Measuring Calculator Speed

Two weeks ago, my summer school Algebra 2 students were exploring sequences and series.  A problem I thought would be a routine check on students’ ability to compute the sum of a finite arithmetic series morphed into an experimental measure of the computational speed of the TI-Nspire CX handheld calculator.  This experiment can be replicated on any calculator that can compute sums of arithmetic series.

PHILOSOPHY

Teaching this topic in prior years, I’ve found that sometimes students have found series sums by actually adding all of the individual sequence terms.  Some former students have solved problems involving  addition of more than 50 terms, in sequence order, to find their sums.  That’s a valid, but computationally painful approach. I wanted my students to practice less brute-force series manipulations.  Despite my intentions, we ended up measuring brute-force anyway!

Readers of this ‘blog hopefully know that I’m not at all a fan of memorizing formulas.  One of my class mantras is

“Memorize as little as possible.  Use what you know as broadly as possible.”

Formulas can be mis-remembered and typically apply only in very particular scenarios.  Learning WHY a procedure works allows you to apply or adapt it to any situation.

THE PROBLEM I POSED AND STUDENT RESPONSES

Not wanting students to add terms, I allowed use of their Nspire handheld calculators and asked a question that couldn’t feasibly be solved without technological assistance.

The first two terms of a sequence are t_1=3 and t_2=6.  Another term farther down the sequence is t_k=25165824.

A)  If the sequence is arithmetic, what is k?

B)  Compute \sum_{n=1}^{k}t_n where t_n is the arithmetic sequence defined above, and k is the number you computed in part A.

Part A was easy.  They quickly recognized the terms were multiples of 3, so t_k=25165824=3\cdot k, or k=8388608.

For Part B, I expected students to use the Gaussian approach to summing long arithmetic series that we had explored/discovered the day before.   For arithmetic series, rearrange the terms in pairs:  the first with last, the second with next-to-last, the third with next-to-next-to-last, etc..  Each such pair will have a constant sum, so the sum of any arithmetic series can be computed by multiplying that constant sum by the number of pairs.

Unfortunately, I think I led my students astray by phrasing part B in summation notation.  They were working in pairs and (unexpectedly for me) every partnership tried to answer part B by entering \sum_{n=1}^{838860}(3n) into their calculators.  All became frustrated when their calculators appeared to freeze.  That’s when the fun began.

Multiple groups began reporting identical calculator “freezes”; it took me a few moments to realize what what happening.  That’s when I reminded students what I say at the start of every course:  Their graphing calculator will become their best, most loyal, hardworking, non-judgemental mathematical friend, but you should have some concept of what you are asking it to do.  Whatever you ask, the calculator will diligently attempt to answer until it finds a solution or runs out of energy, no matter how long it takes.  In this case, the students had asked their calculators to compute values of 8,388,608 terms and add them all up.  The machines hadn’t frozen; they were diligently computing and adding 8+ million terms, just as requested.  Nice calculator friends!

A few “Oh”s sounded around the room as they recognized the enormity of the task they had absentmindedly asked of their machines.  When I asked if there was another way to get the answer, most remembered what I had hoped they’d use in the first place.  Using a partner’s machine, they used Gauss’s approach to find \sum_{n=1}^{8388608}(3n)=(3+25165824)\cdot (8388608/2)=105553128849408 in an imperceptable fraction of a second.  Nice connections happened when, minutes later, the hard-working Nspires returned the same 15-digit result by the computationally painful approach.  My question phrasing hadn’t eliminated the term-by-term addition I’d hoped to avoid, but I did unintentionally create reinforcement of a concept.  Better yet, I got an idea for a data analysis lab.

LINEAR TIME

They had some fundamental understanding that their calculators were “fast”, but couldn’t quantify what “fast” meant.  The question I posed them the next day was to compute \sum_{n=1}^k(3n) for various values of k, record the amount of time it took for the Nspire to return a solution, determine any pattern, and make predictions.

Recognizing the machine’s speed, one group said “K needs to be a large number, otherwise the calculator would be done before you even started to time.”  Here’s their data.

