# Monthly Archives: March 2012

## A little-known series test

I just encountered this one on March 26, 2012 AP Calculus EDG courtesy of Doug Kuhlmann with the original article from the Mathematics Magazine, Vol 57 No.4 September 1984.  The article is available here.

I thought it was cool enough to repost here. The wording is slightly reworked and LaTeXed from Doug’s AP Calculus EDG posting.

Suppose f is a function such that $f(\frac{1}{n})=a_n$ and $f''(0)$ exists.  Then $\sum_{n=1}^{\infty}a_n$ converges absolutely if $f(0)=0$ and $f'(0)=0$ and diverges otherwise.

Here are two examples:

1. If $a_n=\frac{1}{n^2}$ then $f(x)=x^2$.  In this case, $f(0)=0$ and $f'(0)=0$, therefore $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges.
2. If $a_n=\frac{1}{n}$ then $f(x)=x$.  Since $f'(0)=1$, therefore $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges.

## Developing fractional understanding

Here’s another installment of my infrequent commentaries on my eldest daughter’s math development, this time on an unexpected result when we were talking about equivalent fractions.

I’ve been a firm believer in keeping challenging ideas, games, etc. around my children at all times, some of them intentionally beyond what they’re developmentally ready to handle.  Sometimes I’ll ask leading questions to see if there is any interest; sometimes they pick up an idea or toy again and create their own play rules or ask me to explain how it works.  My ground rule is that they should PLAY.  If my explanations ever bore them, they are welcome to drop it at any time and move on to something more interesting.  Not only does this keep with my mantra that learning should be fun (even if it involves work), but I believe it helps them see that mathematics (and anything else they learn) is about enjoyment and pushing yourself to discover more than what exists within the current boundaries of your understanding.

Several weeks ago

With that philosophy in mind, my 2nd grade daughter had been helping me make some fresh bread one afternoon when we needed a cup of one ingredient, but our cup measurer was dirty.  Whether from school or one of our earlier conversations, she responded something like, ‘No problem.  Just use 2 half-cups.‘  Maybe she actually said  ‘3 third-cups,’ but she clearly had the beginnings of fractions down, so I jumped.

So if I needed a half cup of something and my half-cup measurer was dirty, what else could I use?

It took some conversations, but eventually she drew a circle with a line through the center and shaded one side.  When she drew another line through the center roughly bisecting the original sectors, she declared, “Look, Dad.  Cutting each part in half doesn’t change what you have, it just cuts it into more pieces.  So, $\frac{1}{2}$ must be the same as $\frac{2}{4}$.”

That was a pretty cool moment of discovery for her, but then she upped the ante.  After some additional thought, she noticed that both parts of the fraction had been doubled, so she applied her rule again and asked if $\frac{1}{2}$ would also be the same as $\frac{4}{8}$.  Another drawing confirmed her discovery which led to gleeful proclamations that $\frac{1}{2}=\frac{2}{4}=\frac{4}{8}=\frac{8}{16}=\frac{16}{32}$. She knew she could go on, but what was the point?

I asked if doubling was the only way she could make equivalent fractions, which led to $\frac{1}{2}=\frac{3}{6}=\frac{9}{18}$. We didn’t go any further that day, but we had already “cooked up” far more than I had anticipated.  Off the top of my head, I don’t recall how elementary school curricula deal with the scope and sequence of teaching equivalent fractions, but it will be difficult for anyone to convince me that my daughter could have had a better experience or more fun learning.

Last weekend

I was putting my daughter to bed and for some reason she asked how much of a year 9 months was.  Being the teacher, I responded to her question with another question:  “Well, how many months are in a year?

So 9 months is $\frac{9}{12}$ of a year?” She asked.

Good job.  Can you think of any other smaller fractions that might be the same as $\frac{9}{12}$?” It was an innocent question, I thought, trying to get her to take our fraction doubling-trebling idea from earlier in reverse. If she didn’t get it, no big deal, but it was certainly worth asking.  That’s when I got surprised.

Almost immediately, she said, “Four and a half sixths.  Is that right?

