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Math Help - liouville's theorem?

1. liouville's theorem?

Suppose f is an entire function that satisfies $|f(z)|\leq M(1+|z|^m)$ for some constant M and positive integer m. Show that f is a polynomial of at most degree m.

How should I approach this? Do I try to bound $|\frac{f(z)} {1+z^m}|$, and then apply Liouville's theorem? The expression isn't necessarily entire though.

2. Use Cauchy estimates to show that

$\displaystyle \bigg|\frac{\partial^{m+k} }{\partial z^{m+k}}f(z)\bigg|_{z=0}=0$

This will show that the power series of f about zero has only finitely many terms and is therefore a polynomial.

3. Does this work?

For $0\leq |z|\leq R$, $f(z)\leq M(1+ R^m)$.

Then I use the Cauchy estimate:

For $k>0$, $|f^{(m+k)}(0)|\leq \frac{(m+k)!M(1+R^m)}{R^{m+k}}$. Then I use the trick from Liouville's theorem and take R to infinity, so $f^{(m+k)}(0)=0$. From the power series at 0, f has at most degree m.

4. Yep that is the idea!


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# Tikzmark between syntactic tree and matrix?

I am trying to draw a line between the S node and the matrix, but have not been able to get it to work. What am I doing wrong?

MWE:

\documentclass[a4paper,12pt]{article}

\usepackage{fontspec,tikz, tikz-qtree, multicol,avm,array}

\usetikzlibrary{matrix, positioning, trees, calc, arrows, fit,tikzmark,positioning}

\begin{document}

\begin{multicols}{2}

\begin{tikzpicture}
\begin{scope}
\Tree [.\tikzmark{S}{S} [.NP [.Det the ]
[.N cat ] ]
[.VP [.V sat ]
[.PP [.P on ]
[.NP [.Det the ]
[.N; mat ] ] ] ] ]
\end{scope}
\end{tikzpicture}

\columnbreak

\begin{avm}
\scriptsize
${} \tikzmark{PRED}{PRED} avoid <SUBJ, OBJ>'; TNS \neq PAST \cr TOPIC \[ PRED kind<COMP>' \cr DEF +; LOC FAR; NUM SG \cr COMP \[ PRED of <OBJ>' \cr OBJ \[ PRED cake'$ \] \]\tikzmark{topic} \cr
SUBJ $PRED pro'; NUM SG; PERS 1; CASE NOM$ \cr
OBJ $\qquad$\tikzmark{object} \cr
ADJ $PRED usually'$ \]
\end{avm}

\begin{tikzpicture}[remember picture,overlay]
\draw[->] (pic cs:S)--(pic cs:PRED);
\draw[-] (pic cs:topic) to[out=0,in=0,looseness=2]  (pic cs:object);
\end{tikzpicture}
\end{multicols}

\end{document}

• Please don't ask the same question twice. Instead, edit your original question to include the additional information. I've marked your other question as a potential duplicate but this is messy: this one should be the duplicate as it is newer, but it is also a much better question and now has an answer. So... Not sure what to do with it. Perhaps you could delete your other question? [Note that editing a question moves it back to the top of the front page of the site i.e. top of the queue of active questions. So you don't need to ask a new one to ensure it gets attention.] – cfr Dec 22 '14 at 17:35

This is due to the way that tikz-qtree works. The long explanation can be found at my answer to \draw alignment when connecting nodes using tikz-qtree, getting line centered

Cutting-and-pasting my code from that answer is half of the solution here. The other half is to remove the \tikzmark from the tree. Unfortunately, \tikzmark suffers from the same problem as the tikz-qtree (as explained in that answer) so it would need a similar adjustment. Fortunately, this isn't necessary. Since a tikz-qtree is inside a tikzpicture environment, we can use the ordinary tikz node labelling system to refer to nodes in a tikzpicture. (tikzmark was developed for outside a tikzpicture and actually shouldn't be used inside as it then suffers from the nested tikzpicture syndrom.)

