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Recipe for a Leibniz Quarter Pi

Ingredient: $\frac{1}{1+x^2}$

Divide gently in a long division sauce pan:

$\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8-x^{10}...$

Integrate briskly over a low flame:

$\int (1-x^2+x^4-x^6+x^8-x^{10}...)dx=$

$=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+...$

Evaluate for $x=1$ and let stand at room temperature for 1000 terms for accuracy to three decimal places.

$=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}...$

Add lemon zest to taste.

But this doesn’t look like a pi! or taste like a pi!

or a quarter pi!

Ah, but if we return to the original integral

$\int \frac{1}{1+x^2}dx$

and make the trigonometric substitution $x=tan \theta$ so that $dx=sec^2 \theta d \theta$ then:

$\int \frac{1}{1+x^2}dx=\int \frac{1}{1+tan^2 \theta}sec^2 \theta d \theta$

and

$\int \frac{1}{1+tan^2 \theta}sec^2 \theta d \theta=\int \frac{1}{sec^2 \theta}sec^2 \theta d \theta$

$=\int d \theta=\theta$

which, if we return to the original substitution $x=tan \theta$, we see that $\tan^{-1} x=\theta$

So, $\int \frac{1}{1+x^2}dx=\tan^{-1}x$, which means that:

$\tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+...$

and since $\tan^{-1}(1)=\frac{\pi}{4}$, then

$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}...$

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The Many Solutions of the Population Equation

I never studied the logistic equation as a student.  I first encountered this relationship as an instructor in one of the College Algebra textbooks I was reviewing and/or teaching from and was intrigued by the application of this “growth with constraints” model to a natural resource.  In researching applying the logistic model to natural resource consumption, I immediately ran into M. King Hubbard’s work on Peak Oil.

Then, when I was teaching integral calculus last winter, we began a unit on separable differential equations.  I was poking around looking for good application problems that would utilize separable ODEs and ran into the fundamental population differential relationship $\frac{dP}{dt}=kP$, followed by the relationship I had used for the logistic $\frac{dP}{dt}=kP(1-\frac{P}{N})$ with $N$ defined as the “carrying capacity” or maximum growth for the population.

We went through the procedures for solving each of these ODEs (as well as the continuous mixing problems which follow a similar pattern) and then we moved on.

This year when I was teaching this topic again I was reminded of a paper my thesis advisor had given to me back in 2003 about the application of differential equations to modeling fish populations.  I was intrigued by the profusion of models that could be generated by changing the constraints for a given relationship.

Since I didn’t teach integral calculus for another 10 years after I read that paper, I had essentially forgotten most of the equations, formulas and relationships that generated the graphs that had stuck with me.  This year, while covering the separable ODEs with their applications, I began to look into the application of these relationships to fish populations and population in general.

I found two great resources that go through the set up of these relationships in a very clear manner, and each of them includes wonderful graphs showing the multiple solutions that result when the same differential relationship is solved with different initial conditions.

The opening section of Robert Borelli and Courtney Coleman’s Differential Equations: A Modeling Approach can be read here.

A student project from James Madison University written by Bailey Steinworth, Yuhui Wang and Xing Zhang can be read here.

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Derivation of the Cross Product

A few weeks ago, my colleague who teaches Physics asked me about the derivation and justification of the cross-product formula.

The cross product for two vectors will find a third vector that is perpendicular to the original two vectors given.

The given vectors are assumed to be perpendicular (orthogonal) to the vector that will result from the cross product.  This means that the dot product of each of the original vectors with the new vector will be zero.

So, given two vectors $a=\begin{bmatrix} a_1\\ a_2\\ a_3 \end{bmatrix}$

and $b=\begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix}$

we want to find a third vector $n=\begin{bmatrix} n_1\\ n_2\\ n_3 \end{bmatrix}$

so that $n$ is perpendicular to both $a$ and $b$

As I mentioned above this means that we want the dot product of $n$ with each of the two original vectors to be zero.

