## Tuesday, April 10, 2007

### Bringing a 'Series' of Wonders to the Calculus Classroom

The following may drive away most casual readers but it does describe what I try to do every day. One of my goals in starting this blog was to enable a dialogue for effective instructional strategies. My focus has generally been on middle and secondary school curriculum up to Algebra 2, bordering on Precalculus. Today I am sharing a different experience. I hope some of you will appreciate it beyond its technical aspects. Similar developments can be found in some textbooks and experienced teachers already do most of this but as this scenario is fresh in my mind, I thought I'd re-play it for you...

Although most Advanced Placement Calculus (BC) teachers are completing or have already completed the unit on infinite series, I would like to offer a view that I hope brings a sense of 'shock and awe' to the student of the 21st century who rarely has the time to stop and appreciate the beauty of our subject. To those who have been teaching this for a while, you may not quite feel this. However, I still get goosebumps when I observe student reactions as this unfolds in front of their eyes...

Assume that students already have a basic understanding of infinite series, the infinite geometric series in particular.

Consider the following three infinite geometric series:

1+1/2+1/4+1/8+... = 1/(1-1/2) = 2
1+1/3+1/9+1/27+... = 1/(1-1/3) = 3/2
1-1/4+1/16-1/64+... = 1/(1-(-1/4)) = 4/5

Just a collection of simple geometric series, boys and girls?
Genius is looking at an ordinary collection of objects and seeing something different. Some mathematician or mathematicians (research this and report back with their bios!) may have considered a reverse view of these series. Instead of the goal being a formula for the sum of the series, perhaps the goal was to represent a function in a different way. Step back into history...

Consider the general formula for the sum of all these series: 1/(1-r) provided r is between -1 and 1. Replace r by x, the variable we usually use for functions, and we can write:
1 + x + x2 + x3 +... = 1/(1-x) provided x is between -1 and 1.

The 'polynomial' of infinite degree on the left is known as a power series in x. As long as x is between -1 and 1 (the interval of convergence), this 'equation' makes sense and allows us to use algebraic and calculus operations to represent other related functions. Think of how one might have felt when 'discovering' this and I'm just speculating here. The rational function 1/(1-x) is being represented by some kind of polynomial that never ends. Even though x= 1 is not in the interval of convergence, substituting leads to 1/0 = 1 + 1 + 1 + 1 +.... Hmm....
Let's try substitution on this representation.
Replace x by -x2:
(You can show the domain is unchanged)
1/(1 - (-x2)) = 1 + (-x2) + (-x2)2 + (-x2)3 + ... OR
1/(1 + x2) = 1 - x2 + x4 - x6 +...

Ok, let's integrate both sides (assuming it's legal to do so):
tan-1(x) = x - (1/3)x3 + (1/5)x5 - (1/7)x7 + ... + C
Replacing x by 0, we see that C = 0.
Now, you'll have to accept this for the moment (to be proved later), equality holds for x = 1, even though 1 was not in the original interval of convergence! It is not unusual when integrating a power series to see the domain include one or both endpoints even though the original function excluded them!

Thus, tan-1(1) = 1 - 1/3 + 1/5 - 1/7 + ...
Anyone recognize the left-hand side?
The bell rings...

Eric Jablow said...

You might also want to give equal time to −1 in the first series, leading to log 2. Thinking historically, you may want to discuss summability and early attempts to find meaning for divergent series.

Raman said...

thanks ... that was calm and elegant ... all the things we like about math

Dave Marain said...

raman--
thanks!

eric--
I do both of those (including the issue of 1-1+1-1+1-1+... and how one could make a case for various 'sums' in the absence of a rigorous definition, Cesaro sums and the like) but I didn't describe EVERY teachable moment in this post. Unfortunately, the sheer beauty of this topic gets crunched by the deadline for AP's. Since this is my favorite unit, I do much more than is required.

Nick said...

when i took calculus BC my teacher did not go too much in depth about sequences and series (which happened to be my favorite subject at the time) she just basically went over the different test we needed to know. But I learned more interesting things on the subject outside to class. Also you might want to note that the method you showed for finding pi over 4 converges very slowly.

Dave Marain said...

nick--
Today we did just that! I asked them to find the sum of the first several terms and the students recognized that the approximation was poor; since the TI-83/84 can find the sum of a series of a finite number of terms (up to about 1000 terms) they could see quite clearly how slowly the partial sums converged. I suggested they research other series representations for pi on MathWorld and Wikipedia that are far more useful, but this demonstration had a telling effect on them anyway. I know that my passion for this topic can be infectious.
I appreciate the interest this posting is generating. It tells me we could do so much more with this topic if we had the time!

Totally_clueless said...

By using x^2/3 instead of x^2, you can come up with a series for pi/6 which, on first glance, is quite different from the one you get by just multiplying the current series by 2/3.

TC

Dave Marain said...

Nice, tc!
Here's another variation that leads to an interesting sidebar:
Consider f(x) = 1/(x+2)
Method I: Rewrite as (1/2)/(1+(x/2)), which leads to the infinite geometric series:
(1/2)[1-(x/2)+(x^2)/4-(x^3)/8+...]=
(1/2)[1-(1/2)x +(1/4)x^2 - (1/8)x^3
+...]
The interval of convergence for this power series in x would be
-2 < x < 2.

Method II: Rewrite f(x) as
1/(1+(x+1)) which leads to the series:
1 - (x+1) + (x+1)^2 - (x+1)^3 -...
with interval of convergence
-2 < x < 0.
To a student this seems paradoxical, that is, 2 different series with 2 different intervals of convergence for the same function.

Let's evaluate both series when
x = -1 which is common to both intervals.
The first series gives
(1/2)[1+1/2+1/4+1/8+1/16+...] =
(1/2)(2) = 1.

The second series also evaluates to 1 since it's a power series around x = -1! Perhaps not paradoxical to us, but interesting nonetheless. When teaching a topic like this in depth these are questions that naturally arise. We need to think about these ideas ourselves...

Eric Jablow said...

In complex analysis, we have this example:

Consider …+z^{−n} + … + z^{−1} + 1 + z + z^{2} + … + z^{n} + …

The terms with nonnegative exponents converge to 1/(1 − z) in some domain. Replace z with 1/z and multiply by 1/z to get that the terms with negative exponents converge to −1/(1 − z) in some domain. The combined sum would be zero, except that the two domains are disjoint.