# Find $\lim_{n\to\infty} \frac{n^2}{2^n}$

$$\lim_{n\to\infty}\frac{n^2}{2^n}$$

Do you have some tips so I could solve this problem, without the use of L'Hôpital's rule?
Indeed, we didn't see formally L'Hôpital's rule, nor Taylor series so I'm supposed to do this without such "tools".

I've tried using the fact that $n^2 = e^{2\log(n)}$ and $2^n=e^{n\log(2)}$ but didn't manage to eliminate my indeterminate form.

• At least write down what you get after applying l'Hoptial's rule once. Then apply it again. – Alex R. Oct 23 '13 at 17:35
• It look as if $n$ may range over positive integers. In that case, use the fact that by the Binomial Theorem, $(1+1)^n=1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}+\cdots\gt \frac{n(n-1)(n-2)}{6}$. – André Nicolas Oct 23 '13 at 17:40
• What, if anything, do you know about the comparative growth rates of $n$ and $\log(n)$? If you happen to know that $$\lim_{n\to\infty}\bigl[n-\log(n)\bigr]=\infty,$$ then since $$\frac{n^2}{2^n}=\frac{e^{2\log(n)}}{e^{n\log(2)}}=e^{2\log(n)-n\log(2)},$$ and since $\log(2)>0,$ it follows that the limit you're looking for is $0$, and even gives you a start on how it can be proven. – Cameron Buie Oct 23 '13 at 17:53

There are many ways to do that. For example, you can prove that for $n \ge 10$ we have $$2^n > n^3.$$ You can show that using the induction argument.

Regarding to analysis tag, you seem to be familiar to series so let us consider the following series: $$\sum(n^2/2^n)$$ it is not hard seeing that this series is convergent and so the $n-$th term approaches to zero while $n\to\infty$. Clearly, this way needn't using L'Hopital's rule.

• @Amzoti: Thanks for your support my dear friend. I hope you have a good time ahead and a peaceful slumber. :-) – mrs Oct 24 '13 at 1:48
• You are welcome my friend! I appreciate all of the hard work you and amWhy put into the site helping people become better problem solvers! Time to watch a movie and relax as it has been a difficult month at work and preparing two talks! Have a great night! Regards – Amzoti Oct 24 '13 at 1:49

I noticed the OP had wanted L'Hôpital's rule off the table only after I posted my previous reply. I apologize for my inattention. Here is another answer:

Let $a_n=n^2/2^n$ for all $n\in\mathbb{Z}_+$. You can easily compute that \begin{align*} a_{n+1}-a_n=-\frac{n(n-2)-1}{2^{n+1}}, \end{align*} which is negative for $n\geq3$. Therefore, the sequence is eventually decreasing and bounded below (by zero), so it must have a limit.

Obviously, this limit cannot be negative, since $a_n>0$ for all $n$. If we show it cannot be positive, either, then we can conclude that the limit must be zero.

Suppose that the limit is positive: $\lim_{n\to\infty} a_n=c>0$. Since the sequence is decreasing (after omitting $a_1$ and $a_2$, which does not affect convergence properties), it is also true that \begin{align*} c=\lim_{n\to\infty} a_n=\inf_{n\in\mathbb{Z}_+\setminus\{1,2\}}a_n. \end{align*} Then, for $n$ large enough and beyond, we have that \begin{align*} (*)\quad c\leq a_n\leq\frac{3}{2}c. \end{align*}

Therefore, \begin{align*} (**)\quad a_{n+1}=a_{n}\times\frac{a_{n+1}}{a_n}=a_n\times\frac{(n+1)^2}{2n^2}\leq\frac{3}{2}c\times \frac{(n+1)^2}{2n^2}, \end{align*}

where the last inequality follows from $(*)$. But $(n+1)^2/(2 n^2)$ converges to $1/2$ as $n\to\infty$, so that for $n$ sufficiently large and beyond, \begin{align*} \frac{(n+1)^2}{2n^2}\leq\frac{5}{8}. \end{align*} For such values of $n$, $(**)$ implies that \begin{align*} a_{n+1}\leq\frac{3}{2}c\times\frac{(n+1)^2}{2n^2}\leq\frac{15}{16}c<c, \end{align*} where the last inequality is strict because $c$ is positive. But this contradicts $c$ being the infimum of the values of the sequence. The limit thus cannot be positive.

• +1. I appreciate your effort of writing another answer after I pointed out that L'H was off-limits. :) – Lord_Farin Oct 23 '13 at 20:13
• Thanks, @Lord_Farin. :-) – triple_sec Oct 23 '13 at 20:37


# Unique combinations of $n$ non-unique elements in $k$ non-unique buckets with $c$ capacity

I have $n$ non-unique elements, and I have $k$ unordered buckets that can hold anywhere from $0$ to $c$ elements, such that $c * k \geq n$. I would like to find all possible combinations.

For example, given $n=10$, $k=4$, and $c=4$, there are 7 possible distributions:

• 3322
• 3331
• 4222
• 4321
• 4330
• 4411
• 4420

where "3322", for example, means that two buckets have three elements each and the other two buckets have two elements each.

Another way to look at it, is that I want to find unique combinations of $k$ numbers less than or equal to $c$ such that their sum is equal to $n$.

Ideally I'd like an algorithm to be able to generate a list of all acceptable combinations, but knowing a formula for finding the number of combinations given arbitrary $n$, $k$, and $c$ would be helpful. Other answers generally assume buckets have to have a minimum capacity of 1, or they deal with unique elements to some degree, which are not the case here.

First, it is important to note that all solutions can be arranged in descending sorted order. This is useful for the algorithm I will describe, such that it only searches for sorted answers.

Essentially, we start at the first bucket and assign it some item count $0\leq i\leq c$. Now, we realize that assigning the remaining items to the remaining buckets is essentially the same problem, except now we have $n-i$ items, $k-1$ buckets, and each remaining bucket can now only hold up to $i$ items, because if it were to hold anymore, the solution would be unsorted.

Only when we reach the end (no buckets and no items left) do we have a proper solution. This algorithm can also be sped up by realizing that if $\frac{n-i}{k-1}>i$, there is no continuing from this point, as the remaining items can not fit in the remaining buckets without going over $i$.

Below is a solution in Python,

def buckets(n,k,c,solution=[]):
total = 0
for i in range(min(c,n),-1,-1):
if k-1 != 0 and float(n-i)/(k-1) > i:
#Can't fit remaining items
break
solution.append(i)
if n-i == 0 and k-1 == 0:
#We've reached the end of the buckets and have a solution
print(solution)
solution.pop()
return 1
elif k-1 != 0:
#If there's still buckets left, try assigning them
total += buckets(n-i,k-1,i,solution)
#Done checking this bucket assignment
solution.pop()

buckets(10,4,4)

[4, 4, 2, 0]

[4, 4, 1, 1]

[4, 3, 3, 0]

[4, 3, 2, 1]

[4, 2, 2, 2]

[3, 3, 3, 1]

[3, 3, 2, 2]

• Thanks so much for this - it works completely. Jun 1, 2017 at 17:54


# Matrix exponential of a simple bidiagonal matrix

I am interested in finding an expression (closed form or recursive) for the matrix exponential of this banded matrix:

$$\begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & a_1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & a_2 & 1 & \cdots & 0 & 0 \\ 0 & 0 & 0 & a_3 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots &\ddots&\vdots&\vdots \\ 0 & 0 & 0 & 0 & \cdots & a_{n-1} & 1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & a_n \end{pmatrix}$$

For simplicity, assume that $a_k>0~\forall k$. I am quite certain one must exist having played around with it for a while, and looking at the solution for small values of $n$.

