Off on a Tangent
A Fortnightly Electronic Newsletter from the Hope College Department of Mathematics
   February 20, 2008 Vol. 6, No. 9  
http://www.math.hope.edu/newsletter.html


Philosophy of Mathematics Colloquium Today

Title:
An Augustinian Perspective on the Philosophy of Mathematics
Speaker:
Prof. James Bradley, Calvin College
Time:
Thursday, February 21 at 4:00 p.m.
Place:
VWF 238

  Abstract:  From a Christian perspective, both modern and post-modern approaches to the philosophy of mathematics have significant shortcomings. We will explore an alternative. We will look at Augustine of Hippo's views on the four classical themes of the philosophy of mathematicsthe ontology of mathematical objects, their epistemology, the nature of truth in mathematics, and how we account for the effectiveness of mathematics in describing the natural world. We will then trace what has happened to Augustine's perspective in the roughly 1600 years since it was written concluding with a discussion of some spiritual and intellectual problems with the currently dominant secular perspective.

Mime-matics

Title:

Mime-matics
Speakers:

Prof. Tim Chartier, Davidson College
Time:

Friday, February 29 at 3:30 p.m.
Place:

Martha Miller Center 135


Next week's colloquium will not only be at a different time and in a different building, it will also be very different from a tradition talk.  Tim Chartier will combine mathematics and mime with his performance of Mime-matics.  Here are some details.

Abstract: In Mime-matics, Tim Chartier explores mathematical ideas through the art of mime. Whether creating an illusion of an invisible wall, wearing a mask covered with geometric shapes or pulling on an invisible rope, Dr. Chartier delves into mathematical concepts such as estimation, tiling, and infinity. Through Mime-matics, audiences encounter math through the entertaining style of a performing artist who has performed at local, national and international settings.

Biography: Dr. Tim Chartier received both a B.S. degree in applied mathematics and a M.S. degree in computational mathematics from Western Michigan University. After doctoral work in applied mathematics at the University of Colorado at Boulder and a VIGRE postdoctoral position at the University of Washington, he arrived at Davidson College in 2003. Professor Chartier's research in numerical analysis and partial differential equations, sometimes in collaboration with Lawrence Livermore National Laboratory and Los Alamos National Laboratory, has been supported by the Department of Energy. Tim Chartier is also a 2007 recipient of the Henry L. Alder Award for Distinguished Teaching by a Beginning College or University Mathematics Faculty Member from the Mathematical Association of America. As an artist, Tim Chartier's training includes master classes with Marcel Marceau.


Scholarship for students in mathematics education available

The Michigan Council of Teachers of Mathematics gives out a number of scholarships annually to students in mathematics education (both elementary and secondary) through the Miriam Schaefer Scholarship Award Program.  Applications are due April 1, 2008.  You can find out more information about the scholarships, including the application process at http://mictm.org/schols_awards_miriam.html.






Math Club T-shirts

An announcement from the Hope College Math Club:

Thanks to everyone who submitted a design for the 2008 Math T-shirt contest!  It was hard to choose a winner among so many excellent and creative designs, but we did!  The 2008 Math T-shirts will feature the slogan: HOW'S MY DERIVING?  1-800-MATH-CLUB.  We'll be taking orders in classes and other locations soon, so please keep an eye out for your opportunity to get this spring's hottest fashion statement -- the Hope College Mathematics T-shirt!

And, as always, we invite you to come to our regular meetings.  The next meeting will be at 7:00 p.m. in VZN 274 on Wednesday, February 27.  If you'd like to be on the email list for Math Club, please email Dr. Pearson at pearson@hope.edu.


The Problem of the Fortnight

Solve the equation

(ln x)2 - 2.5(ln x)(ln(4x-5)) + (ln(4x-5))2 = 0

where x and all expressions in the equation are real.

Attach your solution (not just the answer!) to a small natual log (i.e. a stick) and
drop it by Dr. Pearson's office (VWF 212) by noon on Friday, February 29.  Be sure to write your name, the name(s) of your professor(s), and your math class(es) on your solution (e.g. Al G. Bragh, Prof. Basey, Math 172).  Good luck, and have fun!