\documentclass[a4paper,12pt]{article}
%\url{https://tex.stackexchange.com/q/219072/86}

\usepackage{fontspec,tikz, tikz-qtree, multicol,avm,array}

\usetikzlibrary{matrix, positioning, trees, calc, arrows, fit,tikzmark,positioning}
\makeatletter

\def\unwind@subpic#1{%
% is #1 the current picture?
\edef\subpicid{#1}%
\ifx\subpicid\pgfpictureid
% yes, we're done
\else
% does #1 have a parent picture?
\expandafter\ifx\csname pgf@sh@pi@#1\endcsname\relax
% no, the original node was not inside the current picture
\pgf@xa=\pgf@x
\pgf@ya=\pgf@y
\pgfsys@getposition{\pgfpictureid}\pgf@shape@current@pos
\pgf@process{\pgfpointorigin\pgf@shape@current@pos}%
\pgf@process{\pgfpointorigin\subpic@parent@pos}%
\pgf@x=\pgf@xa
\pgf@y=\pgf@ya
\else
% yes, apply transform, save picture location, and move up to parent picture
\pgfsys@getposition{\csname pgf@sh@pi@#1\endcsname}\subpic@parent@pos%
{%
\pgfsettransform{\csname pgf@sh@nt@#1\endcsname}%
\pgf@pos@transform{\pgf@x}{\pgf@y}%
\global\pgf@x=\pgf@x
\global\pgf@y=\pgf@y
}%
\unwind@subpic{\csname pgf@sh@pi@#1\endcsname}%
\fi
\fi
}

\def\pgf@shape@interpictureshift#1{%
\def\subpic@parent@pos{\pgfpointorigin}%
\unwind@subpic{\csname pgf@sh@pi@#1\endcsname}%
}

\makeatother
\begin{document}

\begin{multicols}{2}

\begin{tikzpicture}[remember picture]
\begin{scope}
\Tree [. \node (S) {S}; [.NP [.Det the ]
[.N cat ] ]
[.VP [.V sat ]
[.PP [.P on ]
[.NP [.Det the ]
[.N; mat ] ] ] ] ]
\end{scope}
\end{tikzpicture}

\columnbreak

\begin{avm}
\scriptsize
${} \tikzmark{PRED}{PRED} avoid <SUBJ, OBJ>'; TNS \neq PAST \cr TOPIC \[ PRED kind<COMP>' \cr DEF +; LOC FAR; NUM SG \cr COMP \[ PRED of <OBJ>' \cr OBJ \[ PRED cake'$ \] \]\tikzmark{topic} \cr
SUBJ $PRED pro'; NUM SG; PERS 1; CASE NOM$ \cr
OBJ $\qquad$\tikzmark{object} \cr
ADJ $PRED usually'$ \]
\end{avm}

\begin{tikzpicture}[remember picture,overlay]
\draw[->] (S)--(pic cs:PRED);
\draw[-] (pic cs:topic) to[out=0,in=0,looseness=2]  (pic cs:object);
\end{tikzpicture}
\end{multicols}

\end{document}

It works after you switch to pdfLaTeX. For this, it's required to remove fontspec from the loaded packages.

With XeLaTeX I noticed the issue persisting.

• Is there any way to solve this problem without losing fontspec, which I need for polyglossia? – Namenlos Dec 22 '14 at 13:12
• @PRQ Would babel instead of polyglossia be an option? – Stefan Kottwitz Dec 22 '14 at 14:35
• @PRQ Have you tried LuaLaTeX? (Or babel as Stefan Kottwitz suggested.) – cfr Dec 22 '14 at 17:37
• I need to be able to input Greek characters, and babel requires too much work given the number of sentence I need to input. I have never tried LuaLaTeX, but will look into it. – Namenlos Dec 22 '14 at 17:41

format: plain


text:
# A covering space of a Hausdorff space is Hausdorff

Let $p:Y\to X$ be a covering space. If $X$ is Hausdorff, so is $Y$.

Hello,

I have a question to this task. I want to show that $Y$ is a Hausdorff space. Hence for $y_1, y_2\in Y$ with $y_1\neq y_2$ have disjoint neighbourhoods.

