A nice article on recent progress on solving the twin prime conjecture

The twin prime conjecture (see here, here and here for more information) asserts that there are infinitely many primes that have a difference of 2. For example:

3 and 5 are twin primes;

5 and 7 are twin primes;

11 and 13 are twin primes;

17 and 19 are twin primes;

29 and 31 are twin primes; etc.

While most mathematicians believe the twin prime conjecture is correct, an explicit proof has not been found. Indeed, this has been one of the most popular unsolved problems in mathematics — not necessarily because it’s important, but for the curiosity that a conjecture so simply stated has eluded conquest by the world’s best mathematicians.

Still, research continues, and some major progress has been made in the past few years. (I like sharing this story with my students to convince them that not everything that can be known about mathematics has been figure out yet — a misconception encouraged by the structure of the secondary curriculum — and that research continues to this day.) Specifically, it was recently shown that, for some integer N that is less than 70 million, there are infinitely many pairs of primes that differ by N.

http://video.newyorker.com/watch/annals-of-ideas-yitang-zhang-s-discovery-2015-01-28

http://www.newyorker.com/magazine/2015/02/02/pursuit-beauty

For more on recent progress:

 

 

 

 

 

 

My Favorite One-Liners: Part 100

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Today’s quip is one that I’ll use surprisingly often:

If you ever meet a mathematician at a bar, ask him or her, “What is your favorite application of the Cauchy-Schwartz inequality?”

The point is that the Cauchy-Schwartz inequality arises surprisingly often in the undergraduate mathematics curriculum, and so I make a point to highlight it when I use it. For example, off the top of my head:

1. In trigonometry, the Cauchy-Schwartz inequality states that

|{\bf u} \cdot {\bf v}| \le \; \parallel \!\! {\bf u} \!\! \parallel \cdot \parallel \!\! {\bf v} \!\! \parallel

for all vectors {\bf u} and {\bf v}. Consequently,

-1 \le \displaystyle \frac{ {\bf u} \cdot {\bf v} } {\parallel \!\! {\bf u} \!\! \parallel \cdot \parallel \!\! {\bf v} \!\! \parallel} \le 1,

which means that the angle

\theta = \cos^{-1} \left( \displaystyle \frac{ {\bf u} \cdot {\bf v} } {\parallel \!\! {\bf u} \!\! \parallel \cdot \parallel \!\! {\bf v} \!\! \parallel} \right)

is defined. This is the measure of the angle between the two vectors {\bf u} and {\bf v}.

2. In probability and statistics, the standard deviation of a random variable X is defined as

\hbox{SD}(X) = \sqrt{E(X^2) - [E(X)]^2}.

The Cauchy-Schwartz inequality assures that the quantity under the square root is nonnegative, so that the standard deviation is actually defined. Also, the Cauchy-Schwartz inequality can be used to show that \hbox{SD}(X) = 0 implies that X is a constant almost surely.

3. Also in probability and statistics, the correlation between two random variables X and Y must satisfy

-1 \le \hbox{Corr}(X,Y) \le 1.

Furthermore, if \hbox{Corr}(X,Y)=1, then Y= aX +b for some constants a and b, where a > 0. On the other hand, if \hbox{Corr}(X,Y)=-1, if \hbox{Corr}(X,Y)=1, then Y= aX +b for some constants a and b, where a < 0.

Since I’m a mathematician, I guess my favorite application of the Cauchy-Schwartz inequality appears in my first professional article, where the inequality was used to confirm some new bounds that I derived with my graduate adviser.

My Favorite One-Liners: Part 99

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Today’s quip is a light-hearted one-liner that I’ll use to lighten the mood when in the middle of a complex calculation, like the following limit problem from calculus:

Let f(x) = 11-4x. Find \delta so that |f(x) - 3| < \epsilon whenever $|x-2| < \delta$.

The solution of this problem requires isolating x in the above inequality:

|(11-4x) - 3| < \epsilon

|8-4x| < \epsilon

-\epsilon < 8 - 4x < \epsilon

-8-\epsilon < -4x < -8 + \epsilon

At this point, the next step is dividing by -4. So, I’ll ask my class,

When we divide by -4, what happens to the crocodiles?

