## Archive for the ‘Uncategorized’ Category

### The 2008 Edublog Winners!

December 22, 2008

They were announced this weekend here.  There are plenty of great blogs on the list, including the winner of the Best Group Blog was SCC English:  a blog of the English Department from St. Columbia’s College in Dublin, Ireland.  One other favorite of mine is Extreme Biology:  a class blog where the contrubutors are primarily 9th grade students and AP biology students.  (And now I have to make a conscious choice to stop listing favorites or I might as well just reproduce the entire winers list!)  Congratulations to all!

### Holiday Shopping (to buy)

December 16, 2008

What to get the math folk on your holiday list? There are already a few suggestions on holiday lists (e.g. Walking Randomly has some great books listed), so we’ll just add to the fray.

Originally this was going to combine store-bought and homemade presents, but what with the hecticness (that’s a word, right?) of this time of year, the homemade math things are only partially made. Godzilla has grand plans to show a couple in the next day or so, but for today it’s a sample of What’s On The Internet.

You’re probably first wondering about ornaments. These could hang in your window, on a tree, or even be attached to some other lovely gift. There’s a 72 pages of math ornaments here from The Library of Math (e.g. a round ornament with HO3 on it). Cost: $7.49-$9.99. They appear to have roughly a million other objects (buttons, shirts, etc.) with the same kinds of mathy statements, and that’s a lot of browsing.

But if it doesn’t appeal? Perhaps you’d like to do some baking. In that case, of course, you’ll need some cookie cutters shaped like numbers ($23.95 for the whole set of digits) or like the symbol π (at$12.95).

Or a pie dish with the digits of pi. Cost: $24.95. Or maybe you want an original shirt. Then this isn’t the one for you, because TwoPi already created the design from a public domain photo of Iwan Stranski and put it on CafePress so he could order himself a copy. But you can be one of only a few people who has a shirt or mug with the following image on it. Cost:$17.99 — $21.99, or$11.99 for the mug.

(Note: We find the whole money thing confusing, what with this being a group blog and all, so to make it simple we don’t get any profit from this. Which, sadly, doesn’t make the shirts actually free.)

(Another note: Since TwoPi originally uploaded this so he could order one for himself, he put our address http://threesixty360.wordpress.com on the back of some of the shirts.)

Coming up next: things a lot closer to free!

### We’ve Been Nominated!

December 3, 2008

The polls are open for the 2008 EduBlog Awards, and 360 has been nominated in the “Best Group Blog” category!  Thank you to Maria and Mike and anyone else who nominated us!  Head over and vote in up to 16 categories.

### Generating Pythagorean Triples

November 15, 2008

A recent post at jd2718 noted that for all Pythagorean triples $(a, b, c)$, that is to say, for all positive integer solutions to $a^2 + b^2 = c^2$, it turns out that 60 divides the product $abc$.

In the course of exploring proofs of that result, I suggested a proof using the representation $(2uv, u^2-v^2, u^2+v^2)$, which generates all primitive Pythagorean triples (triples which have no common factors).  Proving that 60 divides their product proves the general result, as every Pythagorean triple has the form $(ka, kb, kc)$, where k is a positive integer and $(a, b, c)$ is primitive.

In the ensuing discussion, a question arose as to how we know that all possible Pythagorean triples can be found in this way.   Since it is easier for me to post LaTeX code to this blog rather than in the comments on the jd2718 blog, I’ll present the proof here.

Suppose that $a^2 + b^2 = c^2$, with positive integers $a, b, c$ having no common factors.  This implies that exactly one of a and  b is even (since if both are even, then so is c, leading to a contradiction, while if both a and b are odd, then both $a^2$ and $b^2$ are congruent to 1 mod 4, while $c^2$ would be congruent to 0 mod 4, not 2 mod 4).

So without loss of generality, assume that a is even.  Thus $a^2 = (c-b)(c+b)$ is also even, and since $(c-b)$ and $(c+b)$ differ by an even number, both of those factors must be even.