Has anyone seen this structure before? Does it have a name? Do you know if there is a solution published somewhere?

If you go ahead and compute the answer for small values of $n$, you get:

$$\exp\left( \begin{array}{cc} 0 & 1 \\ 0 & a_1 \\ \end{array} \right) = \left( \begin{array}{cc} 1 & \frac{-1+e^{a_1}}{a_1} \\ 0 & e^{a_1} \\ \end{array} \right)$$

$$\exp \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & a_1 & 1 \\ 0 & 0 & a_2 \\ \end{array} \right) = \left( \begin{array}{ccc} 1 & \frac{-1+e^{a_1}}{a_1} & \frac{-e^{a_2} a_1+a_1+e^{a_1} a_2-a_2}{a_1 \left(a_1-a_2\right) a_2} \\ 0 & e^{a_1} & \frac{e^{a_1}-e^{a_2}}{a_1-a_2} \\ 0 & 0 & e^{a_2} \\ \end{array} \right)$$

Unfortunately, $n=3$ is to large to print here, but a pattern remains.

It suffices to diagonalize $A$ (if the $(a_i)$ are pairwise distinct). I change slightly the notation $A=D+J_n$ where $D=diag(a_1,\cdots,a_n)$ and $J_n$ is the nilpotent Jordan block of dimension $n$. $A=PDP^{-1}$ where the first row of $P$ is

$1,-\dfrac{1}{a_1-a_2},\dfrac{1}{(a_1-a_3)(a_2-a_3)},-\dfrac{1}{(a_1-a_4)(a_2-a_4)(a_3-a_4)},\cdots$

We obtain the other rows of $P$ by circular permutation along the diagonals. Finally $e^A=Pdiag(e^{a_1},\cdots,e^{a_n})P^{-1}$ where the first row of $P^{-1}$ is

$1,\dfrac{1}{a_1-a_2},\dfrac{1}{(a_1-a_2)(a_1-a_3)},\dfrac{1}{(a_1-a_2)(a_1-a_3)(a_1-a_4)},\cdots$

We obtain the other rows of $P^{-1}$ by circular permutation along the diagonals.

• Cool thanks very much, this looks promising. However, I'm not sure exactly what you mean by "circular permutation along the diagonals". Do you mean just "circular permutation" or (equivalently) do you mean that the diagonals, when wrapped, will be constant? When I try this I don't get $PP^{-1}=1$ – Ian Hincks May 8 '14 at 20:28
• Also, can you point me in the direction of how you arrived at this conclusion? – Ian Hincks May 8 '14 at 20:29
• Ah, okay, I figured out that "circular permutation along the diagonal" means that, for example, the second row of $P$ will be $(0,1,-(a_2-a_3)^{-1},((a_2-a_4)(a_3-a_4))^{-1},...)$. – Ian Hincks May 8 '14 at 21:12
• Yes Ian, the second row is as you write it. Let $A_n$ be the matrix of dimension $n$. Then $e^{A_n}$ is the submatrix of $e^{A_{n+1}}$ that is constituted by its first $n$ rows and its first $n$ columns. – user91684 May 9 '14 at 10:36

This response is very late but perhaps still useful to others who come across your question whilst browsing (as I did!).

Your banded (or upper bidiagonal) matrix is what Optiz termed a $\textit{Steigungsmatrix }$ (maybe ... "gradient matrix"). See McCurdy et al (1984), Mathematics of Computation, 43, 501-528. For any $n>0$, all you need is to apply $\textit{Opitz's formula}$. You will need a little knowledge of functions of matrices and divided differences to apply it but it will be worth the effort.

If we denote your matrix by A and its elements by $a_{ij}$ then, using Opitz's theorem, the matrix F defined as F = $g$(A) for a function $g$(.), has entries $f_{ij}$ as follows: $f_{ij}=0$ for $i>j$, $f_{ij} = g(a_{ii})$ for $i=j$, and $f_{ij} = g[a_{ii},a_{i+1,i+1},..,a_{jj}]$ for $i<j$ .

The notation $g[x_0,x_1,...,x_n]$ is the nth divided difference defined recursively as $$g[x_0,x_1,...,x_{(n-1)},x_n] = \frac{g[x_1,...,x_n]-g[x_0,...,x_{(n-1)}]}{x_n-x_0},$$ $g[x_0,x_1] = (g[x_1]-g[x_0])/(x_1-x_0)$ and $g[x] = g(x)$.

Setting $g$(.) to be the exponential function, I get your answers for n=1 and n=2.

The above assumes distinct diagonal entries. If there are repeats in the diagonal terms then you should use the appropriate "confluent" form of the divided difference.


# Prove that the sum of pythagorean triples is always even

Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even

My Attempt, Case by case analysis:

Case 1: a is odd, b is odd. From the first equation,

$odd^2 + odd^2 = c^2$

$odd + odd = c^2 \implies c^2 = even$

Squaring a number does not change its congruence mod 2.

Therefore c is even

$a + b + c = odd + odd + even = even$

Case 2: a is even, b is even. Similar to above

$even^2 + even^2 = c^2 \implies c$ is even

$a + b + c = even + even + even = even$

Case 3: One of a and b is odd, the other is even Without loss of generality, we label a as odd, and b as even

$odd^2 + even^2 = c^2 \implies odd + even = c^2 = odd$

Therefore c is odd

$a + b + c = odd + even + odd = even$

We have exhausted every possible case, and each shows $a + b + c$ is even. QED

Follow Up: Is there a proof that doesn't rely on case by case analysis? Can the above be written in a simpler way?

• In fact there are no Pythagorean triples where the legs ($a$ and $b$) are both odd Feb 27, 2017 at 16:47
• Even though there are shorter proofs, there is something pleasingly straightforward about your proof. Mar 1, 2017 at 13:48

Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$

• Well that makes it easy, props for the quick answer. Feb 26, 2017 at 21:55
• You are welcome, @spyr03. Feb 26, 2017 at 21:55
• $a^2+b^2+c^2\equiv 2(a^2+b^2)\pmod 2$ Could you explain that step? Feb 26, 2017 at 21:55
• @mrnovice $c^2 = a ^2 + b^2\implies a^2+b^2+c^2 = 2(a^2+b^2)$. Feb 26, 2017 at 21:56
• @mrnovice, not at all, the way I had written the answer was missing the context, you've asked a valid question. Feb 26, 2017 at 21:59

Also, Pythagorean triples have a well defined structure: $$a=k(m^{2}-n^{2}),\ \,b=k(2mn),\ \,c=k(m^{2}+n^{2})$$ and $$a+b+c=2k(mn+m^2)$$

Hint

Write $a+b+c=k$, so

$$a^2+b^2=(a+b)^2-2ab= (k-c)^2-2ab=c^2 → k^2-2(kc+ab)=0→k^2=2(kc+ab)$$

Notice that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=2c^2+2(ab+bc+ac)$, so the square of $a+b+c$ is even and thus $a+b+c$ is also even.

