby implementation which takes a zip approach. You can see the second example running online. Edit: Using readlines saves 12 characters. ### Ruby 275265263261258254244243214212 207 H=0...gets.to_i W=0...gets.to_i G=readlines z=->x,y{(H===y&&W===x&&' '!=G[y][x])?1:0} H.map{|j|W.map{|i|l=G[j][i] G[j][i]="@V;#{l}>Fb#{l}<Yd#{l*5}"[z[i+1,j]*8+z[i-1,j]*4+z[i,j+1]*2+z[i,j-1]]if' '!=l}} puts G  Sample 1: http://ideone.com/PfNMA Sample 2: http://ideone.com/sWijD • 1) Parenthesis around the range definition are not necessary. 2) 0..h-1 can be written as 0...h. 3) G=[];h.times{G<<gets} can be written as G=readlines, similarly as in your C# code. 4) After step 3. variable h becomes useless, h and w's values are used only once, so h=gets.to_i;w=gets.to_i;H=(0..h-1);W=(0..w-1) can be written as H=0...gets.to_i;W=0...gets.to_i. 5) In this case and can be written as &&, which not needs the spaces around. 6) You have an extra ; and you counted the newline at the end of file, which are not necessary. This mean 214 chars: ideone.com/CiW0l – manatwork Jun 2 '12 at 11:08 • Wow, thanks! I knew there were improvements to be made, but never thought there were that many. I had tried H=0..gets.to_i when I wrote the code, but it didn't seem to work (obviously it must've been for other reasons). – Cristian Lupascu Jun 2 '12 at 11:52 • Still two points where you can reduce by at least 7 chars: 1) You can use map instead of each 2) z=->... instead of def z...end. – Howard Jun 4 '12 at 11:31 • @Howard Thanks, I've applied the map instead of each change. For the lambda syntax, however, I think that would require the usages of z to be of the form z.call(args) instead of z(args), thus adding a bit to the character count. Please let me know if I'm missing something. – Cristian Lupascu Jun 4 '12 at 12:35 • @Howard Nevermind, I just found out what I was missing. I'll update to use the lambda [] usage. – Cristian Lupascu Jun 4 '12 at 12:37 Javascript, 410 characters: function(t){m={"10110":"b","11100":"d","01101":"Y","00111":"F","10100":";","00101":"V","00110":">","01100":"<","00100":"@"},d="join",l="length",t=t.split('\n').splice(2),t=t.map(function(x)x.split('')),f=function(i,j)t[i]?(t[i][j]||' ')==' '?0:1:0;for(o=t[l];o--;){for(p=t[o][l];p--;){y=[f(o+1,p),f(o,p+1),f(o,p),f(o,p-1),f(o-1,p)],t[o][p]=m[y[d]('')]||t[o][p]}}t=t.map(function(x)x[d](''))[d]('\n');return t;}  ungolfed: function(t){ m={ "10110":"b", "11100":"d", "01101":"Y", "00111":"F", "10100":";", "00101":"V", "00110":">", "01100":"<", "00100":"@" }, d="join", l="length", t=t.split('\n').splice(2), t=t.map(function(x) x.split('')), f=function(i,j) t[i]?(t[i][j]||' ')==' '?0:1:0; for(o=t[l];o--;){ for(p=t[o][l];p--;){ y=[f(o+1,p),f(o,p+1),f(o,p),f(o,p-1),f(o-1,p)], t[o][p]=m[y[d]('')]||t[o][p] } } t=t.map(function(x)x[d](''))[d]('\n'); return t; }  Original, 440 characters: function (t){m={"10110":"b","11100":"d","01101":"Y","00111":"F","10100":";","00101":"V","00110":">","01100":"<","00100":"@"},s="split",d="join",l="length",t=t[s]('\n').splice(2),t=t.map(function(x) x[s]('')),f=function(i,j)i<0||i>=t[l]?0:(j<0||j>=t[i][l]?0:t[i][j]==' '?0:1);for(o=t[l];o--;){for(p=t[o][l];p--;){y=[f(o+1,p),f(o,p+1),f(o,p),f(o,p-1),f(o-1,p)],h=m[y[d]('')];if(h){t[o][p]=h}}}t=t.map(function(x) x[d](''))[d]('\n');return t;}  N.B. I've assumed that the first two input lines are actually irrelevant and the size of the following lines are a correct. I'm also reckoning I might be able to chop a few more chars off this when I get a chance! • Replace the declaration of m to m={22:"b",28:"d",13:"Y",7:"F",20:";",5:"V",6:">",12:"<",4:"@"} then convert m's subscript with parseInt(): m[parseInt(y[d](''),2)]. This reduces the size to 373 characters. – manatwork Jun 2 '12 at 16:29 ## Python, 259 chars H=input() W=input()+1 I=' '.join(raw_input()for i in' '*H) for i in range(H):print''.join(map(lambda(s,a,b,c,d):(s*5+'dY<'+s+'bF>'+s+';V@'+' '*16)[16*(s==' ')+8*(a==' ')+4*(b==' ')+2*(c==' ')+(d==' ')],zip(I,I[1:]+' ',' '+I,I[W:]+' '*W,' '*W+I))[i*W:i*W+W-1])  The program reads the input into a single string I (with spaces separating the lines), zips up a list of 5-tuples containing the character and its four surrounding characters, then looks up the result character using string indexing. ## PHP - 359330282268 257 characters <?php$i=fgets(STDIN)+0;$w=fgets(STDIN)+1;$s='';$m='@<;d>0b0VY00F000'; for(;$i--;)$s.=fgets(STDIN); for(;++$i<strlen($s);){$b=trim($s[$i])?0:15;
foreach(array($i+1,$i+$w,$i-1,$i-$w)as$k=>$x)
$b|=pow(2,$k)*(isset($s[$x])&&trim($s[$x]));
echo $m[$b]?$m[$b]:$s[$i];}

• @PhiNotPi It wasn't working because my locally saved test files had windows-style EOL \r\n. I updated my code to work with unix style EOL \n. – Rusty Fausak Apr 12 '12 at 22:40
• Ok, it seems to be working now. – PhiNotPi Apr 12 '12 at 22:47

