|Off on a Tangent
|A Fortnightly Electronic Newsletter
from the Hope College Department of Mathematics
Mathematics Colloquium Today
Perspective on the Philosophy of Mathematics
|Prof. James Bradley,
21 at 4:00 p.m.
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 mathematics—the 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.
|Prof. Tim Chartier,
|Friday, February 29 at 3:30
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
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
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.
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
An announcement from the Hope College
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
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
of the Fortnight
Solve the equation
- 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
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!
of the Fortnight
The previous POF was: Compute the
(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.
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
Solve for x:
x(9/5) + 6/x = 78/10
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.)
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?)
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?
Frank from Massachusetts
Dear Frank from Massachusetts,
The problem you were referring to is
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
each i = 1, 2, 3, ... Find the total length of the zigzag path A1B2A2B2A3B3A4B4
... Give your answer in closed form.
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 + ...
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+1Bi. This together with the given information means
that the total length of the zigzag path A1B2A2B2A3B3A4B4 ....
Thanks for your
If anyone else has an new (or old) question for me, don't hesitate to send
me an email at firstname.lastname@example.org.
I'm very well acquainted, too, with matters mathematical
~ from the Pirates of
Penzance by Gilbert and Sullivan
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.