$c^2 = a^2 + b^2 = (a+b)^2 - 2ab$.

$2ab = (a+b)^2 - c^2$

$2ab = (a+b+c)(a+b-c)$

Let $n = a+b+c$, and the above becomes:

$2ab = n(n-2c)$

So the right-hand side must be even, but since $n-2c$ is odd when $n$ is odd, $n$ must be even.

• Nice explanation, I never would have thought about writing $2ab = (a + b + c)(a + b -c)$. You could have finished with $2ab = n^2 - 2cn$ and thus n^2 must be even, therefore n is even. Feb 28, 2017 at 23:29

Consider $(a+b+c)^2$

Which is $a^2 + b^2 + c^2 + 2(ab+bc+ca)$

Since $c^2 = a^2 + b^2$ (c being the hypotenuse), $(a+b+c)^2 = 2(c^2 + ab + bc + ca)$ - which is an even number.

and since squares of odd is odd and evens is even $a+b+c$ has to be even.


Codecademy Python Math Symbols | interviewingthecrisis.org

# Writing mathematical expressions - MatplotlibPython plotting.

Codecademy is the easiest way to learn how to code. It's interactive, fun, and you can do it with your friends. Python Comparison Operators. These operators compare the values on either sides of them and decide the relation among them. They are also called Relational operators. Python 3.x: small greek letters are coded from 945 to 969 so,alpha is chr945, omega is chr969 so just type printchr945 the list of small greek letters in a list. :mortar_board:exercise answers. Contribute to ummahusla/Codecademy-Exercise-Answers development by creating an account on GitHub. Many programming languages include libraries to do more complicated math. You can do statistics, numerical analysis or handle big numbers. One topic many programming languages have difficulty with is symbolic math. If you use Python though, you have access to sympy, the symbolic math library. Sympy.

Learn about lists, a data structure in Python used to store ordered groups of data. :mortar_board:exercise answers. Contribute to ummahusla/Codecademy-Exercise-Answers development by creating an account on GitHub. Python 3.x: small greek letters are coded from 945 to 969 so,alpha is chr945, omega is chr969 so just type printchr945 the list of small greek letters in a list.

Python Arithmetic Operators Example - Assume variable a holds 10 and variable b holds 20, then −. Numeric and Mathematical Modules¶ The modules described in this chapter provide numeric and math-related functions and data types. The numbers module defines an abstract hierarchy of numeric types. There is nothing special about multiplication in Python; we just use an asterisk in the same way we did with addition, subtraction, and division. print2 2 I’m sure it comes as no surprise to you that the above outputs 4. Note that if we multiply two integers, the result is of the data type int. Jupyter notebook recognizes LaTeX code written in markdown cells and renders the symbols in the browser using the MathJax JavaScript library. Mathematics Inline and Display Enclose LaTeX code in dollar signs $.$ to display math inline. What are operators in python? Operators are special symbols in Python that carry out arithmetic or logical computation. The value that the operator operates on is called the operand.

x = Symbol'x' y = Symbol'y' A = Matrix[[1,x], [y,1]] Once a matrix is created, you can operate on it. There are functions to do dot products, cross products or calculate determinants. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions area hyperbolic functions, even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. math.floor x ¶ Return the floor of x as a float, the largest integer value less than or equal to x. math.fmod x, y ¶ Return fmodx, y, as defined by the platform C library. Codecademy-Exercise-Answers / Language Skills / Python / Unit 01 Python Syntax / Edvins Antonovs Subfolder rename & ascending ordering for python Latest commit 5fc7ead Jul 14, 2016.

from sympy.plotting import plot from sympy import Symbol x = Symbol 'x' p = plot 2 x3, 3 x1, legend = True, show = False p [0]. line_color = 'b' p [1]. line_color = 'r' p. show Sets Beside FiniteSet which I exemplify below, sympy also includes support for infinite sets and intervals. Ask questions, get help with an exercise, and chat about your Codecademy coursework here. “When one teaches, two learn” — Robert Heinlein.

Codecademy: Python 1. Python Syntax print "Welcome to Python!" my_int = 7 my_float = 1.23 my_bool = True my_int = 7 my_int = 3 print my_int def spam: eggs = 12 return eggs print spamjust a comment """ first comment second comment third comment """ count_to = 12 count_to = 5 - 2 ni = 2 10 ni = 20 / 4 eggs = 10 2 spam = 3 % 2 2. Strings and Console Output 'Help! Help! I\'m being. Yes, you can do symbolic math in Python! The library to take a look at is SymPy. Its official website is. This article is not a SymPy tutorial, as I only want to walk you through some examples to show you the kinds of things that it can do. Programming reference for Ruby. Codecademy is the easiest way to learn how to code. It's interactive, fun, and you can do it with your friends.

Python-based: SymPy is written entirely in Python and uses Python for its language. Lightweight: SymPy only depends on mpmath, a pure Python library for arbitrary floating point arithmetic, making it easy to use. A library: Beyond use as an interactive tool, SymPy can be embedded in other applications and extended with custom functions. Definition of codecademy in thedictionary. Meaning of codecademy. What does codecademy mean? Information and translations of codecademy in the most comprehensive dictionary definitions resource on the web.

Yingxie Python notes from codecademy Chap 1 Python Syntax 13‟ Intro: Python can be used to create web apps, games, even a search engine. 1.1 Variables and Data Types. Originally Answered: is there much math in python? This is a common misconception because there are many math people who also program. In reality, programming is the process of telling a computer to do something-- like telling your friend to do something. List of all math symbols and meaning - equality, inequality, parentheses, plus, minus, times, division, power, square root, percent, per mille,. RapidTables Home › Math › Math symbols › Math symbols. SymPy is a Python library for symbolic computation. It provides computer algebra capabilities either as a standalone application, as a library to other applications, or live on the web as SymPy Live or SymPy. Math Symbols Explained with Python August 03, 2019 When working with Machine Learning projects, you will come across a wide variety of equations that you need to implement in code.

Introduction¶ SymPy is a Python library for symbolic mathematics. It aims to become a full-featured computer algebra system CAS while keeping the code as simple as possible in order to be comprehensible and easily extensible.You can find out which functions and attributesare defined in a module. import math dir mathIf you have a Python script named math.py in the samefolder as your current script, the file math.py willbe loaded instead of the built-in Python module.

08.01.2013 · What is this? As many of you have guessed by now, Sandra is an A.I. meant to be a repository for your questions. Right now, her vocabulary is limited. The symbolic math in sympy is pretty good. It is not up to the capability of Maple or Mathematica, but neither is Matlab but it continues to be developed, and could be helpful in some situations. It is not up to the capability of Maple or Mathematica, but neither is Matlab but it continues to be developed, and could be helpful in some situations. Learn codecademy with free interactive flashcards. Choose from 148 different sets of codecademy flashcards on Quizlet. Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step.

Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. mpmath is a free BSD licensed Python library for real and complex floating-point arithmetic with arbitrary precision. It has been developed by Fredrik Johansson. Embedding Python in LaTeX by Kjell Magne Fauske, Friday, October 24, 2008 Comments: 19 Recently, while browsing the archives of the matplotlib mailing list, I stumbled upon the small python.sty package written by Martin R. Ehmsen. Insights When One Teaches, Two Learn: Meet Codecademy’s Community. We believe “when one teaches, two learn.” To make that belief a reality for learners, we host an online community with tens of. Chapter 5 Operators and Expressions The purspose of computing is insight, not numbers. Richard Hamming Chapter Objectives Learn Python’s arithmetic, string, relational, logical, bitwise operators.

Plotly's Python library is free and open source! Get started by downloading the client and reading the primer. You can set up Plotly to work in online or offline mode, or in jupyter notebooks. The Percent Sign % is a interesting beast in the Python language. It does something called string formatting and it’s a mathematic operator as well. Whatever math is. Let’s find out what the POWER of the % is first, we’ll learn by finding out what NOT to do. Unicode characters are very useful for engineers. A couple commonly used symbols in engineers include Omega and Delta. We can print these in python using unicode characters. Kenneth Tilton I am surprised you do not include the numeric character codes. kt --"The best Algebra tutorial program I have seen. in. ALT Codes - Alt Codes for Maths / Mathematics Welcome to Useful Shortcuts, THE Alt Code resource! If you are already familiar with using alt codes, simply select.

Jupyter Notebook. From the official webpage: The Jupyter Notebook is a web application that allows you to create and share documents that contain live. In this post, I am gonna show you how to write Mathematic symbols in markdown. since I am writing blog post that hosted by Github with Editor Atom, and use plugin markdown-preview-plus and mathjax-wrapper, and use mathjax Javascript display the math symbols on the web page. The type of code that best shows off the power of computers involves math. I wonder whether any coder can succeed without mastery of basic math.

1. Learn Python, a powerful language used by sites like YouTube and Dropbox. Learn the fundamentals of programming to build web apps and manipulate data. Master Python loops to deepen your knowledge. Learn the fundamentals of programming to build web apps and manipulate data.
2. Any text element can use math text. You should use raw strings precede the quotes with an 'r', and surround the math text with dollar signs \$, as in TeX. Regular text and mathtext can be interleaved within the same string.
3. What is SymPy? SymPy is a Python library for symbolic mathematics. It aims to be an alternative to systems such as Mathematica or Maple while keeping the code as.
4. I want to make a dictionary in Python, but with math symbols because then i want to plot that with the symbols. So, i have something like this.

The Python Discord. News about the dynamic, interpreted, interactive, object-oriented, extensible programming language Python. If you are about to ask a "how do I do this in python" question, please try r/learnpython, the Python discord, or the python IRC channel on FreeNode. Note that wasysym also deﬁnes a \lightning symbol. The diﬀerence—other than “ ” vs. The diﬀerence—other than “ ” vs. “ ”—is that the stmaryrd version above is limited to math mode. An online LaTeX editor that's easy to use. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more.

Diese Liste mathematischer Symbole zeigt eine Auswahl der gebräuchlichsten Symbole, die in moderner mathematischer Notation innerhalb von Formeln verwendet werden.


# Number of ways to distribute items (with limit of min and max at each location)

There are a total of $n$ objects in stock. You must keep at least $c_i$ objects in stock at location $i$. Assume $n \ge \sum c_i$.

a.) How many different inventories $(x_1,x_2,\dots,x_r)$ are there?

b.) Each location can store a maximum of $d_i$. Let $r = 2$ and count the number of different invetories $(x_1,x_2)$ with $c_1 \le x_1 \le d_1$ and $c_2 \le x_2 \le d_2$.

My answer for part a: We have no choice over where $c_1 + c_2 + \dots + c_r = c_t$ objects are placed. The ramaining $n-c_t$ objects can be spread across $r$ locations. So, there are $$\binom{(n-c_t) + r - 1}{r-1}$$ different inventories.

My incomplete answer for part b: After placing minimum required objects at each location, we are left with $n-c_1-c_2$ objects to distribute.

The remaining objects can be distributed in $$\binom{(n-c_1-c_2)+2-1}{2-1} = (n-c_1-c_2)+1.$$

Now from the number above, I need to subtract the number of combinations that lead to overflows in the two locations.

[---Not sure if the rest is correct---]

Number of ways location 1 overflows:
Location 1 overflows when $x_1 > d_1$ and $c_2 \le x_2 \le d_2$: $$\binom{(n-(d_1+1)-c_2)+2-1}{2-1} = n-d_1-c_2$$

Number of ways location 2 overflows:
(same reasoning as above) $$n-d_2-c_1$$

Total number of inventories is $$(n-c_1-c_2+1)-[(n-d_1-c_2)+(n-d_2-c_1)] = (d_1+d_2+1)-n$$

Total number of inventories $= (d_1+d_2+1)-n$ doesn't make sense.

-

Your answer to part (a) is fine. In (b) you’ve overlooked the possibility of overflow in both inventories simultaneously.

You got off to a fine start: there are $$I_0 = n-c_1-c_2+1$$ inventories that meet the minimum requirements. Then you want get the number of inventories that overflow location $1$ but meet the minimum requirement at location $2$; your answer of $n-d_1-c_2$ is correct only if $n\ge d_1+c_1$. If $n < d_1+c_1$, it’s not possible to overflow at location $1$ and meet the minimum requirement at location $2$. Thus, the correct result for this step is $$I_1 = \max\{n-d_1-c_2,0\}$$ inventories that overflow at location $1$ and meet the minimum at location $2$. Similarly, there are $$I_2=\max\{n-d_2-c_1,0\}$$ inventories that meet the minimum at location $1$ but overflow location $2$.

However, there may be inventories that overflow at both locations. When you subtract off the inventories that overflow at location $1$ and then those that overflow at location $2$, the inventories that overflow at both locations get subtracted twice, so you have to add them back in.

The basic calculation gives $$\binom{n-(d_1+1)-(d_2+1)+2-1}{2-1} = n-d_1-d_2-1$$ of these inventories, but again this is true only if $n\ge d_1+d_2+1$; if $n < d_1+d_2+1$, it isn’t possible to overflow both locations. Thus, the correct figure is $$I_3=\max\{n-d_1-d_2-1,0\}.$$

The final result is then given by $$I_0-I_1-I_2+I_3.$$

Note that if $n \ge d_1+d_2+1$, this evaluates to $0$. This makes perfectly good sense: in that case it’s impossible to meet the requirements, and there are no acceptable inventories.