# Python, 246 241

H=input();W=1+input()
S=' '
o=W*S
F=o+'\n'.join((raw_input()+o)[:W-1]for k in range(H))+o
print ''.join((16*x+'@;<d>b'+2*x+'V'+x+'Y'+x+'F'+3*x)[
16*(x>S)|8*(a>S)|4*(l>S)|2*(r>S)|(b>S)]for
x,a,l,r,b in zip(F[W:-W],F,F[W-1:],F[W+1:],F[2*W:]))

WC and test on sample 2, diffed with the output of the Ruby solution at the top:

t:~$wc trans.py && python trans.py < lala2 > o && diff -q o ruby_out2_sample 2 11 241 trans.py t:~$

## C# 591 563

string A(string t){var s=new StringReader(t);var h=int.Parse(s.ReadLine());var w=int.Parse(s.ReadLine());var lines=s.ReadToEnd().Split(new[]{"\r\n"},StringSplitOptions.None).Select(x=>x.ToCharArray()).ToArray();for(var i=0;i<h;i++)for(var j=0;j<w;j++){var c=lines[i][j];if(c==' ')continue;var n=(i>0?(lines[i-1][j]!=' '?1:0):0)+(i<h-1?(lines[i+1][j]!=' '?2:0):0)+(j>0?(lines[i][j-1]!=' '?4:0):0)+(j<w-1?(lines[i][j+1]!=' '?8:0):0);lines[i][j]=new[]{'@','V',';',c,'>','F','b',c,'<','Y','d',c,c,c,c,c}[n];}return string.Join("\r\n",lines.Select(l=>new string(l)));}


text:
# Do averaged binary symmetric channels maximize mutual information?

This is a refined version of Do binary symmetric channels maximize mutual information?, which was answered negatively.

Let the random variables $(X, Y)$ be a doubly symmetric binary source with parameter $0 \le p \le 1/2$, i.e., $X,Y \sim \text{Bernoulli}(1/2)$ and $P(X \neq Y) = p$.

Define the two regions $\mathcal{A}, \mathcal{B} \subseteq \mathbb{R}^3$:

• $\mathcal{A}$ consists of all points $(R_0, R_1, R_2)$ such that there exist binary random variables $U,V$, satisfying the Markov chain $U - X - Y - V$ and \begin{align} R_1 &\ge \mathrm{I}(U;X) , \\ R_2 &\ge \mathrm{I}(Y;V) , \\ R_0 &\le \mathrm{I}(U;V) , \end{align} where $\mathrm{I}(\cdot;\cdot)$ is mutual information.

• $\mathcal{B}$ consists of all points $(R_0, R_1, R_2)$ such that there exist probabilities $a,b \in [0,1]$ with \begin{align} R_1 &\ge 1 - \mathrm{H}(a * p) , \\ R_2 &\ge 1 - \mathrm{H}(b * p) , \\ R_0 &\le 1 - \mathrm{H}(a * p *b) , \end{align} where $a*b := a(1-b)+(1-a)b$ is binary convolution and $\mathrm{H}()$ is the binary entropy function.

My Question: Is $\mathrm{conv}(\mathcal A) = \mathrm{conv}(\mathcal B)$, where $\mathrm{conv}()$ denotes the convex hull?

The definitions of $\mathcal{A}$ and $\mathcal{B}$ are almost the same, except that the channels $X \to U$ and $Y \to V$ are required to be symmetric for $\mathcal{B}$ and hence $\mathcal B \subseteq \mathcal A$.
The question Do binary symmetric channels maximize mutual information? asked whether $\mathcal A = \mathcal B$ and a counterexample was provided for $p=0$, which does not seem to apply here.

format: plain


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Resources tagged with: Resourceful

Filter by: Content type:
Age range:
Challenge level:

There are 59 NRICH Mathematical resources connected to Resourceful, you may find related items under Habits of Mind.

Broad Topics > Habits of Mind > Resourceful

Box Plot Match

Age 14 to 16 Challenge Level:

Match the cumulative frequency curves with their corresponding box plots.

Perception Versus Reality

Age 14 to 18 Challenge Level:

Infographics are a powerful way of communicating statistical information. Can you come up with your own?

Shifting Times Tables

Age 11 to 14 Challenge Level:

Can you find a way to identify times tables after they have been shifted up or down?

Tower of Hanoi

Age 11 to 14 Challenge Level:

The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.

How Much Can We Spend?

Age 11 to 14 Challenge Level:

A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?

Dicey Operations

Age 11 to 14 Challenge Level:

Who said that adding, subtracting, multiplying and dividing couldn't be fun?

Nice or Nasty

Age 7 to 14 Challenge Level:

There are nasty versions of this dice game but we'll start with the nice ones...

The Remainders Game

Age 7 to 14 Challenge Level:

Play this game and see if you can figure out the computer's chosen number.

Treasure Hunt

Age 7 to 14 Challenge Level:

Can you find a reliable strategy for choosing coordinates that will locate the treasure in the minimum number of guesses?

M, M and M

Age 11 to 14 Challenge Level:

If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?

Which Spinners?

Age 14 to 18 Challenge Level:

Can you work out which spinners were used to generate the frequency charts?

Difference Sudoku

Age 14 to 16 Challenge Level:

Use the differences to find the solution to this Sudoku.

A Chance to Win?

Age 11 to 14 Challenge Level:

Imagine you were given the chance to win some money... and imagine you had nothing to lose...

Transformation Game

Age 11 to 14 Challenge Level:

Why not challenge a friend to play this transformation game?

Substitution Cipher

Age 11 to 14 Challenge Level:

Find the frequency distribution for ordinary English, and use it to help you crack the code.

Product Sudoku

Age 11 to 14 Challenge Level:

The clues for this Sudoku are the product of the numbers in adjacent squares.

Semi-regular Tessellations

Age 11 to 16 Challenge Level:

Semi-regular tessellations combine two or more different regular polygons to fill the plane. Can you find all the semi-regular tessellations?

Cosy Corner

Age 11 to 14 Challenge Level:

Six balls are shaken. You win if at least one red ball ends in a corner. What is the probability of winning?

Two's Company

Age 11 to 14 Challenge Level:

Seven balls are shaken. You win if the two blue balls end up touching. What is the probability of winning?

