A series expansion which occurs often in applications is the binomial expansion. In this article we will discuss about

• What is binomial expansion?
• How to arrive at that expansion?
• Where it is used?
• Its relation with pascal’s triangle
• Issac newton’s approach to binomial expansion
• A ‘C’ program to calculate the binomial coefficients

A "Binomial" is a polynomial expression with two terms, like x+y, x+y^2 etc

"Binomial expansion" is the expansion of the expressions like (x+y)^3, (x+y)^6, where the entire binomial is raised to some power.

First, we’ll look at the expansion of "generic" binomial x+y, and its powers (x+y)^2, (x+y)^3, …     (x+y)^n.

————–(1)

We can see some kind of patterns in the powers of x and y and the coefficents from the above expansions.

Lets first look at the powers of x and y ,

Power of (x,y) in the (k)th term

For example in,

The power of x and y in the 2nd term (K) is 1 and 1 respectively.

Similary, the power of x and y for equations (1) is given in Table (1)

Table1: Power of (x,y) in the (k)th term

From Table(1) it is clear that (x+y)^n will have n+1 terms, the ‘k’ th term will have a (n-k+1)th power of x, and (k-1)th power of y.

Now lets look at the coefficients for equation (1) 

The coefficients in table 2 can be written as shown in figure 1

This is called pascal’s triangle and it is easy to generate any rows of the triangle. Just put 1’s along the sides of the triangle and add the pair of values from the previous row to get the next element. For example to get value 3 in the third row we need to add the pair of values from the previous rows, which happens to be 1 and 2.

Now lets try to expand (x+y)^n. If B(n,k) is the kth entry in the nth row of the pascal triangle. Then

——(2)

Its summation equivalent is

—————(3)

If we start k from ‘0′ instead of ‘1′ eqn (3) becomes

—————– (4)

C(n,k) in equation (4) is called as binomial coefficient and its value is

          ————— (5)

The above expansion was known long before the times of newton (1642-1727). However i like to share an  interesting information on how he arrived at the above expansion while he was investigating on calculus.

We all knew that Wallis laid the foundation of modern integral calculus by finding the area under the parabola (Refer previous article- Wallis method of integration). In the similar way Wallis also determined the area of quadrature of circle which is,

————-(6)

The work of Wallis was continued by Newton who replaced the fixed upper limit of unity in equation (6) by x and obtained the below results

———–(7)

After noticing some kind of patterns in equation (7) he concluded that

and by differentiating on both sides he discovered the series

If -x^2 =x, the binomial theorem as given in equation(4) is revealed.

 ——(8)

and if we put,

Equation (8) becomes,

 ——– (9)

and in DSP, equation(9) is the Z-Transform of the right sided signal.

Further this expansion is very useful if we expect some terms to cancel out in our derivations!

Implementation of Binomial Coefficient:

To calculate binomial coefficients we just need to implement the below equation,

 ————-(10)

However it is not easy as it looks, since "n!" is highly prone to overflow errors.

There are two ways to solve this problem

1. a) Determine the factorial after taking log on both sides of the above equation.

          b) To get the actual output take the exponential of step (a)

    2.    Recursive implemetation of Equation (10)

Recursive implementation to calculate binomial coefficent can be downloaded here

References:

1. A.Antoniou, On the Roots of Digital signal processing – Part 1
2. Wikepedia, Binomial coefficent

From the previous articles, it should be clear that archimedes proved the results using geometry alone. For the sake of simplicity and understanding we have used  geometric series, harmonic mean terminologies. The shift from geometrical thinking to arithmetic happend only during the early 1600s by cavalieri and John wallis. They developed the techniques for finding the areas of various geometric figures by extending the archimedes method using arithmetics and Algebra.

Wallis method of Integration:

Wallis was intrested to find the area under the parabola y=x^2

For that he divided the area under parabola into rectangles as shown in figure 1. He noticed that

Area abcfa = k^2 * ε

and Area abdea = n^2 * ε

Therefore if we add up the areas of the rectangles in the above figure we will be closer to the area under the parabola which is given by

Where,

= The area of the rectangle ABCDA

Till this point it is a geometrical work by wallis. However he generalized the above relation analytically using the principle of induction as follows.

Infact, before this work, Wallis had observed in general that,

———- (2)

Sub of (2) in (1) gives

             ———- (3)

Wallis is the person who proposed the symbol for infinity (∞) and also said that the above relation can be better approximated as ‘n’ tends to infinity

 ————– (4)

This work can be much appreciated with the help of an example

To estimate the area under the parabola y = x^2 from 0 to 1, we will draw rectangles from left to right of the curve as shown in figure 2. 

