## Archive for the ‘Miscellaneous’ Category

### Average percentage

July 21, 2009

There was a show on TV late last night, but I don’t know what the show was or, for that matter, what channel it was on.  Two men and two women had been trying to lose weight in some boys vs. girls competition, so they added up the weight loss for each pair, divided by their previous weight, and compared.  They guys had lost more weight, but the gals had lost a higher percentage so they won.

Comparing percentages instead of absolute amounts made sense, but I’m not sure what the fairest way to make that comparison is.  Suppose you had a team with a 100 and a 200 pound person, and another team with two 150 pound people.   If each person lost 5% of their body weight then the total for each team would be 15 pounds, or 5%.

But what if someone lost 1 pound?  If it were the 100 pound person, that seems more significant than the 200 pound person, but they’re counted the same because those two are on the same team.  Likewise, if the 200 pound person lost an extra 10 pounds and one of the 150-pound people lost 9 pounds, the first team would win even though the second individual lost a higher individual percentage because it’s the team total that is used for the percentage, and both teams weigh 300 pounds.

I was thinking that it might make more sense to compute the percentages and then average them.  The problem is that it might be easier for a 200 pound person to lose even 5% of their weight than a 100 pound person so even that is imperfect — ease and health of weight loss don’t quite follow a linear scale.  But more importantly, I suspect, it might just be one step too confusing for a late night show.

### Margin of Error in packaging

July 7, 2009

Hello July!  It turns out that the family reunion we’ve been at didn’t have internet access, so instead of taking breaks to write about math we had to spend the entire time relaxing.

More multiplication tomorrow, but for now here’s a picture from the folk at Evil Mad Scientist Laboratories (some rights reserved on the picture).  Notice that the Margin of Error is included in the net weight!  It makes me want to run out and see if I can find similar examples.

### Calculus Demonstration: 3D printing

June 18, 2009

I love the concept of 3D printing.  Of course, I also really enjoy teaching Calc II when we talk about slicing and shell formulas and volumes of revolution, because I remember the AHA! moment when I suddenly put together what the formulas were all describing and it all made perfect sense.  (Sadly, this moment came when I was a senior studying for comps, several years after actually taking the class, but still, AHA! moments are glorious.)

In 3D printing, objects are built from the bottom up, cross section by cross section, the same way you’re supposed to envision the pieces when you calculate volumes by slicing.  This article in the Christian Science Monitor last week likens it to building with legos, although my experience with legos is that separate sections are constructed and then put together (you build the walls, then add the furniture, then the roof); that concept might work with printing too, where you print separate components and then put them together.  And what’s amazing is that you can print some pretty complicated things with moving parts.

So what is used for the printing?  The article above describes a layer of powder being put down and the printing is actually done by spraying glue instead of ink.  Wikipedia also describes printers that build with a liquid gel.  But my favorite is printing done with candy.

That’s right:  candy.  Not surprisingly, the CandyFab 6000 and it’s earlier prototypes are made by the folk at Evil Mad Scientist Laboratories.  Here’s an example of 3D printing:

Isn’t that amazing?  Here’s the machine it was done on:

You can see it making some 3D Candy above.  They keep making improvements, as you can see in the two dodecahedrons below (printed at different times):

Here’s another fancy piece o’ printing:

And finally, here’s a Möbius Strip.  I thought it was a cake at first but no, it’s eight pounds of sugar.

Now that’s some pretty neat printing.  You can even see the “slices” on the surface.

All photos are licensed under Creative Commons; the photos link to their home on Flickr, and you can find even more here, plus more information here.

### The Wolfram Alpha Bandwagon

May 18, 2009

I knew that Wolfram Alpha was coming, but couldn’t quite figure out what it was so I didn’t keep too close an eye out for it.  Then I read on Teaching College Mathematics and the Number Warrior that it was up, and I was pretty impressed with the screen shots.    Since then, I’ve been playing around with it, and I’m impressed.

It solves problems:

(See how you can switch from exact forms to decimal approximations?  With series, you can even tell it to use more terms.)

It gives you data:

(My favorite part is the info at the bottom, about population density and population growth.  I can see those as being useful for writing problems in stats or calculus.  I wasn’t able to get it to predict the population in a given year or predict when it would be a certain population, except once accidentally when it said the US population would be something like 4 quadrillion some year in the distant future.)