Odds and Evens

Age 11 to 14 Challenge Level:

Are these games fair? How can you tell?

Property Chart

Age 11 to 14 Challenge Level:

A game in which players take it in turns to try to draw quadrilaterals (or triangles) with particular properties. Is it possible to fill the game grid?

Shapely Pairs

Age 11 to 14 Challenge Level:

A game in which players take it in turns to turn up two cards. If they can draw a triangle which satisfies both properties they win the pair of cards. And a few challenging questions to follow...

Age 11 to 14 Challenge Level:

A game for 2 or more people, based on the traditional card game Rummy. Players aim to make two `tricks', where each trick has to consist of a picture of a shape, a name that describes that shape, and. . . .

Arithmagons

Age 14 to 16 Challenge Level:

Can you find the values at the vertices when you know the values on the edges?

Stars

Age 11 to 14 Challenge Level:

Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit?

Isosceles Triangles

Age 11 to 14 Challenge Level:

Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?

Square It

Age 11 to 16 Challenge Level:

Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.

On the Edge

Age 11 to 14 Challenge Level:

If you move the tiles around, can you make squares with different coloured edges?

Cuboids

Age 11 to 14 Challenge Level:

Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?

The Spider and the Fly

Age 14 to 16 Challenge Level:

A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly?

Special Numbers

Age 11 to 14 Challenge Level:

My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?

Hexy-metry

Age 14 to 16 Challenge Level:

A hexagon, with sides alternately a and b units in length, is inscribed in a circle. How big is the radius of the circle?

Age 11 to 14 Challenge Level:

How many different symmetrical shapes can you make by shading triangles or squares?

Triangles to Tetrahedra

Age 11 to 14 Challenge Level:

Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?

In a Box

Age 14 to 16 Challenge Level:

Chris and Jo put two red and four blue ribbons in a box. They each pick a ribbon from the box without looking. Jo wins if the two ribbons are the same colour. Is the game fair?

Attractive Tablecloths

Age 14 to 16 Challenge Level:

Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?

Marbles in a Box

Age 11 to 16 Challenge Level:

How many winning lines can you make in a three-dimensional version of noughts and crosses?

Two and Two

Age 11 to 14 Challenge Level:

How many solutions can you find to this sum? Each of the different letters stands for a different number.

Where Can We Visit?

Age 11 to 14 Challenge Level:

Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?

Funny Factorisation

Age 11 to 14 Challenge Level:

Using the digits 1 to 9, the number 4396 can be written as the product of two numbers. Can you find the factors?

How Old Am I?

Age 14 to 16 Challenge Level:

In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?

Dozens

Age 7 to 14 Challenge Level:

Do you know a quick way to check if a number is a multiple of two? How about three, four or six?

Gabriel's Problem

Age 11 to 14 Challenge Level:

Gabriel multiplied together some numbers and then erased them. Can you figure out where each number was?

Unequal Averages

Age 11 to 14 Challenge Level:

Play around with sets of five numbers and see what you can discover about different types of average...

Reversals

Age 11 to 14 Challenge Level:

Where should you start, if you want to finish back where you started?

Age 11 to 14 Challenge Level:

What happens when you add a three digit number to its reverse?

Cryptarithms

Age 11 to 14 Challenge Level:

Can you crack these cryptarithms?

Wipeout

Age 11 to 14 Challenge Level:

Can you do a little mathematical detective work to figure out which number has been wiped out?

Constructing Triangles

Age 11 to 14 Challenge Level:

Generate three random numbers to determine the side lengths of a triangle. What triangles can you draw?

Finding Factors

Age 14 to 16 Challenge Level:

Can you find the hidden factors which multiply together to produce each quadratic expression?


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# elliptic regularity of Neumann problem on Square

I asked a similar question the other day, but I will be more precise now.
Consider $\Omega:=(0,1 ) \times (0,1)$ and consider
$$- u_{xx}(x,y) - u_{yy}(x,y) + a(x) u_x + b(y) u_y + u = f(x,y) \mbox{ in } \Omega$$ with $\partial_\nu u=0$ on $\partial \Omega$ (assume one has some sort of variational solution $u \in H^1(\Omega)$. Suppose $a(x),b(y)$ are smooth in $\Omega$ and assume $f$ is Holder continuous on $\Omega$. \

By looking at the even extension of $u$ (across the various boundaries and which then makes one look at odd extensions of $a$ and $b$ which have jump discontinuities) i seem to be able to prove that $u \in C^{2,\alpha}( \overline{\Omega})$. I am curious whether this result is true or ? So my question is: does this regularity seem correct or is it known one does not have this regularity at the corners? (the result suprised me a bit but I am also not familiar with what to expect with Neumann boundary conditions).
thanks a bunch for any comments.


text:
# Kepler-46

Observation data Epoch J2000      Equinox J2000 Constellation Artist's conception of the Kepler-46 system showing the two planets. Planet b transits its star, the same method by which it was detected. Lyra 19h 17m 04.4929s[1] 42° 36′ 15.041″[1] 13.814[2] 13.436[2] 13.347[2] Proper motion (μ) RA: 3.156±0.046[1] mas/yr Dec.: 0.894±0.053[1] mas/yr Parallax (π) 1.2586 ± 0.0277[1] mas Distance 2,590 ± 60 ly (790 ± 20 pc) Absolute bolometricmagnitude (Mbol) 5.18 (predicted)[note 1] Mass 0.902+0.040−0.038[3] M☉ Radius 0.938+0.038−0.039[3] R☉ Temperature 5155±150[3] K Metallicity [Fe/H] 0.41±0.10[3][note 2] dex Rotation 27.859±0.075 days[4] Age 9.7+3.7−3.5[3] Gyr 2MASS J19170449+4236150, KIC 7109675[2] SIMBAD data KIC data

Kepler-46, previously designated KOI-872, is a star located in the constellation Lyra. Observed since 2009 by the Kepler space observatory, it has since been found to possess a planetary system consisting of at least two planets and while it has a similar mass to the Sun (90%) it is significantly older at ten billion years.[3]

Kepler-46 b (previously KOI-872.01), was the first planet discovered in the system. It was found through detailed analysis of Kepler space observatory data. An additional planet, Kepler-46 c, was discovered by an outside group using Kepler public data through analysis of transit timing variations. While only one additional planet was confirmed by the analysis, the study revealed the potential existence of either an unconfirmed planet KOI-872.03 (KOI-872 d)[5]. Validation by multiplicity method allowed to confirm the existence of this planet which was then renamed Kepler-46d.

