Calculus: Linear Approximations, II

As I mentioned last week, I am a fan of emphasizing the idea of a derivative as a linear approximation.  I ended that discussion by using this method to find the derivative of \tan(x).   Today, we’ll look at some more examples, and then derive the product, quotient and chain rules.

Differentiating \sec(x) is particularly nice using this method.  We first approximate

\sec(x+h)=\dfrac1{\cos(x+h)}\approx\dfrac1{\cos(x)-h\sin(x)}.

Then we factor out a \cos(x) from the denominator, giving

\sec(x+h)\approx\dfrac1{\cos(x)(1-h\tan(x))}.

As we did at the end of last week’s post, we can make h as small as we like, and so approximate by considering 1/(1-h\tan(x)) as the sum of an infinite series:

\dfrac1{1-h\tan(x)}\approx1+h\tan(x).

Finally, we have

\sec(x+h)\approx\dfrac{1+h\tan(x)}{\cos(x)}=\sec(x)+h\sec(x)\tan(x),

which gives the derivative of \sec(x) as \sec(x)\tan(x).

We’ll look at one more example involving approximating with geometric series before moving on to the product, quotient, and chain rules.  Consider differentiating x^{-n}. We first factor the denominator:

\dfrac1{(x+h)^n}=\dfrac1{x^n(1+h/x)^n}.

Now approximate

\dfrac1{1+h/x}\approx1-\dfrac hx,

so that, to first order,

\dfrac1{(1+h/x)^n}\approx \left(1-\dfrac hx\right)^{\!\!n}\approx 1-\dfrac{nh}x.

This finally results in

\dfrac1{(x+h)^n}\approx \dfrac1{x^n}\left(1-\dfrac{nh}x\right)=\dfrac1{x^n}+h\dfrac{-n}{x^{n+1}},

giving us the correct derivative.

Now let’s move on to the product rule:

(fg)'(x)=f(x)g'(x)+f'(x)g(x).

Here, and for the rest of this discussion, we assume that all functions have the necessary differentiability.

We want to approximate f(x+h)g(x+h), so we replace each factor with its linear approximation:

f(x+h)g(x+h)\approx (f(x)+hf'(x))(g(x)+hg'(x)).

Now expand and keep only the first-order terms:

f(x+h)g(x+h)\approx f(x)g(x)+h(f(x)g'(x)+f'(x)g(x)).

And there’s the product rule — just read off the coefficient of h.

There is a compelling reason to use this method.  The traditional proof begins by evaluating

\displaystyle\lim_{h\to0}\dfrac{f(x+h)g(x+h)-f(x)g(x)}h.

The next step?  Just add and subtract f(x)g(x+h) (or perhaps f(x+h)g(x)).  I have found that there is just no way to convincingly motivate this step.  Yes, those of us who have seen it crop up in various forms know to try such tricks, but the typical first-time student of calculus is mystified by that mysterious step.  Using linear approximations, there is absolutely no mystery at all.

The quotient rule is next:

\left(\dfrac fg\right)^{\!\!\!'}\!(x)=\dfrac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}.

First approximate

\dfrac{f(x+h)}{g(x+h)}\approx\dfrac{f(x)+hf'(x)}{g(x)+hg'(x)}.

Now since h is small, we approximate

\dfrac1{g(x)+hg'(x)}\approx\dfrac1{g(x)}\left(1-h\dfrac{g'(x)}{g(x)}\right),

so that

\dfrac{f(x+h)}{g(x+h)}\approx(f(x)+hf'(x))\cdot\dfrac1{g(x)}\left(1-h\dfrac{g'(x)}{g(x)}\right).

Multiplying out and keeping just the first-order terms results in

\dfrac{f(x+h)}{g(x+h)}\approx f(x)g(x)+h\dfrac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}.

Voila!  The quotient rule.  Now usual proofs involve (1) using the product rule with f(x) and 1/g(x), but note that this involves using the chain rule to differentiate 1/g(x);  or (2) the mysterious “adding and subtracting the same expression” in the numerator.  Using linear approximations avoids both.

The chain rule is almost ridiculously easy to prove using linear approximations.  Begin by approximating

f(g(x+h))\approx f(g(x)+hg'(x)).

Note that we’re replacing the argument to a function with its linear approximation, but since we assume that f is differentiable, it is also continuous, so this poses no real problem.  Yes, perhaps there is a little hand-waving here, but in my opinion, no rigor is really lost.

