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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.

## 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!

## Combinatorics

We were covering some elementary combinatorics in the Pre-Calculus course this past week and were discussing some problems from VK Balakrishnan’s Discrete Mathematics textbook.  I took a Discrete Math course at University of Maine in 2000 from Balakrishnan and really enjoyed the class.

I needed to make a quiz question for the Pre-Calculus class and tried to remember some of the more interesting ways Balakrishnan had constructed his problems and came up with this one:

Seven people (A, B, C, D, E, F, G) want to ride into Portland but the car will only hold four.  Person “A” owns the car and will always be driving.  People “F” and “G” don’t get along and won’t ride together.  If it makes no difference where in the car people sit, how many ways are there for four people to ride into Portland together?

## Confounding Calculus

Another fun Calculus problem we ran into the other day involved using the quotient rule versus using the product rule.  This problem also came from the Briggs and Cochran book in the section on Implicit Differentiation (pg. 162 #15).  So, the problem this time started out as

Use implicit differentiation to find $\frac {dy}{dx}$.

$x^3=\frac {x+y}{x-y}$

So, if we start out by differentiating the expressions on each side, we get

$3x^2=\frac {(x-y)(1+y')-(x+y)(1-y')}{(x-y)^2}$

If we expand and simplify a little -

$3x^2=\frac {(x+xy'-y-yy')-(x-xy'+y-yy')}{(x-y)^2}$

$3x^2=\frac {x+xy'-y-yy'-x+xy'-y+yy'}{(x-y)^2}$

$3x^2=\frac {2xy'-2y}{(x-y)^2}$

So then

$3x^2(x-y)^2=2xy'-2y$

and

$\frac{3x^2(x-y)^2+2y}{2x}=y'$

This is the answer that appears in the text – but, if we assume that $x\neq y$, then we can clear the denominator in the original problem and use implicit differentiation on the result.

$x^3=\frac {x+y}{x-y}$

$x^3(x-y)=x+y$

$x^4-x^3y=x+y$

Then, differentiating

$4x^3-(3x^2y+x^3y')=1+y'$

or

$4x^3-3x^2y-x^3y'=1+y'$

Then, to isolate $y'$:

$4x^3-3x^2y-1=x^3y'+y'$

$4x^3-3x^2y-1=y'(x^3+1)$

and

$\frac{4x^3-3x^2y-1}{x^3+1}=y'$

I spent some time playing around with these, and believe me, they are not the same.  I graphed them and here is what the first answer $\frac{3x^2(x-y)^2+2y}{2x}=y'$ looks like:

And here is the second version, $\frac{4x^3-3x^2y-1}{x^3+1}=y'$

I tried plugging values for $x$ and $y$ into these different expressions for the derivative of the same function to see if I got the same answer.  This led me to realize the solution to the conundrum!

I started by plugging in some simple values for $x$ and $y$ like $x=1, y=1$ but quickly realized that this wasn’t valid because $x\neq y$.  So then I tried $x=1, y=0$ and for these values, the two expressions were the same!

$\frac{3(1)(1-0)^2+2(0)}{2(1)}=y'$

$\frac{3}{2}=y'$

and

$\frac{4(1)^3-3(1)^2(0)-1}{(1)^3+1}=y'$

$\frac{4-1}{1+1}=y'$

$\frac{3}{2}=y'$

These are the same – because the point (1,0) satisfies the original function: $x^3=\frac {x+y}{x-y}$.

So I got out the Excel spreadsheet and found a series of points that satisfied the original function and tried a few of these points in the two expressions for the derivative and for each one the value of the two expressions is the same!

So next I was determined to show that assuming the original function was true would show the equality of the two different expressions for the derivative.

So, assuming $x^3=\frac {x+y}{x-y}$ we need to turn $\frac{4x^3-3x^2y-1}{x^3+1}=y'$ into $\frac{3x^2(x-y)^2+2y}{2x}=y'$.