Furthermore, exactly one of $(c-b)$ and $(c+b)$ is 2x(an odd number), since if both factors are divisible by 4, it would follow that both b and c would be even (since b and c are $\frac{ (c+b)\pm (c-b)}{2}$.)   Likewise, any odd common factor of $(c-b)$ and $(c+b)$ would also then be a common factor of both b and c, from which it follows that

• $(c+b)$ and $(c-b)$ have no odd common factors
• one of these expressions is of the form $2 u^2$ (for some odd u)
• and hence the other expression is of the form $2 v^2$, where $v^2 = \frac{a^2}{4u^2}$, where u and v are positive integers

It now follows that $c = u^2 + v^2$, and $b = u^2 - v^2$.  We can find a using $a^2=(c+b)(c-b) = (2u^2)(2v^2)$, and thus $a = 2uv$.

Conclusion:  every primitive Pythagorean triple has the form $(2uv, u^2-v^2, u^2+v^2)$, for positive integers u and v.

### A New Approach to Science?

October 28, 2008

Once again we turn to the writers at Robot Chicken to illuminate math and physics (also here if YouTube takes it down):

### Using calculus to generate the quadratic formula

October 19, 2008

I get (unreasonable considerable) joy in finding fancy new proofs of elementary results, proofs that might come under the heading of mathematics made difficult or obscure. (I’ve never seen Linderholm’s book, but suspect it would appeal to a twisted part of my psyche.) I don’t quite understand the psychology behind this liking, but there it is nonetheless.

One of my favorite examples follows: solving quadratic equations using integration to complete the square.

Suppose $f(x)=ax^2+bx+c$, and we want to find the solutions to $f(x)=0$.

Note that $f'(x) = 2ax+b$ and $f(0) = c$, and thus it must be that $f(x) = c+ \int_0^x 2at+b \; dt$.

We compute this antiderivative using the change of variables $w=2at+b, \quad dw = 2a\;dt$, which leads to $c + \int_b^{2x+b} \frac1{2a} \; w \; dw$. This last expression is equal to $c + \left( \frac1{4a} (2ax+b)^2 - \frac{b^2}{4a} \right)$.

Thus the roots of $f(x) = 0$ are found by solving the equation $c + \frac{ (2ax+b)^2 - b^2 }{4a} = 0$.

This leads to $(2ax+b)^2 -b^2 = -4ac$, and thus $2ax+b = \pm \sqrt{b^2 - 4ac}$, and so finally we arrive at the roots $x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$.

Forthcoming: Using Fubini’s Theorem to prove Integration by Parts

### Today in Pi

October 7, 2008

Happy Decimal Digits 23,913,007 through 23,913,015 of Pi day! Yes, at the 23,913,007th digit (not counting the 3) is the string 10/07/2008 [precisely, ...95510072008255...]. Apparently being only 24 million digits in in a bit of a lucky stroke: tomorrow’s date of 10/08/2008 first appears 100 million digits later, at the 124,023,083th place. (See Pi-Search for more such fun.)

Then I stared wondering about e. Sadly, being the less famous relation of π has its drawbacks and e doesn’t have its own search page. There is, however, a list of the first two million digits of e, which includes 1007 for 10/07, but not 10072008. Bummer we weren’t writing this a year ago: the string 10/7/2007 manages to squeeze in towards the end.

(The string 360 is much more popular: it starts at the 285th decimal digit in the expansion of pi, and occurs almost immediately in e: 2.7182818284590452353602874…)

Suppose you want to look for a name instead? Dr. Mike has translated 31,415,929 digits of pi and 27,182,818 digits of e into Base 27, to represent the 26 letters and punctuation. The laws of probability make long names hard to find, and even with all those digits GODZILLA doesn’t appear in the translated numbers (nor does MOTHRA), though a lot of other words do, as he explains at the bottom of each page. Interestingly, Dr. Mike used $\sum \frac{1}{n!}$ in base 27 to estimate the digits of e. Hooray for Calc II and series convergence!

### Fibonacci mileage

September 21, 2008

I ran across a little tidbit about the Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21,34, 55, 89,…. It’s a mere coincidence, but still pretty cool:

As the list of numbers goes on, the ratio of successive terms gets closer and closer to the Golden Ratio ø≈1.618. And another number that’s close to 1.618 is the number of kilometers in a mile (1.609). This means that Fn miles is approximately Fn+1 kilometers, where Fn and Fn+1 are successive Fibonacci numbers.