Since g is differentiable, then g'(x) exists, and so we can make hg'(x) as small as we like, so the “hg'(x)” term acts like the “h” term in our linear approximation.  Additionally, the “g(x)” term acts like the “x” term, resulting in

f(g(x+h)\approx f(g(x))+hg'(x)f'(g(x)).

Reading off the coefficient of h gives the chain rule:

(f\circ g)'(x)=f'(g(x))g'(x).

So I’ve said my piece.  By this time, you’re either convinced that using linear approximations is a good idea, or you’re not.  But I think these methods reflect more accurately the intuition behind the calculations — and reflect what mathematicians do in practice.

In addition, using linear approximations involves more than just mechanically applying formulas.  If all you ever do is apply the product, quotient, and chain rules, it’s just mechanics.  Using linear approximations requires a bit more understanding of what’s really going on underneath the hood, as it were.

If you find more neat examples of differentiation using this method, please comment!  I know I’d be interested, and I’m sure others would as well.

In my next installment (or two or three) in this calculus series, I’ll talk about one of my favorite topics — hyperbolic trigonometry.

Calculus: Linear Approximations, I

Last week’s post on the Geometry of Polynomials generated a lot of interest from folks who are interested in or teach calculus.  So I thought I’d start a thread about other ideas related to teaching calculus.

This idea is certainly not new.  But I think it is sorely underexploited in the calculus classroom.  I like it because it reinforces the idea of derivative as linear approximation.

The main idea is to rewrite

\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}h=f'(x)

as

f(x+h)\approx f(x)+hf'(x),

with the note that this approximation is valid when h\approx0.  Writing the limit in this way, we see that f(x+h), as a function of h, is linear in h in the sense of the limit in the definition actually existing — meaning there is a good linear approximation to f at x.

Moreover, in this sense, if

f(x+h)\approx f(x)+hg(x),

then it must be the case that f'(x)=g(x).  This is not difficult to prove.

Let’s look at a simple example, like finding the derivative of f(x)=x^2.  It’s easy to see that

f(x+h)=(x+h)^2=x^2+h(2x)+h^2.

So it’s easy to read off the derivative: ignore higher-order terms in h, and then look at the coefficient of h as a function of x.

Note that this is perfectly rigorous.  It should be clear that ignoring higher-order terms in h is fine since when taking the limit as in the definition, only one h divides out, meaning those terms contribute 0 to the limit.  So the coefficient of h will be the only term to survive the limit process.

Also note that this is nothing more than a rearrangement of the algebra necessary to compute the derivative using the usual definition.  I just find it is more intuitive, and less cumbersome notationally.  But every step taken can be justified rigorously.

Moreover, this method is the one commonly used in more advanced mathematics, where  functions take vectors as input.  So if

f({\bf v})={\bf v}\cdot{\bf v},

we compute

f({\bf u}+h{\bf v})={\bf u}\cdot{\bf u}+2h{\bf u}\cdot{\bf v}+h^2{\bf v}\cdot{\bf v},

and read off

\nabla_{\bf v}f({\bf u})=2{\bf u}\cdot{\bf v}.

I don’t want to go into more details here, since such calculations don’t occur in beginning calculus courses.  I just want to point out that this way of computing derivatives is in fact a natural one, but one which you don’t usually encounter until graduate-level courses.

Let’s take a look at another example:  the derivative of f(x)=\sin(x), and see how it looks using this rewrite.  We first write

\sin(x+h)=\sin(x)\cos(h)+\cos(x)\sin(h).

Now replace all functions of h with their linear approximations.  Since \cos(h)\approx1 and \sin(h)\approx h near h=0, we have

\sin(x+h)\approx\sin(x)+h\cos(x).

This immediately gives that \cos(x) is the derivative of \sin(x).

Now the approximation \cos(h)\approx1 is easy to justify geometrically by looking at the graph of \cos(x).  But how do we justify the approximation \sin(h)\approx h?

Of course there is no getting around this.  The limit

\displaystyle\lim_{h\to0}\dfrac{\sin(h)}h

is the one difficult calculation in computing the derivative of \sin(x).  So then you’ve got to provide your favorite proof of this limit, and then move on.  But this approximation helps to illustrate the essential point:  the differentiability of \sin(x) at x=0 does, in a real sense, imply the differentiability of \sin(x) everywhere else.

