Mr Honner

Speed Limit Infimum

I claim that this is the minimum maximum speed. Have you ever seen anything lower?

By MrHonner, 13 yearsJune 7, 2013 ago

Math Quiz — NYT Learning Network

Through Math for America, I am part of an ongoing collaboration with the New York Times Learning Network. My latest contribution, a Test Yourself quiz-question, can be found here

Test Yourself Math — June 3, 2013

This question is related to California’s projected $4.4 billion budget surplus. What is the per capita share of that amount?

By MrHonner, 13 yearsJune 4, 2013 ago

Application Statistics

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.

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, 13 yearsMay 28, 2013 ago

Appreciation Photography

Math Photo: Cylindrical Cone

By MrHonner, 13 yearsMay 26, 2013 ago

Appreciation Geometry

Decomposing Functions into Even and Odd Parts

When 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

Derivatives of Even Functions

By MrHonner, 13 yearsMay 21, 2013 ago

Speed Limit Infimum

Math Quiz — NYT Learning Network

How Old is the Oldest Person You Know?

Math Photo: Cylindrical Cone

Decomposing Functions into Even and Odd Parts

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