Trying out a challenging problem shared by Amy Hogan

Saw a neat tweet from Amy Hogan yesterday:

I found a series that gives the solution, but don’t know a closed form for the solution. Still, though, I thought sharing this problem with kids would be fun – some of the basic ideas you need to start down the road to the solution are accessible to kids. I’m really happy with how the discussion went today.

We started out by reading the problem and I asked the boys a few questions to make sure they understood how the game worked:

Next we started talking about the solution. My younger son had a little bit of a misconception initially, but the problem started to make a bit more sense for him after my older son suggested a different approach.

Still, it was interesting to hear the boys discuss whether the probability of winning in exactly 3 tosses was 1/4 or 1/8.

Now we tried to figure out how likely it would be to win with a string of exactly three heads at the end (so your final four flips would be THHH).

Finally, we tried to count the ways that you could win with final flips of THHHH – so winning by flipping four heads in a row. Their guess in the last video was that there would be 3 cases – but they realized fairly early on that there would be more than 3.

After we finished I showed them my code in Mathematica that calculated the sum that gives the final answer.  To 10 decimal places the answer is 0.7112119049.

Thank to Amy Hogan for sharing this problem – even though the exact answer is a little out of reach for young kids, it is still a terrific problem for them to study.

Advertisements

Playing with Polynomials

We’ll be doing a little bit of review work in the Integrated CME Project III book. Today my son came across an interesting problem about trying to (sort of) match two polynomials. He came up with a nice solution this morning and we talked about the problem when he got home from school today.

The problem goes like this:

Find a polynomial that agrees with x^3 - x at x = 1, 2, and 3, and has a value of 0 at x = 4.

Here’s my son talking through his solution:

After he finished his explanation, I showed him my solution to the problem:

To wrap up we went to Mathematical to look at both solutions and also so that I could show him a little surprise:

So, a nice start to this review project. It’ll be fun to work through the book over the summer.

Paula Beardell Krieg’s 72 degree question

A few weeks ago I got this question from Paula Beardell Krieg on Twitter:

Today I went through this problem with the boys – the difficulty of this exercise surprised me a bit. They really struggled to see how you could tell if an angle was 72 degrees.

Here’s the introduction. The boys noticed a few things about the picture and got some ideas with how to proceed:

Next we drew the two squares on a piece of paper and I let the boys explore the question. Here they struggled to make much progress beyond the things that they noticed in the first part of the project:

The thing giving them trouble was that they didn’t know any relationships between angles in a right triangle with a 72 degree angle. That left them completely stuck. Eventually they decided to measure the squares and found that they had something that looked like a triangle with side 1, 3, and \sqrt{10}.

Next we explored some of the ideas around 1, 3, \sqrt{10} triangles. After a little nudging from me they decided to measure the angles with a protractor.

Now I showed them my solution and let them see where the 1, 3, \sqrt{10} triangle comes up:

Finally, I let them play with two sets of triangles that I printed overnight. Two of these triangles are right triangles with 72 and 18 degree angles, and two of them are 1, 3, \sqrt{10} triangles. The question is -> are all 4 triangles the same?

Here are pictures (to scale) of the two triangles. You can see how similar they are.

First, the right triangle with a 72 degree angle:

72 degree picture

Second, the 1, 3, \sqrt{10} triangle:

one three root 10 picture.jpg

Tomorrow we’ll explore a second similarity between these two triangles. I found it playing around while I was making the triangles yesterday 🙂

Working through an AMC 8 geometry problem

My younger son was working through the 1989 American Junior High Mathematics Exam this morning and got stuck on this problem:

Screen Shot 2017-07-14 at 8.33.41 AM copy.jpg

Here’s a link to the entire exam on Art of Problem Solving’s website:

The 1989 AJHME on Art of Problem Solving’s website

I thought this problem would make for a nice project since there are a couple of good mathematical ideas in it, so we sat down to talk about it. My younger son talked through his approach first:

My older son went next and had a different approach:

To wrap up we talked about how the answer would change if the problem was set up with a slightly different arrangement of the cubes and the boys found their way to an important idea in geometry:

I’m really happy that the old AMC problems are available – they are a wonderful resource to use to find challenging but accessible problems for kids.

