Sep 072016
 

The struggle to understand why we teach K-12 mathematics in the order we do, and the content we do is real. I have wondered about this for a long time, and really have never found a good answer.

I threw out the idea of teaching y=mx+b as the only way to write lines (even though the district materials at the time said it was all we needed). I took a lot of heat for that decision from some people. I was told I was completely wrong; by teachers. I stuck to my guns because y=mx+b is a stupid way to teach lines. And in the end, I was told by other teachers that I influenced them to change too.

But really, K-12 mathematics education is nothing like this:

Mathematics as human pursuit

Think of Lockhart’s Lament.  You read Paul’s words, and you are hit by the poetry he sees in math. It is also 25 pages long. I read somewhere that Lockhart’s Lament is the the most powerful and often cited mathematics education document that is never acted upon. What does that say about us, as educators, who cite it?

Lockhart is passionate about math education, and he feels that the current state (in 2002) of math education is in trouble. His words may be as apt today as it was then. On page 2 he writes,

In fact, if I had to design a mechanism for the express purpose of destroying a child’s natural curiosity and love of pattern-making, I couldn’t possibly do as good a job as is currently being done— I simply wouldn’t have the imagination to come up with the kind of senseless, soul-crushing ideas that constitute contemporary mathematics education.

How much impact has Lockhart had on mathematics education? Often cited, rarely used or implemented. And yet, my Twitter feed and Facebook still have things like this pop up regularly.

Mathmatician is like a painter G H Hardy

What beautiful words representing fantastic ideals. Are you starting to see the cognitive dissonance I am feeling today? Too bad none of these ideals are found in our textbooks or our standards (and don’t get me wrong, I am not hating on the CCSS-M here). In fact, much of school mathematics is exactly how Seymour Papert described it here.

Papert - outwitting teachers as school goal

It is mindless, repetitive, and dissociated.

So as I was thinking of the question of “Why?”, I stumbled upon this article. Why We Learn Math Lessons That Date Back 500 Years? on NPR. To find out the answer is pretty much, “Because we always have,” is sad, disappointing and frustrating. We have taught it this way for the last 500 years, so we will continue to teach it this way for the foreseeable future.

I call B.S.

Seriously. We need to rethink how we teach math in a substantive manner.

We are part of a system that is not allowing learners to find the joy of mathematics, but the drudgery of mathematics and of learning. And this is not new. Not by any means. Edward Cubberly, Dean of the Stanford University School of Education around 1900) said,

Our schools are, in a sense, factories, in which the raw products (children) are to be shaped and fashioned into products to meet the various demands of life. The specifications for manufacturing come from the demands of twentieth-century civilization, and it is the business of the school to build its pupils according to the specifications laid down.

The fact that the specifications of education haven’t changed in hundreds of years is a problem (see the NPR article). It may even be THE problem. I am not so confident to claim that for sure, but it is definitely A problem.

At what point do we, as teacher leaders, rise up and demand this change. We see the damage. We see the issues. We must start demanding the curriculum be changed to meet the needs of our learners. I am not sure that the CCSS-M is that change. It seems like it is codifying the 500 year old problems that we are currently doing.

But it doesn’t not have to.

The Modeling Standard is gold. It is also 1 single page in the entire document.

I will just end this rant with that thought. Oh, and this thought. No more broccoli flavored ice cream.

textbook math is like broccoli ice cream

Jan 172015
 

Nothing annoys me more in teaching math than a bunch of rules to memorize, and rational function come with their own complete set of rules to memorize. I really find that annoying, and I have been on a personal quest to make sense of algebra through a combined set of understandings that will bring comprehension, not rule following.

I have found that in large part through the (h,k) form of the algebraic functions (and here too). Not just a little, but the (h,k) form now drives my entire instruction to the point where my learners are asking me first “how do we undo this” instead of “what chapter is this” as we are learning the math.

So, rational functions. How do the “rules” of horizontal asymptotes fit for rational functions. I really struggled with this the first year I was working on the translations and (h,k) ideas, but this year it all fell into place.

Lets take two functions, f(x) and g(x) where the highest degree is m for the numerator and n for the denominator (just keeping things in alphabetical order).

The rules that everyone knows and hates:

If m=n, then horizontal asymptote is: y=a/b where a and b are the leading coefficients of the numerator and denominator.
If m>n, then there is no H asymptote [or some books say if m=n+1 then there is a slant asymptote]
if m<n, then H asymptote is: y=0.