Some purists out there might complain that $\frac{4.5}{6}$ isn’t “proper,” but I’ve seen far too many situations where rigid insistence on proper form served instead to stifle creativity far more than to enhance understanding or to encourage deeper exploration or creativity.  I praised the heck out of her solution, letting her know that she had just made a fraction “smaller” for the first time (that I knew of).”  I didn’t mention that her proposed numerator wasn’t whole.  It didn’t matter.

How did you do that?

Easy,” she said, as if her answer would have been obvious to anyone.  “If you can double the parts of a fraction, why can’t you halve them, too?

Why not, indeed?  No matter what they end up looking like.

I asked if there were any other smaller, equivalent fractions.  That took lots more thought and time than I expected, certainly more than her nearly instantaneous $\frac{4.5}{6}$ had required.  Eventually, she asked if I could hold out 2 fingers beside her 10 so that she could look at 12 fingers together.  A little more thought led to her grouping the 12 fingers into 4 equal groups and a claim that $\frac{9}{12}=\frac{3}{4}$.  I still don’t completely understand her long explanation, but it wasn’t as clear (to me, anyway) that she was saying that 3 of her sub-groups contained the original 9 fingers (or months), and so 9 of an original 12 individual months was the same as 3 of 4 equal sub-groups of months.

When she woke the next morning, she asked if we could play any more fraction games

Conclusion

As students of all ages are learning, we need to reserve space for them to think and be creative.  They should be allowed to give correct mathematical solutions, even if those answers aren’t the arithmetic solutions our “trained” minds expect.  For me, I expected my daughter to say $\frac{9}{12}=\frac{3}{4}$, if she answered at all.  I was far more delighted to hear $\frac{9}{12}=\frac{4.5}{6}$ than she will ever know, and I’m convinced that she’s becoming more confident and capable because she was allowed to do so.

## Running into Math

Here’s a real-world math problem I just found.

For the last two years, the AJC Peachtree Road Race in Atlanta, the “World’s Largest 10K” (it happens every July 4th), has been using a lottery system to determine which non-invited runners get race numbers.

To accommodate those who would like to participate in the AJC Peachtree Road Race with their family and friends, the lottery registration system allows groups of up to 10 people to enter the lottery as a “Group”.  During the selection process, if a “Group” is selected everyone in the group will receive an entry.  If a “Group” is not selected through the lottery, no one in the group will receive entry into the event.  Those entering the lottery as a group have an equal chance of getting into the event as those entering as individuals (source, emphasis added).

Assume a full group of 10 runners enters as a group.  If any 1 runner in the group is selected in the lottery, every runner in the group gets a race number even if no one else in the group is chosen.  On the surface, this seems like it ought to give a runner a better chance of getting a lottery number if entering as a group.  But … the organizers claim that individuals seeking race numbers have an equal probability of getting into the race whether entering solo or in a group.  So how do they do it?

I didn’t find this problem at the right point in my class’ curriculum sequence this year (I get that I raise lots of rightfully debatable curriculum & teaching issues here), but maybe it will work for one of you.  Even so, I’m trying to create a 10-15 minute gap in an upcoming class to give this problem as a “cool (or real) math moment” that I have from time to time in my courses.  If I can get some student results, I’ll post them here.  I’ll provide links/posts from here to any pages or tweets that tackle this.  Enjoy.

## An unexpected lesson from technology

This discovery happened a few years ago, but I’ve just started ‘blogging, so I guess it’s time to share this for the “first” time.

I forget whether my calculus class at the time was using the first version of the TI-Nspire CAS or if we were still on the TI-89, but I had planned a very brief introduction to the CAS syntax for computing symbolic derivatives, but my 5-minute introduction in the first week of introducing algebraic rules of derivatives ended up with my students discovering antiderivative rules simply because they had technology tools which allowed them to explore beyond where their teacher had intended them to go.

They had absolutely no problem computing algebraic derivatives of power functions, so the following example was used not to demonstrate the power of CAS, but to give easily confirmed outputs.  I asked them for the derivative of $x^5$, and their CAS gave the top line of the image below.

(In case there are readers who are TI-Nspire CAS users who don’t know the shortcut for computing higher order derivatives, use the left arrow to place the cursor after the dx in the “denominator” of $\frac{d}{dx}$ and press the carot (^) key.  Then type the integer of the derivative you want.)