It also converts units.  I know that you can do this easily on Google, but this gives you a whole selection and you can pick the one that you like best.

And you get to look up cool stuff, like cities, movies, colleges, and names:

So all in all, it seems like it’s a combination of many of the things I like about Wolfram’s MathWorld, Wikipedia, and Google.  It doesn’t supplant any of them, but it’s quite user-friendly and I’m looking forward to seeing what else it does.

### Big Bills in the US

May 3, 2009

After reading Denise’s post on large amounts of money, I started to wonder what the largest bill in circulation is.    I started with the US.

The largest bill I knew of was the $100 bill. Ben Franklin invented the carriage odometer. I figured there were a few bills above that, getting into the$1000s, but it turns out that this is the largest denomination in circulation in the US.

There used to be a $500 bill. A 1918 version had John Marshall on it, but the 1928 and 1934 runs had William McKinley. The man who shot William McKinley was executed in Auburn, not far from here. And there used to be a$1000 bill, in 1918 with Alexander Hamilton and in 1928 with Grover Cleveland.  We have two cards for Grover Cleveland in our Deck of US Presidents since he served non-consecutive terms.

There’s a $5000 bill with James Madison. His wife’s sister’s husband’s uncle was George Washington. There’s even a$10,000 bill with Salmon Portland Chase.  He was Secretary of the Treasury in 1862 when the first bills were printed and he put his own picture on some early bills although this one was designed after he died.

That’s it for bills, although there was a $100,000 gold certificate with Woodrow Wilson. In 1915 Woodrow Wilson became the first sitting president to go to a World Series baseball game; a year later he thew the opening ball. “So what happened to all that fine money?” you may be wondering. Well, bills just weren’t all that popular. OK, I’m sure they were popular in the sense that no one would turn them down, but there weren’t a lot of them in circulation. The$100,000 bill [with a buying power today over over \$1.5 million] was never in public circulation — it was only for Official Transactions between Federal Reserve Banks [and apparently it’s illegal to for collectors to even have one], and the others haven’t been printed in more than 60 years.  Indeed, in 1969 the Federal Reserve said something along the lines of “Enough of this” and discontinued them.  I think that means that they stopped giving them out as change, although they’re still officially legal tender so those of you who have them in your wallet are still OK.

### Tetrahedral Pyramids (part 2 of 3)

April 24, 2009

The first post in this series, Pascal’s Pyramid (part 1 of 3), explored number patterns that arise in a 3 dimensional version of Pascal’s Triangle: a pyramid with square cross sections.  Another way to do a 3D version of Pascal’s Triangle would use triangular cross sections.   If you think of building a tetrahedron by stacking oranges in a pile, each orange is in contact with (up to) three oranges that lie in the level above it.

As before, we set the top number in the pyramid to be 1, and assume that at the lower levels, each number will be the sum of the (up to) three adjacent numbers on the previous level.

Also as before, we know to expect the binomial coefficients to appear on each of the three triangular faces of our new pyramid.

In the interior, the numbers generated satisfy the recursion

f(n+1, a+1, b+1) = f(n, a, b) + f(n, a, b+1)+ f(n, a+1, b+1)

where f(0,0,0) = 1, and f(n,a,b) is understood to be zero if any of the parameters go out of range (that is, if a or b is either negative or greater than n, or if b>a).

The resulting pyramid:

And for example, the numbers on the seventh level from the top are:

Examining this triangle, we recognize an odd variation on Pascal’s triangle:  each row is a multiple of a row of the usual Pascal triangle, where the multiplier is the left-most entry in the row.  Since those terms are also terms on an outer face of the pyramid, we know they too are binomial coefficients, and we are led to conjecture that  $f(n, a, b) = { n \choose a } {a \choose b}$.

As before, we can prove this result by induction on n.  The base case, when n=0, works, since ${0 \choose 0} {0 \choose 0} = 1$.