So computing derivatives in this way doesn’t save any of the hard work, but I think it makes the work a bit more transparent.  And as we continually replace functions of h with their linear approximations, this aspect of the derivative is regularly being emphasized.

How would we use this technique to differentiate f(x)=\sqrt x?  We need

\sqrt{x+h}\approx\sqrt x+hf'(x),

and so

x+h\approx \left(\sqrt x+hf'(x)\right)^2\approx x+2h\sqrt xf'(x).

Since the coefficient of h on the left is 1, so must be the coefficient on the right, so that

2\sqrt xf'(x)=1.

As a last example for this week, consider taking the derivative of f(x)=\tan(x).  Then we have

\tan(x+h)=\dfrac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}.

Now since \sin(h)\approx h and \cos(h)\approx 1, we have \tan(h)\approx h, and so we can replace to get

\tan(x+h)\approx\dfrac{\tan(x)+h}{1-h\tan(x)}.

Now what do we do?  Since we’re considering h near 0, then h\tan(x) is small (as small as we like), and so we can consider

\dfrac1{1-h\tan(x)}

as the sum of the infinite geometric series

\dfrac1{1-h\tan(x)}=1+h\tan(x)+h^2\tan^2(x)+\cdots

Replacing, with the linear approximation to this sum, we get

\tan(x+h)\approx(\tan(x)+h)(1+h\tan(x)),

and so

\tan(x+h)\approx\tan(x)+h(1+\tan^2(x)).

This give the derivative of \tan(x) to be

1+\tan^2(x)=\sec^2(x).

Neat!

Now this method takes a bit more work than just using the quotient rule (as usually done).  But using the quotient rule is a purely mechanical process; this way, we are constantly thinking, “How do I replace this expression with a good linear approximation?”  Perhaps more is learned this way?

There are more interesting examples using this geometric series idea.  We’ll look at a few more next time, and then use this idea to prove the product, quotient, and chain rules.  Until then!

The Geometry of Polynomials

I recently needed to make a short demo lecture, and I thought I’d share it with you.  I’m sure I’m not the first one to notice this, but I hadn’t seen it before and I thought it was an interesting way to look at the behavior of polynomials where they cross the x-axis.

The idea is to give a geometrical meaning to an algebraic procedure:  factoring polynomials.  What is the geometry of the different factors of a polynomial?

Let’s look at an example in some detail:  f(x)=2(x-4)(x-1)^2.poly0b

Now let’s start looking at the behavior near the roots of this polynomial.

poly0c

Near x=1, the graph of the cubic looks like a parabola — and that may not be so surprising given that the factor (x-1) occurs quadratically.

poly0d

And near x=4, the graph passes through the x-axis like a line — and we see a linear factor of (x-4) in our polynomial.

But which parabola, and which line?  It’s actually pretty easy to figure out.  Here is an annotated slide which illustrates the idea.

Day137poly1

All you need to do is set aside the quadratic factor of (x-1)^2, and substitute the root, x=1, in the remaining terms of the polynomial, then simplify.  In this example, we see that the cubic behaves like the parabola y=-6(x-1)^2 near the root x=1. Note the scales on the axes; if they were the same, the parabola would have appeared much narrower.

We perform a similar calculation at the root x=4.

Day137poly2

Just isolate the linear factor (x-4), substitute x=4 in the remaining terms of the polynomial, and then simplify.  Thus, the line y=18(x-4) best describes the behavior of the graph of the polynomial as it passes through the x-axis.  Again, note the scale on the axes.

We can actually use this idea to help us sketch graphs of polynomials when they’re in factored form.  Consider the polynomial f(x)=x(x+1)^2(x-2)^3.  Begin by sketching the three approximations near the roots of the polynomial.  This slide also shows the calculation for the cubic approximation.

Day137poly3.png

Now you can begin sketching the graph, starting from the left, being careful to closely follow the parabola as you bounce off the x-axis at x=-1.

poly1d

Continue, following the red line as you pass through the origin, and then the cubic as you pass through x=2.  Of course you’d need to plot a few points to know just where to start and end; this just shows how you would use the approximations near the roots to help you sketch a graph of a polynomial.

poly1f

Why does this work?  It is not difficult to see, but here we need a little calculus.  Let’s look, in general, at the behavior of f(x)=p(x)(x-a)^n near the root x=a.  Given what we’ve just been observing, we’d guess that the best approximation near x=a would just be y=p(a)(x-a)^n.