NspireTime1

They graphed the first 5 values on a second Nspire and used the results to estimate how long it would take their first machine to compute the even more monumental task of adding up the first 50 million terms of the series–a task they had set their “loyal mathematical friend” to computing while they calculated their estimate.

NspireTime2

Some claimed to be initially surprised that the data was so linear.  With some additional thought, they realized that every time k increased by 1, the Nspire had to do 2 additional computations:  one multiplication and one addition–a perfectly linear pattern.  They used a regression to find a quick linear model and checked residuals to make sure nothing strange was lurking in the background.

NspireTime4

The lack of pattern and maximum residual magnitude of about 0.30 seconds over times as long as 390 seconds completely dispelled any remaining doubts of underlying linearity.  Using the linear regression, they estimated their first Nspire would be working for 32 minutes 29 seconds.

NspireTime3

They looked at the calculator at 32 minutes, noted that it was still running, and unfortunately were briefly distracted.  When they looked back at 32 minutes, 48 seconds, the calculator had stopped.  It wasn’t worth it to them to re-run the experiment.  They were VERY IMPRESSED that even with the error, their estimate was off just 19 seconds (arguably up to 29 seconds off if the machine had stopped running right after their 32 minute observation).

HOW FAST IS YOUR NSPIRE?

The units of the linear regression slope (0.000039) were seconds per k.  Reciprocating gave approximately 25,657 computed and summed values of k per second.  As every increase in k required the calculator to multiply the next term number by 3 and add that new term value to the existing sum, each k represented 2 Nspire calculations.  Doubling the last result meant their Nspire was performing about 51,314 calculations per second when calculating the sum of an arithmetic series.

NspireTime5

My students were impressed by the speed, the lurking linear function, and their ability to predict computation times within seconds for very long arithmetic series calculations.

Not a bad diversion from unexpected student work, I thought.

Infinite Ways to an Infinite Geometric Sum

One of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to \displaystyle \frac{g}{1-r} for first term g and common ratio r when \left| r \right| < 1.  I was glad she was dissatisfied with blind use of a formula and dove into a familiar (to me) derivation.  In the end, she shook me free from my routine just as she made sure she didn’t fall into her own.

STANDARD INFINITE GEOMETRIC SUM DERIVATION

My standard explanation starts with a generic infinite geometric series.

S = g+g\cdot r+g\cdot r^2+g\cdot r^3+...  (1)

We can reason this series converges iff \left| r \right| <1 (see Footnote 1 for an explanation).  Assume this is true for (1).  Notice the terms on the right keep multiplying by r.

The annoying part of summing any infinite series is the ellipsis (…).  Any finite number of terms always has a finite sum, but that simply written, but vague ellipsis is logically difficult.  In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series.  You can accomplish this by continuing the pattern on the right:  multiplying both sides by r

r\cdot S = r\cdot \left( g+g\cdot r+g\cdot r^2+... \right)

r\cdot S = g\cdot r+g\cdot r^2+g\cdot r^3+...  (2)

This seems to make make the right side of (2) identical to the right side of (1) except for the leading g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical.  After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.

(1)-(2) = S-S\cdot r = S\cdot (1-r)=g

\displaystyle S=\frac{g}{1-r}

STUDENT PREFERENCES

I despise giving any formula to any of my classes without at least exploring its genesis.  I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.

In my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number prefer the quick “multiply-by-r-and-subtract” approach used to derive the summation formula.  For many, apparently, the dynamic manipulation is more meaningful than a static rule.  It’s very cool to watch student preferences at play.

K’s VARIATION

K understood the proof, and then asked a question I hadn’t thought to ask.  Why did we have to multiply by r?  Could multiplication by r^2 also determine the summation formula?

I had three nearly simultaneous thoughts followed quickly by a fourth.  First, why hadn’t I ever thought to ask that?  Second, geometric series for \left| r \right|<1 are absolutely convergent, so K’s suggestion should work.  Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “r^2 formula” looked like, it had to be algebraically equivalent to the standard form.  While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K’s question and the equivalence from Thought #3.

Multiplying (1) by r^2 gives

r^2 \cdot S = g\cdot r^2 + g\cdot r^3 + ... (3)

and the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to get the inevitable result.