If we assume (for some specific value of n)  that each $f(n, a, b) = {n \choose a} { a \choose b}$, then it follows that

$f(n+1, a+1, b+1) = {n \choose a}{a \choose b} + {n \choose a}{a \choose {b+1}}+ {n \choose {a+1}}{{a+1} \choose {b+1}}$

$= {n \choose a} \bigg( {a \choose b} + {a \choose {b+1}} \bigg) + {n \choose {a+1}}{{a+1}\choose {b+1}}$

$= {n \choose a}{{a+1} \choose {b+1}} + {n \choose {a+1}}{{a+1} \choose {b+1}}$

$= {{n+1} \choose {a+1}}{{a+1}\choose {b+1}}$

And thus we have our identity by induction on n.

This suggests a direction for exploration:  what interesting things do we know about Pascal’s Triangle, that might generalize to three dimensions?

Coming up:  Sierpinski’s Pascal Pyramid (3/3)

### Pascal’s Pyramid (part 1 of 3)

April 20, 2009

In an earlier post, Ξ had mentioned a question from Who Wants To Be A Millionaire that made reference to “a famous “pyramid” of numbers that starts with the number one on top”.  The intended answer was “Pascal’s Triangle” (my emphasis).

That got me wondering what a three dimensional analogue of the Pascal Triangle might look like.

Today we’ll explore a Pascal-like construction of a square pyramid.   The top-most point on the pyramid will be assigned the value 1.  As we move down the pyramid,   let’s assign each point a value  by taking the sum of the 4 terms directly above and adjacent to each point.

If you think about the 4 outer faces, each point along a face only has two points that lie above it.  Thus if we restrict our attention just to one outer face, our construction is identical to the two dimensional Pascal triangle, and for  each face we should get the familiar binomial coefficients.

One can describe  how a given entry depends on those above it:  If we let f(n, a, b) be the entry in the nth layer down from the top, in row a and column b, then the entries that lie above will be in layer n-1, their row will be a or a-1, and their column will be b or b-1.  Thus

f (n, a, b) = f(n-1, a-1,b-1) + f(n-1, a, b-1) + f(n-1, a-1, b) + f(n-1, a, b)

with f(0,0,0)=1, and f understood to be zero if any of the parameters get out of range (e.g. if a or b > n, or is negative).

Numerically, we get the following pattern (for 0 ≤ n ≤ 6):

For example, the entries in the fifth layer are:

Each entry appears to be the product of the first entry in its row, with the first entry in its column.  Since we know the row and column values are binomial coefficients, we conjecture that $f(n,a,b) = {n \choose a} {n \choose b}$.

To prove this, we start by verifying that when n=0, the product of the binomial coefficients is equal to 1.  Proceeding inductively, we assume that the result holds for n-1, and we compute f(n,a,b):

${ {n-1} \choose {a-1}} {{n-1} \choose {b-1}} + {{n-1} \choose {a}} { {n-1} \choose {b-1}} + {{n-1} \choose {a-1}}{{n-1} \choose {b}} + {{n-1} \choose {a}} {{n-1} \choose {b}}$

$= \bigg( {{n-1} \choose {a-1} } + {{n-1} \choose {a}} \bigg) \bigg( {{n-1} \choose {b-1}} + {{n-1} \choose {b}} \bigg)$

$= { {n}\choose {a}} {{n} \choose {b}}$

QED by induction; each entry in Pascal’s Square Pyramid is a product of two binomial coefficients, and is equal to the sum of the four terms immediately above it.

Coming upTetrahedral pyramids (2/3);  Sierpinski’s Pascal Pyramid (3/3)

### Inversions

March 28, 2009

Dave Richeson over at Division by Zero has a post today about ambigrams.  Ambigrams, also known as inversions, are words that have some sort of symmetry.  In the post he shares some of the ambigrams he’s created.

I am, sadly, not so creative.   Fortunately, there are some computer programs that do the work for me!  FlipScript (again from Dave’s article) does a nice job, but they’re strict about copyright (you have to buy the images to be able to show them; still, many are worth buying for a special occasion).

Ambigram.Matic has a more public generator.  The script isn’t as fancy, but they write that the intent isn’t to be a finished product but just a starting place.  Here’s one they came up with:

Not too shabby!  Be sure to take Dave’s warning seriously about these sites, though — they can be amazing time sinks.

### Universal sets and the Russell paradox

March 18, 2009

I was working with some students on set theory recently, and we were momentarily puzzled by their textbook’s definition of subset:

Let A and B be two sets contained in some universal set U. […] The set A is a subset of a set B if each element of A is an element of B….  More formally, A is a subset of B provided that for all x ε U, if x ε A then x ε B.