I wanted my students to compute the 2nd and 3rd derivatives and confirm the power rule which they did with the following screen.

That was the extent of what I wanted at the time–to establish that a CAS could quickly and easily confirm algebraic results whether or not a “teacher” was present.  Students could create as many practice problems as were appropriate for themselves and get their solutions confirmed immediately by a non-judgmental expert.  Of course, one of my students began to explore in ways my “trained” mind had long ago learned not to do.  In my earlier days of CAS, I had forgotten the unboundedness of mathematical exploration.

Shortly after my syntax 5-minutes had passed and I had confirmed everyone could handle it, a young man called me to his desk to show me the following.

He understood what a 1st or 2nd derivative was, but what in the world was a negative 1st derivative?  Rather than answering, I posed it to the class who pondered a few moments before recognizing that “underivatives” (as they called them in that moment) of power functions added one to the current exponent before dividing by the new exponent.  They had discovered and explained (at least algebraically) antiderivatives long before I had intended.  Technology actually inspired and extended my students’ learning!

Then I asked them what the CAS would give if we asked it for a 0th derivative.  It was another great technology-inspired discussion.

I really need to explore more about the connections between mathematics as a language and the parallel language of technology.

## Math doesn’t happen the way it’s printed

Learning is messy.  Real math is messy.

If you think about it, there shouldn’t be any wonder why students learning a math idea for the first time get frustrated when the work they produce looks almost nothing like the pristine, sharp, and short solutions provided by textbooks and their classroom teachers.

Mathematics is discovered through pattern recognition or trial-and-error or pushing the boundaries of a situation or equation to explore those great “What if…” questions.  Discovery usually is just plain messy, but by the time mathematical writing gets into publication, it has been edited and refined far beyond its messy origins.  We sweep under the publishing rug all of the mistakes and dead ends that taught us so much, delivering our final results all dressed up in tight, pithy expressions.

My latest reminder of this was an exploration of the product rule in my calculus classes this week.  I’ll give a short-and-sweet “textbook” proof of this at the end of this post, but I’ll start with their first exploration of discovering a rule.

At the start of yesterday’s class, my students understood

• the definition of the derivative,
• if a function is differentiable at $x=a$, then sufficiently zooming in on the function at that point essentially shows the tangent line to the function at that point,
• the derivative rule of power functions, and
• if a function was horizontally translated then its derivative experienced the same horizontal translation.

So how do you get the product rule for some $y=f(x)*g(x)$ from that?

Assuming f and g are differentiable at some point $x=a$, then $f(x) \approx f'(a)*(x-a)+f(a)$ and $g(x) \approx g'(a)*(x-a)+g(a)$ for values of x near $x=a$.  Therefore,

$\frac{d}{dx}(f(x)*g(x)) \approx \frac{d}{dx}([f'(a)*(x-a)+f(a)][g'(a)*(x-a)+g(a)])$
$\approx \frac{d}{dx}(f'(a)g'(a)(x-a)^2+(f'(a)g(a)+f(a)g'(a))(x-a)+f(a)g(a))$
$\approx 2f'(a)g'(a)(x-a)+(f'(a)g(a)+f(a)g'(a))+0$

The derivatives of $(x-a)^2$ and $(x-a)$ are the translated derivatives of $x^2$ and $x$–possibilities with their early knowledge without any need for the chain rule.

Therefore, $\frac{d}{dx}(f(x)*g(x))|_{x=a}=$
$=2f'(a)g'(a)(a-a)+(f'(a)g(a)+f(a)g'(a))$
$=f'(a)g(a)+f(a)g'(a)$, the product rule!

It was not the easiest or cleanest of early investigations for my students yesterday.  AND there was lots of potential fudging around the concept of local linearity. BUT … now that the rule has been named, it can be proved  more formally, something we attempted today.
By definition of the derivative,
$\displaystyle \frac{d}{dx}(f(x)*g(x))=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}$
But if an $f'(x)$ term is to emerge from this, then the difference quotient must contain an isolated $\frac{f(x+h)-f(x)}{h}$ term.  The only $f(x+h)$ in the original derivative expression also contains an $g(x+h)$ term, so subtracting $f(x)g(x+h)$ will allow the $g(x+h)$ to factor out.  To balance that subtraction, $f(x)g(x+h)$ also needs to be added back.