(1)-(3) = S-S\cdot r^2 = g+g\cdot r

S\cdot \left( 1-r^2 \right) = g\cdot (1+r)

\displaystyle S=\frac{g\cdot (1+r)}{1-r^2} = \frac{g\cdot (1+r)}{(1+r)(1-r)} = \frac{g}{1-r}

That was cool, but this success meant that there were surely many more options.

EXTENDING

Why stop at multiplying by r or r^2?  Why not multiply both sides of (1) by a generic r^N for any natural number N?   That would give

r^N \cdot S = g\cdot r^N + g\cdot r^{N+1} + ... (4)

where the ellipses of (1) and (4) are again identical by the method of Footnote 2.  Subtracting (4) from (1) gives

(1)-(4) = S-S\cdot r^N = g+g\cdot r + g\cdot r^2+...+ g\cdot r^{N-1}

S\cdot \left( 1-r^N \right) = g\cdot \left( 1+r+r^2+...+r^{N-1} \right)  (5)

There are two ways to proceed from (5).  You could recognize the right side as a finite geometric sum with first term 1 and ratio r.  Substituting that formula and dividing by \left( 1-r^N \right) would give the general result.

Alternatively, I could see students exploring \left( 1-r^N \right), and discovering by hand or by CAS that (1-r) is always a factor.  I got the following TI-Nspire CAS result in about 10-15 seconds, clearly suggesting that

1-r^N = (1-r)\left( 1+r+r^2+...+r^{N-1} \right).  (6)

geometric1

Math induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS.  From there, dividing both sides of (5) by \left( 1-r^N \right) gives the generic result.

\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right)}{\left( 1-r^N \right)}

\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right) }{(1-r) \cdot \left( 1+r+r^2+...+r^{N-1} \right)} = \frac{g}{1-r}

In the end, K helped me see there wasn’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways.  Nice.

FOOTNOTES

1) RESTRICTING r:  Obviously an infinite geometric series diverges for \left| r \right| >1 because that would make g\cdot r^n \rightarrow \infty as n\rightarrow \infty, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum.

For r=1, the sum converges iff g=0 (a rather boring series). If g \ne 0 , you get a sum of an infinite number of some nonzero quantity, and that is always infinite, no matter how small or large the nonzero quantity.

The last case, r=-1, is more subtle.  For g \ne 0, this terms of this series alternate between positive and negative g, making the partial sums of the series add to either g or 0, depending on whether you have summed an even or an odd number of terms.  Since the partial sums alternate, the overall sum is divergent.  Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.

2) NOT ALL INFINITIES ARE THE SAME:  There are two ways to show two groups are the same size.  The obvious way is to count the elements in each group and find out there is the same number of elements in each, but this works only if you have a finite group size.  Alternatively, you could a) match every element in group 1 with a unique element from group 2, and b) match every element in group 2 with a unique element from group 1.  It is important to do both steps here to show that there are no left-over, unpaired elements in either group.

So do the ellipses in (1) and (2) represent the same sets?  As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant.  For the second approach using pairing, we need to compare individual elements.  For every element in the ellipsis of (1), obviously there is an “partner” in (2) as the multiplication of (1) by r visually shifts all of the terms of the series right one position, creating the necessary matches.

Students often are troubled by the second matching as it appears the ellipsis in (2) contains an “extra term” from the right shift.  But, for every specific term you identify in (2), its identical twin exists in (1).  In the weirdness of infinity, that “extra term” appears to have been absorbed without changing the “size” of the infinity.

Since there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero.

CAS Presentations at USACAS-9

I had two presentations at last Saturday’s USACAS-9 conference at Hawken School in Cleveland, OH.  Following are outline descriptions of the two sessions with links to the PowerPoint, pdf, and .tns files I used.  I’m also adding all of this information to the Conference Presentations tab of this ‘blog.

Powerful Student Proofs

This session started with a brief introduction to a lab that first caught my eye at the first USACAS conference years ago.

You know how the graph of y=ax^2+bx+c behaves when you vary a and c, but what happens when you change b?

I ‘blogged on this problem here and here.  In the session, we used TI-Nspire file QuadExplore.