In other words, if you’re driving at 13 miles per hour, that’s approximately 21 kilometers per hour (actual: 20.9). If you’re driving at 21 miles per hour, that’s approximately 34 kilometers per hour (actual: 33.8).

This way cool observation was found at Futility Closet, found via the Evil Mad Scientist Laboratories August Linkdump.

### Mathematician of the week: Johann Lambert

August 27, 2008

Johann Lambert was born on August 26, 1728. (His father, Lukas Lambert, was a tailor, and was not involved in the discovery of a primality test for Mersenne numbers.)

By the time he was a teenager, Lambert found himself working full-time for his father, while pursuing his academic studies at night. Over the ensuing decade, Lambert held a number of jobs while teaching himself mathematics and science. He first attracted attention on the academic stage with an article on heat, published in 1755. A mere three years later Lambert published a book on the passage of light through various media.

Lambert struggled throughout his life to find academic employment, but in spite of that he was quite prolific, publishing over 150 articles and books by the time of his death at the age of 49.

He had two particularly significant mathematical contributions:

• His Theorie der Parallellinien (1766) was a study of the consequences of the negation of the parallel postulate, including the correlation between the sum of the angles of a triangle and its area in a non-euclidean geometry.
• In 1768, Lambert gave the first rigorous proof of the irrationality of π. Specifically, he showed that if x is a nonzero rational number, then both $e^x$ and $\tan x$ must be irrational, using the continued fraction representations of those two functions.

### Happy Pi Day!

July 22, 2008

Happy Pi Day!!!

22 July is celebrated throughout (much of) the world as Pi Day, for the ratio 22/7 is a reasonably accurate rational approximation to the number π.

Pi Day is also celebrated on March 14, in those parts of the world who would abbreviate today’s date ( July 22, 2008 ) as 7/22/2008, since March 14 becomes 3/14 under such a scheme. According to Wikipedia (“So you know it’s true!”™), only a handful of countries follow this scheme. Most would abbreviate using either a little-endian scheme ( 22/7/2008 ) or a big-endian scheme ( 2008/7/22 ). The amount of space on Wikipedia devoted to a flamewar discussion about the relative merits of each scheme is astounding.

There are many days when I’m happy to be a mathematician, and not a copy editor for an international open content network based encyclopedia.

### How fast was Godzilla running?

June 11, 2008

A bunch of fossilized dinosaur footprints in the Mideast were recently discovered by Mohammed Al-Daheri who, incidentally, is a journalist (so all you folk who wanted to be paleontologists but didn’t get around to it, it’s not too late to make great discoveries!). The footprints looked like these ones:

but were in Yemen, and not in the Science Museum in Logroño, Spain (where jynus took this picture). National Geographic has a neat article about the footprints with even neater pictures here. Part of the article states, “The paleontologists were also able to infer the size, age, and speed of the sauropods based on their prints.”

Size of the sauropod is presumably related to the size of the footprints, and age might be related to size, but how did paleontologists determine the speed? To demonstrate, we’ll use footprints to determine how fast Godzilla was moving in the picture below:

The first thing we can do from the footprint is to determine the Godzilla’s leg length. From fossils and stuff “we” know that most dinosaur hind feet were ¼ of the size of their leg length, so in theory we should be able to measure the footprint and multiply by 4. It turns out, though, that Godzilla has disproportionally large feet, so we’ll cheat and measure his leg length exactly. Our Godzilla’s leg length is 6.5″, which we’ll translate to 0.1651 meters so we can be all metricy like the scientists.

Next we have to measure the stride length.

Exactly 8 inches, which is 0.2032 meters.

Now we find the relative stride length with the following formula:
Relative Stride Length (RSL) = (Stride Length)/(Leg Length).

Godzilla’s Relative Stride Length is about 1.23.

Now it gets complicated. A famous scientist guy in the Biology department at Leeds University, R.M. Alexander, published something about something called Dimensionless Speed. (See how well I understand it? I think it takes into account that similar animals move in a similar way, regardless of size. ) He came up with the following equation:

Speed = (Dimensionless Speed)√(leg length ·g)

The constant “g” is that familiar gravitational constant 9.80665 meters per second per second. Bet you never expected to see that here, right?