A terrific example for calculus students from Nassim Taleb and Alexander Bogomolny

I saw a wonderful exchange on twitter yesterday on a problem posted by Alexander Bogomolny:

At first this problem didn’t really jump off the page as a good first year calculus example, but then I as the solution that Nassim Taleb posted:

I’m a tiny bit time constrained this morning and can’t get the Taleb tweet to embed right, so here’s the solution a second time just in case the embedding remains broken:

Taleb

So, Taleb reduces the difficult-looking limit and sums to two integrals. The ideas underlying this reduction are both beautiful on their own and fundamental in calculus.

A few questions that I think would be worth discussing with calculus students are:

(1) [this one was discussed in the twitter thread] Why did the integrals start at 0, and does that matter?

(2) Why is ratio of the integrals equal to the ratio of the sums? This answer to this question is related to the answer to (1). It is also an excellent way to reinforce some of the main ideas behind Riemann sums.

(3) Probably less mathematically interesting, but a good challenge exercise for students is evaluating Taleb’s integral formulation of the problem using l’Hospital’s rule. I say “less mathematically interesting” because you have to evaluate the same integrals in both approaches, but the approach via l’Hospital’s rule allows you to discuss the Fundamental Theorem of Calculus and also review the chain rule. The arithmetic here requires you to be extra careful, but I think the other ideas outweigh the annoying arithmetic.

Too bad that the school year is over – but this is a great example to keep in your back pocket for next year’s calculus classes!

A night with Cut the Knot, Nassim Taleb, and some Supernova

Please note the correction at the bottom of the post

A further correction – there is still an error. Ugh. This approach may not work, unfortunately . . .

Saw a neat problem from Alexander Bogomolny earlier today:

I actually missed the problem when it was initial posted, but saw it via Nassim Taleb’s solution:

The problem sort of gnawed at me all day and I figured it was in the maybe 1 in 10 problems that Bogomolny posts that I might be able to solve.

So, tonight I poured a glass of Supernova and gave it a go

One thing on my mind all day with this problem was Jensen’s inequality. What I would *love* to be able to do is say that by Jensen’s inequality:

(1/3) \sqrt{x^2 + 3} + (1/3) \sqrt{y^2 + 3} + (1/3) \sqrt{xy + 3}

\geq \sqrt{ (1/3)( x^2 + y^2 + xy) + 3}

Which is easily seen to be \geq 2 because of the constraint x + y = 2. That work would show the original inequality was \geq 6.

The approach has a tiny bit of merit since \sqrt{x^2 + 3} is concave up for x between 0 and 2 -> here’s a little Mathematica plot showing that the second derivative is indeed positive on 0 to 2:

Plot1

But . . . the problem is that folding in the 3rd term in the sum is stretching the rules of Jensen’s inequality a bit, I think, since it is not of the form \sqrt{a^2 + 3}.

With the first two terms, though, applying Jensen’s inequality seems ok, but I now need (1/2)’s instead of (1/3)’s since there are only two terms. So, I’ll use Jensen’s on the first two terms only and try to show that

(1/2) \sqrt{x^2 + 3} + (1/2) \sqrt{y^2 + 3} + (1/2) \sqrt{xy + 3} \geq 3

By Jensen’s inequality this new sum is

\geq \sqrt{ \frac{x^2 + y^2}{2} + 3} + (1/2) \sqrt{xy + 3}

A bit of algebra and the fact that x + y = 2 allows us to simplify this expression to:

\sqrt{5 - xy} + (1/2) \sqrt{xy + 3}

and then further to:

\sqrt{ (x - 1)^2 + 4} + (1/2) \sqrt{4 - (x - 1)^2}

Now we are just down to a fairly straightforward calculus problem, and I’ll let Mathematica do the heavy lifting since the algebra isn’t that interesting:

Plot2

We can see visually that the minimum occurs at x = 1 from the plot, and the plot of the derivative further confirms that there is only one critical point. The value of the last expression at x = 1 is indeed 3 as we were hoping.