Okay, I hate these. I really wanted to understand why, and I fully understood when I explored how to get any rational function into the (h,k) form. How do you do that, you ask? Simple. You do the long division and rewrite the equation in the new form.

First off, though, we need some functions to explore. I have a Desmos file with 1600 different possible rational functions:
 Seriously, 1600 possible functions. 40 for numerator and same 40 for denominator.

I tried typing it all out, but failed, so I wrote it out and took a picture:

2015-01-17 16.05.06

What we see is that the ‘k’ value is always the horizontal asymptote. What we also see, is that there is ALWAYS an asymptote when m>n, and sometimes it is a linear slant. It also, can be a quadratic slant, or cubic slant. What is important is that the horizontal asymptote is a way to discuss the END BEHAVIOR of the curve. If we have a slant asymptote, what is happening is the original function is approaching the value of another function instead of a constant.

Rock my world.

So, 2x^4 +3x^3-2x^2 + 5 divided by 2x^2+4x-2 gives us a ‘k’ of x^2 -.5x +1. The “slant” asymptote is a quadratic function.

2015-01-17 17.14.21Here is the math:
 and the Desmos file.

What is amazing here is the long division and putting the function into (h,k) form means you do not have to remember ANY rules with rational functions. It also means there is a reason to teach long division of functions as well.

If our goal is to create a unified, sense-making structure in algebra, this is how it is done.

Let me know if I have made a mistake somewhere or there are flaws in my thinking. This is one piece of the larger structure I am seeing with this approach to algebra, and I really want to push the envelop and limits of of the method.
At this point, what I see is that the “rules” of horizontal asymptotes are nothing more than tricks. The math is the long division and rewriting the function into the (h,k) form to show the translations, and reflection.

In addition, if you look at the functions I used in the explanation above (the first picture I used), you will see that only when the function is put in (h,k) for does the reason for the reflection show up. If the function is left in standard form, the reflection is hidden.

Nix the Tricks! This is the reason.

Nov 072014
 

This post is a branch of yesterday’s post on polynomials. As were are teaching this year, I am giving them some tough, torturous problems that are about incorporating all the math from previous units into one problem each time.

Yes, I want every problem to be a review of the previous year’s materials. And it is working so far. Another teacher and were looking at this problem set at lunch and we were discussing whether or not we had to have pairs of parenthesis that were conjugates.

problemset

I mean, I did give them pairs that were conjugates in each of these, but do I have to?

No. And not just no, but why should I?

Let’s think of a polynomial differently. A polynomials is nothing more that a series of “lines” that are multiplied. Those “lines” can be real lines  (x+2) or irrational lines (x+2sqrt(3)) or even imaginary “lines” (x+2i). When we take these “lines” and multiply them together we get a polynomial function.

cubic1 or cubic2

or even! cubic3. Yes, I had to do this in the Nspire software because this last graph breaks Desmos. Desmos cannot graph the imaginary numbers of the intercept form of a polynomial. #sadness

But notice that in each of these graphs I used the conjugates. Did I have to?

No.

cubic4

Not at all. Here is a polynomial function, perfectly formed, that is composed of three lines multiplied together. Can I do it with three different imaginary numbers? No. That is where we MUST use conjugates. Only in that one, special case.

So the question is why do some teachers think that conjugates are required in ALL cases of polynomials? The reason is very simple. Only when we have conjugates can we form a standard form function that can be solved with the quadratic formula.

That’s it. That is the only reason.

I was pleasantly surprised to find this series of old posts by Mr. Chase this week as well:

Do Irrational Roots come in Pairs Part 1

Do Irrational Roots come in Pairs Part 2

Do Irrational Roots come in Pairs Part 3

He approaches it from the standard form side, but I think it is far easier to see why it is perfectly okay from the intercept form side of a polynomial.

The real issue is do you have to have conjugates of imaginary numbers in a polynomial. I think, and I do emphasize the THINK, that we do. If we have an imaginary number in the standard form polynomial then we can not graph it on the real plane.  I have been trying to think of a work around (reminiscent of what I did with the quadratics here and here.) Still wrapping my head around it. Not sure if I can see a way out.

 

How does this relate to my class? Well, my learners are doing problems like the fourth picture. That counts.