I want to prove this:

Let $p:Y\to X$ be a covering space. $y_1, y_2\in Y$, with $y_1\neq y_2$. $x_1:=p(y_1)$ and $x_2:=p(y_2)$. Then exist open sets $x_1\in U_1\subseteq X$ and $x_2\in U_2\subseteq X$, such that $p^{-1}(U_i)$ with $i=1,2$ is the disjoint union of open sets.

Hence there are open sets $y_1\in V_1\subseteq p^{-1}(U_1)$ and $y_2\in V_2\subseteq p^{-1}(U_2)$.

Claim: $V_1\cap V_2=\emptyset$.

I do not think, that this works... Is this approach any good? Do you have tips on how I can do better? Thanks in advance.

• Hint: You never used the hypothesis that $X$ is Hausdorff. – Lee Mosher Jun 13 '16 at 22:11

You need to break your proof up into two parts.

Suppose $x_1 \neq x_2$. We can choose $U_1$ and $U_2$ to be disjoint, and so $p^{-1}(U_1)$ and $p^{-1}(U_2)$ will be disjoint (if they had a point $z$ in common, $p(z) \in U_1\cap U_2 = \emptyset$ which is impossible). Now you can take $V_1 = p^{-1}(U)$ and $V_2 = p^{-1}(U_2)$.

Suppose $x_1 = x_2$; call this point $x$. Let $U$ be an evenly covered open neighbourhood of $x \in X$, then $p^{-1}(U) \cong \bigsqcup_{\alpha} U_{\alpha}$ where $p|_{U_{\alpha}} : U_{\alpha} \to U$ is a homeomorphism. Suppose $y_1 \in U_{\beta_1}$ and $y_2 \in U_{\beta_2}$, then it follows that $\beta_1 \neq \beta_2$. To see this, suppose $\beta_1 = \beta_2$; call this value $\beta$. As $p|_{U_{\beta}}$ is a homeomorphism and $p|_{U_{\beta}}(y_1) = p|_{U_{\beta}}(y_2)$, we have $y_1 = y_2$; this is a contradiction as $y_1$ and $y_2$ were assumed to be distinct. As $\beta_1 \neq \beta_2$, we can take $V_1 = U_{\beta_1}$ and $V_2 = U_{\beta_2}$.

• @Error404: Right, I overlooked that. I will edit accordingly. – Michael Albanese Sep 10 '18 at 13:36

Hint:

First case $x_1\neq x_2$, since $X$ is separated, there exists $U_1,U_2$ open neighborhoods of respectively $x_1$ and $x_2$ such that $U_1\cap U_2$ is empty. $p^{-1}(U_1)\cap p^{-1}(U_2)$ is empty and $y_1\in p^{-1}(U_1), y_2\in p^{-1}(U_2)$.

Second case $x_1=x_2=x$, since $p$ is a covering, there exists a neighborhoo $U$ of $x$ such that for every $y\in p^{-1}(x)$, there exists a neighborhood $V_y$ such that the restriction of $p$ to $V_y$ is an homeomorphism $p_{\mid V_y}:V_y\rightarrow U$, and $V_{y_1}\cap V_{y_2}$ is empty by definition.


text:
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 July 7th, 2013, 12:53 PM #1 Senior Member   Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0 Check my work please. Thank you Solve each equation for x. $e^{7-4x}= 6$ Here is my work. $e^{7-4x} -6= 0$ $\ln(e^{7-4x})\ln(-6)= 0$ $7 - 4x - \ln(6)= 0$ add + 4x $7 - \ln(6)= 4x$ $\dfrac{7 - \ln(6)}{4}= \dfrac{4x}{4}$ $\dfrac{7 - \ln(6)}{4}= x$ but the book is saying the answer is $\dfrac{1}{4} (7 - \ln 6)$ <-- notice they don't put the 6 in parentheses like I did ln(6) is there any significance in this? or no is 1/4 * somethinig the same as something / by 4 ? Sorry if this sounds like an elementary type question. I just want to understand if my answer is correct or not?
 July 7th, 2013, 01:29 PM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Check my work please. Thank you I'm amazed you can manipulate exponentials and logs and not recognise that your answer is the same as the book answer. You made things more difficult for yourself by bringing the 6 across to the LHS. You should have left it where it was! After that, your working goes off the rails a bit. The third line doesn't follow from the second. I'd leave the 6 on the RHS and try again!
 July 7th, 2013, 01:37 PM #3 Senior Member   Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0 Re: Check my work please. Thank you Yea Pero, well I'm glad your a genius. When things don't look the same, I get confused. Like for example, why are they not saying ln(6) in the book? They are just saying ln 6. There has got to be a difference, like is it LN OF 6 or is it LN * 6?
July 8th, 2013, 04:44 AM   #4
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Quote:
 Originally Posted by Shamieh $\ln(e^{7-4x})\ln(-6)= 0$
That should have been $\ln(e^{7-4x})\,-\,\ln(6)\,=\,0.$