## Planetary system

Planet b is a gas giant planet with a mass slightly less than that of Jupiter.[3] The second planet in the system was among the first to be discovered through the method of transit timing variations, and through its confirmation of KOI-872 c with a 99% confidence level has shown that the method of detection may be used to detect future extrasolar planets and, possibly, extrasolar moons.[6] This second planet exerted a gravitational force on the first planet, orbiting its host star in just 34 days. While this usually occurs on an extremely regular schedule, additional planets within the system can disrupt the time of the transit, and these disruptions can indicate the presence of a planet, even if the disrupting planet does not pass in front of the host star itself.[6]

The data show that Kepler-46 c is an approximately Saturn-mass object with an orbital period of 57 days.[6] As the planet does not itself transit its host star, there is no way of knowing its size (probably a similar size to its sibling). The measurements also suggest the existence of another planet orbiting with a period of about 6.8 days,[5] this planet was later confirmed.[7]

The method in which the planet was detected is similar to the way that the planet Neptune was discovered, in which the newly discovered planet is known by its pull on another which is already known to exist.[8]

The Kepler-46 planetary system[5][3]
Companion
(in order from star)
Mass Semimajor axis
(AU)
Orbital period
(days)
b 0.885+0.374
−0.343
MJ
0.1971±0.0001 33.648+0.004
−0.005
0.0321+0.0069
−0.0078
89.04±0.14° 0.810+0.035
−0.36
RJ
c 0.362±0.016 MJ 0.2811±0.0003 57.325+0.116
−0.098
0.0354+0.0057
−0.0059
88.66+0.26
−0.27
°
d 0.0679±0.0035 6.76671+0.00013
−0.00012
0(assumed) 88.55+0.49
−0.69
°
0.1510+0.0094
−0.0098
RJ

## References

1. Brown, A. G. A.; et al. (Gaia collaboration) (August 2018). "Gaia Data Release 2: Summary of the contents and survey properties". Astronomy & Astrophysics. 616. A1. arXiv:. Bibcode:. doi:. Gaia Data Release 2 Vizier catalog entry
2. ^ a b c d "Kepler-46". SIMBAD. Centre de données astronomiques de Strasbourg. Retrieved May 16, 2012.
3. Saad-Olivera, Ximena; et al. (2017). "Masses of Kepler-46b, c from Transit Timing Variations". The Astronomical Journal. 153 (4). 198. arXiv:. Bibcode:2017AJ....153..198S. doi:10.3847/1538-3881/aa64e0.
4. ^ McQuillan, A.; Mazeh, T.; Aigrain, S. (2013). "Stellar Rotation Periods of The Kepler objects of Interest: A Dearth of Close-In Planets Around Fast Rotators". The Astrophysical Journal Letters. 775 (1). L11. arXiv:. Bibcode:2013ApJ...775L..11M. doi:10.1088/2041-8205/775/1/L11.
5. ^ a b c Nesvorny, D.; et al. (2012). "The Detection and Characterization of a Nontransiting Planet by Transit Timing Variations". Science. 336 (6085): 1133. arXiv:. Bibcode:2012Sci...336.1133N. doi:10.1126/science.1221141.
6. ^ a b c Moskowitz, Clara (May 10, 2012). "Hidden Alien Planet Revealed by Its Own Gravity". Space.com. Space.com. Retrieved May 10, 2012.
7. ^ Rowe, Jason F.; et al. (2014). "Validation of Kepler's Multiple Planet Candidates. III. Light Curve Analysis and Announcement of Hundreds of New Multi-planet Systems". The Astrophysical Journal. 784. 45. arXiv:. Bibcode:2014ApJ...784...45R. doi:10.1088/0004-637X/784/1/45.
8. ^ Crockett, Christopher (May 12, 2012). "New planet found in distant solar system by its tug on another world". EarthSky. Earthsky Communications. Retrieved May 19, 2012.

## Notes

1. ^ Figure based on the following equations, which calculated bolometric (total) luminosity across all spectra based on effective temperature: ${\displaystyle m_{\rm {star}}=m_{\rm {sun}}-2.5\log _{10}\left({L_{\rm {star}} \over L_{\odot }}\cdot \left({\frac {d_{\rm {sun}}}{d_{\rm {star}}}}\right)^{2}\right)}$ (cf. Luminosity) and ${\displaystyle M=m_{\rm {star}}-5((\log _{10}{D_{L}})-1)\!\,}$ (cf. Absolute magnitude)
2. ^ This measurement indicates the log10 of the relative abundance of iron in the measured star to that of the Sun.


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## 35.38 Descending quasi-affine morphisms

In this section we show that “quasi-affine morphisms satisfy descent for fpqc-coverings”. Here is the formal statement.

Lemma 35.38.1. Let $S$ be a scheme. Let $\{ X_ i \to S\} _{i\in I}$ be an fpqc covering, see Topologies, Definition 34.9.1. Let $(V_ i/X_ i, \varphi _{ij})$ be a descent datum relative to $\{ X_ i \to S\}$. If each morphism $V_ i \to X_ i$ is quasi-affine, then the descent datum is effective.

Proof. Being quasi-affine is a property of morphisms of schemes which is preserved under any base change, see Morphisms, Lemmas 29.13.3 and 29.13.5. Hence Lemma 35.36.2 applies and it suffices to prove the statement of the lemma in case the fpqc-covering is given by a single $\{ X \to S\}$ flat surjective morphism of affines. Say $X = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$ so that $R \to A$ is a faithfully flat ring map. Let $(V, \varphi )$ be a descent datum relative to $X$ over $S$ and assume that $\pi : V \to X$ is quasi-affine.