This usually gets the desired laugh out of the middle-school rule about how the insatiable “crocodiles” of an inequality always point to the larger quantity, leading to the next step:

2 + \displaystyle \frac{\epsilon}{4} > x > 2 - \displaystyle \frac{\epsilon}{4},

so that

\delta = \min \left( \left[ 2 + \displaystyle \frac{\epsilon}{4} \right] - 2, 2 - \left[2 - \displaystyle \frac{\epsilon}{4} \right] \right) = \displaystyle \frac{\epsilon}{4}.

Formally completing the proof requires starting with |x-2| < \displaystyle \frac{\epsilon}{4} and ending with |f(x) - 3| < \epsilon.

My Favorite One-Liners: Part 88

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

In the first few weeks of my calculus class, after introducing the definition of a derivative,

\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h},

I’ll use the following steps to guide my students to find the derivatives of polynomials.

  1. If f(x) = c, a constant, then \displaystyle \frac{d}{dx} (c) = 0.
  2. If f(x) and g(x) are both differentiable, then (f+g)'(x) = f'(x) + g'(x).
  3.  If f(x) is differentiable and c is a constant, then (cf)'(x) = c f'(x).
  4. If f(x) = x^n, where n is a nonnegative integer, then f'(x) = n x^{n-1}.
  5. If f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 is a polynomial, then f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1.

After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.

Example 1. Let A(r) = \pi r^2. Notice I’ve changed the variable from x to r, but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.)

What’s the derivative? Remember, \pi is just a constant. So A'(r) = \pi \cdot 2r = 2\pi r.

Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Generally, students start waking up even though it’s near the end of class. I continue:

Example 2. Now let’s try V(r) = \displaystyle \frac{4}{3} \pi r^3. Does this remind you of anything? (Students answer: the volume of a sphere.)

What’s the derivative? Again, \displaystyle \frac{4}{3} \pi is just a constant. So V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2.

Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)

By now, I’ve really got my students’ attention with this unexpected connection between these formulas from high school geometry. If I’ve timed things right, I’ll say the following with about 30-60 seconds left in class:

Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (Students: stunned silence.)

This is what’s known as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (Students groan, as they really want to know the answer immediately.) Class is dismissed.

If you’d like to see the answer, see my previous post on this topic.

My Favorite One-Liners: Part 50

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s today’s one-liner: “To prove that two things are equal, show that the difference is zero.” This principle is surprisingly handy in the secondary mathematics curriculum. For example, it is the basis for the proof of the Mean Value Theorem, one of the most important theorems in calculus that serves as the basis for curve sketching and the uniqueness of antiderivatives (up to a constant).

And I have a great story that goes along with this principle, from 30 years ago.

I forget the exact question out of Apostol’s calculus, but there was some equation that I had to prove on my weekly homework assignment that, for the life of me, I just couldn’t get. And for no good reason, I had a flash of insight: subtract the left- and right-hand sides. While it was very difficult to turn the left side into the right side, it turned out that, for this particular problem, was very easy to show that the difference was zero. (Again, I wish I could remember exactly which question this was so that I could show this technique and this particular example to my own students.)

So I finished my homework, and I went outside to a local basketball court and worked on my jump shot.

Later that week, I went to class, and there was a great buzz in the air. It took ten seconds to realize that everyone was up in arms about how to do this particular problem. Despite the intervening 30 years, I remember the scene as clear as a bell. I can still hear one of my classmates ask me, “Quintanilla, did you get that one?”

I said with great pride, “Yeah, I got it.” And I showed them my work.

And, either before then or since then, I’ve never heard the intensity of the cussing that followed.

Truth be told, probably the only reason that I remember this story from my adolescence is that I usually was the one who had to ask for help on the hardest homework problems in that Honors Calculus class. This may have been the one time in that entire two-year calculus sequence that I actually figured out a homework problem that had stumped everybody else.