Strictly speaking, one should minimize ambiguity by writing ln(6) rather than ln 6, but the rules are often relaxed where the intention is reasonably clear. For example, cos 2A is assumed to mean cos(2A). When using LaTeX, you should type \ln instead of ln, so that the function name is recognized and not rendered in italics. I amended your post to do this for you.

 July 8th, 2013, 11:57 AM #5 Senior Member   Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0 Re: Check my work please. Thank you Ahh, cool I understand now Skipjack. Thank you, that was the answer I was looking for.
July 8th, 2013, 08:19 PM   #6
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Re: Check my work please. Thank you

Quote:
 Originally Posted by Shamieh $e^{7-4x}= 6$
REMEMBER: if a^p = b, then p = log(b) / log(a)
So:
7 - 4x = log(6) ; almost finished!

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# Dave Radcliffe’s polynomial activity part 2

Last week I saw some really neat tweets from Dave Radcliffe. For example:

Those tweets led to a fun project yesterday:

Dave Radcliffe’s polynomial activity day 1

Today I had each of the boys explore $(1 + x + x^2)^n$ mod 2 and mod 3. This is a harder exploration to do by hand (and made harder because I was out this morning and they worked on it alone). Still, it was interesting to hear what they had to day.

My younger son chose the more complicated activity of looking at the powers mod 3. Here’s what he found:

We then went to the computer to check if any of the patterns he thought were there would continue. He had some ideas but unluckily none of them worked. We’ll play more later to see if we can crack the code on the patterns:

Next I talked to my older son. He looked at powers of the polynomial $(1 + x + x^2)$ mod 2.

Here’s what he noticed:

He didn’t have any conjectures, so I showed him the picture that Dave Radcliffe tweeted and that led to him seeing some additional patterns in what he’d written down on the sheet of paper:

So, I’m glad I saw Dave’s tweets because this project is a great computer math exercise. Exploring powers of these polynomials would have been next to impossible without the computer help, but with the computer help we were able to explore a few patterns. It’ll be fun to try to find ways to explore the patterns a bit more and see what we can find.

# Dave Radcliffe’s polynomial activity day 1

Saw this really fun tweet from Dave Radcliffe yesterday:

This looked like a fun project for kids, though it wasn’t obvious how to get started. It turns out that Mathematica has a handy function called PolynomialMod[] that tells you what a polynomial looks like modulo an integer – so that made life easier!

I decided that for today’s project we’d explore $(1 + x)^n$ using Mathematica and see what patterns we could find. The introduction to today’s project involved introducing basic polynomial multiplication. Luckily, a natural way to multiply polynomials looks a lot like multiplying 2-digit numbers. I used that connection to introduce the project:

After the introduction I had the boys play on Mathematica and compute various powers of $(1 + x)^n$ starting with $(1 + x)^0$. We got a little confused between Fibonacci numbers and Pascal’s triangle, but here is what they saw:

For the last part of the project today we used PolynomialMod[] to look at the various powers of $(1 + x)^n$ in mod 2. I wanted to get them used to this Mathematica function to make it easier to explore $(1 + x + x^2)^n$ mod 2 tomorrow. After they explored the powers of $(1 + x)^n$ mod 2 up to n = 8, we talked about patterns in the numbers:

So, a fun little computer math project. It was fun to hear the kids talk about the patterns and also fun to talk about some basic ideas like polynomial multiplication and modular arithmetic. Definitely excited to explore some of the more complicated patters tomorrow.