Problem Solvers of the Fortnight

The previous POF was: Compute the integral

∫  (x6 + x3) (x3 + 2)1/3 dx

Congratulations to: Chriss Hall, Josh Kinder, Brendan Krueger, Kristian Cunningham, Beth Heisel, Emily West, Eric O'Brien, Eric Lunderberg, Andrea Eddy, Kelsey Bos, Dan Waldo, Ron Radcliffe, Jill Immink, Laura Smallegan, James, Daly, Laura Shears, Kimberly Klask, Elvis, Thao Le, AuSable Schweibert, Mark Pannagio, Megan Pearson, Luke Wendt, Hannah Kasperson, Chris Ploch, Jenny Rautiola, Jessica Clouse, Zach Mitchell, Nate Poel and Stephanie Pasek for determining that the correct answer is 1/8 (x6 + 2x3)4/3 + C.


Special Challenge Problem from Ryan

A few days ago, the following problem appeared in neatly written cursive handwriting outside Dr. Stephenson's door.  We'll put it to you as a special challenge problem.

Solve for x:

x(9/5) + 6/x = 78/10

The problem was posed by Dr. Stephenson's son Ryan, who's a fifth grader.  Although this special challenge problem will not count as a Problem of the Fortnight, we encourage you to drop a solution in the envelope Ryan has created on the bulletin board outside his dad's office.  Ryan will choose a winner randomly from the correct solutions he receives.  We at Off on a Tangent will keep you posted about this special challenge problem contest and announce the winner when Ryan selects one.  (Note: There isn't a due date on the problem, so we encourage you to submit your solution soon!  We'll try to encourage Ryan to accept solutions through the end of February.)

Hello, Hello Mr. Columbian Pi

While some of us have trouble remembering what our office numbers are, Jaime Garcia Serrano (aka the Human Calculator), from Colombia, recited by memory and for hours, random sections of the number pi taken to 150,000 decimal places.  He completed this feat last month at the University of Computense of Madrid.  He is shown here posing with the number pi displayed on a screen and on his face at the University Computense of Madrid.  (Editor's note:  Do you think he ever forgets where he put his car keys?)





Ask Elvis

Hello Elvis,

The Internet is a wonderful thing. Your "Off on a Tangent" newsletter from November 1, 2006 is still out there and I stumbled upon it. I must know if I am correct in my answer to the Problem of the Fortnight: the answer is infinity because the segments never reach zero in length. Correct?

Thanks,
Frank from Massachusetts

Dear Frank from Massachusetts,

The problem you were referring to is as follows:
In the figure below angle AOB has a measure of 15 degrees and the length of segment A1B1 is 4.  Segment AiBi is perpendicular to OB for each i = 1, 2, 3, ...  The lengths of segment AiBi is the same as Ai+1Bi for each i = 1, 2, 3, ...  Find the total length of the zigzag path A1B2A2B2A3B3A4B4 ...  Give your answer in closed form.




One of the interesting things about mathematics is that the sum of an infinite number of numbers (with none of them being zero) can actually be a finite number.  For example, you  probably know that 1/3 can be written in decimal form as 0.3333...  We can also think of this decimal as sum of numbers, namely 

0.3 + 0.03 + 0.003 + 0.0003 + ... 

While the numbers in this infinitely long list never reach zero, its sum is certainly finite.  This works the same in our old problem of the fortnight.  It is true that that the segments never reach zero in length, but the length approaches zero.  The following is our original solution to this problem.

The triangles Ai+1BiBi+1 are all 30-60-90 degree triangles.  This means that the lengths of segment Ai+1Bi+1 = sqrt(3)/2 times the length of segment Ai+1BiThis together with the given information means that the total length of the zigzag path A1B2A2B2A3B3A4B4 .... is


Thanks for your question Frank! 

Your Pal,

Elvis


If anyone else has an new (or old) question for me, don't hesitate to send me an email at elvis@hope.edu.





I'm very well acquainted, too, with matters mathematical
I understand equations, both the simple and quadratical
About binomial theorem I'm teeming with a lot o' news
With many cheerful facts about the square of the hypotenuse.

~ from the Pirates of Penzance by Gilbert and Sullivan (1879)

Off on a Tangent