So now we just need to figure out the Dimensionless Speed, and we do that using the Relative Stride Length. There’s a graph in Figure 2 of this article, but it didn’t have a category for “Giant Monsters”. If Godzilla were like a human, then a Relative Stride Length of 1.23 would correspond to a Dimensionless Speed of about 0.3. On the other hand, in this activity (an outline of doing a lesson like this in High School Classes) there is a chart of four dinosaurs’ RSL and DS, and plugging those numbers into my calculator gives:

(Dimensionless Speed) = 0.6425(Relative Stride Length)-0.4679

(with a correlation r=.9997, so it’s pretty good).

If we treat Godzilla like a dinosaur, then plugging in his RSL of 1.2308, we get a Dimensionless Speed of 0.3228, which is close to the 0.3 we estimated above. So let’s use that.

Now we plug in DS=0.3228, leg length = 0.1651 meters, and g=9.80665 into
Speed = (Dimensionless Speed)√(leg length·g)

and we find that Godzilla’s speed is a whopping 0.41 meters per second! That’s 41 centimeters per second, which is pretty fast for a guy whose barely a foot tall. To put it into perspective, if you multiply by the scaling factor of 133.3 (from when we calculated Godzilla’s weight), it’s equivalent to the original 50 meter Godzilla running at just over 54 meters per second, or just over 122 miles per hour! Boy, you’d never know that he could run that fast, could you?

Makes you wonder just how fast those Yemeni sauropods were booking, doesn’t it?

I’m thinking that if I’d done the calculations before writing all this up, Godzilla’s footprints would have been placed a little closer together. Thanks to Bill Korth, who demonstrated this method of figuring out dinosaur speeds at a teacher workshop. He found it in Dinosaurs: The Textbook (3rd ed.) by Spencer G. Lucas.

### Mathematician of the week: Paul Guldin

June 8, 2008

Paul Guldin spend much of his teaching career at Jesuit colleges in Rome and Graz, Austria. He is remembered for Guldin’s Theorem: If any plane figure revolve about an external axis in its plane, the volume of the solid body so generated is equal to the product of the area of the figure and the distance traveled by the center of gravity of the figure.

Guldin published this result in his De centro gravitatis (1635-1641), and the result is more widely referred to as Pappus’s Theorem, having been anticipated by Pappus of Alexandria circa 320 CE.

Ivor Bulmer-Thomas writes about the question of whether Guldin had known of Pappus’s work in an article in Isis (vol 75, 1984).  He argues convincingly that Guldin is almost certain to have read Pappus’s work in translation in 1618, and probably unconsciously plagiarized it some 20 years later when completing his own work on centers of gravity.  Bulmer-Thomas’s article debunks a longstanding error in historiography:  Moncula’s 1758 history of mathematics erroneously stated that Pappus’s Theorem had been omitted from the first Latin translation of his work, and consequently Guldin’s discovery must have been arrived at independently.  Moncula corrected this error in the second edition of his history (1799), but the incorrect account has gotten wider dissemination (from Paul Ver Eecke [editor of a modern collection of Pappus's writings], Carl Boyer’s History of Mathematics, and even the Dictionary of Scientific Biography‘s entry for Guldin, for example).

Moral:  as much as possible, check and recheck the claims of your secondary sources, even if they are highly regarded and generally known to be sound.

Mathematicians with birth or death anniversaries during the week of June 8 through June 14:

June 8: Birthdays of Giovanni Cassini [1625] (discovered Cassini gap in Saturn’s rings; introduced use of Jovian moon observations to determine longitude), Caspar Wessel [1745] (“Argand” diagram), and Charlotte Angas Scott [1858] (algebraic geometer; first head of math dept at Bryn Mawr)

June 9: Birthday of John Littlewood [1885] (worked with Hardy on theory of functions); death anniversaries of John Machin [1751] (use of series to compute over 100 digits of π) and Vivienne Malone-Mayes [1995] (asymptotic analysis; fifth African-American woman to earn a PhD in mathematics [in 1966])