Next, we explored briefly the same review of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors.  In the session, we used TI-Nspire file Intro Polar.

Having a complete grasp of polar graphs of limacons, cardioids, rose curves, and hybrids of these, I investigated what would happen for curves of the family r=cos \left( \frac{\theta}{k} \right).  Curiously, for k=3, I encountered a curve that looked like a horizontal translation of limacons–something that just shouldn’t happen within polar coordinates.

Polar1

One of my former students, Sara, used a CAS to convert a polar curve to Cartesian, translate the curve, and convert back to polar.  She then identified and solved a trig identity to confirm what the graph suggested.  A complete description of Sara’s proof is below.  I originally ‘blogged on Sara’s work here which was a much more elegant solution to the problem than my initial attempt.  It’s always cool when a student’s work is better than her teacher’s!  I used TI-Nspire file Polar Fractions in Saturday’s session.

The last example presented itself when I created a document to model the family of conic curves resulting from manipulating the coefficients of Ax^2+Bxy+Cy^2+Dx+Ey+F=0.  After I created  dynamic points for the foci of the conics, something unusual happened when the E parameter for horizontal ellipses and hyperbolas varied.

Image5b

Image5a

The foci for hyperbolas followed an ellipse, and the locus of elliptical foci appeared to be a hyperbola.  Another former student, Lilly, proved this property to be true.  A detailed explanation of Lilly’s proof is below.  We were fortunate to have Lilly’s work published in the Mathematics Teacher in May, 2014.

To demonstrate this final part of the session, I used TI-Nspire file Hidden Conic Behavior.

Here is my PowerPoint file for Powerful Student Proofs.  A more detailed sketch of the session and the student proofs is below.

Bending Asymptotes, Bouncing Off Infinity, and Going Beyond

The basic proposal was that adding the Reciprocal transformation to the palette of constant dilations and translations dramatically simplified understanding of the behavior of rational functions around even and odd vertical asymptotes (bouncing off and passing through infinity).  Just like lead coefficients of polynomials determine their end behavior, so, too, do the lead coefficients of proper rational expressions define the end behavior of rational functions.

Extending the idea of reciprocating and transforming functions, you can quickly explain exponential decay from exponential growth, derive the graphs of y=\frac{1}{x} and y=\frac{1}{x^2}, and completely explain why logistic functions behave the way they do.

We finished with a quick exploration of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors.

I used TI-Nspire Bending and Intro Polar files in the demonstration.  Here is my outline PowerPoint file for Bending Asymptotes.

Graphing Ratios and Proportions

Last week, some colleagues and I were pondering the difficulties many middle school students have solving ratio and proportion problems.  Here are a few thoughts we developed to address this and what we think might be an uncommon graphical extension (for most) as a different way to solve.

For context, consider the equation \displaystyle \frac{x}{6} = \frac{3}{4}.

(UNFORTUNATE) STANDARD METHOD:

The default procedure most textbooks and students employ is cross-multiplication.   Using this, a student would get

\displaystyle 4x=18 \longrightarrow x = \frac{18}{4} = \frac{9}{2}

While this delivers a quick solution, we sadly noted that far too many students don’t really seem to know why the procedure works.  From my purist mathematical perspective, the cross-multiplication procedure may be an efficient algorithm, but cross-multiplication isn’t actually a mathematical function.  Cross-multiplication may be the result, but it isn’t what happens.

METHOD 2:

In every math class I teach at every grade level, my mantra is to memorize as little as possible and to use what you know as broadly as possible.  To avoid learning unnecessary, isolated procedures (like cross-multiplication), I propose “fraction-clearing”–multiplying both sides of an equation by common denominatoras a universal technique in any equation involving fractions.  As students’ mathematical and symbolic sophistication grows, fraction-clearing may occasionally yield to other techniques, but it is a solid, widely-applicable approach for developing algebraic thinking.

From the original equation, multiply both sides by common denominator, handle all of the divisions first, and clean up.  For our example, the common denominator 24 will do the trick.