My Favorite One-Liners: Part 46

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Today’s one-liner is something I’ll use after completing some monumental calculation. For example, if z, w \in \mathbb{C}, the proof of the triangle inequality is no joke, as it requires the following as lemmas:

  • \overline{z + w} = \overline{z} + \overline{w}
  • \overline{zw} = \overline{z} \cdot \overline{w}
  • z + \overline{z} = 2 \hbox{Re}(z)
  • |\hbox{Re}(z)| \le |z|
  • |z|^2 = z \cdot \overline{z}
  • \overline{~\overline{z}~} = z
  • |\overline{z}| = |z|
  • |z \cdot w| = |z| \cdot |w|

With all that as prelude, we have

|z+w|^2 = (z + w) \cdot \overline{z+w}

= (z+w) (\overline{z} + \overline{w})

= z \cdot \overline{z} + z \cdot \overline{w} + \overline{z} \cdot w + w \cdot \overline{w}

= |z|^2 + z \cdot \overline{w} + \overline{z} \cdot w + |w|^2

= |z|^2  + z \cdot \overline{w} + \overline{z} \cdot \overline{~\overline{w}~} + |w|^2

= |z|^2 + z \cdot \overline{w} + \overline{z \cdot \overline{w}} + |w|^2

= |z|^2 + 2 \hbox{Re}(z \cdot \overline{w}) + |w|^2

\le |z|^2 + 2 |z \cdot \overline{w}| + |w|^2

= |z|^2 + 2 |z| \cdot |\overline{w}| + |w|^2

= |z|^2 + 2 |z| \cdot |w| + |w|^2

= (|z| + |w|)^2

In other words,

|z+w|^2 \le (|z| + |w|)^2.

Since |z+w| and |z| + |w| are both positive, we can conclude that

|z+w| \le |z| + |w|.

QED

In my experience, that’s a lot for students to absorb all at once when seeing it for the first time. So I try to celebrate this accomplishment:

Anybody ever watch “Home Improvement”? This is a Binford 6100 “more power” mathematical proof. Grunt with me: RUH-RUH-RUH-RUH!!!

My Favorite One-Liners: Part 43

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. q Q

Years ago, my first class of students decided to call me “Dr. Q” instead of “Dr. Quintanilla,” and the name has stuck ever since. And I’ll occasionally use this to my advantage when choosing names of variables. For example, here’s a typical proof by induction involving divisibility.

Theorem: If n \ge 1 is a positive integer, then 5^n - 1 is a multiple of 4.

Proof. By induction on n.

n = 1: 5^1 - 1 = 4, which is clearly a multiple of 4.

n: Assume that 5^n - 1 is a multiple of 4.

At this point in the calculation, I ask how I can write this statement as an equation. Eventually, somebody will volunteer that if 5^n-1 is a multiple of 4, then 5^n-1 is equal to 4 times something. At which point, I’ll volunteer:

Yes, so let’s name that something with a variable. Naturally, we should choose something important, something regal, something majestic… so let’s choose the letter q. (Groans and laughter.) It’s good to be the king.

So the proof continues:

n: Assume that 5^n - 1 = 4q, where q is an integer.

n+1. We wish to show that 5^{n+1} - 1 is also a multiple of 4.

At this point, I’ll ask my class how we should write this. Naturally, I give them no choice in the matter:

We wish to show that 5^{n+1} - 1 = 4Q, where Q is some (possibly different) integer.

Then we continue the proof:

5^{n+1} - 1 = 5^n 5^1 - 1

= 5 \times 5^n - 1

= 5 \times (4q + 1) - 1 by the induction hypothesis

= 20q + 5 - 1

= 20q + 4

= 4(5q + 1).

So if we let Q = 5q +1, then 5^{n+1} - 1 = 4Q, where Q is an integer because q is also an integer.

QED

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On the flip side of braggadocio, the formula for the binomial distribution is

P(X = k) = \displaystyle {n \choose k} p^k q^{n-k},

where X is the number of successes in n independent and identically distributed trials, where p represents the probability of success on any one trial, and (to my shame) q is the probability of failure.

 

 

My Favorite One-Liners: Part 13

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a story that I’ll tell my students when, for the first time in a semester, I’m about to use a previous theorem to make a major step in proving a theorem. For example, I may have just finished the proof of

\hbox{Var}(X+Y) = \hbox{Var}(X) + \hbox{Var}(Y),

where X and Y are independent random variables, and I’m about to prove that

\hbox{Var}(X-Y) = \hbox{Var}(X) + \hbox{Var}(Y).

While this can be done by starting from scratch and using the definition of variance, the easiest thing to do is to write

\hbox{Var}(X-Y) = \hbox{Var}(X+[-Y]) = \hbox{Var}(X) + \hbox{Var}(-Y),

thus using the result of the first theorem to prove the next theorem.

