How Old is the Oldest Person You Know?

The Prudential commercial that aired during Super Bowl 47 features what Steven Strogatz calls the most viewed histogram of all time.

According to the commercial people were asked the age of the oldest person they know, and their answers were plotted.  The resulting histogram is somewhat “normal” looking, and the average age is in the low 90s.

prudential histogram

The commercial’s message is clear:  “Look at how old people get!  You need to be better prepared for your retirement!  Come see a Prudential representative today.”

This is a good example of the subtle ways mathematics can be used to manipulate the opinions of the quantitatively unsophisticated.

The above histogram is intentionally designed to mislead viewers into thinking they may be significantly unprepared for retirement.  The average life expectancy in the US is around 78 years, but this number may not be shocking enough for advertisitng purposes.  So instead of life expectancy, Prudential used age of the oldest person you know, a data set whose average is about 15 years higher.

Showing a histogram that suggests people are likely to live into their 90s might motivate some viewers to head down to their local Prudential office, worried that they aren’t properly prepared for retirement.  But the data on display here isn’t really relevant, and the difference is so subtle that most people won’t notice the distinction.  In reality, the age of the oldest person you know has very little to do with how long you will live.

Imagine asking each member of a large group to name the salary of the highest-paid person they know.  The average of these responses, the average highest-known-salary, will almost certainly be much higher than the average salary of the people in the group.  It would be ridiculous to try to estimate the average salary of the group by looking at the average highest-known-salary, but in a sense, that is exactly what Prudential is doing in this commercial.

The fact that they are doing it intentionally to further their interests provides yet another example of  the vital need for quantitative literacy in today’s world.

By MrHonner, ago

Decomposing Functions into Even and Odd Parts

even and odd function decompositionWhen it comes to functions, the concepts even and odd have always been important to me as a teacher. Connecting the algebraic and geometric representations of mathematical ideas is a primary goal in my classroom, and these concepts provide great opportunities to do that.

Algebraically, a function is even if f(-x) = f(x), and this condition manifests itself geometrically as symmetry with respect to the y-axis in the graph of y = f(x). A function is odd if f(-x) = -f(x), and geometrically this means that the graph of y = f(x) is symmteric with respect to the origin. Knowing a function is even or odd provides a wealth of information to work with, and can make solving some problems trivially easy.

But it wasn’t until recently that I learned the following amazing fact: Functions can essentially be uniquely decomposed into even and odd parts!

Claim:  Let f(x) be a non-zero, real-valued function whose domain is symmetric about the origin; that is, f(x) exists implies f(-x) exists. Then f(x) can be uniquely expressed as the sum of an even function and an odd function.

Proof:  For any function f(x) \neq 0, define the functions a(x) and b(x) in the following way:

a(x) = \frac{f(x)+f(-x)}{2}   and   b(x) = \frac{f(x)-f(-x)}{2}

First, we see that

a(x) + b(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2} = \frac{2f(x)}{2} = f(x).

Next, since a(-x) = \frac{f(-x)+f(x)}{2} = a(x), we have that a(x) is even.

Similarly, since b(-x) = \frac{f(-x)-f(x)}{2} = -\frac{(f(x)-f(-x))}{2} = -b(x), we have that b(x) is odd. Thus, f(x) can be expressed as the sum of an even function and an odd function.

Now, suppose f(x) could written as the sum of an even and an odd function in two ways:

f(x) = a_{1}(x) + b_{1}(x) = a_{2}(x) + b_{2}(x)

A little algebra gives us

a_{1}(x) - a_{2}(x) = b_{2}(x) - b_{1}(x)

Since the sum of even functions is even and the sum of odd functions is odd, we have an even function, a_{1}(x) - a_{2}(x), equal to an odd function, b_{2}(x) - b_{1}(x). The only function that is both even and odd is the zero function (another fun proof!), therefore

a_{1}(x) - a_{2}(x) = b_{2}(x) - b_{1}(x) = 0

and so

a_{1}(x) = a_{2}(x)

b_{1}(x) = b_{2}(x)

Thus, this representation of f(x) is unique. (Note: since 0 is both even and odd, we can consider f(x) = f(x) + 0 to be the unique decomposition in case f(x) is itself even or odd.)

I was fortunate to encounter this unfamiliar fact at a time when hyperbolic trig functions were on my mind, which made it obvious to me where the hyperbolic sine and cosine functions come from: They are the even and odd parts of e^x!

e^x = \frac{e^{x} - e^{-x}}{2} + \frac{e^{x} + e^{-x}}{2} = sinh(x) + cosh(x)

I also used this fact in a fun but inefficient proof that the derivative of an even function is an odd function.

Are there are other cool consequences of this unique decomposition of functions?

Related Posts

By MrHonner, ago

Math Art: Building Sines

This is Building Sines, one of the pieces I will have on display at the 2013 Bridges Math and Art conference in Enschede, the Netherlands.

Building SinesBuilding Sines, by Patrick Honner

Inspired by the mathematicians, computer scientists, and artists of the Bridges organization, I have started writing computer programs to alter and transform my photography in mathematical ways.  Here, a custom Python script was used to smoothly compress the original image along a vertical sine wave, creating an interesting visual effect.

You can see the other pieces I will have on display here, and you can peruse the entire Bridges 2013 gallery here.  I will also be presenting a short paper at the conference about my work and its potential as a project-based learning activity for teachers and students.

 

By MrHonner, ago
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