\displaystyle 24 \cdot \frac{x}{6} = 24 \cdot \frac{3}{4}

4 \cdot x = 6 \cdot 3

\displaystyle x = \frac{9}{2}

Notice that the middle line is precisely the result of cross-multiplication.  Fraction-clearing is the procedure behind cross-multiplication and explains exactly why it works:  You have an equation and apply the same operation (in our case, multiplying by 24) to both sides.

As an aside, I’d help students see that multiplying by any common denominator would do the trick (for our example, 12, 24, 36, 48, … all work), but the least common denominator (12) produces the smallest products in line 2, potentially simplifying any remaining algebra.  Since many approaches work, I believe students should be free to use ANY common denominator they want.   Eventually, they’ll convince themselves that the LCD is just more efficient, but there’s absolutely no need to demand that of students from the outset.

METHOD 3:

Remember that every equation compares two expressions that have the same measure, size, value, whatever.  But fractions with differing denominators (like our given equation) are difficult to compare.  Rewrite the expressions with the same “units” (denominators) to simplify comparisons.

Fourths and sixths can both be rewritten in twelfths.  Then, since the two different expressions of twelfths are equivalent, their numerators must be equivalent, leading to our results from above.

\displaystyle \frac{2}{2} \cdot \frac{x}{6} = \frac{3}{3} \cdot \frac{3}{4}

\displaystyle \frac{2x}{12} = \frac {9}{12}

2x=9

\displaystyle x = \frac{9}{2}

I find this approach more appealing as the two fractions never actually interact.  Fewer moving pieces makes this approach feel much cleaner.

UNCOMMON(?) METHOD 4:  Graphing

A fundamental mathematics concept (for me) is the Rule of 4 from the calculus reform movement of the 1990s.  That is, mathematical ideas can be represented numerically, algebraically, graphically, and verbally.  [I’d extend this to a Rule of 5 to include computer/CAS representations, but that’s another post.]  If you have difficulty understanding an idea in one representation, try translating it into a different representation and you might gain additional insights, or even a solution.  At a minimum, the act of translating the idea deepens your understanding.

One problem many students have with ratios is that teachers almost exclusively teach them as an algebraic technique–just as I have done in the first three methods above.  In my conversation this week, I finally recognized this weakness and wondered how I could solve ratios using one of the missing Rules: graphically.  Since equivalent fractions could be seen as different representations of the slope of a line through the origin, I had my answer.

Students learning ratios and proportions may not seen slope yet and may or may not have seen an xy-coordinate grid, so I’d avoid initial use of any formal terminology.  I labeled my vertical axis “Top,” and the horizontal “Bottom”.  More formal names are fine, but unnecessary.  While I suspect most students might think “top” makes more sense for a vertical axis and “bottom” for the horizontal, it really doesn’t matter which axis receives which label.

In the purely numeric fraction in our given problem, \displaystyle \frac{x}{6} = \frac{3}{4}, “3” is on top, and “4” is on the bottom.  Put a point at the place where these two values meet.  Finally draw a line connecting your point and the origin.

ratio1

The other fraction has a “6” in the denominator.  Locate 6 on the “bottom axis”, trace to the line, and from there over to the “top axis” to find the top value of 4.5.

ratio2

  Admittedly, the 4.5 solution would have been a rough guess without the earlier solutions, but the graphical method would have given me a spectacular estimate.  If the graph grid was scaled by 0.5s instead of by 1s and the line was drawn very carefully, this graph could have given an exact answer.  In general, solutions with integer-valued unknowns should solve exactly, but very solid approximations would always result.

CONCLUSION:

Even before algebraic representations of lines are introduced, students can leverage the essence of that concept to answer proportion problems.  Serendipitously, the graphical approach also sets the stage for later discussions of the coordinate plane, slope, and linear functions.  I could also see using this approach as the cornerstone of future class conversations and discoveries leading to those generalizations.

I suspect that students who struggle with mathematical notation might find greater understanding with the graphical/visual approach.  Eventually, symbolic manipulation skills will be required, but there is no need for any teacher to expect early algebra learners to be instant masters of abstract notation.

Student Quadratic Creativity

I’m teaching Algebra 2 this summer for my school.  In a recent test on quadratic functions, I gave a question I thought would be a little different, but still reachable for those willing to make connections or exert a little creativity.