Just what does “best approximation” mean?  One way to think about approximating, calculuswise, is matching derivatives — just think of Maclaurin or Taylor series.  My claim is that the first n derivatives of f(x)=p(x)(x-a)^n and y=p(a)(x-a)^n match at x=a.

First, observe that the first n-1 derivatives of both of these functions at x=a must be 0.  This is because (x-a) will always be a factor — since at most n-1 derivatives are taken, there is no way for the (x-a)^n term to completely “disappear.”

But what happens when the nth derivative is taken?  Clearly, the nth derivative of p(a)(x-a)^n at x=a is just n!p(a).  What about the nth derivative of f(x)=p(x)(x-a)^n?

Thinking about the product rule in general, we see that the form of the nth derivative must be f^{(n)}(x)=n!p(x)+ (x-a)(\text{terms involving derivatives of } p(x)). When a derivative of p(x) is taken, that means one factor of (x-a) survives.

So when we take f^{(n)}(a), we also get n!p(a).  This makes the nth derivatives match as well.  And since the first n derivatives of p(x)(x-a)^n and p(a)(x-a)^n match, we see that p(a)(x-a)^n is the best nth degree approximation near the root x=a.

I might call this observation the geometry of polynomials. Well, perhaps not the entire geometry of polynomials….  But I find that any time algebra can be illustrated graphically, students’ understanding gets just a little deeper.

Those who have been reading my blog for a while will be unsurprised at my geometrical approach to algebra (or my geometrical approach to anything, for that matter).  Of course a lot of algebra was invented just to describe geometry — take the Cartesian coordinate plane, for instance.  So it’s time for algebra to reclaim its geometrical heritage.  I shall continue to be part of this important endeavor, for however long it takes….

The Puzzle Archives, II

This week, I’ll continue with some more problems from the contests for the 2014 conference of the International Group for Mathematical Creativity and Giftedness.  We’ll look at problems from the Intermediate Contest today.  Recall that the first three problems on all contests were the same; you can find them here.

The first problem I’ll share is a “ball and urn” problem.  These are a staple of mathematical contests everywhere.

You have 20 identical red balls and 14 identical green balls. You wish to put them into two baskets — one brown basket, and one yellow basket. In how many different ways can you do this if the number of green balls in either basket is less than the number of red balls?

Another popular puzzle idea is to write a problem or two which involve the year of the contest — in this case, 2014.

A positive integer is said to be fortunate if it is either divisible by 14, or contains the two adjacent digits “14” (in that order). How many fortunate integers n are there between 1 and 2014, inclusive?

The other two problems from the contest I’ll share with you today are from other contests shared with me by my colleagues.

In the figure below, the perimeters of three rectangles are given. You also know that the shaded rectangle is in fact a square. What is the perimeter of the rectangle in the lower left-hand corner?

Day136problem

I very much like this last problem.  It’s one of those problems that when you first look at it, it seems totally impossible — how could you consider all multiples of 23?  Nonetheless, there is a way to look at it and find the correct solution.  Can you find it?

Multiples of 23 have various digit sums. For example, 46 has digit sum 10, while 8 x 23 = 184 has digit sum 13. What is the smallest possible digit sum among all multiples of 23?

You can read more to see the solutions to these puzzles.  Enjoy!

Continue reading The Puzzle Archives, II

Bay Area Mathematical Artists, VI

As I mentioned last time, this meeting took place at Santa Clara University.  As we have several participants in the South Bay area, many appreciated the shorter drive…it turns out this was the most well-attended event to date.  Even better, thanks to Frank, the Mathematics and Computer Science Department at Santa Clara University provided wonderful pastries, coffee, and juice for all!

Our first speaker was Frank A. Farris, our host at Santa Clara University.  (Recall that last month, he presented a brief preview of his talk.)  His talk was about introducing a sound element into his wallpaper patterns.

In order to do this, he used frequencies based on the spectrum of hexagonal and square grids.  It’s not important to know what this means — the main idea is that you get frequencies that are not found in western music.