June 10: birthdays of Mohammad Abu’l-Wafa [940] (introduction of tangent function; mathematical astronomy) and Pierre Duhem [1861] (mathematical physics); deaths of André-Marie Ampère [1836] (PDEs, chemistry, and physics) and Antonio Cremona [1903] (introduced use of graphical methods in study of statics)

June 11: Birthday of Hilda Hudson [1881] (algebraic geometry); death of Wilhelm Meyer [1934] (invariant theory)

June 12: Birthdays of Paul Guldin [1577] (Guldin’s Theorem) and Vladimir Arnold [1937] (solution of Hilbert’s 13th problem; 2001 recipient of the Wolf Prize); death of Jean Frenet [1900] (differential geometry)

June 13: Birthdays of James Clerk Maxwell [1831] (study of electricity and magnetism), Ernst Steinitz [1871] (abstract algebra), William Gosset [1876] (“Student”‘s t-distribution), and John Nash [1928] (game theory)

June 14: Birthdays of Nilakantha Somayaji [1444] (mathematical astronomy, series for arctangent), Andrei Andreyevich Markov [1856] (probability theory), Alonzo Church [1903] (computability; lambda calculus), and Atle Selberg [1917] (zeta function); death of Colin Maclaurin [1746] (analysis)

Sources:  Ivor Bulmer-Thomas, “Guldin’s Theorem-Or Pappus’s?”  Isis 75 (1984) 348-352 (for information on Guldin’s Theorem and the controversy over its provenance);   the MacTutor History of Mathematics website (for dates and biographical data generally; for the image of Paul Gulden)

### Hard Drive Failure

June 4, 2008

One of my stats students stopped by this afternoon with an interesting question: she needed a new hard drive, and was looking at this one, which boasted an Annual Failure Rate of 0.34%. What she was wondering was how to calculate the chances that the hard drive would fail within, say, 5 or 10 years. Her first guess had been to multiply 0.34% by that number of years, but she didn’t think that was actually the way to go.

My first thought was to break it down into the chances that the hard drive would fail in one year [(0.0034)] and then add on the chances that it didn’t fail the first year but did the second [(0.9966)(0.0034)] and then add on the chances that it didn’t fail the first two years but did the third [(0.9966)2(0.0034)] etc., all the way up to adding on the chances that it didn’t fail the first (N-1) years but did in the Nth year [(0.9966)N-1(0.0034)].

Then I added them all up, and factored out the (0.0034), giving (0.0034)[1+(0.9966)+(0.9966)2+...+(0.9966)N-1]. But 1+x+x2+…+xN-1=(1-xN)/(1-x), so that equation simplified to (0.0034)[1-(0.9966)N]/[1-0.9966],
which further simplified to 1-(0.9966)N. And when I thought about it, that made sense: (0.9966)N is the probability that the hard drive does NOT fail anytime in the first N years, so the probability that it does fail would be 1-(0.9966)N.

Then we started looking at a few actual numbers. The likelihood that the hard drive fails within one year is 0.34%, that it fails within two years is 0.68%, that it fails within three years is 1.02%. Indeed, since 0.9966 is so close to 1 it has a very small effect so the student’s initial thought of N·(0.34%) turns out to be a pretty good estimate for quite a few years [at 10 years out the theoretical probability would be 3.35% compared to 3.4%, and 20 years out the theoretical probability is 6.59% compared to 6.8%.]

This lead me to wondering about another piece of data she’d pointed out: the Mean Time Before Failure (MTBT) was listed as 700,000 hours, which translates to about 79.85 years of continuous play. And where do they get THAT number from, I ask you? Certainly not experimental data. A later thought I had was that it was the number of years before the likelihood of disk failure was 50%, but 1-(0.9966)79.85≈0.238, so there should be less than a 25% chance that a given hard drive fails within 79.85 years. Indeed, 1-(0.9966)N=.5 when (0.9966)N=.5, which happens when N=ln(.5)/ln(0.9966)≈203 years. Hey, why aren’t stores advertising that data? Shouldn’t the Antiques Roadshow have featured a few Revolutionary War hard drives, bought for 25¢ at the neighborhood garage sale?