Write a system of quadratic functions that has exactly one solution:  (1,1).

Their handheld graphing calculators were allowed.  Some students definitely had difficulty with the challenge, some gave a version of the answer I expected, and one adopted a form I knew was possible, but doubted anyone would actually find during a test situation.

I show my students’ solutions below.  But before you read on, can you give your own solution?

SOLUTION ALERT!  Don’t read further if you want to find your own solution.


WHAT I EXPECTED

We’ve had many discussions in class about the power of the Rule of 4–that math ideas can be expressed numerically, graphically, algebraically, and verbally.  When you get stumped in one representation, being able to shift to a different form is often helpful.  That could mean a different algebraic representation, or a different Rule of 4 representation altogether.

The question is phrased verbally asking for an algebraic answer.  But it asks about a solution to a system of equations.  I hoped my students would recall that the graphical version of a system solution is equivalent to the point(s) where the graphs of the equations intersected.  In my mind, the easiest way to do this is to write quadratic functions with coincident vertices.  And this is most easily done in vertex form.  The cleanest answer I ever got to this question was

quad1

A graphical representation verifies the solution.

quad3

Another student recognized that if two parabolas shared a vertex, but had different “slopes”, their only possible point of intersection was exactly the one the question required.  Here’s a graphical version of her answer.

quad2

From these two, you can see that there is actually an infinite number of correct solutions.  And I was asking them for just one of these!  :)

WHAT I KNEW, BUT DIDN’T EXPECT

Another way to solve this question makes use of the geometry of quadratic graphs.  If two quadratics have the same leading coefficients, they are the same graph, intersect exactly once, or never intersect.  This is a very non-trivial idea for most students.  While I’m not convinced the author of the following solution had this in mind when he answered the question, his solution works because of that fact.  Here’s what J wrote on last week’s test and its graph.

quad4

quad5

J used more equations than he needed, but had he restricted himself to just two equations, I’m not sure the lovely pattern would have been so obvious.

This is a very different (and super cool) answer than what I expected my students to produce.  Lesson re-learned:  Challenge your students, give them room to express creativity and individuality, and be prepared to be amazed by them.

NEXT STEPS

J’s answer actually opens the door to other avenues of exploration.

  1. Can you generalize the form of all of J’s equations, essentially defining a family of quadratics?  Can you prove that all members of your generalization satisfy the question posed and that no other answers are possible?
  2. Can you find forms of other generalized families of quadratic functions whose only solution is (1,1)?
  3. Notice that there were two types of solutions above:  A) those with coincident vertices and different lead coefficients and B) those with identical lead coefficients and different vertices.  Are these the only types of quadratics that can answer this question?  That is, is there a system of quadratics with (1,1) as the only solution that have identical vertices and lead coefficients?  Could both be different and (1,1) be the only solution?
  4. If I relax the requirement that the quadratics be functions, what other types of quadratics are possible?  [This could be a very nice calculus question!]

For my part, I’m returning to some of these questions this week to stretch and explore my student’s creativity and problem-solving.

I’d love to hear what you or your students discover.

Innumeracy and Sharks

Here’s a brief snippet from a conversation about the recent spate of shark attacks in North Carolina as I heard yesterday morning (approx 6AM, 7/4/15) on CNN.

George Burgess (Director, Florida Program for Shark Research):  “One thing is going to happen and that is there are going to be more [shark] attacks year in and year out simply because the human population continues to rise and with it a concurrent interest in aquatic recreation.  So one of the few things I, as a scientist, can predict with some certainty is more attacks in the future because there’s more people.”

Alison Kosik (CNN anchor):  “That is scary and I just started surfing so I may dial that back a bit.”

This marks another great teaching moment spinning out of innumeracy in the media.  I plan to drop just those two paragraphs on my classes when school restarts this fall and open the discussion.  I wonder how many will question the implied, but irrational probability in Kosik’s reply.

TOO MUCH COVERAGE?

Burgess argued elsewhere that

Increased documentation of the incidents may also make people believe attacks are more prevalent.  (Source here.)

It’s certainly plausible that some people think shark attacks are more common than they really are.  But that begs the question of just how nervous a swimmer should be.