And so I have a little story that I tell students about this principle. I think I was 13 when I first heard this one, and obviously it’s stuck with me over the years.

At MIT, there’s a two-part entrance exam to determine who will be the engineers and who will be the mathematicians. For the first part of the exam, students are led one at a time into a kitchen. There’s an empty pot on the floor, a sink, and a stove. The assignment is to boil water. Everyone does exactly the same thing: they fill the pot with water, place it on the stove, and then turn the stove on. Everyone passes.

For the second part of the exam, students are led one at a time again into the kitchen. This time, there’s a pot full of water sitting on the stove. The assignment, once again, is to boil water. Nearly everyone simply turns on the stove. These students are led off to become engineers. The mathematicians are ones who take the pot off the stove, dump the water into the sink, and place the empty pot on the floor… thereby reducing to the original problem, which had already been solved.

Engaging students: Completing the square

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Deborah Duddy. Her topic, from Algebra: completing the square.

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What interesting word problems using this topic can your students do now?

Applying what is learned in the class is very vital in fact it is a process TEKS that teachers need to use to maximize student’s understanding. “When are we going to use this in real life?” and “Why do we need to know this?” are questions that students ask on a daily basis. Connecting material to the real world helps engage students and develops critical thinking. Describing a path of a ball, how far an item can be tossed in the air and how to maximize profits for a company are just some examples of how quadratics can be used in the real world.

One important event happens during high school; students receive their driver’s license. In their written driver’s test, students must know the distance needed to stop a car at certain speed limits. Using an example like the one below will be interesting for the students and help connect lesson material and real life. 

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How could you as a teacher create an activity or project that involves your topic?

To begin class and get students involved with their learning, the class will participate in an activity. Each pair of students will have two different cards such as (x+2)^2 and x^2+4x+4, and any variations of these problems. They can only look at the (x+2)^2 card. Students will work out the problem on paper. Students will be asked to remember how to find the area of a square and then set up a square with the dimensions matching the first card. From there, the pairs would use algebra tiles (after knowing what each tile stands for) and attempt to “complete the square”. This activity will be used as an engage and a beginning explore for the students. This activity will help students see completing a square geometrically.

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How does this topic extend what your students should have learned in previous courses?

Completing the square is another way of solving/factoring the equation. The process of completing the square is to turn a basic quadratic   equation of ax^2 + bx + c = 0 into a(x-h)^2 + k = 0 where (h,k) is  the vertex of the parabola. Therefore this process is very beneficial because it helps students graph the quadratic equation given. In order to find h and k, students should be able to factor, square a term, find the square root and manipulate the equation.

In solving the equation by completing the square is to subtract the constant off the left side and onto the right side. Then students take the coefficient off the x-term divide it then square it. Students then add this number to both sides of the equations. By simplifying the right side of the equation, students give the perfect square. Then solve the equation left by taking the square root of both sides and determining x.

 

References:

http://www.classzone.com/eservices/home/pdf/student/LA205EBD.pdf

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 18

The Riemann Hypothesis (see here, here, and here) is perhaps the most famous (and also most important) unsolved problems in mathematics. Gamma (page 207) provides a way of writing down this conjecture in a form that only uses notation that is commonly taught in high school:

If \displaystyle \sum_{r=1}^\infty \frac{(-1)^r}{r^a} \cos(b \ln r) = 0 and \displaystyle \sum_{r=1}^\infty \frac{(-1)^r}{r^a} \sin(b \ln r) = 0 for some pair of real numbers a and b, then a = \frac{1}{2}.

As noted in the book, “It seems extraordinary that the most famous unsolved problem in the whole of mathematics can be phrased so that it involves the simplest of mathematical ideas: summation, trigonometry, logarithms, and [square roots].”

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When I researching for my series of posts on conditional convergence, especially examples related to the constant \gamma, the reference Gamma: Exploring Euler’s Constant by Julian Havil kept popping up. Finally, I decided to splurge for the book, expecting a decent popular account of this number. After all, I’m a professional mathematician, and I took a graduate level class in analytic number theory. In short, I don’t expect to learn a whole lot when reading a popular science book other than perhaps some new pedagogical insights.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.