But then when I was showing a rough draft of this post to TwoPi, he pointed out that that’s the median, not the mean. The mean would be an expected value, which would be the infinite sum ΣN(0.0034)(0.9966)N-1, which is (0.0034)·ΣN(0.9966)N-1. Thinking back to Calc II, the geometric series ΣxN converges to 1/(1-x) for |x|<1; taking the derivative of both sides yields ΣNxN-1 converging to 1/(1-x)2 when |x|<1. This means that (0.0034)·ΣN(0.9966)N-1 just equals (0.0034)·1/(1-0.9966)2, or 1/(0.0034)≈294 years.

OK, all that work and it didn’t clear up anything except to indicate that George Washington could have gotten a hard drive for his birthday and it might still be working.

All in all, I think we got a good handle on the theory and absolutely none on the reality. She sent me a link later this afternoon to this article on a study by Carnegie-Mellon that indicated that the Mean Time to Failure Rates were greatly exaggerated by manufacturers (ya think?), up to 15 times as long as they should be, because they failed to take into account that the #1 cause of failure is age: the likelihood that a hard drive will fail in its 10th year is a lot higher than the likelihood that it will fail in its 2nd year.

In other words, it doesn’t really matter if they say the drive will last 80 years or 800 years, because it won’t. But it’s still worthwhile to buy a new hard drive when the old one goes belly up.

Thanks to Anya for the idea and for the links in this post!

### Carnival Submissions Welcome!

May 28, 2008

Just a reminder that in two days (Friday May 30), we’ll be hosting the 34th Carnival of Mathematics!!  Submissions are welcome via the Official Submit Form, via the comments below, and via email to hlewis5 followed by the @ and then ending with naz.edu (put something like “Carnival” in the subject line if you think of it).

In theory we’ll be putting it together tonight and tomorrow.  In reality, we’ll be working on it tomorrow night and Friday morning, so anything submitted anytime Thursday will certainly be included.

### How much is enough?

May 10, 2008

If you notice a pattern, how many times do you have to check that it works before being certain. Six? Twenty? Two thousand?

Two favorite examples of mine that demonstrate that patterns can continue for a long time before going awry:

The first example is the polynomial f(n)=n2+41n+41. If you plug in any whole number for n from 1 to 40, then f(n) is a prime number: f(1)=83, f(2)=127, all the way to f(40)=3281. But f(41)=3403=41·83 is not prime. So something that works 40 times in a row might fail.

The second example are the cyclotomic polynomials. Look at polynomials of the form xn-1 that have been factored:

x-1=(x-1)
x2-1=(x-1)(x+1)
x3-1=(x-1)(x2+x+1)
x4-1=(x-1)(x+1)(x2+1)
x5-1=(x-1)(x4+x3+x2+x+1)
x6-1=(x-1)(x+1)(x2+x+1)(x2-x+1)

The last polynomial in each case is the cyclotomic polynomial of order n. [It has a much more technical definition using imaginary numbers and the product of primitive roots of unity]. And at first glance it looks like the coefficients are 0, 1, or -1. Even at second glance or sixth glance, since it’s true for the first 104 cyclotomic polynomials. But not the 105th.

x105-1=(x-1)•(x2+x+1)•(x4+x3+x2+x+1)•
(x6+x5+x4+x3+x2+x+1)•(x8-x7+x5-x4+x3-x+1)•
(x12-x11+x9-x8+x6-x4+x3-x+1)•
(x24-x23+x19-x18+x17-x16+x14-x13+x12-x11
+x10-x8+x7-x6+x5-x+1)•
(x48+x47+x46-x43-x42-2x41-x40-x39+x36+x35
+x34+x33+x32+x31-x28-x26-x24-x22-x20+x17+x16
+x15+x14+x13+x12-x9-x8-2x7-x6-x5+x2+x+1)

See those two coefficients of 2 in that last polynomial? So the pattern of coefficients being only 0, 1, or -1 fails. Interestingly, the reason for this initial failure occurring so late in the game is that 105 is the smallest number that has three distinct odd prime factors (105=3·5·7). The integer 385 is also a product of three distinct odd primes (385=3·5·11) and the 385th cyclotomic polynomial is the first one to have a 3 as a coefficient (see Wolfram Mathworld).

Patterns. You just can’t trust them.