Frank’s idea was to take his wallpaper patterns, and add music to them using these non-traditional frequencies.  Here is a screenshot from one of his musical movies:

Day135AmaryllisPMG

Frank was really excited to let us know that the San Jose Chamber Orchestra commissioned work by composer William Susman to accompany his moving wallpaper patterns.  The concert will take place in a few weeks; here is the announcement, so you are welcome to go listen for yourself!

Day135susman_farris invite_18

Frank has extensive information about his work on his website http://math.scu.edu/~ffarris/, and even software you can download to make your very own wallpaper patterns.  Feel free to email him with any questions you might have at ffarris@scu.edu.

The second talk, Salvador Dali — Old and New, was given by Tom Banchoff, retired from Brown University.  He fascinated us with the story of his long acquaintance with Salvador Dali.  It all began with an interview in 1975 with the Washington Post about Tom’s work in visualizing the fourth dimension.

He was surprised to see that the day after the interview, the article Visual Images And Shadows From The Fourth Dimension in the next day’s Post, as well as a picture of Dali’s Corpus Hypercubus (1954).

Dali_Crucifixion_hypercube
Salvador Dali’s Crucifixion (Corpus Hypercubus)

But Tom was aware that Dali was very particular about giving permission to use his work in print, and knew that the Post didn’t have time to get this permission in such a short time frame.

The inevitable call came from New York — Dali wanted to meet Tom.  He wondered whether Dali was simply perturbed that a photo of his work was used without permission — but luckily, that was not the reason for setting up the meeting at all.  Dali was interested in creating stereoscopic oil paintings, and stereoscopic images were mentioned in the Post article.

Thus began Tom’s long affiliation with Dali.  He mentioned meeting Dali eight or nine times in New York (Dali came to New York every Spring to work), three times in Spain, and once in France.  Tom remarked that Dali was the most fascinating person he’d ever met — and that includes mathematicians!

Day135Dali
Tom with Salvador Dali.

Then Tom proceeded to discuss the genesis of Corpus Hypercubus.  His own work included collaboration with Charles Strauss at Brown University, which included rendering graphics to help visualize the fourth dimension — but this was back in the 1960’s, when computer technology was at its infancy.  It was a lot more challenging then than it would be today to create the same videos.

Day135net
Tom’s rendition of a hypercube net.

He also spent some time discussing a net for the hypercube, since a hypercube net is the geometrical basis for Dali’s Corpus Hypercubus.  What makes understanding the fourth dimension difficult is imagining how this net goes together.

It is not hard to imagine folding a flat net of six squares to make a cube — but in order to do that, we need to fold some of the squares up through the third dimension.  But to fold the hypercube net to make a hypercube without distorting the cubes requires folding the cubes up into a fourth spatial dimension.

This is difficult to imagine!  Needless to say, this was a very interesting discussion, and challenged participants to definitely think outside the box.

Tom remarked that Dali’s interest in the hypercube was inspired by the work of Juan de Herrera (1530-1597), who was in turn inspired by Ramon Lull (1236-1315).

Tom also mentioned an unusual project Dali was interested in near the end of his career.  He wanted to design a horse that when looked at straight on, looks like a front view of a horse.  But when looked from the side, it’s 300 meters long!  For more information, feel free to email Tom at banchoff@math.brown.edu.

Day135horse
Dali’s sketch of the horse.

Suffice it to say that we all enjoyed Frank’s and Tom’s presentations.  The change of venue was welcome, and we hope to be at Santa Clara again in the future.

Following the talks, Frank generously invited us to his home for a potluck dinner!  He provided lasagna and eggplant parmigiana, while the rest of us provided appetizers, salads, side dishes, and desserts.

As usual, the conversation was quite lively!  We talked for well over two hours, but many of us had a bit of a drive, so we eventually needed to make our collective ways home.

Next time, on April 7, we’ll be back at the University of San Francisco.  At this meeting, we’ll go back to shorter talks in order to give several participants a chance to participate.  Stay tuned for a summary of next month’s talks!

Art Exhibition: Golden Section 2018

The Regional Meeting of the Golden Section of the Mathematical Association of America was held at California State University, East Bay.  The local organizer was Shirley Yap, fellow mathematical artist, who deserves kudos for the monumental amount of work it takes to organize a conference like this!  I helped out by organizing this year’s Art Exhibition.