What threw us was the reference to a universal set U.  Why bother with that?  Why wouldn’t we just quantify over all x in A, instead of all x in U?

After a bit of thought, I realized there were a few reasons:

1. This makes defining set equality a bit cleaner:  A=B provided for each x in U, x ε A iff x ε B.  (If we didn’t have a universal set U to refer to, we’d presumably have to do two separate if-then statements, one quantifying over A, the other over B.)
2. It places the formal discussion of set theory in the familiar setting of Venn diagrams.
3. It sets the stage for dealing with set operations (union, intersection, and especially complements).

There is a 4th reason:  a desire to avoid impredicative definitions (according to the Oxford Dictionary of Philosophy, “Term coined by Poincaré ; for a kind of definition in which a member of a set is defined in a way that presupposes the set taken as a whole”;  more loosely, definitions that can potentially be self-referential).

Sidestepping the technicalities,   I wondered if we could easily craft a compelling reason to work within a universal set: what are the consequences of ignoring it?  The obvious place to worry is in computing set complements.  And thinking about self-reference, I wondered what could go wrong in building Øc without reference to a universal set?

If we call that complement M (for “monster”, or “massively huge”, or…), we see a set that contains everything that isn’t in the empty set.   “Cool!”, you’re thinking — that’s a great choice for a universal set U!

Well, hold on a bit.  This M has lots of stuff in it.  17, and {17}, and cool mathy stuff like that.  But it also has my old 82 Impala, and every sock I’ve ever lost, and every subset of the set of all socks that I’ve lost.   Everything you can imagine is an element of M.

For that matter, M even contains itself as an element, since M contains every thing as an element.

Once you see that this monster has the bizarre property that MεM, you probably begin to worry.  Maybe we want a smaller universe, that only has the well-behaved stuff, those things in M that aren’t elements of themselves.

Thus we define $R := \{ x \in M \quad : \quad x \not \in x \}$, the set of reasonable objects.  This set is a better candidate for our universe: it avoids those strange objects that are members of themselves.  (It still has my 82 Impala, though.)

B ut now, is R itself a reasonable object?  Is R ε R ?  If R is a reasonable object, then it should be a member of R, which is a very unreasonable thing for it to do. (That is, if R ε R, then R doesn’t satisfy the membership criteria for R, and shouldn’t be a member of R.)

So apparently then R is unreasonable.  But then we’d expect $R \not \in R$, which is to say that R is a reasonable object, and thus should be a member of R.

In this way, we’re led to a variation on the Russell Paradox.
Apparently building Øc without first pinning down our universe is a bad idea.

The image above is a public-domain harbor map, showing the canning docks in Liverpool England.

Bertrand Russell  first discovered the self-referential paradox of building a set of all sets that are not members of themselves.   He was not (so far as I am aware) related to Baron Russell of Liverpool (Sir Edward Russell), but in any event I shall henceforth look at that image and see “Russell’s Pair of Docks”.

### a and 1/a

February 18, 2009

In the last Carnival of Mathematics, two of the number facts were:

• 1/49 is the sum of the series 0.02+0.0004+0.000008+…
• 49 is the sum  the series 0.98+0.982+0.983+…

(Incidentally, I originally tried to put parentheses around the 0.98, but having an 8 and ) next to each other made the end format as 8) and that looked pretty funny.)

It turns out that it’s not a coincidence that the ratios in the two geometric series (0.02 and 0.98 ) add to 1.  As proof, suppose that:

$\frac{1}{a}=x+x^2+x^3+...$

The first term and ratio of this geometric series both equal $x$ and so the series sums to $\frac{x}{1-x}$.  But we said this series was equal to  $\frac{1}{a}$ so it follows that

$a=\frac{1-x}{x}=\frac{(1-x)}{1-(1-x)}$, which is the sum of the geometric series whose first term and ratio both equal $(1-x)$.  In other words,

$a = (1-x)+(1-x)^2+(1-x)^3...$

Not the most exciting fact in the world, but still intriguing.