Jun 102013
 

This post is a little personal venting along with some serious math. I am in an Algebra 2 training today, and the instructor is doing some long division for finding slant asymptotes. Okay, nothing wrong with that, but I ask why we aren’t using synthetic division when we just finished using synthetic division in the previous example when the WHOLE room (minus 2 pp) yells it me, “Because it can’t be done.” When I insist it can work, they challenge me to put it on the board, they just don’t believe it. The session ends before I can.

The frustrating thing is, the entire room of math teachers with the exception of 2 others are completely and totally wrong. At least 2 other people had my back.

As evidence, see this paper by

Okay, I admit, that paper is kind of rough reading. How about some easier stuff then.

From Pat’s blog, we have the following articles: Link 1, Link 2, Link 3, and Link 4.

And, just to prove to myself that Pat is correct, (and he is, don’t doubt it) I tried a problem side by side with the long division I know works:

image

Really, it works. I especially like the fact that synthetic division does not always need a 1 in front of the leading term.

image

Oops, I just did another synthetic division that is impossible!

I think I need to do four more so that I can be like the White Queen and do six impossible things before breakfast.

And yes, I am getting to the training tomorrow early so these can be on the board when everyone walks in. Don’t tell me something is impossible unless it truly is. Calling something impossible because you don’t want to think or learn about it is kind of like saying the quadratic equation y = x2 – x + 7 has no solutions. It has solutions, they just aren’t nice and easy like you are used to.

—————–

Special thanks to Pat, who introduced me to this in 2009 in his blog.

Jul 292012
 

Many people ask me why I ride my motorcycle long distances in the summer. This summer I traveled from Reno, NV to St. Louis, MO. It was around 4000 miles, round trip, and brutally hot for a couple of states worth of riding.

But, that traveling allows me one single thing I rarely get. Time away from all distractions. It worked. I thought long and hard about the problem I talked about last post; A visual representation to imaginary solutions of quadratics. Somewhere in Wyoming I had the idea on how to prove it. By the time I hit Utah, I had the solution worked out in my head, and I needed to jot some notes. It honestly took me several hours to type up the solution, and without further ado, here it is.

graph5

The Goal:

To prove that in a general case, the circle that is created by reflecting a parabola with imaginary roots (the orange one) about its vertex (the black one) will have as its radius the value of the imaginary roots of the original.

We will begin with clip_image002[12] as our initial equation, with one requirement that the discriminant is negative;  clip_image004[10]. This will ensure that our initial quadratic equation has imaginary roots and the parabola exists above the x axis as shown.

 

Now, we need to reflect this equation around the vertex, but just adding a negative sign in front of the “a” will not do it. If we add that sign in and make it “-a” it will reflect around the x-axis, not the vertex. Therefore, we are going to need to complete the square, get the original equation in vertex form and then add the minus sign to reflect.

 

Given equation                                                                       clip_image002[13]

 

First, divide all terms by “a” and set the y = 0                              clip_image006 

This gives us a first coefficient of 1, which makes

Completing the square possible. Next, we will complete

the square by using clip_image008 and its square.                        clip_image010

 

now that the perfect square trinomial has been constructed       clip_image012

we can factor the trinomial into vertex form.

 

The center of the circle above can be clearly seen in this form, and is: clip_image014 We will need this later.

 

Now we need to solve the reflected parabola for x.               clip_image016

 

Add & Subtract the constant terms from both sides to get:          clip_image018

 

Move the negative sign from the right to the left side:                 clip_image020

 

Take the square root of both sides:                                       clip_image022

 

Finally subtract the constant term from both sides:                  clip_image024


Notice that we have essentially derived a version of the quadratic formula. It doesn’t look exactly like the standard version we all memorize, but it is the same, with one important difference. There is a sign change to the terms inside the radical sign! That will be very important.

 

This formula gives us where the reflected parabola crosses the x-axis, so we now have 2 points on the circle, the plus and minus, and the center of the circle.

 

The final step of the proof is to show that the radius of the circle, or to put it in another way, the distance from the center of the circle to one of the roots of the reflected parabola, is identical to the imaginary part of the solution / roots of the non-inverted parabola. So, onward to the distance formula.

 

We need to find the distance from clip_image014[1] to clip_image026.

 

Distance formula:                                            clip_image028

Insert the point values for x and y       clip_image030


Using just 1 of the 2 values for the + or -.