# A neat problem Dan Anderson shared with us

We’ve spent the last couple of days studying divisors of integers – mainly the number of divisors and the sum of those divisors. This topic came to us via a “things you should know for math contests” list that math team at my older son’s school gave to the kids.

We’ve used Mathematica to help us get a feel for these topics and that computer work (I assume) prompted Dan to share this problem with us:

The problem is: Find the first triangular number with more than 500 divisors.

I asked the boys if they wanted to try to tackle this problem, and they wanted to give it a try. So . . . off we went:

Once the kids understood the problem, I thought it would be useful to spend some time talking about how we could approach the problem. The boys had some pretty good ideas:

The one thing I wanted to spend some extra time on was an alternate way to calculate the triangular numbers. The method that the boys proposed was actually fine, but it seemed like an extra couple of minutes talking about a different approach would be time well spent:

Now we went to the computer to implement our plan. We found that the 12,375th triangular number, 76,576,500, was the first triangular number with more than 500 divisors:

The boys were a little surprised to learn that the first triangular number with more than 500 divisors was smaller than the one with exactly 500 divisors. In fact, we didn’t even find one with 500 divisors yet. In the next part of the project we looked for that number. We did find that number, but it was much larger than we expected – the 1,569,375th triangular number is 1,231,469,730,000, which has exactly 500 divisors!

We wrapped up by looking at 5 of the triangular numbers with exactly 500 factors. They all shared a common factor of 16. We decided to look to see if there was an odd triangular number with exactly 500 factors. As of now (3 hours) after finishing up the project, the computer has not found one.

So, a really fun computer project with the boys. Thanks to Dan Anderson for providing this challenging problem!

# A scary approach and a not-so-scary approach to a challenging math problem

Today was the last day of math projects for at least a week because of some vacation and work travel. Rather than jumping ahead in our book I thought it would be neat to show them an interesting identity in Pascal’s triangle.

Prove that for any integer n greater than 0:

${n \choose 0}^2 + {n \choose 1}^2 + \ldots {n \choose n}^2 = {2n \choose n}$

Instead of taking the scary approach, we started today’s problem by talking about counting paths in a square grid. We explored this type of question for the first time in yesterday’s project:

What learning math sometimes looks like: Counting paths in a grid

Here’s what the boys thought about counting these paths today followed by their approach to counting a few subsets of the total paths:

In the last video we started counting paths that passed through specific points in the diagonal of the grid. Here we finish off that calculation to find an interesting, though fairly complicated-looking expression:

For the last part of the project we take a look at what this identity means in terms of Pascal’s triangle. The fairly easy to see relationship here is a nice surprise! Once the boys see the surprise, they are able to find other cases of this identity in Pascal’s triangle

So, a fun project before we take a break for a week. The connections with counting and Pascal’s triangle really are amazing 🙂

# Introduction to counting with combinations

This morning we started looking at combinations and “choosing” numbers. I think this is one of those topics that seems much easier after you’ve learned it than it seemed when you were learning it. After a quick introduction to the idea of choosing number, the boys worked on an example problem that involved counting the number of games in a round robin tournament:

The were able to work through that problem, but I thought that thinking through it again would still be helpful, so we reviewed the problem from start to finish.

After that quick review we began to talk about choosing numbers. I let the boys explain their ideas about these numbers. One fun thing that happened in this part of our project is the boys discovered that there is a little bit of symmetry in these numbers:

Now we looked at patterns that arise in the choosing numbers. There’s a little trick that I mistakenly thought we’d already talked about – that 0! = 1. After telling them that was simply a definition, we moved on to finding a fun pattern that comes up in the choosing numbers – Pascal’s triangle!

Finally, as a way of confirming the connection to Pascal’s triangle, we looked to see if the addition relationship between two rows of the triangle also shows up in the choosing numbers. This is one of the first examples of a combinatorial proof that the boys have seen!

So, a really fun project showing that counting has some surprising connections to other topics in math. It was fun to hear their ideas (and their surprise) when they found the connection to Pascal’s triangle. Showing them a basic combinatorial proof at the end was fun, too – those proofs can be absolutely amazing!