-

Unless I am reading the problem wrong, this is the same as a couple of other problems that were asked in the last couple of days. You are given a total and want to find the number of ways to make that total $\left(\displaystyle\sum_{i=1}^rx_i=n\right)$ given a range constraint on each summand $(c_i\le x_i\le d_i)$. The answer is the coefficient of $x^n$ in $$\prod_{i=1}^r\frac{x^{d_i+1}-x^{c_i}}{x-1}\tag{1}$$ We can apply this to the special cases you have listed.

a) you let $d_i\to\infty$. In that case, $(1)$ becomes $\displaystyle\prod_{i=1}^r\frac{x^{c_i}}{1-x}=x^{\sum_{i=1}^rc_i}\frac{1}{(1-x)^r}$, so the answer becomes the coefficient of $x^{n-\sum_{i=1}^rc_i}$ in $\frac{1}{(1-x)^r}$. \begin{align} \frac{1}{(1-x)^r} &=\sum_{i=0}^\infty\binom{-r}{i}(-x)^i\\ &=\sum_{i=0}^\infty\binom{r+i-1}{i}x^i\\ &=\sum_{i=0}^\infty\binom{r-1+i}{r-1}x^i \end{align} Therefore, the answer is $\displaystyle\binom{r-1+n-\sum_{i=1}^rc_i}{r-1}$.
b) you set $r=2$. In this case, $(1)$ becomes $$\left(x^{c_1}+x^{c_1+1}+\dots+x^{d_1}\right)\left(x^{c_2}+x^{c_2+1}+\dots+x^{d_2}\right)\tag{2}$$ and the answer is the coefficient of $x^n$ in $(2)$.

Suppose $d_1-c_1\le d_2-c_2$, then the answer is $$\text{the coefficient of }x^n\text{ in }(2)=\left\{\begin{array}{}n-c_1-c_2+1&\text{for }c_1+c_2\le n\le d_1+c_2\\d_1-c_1+1&\text{for }d_1+c_2+1\le n\le c_1+d_2-1\\d_1+d_2-n+1&\text{for }c_1+d_2\le n\le d_1+d_2\\0&\text{otherwise}\end{array}\right.$$

-
Please show the closed form, I can't get the answer. –  jamesio Oct 3 '11 at 17:04
@jamesio: I have included the gory details. –  robjohn Oct 3 '11 at 17:53


# Integral substitution

## Main Question or Discussion Point

How do I evaluate the integral $$\int_a^b$$ f(x4) dx = $$\int_y^z$$ f(u) dx/du du, where u=x4
y=a4
z=b4

## Answers and Replies

HallsofIvy
Homework Helper
How do I evaluate the integral $$\int_a^b$$ f(x4) dx = $$\int_y^z$$ f(u) dx/du du, where u=x4
y=a4
z=b4
How you integrate the function f depends strongly on what f is, don't you think?

f is $$\int_-1^1$$ 1+x4 dx

f is $$\int_1^1$$ 1+x4 dx
meant to be a -1 at the bottom of the integral.
How could I do this then.

jgens
Gold Member
Well, if your integral is simply

$$\int_{-1}^1 1 + x^4\, dx$$

Then it's pretty simple. Just seperate the integrand so that,

$$\int_{-1}^1 \,dx + \int_{-1}^1 x^4 \, dx$$

But, based on the context of the question I'm not sure that's what you're looking for. Perhaps you could repost exactly what you mean clearly?

Sure. Consider the integral $$\int_{-1}^1$$ 1+x4 dx
How do I evaluate this integral using the substitution u=x4 and the formula:
$$\int_a^b$$ f(x4) dx = $$\int_{a^4}^{b^4}$$ f(u) dx/du du, where u=x4

jgens
Gold Member
Why would you use that method to evaluate the integral? The integral you have posted can easily be evaluated using the power rule for integration.

I thought this was a more interesting way but just don't know how to get started.

Also I someone could help me with this problem.
How could I evaluate the integral above using u=x4, but seperating it into 2 integrals.

Cyosis
Homework Helper
Taking detours isn't a more interesting way, but here goes.
$u=x^4 \Rightarrow du=4x^3dx$

$$\int 1+x^4 dx=\int \frac{1+u}{4(u^{3/4})} du$$

Making life a lot harder!

Last edited:
Gib Z
Homework Helper
Personally I wouldn't discourage him, I find it a good way to learn how to compute integrals and what methods to use can be supplemented by trying other methods that come to mind, and seeing why they may not be as efficient. Once he computes that integral Cyosis got him, I think he will be able to see in future why it may not be wise to compute an integral as such.

Cyosis
Homework Helper
You raise a very good point. I guess it's all too easy to see if something is going to be inefficient or not when you have some experience with the topic at hand.

Note: I didn't take the limits into account so you still have to do that yourself.

Why would you use that method to evaluate the integral? The integral you have posted can easily be evaluated using the power rule for integration.
You are too harsh. Sure, for this function there are easy ways to do the integral. But for other functions f maybe you really should do the substitution. And why not learn to do the substitution on simple problems that can be evaluated in other ways? Then, for example, you can check your answers and see when you make a mistake.


# Synthetic Geometry Session Followup Questions and Problem Summary

In our first session, we spent some time on this problem: Let $ABC$ be a triangle in which $\angle A = 60^{\circ}$ Let $BE$ and $CF$ be the angle bisectors with $E$ on $AC$ and $F$ on $AB$. Let $M$ be the reflection of $A$ in the line $EF$. Prove that $M$ lies on $BC$

1. $AEIF$ is cyclic because $\angle BIC=90+\frac {\angle A}{2}=120^{\circ}\implies \angle FIE+\angle A=180^{\circ}$. The cyclic result only occurs when $\angle A=60^{\circ}$, and it gives us information about the angles of $\triangle AEF$ and consequently those of $\triangle FME$

2. $I$ is the circumcenter of $MEF$ because $IE=IF$ and $\angle FIE=2\angle FME$. This result relates $M$ with $I$ and gives us more angle relations.

3. $BMIF$ is cyclic, which we proved by using the above properties to angle chase $\angle FBI=\angle FMI$. Symmetrically $CMIE$ is cyclic, so $\angle IMB+\angle IMC=\angle AFI+\angle AEI=180^{\circ}$, which implies $B,M,C$ are collinear and we are done.

The discovery of these properties makes the problem transparent, meaning we know what makes our main result true. What this also means is that we can probably find simpler solutions. Indeed, there are many ways to solve this problem:

Using angle bisector theorem and Miquel point (Avoids tedious angle chasing):

Since $AEIF$ is a cyclic quadrilateral, we know the miqual point of its complete correspondence lies on $BC$. This means the circumcircles of $BFI,CEI, ABE,ACF$ all have a common point on $BC$, we denote the point $M'$. Because $ACM'F$ is cyclic, $\frac {FM'}{BF}=\frac {AC}{BC}=\frac {AF}{BF}$; therefore $AF=FM'$ and similarly $AE=EM'$. This is enough to establish $M=M'$ and $M$ lies on $BC$

Utilizing $EF$ as the axis of symmetry:

Let $EF\cap BC=P$. It suffices to prove $PE$ bisects $\angle APB$. We can achieve this with our first property.

Now I will propose a few follow up questions which you guys should now easily answer:

We keep the notations of our main problem:

1. $DBC$ is equilateral such that $D,A$ are on opposite sides of $BC$. Prove that $DE=DF$

2. Through $I$ construct the perpendicular to $EF$ which intersects $EF,BC$ at $X,Y$. Prove that $2IX=IY$

3. $O$ is the circumcenter of $AEF$, Prove that $OM\perp BC$.