So, Jensen’s inequality, a bit of calculus, and a nice glass of scotch shows that the original inequality is indeed true.

Thanks to Alexander Bogomolny for the problem, and to Nassim Taleb for his solution that got me thinking about the problem.


Correction

I received a note from Alexander Bogomoly over night. He spotted an error in the calculation:

[/embed]https://twitter.com/CutTheKnotMath/status/873413279749722113[/embed]

and I thought my kids having trouble sleeping and waking me up at 5:00 am today was a bad start to the day!

But it seems that I’ve gotten very lucky – both learning from my carelessness in applying Jensens inequality and that the path forward from Bogomolny’s correction is easier than the path I actually took.

Starting here – we wish to show that:

(1/2) \sqrt{x^2 + 3} + (1/2) \sqrt{y^2 + 3} + (1/2) \sqrt{xy + 3} \geq 3

The correction shows that the expression on the left hand side is \geq than

\sqrt{ (\frac{x + y}{2})^2 + 3} + (1/2) \sqrt{xy + 3}

but since x + y = 2, the first piece of this expression is equal to 2 and the 2nd expression simplifies as before. So we are left with

2 + (1/2) \sqrt{4 - (x - 1)^2}

and this expression has a maximum of 3 at x = 1.

That means that the expression we were trying to show to be greater than 3 is indeed greater than 3, and the expression in the original tweet is greater than 6.

I’m grateful to Alexander Bogomolny for spotting the error in my original argument.

A hand waving approach to a problem posted by Cut the Knot

Saw this tweet yesterday:

It was a fun problem to think about and the two solutions on the site use the Stolz-Cesaro Lemma, which is basically l’Hospital’s rule for sums.

Through the various Christmas preparations yesterday I was wondering if there was a simple way to see why the limit exists in the first place. What follows below isn’t a rigorous proof (or even close to one!) but instead how I convinced myself that the limit probably does exist.

Since seeing Tim Gowers “live blog” his solution to an old IMO problem, I’ve been interested in occasionally sharing the solution process rather than polished solutions to problems. Two examples of problems I’ve used for that idea are below:

A Challenge / Plea to math folks

A challenge relating to a few problems giving my son trouble

So, for the problem at hand, here’s my “hand waving” approach to convincing myself that the limit even existed:

We know that:

\lim_{x\to\infty} E_n = \lim_{x\to\infty} (1 + \frac{1}{n})^n = e

and that

\lim_{x\to\infty} H_n = \lim_{x\to\infty} (1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}) \approx \ln{n} + \gamma

So I’ll approximate the difference we are considering like this for large n:

(E_{n+1})^{H_{n+1}} - (E_n)^{H_n}

\approx e^{H_{n+1}} - e^{H_n}

\approx (e^{H_n})(e^{\frac{1}{n+1}} - 1)

\approx (e^{\ln{n} + \gamma}) (\frac{1}{n+1} + \frac{1}{2(x+1)^2} + \ldots )

\approx (e^{\gamma}) * n * (\frac{1}{n+1} + \frac{1}{2(n+1)^2} + \ldots )

\approx e^{\gamma}

The fact that this hand waving approach arrives at the “right” answer is just a coincidence as I’m playing pretty fast and loose with limit rules. But, at least I now have some indication that this strange (and lovely!) \infty - \infty limit might actually exist.

Just for fun here’s what the expression looks like for n up to 1,000,000:

CuttheKnot.jpg

Definitely a fun little problem to noodle over 🙂