# Counting, Pascal’s triangle, and binary numbers

In yesterday’s counting project the boys noticed a connection between a counting problem and binary numbers. Here’s that project:

Revisiting an AMC 10 Counting Problem

For today’s project I wanted to explore that connection an a little more depth. To start off we looked at the connection between counting arrangements and Pascal’s triangle:

In the first part of this project we saw a connection between counting pairings of tourists and guides has a interesting connection with Pascal’s triangle. Here we look more carefully at that connection by trying to understand how the rule that tells you how to construct the rows of Pascal’s triangle also shows up when you count these pairings.

We explored the connection here in two parts. In this first part we show the 6 ways that you can pair 4 tourists with 2 guides when each guide has 2 tourists. We also show the 3 ways to pair 3 people with 2 guides where the first guide gets 1 person and the 3 ways to pair 3 people with 2 guides where the first guide gets 2 people.

Now we are ready to find the connection between the two lists me made in the prior video. That connection is important because it shows that the same addition rule that gives the rows of Pascal’s triangle also applies to counting arrangements of certain sets, and therefore helps you understand why Pascal’s triangle helps you count those arrangements.

In the last part of the project we explore the connection between binary numbers and Pascal’s triangle. We do this using an example of 5 digit binary numbers (from 00000 to 11111). This connection allows you to see that the rows of Pascal’s triangle always add up to be a power of 2.

So, a nice little project showing some fun connections between Pascal’s triangle, counting, and binary numbers. Some of these connections are pretty deep and I certainly don’t expect that the boys will have understood every detail from this project. They did seem to have fun with it, though, and their understanding seems to have come a long way from when we worked through the AMC 10 problem earlier this week.

# Talking through Dan Anderson’s mod 2 Pascal’s Triangle

Saw this tweet from Dan Anderson today:

I didn’t have the best math talks with the boys this morning and was sort of bummed out about it all day. Because the problem Dan posted has a really surprising and fun math connection in the solution I thought that going through it would help us end the day on a better note than we how we started it.

Slightly unfortunately we were a little pressed for time, but it was still a nice little project.

We started by just talking about Dan’s problem and about Pascal’s triangle in general by taking an initial look to see if we can find any pattern in the number of odd numbers in each row. The boys noticed a few interesting patterns.

We started this next section by talking in a little more detail about some of the patterns. My younger son noticed that there were a couple of patterns relating to the rows whose number was a power of 2.

The boys used the ideas they found here to take a guess at the number of odd numbers in rows 2048 and 2047, but weren’t sure how to get back to row 2015, yet.

This next section is where I’m a little bummed that we were pressed for time. I wanted them to brain storm about other things that were related to powers of 2, but they got a little stuck.

I used the idea from the picture in Dan’s tweet to write Pascal’s triangle mod 2. Writing out the triangle that way made my older son think about binary. They also were surprised to see something that looked like Sierpinski’s triangle emerging on the board!

From there we wrote out the row numbers in binary and looked for a connection to the number of odd numbers in each row. The surprise in this problem is that there is indeed a connection!

With the clock ticking down to when they had to get out the door, we went to Wolfram Alpha to see what 2015 was in binary. From there they boys guessed that that the number of odd numbers in row 2015 of Pascal’s triangle would be 1024. Wolfram Alpha confirmed this conjecture.

So, though I wish I would have had more time, this was still a fun little project. I’m always trying to help the boys see fun connections in math. Here I really had to show them the connection rather than letting them discover it, but that’s going to happen every now and then. Happy to end the day with this fun project.

# Angry Birds and Snap Cubes: Using Bryna Kra’s MoMath public lecture to talk math with kids

Last night I watched Bryna Kra’s public lecture at the Museum of Math:

I’m not quite sure how to talk through some of the simple dynamical system ideas in the lecture, but the earlier material about patterns and the pigeonhole principle are definitely fun topics to talk about with kids. We used our collection of snap cubes and Angry Bird stuffed animals as props 🙂

In the first part of the talk we introduce the pigeonhole principle and talk through a simple pattern with only single blocks based on one of the elementary patterns Kra uses in her talk. This simple pattern allows us to get a little bit of practice identifying the “pigeons” and the “pigeonholes” in a problem:

In the second talk we look at a slightly more complicated pattern – patterns you get with two blocks rather than one. For this pattern we consider the order of the birds to be important – so a (red, blue) group is different than a (blue, red) group. The example we look at in the last part of today’s talk will consider those two groups to be the same.