 

Simplify the subtractions:                                               clip_image032

 

Finally, square the inside term leaving the following:             clip_image034

 

This leaves us with a pseudo-determinant of:                     clip_image036

 

 

However, in setting up the problem initially, we stipulated that the determinant clip_image038 would be negative. If that is true, then the value of inside the radical sign in our last step must be positive!

 

[And yes, I am cheating. I am leaving it to the reader to show that the way it is written above in the last step as the “pseudo-determinant” and the regular determinant are essentially equivalent.]

 

Not only that, but the value of clip_image040 which is from our inverted quadratic, is the same value but opposite sign of the more familiar clip_image042 from the quadratic equation.  If clip_image038[1] is negative, our inverted quadratic will be positive with the same value (oh, and it works in reverse too!)

 

There, I now proved that the reflecting a parabola with imaginary roots around its vertex will allow you to calculate the imaginary part of the complex answer as the radius of a circle created by the reflection.

 

QED.

Jul 272012
 

For today’s #myfavfriday I am presenting an idea that has been percolating in my head for a while. If you want to know what a #myfavfriday is, then see Druinok’s blog here.

MyFavFri-t

Learners have a devil of a time with quadratics. Afterall, there can be 2 solutions, 1 solution, or no solutions in Algebra 1, and then in Algebra 2, we come at them with the fact that those equations with no solution really do have 2 solutions after all, they are just “imaginary” (could there be a worse name for them, really? Thanks a lot Descarte.”)

But I came across a picture on some site one day, and it has stuck with me. I never bookmarked it, or wrote down the site, so it is lost to me (and I have searched hard for it) but the work blew my mind, and as I have shown it to learners, they have at least gotten a sense that the “imaginary” really does have meaning.

Let’s begin with 2 equations and graphs that are simple, straight-forward and make sense. [all images are clickable to see full size]

graph1and graph2

The equations are y = x^2 – 4x + 3 and y = x^2 – 4x + 4.

A simple change of one number changes the number of solutions from 2 distinct to 2 repeating solutions, and learners don’t have a problem with that idea, generally. Then comes this bad boy.

graph3 y = x^2 –4x + 6

Now they have to do the whole Quadratic formula on it to get the solution, and the solution has those i thingies in it, which makes them all confused and irritated until they wrap their heads around it. And why does it still have 2 solutions? It doesn’t touch at all!

But wait! We can play a game with this quadratic function. What if we reflect the parabola around the vertex in the downward direction? Then we end up with something that looks like this:

graph4a To do this reflection, we first had to complete the square on the original equation to get y = (x-2)^2 + 2. Now, with this equation, we can put the – sign into the equation and get the reflection, y = -(x-2)^2 + 2.

But hold on, see those 2 points where it crosses the X-axis? And see the Axis of Symmetry that goes through both equations? If we use those three points as definitions for a circle, we get the following graph and equation.

graph5 (x-2)^2 + y^2 = 2

Guess what the solution to the quadratic equation y = x^2 –4x + 6 is. If you guessed 2 + root(2)i and 2 – root(2)i  then you are absolutely correct.

The real number part of the complex solution of a quadratic with two imaginary roots is the X value of the Axis of Symmetry, and the imaginary part of the solution is the radius of the circle created by the center and endpoints created when the inverted parabola crosses the X-Axis!

Okay, mind blown. Why? How could I prove this?

Aha! now come into play the hours I spend on a motorcycle every summer. How could I PROVE that this will always work? I have the proof. I am working it up, but it is a pain to type. That, I think, will be the focus of a future, #myfavfriday!

[And I really need to look in to a LaTex module for my blog if I am going to do math. The equations look horrible.]

Edit: 29 July 2012: I proved this assertion, at least to my satisfaction in a followup post: http://blog.mrwaddell.net/archives/348

Edit: 4 August 2012: I found, stuffed in the bottom of my backpack, a rumpled piece of paper with this link on it. I think I did this page justice with my treatment. I wish I had found the page before I spent hours thinking about how to prove it, it gives the suggestion right there at the bottom!

Edit: 18 Dec 2012: @Mythagon posted this picture on Twitter. It is a great visual of what is discussed above, and clearly shows why the rotation is so important.

From: Teaching Mathematics, 2nd edition by m. Sobel and E. Maletsky

Edit: 27 Sep 2015: Wow, a long time since the original post, however I still come back to this every year. Love it. Now, Luke Walsh, aka @LukeSelfwalker added this to the mix. Love it. Click it for the live Desmos file.