[As a further aside, if $a$ is an integer then it turns out that $x=\frac{1}{a+1}$.  For example, $\frac{1}{4}=(\frac{1}{5})+(\frac{1}{5})^2+(\frac{1}{5})^3...$ and therefore $4=(\frac{4}{5})+(\frac{4}{5})^2+(\frac{4}{5})^3...$.  I’m not sure if this makes the series more or less interesting, so I’ll pretend the answer is more.]

Blame Credit for this post actually goes to TwoPi, who first came up with the sums for 49 and 1/49 and who noticed the pattern of adding to 1.  Credit for the photo goes to Arjan Dice; it’s published here on Wikipedia.

### Predicting the Super Bowl

February 1, 2009

It is probably bad form to start off with a disclaimer, but here it goes anyways:  I don’t follow the NFL.  I am aware that the Steelers and the Cardinals are in the Super Bowl, and I am aware that the Cardinals are not from St. Louis, where the Rams play now that they’ve left LA, and the Raiders aren’t in LA anymore either.  (Nor are the Chargers.)  But I’d be hard pressed to name any of the current players for the Steelers or Cardinals, so any opinion I might have to offer about the game will be based on other data.

I am going to a Super Bowl party this afternoon, and there is likely to be discussion (or perhaps -ahem- a “party game” with prizes and such) focused on the final score of the contest.

What strategy would one employ in order to be most likely to predict the final outcome?

After perusing the final scores of 12,595 NFL games (from 1922 to the present), I’ve found a few patterns:

• The most frequently occurring scores for an individual team are:
• 17 points  (7.2% of all scores),
• 14 (6.3%),
• 7 (6.1%),
• 24 (6.0%), and
• 10 (5.9%).
• The least frequently occurring total scores [under 20]  are:
• 1 point (which never can happen under NFL rules),
• 4 points (which happened once:  Racine 10, Chicago 4, on November 25, 1923),
• 5 points (18 times in NFL history),
• 2 points (31 times), and
• 8 points (36 times)
• It is possible for a team to score exactly 1 point under NCAA Football rules, but I haven’t had a chance to search through that ocean of data to see if it has ever happened.  [If a team is attempting a PAT, and a safety occurs, the defending team gets 1 point under the NCAA rulebook [see page FR-108.]
• The most frequently occurring final scores are:
• 20 – 17 (211 times),
• 17-14 (162 times),
• 27-24 (153 times),
• 13-10 (141 times), and
• 24-17 (125 times).
• Modulo 10, the most frequently occurring team scores are:
• 0 (19.3%),
• 7 (19.1%),
• 4 (15.1%), and
• 3 (12.3%);
• The least frequently occurring team score modulo ten is 2 (3.1%).
• Modulo 10, the most frequently occurring pairs of scores are:
• {0,7} (8.1% of all games), followed by
• {4,7} (6.8%),
• {0,3} (5.8%),
• {0,4} (5.2%), and
• {3,7} (4.5%).
• The least likely combinations modulo 10 are:
• {2,2} (0.024% = 3 times in NFL history), followed by
• {9,9} (0.10%),
• {5, 5} (0.14%),
• {5, 9} (0.25%), and
• {2, 8} (0.26%)

My own Super Bowl prediction?  Cool ads and lots of fun chatting with friends.

### What’s the (Geometric) Pattern?

January 26, 2009

To go along with yesterday’s post, here is another fun “Find the Pattern” problem.  I got this from Helen Timberlake, but I’m not sure if it was her creation or not.

### My Presidential Age

January 20, 2009

I celebrated a Presidential Age for the last half of 2008.    By that, I mean I was 43 years old during the term in office of the USA’s 43rd President.   And in fact, later this coming year I will turn 44, and have a second Presidential Age.

In this post, I want to explore some of the properties of this notion of “Presidential Age”.  In particular:

• Everyone who lives long enough should expect to experience a Presidential Age of their own,
• only some people are likely to experience two, and
• experiencing three or more is rare but possible, and has happened in our nation’s history.

In what follows, I want to prove the first statement, explore the likelihood of the second, and give examples of the third.

### Everyone who lives long enough will experience a Presidential Age.

Assumptions:  A)  People live arbitrarily long lives, and B) over the course of your life, the mean length of a Presidential term in office will be greater than 1 year.