MEDIA MANIPULATION

CNN–like almost all mass media, but not nearly as bad as some–shamelessly hyper-focuses on catchy news banners, and what could be catchier than something like ‘Shark attacks spike just as tourists crowd beaches on busy July 4th weekend”?  Was Kosik reading a prepared script that distorts the underlying probability, or was she showing signs of innumeracy? I hope it’s not both, but neither is good.

IRRATIONAL PROBABILITY

So just how uncommon is a shark attack?  In a few minutes of Web research, I found that there were 25 shark attacks in North Carolina from 2005-2014.  There was at least one every year with a maximum of 5 attacks in 2010 (source).  So this year’s 7 attacks is certainly unusually high from the recent annual average of 2.5, but John Allen Paulos reminded us in Innumeracy that [in this case about 3 times] a very small probability, is still a very small probability.

In another place, Burgess noted

“It’s amazing, given the billions of hours humans spend in the water, how uncommon attacks are,” Burgess said, “but that doesn’t make you feel better if you’re one of them.”  (Source here.)

18.9% of NC visitors went to the beach (source) .  In 2012, there were approximately 45.4 million visitors to NC (source).  To overestimate the number of beachgoers, Let’s say 19% of 46 million visitors, or 8.7 million people, went to NC beaches.  Seriously underestimating the number of beachgoers who enter the ocean, assume only 1 in 8 beachgoers entered the ocean.  That’s still a very small 7 attacks out of 1 million people in the ocean.  Because beachgoers almost always enter the ocean at some point (in my experiences), the average likely is much closer to 2 or fewer attacks per million.

To put that in perspective, 110,406 people were injured in car accidents in 2012 in NC (source).  The probability of getting injured driving to the beach is many orders of magnitude larger than the likelihood of ever being attacked by a shark.

CONCLUSIONS AND READING SUGGESTIONS

Alison Kosik should keep up her surfing.

If you made it to a NC beach safely, enjoy the swim.  It’s safer than your trip there was or your trip home is going to be.  But even those trips are reasonably safe.

I certainly am not diminishing the anguish of accident victims (shark, auto, or otherwise), but accidents happen.  But don’t make too much of one either.  Be intelligent, be reasonable, and enjoy life.

In the end, I hope my students learn to question facts and probabilities.  I hope they always question “How reasonable is what I’m being told?”

Here’s a much more balanced article on shark attacks from NPR:
Don’t Blame the Sharks For ‘Perfect Storm’ of Attacks In North Carolina.

Book suggestions:
1)  Innumeracy, John Allen Paulos
2) Predictably Irrational, Dan Ariely

Next Steps from a Triangle

Watching the news a couple mornings ago, an impossible triangle appeared on the screen.  Hopefully some readers might be able to turn some first ideas a colleague and I had into a great applied geometry lesson.  What follows are some teacher thoughts.  My colleagues and I hope to cultivate classes where students become curious enough to raise some of these questions themselves.

CNN_triangle

WHAT’S WRONG?

At first glance, the labeling seems off.  In Euclidean geometry, the Triangle Inequality says the sum of the lengths of any two sides of a triangle must exceed the length of the third side.  Unfortunately, the shorter two sides sum to 34 miles, so the longest side of 40 miles seems physically impossible.  Someone must have made a typo.  Right?

But to dismiss this as a simple typo would be to miss out on some spectacular mathematical conversations.  I’m also a big fan of taking problems or situations with prima facie flaws and trying to recover either the problem or some aspects of it (see two of previous posts here and here).

WHAT DOES APPROXIMATELY MEAN?

Without confirming any actual map distances, I first was drawn to the vagueness of the approximated side lengths.  Was it possible that this triangle was actually correct under some level of round-off adjustment?  Hopefully, students would try to determine the degree of rounding the graphic creator used.  Two sides are rounded to a multiple of 10, but the left side appears rounded to a nearest integer with two significant digits.  Assuming the image creator was consistent (is that reasonable?), that last side suggests the sides were rounded to the nearest integer.  That means the largest the left side could be would be 14.5 miles and the bottom side 20.5 miles.  Unfortunately, that means the third side can be no longer than 14.5+20.5=35 miles.  Still not enough to justify the 40 miles, but this does open one possible save.