Using Equation (3) the area under the parabola in figure (2) is estimated as

= 0.375

Where, n = 4 (Number of rectangles) and  = 1 ( Area of rectangle ABCDA )

If ‘n’ equals infinity equation (4) is used

Which gives Ap = 0.33333

This process is the same as, 

Wallis method of Integration is the base of modern integration.

References:

1. Jacqueline A.stedall, The Discovery of wonders: Reading between the lines of John wallis’s Arithmetica infinitorium, Springer-verlag, 2001
2. A.Antoniou, On the roots of digital signal processing – part 1

In the last article (Dated: 1 July, 2008) we had discussed about the origination of  successive approximation technique, discretization process and its relation with interpolation. Another intellectual achievement of archimedes is the invention of integration. 

In this article,

• We will discuss about the origination of Integral calculus by explaining his method (geometrical proof) to calculate the area enclosed by a parabola
• We will solve an example to gain more knowledge on the process. In this example, the area enclosed by a parabola is determined using modern integral calculus and the result is compared with Archimedes method of calculation.
• The relationship between geometric series and integration will also be exploited using Archimedes method

In a letter (Quadrature of parabola) to his friend Dositheus, archimedes wrote that the total area under the parabola [Fig 1] is 4/3 of the area of the blue triangle (i.e).

Area under the parabola = 4/3 * Area of the blue triangle

To prove this archimedes first determined that blue triangle has 8 times the area of each green triangle, each green triangle has 8 times the area of each yellow triangle and so on

(i.e)  He proved that Δ ACB = 8 * Δ ADC [Refer Fig 2]

Proof:

In Figure 2, M is the midpoint of the chord AB. A property of parabolas in general is that if AM is parallel to DE then

AM^2 / DE^2 = MC/EC      ———– (1)

Archimedes refers to the classical work on conics by Euclid and aristaeus for the derivation of this property [ Refer: The Works of Archimedes, Page:235  ]

Further in figure 2, P is the midpoint of AM

Therefore, AM = 2 * DE      ———– (2)

Substitute (2) in (1)

MC/EC = AM^2 / DE^2 = (2*DE)^2/DE^2 = 4 * DE^2/DE^2 =4

Which implies MC = 4*EC   ————— (3)

We know that, MC = ME+EC

Therefore ME = 3*EC , EC = ME/3 ———- (4)

Substituting (4) in (3) Gives

MC = 4/3 * ME = 4/3 * PD     [Since PD = ME]

Δ ADC is similar to Δ DCW, Hence MC = 2 * PW

PW = 2/3 * PD [ Since MC = 4/3 * PD] and PW = 2 * WD.

Therefore Δ ACP = 2 * Δ ADC, Δ ACM = 4 * Δ ADC and Δ ACB = 8 * Δ ADC

=> Δ ADC = (1/8 )* Δ ACB

The triangle formed by the chord CB also has an eight of the area of Δ ACB

Similarly we can inscribe 4 yellow triangles which has an eight of the area of the blue triangle. This process continues indefinetely

Assuming that the blue triangle in Figure 1 has area 1, the total area under the parabola is an infinite sum

1 + 2*( 1/8 ) + 4*(1/8)^2 + 8*(1/8)^3 + ………

Here the first term represents the area of the blue triangle, the second term the area of two green triangles, the third term the area of four yellow triangles and so on. Simplifying the ablove fraction gives

1+1/4+1/16+1/64 + ……

This is a Geometric ratio with common ratio 1/4. The sum of the geometric series is given by 1/(1-r) where ‘r’ is the common ratio

=> 1/1-(1/4) = 4/3

In general we can say that the area under the parabola is 4/3 * area of the blue triangle.

Example:

To gain more knowledge into the proof we will find the area under parabola y = x^2 +1 intersected by a line at y=10 

Archimedes method:

We know by archimedes method, Area under the parabola = 4/3 * Area of Δ ABC

From Figure 3, Area of Δ ABC = 1/2 (b * h)

h=9, b= abs(-3) + abs(3) = 6

=(1/2 )* 54 => Area of Δ ABC = 27

Therefore area under the parabola = 4/3*(27) = 36

Modern (or) well known Integral calculus:

Where f(x) = y-(x^2)

as we had expected.