Roughly speaking, the reason why everyone gets a PA is the fact that your age increases by 1 each year, while the President Number increases more slowly than that on average, so eventually your age must become the larger of the two quantities.  (Granted, in a few hundred years, the Presidential numbers will be larger than 100, and it will be harder for people to live long enough to experience their PA, but the general principle still holds.)

Making the previous paragraph more rigorous is at first glance non-trivial, as the two functions involved (your age and the Presidential number) are not continuous, so the Intermediate Value Theorem from the calculus doesn’t apply.  And in fact you can have  rational valued functions where f(a) < g(a) for all small enough a, and f(b) > g(b) for all large b, without f(x) = g(x) at any intervening point.  (See the comments to a puzzle on winning percentages from jd2718’s blog for one such example.)

For the Age and Prez functions, though, it is significant that they are integer-valued, their values only increase, and they always increase by exactly 1.  Thus the only way my Age can first  become strictly larger than the Prez number is if I have a birthday (so my Age jumps by 1) on a day when there is no change in administration (so the Prez number stays unchanged), and hence just prior to my birthday my Age must have actually matched the Prez number.

### Getting more than one Presidential Age

Clearly lots of people will have more than one Presidential Age.  Anyone who turned 43 between January 21, 2007 and January 19, 2009 will have been 43 years old during the GWB era, and will be 44 years old during the BHO era.  But those who turned 43 earlier during the Bush presidency will have missed their chance, having turned 45 before President Obama took the oath of office.

The likelihood of getting a second Presidential Age depends, then, on the length of the term of the relevant President.  If all Presidents served their full term in office, then in order to get more than one Presidential Age, your Age and Prez numbers must first coincide within two years of the end of that President’s term, and we’d expect this to occur between 1/2 and 1/4 of the time (depending on whether that president served one or two terms).

Reality is of course not so simple as all that, as some Presidents served less than their full term in office.  Some examples:

### Nixon-Ford-Carter

Richard Nixon was our nation’s 37th President.  He resigned the presidency on August 9, 1974, and was succeeded by Gerald Ford.  Jimmy Carter followed as the 39th President on January 20, 1977.

An individual whose was born between 1/21/37 and 8/9/37 would have been 37 during the last months of the Nixon administration, 38 during Ford’s, and 39 at the beginning of Carter’s term in office.

### Van Buren – Harrison – Tyler

A more extreme example occurs in the 1840s, with the death of William Henry Harrison just one month after taking office.  In that case, anyone who was nine years old between March 4, 1841 and April 4, 1841 would have three Presidential Ages (8, 9, 10).

### More than three?

In principle, it is possible to have more than three Presidential Ages, although for that to happen would require  consecutive abbreviated terms in office, something our nation has never experienced, and hopefully never will.

### Temps fail the Intermediate Value Theorem

January 15, 2009

Rochester’s in the middle of a cold spell right now, with highs in the single digits.   And apparently sudden jumps in temperature.  Check out today’s paper:

On which day are they predicting that it will be 11°F  (or -12°C)?

### Math and the Fiber Arts

January 10, 2009

We’re back from the Joint Mathematics Meetings in DC!  There were lots of great books at the Exhibits, lots of great rocks at the Smithsonian (more on that tomorrow), and lots of great talks at the conference.

One of my favorite sessions was the AMS Special Session on Mathematics and the Fiber Arts.  You can see the complete session (with links to home pages and talk summaries) here, but a few of the speakers also have web pages about their topics.

sara-marie belcastro talked  about braid words that describe the twisting that occurs when making helix stripe pattern.  She has an extensive page on mathematical knitting.

Irena Swanson showed how to make semi-regular tessellations while quilting — not just by sewing the pieces together, but by making use of some intriguing shortcuts in pieces and cutting.  She has a page with some of her mathematical quilts.

Mary Shepherd used cross stitching to illustrate cosets in group theory.  She has a web page about using mathematics to design cross-stitching and symmetry which also has some of the coset information from the talk.  It also shows her using cross stitch to show Frieze Patterns and most of the wallpaper patterns (but not all, because those that have a rotation of 60° can’t be put on the grid pattern).

Good times,  good times.  Now I just have to switch my mind away from knitting and back to classes!

The photo above is of two hyperbolic planes that I knit crocheted last year for my geometry class.