According to Morphisms, Lemma 29.13.3 this means that

$V \longrightarrow \underline{\mathop{\mathrm{Spec}}}_ X(\pi _*\mathcal{O}_ V) = W$

is a quasi-compact open immersion of schemes over $X$. The projections $\text{pr}_ i : X \times _ S X \to X$ are flat and hence we have

$\text{pr}_0^*\pi _*\mathcal{O}_ V = (\pi \times \text{id}_ X)_*\mathcal{O}_{V \times _ S X}, \quad \text{pr}_1^*\pi _*\mathcal{O}_ V = (\text{id}_ X \times \pi )_*\mathcal{O}_{X \times _ S V}$

by flat base change (Cohomology of Schemes, Lemma 30.5.2). Thus the isomorphism $\varphi : V \times _ S X \to X \times _ S V$ (which is an isomorphism over $X \times _ S X$) induces an isomorphism of quasi-coherent sheaves of algebras

$\varphi ^\sharp : \text{pr}_0^*\pi _*\mathcal{O}_ V \longrightarrow \text{pr}_1^*\pi _*\mathcal{O}_ V$

on $X \times _ S X$. The cocycle condition for $\varphi$ implies the cocycle condition for $\varphi ^\sharp$. Another way to say this is that it produces a descent datum $\varphi '$ on the affine scheme $W$ relative to $X$ over $S$, which moreover has the property that the morphism $V \to W$ is a morphism of descent data. Hence by Lemma 35.37.1 (or by effectivity of descent for quasi-coherent algebras) we obtain a scheme $U' \to S$ with an isomorphism $(W, \varphi ') \cong (X \times _ S U', can)$ of descent data. We note in passing that $U'$ is affine by Lemma 35.23.18.

And now we can think of $V$ as a (quasi-compact) open $V \subset X \times _ S U'$ with the property that it is stable under the descent datum

$can : X \times _ S U' \times _ S X \to X \times _ S X \times _ S U', (x_0, u', x_1) \mapsto (x_0, x_1, u').$

In other words $(x_0, u') \in V \Rightarrow (x_1, u') \in V$ for any $x_0, x_1, u'$ mapping to the same point of $S$. Because $X \to S$ is surjective we immediately find that $V$ is the inverse image of a subset $U \subset U'$ under the morphism $X \times _ S U' \to U'$. Because $X \to S$ is quasi-compact, flat and surjective also $X \times _ S U' \to U'$ is quasi-compact flat and surjective. Hence by Morphisms, Lemma 29.25.12 this subset $U \subset U'$ is open and we win. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


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# Practice Problems for Module 3

Sections 6.3 and 6.4

Directions. The following are review problems. We recommend you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it. You can also follow the link for the full video page to find related videos.
1. Set up an integral representing the volume of the solid obtained by rotating the region bounded by the following curves about the line $$x=5$$: $$y=3x−x^2$$ and $$y=3x−9$$.

An integral giving the volume is $$\displaystyle \int_{-3}^3 \! 2\pi \left(9-x^2\right)(5-x)\, dx$$.
Video Errata: Speaker said "Actually this is the volume of the shell" and wrote $$V$$, which is not true. It is the surface area!

To see the full video page and find related videos, click the following link.

2. Set up an integral representing the volume of the solid obtained by rotating the region bounded by the following curves about the line $$y = 5$$: $$y =\sqrt{x−4}$$, the $$x$$-axis and $$x = 8$$.

An integral giving the volume is $$\displaystyle \int_{0}^2 \! 2\pi (5-y)\left(4-y^2\right)\, dy$$.

To see the full video page and find related videos, click the following link.

3. Suppose a spring has a natural length of 3 ft and it takes 10 ft-lb to stretch it from 5 ft to 8 ft.
1. How much work is required to stretch the spring from 4 ft to 7 ft?
2. How far beyond its natural length would a force of 3 lb keep the spring stretched?

1. 150/21 ft-lb
2. 63/20 ft

To see the full video page and find related videos, click the following link.

4. A hemispherical tank has a radius of 10 m with a 2 m spout at the top of the tank. The tank is filled with water to a depth of 7 m. Set up an integral that would compute the work required to pump all the water out of the spout. Use the fact that the weight density of water is 9800 N/m$$^3$$.

The total work is given by $$\displaystyle \int_3^{10} 9800\pi \left( 100-y^2\right) (y+2) \, dy$$.

To see the full video page and find related videos, click the following link.

5. A trough with isosceles triangles as its ends is filled with water to a depth of 2 ft. The tank is 4 ft tall, 6 ft across at the top, 10 ft long and has a 3 ft spout. Set up an integral that would compute the work required to pump all the water out of the spout. Use the fact that the weight density of water is 62.5 lb/ft$$^3$$.

The total work is given by $$\displaystyle \int_2^{4} 62.5 \left[ 15(4-y)\right](y+2) \, dy$$.
Video Errata: When labeling the picture on the left (the one in 3D), the speaker said to start the axes at the top of the water. This is not correct: start it at the top of the tank as he did in the right-hand drawing, looking at the tank from the side. Also, the speaker used the wrong weight density: since we are working in ft, it needs to be 62.5 ft/lb$$^3$$, NOT 9800 N/m$$^3$$.

To see the full video page and find related videos, click the following link.


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# Limit question: from proof that t-distribution approaches N(0,1)

## Homework Statement

I'm trying to prove that the random variable with t distribution with k degrees of freedom will have N(0,1) as k→∞ the two equations below are derived from the pdf of the t-distribution:

2. Homework Equations which I can't get my head around

$$lim_{k\to\infty}((1+\frac{x^2}{k})^k)^{-0.5} = e^{-\frac{x}{2}}$$

## The Attempt at a Solution

I have proved $$lim_{k\to\infty}((1+\frac{x^2}{k})^{-0.5}=1$$, which I admit wasn't too hard.

The statement involving the limit of gamma function was given to us as "knowing that it can be shown" but I would appreciate if someone was willing to explain it/point me to the source where I can find it.

## Answers and Replies

SammyS
Staff Emeritus
Homework Helper
Gold Member
Do you know that $\displaystyle \lim_{t \to \infty } \left ( 1 + \frac{1}{t} \right )^{t}=e\ ?$

BTW: Either you should have x2 in your exponent, or you should have x, rather than x2 in your limit expression.

Do you know that $\displaystyle \lim_{t \to \infty } \left ( 1 + \frac{1}{t} \right )^{t}=e\ ?$

BTW: Either you should have x2 in your exponent, or you should have x, rather than x2 in your limit expression.