This year, I thought I’d give you a virtual tour of the exhibit!  So I asked contributing artists to submit their own personal statement about their work and/or mathematical art in general, as well as an image of one of their displayed artworks.  I’ll let the artists speak for themselves….  (By the way, the order the artists are presented in is the order in which they sent me their information.  There is no ranking implicit in the order.)

Shirley Yap (shirley.yap@csueastbay.edu)

Day134test7

I created this image out of golden spirals.   While working on a math demonstration for my students, I unpacked a roll of netting. During the unraveling process, I had a vision of a two-dimensional golden spiral unravelling, which I tried to recreate with code.  I wanted to viewer to not just witness the unraveling, but also be inside the web of the fabric.  I often create code that has a lot of randomness in it, so that it captures a moment in time that can never be recreated.

Frank A. Farris (ffarris@scu.edu and math.scu.edu/~ffarris)

Day134FiloliSubmissionSize

Mathematicians can feel lonely to find ourselves face to face with the most beautiful thoughts humans have ever known, only to realize that communicating our experience is unreasonably difficult. I have found comfort in visual art, digitally computed images that are the best I can do (short of giving an hour lecture) to say, ‘This is the beauty of mathematics.’

David Honda (snaporigami@gmail.com and  snaporigami.weebly.com)

Day134p1019930_orig

I’m primarily a middle school mathematics teacher with one of my hobbies being Origami and other paper crafts. Some years back I became interested in the work of Heinz Strobl which uses joined, folded strips of paper to create various structures. Much like unit origami, the structures are held together solely by the folds, no adhesives. My interest soon became an obsession and I’ve been neck-deep in little strips of paper ever since. Lately I’ve been exploring concepts in Topology.  This particular work is my attempt at a Klein Bottle.

Dan Bach (www.dansmath.com, art@dansmath.com, and @dansmath)

Day134Primebeadssmall
Prime Bead Spirals

 “I’m a career college mathematics teacher, now an interactive book author and 3D math artist. I like to illustrate number theory and vector calculus principles with surprising and colorful images, using a software palette of Mathematica, Cheetah3D, and iBooks Author. My math art encourages viewers to think, notice, and wonder. And hopefully say, ‘That’s cool! That’s math?’ “

Day134Shuffling08_20180219_151104_128c_2p_o2x
This is a visualization of seven repeated perfect out-shuffles of a deck with 128 cards. The horizontal lines represent the particular orders of the cards throughout the shuffling, and the vertical curves represent the path each card takes from start to finish. The curves are colored from black to white in order to show the mechanics of the shuffling. The dots are colored from black to red to black in order to show that each perfect out-shuffle preserves the so-called “stay-stack principle”. Notice that the order of cards returns to the original order after seven shuffles.

Linda Beverly (lbeverly@horizon.csueastbay.edu and https://mathcsresearch.wordpress.com/)

Day134watercolor_frame_17_k128-8_by_10_webpage_quality
Watercolors

This image is an embedding of a photograph of a set of watercolors. The embedding was performed using Locally Linear Embedding (LLE), a nonlinear dimensionality reduction technique, introduced by Saul & Roweis in 2000. LLE is an unsupervised machine learning algorithm.

I am pursuing a double major in Mathematics and Computer Science. I am currently working on research in geometric dimensionality reduction and unsupervised machine learning. This work will be extended into neural networks and deep learning. I enjoy seeking out interesting intersections between mathematics, computer science, and art.

Jason Herschel (jherschel@gmail.com)
Day134JasonHerschel-100pingpongballswLEDs

Undulating organic light show with minimal code generated by a 16 MHz processor calculating color and brightness values through a Perlin noise algorithm. Blobs appear to grow, shrink, and drift relaxingly over the LED grid. 100 ping pong balls covering 100 individually addressable LEDs on poster board with Arduino Nano v.3 controller and battery pack. 

Paul Gonzalez-Becerra (pgonzbecer@gmail.com and http://www.pgonzbecer.com/)
Day134perlin-2
Programming is my art. I might not be a good “designer”, but I am a good developer where I am able to take a structured approach on art. I specialize in computer graphics, thus my understanding of the mathematics behind geometry, 3D models, and 2D sprites are better than my ability to free-form draw them.