From there, the proof becomes$\displaystyle \frac{d}{dx}(f(x)*g(x))=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}$
$\displaystyle =\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}$
$\displaystyle =\lim_{h\to 0}g(x+h)*\frac{f(x+h)-f(x)}{h}+f(x)*\frac{g(x+h)-g(x)}{h}$
$\displaystyle =f'(x)g(x)+f(x)g'(x)$

It’s pretty, but the addition and subtraction of $f(x)g(x+h)$ is completely black box or black magic if you don’t have a reason to do it.  Hopefully, the reasoning above provides such a reason, but it is a result of deep reasoning.  But that’s the reality of most published mathematics, and THAT is what students see as their production expectation the first time they try.  And THAT is one reason why many students get frustrated with mathematics.  The work you produce when you are learning is rarely (if ever) so pretty.

Students need room to be creative, they need room to experiment, and they need to know the math they produce doesn’t need to look pretty when first created.

## CAS Derivatives

Here’s a fun problem from my calculus class today, enhanced by CAS. As a set-up, our last unit focused on interpreting the meaning of derivatives with multiple interpretations of the definition of the derivative as the only algebraic work they’ve done.  In that unit, the students discovered that vertical translations on functions didn’t change their derivatives, and horizontal translations on functions changed the corresponding derivatives by the same horizontal translation.  From their work with derivatives of power functions using a definition of the derivative, they hypothesized and proved $\frac{d}{dx}(x^n)=n*x^{n-1}$ for natural (and a few other) values of n.  Knowing nothing else, I posed this.

Use your CAS, determine the derivatives of $y=ln(x)$, $y=ln(2x)$, $y=ln(3x)$, and $y=ln(4x)$.  Use your results to hypothesize the derivative of $y=ln(n*x)$.  Justify your claim.

The following image from the first part shows that the pattern is easy to spot.

Unfortunately, I posed the problem with only 10 minutes remaining in class, but the students clearly knew $\frac{d}{dx}(ln(n*x))=\frac{1}{x}$, but the looming question wasWHY.  With a couple minutes to spare, one guessed rules of logarithms might apply, but not having used them since their first semester exam, he didn’t recall the property.  Some colleagues may argue that I should have insisted on my students having those rules memorized in advance, but I firmly believe that this particular problem actually gave a reason for my students to relearn their logarithm properties.

I let the awkward moment hang there until another called out with glee,
$ln(n*x)=ln(n)+ln(x)$“, to which a third exclaimed, “and $ln(n)$ is a constant, making the derivative of $ln(n*x)$ the same as the derivative of
$ln(x)$,” clearly using her understanding of the effect of translations on functions and their derivatives that she learned in the last unit.

Two other nice ideas emerged:

1. They thought it convenient that $\frac{d}{dx}(ln(n*x))=\frac{1}{x}$, but now really want to know why.
2. A few observant ones noticed that $ln(x)$ and $\frac{1}{x}$ have different domains.  To these, I pointed out the warning at the bottom of the CAS screen above that most had completely overlooked when getting their initial answers.

These questions still linger for the class, but I argue that the use of CAS in my calculus class this afternoon

• emphasized the importance of logarithm properties in a more meaningful way than any pleas from me could have done,
• left a need in many to discover why $\frac{d}{dx}(ln(n*x))=\frac{1}{x}$, and
• raised domain issues that ultimately will lead to a deeper understanding for the existence of the absolute value in $\int{\frac{1}{x}dx}=ln(|x|)$.

It was a good end to our first day back after Spring Break and left my students wanting to know more.  What a great place to leave a group of students, thanks to CAS.

## Screencasting some calculus

OK, no CAS here, but the power of screencasting allows one to add a voice-over to problems that are difficult to capture in the linear text that predominates Web-based writing.  The following two problems were requests from March 6-7, 2012 on the AP Calculus EDG.  My solutions are screencasts using the ShowMe Whiteboard app on my iPad.  Thanks to my former student (now technology consultant) Beth Holland for the lead to that great app.