Hi Sammy, I know the definition of e but the limit is still not trivial to me. Is the equation below some sort of a rule? Thanks!

$$lim_{k\to\infty}(1+\frac{b}{k})^k= e^b$$

You're right. I added ^2


text:
## Files in this item

FilesDescriptionFormat

application/pdf

9512477.pdf (5MB)
(no description provided)PDF

## Description

 Title: Growth and characterization of centimeter-sized crystals of pyridine dodecasil-3C Author(s): Marquart, Todd A. Doctoral Committee Chair(s): Klemperer, Walter G. Department / Program: Chemistry Discipline: Chemistry Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Chemistry, Inorganic Abstract: Hydrothermal seeded growth was used to grow centimeter-sized crystals of the all-silica zeolite ZSM-39, also known as the clathrasil dodecasil-3C (D3C). Through careful optimization of reaction conditions coupled with judicious choice of silica sources, conditions were found that minimize the effects of Ostwald's rule of successive transformations allowing for the growth of large, high-purity crystals of the metastable zeolite. The twinning present in pyridine D3C crystals was removed by thermal methods. X-ray single crystal studies of detwinned crystals coupled with data from optical crystallography studies were used to determine the structure at ambient temperature. The high-temperature structure was also determined by single crystal X-ray diffraction methods carried out at temperatures above the phase transformation at 161$\sp\circ$C. Twinning in both the ambient and high-temperature phases was examined. In order to examine the position and motion of the guest molecule, a combination of computer simulations and $\sp2$H NMR studies was undertaken. Issue Date: 1994 Type: Text Language: English URI: http://hdl.handle.net/2142/19378 Rights Information: Copyright 1994 Marquart, Todd A. Date Available in IDEALS: 2011-05-07 Identifier in Online Catalog: AAI9512477 OCLC Identifier: (UMI)AAI9512477


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# How is angular momentum related to torque?

Apr 28, 2018

\vec{\tau} = \frac{d\vec{L}}{dt};
$\setminus \vec{L}$ - Angular Momentum; $\setminus \vec{\setminus \tau}$ - Torque;

#### Explanation:

Torque is the rotational equivalent of force and Angular Momentum is the rotational equivalent of Translational Momentum.

Newton's second law relates Translational Momentum to Force,
$\setminus \vec{F} = \frac{d \setminus \vec{p}}{\mathrm{dt}}$

This can be extended to rotational motion as follows,
$\setminus \vec{\setminus \tau} = \frac{d \setminus \vec{L}}{\mathrm{dt}}$.

So Torque is the rate of change of Angular Momentum.


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# Is the “recovery phrase” generated by a wallet app platform independent?

I recently started a wallet with Coinbase Wallet, and I was asked to backup a 12 word "recovery phrase" that is supposed to be able to recover my wallet in the event that I delete the app from my phone. I have previously used other wallet apps, which often give "recovery phrases" of varying word count. I read online that the phrase is passed through an algorithm and used to recover a key or a group of keys.

I am now curious if these recovery phrases are platform independent (meaning you can recover your keys just by way of knowing the phrase), or if the app must exist with a specific algorithm in order to recover keys (your keys are lost if the recovery implementation is destroyed, e.g. Coinbase shuts down operation and removes their app).

The mnemonic words that is given to you by the wallets is a representation of entropy. This entropy is then passed through PBKDF2 hashing function to generate seed. The industry standard for this conversion of mnemonic to seed is documented in BIP 39 and most wallets follow that method. The seed is then passed to HMAC-SHA512 to get a master extended private key.


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# Statistics

## Getting started

### Typing your data in the table

When you enter the Statistics app, you must type your data in a two-column table. You can add up to 3 data tables.

• In the first column (Values), enter the values of your statistical data.
• In the second column (Sizes), enter the frequencies/sizes associated with each value in your dataset, i. e. the number of occurrences of each value.

When you fill in the first column, the second column is automatically filled in with the value $1$. This means that each of the values in your statistical data appears only once in the set. Change the sizes values in the second column if the values in your statistical set appear more than once.

Like, for example:
Let’s consider the following statistical data: $1$, $1$, $1$, $2$, $3$, $3$, $4$, $5$, $5$.
To enter this statistical data in the table, proceed as follows.

Values V1 Sizes N1
$1$ $3$
$2$ $1$
$3$ $2$
$4$ $1$
$5$ $2$

You can also enter frequencies in the Sizes column.

### Plotting data as a histogram

Once you have typed your data into the table in the Data tab, you can plot it as a histogram.

1. Select the Histogram tab at the top of the screen.
2. Confirm by pressing OK.

You then see the histogram that displays your data.

### Plotting data as a box plot

Once you have typed your data into the table in the Data tab, you can plot it as a box plot.

1. Select the Box tab at the top of the screen.
2. Confirm by pressing OK.

You then see the box plot that displays your data.

### Displaying statistical variables

Once you have typed your data into the table in the Data tab, you can display the statistical variables: mean, standard deviation, median,…

1. Select the Stats tab at the top of the screen.
2. Confirm by pressing OK.

You then see the table of statistical variables.

## Using the Data tab

### Delete a value from the data table

You can delete a row from the table by selecting a cell in that row and pressing del.

You can change the content of a cell by selecting it and typing a new value with the keyboard.

### Clearing a column of the table

You can delete all the values in a column of the table.

1. Select the name of the column you want to clear. Confirm by pressing OK.
2. The column options menu opens. Select Clear column and confirm with OK.

Clearing the Values column also clears the Sizes column.
Clearing the Sizes column fills this column with the value $1$.

### Generate a list with a formula

You can create a column of the data table using a formula involving another column.

1. Select the name of the column you want to fill. Confirm by pressing OK.
2. The column options menu opens. Select Fill with formula and confirm with OK.
3. Type your formula using the name of another column. For example, if you want column V2 to be filled with the values of V1 divided by 2, write V1/2 in the text field at the bottom of the screen. Then press OK.

To enter a capital letter, press shift then alpha then the letter to display.

## Using the Histogram tab

### Moving the cursor in the histogram

When you are in the Histogram tab, you can read the numbers and frequencies in the bar at the bottom of the screen for each rectangle. The intervals represented by the rectangles are also displayed.