First I substituted $\frac{x+y}{x-y}$ in for $x^3$, so:

$\frac{4x^3-3x^2y-1}{x^3+1}=y'$

$\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{x+y}{x-y}+1}=y'$

$\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{x+y}{x-y}+\frac{x-y}{x-y}}=y'$

or

$\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{2x}{x-y}}=y'$

Then I cleared the fraction in the numerator and denominator:

$(\frac{x-y}{x-y})\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{2x}{x-y}}=y'$

$\frac{4(x+y)-3x^2y(x-y)-1(x-y)}{2x}=y'$

OK, so this is

$\frac{3x^2y^2-3x^3y+4x+4y-x+y}{2x}=y'$

or

$\frac{3x^2y^2-3x^3y+3x+5y}{2x}=y'$

I saw that where I was headed with this had a $2y$ and a $3x^2$ in it, so I moved a few things around and ended up with:

$\frac{3x^2y^2-3x^3y+3(x+y)+2y}{2x}=y'$

Going back to the original function $x^3=\frac {x+y}{x-y}$ I saw that I could substitute $x^3(x-y)$ in for the $(x+y)$:

$\frac{3x^2y^2-3x^3y+3(x^3(x-y))+2y}{2x}=y'$

$\frac{3x^2y^2-3x^3y+3(x^4-x^3y)+2y}{2x}=y'$

$\frac{3x^2y^2-3x^3y+3x^4-3x^3y+2y}{2x}=y'$

Factor out $3x^2$ from the everything except the $2y$ and this is starting to look good:

$\frac{3x^2(y^2-xy+x^2-xy)+2y}{2x}=y'$

And there it is:

$\frac{3x^2(x-y)^2+2y}{2x}=y'$

- sigh of relief -

## Inequalities – Graphical and Algebraic

I’m teaching a Calculus course this year and came across an interesting problem the other day.  We were working from the Briggs & Cochran Calculus book and discussing the quotient rule.  Problem #40 on page 127 says given the function:

$f(t)=\frac {3t^2}{t^2+1}$

a) Find the values of $t$ for which the slope of the curve is zero.

b) Does the graph of $f(t)$ have a slope of $3$ at any point?

OK – so we need to apply the quotient rule and find $f'(t)$

$f'(t)= \frac{(t^2+1)(6t)-(3t^2)(2t)}{(t^2+1)^2}$

OR

$f'(t)= \frac{6t^3+6t-6t^3}{(t^2+1)^2}$

$f'(t)= \frac{6t}{(t^2+1)^2}$

OK – so part (a) is easy enough.  The derivative is equal to zero when $x=0$.  But what about part (b) ?  We can graph the derivative and see that it’s always less than $3$, but what about algebraically?

The problem can be posed in a number of different ways.  I chose to set it up as:

$3> \frac{6t}{(t^2+1)^2}$

so

$3(t^2+1)^2 > 6t$

and

$(t^2+1)^2 > 2t$

Again, we can graph this and SEE clearly that the graph of $(t^2+1)^2$ is definitely above the graph of $2t$, but I like to know what’s happening algebraically.  In fact, I used to give my College Algebra students algebraic inequalities as an introductory project, so I really wanted to figure this out.

Of course, this came up in the middle of class, so we looked at the graph and I began to consider the algebraic reasoning, but didn’t get far, so I told the class that I would think about it while they had their quiz – but still to no avail.  I worked on it during my free time yesterday and made some progress, then when I came in this morning it all fell together:

$t=0$

$\frac{6t}{(t^2+1)^2}=0$

OK, done there.

$t<0$

$6t<0$

$(t^2+1)^2>0$

$\frac{6t}{(t^2+1)^2}<0$

Again – ok, done.  What was the sticking point for me yesterday were the positive values of $t$. So:

$t>0$

$(t-1)^2\geq0$

$t^2-2t+1\geq0$

So, $t^2+1\geq2t$

Also, $t^2+1>1$

So, $(t^2+1)(t^2+1)>1(t^2+1)\geq2t$

So now we know that $(t^2+1)^2>2t$ and can work our way back to

$3> \frac{6t}{(t^2+1)^2}$

For some reason, I find this beautiful!