But what if all three sides were measured to the nearest 10 instead of my assumed ones place?  In this case the sides would be approximately 10, 20, and 40.  Again, this looks bad at first, but a 10 could have been rounded from a 14.9, a 20 from a 24.9, making the third side a possible 14.9+24.9=39.8, completely justifying a third side of 40.    This wasn’t the given labeling, but it would have potentially saved the graphic’s legitimacy.

GEOMETRY ALTERNATIVE

Is there another way the triangle might be correct?  Rarely do pre-collegiate geometry classes explore anything beyond Euclidean geometry.  One of my colleagues, Steve, proposed spherical geometry:

Does the fact that the earth is round play a part in these seemingly wrong values (it turns out “not really”… Although it’s not immediately clear, the only way to violate the triangle inequality in spherical geometry is to connect point the long way around the earth. And based on my admittedly poor geographical knowledge, I’m pretty sure that’s not the case here!)

SHORTEST DISTANCE

Perhaps students eventually realize that the distances involved are especially small relative to the Earth’s surface, so they might conclude that the Euclidean geometry approximation in the graphic is likely fine.

Then again, why is the image drawn “as the crow flies”?  The difficult mountainous terrain in upstate New York make surface distances much longer than air distances between the same points.  Steve asked,

in the context of this problem (known location of escaped prisoners), why is the shortest distance between these points being shown? Wouldn’t the walking/driving distance by paths be more relevant?  (Unless the prisoners had access to a gyrocopter…)

The value of a Euclidean triangle drawn over mountainous terrain has become questionable, at best.

FROM PERIMETER TO AREA

I suspect the triangle awkwardly tried to show the distances the escapees might have traveled.  Potentially interesting, but when searching for a missing person in the mountains–the police and news focus at the time of the graphic–you don’t walk the perimeter of the suspected zone, you have to explore the area inside.

A day later, I saw the search area around Malone, NY shown as a perfect circle.  (I wish I had grabbed that image, too.).  Around the same time, the news reported that the search area was 22 square miles.

  • Was the authorities’ 22 measure an approximation of a circle’s area, a polygon based on surface roads, or some other shape?
  • Going back to the idea of a spherical triangle, Steve hoped students would ask if they could “compute that from just knowing the side lengths? Is there a spherical Herons Formula?”
  • If the search area was a more complicated shape, could you determine its area through some sort of decomposition into simpler shapes?  Would spherical geometry change how you approach that question?  Steve wondered if any students would ask, “Could we compute that from just knowing the side lengths? Is there a spherical Herons Formula?
  • At one point near the end of the search, I hear there were about 1400 police officers in the immediate vicinity searching for the escapee.  If you were directing the search for a prison escapee or a lost hiker, how would you deploy those officers?  How long would it take them to explore the entire search zone?  How would the shape of the potential search zone affect your deployment plan?
  • If you spread out the searchers in an area, what is the probability that an escapee or missing person could avoid detection?  How would you compute such a probability?
  • Ultimately, I propose that Euclidean or spherical approximations seriously underestimated the actual surface area?  The dense mountainous terrain significantly complicated this search.  Could students extrapolate a given search area shape to different terrains?  How would the number of necessary searchers change with different terrains?
  • I think there are some lovely openings to fractal measures of surface roughness in the questions in the last bullet point.

ERROR ANALYSIS

Ultimately, we hope students would ask

  • What caused the graphic’s errors?  Based on analyses above and some Google mapping, we think “a liberal interpretation of the “approximately” label on each leg might actually be the culprit.”  What do the triangle inequality violations suggest about round-off errors or the use of significant digits?
  • The map appeared to be another iteration of a map used a few days earlier.  Is it possible that compounded rounding errors were partially to blame?
  • Surely the image’s designer new the triangle was an oversimplification of the reality.  Assuming so, why was this graphic used anyway?  Does it have any news value?  Could you design a more meaningful infographic?

APPRECIATION

Many thanks to Steve Earth for his multiple comments and thoughts that helped fill out this post.