To move the selection to another rectangle in the histogram, use the Left arrow and Right arrow keys.

To move to another data set, use the Up arrow and Down arrow keys.

### Setting histogram parameters

You can change the width of the histogram rectangles (bin width) and the start value of the data set.

1. Press OK.
2. The histogram settings menu opens. Enter the values for the width of rectangles and the start of the set. Confirm by selecting the Validate button and pressing the OK key.

## Using the Box tab

In the Box tab, you can read the statistical variables below the box plot:

• Minimum
• First quartile
• Median
• Third quartile
• Maximum

To move the cursor, use the Left arrow and Right arrow keys.

To move to another data set, use the Up arrow and Down arrow keys.

## Using the Stats tab

The Stats tab displays the statistical variables calculated using the data in the Data tab:

• Total size
• Minimum
• Maximum
• Range
• Mean
• Standard deviation
• Variance
• First quartile
• Third quartile
• Median
• Interquartile range
• Sum
• Sum of squares
• Sample standard deviation


text:
Groupe d’études et de recherche en analyse des décisions

# Sensitivity of Trust-Region Algorithms on their Parameters

## Nick Gould, Dominique Orban, Annick Sartenaer et Philippe Toint

In this paper, we examine the sensitivity of trust-region algorithms on the param- eters related to the step acceptance and update of the trust region. We show, in the context of unconstrained programming, that the numerical efficiency of these algo- rithms can easily be improved by choosing appropriate parameters. Recommended ranges of values for these parameters are exhibited on the basis of extensive numerical tests.

, 18 pages


text:
Optional Donation Option 1 $1.00 USD Option 2$2.00 USD Option 3 $5.00 USD Option 4$10.00 USD Option 5 \$25.00 USD

## How many zeros does a polynomial function have?

A polynomial function of degree $$n$$ has $$n$$ zeros, provided multiple zeros are counted more than once and provided complex zeros are counted.

Keep in mind that any complex zeros of a function are not considered to be part of the domain of the function, since only real numbers domains are being considered. It is only for the purpose of counting zeros that we consider complex and multiple zeros.

## Exercise 2.5.1

Let $$f ( x ) = x^2 + 1$$.

Use the quadratic formula to show that the imaginary numbers $$i$$ and $$- i$$ are the two zeros of $$f$$. Verify this using synthetic division.

See Solution

## Exercise 2.5.2

Let $$f ( x ) = x^2 + 2 x + 1$$.

Factor $$f ( x ) = x^2 + 2 x + 1$$ and show that it contains the same linear factor twice. Thus it has a double zero.

See Solution

## Exercise 2.5.3

Let $$f ( x ) = 2 x^3 + 3 x^2 + 8 x - 5$$.

This function $$f(x)$$ has one real zero and two complex zeros.

Use the rational zero principle from section 2.3 to list all possible rational zeros. Then use synthetic division from section 2.4 to find a rational zero from among the possible rational zeros. Use synthetic division to factor the function. Use the quadratic formula on the quotient to find the two complex zeros of the function.

See Solution

## Complex zeros occur in pairs.

Notice that in exercises 2.5.1 and 2.5.3, the complex zeros were conjugates of each other. This always happens. We can state the following general principle:

Complex zeros of polynomial functions occur in conjugate pairs.

## Descartes Rule of Signs for positive zeros

Next, we are going to investigate a general principle which tells us how many positive zeros a polynomial function may have.

Consider the polynomial function $$f ( x ) = 2 x^3 + 3 x^2 + 8 x - 5$$. Notice that it is written in standard form for polynomials, that is, terms are written in decreasing order according to their exponents. Now, count the number of changes in sign of the coefficients. There is exactly one such change in sign. Thus, there can be only one positive zero.

Descartes' rule of signs for positive zeros may be stated as follows:

When a polynomial function is written in standard form, the number of changes in sign of the coefficients is the maximum number of positive zeros of the function. The actual number of positive zeros may be less than the maximum by an even amount.

## Exercise 2.5.4

Consider the function $$g ( x ) = 2 x^5 - 8 x^4 + 7 x^3 - 2 x^2 + 5 x - 1$$.

Use the positive zeros principle to show that $$g$$ has either five, three or one positive zeros.

See Solution

## Descartes rule of signs for negative zeros.

We can use we can use Descartes' Rule of Signs for positive zeros to find a Descartes' rule of signs for negative zeros..

Consider the graph of y$$= g ( - x )$$. It will look like the graph of $$y = g ( x )$$ reflected about the $$y$$-axis. Thus there will be a one-to-one correspondence with the negative zeros of $$y = g ( x )$$ and the positive zeros of $$y = g ( - x )$$. That is to say, the number of positive zeros of

$y = g ( - x ) = - 2 x^5 - 8 x^4 - 7 x^3 - 2 x^2 - 5 x - 1$

equals the number of negative zeros of

$y = g ( x ) = 2 x^5 - 8 x^4 + 7 x^3 - 2 x^2 + 5 x - 1$

Because there are no sign changes in the cofficients of $$y = g ( - x )$$, it follows that $$y = g ( x )$$ has no negative zeros.

The following table shows the possible distribution of the five zeros of $$g ( x )$$ :

 Positive $$5$$ $$3$$ $$1$$ Negative $$0$$ $$0$$ $$0$$ Complex $$0$$ $$2$$ $$4$$

In other words, there are only three possible arrangements of the zeros of $$g$$. Either $$g$$ has five positive zeros, no negative and no complex zeros, or it has three positive zeros, no negative and two complex zeros, or it has one positive zero, no negative and four complex zeros.

## Exercise 2.5.5

Draw a chart showing the possible distribution of the zeros of the function

$$p ( x ) = 2 x^5 + 7 x^3 - 2 x^2 + 5 x - 1$$

See Solution

## Bounds on sets of zeros.

Next, let us consider the polynomial function $$f ( x ) = x^4 + 5 x^3 + 3 x^2 + 2 x + 1$$. Since there are no changes of sign in the coefficients, there are no positive zeros.

Now, consider the function$$g ( x ) = ( x - 1 ) ( x^4 + 5 x^3 + 3 x^2 + 2 x + 1 )$$.

