# The Approximation of Pi

by:

Heather Bridges

Note: The class needs to be familiar with the expanding of expressions of the form ( a+b)^n using the binomial theorem.

Isaac Newton developed a way of estimation the area under a curve back in the late 1600's. His ideas were discovered many years later in De Analysi.

Newton found that if y = a x ^ (m/n) was the form of the curve then

.

Have the students to check to see if this works using a familiar curve lik y = x. If we put y = x in the form as stated above then a = 1, m=1, and n=1. So after plugging in the values into Newtons formula we see that the area under the curve is A = 1/2 x^2.

The students should then use another way to find that same area using the formula for a triangle. The will discover that the area is still the same so Newton's formula passes their test.

Newton then went on to add that if the curve is made up of several terms likewise the area will be made up of several terms using the same methods as above.

The students could us a few examples of this like given the curve
the area under the curve is .

Now , the whole idea is for the students to use the binomial theorem and Newton's formula for estimation the area under the curve to develop the approximation for pi.

Prompt the students to name all the things they know about the semicircle given only the information above. The should be able to come up with the radius, the center, the equation, and the area. Divide the students into two groups. Half of the students work to find the area of ABD by Newton's ideas and the other using straight geometry while the teacher provides assistance.

Area by Newton:

Beginning with the equation for the semicircle, adjust it so it is in the form of Newton's or in other words solve for y. The students should then use the binomial theorem to expand the equation. Then using Newton's method they can find a formula for the area under the curve without substituting in a value for x. Then since we know that x = .25 they can plug in the value using a spreadsheet software to find that the Area is approximately equal to .0767.

Area by geometry:

The first step is to just focus on the triangle ADC. The students can use the pythagorean theorem to solve for the length of AD. Then they can find the area of ADC. The tricky part is for the students to pick up on the fact that the portion ABC of the semicircle is one third of it. Using this they can discover that the area = 1/24 *pi. THen they can subtract the area of triangle ADC from the portion ABC to get the area of the portion ABD.

Then come back together to use all the students information

.0767 = (1/24)pi - ( sqrt(3))/32.

Finally we can solve this to find that pi = 3.142.