The boys were able to see the four different patters that we could make with the two birds / blocks. My older son even noticed a connection with Pascal’s triangle which was fun to see. We then talked about how to count the different types of pairs by looking at the number of choices we had for the first bird and for the second bird. That led my younger son to wonder if there would be a total of 9 groups of two birds if we allowed three different birds in the pattern. Pretty fun discussion:

At the end of the last talk my younger son wondered what would happen if we used three different colors of blocks rather than two. I hadn’t planned on discussing that problem, but what the heck! It was interesting to see the kids figure out how to group the blocks to make the 9 pairs. They were also now able to see how the patterns would continue if we varied the colors and/or number of blocks in the pattern. Fun little exercise. Watching this again I wish I would have spent a little time responding to my older son’s comment that there was no connection to Pascal’s triangle pattern anymore – oh well, next time!

Our last project was a slightly different twist on the Pigeonhole principle. We looked at a tournament involving 4 birds in which each game involves 2 birds. The question I had the boys look at was this: If there are 7 total games played in this tournament, show that at least two of the games must involve the same two players.

I liked their approach to solving this problem. Their instinct was to solve the problem by listing out all of the types of games that could happen. If we were at our whiteboard I would have drawn a square with its sides and diagonals, but their list of all of the types of games was good enough for this project. They had a little difficulty identifying the pigeons and pigeonholes here, but that’s ok, it isn’t always so obvious how to make that identification.

So, a fun project based on another MoMath talk. See here for our last project based on a MoMath lecture:

Part 3 of using Terry Tao’s MoMath lecture to talk about math with kids

I think the public lectures at the Museum of Math are a great way for kids to see some amazing math. There will surely be some lectures that are too advanced for young kids, but many of these lectures have ideas in them that are not hard at all for kids to understand. With Bryna Kra’s lecture, the ideas about patterns and the pigeonhole principle are topics that kids can play with and really enjoy. I’m super glad that MoMath is making these lectures available to the public. It is really fun to show kids some ideas that professional mathematicians use in their research, and hopefully also a great way to inspire a new generation of mathematicians!

# Amazing how many things connect to the number 11.

A few weeks ago David Wees asked the following question on Twitter:

The there were many fun answers to David’s question if you click through to the thread on twitter.  There’s another one in this super blog post that Patrick Honner linked last week, too:

The ideas in Vincnet Knight’s post have been kicking around in the back of my mind for several days.  Today I stumbled on an old favorite Ben Orlin post that sort of connected the dots for me:

A Teaching Philosophy I’m Not Ashamed Of

This quote, in particular, speaks volumes:  “Math is big ideas, approached from as many angles as possible.

Funny enough the dots that got connected led to nearly the opposite conclusion in this specific case.  For me the number 11 is a simple, or in Orlin’s words, small idea, but this small idea allows you to approach many different areas of mathematics directly.  It really is an incredible how useful this simple idea is.

We’ve had a lot of discussions where the number 11 proved to be a surprisingly useful starting point.  One interesting connection is Pascal’s triangle.  We discussed how Pascal’s triangle relates to powers of 11 here:

Pascal’s Triangle and Powers of 11

Here’s a sample of that discussion:

Another fun discussion that related to the number 11 came when we were talking about converting numbers between bases:

All about that base: A fun exercise from Art of Problem Solving

Here’s a sample of that discussion:

Finally, although this technically relates to the number 12 (oh so close!) you could easily replicate this exercise for the number 11 and the polynomial $(x + 1)^n$:

Complete This Sentence: Math is _______

A sample of this discussion is here:

So, a lot of thinking on my end inspired by Vincent Knight’s post.  Who would have thought that this one little number could be so interesting!