Note by Xuming Liang
5 years, 5 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

For follow up problem 1 :

To prove $DE=DF$ , it suffices to establish congruence between $\Delta DIF$ and $\Delta DIE$ (not dead yet) . We have $DI=DI$ and we have already established that $IF=IE$ by having $\angle IFE= \angle IEF=30$. So we need to prove a pair of angles congruent. Since we want a good connection between the angles and the discovered stuff , we prefer proving $\angle DIF=\angle DIE$. By cyclicity we have $m\angle BIF=m\angle CIE=60$. So we need to prove $\angle DIB=\angle DIC=\dfrac{\angle BIC}{2}=\dfrac{120}{2}=60$. Now how to achieve this? We have not yet utilized the equilateral triangle yet!

So, consider $\Box BICD$ Can we prove it cyclic? Yes! We have to prove that $\angle EBD+\angle ECD=180$ and its easy.

$\angle IBD+\angle ICD =\angle IBC+\angle CBD+\angle ICD+\angle BCD=\dfrac{B}{2}+60+\dfrac{C}{2} + 60=120+60=180 \\ \Rightarrow \Box BICD \ is \ cyclic!$

So by cyclicity we have $\angle DIB=\angle DIC=60$ which we had to prove.

- 5 years, 5 months ago

Very nice reasoning using congruence! For the cyclic part, I believe you meant to say $B,I,C,D$ are concyclic, which is true because $\angle BIC+\angle BDC=180$. Since $BD=DC$, $ID$ bisects $\angle BIC$, which is what you wanted to prove.

- 5 years, 5 months ago

Sorry , yes it is that $B,I,C,D$ are concyclic. I have edited it and few angles too.

- 5 years, 5 months ago

A fast way to prove that $BICD$ is cyclic is to notice that $\angle BIC = 120^{\circ}$ and $\angle BDC = 60^{\circ}$.

- 5 years, 5 months ago

First we have $\Delta IXF$ as 30-60-90 triangle. So $FI=2IX$. So we have to prove that $FI=IY$. So consider $\Delta BIF$ and $\Delta BIY$ , where we have $\angle FBI=YBI$ , $BI=BI$ , $\angle BIF=\angle BIY=60$ (by angle chase). So they are congruent by ASA test and hence $FI=IY$ is proved.

- 5 years, 5 months ago

I pretty much proved it the same as you. Nice solution.

- 5 years, 5 months ago

Yesterday we did find that $\angle EMC=60$ So to prove our result , we must prove that $\angle OME=30$. Using the fact that $O$ is circumcentre , we have $\angle EOF=\angle EIF=120$ and by some angle chase we have $\angle OEI=\angle OFI=60$ where $\angle OFE=\angle IFE=\angle OEF=\angle IEF=30$. We get a beautioful result that $\Box FOEI$ is a rhombus which has one of its diagonals that is $OI$ equal to its side. Using these facts and that $I$ is circumcenter of $\Delta EMF$ we have $EI=OI=IF=IM$ and thus $\Box EOFM$ is cyclic giving $\angle OFE= \angle OME=30$ which was to be proved :)

- 5 years, 5 months ago

When will be the next discussion?

- 5 years, 5 months ago

Problem 3: Since $M$ is the Miquel point, there exists a spiral similarity centered at $M$ that maps $FI \rightarrow AE$. Denote $M_1, M_2$ as the midpoint of $AF$ and $EI$, respectively. Thus, the spiral similarity maps $FI \rightarrow M_1M_2$. Using the circumcircles of $\triangle BFI$ and $BM_1M_2$, this implies that $M_1M_2BM$ is cyclic. Since $M_1OM_2B$ is cyclic, this implies that $M_1OM_2MB$ is cyclic with diameter $OB$. Thus, $\angle OMB = 90^{\circ}$ and we are done.

- 5 years, 5 months ago

Great solution. This fact is actually true as long as $BICD$ is cyclic.

- 5 years, 5 months ago


A bacteria colony is made of billions of organisms. “If you can't explain it simply, you don't understand it well enough.” —Einstein (, Understanding Discrete vs. Asking for help, clarification, or responding to other answers. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. But if you're asking if it's spatially continuous then the answer is no. Hence, the temperature variable placed against the time variable would be represented by a continuous curve. It … Now that we know the continuous rate is -47.9% per year, we can work out how long until we're at 50%: This is a tricky one: the stock market changes every day, so it seems like it'd continuous, but there isn't an underlying predictable rate. Random variables represent quantities or qualities that randomly change within a population. This can be done regardless of whether the sample size is specified by a discrete unit or a continuous measure. Measurement Scale and Context For example, 8 bits have 256 possible values, and 16 bits have 65536. The natural log works on the ratio between the new and old value: $\frac{\text{new}}{\text{old}}$. Learn more about how features and surfaces can be represented as either discrete or continuous in ArcGIS. It is continuous if you consider the values. What to do to speed up the paper publication process? The measurements themselves are discrete, but how you represent them need not be. What's that Professor? For example, you could represent them as a continuous density surface: Or, as discrete 3D bar charts extruded from the census tracts (in this case a grid): A population count is a point measure. We can't have half a student! Why is population density a continuous data type when it is typically measured for aggregate areas such as census tracts or districts/neighbourhoods (ie, it can't be measured at any point on a surface like gradient or temperature). The population of a city is always a continuous variable as people are always moving to and from the city as well as births & deaths. To simulate a weather system, for example, the tracking occurs continuously as all elements are constantly changing. How to say "garlic", "garlic clove" and "garlic bulb" in Japanese? "12% interest per day" is different than "12% interest per year".). BetterExplained helps 450k monthly readers with friendly, intuitive math lessons (more). (Notice how the rate must be scaled to match the time period. Something areal data is not. If the population is discrete, this distribution is a stick (or bar) graph. ), How To Think With Exponents And Logarithms, Q: Why is e special? It is areal data as @Radar have said. Continuous Growth. Updated my answer. Sample Data Representing a Population Distributions. In Star Trek TNG Episode 11 "The Big Goodbye", why would the people inside of the holodeck "vanish" if the program aborts? The function itself need not be continuous. Ah, but the caveat is the application of a kernel function on the point data. A surface for which each location has a specified or derivable value. For a census tract you can ask the question: How many people exist per unit area? Why was the name of Discovery's most recent episode "Unification III"? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The ratio between new and old was 37/53, so ln(37/53) = -.359 = -35.9% continuous growth over our time period. What's the best way to produce a density map from weighted points in QGIS? For a discrete distribution, probabilities can be assigned to the values in the distribution - for example, "the probability that the web page will have 12 clicks in an hour is 0.15." Argh! Enjoy the article? For a spatial application the consideration of interpolation is important. (Continuous growth requires a smaller rate because of compounding.). Discrete and Continuous Data. We can take them, split them into smaller, more frequent changes, and spread them out. In our case, we grew from 1 to 2, which means our continuous growth rate was ln(2/1) = .693 = 69.3%. $2^n$ (where n is an integer) models discrete scenarios like coin flips or binary digits. Are you telling me the bacteria colony just happens to have a continuous rate of precisely ln(2) over the course of a day? elevation). There are two definitions of continuous data ( at least that I could find online). find the population density of some points in qgis/postgis, Calculating population density from urban raster reclasiffy, How to create a population density heatmap from attributes, Find average density value within a polygon in ArcMap, Sources for population density map of Europe. For example, the half-life of Carbon-14 is 5700 years. These practice problems focus on distinguishing discrete versus continuous random variables. A clear understanding of the difference between discrete and continuous data is critical to the success of any Six Sigma practitioner. +1 This answer addresses both spatial and non-spatial forms of data continuity, which is important considering the forum in which the question was asked. To learn more, see our tips on writing great answers. What are the possible outcomes? (The model gets more complex as you account for how long it takes for cubs to have children of their own.). We see a lot of jumpy changes, and sample them at yearly intervals to see how we're doing. I'd prefer you told me the colony doubled while a grad student stared at a petri dish for 24 hours straight. None of this "wait until we decay by 50% so humans can count it easier" nonsense. Let’s see the definition: Continuous data is information that could be meaningfully divided into finer levels. A continuous distribution is one in which data can take on any value within a specified range (which may be infinite). What does “blaring YMCA — the song” mean? Population ecology is the study of population and how they change over time in relation to the environment including the environmental influences on population density and distribution, age structure and variations in population size. A discrete variable is a number that can be counted. Breeding seasons introduce some delay in the regulative process. The natural log finds the continuous rate behind a result. One of my pet peeves were problems like "A bacteria colony doubles after 24 hours...". Discrete and Continuous Data. Density is an areal measure and in itself implies that there exists a container variable (e.g. All data that are the result of counting are called quantitative discrete data. I visualize change as events along a timeline: Discrete changes happen as distinct green blobs. With discrete growth, we can see change happening after a specific event. Occasionally when analyzing data between groups or variables there are limitations on the validity of measured tests due to the lack of evidence, data, or information because a sample size is too small or has low value. Example: Material X decayed from 53kg to 37kg over 9 months. We can't point to an event and say "It changed here". The radioactive material is changing every instant. Automating calculation of Population Density in Census Tracts? jar that holds water, or in this case a census tract that holds people). Discrete Data. (1.98kg... 1.99kg... 2.00kg. If your system does change continuously, why not provide the continuous rate and write $e^{\ln(2) x}$? Should live sessions be recorded for students when teaching a math course online? Show Answer. This happened over 9 months, so the monthly continuous rate is -35.9/9 = -3.98%. Using $e$ as a base ($e^{\text{rate} \cdot \text{time}}$) implies you want people to think about change that happens at every moment. (Brush up on the number e and the natural logarithm.). I wrote this post because my video on e had questions about how $2^x$ represents "staircase growth". This is similar to saying the average family has 2.3 kids.). That is, the function's domain is an uncountable set.