1. A region is defined by $y=e^x$, $x=1$, $x=8$, and the x-axis.  What is the volume of the solid that results when that region is revolved around the line $y=4$?  Click here for my solution.

As an aside, colleague John Burk (@occam98) pointed me to a cool article, Volumes of Revolution & 3D Modeling, that quite rightly identifies the most challenging part about volumes of revolutions for students:  visualization.  This is another reason why we need to make greater use of screencasting or other technologies to get past the walls of words.

1. A curve is defined parametrically by $x=t^2+1$ and $y=4t^3+3$.  What is the length of that curve for $-1\le t \le 0$?  Click here for my solution.

I find that many students (and some teachers) have difficulty sorting out the different forms of arclength.  My screencast describes very briefly why I believe students should work from the single arclength formula, $\int \sqrt{dx^2+dy^2}$.

## CAS Publications

After this past weekend’s 2012 T3 International Conference, it’s clear that several folks are interested in CAS-related materials, but just don’t know where to look.

I’ve updated the Resources Page on this ‘blog to include some sites, links, publications, and research, but I know there’s LOTS MORE I haven’t included or mentioned.  If you have any CAS-related resources, please email me with a link or reference, and I’ll update the resources page.

Let’s spread the word about the power of CAS to enhance student learning and mathematics exploration.

## An unexpected identity

Typical high school mathematics considers only two types of transformations:  translations & scale changes.  From a broader perspective, a transformation is an operation that changes its input into an output.  A function is also an operation that changes inputs.  Therefore, functions and transformations are the same thing from different perspectives.  This post will explore an unexpected discovery Nurfatimah Merchant and I made when applying the squaring function (transformation) to trigonometric functions, an idea we didn’t fully realize until after the initial publication of PreCalculus Transformed.

When a function is transformed, some points are unchanged (invariant) while others aren’t.  But what makes a point invariant in a transformation?  From a function perspective, point a is invariant under transformation T if $T(a)=a$.  Using this, a squaring transformation is invariant for an input, a, when $a^2=a\Rightarrow a*(a-1)=0 \Rightarrow a=\{0,1\}$.

Therefore, input values of  0 and 1 are invariant under squaring, and all other inputs are changed as follows.

• Negative inputs become positive,
• $a^2 for any $0, and
• $a^2>a$ for any $a>1$.

So what happens when the squaring transformation is applied to the graph of $y=sin(x)$ (the input) to get the graph of $y=(sin(x))^2$ (the output)?  Notice that the output of $sin(x)$ is the input to the squaring transformation, so we are transforming y values.  The invariant points in this case are all points where $y=0$ or $y=1$.  Because squaring transforms all negative inputs into positive outputs, the first image shows a dashed graph of $y=sin(x)$ with the invariant points marked as black points and the negative inputs made positive with the absolute value function.

All non-invariant points on $y=|sin(x)|$ have magnitude<1 and become smaller in magnitude when squared, as noted above.  Because the original x-intercepts of $y=sin(x)$ are all locally linear, squaring these creates local “bounce” x-intercepts on the output function looking locally similar to the graphs of polynomial double roots.  The result is shown below.

While proof that the final output is precisely another sinusoid comes later, the visual image is very compelling.  This looks like a graph of $y=cos(x)$ under a simple scale change ($S_{0.5,-0.5}$) and translation ($T_{0,0.5}$), in that order, giving equation $\displaystyle\frac{y-0.5}{-0.5}=cos(\frac{x}{0.5})$ or $y=\frac{1}{2}-\frac{1}{2}cos(2x)$.  Therefore,

$sin^2(x)=\frac{1}{2}-\frac{1}{2}cos(2x)$.

We later rewrote this equation to get

$cos(2x)=1-2sin^2(x)$.

The initial equation was a nice enough exercise, but what we realized in the rewriting was that we had just “discovered” the half-angle identity for sine and a double-angle identity for cosine using a graphical variation on the squaring transformation!  No manipulation of angle sum identities was required!  (OK, they really are for an honest  proof, but this is pretty compelling evidence.)

Apply the squaring transformation to the graph of $y=cos(x)$ and you get the half-angle identity for cosine and another variation on the double-angle identity for cosine.

We thought this was a nice way to sneak up on trigonometric identities.  Enjoy.