Day134Islamic 8-fold Fractal Flower Median 900x900

All of my work stems from one core impulse: to celebrate the inherent beauty of mathematical forms. Since traveling to India in 2012, I have been particularly focused on blending traditional Islamic motifs with polyhedra and fractals. The results are distinctly Islamic in flavor but with a modern twist.

This piece has global and local 8-fold rotational symmetry around each gold star. Star centers occupy the nodes of 8 fractal quaternary trees, which are pruned at the octant boundaries. The original central star is replaced by an inward progression of the same fractal diminution.

Vince Matsko (vince.matsko@gmail.com,  www.vincematsko.com)

2017-03-14sierpinksi

This piece is based on fractal binary trees. The usual way of creating a binary tree is to move forward, then branch to the left and right some fixed angle as well as shrink, and repeat recursively. Recent work involves specifying the branching by arbitrary affine transformations. In this piece, the affine transformations were chosen so that as the tree grows, nodes are repeatedly visited. The nodes are covered by disks which become smaller with each iteration, accounting for the overlapping circles. The research needed to produce these images was undertaken jointly with Nick Mendler.

I hope you enjoyed this virtual tour of the Art Exhibition at the recent MAA Regional Conference.  The upcoming conference is next March; stay tuned for another virtual tour in about a year!

The Puzzle Archives, I

In going through some folders in my office the other day, I came across some sets of mathematics puzzles I wrote for a conference of the International Group for Mathematical Creativity and Giftedness in 2014.  Teachers of mathematics of all levels attended, from elementary school to university.  The organizing committee (which included me) thought it might be fun to have some mathematical activity that conference attendees could participate in.

So I and my colleagues created three levels of contests — Beginning, Intermediate, and Advanced — since it seemed that it would be difficult to create a single contest that everyone could enjoy.  But I did include three problems that were the same at every level, so all participants could talk about some aspect of the contests with each other.

Participants had a few days to get as many answers as they could, and we even had books for prizes!  Many remarked how much they enjoyed working out these puzzles.

Now this conference took place before I started writing my blog.   I have written several similar contests over the years for various audiences, and so I thought it would be nice to share some of my favorite puzzles from the contests with you.  And so The Puzzle Archives are born!

First, I’ll share the three puzzles common to all three contests.  I needed to create some puzzles which were fun, and didn’t require any specialized mathematical knowledge.  As I’m a fan of cryptarithms and the conference took place in Denver, I created the following puzzle.  Here, no letter stands for the digit “0.”

Day133Puzzle1

For the next puzzle, all you need to do is complete the magic square using the even numbers from 2 to 32.  Each row, column, and diagonal should add up to the same number.  There are two solutions to this puzzle — and so you need to find them both!

Day133Puzzle2

And of course, I had to include one of my favorite types of puzzles, a CrossNumber puzzle.  Remember, no entry in a CrossNumber puzzle can begin with “0.”

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I also included a few geometry problems, staples of any math contest.  For the first one, you need to find the area of the smallest circle you could fit the following figure into.  Both triangles are equilateral; the smaller has side length 1 and the larger has side length 2.

Day133Puzzle4

And for the second one, you need to find the radius of the larger circle.  You are given that the smaller circle has a diameter of 2 units, and the sides of the square are 2 units long.  Moreover, the smaller circle is tangent to the square at the midpoint of its top edge, and is also tangent to the larger circle.

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The last two problems I’ll share from this contest are number puzzles.  The first is a word problem, which I’ll include verbatim from the contest itself.

Tom and Jerry each have a bag of marbles. Tom says, “Hey, Jerry. I have four different colors of marbles in my bag. And the number of each is a different perfect square!” Jerry says, “Wow, Tom! I have four different colors of marbles, too, but the number of each of mine is a different perfect cube!”

If Tom and Jerry have the same total number of marbles, what is the least number of marbles they can have?

And finally, another cryptarithm, but with a twist.  In the following multiplication problem, F, I, N, and D represent different digits, and the x‘s can represent any digit.  Your job is to find the number F I N D. (And yes, you have enough information to solve the puzzle!)

Day133Puzzle5

Happy solving!  You can read more to see the solutions; I didn’t want to just put them at the bottom in case you accidentally saw any answers.  I hope you enjoy this new thread!

(Note:  The FIND puzzle was from a collection of problems shared by a colleague.  The first geometry problem may have come from elsewhere, but after four years, I can’t quite remember….)
Continue reading The Puzzle Archives, I