References:

1.  The Works of Archimedes, Page:235 
2. Gordon swain and Thomas Dence, Archimedes’ Quadrature of the parabola revisited, Mathematics magazine. Vol. 71. No.2 (Apr.1998), pp.123-130  [This article is not freely downloadable. However it is very intresting and worth to pay ]

I wish to start this blog by describing the Archimedes method of calculating the perimeter of circle with a diameter of one unit, which happens to be equal to ∏.  You might ask why to study an old algorithm? or how does a perimeter of circle help us in today’s applications?. You will surely appreciate the below reasons to study this algorithm.

• The connection between the perimeter of a circle to the origin of signal processing can be revealed by looking  at Archimedes method.
• Sine waves have a fundamental period of 2*∏.  Therefore ∏ (Perimeter of circle with a diameter of one unit) becomes vital in signal processing
• Learning history will motivate us to learn advance algorithms.

In this article i will help you to understand the first bulleted point.

Archimedes of Syracuse contributed quite substantially to the theory of mechanics and mathematics. One of his greatest achievement is his method of calculating the perimeter of a circle with a diameter of one unit, which happens to be equal to ∏. He inscribed a hexagon in a circle of diameter of one unit and divided the hexagon into six equilateral triangles as shown in figure 1. Since the diameter of the circle is one unit , the sides of each triangle is half a unit long.

Therefore, perimeter of the inscribed Hexagon = 6 * 1/2 = 3 units

Archimedes concluded that the perimeter of the circle ∏, is larger than 3, the perimeter of hexagon.

a6 = 3 < ∏

Where ‘a6′ is the perimeter of the inscribed hexagon

He then circumscribed the circle by another hexagon as illustrated in figure 2.

To find out the perimeter of the larger hexagon, Pythagoras theorem helped him.

The Length of the sides of the equilateral triangle ABC is calculated as follows:

In right angled triangle ABD,

L^2 = (L/2)^2 + (1/2)^2 [ Pythogoras theorem]

Upon simplification,

L^2 = 1/3 , which implies L = 1/√3

Therefore the perimeter of the larger hexagon is 6*( 1/√3)= 2*√3 = 3.4641

From this Archimedes concluded that the perimeter of circle would be less than 3.4641, i.e.,

a6=3< ∏ <3.4641=b6, where ‘b6′ is the perimeter of the larger hexagon.

Most of us would have stopped at this point. However Archimedes thought about the next step. He doubled the sides of the inside hexagon to obtain a 12 sided polygon and calculated the perimeter of the inside polygon(a12). Similarly he doubled the sides of of the outside hexagon and calculated the perimeter of outside polygon (b12). He found out that the perimeter of outside polygon (b12) is given by the harmonic mean of a6 and b6.

Harmonic mean is defined as the reciprocal of the arithmetic mean of the two numbers, i.e,

b12 = 1/1/2*(1/a6+1/b6) = 2*a6*b6/a6+b6                  ——–  (1)

and a12 is given by the geometric mean of a6 and b12, i.e;

a12 = √a6*b12                                                                ——– (2)

Using equations (1) and (2) we get,

b12=3.2154 and a12=3.1058. Therefore ∏ value should lie between,

3<3.1058<∏<3.2154<3.4641

Obviously we know that a 12-sided polygon will provide a closer bound on the perimeter of circle compared to the 6 sided polygon.

The genius also noted that equations (1) and (2) readily extend to 24, 48, 96 sided polygons. Infact the relation holds good for 2n sided polygons which is given as,

b2n = 2*an*bn / an+bn

and

a2n = √an*b2n

Using these relations the lower and upper bounds of ∏, given in table 1 is obtained

The average value of ∏ for the 96 sided polygon is referred as the archimedes algorithm. It has an erorr of about 0.008%.

Relation to signal processing:

The bounds of ∏ is due to the discrete functions of ‘n’ and there averages are successive approximations of the perimeter of the circle. The successive approximations using geometric and harmonic mean is a method of interpolation and the discrete functions of ‘n’ can be depicted as the discretization process in modern terminology.

Approximation of circle interms of polygon was done independently in china where Tsu Chung-Chi(430-501) found the more precise bound

3.1415926<∏<3.1415927

and said that 22/7 is an inaccurate value of ∏ and termed 355/113 as the accurate value of ∏

Software:

Access the below link to download the ‘C’ program. The program works upto 6144 sided polygons.

Download Archimedes method

References:

1. A.Antoniou, On the Roots of Digital signal processing – Part 1
2. Indexes of Biographies. The Mactutor History of Mathematics Archive