$a \cdot n=a_1n_1+a_2n_2+a_3n_3=0$

and

$b \cdot n=b_1n_1+b_2n_2+b_3n_3=0$

This gives us two equations to work with.  Since we have three variables to solve for $(n_1, n_2, n_3)$, we’ll need another equation to work with.

The website Heaven’s in the backyard introduces a third constraint that the modulus of the cross product vector $n$ be equal to 1.

This creates a third equation $n_1^2+n_2^2+n_3^2=1$ and allows us to solve for $n_1, n_2,$ and $n_3$ in terms of $a_1, a_2, a_3, b_1, b_2$ and $b_3$

As mentioned above the web page Heaven’s in the backyard does a nice job with the derivation of the values for $n_1, n_2,$ and $n_3$ and ends up with the formula $n=\begin{bmatrix} a_2b_3-a_3b_2\\ a_3b_1-a_1b_3\\ a_1b_2-a_2b_1 \end{bmatrix}$.

I don’t see this mentioned in the derivation, but it appears that the $\sqrt{Z}$ term that is factored out and defined to make the derivation work more smoothly is actually the modulus of the cross product vector $n$.

Assuming that the cross product vector has a length of $1$ results in an answer that is multiplied by $\frac{1}{\sqrt{Z}}$, because the actual perpendicular $(n)$ has a modulus equal to $\sqrt{(a_2b_3-a_3b_2)^2+(a_3b_1-a_1b_3)^2+(a_1b_2-a_2b_1)^2}$ which is $\sqrt{Z}$.

After working through some of these ideas, I became interested in where the cross-product came from.

At Wikipedia, they mention that Joseph-Louis Lagrange, the French/Italian mathematician of the late 18th century provided a formula for this in a paper from 1773 that was focused on the properties of a tetrahedron.  The calculations related to the cross-product appear in the first few pages of the paper.  Lagrange posits 9 “quantités” $x, y, z, x', y', z', x'', y'', z''$ and then proceeds through a blizzard of calculations based on these quantities.

If each triplet of values $q=(x, y, z)$, $r=(x', y', z')$ and $s=(x'', y'', z'')$ is considered as the coordinates of a vector, then the first calculations are the cross-products of each vector with each of the others.  In Lagrange’s notation, we could identify each of the cross products as

$r \times s=\begin{bmatrix} \xi\\ \eta\\ \zeta \end{bmatrix}$, $q \times s=\begin{bmatrix} \xi'\\ \eta'\\ \zeta' \end{bmatrix}$, $q \times r=\begin{bmatrix} \xi''\\ \eta''\\ \zeta'' \end{bmatrix}$

On the following page Lagrange identifies the square of the modulus for each of the cross-product vectors as $\alpha$, $\alpha'$ and $\alpha''$

Several pages later Lagrange notes that the dot product of each original vector with the appropriate cross-product produces a zero result.

There are two things that fascinate me about this – (1) the depth of this seemingly simple question – how do you justify the cross-product formula? and (2) what was Lagrange up to in this paper? – what is the purpose of the multitude of calculations that he completes in the paper from 1773 (Solutions analytiques de quelques problèmes sur les pyramides triangulaires)?

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Infinite Series Approximations for Trigonometric functions from 14th century India

The infinite series approximations that have been used for many years to calculate the values of trigonometric functions have traditionally been attributed to Brook Taylor and Colin Maclaurin, European mathematicians of the early 18th century who were building on the work of Newton, Leibniz, James Gregory and Isaac Barrow among others.

However, I recently discovered that they were not the first to use these techniques.  As author George Gheverghese Joseph points out in the previous link, the work of Newton and Leibniz was tremendous, however the Indian development of infinite series approximations for trigonometric functions was equally amazing and important.  In addition, it came nearly 300 years before the European development of these techniques.