## is population discrete or continuous

Backyard System Of Poultry Rearing, Anti American Synonym, Sm Megamall Cyberzone Store Directory, Dollar Tree Pharmacy, Gilchrist County Map, Haier Window Air Conditioner, Roasted Eggplant Baby Led Weaning, How To Make Mcdonald's Habanero Sauce,


The chromatic number of the union of two graphs

Let $$G_n$$ be the graph on the set of all binary strings of length $$n$$ with two strings adjacent whenever they are Hamming distance $$2$$ away from each other, or one of them lies below another one; thus, for instance, $$G_2=K_4$$, and $$G_3$$ has $$25$$ edges. What is the chromatic number of $$G_n$$?

There is a simple, but not quite obvious construction showing that $$\chi(G_n)\le n(n+1)/2+1$$, and I am interested in a matching lower bound. Computations give $$\chi(G_1)=2,\ \chi(G_2)=4,\ \chi(G_3)=5,\ \chi(G_4)=9,$$ $$\chi(G_5)=12,\ \chi(G_6)=16,\ \text{and}\ \chi(G_7)\le 17.$$ (The first five values are easy to compute, the last two are reported by Gordon Royle in the comments.)

An update. The colorings I am interesting in are of a special nature: if the strings $$s$$ and $$t$$ are same-colored, then (identifying strings with sets) also $$s\setminus t$$ and $$t\setminus s$$ are same-colored. Let $$\chi^*(G)$$ denote the smallest number of colors needed to properly color the graph $$G$$ in this special way. Is it true that $$\chi^*(G_n)=(1/2+o(1))n^2$$? That, say, $$\chi^*(G_n)>(1/4+o(1))n^2$$?

• @JoshuaZ: sure; color each vertex with the scalar product of the corresponding string and the vector $(1,2,\dotsc,n)$.
– Seva
Nov 20 at 16:36
• what do you mean by lying below? Nov 21 at 8:48
• @FedorPetrov: one of the corresponding subsets of the $n$-element set being contained in another one.
– Seva
Nov 21 at 9:03
• Here is a nice $17$-colouring of $G_7$ where I am using subsets of $\{0,1,\ldots,6\}$ to represent the vertices. Let $\sigma = (0,1,2,3,4,5,6)$ be a cyclic permutation and let $\tau$ be the complement map on the subsets. Let $F= \{013, 124, 235, 346, 450, 561, 602\}$ be the Fano plane and set $C = F \cup F^\tau$. Then take $D = \{0, 16, 25, 34, 124, 135, 236, 456\}$ and its seven rotations under powers of $\sigma$, and $D^\tau$ and its seven rotations under powers of $\sigma$. Then one colour class for each of $\emptyset$ and the entire set, making a total of 17 classes. Nov 22 at 7:04
• So $G_6$ has chromatic number $16$ - this took quite a long time to compute, almost all of which was spent searching in vain for a $15$-colouring. I think it will be hard to get close to an exact answer. Nov 23 at 2:29

This is a comment, not an answer, but it will be more convenient to post it as an answer. Consider the vertices of $$G_n$$ as subsets of $$[n]=\{1,\dots,n\}$$.

Observation 1. $$\chi(G_n)\ge\left\lfloor\frac{3n}2\right\rfloor+1$$.

This is because $$G_n$$ contains a clique of that size, namely,
$$\varnothing,\{1\},\{2\},\{1,2\},\{1,2,3\},\{1,2,4\},\{1,2,3,4,\},\{1,2,3,4,5\},\dots$$.

Observation 2. The OP noted that $$\chi(G_n)\le\binom{n+1}2+1=\frac{n^2+n+2}2$$. For odd $$n$$ this bound can be improved to $$\chi(G_n)\le\frac{n^2+3}2$$.

The vertices $$\varnothing$$ and $$[n]$$ get their own colors. If $$1\le|X|\le n-1$$, color $$X$$ with the ordered pair $$\left(\sum_{x\in X}x\mod(n+1),\ \left\lceil\frac{|X|}2\right\rceil\right).$$ This is a proper coloring, so (assuming $$n$$ is odd) $$\chi(G_n)\le2+(n+1)\cdot\frac{n-1}2=\frac{n^2+3}2.$$

Observation 3. The clique number $$\omega(G_n)$$ is exactly $$\left\lfloor\frac{3n}2\right\rfloor+1$$.