# A day in the life: building and extending number sense

I’m not entirely sure why but I’ve been spending a lot of time recently thinking about different ways to build up number sense.  About a week ago I started a chapter on similar triangles with my older son, and the problems in that chapter have helped me gain a better understanding of the importance of building “algebra sense” (for lack of a better phrase) too.    I’m surprised how many opportunities there are to focus on both of these topics now that I’m actively paying attention to them.   An odd coincidence today made me want to write up the conversations I had with my kids this morning.

But first I want to back up to a coincidence from yesterday.

As I mentioned above I’ve been studying similar triangles with my older son for a week or so.  The bit of math that seems to be giving him the most difficulty isn’t the geometry, it is working with the ratios that arise in the problems about similar trirangles.   Here’s one of the problems we worked through yesterday just to give an example of the ratios that come up in these problems:

I felt that it would be good to review some of the algebra behind these ratio equations before finishing the similar triangle section and found several sets of practice problems on Khan Academy that provided more or less exactly the review I was looking for.  Here’s one set for example:

Although people have widely differing views about Khan Academy, I think one nice advantage that it has is that the problem sections are great for this type of review.

Interestingly last night Steven Strogatz posted this picture on twitter:

The similarity between the homework I wanted my son to do and the homework assignment in Strogatz’s tweet got me thinking about context.  The motivation to learn more about geometry was enough for my son to understand the purpose of the Khan Academy problems.  Actually, he even asked to do more.   I don’t know the context of the other homework assignment, but do think that without proper context that assignment could seem quite dull.  This coincidence from yesterday reminds me to be careful to be clear about why I’m asking the boys to do the homework I give them.

Now on to today . . . .

This morning my younger son and I were talking about palindromes (section 6.5 in Art of Problem Solving’s Introduction to Number Theory book).  We began with several simple examples – numbers like 11, 454, 34543 – and then he stopped me:

kid:  “I know a long list of palindromes.”
me:  “what is it?”
kid: [ writes the first 4 rows of Pascal’s triangle on the board ]

This example is definitely a fun one for looking at palindromes, but it also turns out to be a great one for building on number sense.  The connection I wanted to focus on was how the rows related to powers of 11, and how that connection seems to break down in the row:  1 5 10 10 5 1.

My first question to him was whether or not this specific row was a palindrome.  He surprised me by saying that although the number you get by putting all of the terms together, namely 15101051, was not a palindrome, you could get a palindrome you looked only at the last digits, so 150051.  Interesting observation.  We’ll have to return to this topic later when we talk about modular arithmetic!

My next question for him was about the powers of 11.  Starting at $11^0$, the powers of 11 are 1, 11, 121, 1331, 14641, and 161,051.  Why did we lose the connection to Pascal’s triangle when we computed $11^5$?  This led to a wonderful conversation about place value and eventually to showing why we did not actually lose the connection to Pascal’s triangle at all.  Really fun, and I think a neat way to talk through place value while getting in a little arithmetic practice, too.

Later in the morning my older son got tripped up on this problem from the 2006 AMC 8:

2006 AMC 8 problem 24

The problem has a really lucky connection to palindromes since an important observation in solving it is that one number is equal to another number multiplied by 101.  Talking through this problem also led to a good conversation about place value.  Luckily the notes from the conversation about Pascal’s triangle and place value happened to still be up on the board when this second conversation took place.

Seeing some of the earlier work that was on the board my older son said that he thought you could make the row 1 5 10 10 5 1 into a palindrome by working in base 11.  Ha – another unexpected response, but also now a wide open door to talk a little about what I’m calling “algebra sense.”

We quickly reviewed the place value conversation I had with my younger son about how the rows connect to powers of 11, but then looked at what happens in base 11.  Surprise –  powers of 12!!  Don’t think he saw that coming 🙂    Now maybe 5 to 10 minutes of conversation about what the polynomials $(x + 1)^n$ and $(x + y)^n$ look like and we’ve quite unexpectedly done some neat work that helps build up familiarity with algebra and algebraic expressions.

So a fun morning.  As I have the goal of working on number sense in the back of my mind, I’m excited to see all of opportunities that come up to work on it.  Algebra sense, too, but Strogatz’s post from yesterday reminds me to be extra careful about context.  It is fun to take advantage of the lucky times like this morning when that context appears almost by magic!