Madhava of Sangamagrama is generally recognized of the founder of the Kerala school of mathematics and astronomy in what is today the state of Kerala in southwest India.  The work of the mathematicians of the Kerala school was based on a desire for accurate trigonometric values for use in navigation.

Madhava lived in the late 1300s and early 1400s and most of his original work has been lost.  However, he is mentioned frequently in the surviving work of later mathematicians from the Kerala school.  Madhava is credited with power series calculations for the sine, cosine, tangent and arctangent, and like Leibniz, he used the arctangent power series to approximate the value of $\pi$ to 13 decimal places.

Victor Katz’ A History of Mathematics (Brief Edition) has a wonderful and detailed derivation of the Kerala school trigonometric series, with diagrams showing how they used the relationships between the angles, radii, chords and arcs in a circle to arrive at these amazing calculations.

Katz also published this derivation in a paper for the MAA (Mathematics Magazine, vol. 68, n. 3, June 1995, pp. 163-174)

The derivation for the infinite series begins on page 169 (pg. 7 in the pdf).

I’ve just begun to unpack this derivation and will post a step-by-step explanation of Katz’ work in the “near” future.

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Deriving the Area of a Circle

During the 1998-99 school year I was teaching math at Ellsworth High School in Maine.  I got to talking with one of the other math teachers there about deriving the area of a circle.  We started approaching the idea from a similar perspective as Archimedes had 2300 years ago in his work with circles and $\pi$.  We inscribed a polygon in a circle and then took the limit of the area of the polygon as the number of sides ($n$) approached infinity.  For the area of the polygon, we divided it into $n$ triangles and used the area formula $A=\frac {1}{2} ab\sin C$.

In this diagram, both $a$ and $b$ are radii of the circle and are labeled as $R$.  The central angle $\theta$ which will be used to find the area of the triangle is actually equal to $\frac {360^\circ}{n}$ or (and this will be VERY important later) $\frac {2\pi}{n}$.

So the area of the polygon is $n$ times the area of the triangle, or:

$n*\frac {1}{2} R*R*\sin \frac{2 \pi}{n}$

For this to be the area of the circle we want to take the limit of the previous expression as $n \to \infty$.

$A=\lim_{n \to \infty} n*\frac {1}{2} R*R*\sin \frac {2 \pi}{n}$

Since the $\frac {1}{2}*R*R$ don’t involve $n$, we can pull them out from the limit.

$A=\frac {1}{2} R^2*\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$

Now comes the calculus – we were just talking about the derivative of the sine function in Calculus I last week, which is what made me think about this.  Remember that:

$\lim_{x \to 0} \frac{\sin x}{x}=1$

So, we need to establish this limit somewhere in our derivation and then replace it with $1$.  The most likely place for this would be in here:

$\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$

but we need to have:

$\lim_{\frac{2 \pi}{n} \to 0} \frac{\sin \frac {2 \pi}{n}}{\frac{2 \pi}{n}}$

OK – so, let’s multiply our origninal expression by $\frac{2 \pi}{2 \pi}$.

$\lim_{n \to \infty} \frac{2 \pi}{2 \pi}*n*\sin \frac {2 \pi}{n}$

or

$\lim_{n \to \infty} 2 \pi*\frac{n}{2 \pi}*\sin \frac {2 \pi}{n}$

which is equivalent to

$\lim_{n \to \infty} 2 \pi*\frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

Let’s bring back the whole expression:

$A=\frac {1}{2} R^2*\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$

$A=\frac {1}{2} R^2*\lim_{n \to \infty} 2 \pi*\frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

$A=\frac {1}{2} R^2*2 \pi \lim_{n \to \infty} \frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

as $n \to \infty$,

$\frac{2 \pi}{n} \to 0$

$A=\frac {1}{2} R^2*2 \pi \lim_{\frac{2 \pi}{n} \to 0} \frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

$A=\frac {1}{2} R^2*2 \pi *1$

or

$A=\pi R^2$

When we were working through this the first time, we kept the $360^\circ$ in the formula and ended up with $A=180^\circ R^2$, which didn’t make any sense until we realized that $180^\circ$ is $\pi$ radians!