## The Pirate’s Treasure

I was remembering today a contest problem I encountered back in 1999-2000 about pirates and buried treasure.  At first, I couldn’t remember the problem, then when I found multiple on-line versions of it (physicsforums.com, mathpages.com, Bradley University, geometer.org, the mathfactor podcast, and University of Georgia), I couldn’t remember how I had solved it!  It apparently appears originally in George Gamow’s One, Two Three,…Infinity.

The essential features of the problem are that two pirates arrive on a desert island on which there are three prominent trees.  Choosing the most prominent tree as their starting place, the pirates walk first to one tree, counting their steps.  When they arrive at the tree, they turn 90 degrees to the right and walk the same number of steps and mark the spot.

Then they return to the starting place and walk to the other tree, again counting steps.  When they arrive at the second tree, they turn 90 degrees to the left and walk the same number of steps and mark that spot.  They then bury their treasure mid-way between the two marked spots.

Upon returning to the island some time later, they find that the most prominent tree that was their starting point is now gone.  Can they still find the treasure?

Once I was sure I had the right problem, I remembered finally that I had originally used a coordinate proof.

The comments in the Math Factor podcast contain a nice solution using coordinate geometry with a nifty ending that makes finding the treasure fairly simple.

The solution in Gamow’s original apparently uses complex numbers to solve it.

The University of Georgia site asks students to find four separate proofs using 1) complex numbers, 2) Euclidean geometry, 3) coordinate geometry and 4) vector algebra.

## Fun Stuff

I was reading Kate Nowak’s blog f(t) yesterday and followed a link to the Phillips Exeter Academy website where they’ve posted a pretty amazing collection of problem sets for all levels of math from Pre-algebra to Calculus (and beyond!).

I’ve only just begun to work on some of these, but I’m having a good time so far.

Can you find a fraction so that the difference between the fraction and its reciprocal is exactly equal to 1?

That is $\frac ab - \frac ba=1$

The problem gave the example that $\frac85 - \frac58 = \frac{39}{40}$

D’oh – off by $\frac{1}{40}$

Then the question asks – Can you find another fraction that gets closer than this?

I approached this from a couple of different ways – first I took the original equation $\frac ab - \frac ba=1$ and created a common denominator to get $\frac{a^2 - b^2}{ab} = 1$ or $a^2 - b^2 = ab$ and $a^2 - ab - b^2 = 0$.

I didn’t pursue this past that point, but did come back to it later.

Next, I broke out the spreadsheet and set it up to take all the numbers from 1-30 and create fractions and their reciprocals from these and subtract them.

Looking at all that, I noticed a few places where the difference between 1 and the $\frac ab - \frac ba$ was smaller than $\frac{1}{40}$.

This happened for $\frac {13}{8} - \frac {8}{13}$ and $\frac {21}{13} - \frac {13}{21}$.  Then it was time for class.

The numbers in the fractions that were getting close to 1 had caught my eye yesterday and this morning when I came in, I sat down with it again and saw that they were all consecutive Fibonacci numbers.  So, I made a new spreadsheet with Fibonacci numbers and the fractions and reciprocals and noticed that the difference $\frac ab - \frac ba$ was approaching 1.

At some point yesterday afternoon I went to Wolfram Alpha and typed in $a^2 - ab - b^2 = 0$ just to see what I would get and it provided a relationship between the two variables that comes from treating one of the variables as a constant so that

$a=\frac b2 (1 \pm \sqrt{5})$

In other words a and b cannot both be integers!  And of course, some of you may have picked up on this sooner than I did – the number that produces the exact value of 1 is phi – The Golden Ratio.

$\phi - \frac {1}{\phi}=1$.