Its only positive zero is 1, since ( x - 1 ) is a factor and since $$x^4 + 5 x^3 + 3 x^2 + 2 x + 1$$ has no positive zeros. Furthermore, since the leading coefficient is positive, the graph of$$y = g ( x )$$ to the right of $$x =1$$ lies entirely above the $$x$$-axis.

Now, consider the polynomial function $$h ( x ) = ( x - 1 ) ( x^4 + 5 x^3 + 3 x^2 + 2 x + 1 ) + 3$$.

Its graph is just the graph of $$y = g ( x )$$ raised upwards by three units. Thus $$y = h ( x )$$ lies entirely above the $$x$$-axis for $$x\ge1$$. So $$1$$ is an upper bound on the set of zeros of the polynomial function $$y = h ( x )$$.

Next, consider the polynomial function $$f ( x ) = - x^4 - 2 x^3 - x^2 - 5 x - 2$$. Since it has no change in the signs of its coefficients, it has no positive zeros. Thus the function $$g ( x ) = ( x - 1 ) ( - x^4 - 2 x^3 - x^2 - 5 x - 2 )$$ has no zeros greater than $$1$$. And since the leading coefficient of $$g$$ is negative, all the graph of $$g$$ to the right of $$1$$ lies below the $$x$$-axis. Thus the graph of a function such as $$h ( x ) = ( x - 1 ) ( - x^4 - 2 x^3 - x^2 - 5 x - 2 ) - 3$$, for example, will also lie entirely below the $$x$$-axis to the right of $$1$$. Thus $$1$$ would be an upper bound on the zeros of $$h$$.

Now what is the point of all this? How can this example be generalized?

In general, if $$c>0$$ and $$f ( x ) = ( x - c ) q ( x ) + r$$ and if $$r$$ and all the coefficients of $$q ( x )$$ are of the same sign, then $$c$$ is an upper bound on the set of zeros of $$f ( x )$$.

Let us see how to use this principle to find an upper bound on the set of zeros of the function$$f ( x ) = x^5 - 2 x^4 + 7 x^3 - x^2 - 12 x - 3$$.

Use synthetic division to rewrite the function in the following form: $$f ( x ) = ( x - 2 ) ( x^4 + 7 x^2 + 13 x + 14 ) + 25$$.

Then, using the principle above, $$2$$ is an upper bound on the set of zeros of $$f$$.

If we are searching for the zeros of a polynomial function by means of synthetic division and find that, when evaluating $$f ( c )$$, for some $$c > 0$$, we get a row of all non-negative (or non-positive) numbers from the synthetic division process, then we know that $$c$$ is an upper bound on the set of zeros of $$f$$.

## Exercise 2.5.6

Use synthetic division to find the smallest positive integer which is an upper bound on the zeros of the function$$f ( x ) = x^3 - 3 x^2 - 3 x + 1$$

See Solution

## Exercise 2.5.7

Find the smallest positive integer which is an upper bound on the zeros of the function $$g ( x ) = - x^3 + 3 x^2 + 3 x - 1$$.

See Solution

## Greatest integral lower bounds on the set of negative zeros

This procedure can be adapted to find negative numbers which are lower bounds on the set of zeros of a polynomial function. Consider the following fact: The number $$- c$$ is a zero of $$y = f ( x )$$ if and only if $$c$$ is a zero of $$y = f ( - x )$$. [Remember that the graph of $$y = f ( - x )$$ is the reflection of the graph of $$y = f ( x )$$ in the $$y$$-axis.] Thus, we can find the negative lower bounds of$$y = f ( x )$$ by finding the positive upper bounds of $$y = f ( - x )$$ and changing the sign.

## Exercise 2.5.8

Find the largest negative integer which is a lower bound on the zeros of the function $$f ( x ) = x^3 - 5 x + 2$$.

See Solution

## Exercise 2.5.9

Find the largest negative integer which is a lower bound on the zeros of the function $$g( x ) = - x^3 + 5 x - 2$$.

See Solution


text:
## Camel Sequences and Quadratic Residues

### Authors: V. Gurvich and Li Sheng

ABSTRACT

Given an even cyclic $(+1,-1)$ sequence $s = (s_0,..., s_{2n-1})$ which consists of $n$ plus ones and $n$ minus ones, let us compute $i + j \ (mod\ 2n)$ for all $n^2$ $(+1,-1)$-pairs $(s_i,s_j)$ and distribute obtained $n^2$ numbers between $2n$ "boxes" $1,2,...,2n-1,2n=0 \ (mod\ 2n)$. The average cardinality of a box is $\frac{n^2}{2n}=\frac{n}{2}$. Some sequences have quite remarkable box-distributions, "almost average everywhere with two big humps". For example,

$n=5, s = (+1+1-1-1+1-1+1-1-1+1),\ B = (3333133330) = (3^4 1 \, 3^4 0)\$

$n=7, s = (+1+1+1-1+1-1-1-1+1+1-1+1-1-1),\ B = (33333363333337) = (3^6 6 \, 3^6 7)$.

In general, given an {\em odd} $n$, the box-distributions $(\lfloor\frac{n}{2}\rfloor^{n-1} \ (n-1) \\lfloor\frac{n}{2}\rfloor^{n-1} \ n)$ and $(\lceil\frac{n}{2}\rceil^{n-1} \ 1 \ \lceil\frac{n}{2}\rceil^{n-1} \ 0)$ as well as the sequences, which generate them, will be called the {\em camel distributions} and {\em camel sequences}, respectively {\em up-camel} and {\em down-camel}. For example, the first sequence above is down-camel, and the second one is up-camel. Camel sequences have applications in extremal graph theory. Here we prove that there are infinitely many 'camels' of both types. More precisely, for every prime $n=4j-1$ we construct an up-camel sequence and for every prime $n=4j+1$ a down-camel one. In both cases these sequences are related to quadratic residues and non-residues modulo $n$. We conjecture that there are no other camel sequences.

Key Words: quadratic residues, camel sequences, maximin, minimax, extremal graph theory, vertex-enumerated graphs.

Paper Available at: ftp://dimacs.rutgers.edu/pub/dimacs/TechnicalReports/TechReports/1999/99-40.ps.gz