We showed in Observation 1 that $$\omega(G_n)\ge\left\lfloor\frac{3n}2\right\rfloor+1$$. Let us prove by induction that $$\omega(G_n)\le\left\lfloor\frac{3n}2\right\rfloor+1$$.

We may assume $$n\ge2$$. Let $$\mathcal C$$ be a maximal clique in $$G_n$$. Let $$\mathcal C_h=\{X\in\mathcal C:|X|=h\}$$. Choose $$h$$ so that $$0\lt h\lt n$$ and $$\mathcal C_h\ne\varnothing$$.

If $$\bigcup\mathcal C_h=[n]$$ and $$\bigcap\mathcal C_h=\varnothing$$, then $$|\mathcal C|=|\mathcal C_h|+2\le n+2\le\left\lfloor\frac{3n}2\right\rfloor+1$$. Otherwise, choose $$X\in\{\bigcup\mathcal C_h,\ \bigcap\mathcal C_h\}$$ so that $$0\lt|X|\lt n$$, and let $$k=|X|$$. Then every element of $$\mathcal C$$ is a subset or superset of $$X$$. Since $$0\lt k\lt n$$, by induction we have $$|\mathcal C|\le\omega(G_k)+\omega(G_{n-k})-1\le\left\lfloor\frac{3k}2\right\rfloor+\left\lfloor\frac{3(n-k)}2\right\rfloor+1\le\left\lfloor\frac{3n}2\right\rfloor+1.$$

• Some more computational snippets: the cliques you found are the unique maximum clique (up to equivalence under the automorphism group) for $G_4$, $G_6$ and $G_8$. For $G_5$, $G_7$ and $G_9$ there are other classes of equally-large cliques, but none larger. Nov 25 at 2:56
• So I liked your original proof that if $|\mathcal{C}_h| = 1$ for some $h$, then the result follows. Did you decide that it was not so obvious to prove that there must be a layer containing just one vertex of the clique? Nov 26 at 5:11


# How to solve for x/z and y/z here?

I got stuck solving these two equations: $$a_1(x/z) + b_1 (y/z) + c_1 = 0$$ and, $$a_2(x/z) + b_2 (y/z) + c_2 = 0$$ for $$x/z$$ and $$y/z$$.

The desired result would be: $${x \over z} = {b_1c_2 - b_2c_1\over a_1b_2 - a_2b_1}$$,

$${y \over z} = {c_1a_2 - c_2a_1\over a_1b_2 - a_2b_1}$$

How do I get there? I keep getting dead ends.

Edit: I basically came this far: $$u = v{-b_1 \over a_1} - {c_1 \over a_1}$$ and, $$u = v{-b_2 \over a_2} - {c_2 \over a_2}$$

And ofcourse $$u = x/z$$, $$v = y/z$$

Now i got here, $$v{-b_1 \over a_1} - {c_1 \over a_1} = v{-b_2 \over a_2} - {c_2 \over a_2}$$

Then I took all the v's to one side and factored v outside: $$v({-b_2 \over a_2} + {b_1 \over a_1}) = {c_1 \over a_1} - {c_2 \over a_2}$$

now divide both sides by that b and a thingy on the left, but now it gets messy...

• Can you show which dead end you run into? It should be simple enough to substitute your proposed solutions into the equations and simplify to show that they work. Jan 30 '16 at 16:08
• Rename $\frac xz=u$ and $\frac yz=v$ and rewrite the initial equations. Jan 30 '16 at 16:08
• Hint: take the last thing, check that it is correct (the second term on the left should be $\frac{b_2}{a_2}$, and the plus on the right should be a minus), then multiply by $a_1a_2$ on both sides then solve for $v$. You just obtained $\frac yz$ as in the solution you give :). Edit You just fixed the sign while I typed. Jan 30 '16 at 16:44
• I suppose I am stuck on how to divide c1/a1 - c2/a2 by that thing on the left now. Any clever way to do that? Jan 30 '16 at 16:56
• See my answer below. I started from a little earlier than that equation, but to make the division, I suggest you multiply both sides by $a_1a_2$, and then divide. The multiplication eliminates all denominators. This is why it is advisable. Jan 30 '16 at 17:04

## 2 Answers

HINT: with $$\frac xz=u$$ and $$\frac yz=v$$ we get $$a_1u+b_1v+c_1=0$$ $$a_2u+b_2v+c_2=0$$ if $$b_1\ne 0$$ we have $$v=-\frac{c_1}{b_1}-\frac{a_1}{b_1}u$$ and we get an equation for $$u$$ $$a_2u+b_2\left(-\frac{c_1}{b_1}-\frac{a_1}{b_1}u\right)+c_2=0$$ can you proceed?

• it seems we get a different thing, did I do something wrong in my original post? I edited it in. Jan 30 '16 at 16:42
• I urge you to decide if you mean to use slash fractions $a/b$ or normal fractions $\frac ab$, because mixing the two can be quite confusing for readers. @TheProgramMAN123 I think you just followed a different path than he did. You solved both for $u$ and then equalled the two expressions, he solved one for $v$ and then substituted the expression in the other one. Jan 30 '16 at 16:48
• If you solve the last equation in this post for $u$, you get the solution in the question, with signs changed both in numerator and in the denominator. Jan 30 '16 at 16:51

Let me just continue your attempt. You got to:

$$\left\{\begin{array}{c} u=-\frac{b_1}{a_1}v-\frac{c_1}{a_1} \\ u=-\frac{b_2}{a_2}v-\frac{c_2}{a_2} \\ \end{array}\right.$$

If we equal these two expressions for $u$ and multiply both sides by $a_1a_2$, we get:

$$-b_1a_2v-c_1a_2=-b_2a_1v-c_2a_1.$$

Bringing the RHS's $v$ term to the left and the LHS's "constant" term to the right, this becomes:

$$v(a_1b_2-a_2b_1)v=a_2c_1-a_1c_1.$$

Yes, I also factored out $v$ from the LHS and reordered the factors in the various $a,b,c$ products. Is this not precisely the solution you gave, that is:

$$\frac yz=v=\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}?$$

Wonderful! To complete, we substitute this into the equation for $u$, one of the two, I mean. To simplify the calculations, we multiply by $a_1$ the first equation of the system at the start of the answer, and then substitute, getting:

$$a_1u=-b_1\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}-c_1.$$

Let us make that RHS into a single fraction:

$$a_1u=\frac{\overline{-b_1a_2c_1}+b_1a_1c_2-c_1a_1b_2+\overline{c_1a_2b_1}}{a_1b_2-b_1a_2}=\frac{a_1(b_1c_2-c_1b_2)}{a_1b_2-b_1a_2}.$$

The overlines simply indicate those two terms cancel. Oh, but if we divide both sides by $a_1$, we get the desired result, don't we? Very good. We are done.

Bottom line: when you have coefficients with fractions, the trick is often to sum all fractions and see if you can multiply/divide to simplify numerators or denominators. Don't be afraid of complicated coefficients: just have patience and do the algebra to the end, and you will get to the desired result :).

• thanks a bunch! I guess the lesson for me here is to remove those fractions as soon as you can. They make things confusing and messy. i guess multiplying both sides by a1a2 is a clever way to do that. the solution really is painfully simple then. Jan 30 '16 at 18:29