Archimedes’ conclusion was that the area of the circle was equal to one-half the radius times the circumference:

$\frac{1}{2} R*2\pi R = \pi R^2$

or, since the mathematicians in the Greek tradition like Archimedes thought in terms of geometry – the area of a circle is the same as the area of triangle with height equal to the radius of the circle and base equal to the circumference of the circle!

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History of Math

I’ve been teaching an on-line History of Math course (with a HUM humanities prefix) this term.

The posts for that course are here.

The most recent post was about the French mathematicians of the 17th century – Viète, Mersenne, Fermat, Descartes and Pascal.

French Mathematics of the 17th century

Francois Viète (1540-1603)

Francois Viète was the son of a lawyer in 16th century France.  He is credited with devising a scheme* in which unknown quantities in algebra would be represented by letters that are vowels and constant quantities would be represented by letters that are consonants.  At the time, the Arabic algebra that had been transferred to Europe over the previous 500 years was based on prose writing – everything was described in words.  After Viète’s initial use of letters for unknowns and constants, René Descartes later began to use letters near the end of the alphabet for unknowns (x, y, z) and letters from the beginning of the alphabet for constants (a, b, c).  This practice continues today.

In 1593, the Dutch ambassador to France said to French King Henry IV that a well-known Dutch mathematician had posed a problem that was beyond the capabilities of ANY French mathematician.  Henry IV passed the problem along to Viète and Viète was able to solve it.

Viète began a correspondence with Roomen, the Dutch mathematician who had posed the problem originally and became one of the first internationally recognized French mathematicians.  He worked mainly in trigonometry, astronomy and the theory of equations.

*This link is a paper written by a college student at Rutgers University in New Jersey.  Papers on other subjects by other students in the same course can be found here.

Marin Mersenne (1588-1648)

Marin Mersenne was a French monk best known for his research into prime numbers.  He also did important research into the musical behavior of a vibrating string, showing that the frequency of the vibration was related to the length, tension, cross section and density of the material.

Mersenne primes are prime numbers of the form $2^p-1$, where p is a prime number itself.  For example

$2^2-1=3$ which is prime

$2^5-1=31$ and so on.

Mersenne was also interested in the work that Copernicus had done on the movement of the heavenly bodies and despite the fact that, as a monk, he was closely tied to the Catholic church, he promoted the heliocentric theory in the 1600′s.

Mersenne was also known as a friend, collaborator and correspondent of many of his contemporaries.  Fermat, Pascal, Descartes, Huygens, Galileo, and Torricelli all corresponded with Mersenne and the exchange of ideas among these scientists promoted the understanding of music, weather and the solar system.

René Descartes (1596-1650)

René Descartes is probably best known for two things.  One is the conclusion “I think therefore I am” (Cogito ergo sum in Latin and Je pense donc je suis in French) and the other is the geometric coordinate system generally known as the Cartesian plane.

Descartes joined the army of Prince Maurice of Nassau in 1619 and was in Bavaria (southern Germany) and Bohemia (Czech Republic) during the beginning of the Thirty Years War.

The importance of the Cartesian Plane is difficult for us to understand today because it is a concept that we are taught at a young age.  Locating objects on a grid by their horizontal and vertical coordinates is so deeply embedded in our culture that it is difficult to imagine a time when it did not exist.

Before Descartes’ grid system took hold, there was Geometry:

and there was Algebra:

(Click on photo for larger view)

…and they were separate fields of endeavor.  The idea that a geometric shape like a parabola could be described by an algebraic formula that expressed the relationship between the curve’s horizontal and vertical components really is a ground-breaking advance.  It is so ground-breaking that once it happened, people began to forget that it hadn’t always been that way.

Once this new method for describing curves was developed, the question of finding the area under a curve was addressed.  This is the general problem of Integral Calculus.  Descartes (among others) saw that, given a polynomial curve $y=x^n$, the area under the curve could be found by applying the formula $A=\frac{x^{n+1}}{n+1}$

These were the rudimentary beginnings of the development of the Calculus that would be devised by Isaac Newton and Gottfried Leibniz in the ensuing years.

Fermat (1601-1665)

Pierre Fermat is also mostly remembered for two important ideas – Fermat’s Last Theorem and Fermat’s Little Theorem.  Fermat’s Last Theorem is a simple elegant statement – that Pythagorean Triples are the only whole number triples possible in an equation of the form $a^n+b^n=c^n$.

Pythagorean Triples are interesting groups of numbers that satisfy the Pythagorean relationship $a^2+b^2=c^2$.  Triples such as {3,4,5} {6,8,10} {8,15,17} {7, 24, 25} can be found that satisfy the equation.  But – Fermat’s Last Theorem says that if the $n$ in the original equation is any number higher than two, then there are no whole number solutions.

It’s true – but very difficult to prove.  Mathematicians tried for 350 years or so to prove this theorem before it was finally accomplished by Andrew Wiles in 1995.

By the way, you can generate Pythagorean Triples using the following formulas:

Pick two numbers $x$ and $y$, with $x>y$

$a=x^2-y^2$

$b=2xy$

$c=x^2+y^2$

Fermat’s Little Theorem is a useful and interesting piece of number theory that says that any prime number $p$ divides evenly into the number $a^{p-1}-1$, where $a$ is any number that doesn’t share any factors with $p$.

Blaise Pascal (1623-1662)

Blaise Pascal was the son of Etienne Pascal, who was a lawyer and amateur mathematician.  Etienne Pascal knew Marin Mersenne and often visited him at his Paris monastery, and when Blaise was a teenager he sometimes accompanied his father on these visits.

Pascal’s first published paper was a work on the conic sections.  He also did research on the composition of the atmosphere and noticed that the atmospheric pressure decreased as the elevation increased.  This led him to believe that beyond the atmosphere there existed a vacuum in which there was no atmospheric pressure.

René Descartes visited Pascal in 1647 and they argued about the existence of a vacuum beyond the atmosphere.  Descartes felt that this was impossible and criticized Pascal, saying that he must have a vacuum in his head.

Pascal is known for the structure of Pascal’s Triangle, which is a series of relationships that had previously been discovered by mathematicians in China and Persia.

Here is Pascal’s version:

Here is the Chinese version:

Here is a version that we often see in textbooks:

Each successive level is created by adding the two numbers above it, so in the 6th row {1,5,10,10,5,1} the 10 is created by adding the 4 and the 6 from the row above it.  These number patterns are actually quite useful in a wide variety of situations.

In raising a binomial to a power like $(x+y)^5$, the coefficients of each term are the same as the numbers from the 6th row:

$(x+y)^5=1x^5+5x^4+10x^3+10x^2+5x+1$

These numbers are also related to Discrete Mathematics and Combinatorics which describes how many ways there are to choose something from a series of possibilities.

There was a lot of great mathematics happening in Italy, England, Holland and Germany during the 17th century, but this collection of French mathematicians spanning nearly 100 years produced a tremendous amount of very important mathematical ideas.

The English, Germans and Swiss would make great contributions to mathematics in the 18th century with Newton, Leibniz, the Bernoullis, Euler and others, while the French would still contribute with the works of Laplace, Lagrange and Legendre.

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Benoit Mandelbrot (1924-2010)

Benoit Mandelbrot

A Greek among Romans

Benoit Mandelbrot passed away last week in Cambridge, Mass. at the age of 85.

Nassim Nicholas Taleb dedicated his book The Black Swan to Mandelbrot, calling him “a Greek among Romans.”

Mandelbrot is probably the most important mathematician of the last 50 years.  His ideas on using mathematics to describe and understand the world around us have influenced many mathematicians and scientists and led to a new understanding of how to apply mathematical ideas.

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