Julia Set Animation Part 5 – degree 6 polynomial Julia sets

This is WuFeng again with another Julia Set animation video. This fractal is generated by representing each pixel as a complex number and then calculating a color value based off of an iterated function system. Here is the another Julia Set animation.

This Julia Set animation is created from a set of 7200 different Julia Sets on the equation z_{n+1} = z_n^6 + c rather then the famous z_{n+1} = z_n^2 + c, and here the values of c = .8 \cos(x) + .8 i \sin (x)

Here are selected images at higher resolutions:

Selected Images

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For more Fractals images visit the Fractal Corner of the Fun Corner.

Pi Day Countdown #10 – Pi and the triangle Part 2

From the previous Pi day post we know that the function arc tangent can be used to calculate Pi, the only problem is now to approximate the value of arc tangent itself. This can be done by using a technique from calculus called Taylor series, which transforms a function into an infinite series of terms that when added produced the exact value desired.

The Taylor series of arctangent is as follows:

arctan(x) = x + x^3 / 3 - x^5 /5 + x^7/7 - x^9/9 ...

However since we know that arctan(1) = \pi /4, we only need to substitute in 1 to get a good approximation of \pi.

arctan(1) = 1 + 1 / 3 - 1 /5 + 1/7 - 1/9 ... = \pi / 4

However since it is also impossible to add up all infinite terms, the more terms that we compute this series for, the better the approximation we get for \pi / 4.

PS: For more Pi Day countdown articles, visit the fun corner

Pi Day Countdown #9 – Pi and the triangle Part 1

The study of the triangles or trigonometry is an ancient branch of geometry studied by the Greeks, Indians and Babylonians. Trigonometry deals with the study of angles and lengths of a triangle, and indirectly the circle, because of this, trigonometry gives a powerful tool to approximate and solve Pi.

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One of the ways that trigonometry can be used to approximate Pi is via equations known as trig formulas. One such formula is called arctangent (arctan(x)) which as a function, will give the degree of any angle, based off of the ratio of the opposite and adjacent sides. arctan(x) however will only work for right triangles and the reference angle cannot be the right angle.

Now since it takes a 360 degree rotation to form a circle and since the circumference of the unit circle is 2 \pi, this means that degrees can also be represented by fractions of 2 \pi. This unit if measure is called a radian. For example 60^{\circ}  is \pi / 3, 45^{\circ} is \pi / 4.

So with proper values, an approximation of arctan(x) will give an approximation of \pi. A good case to start with is arctan(1). This right triangle has an angle of degree 45^{\circ}, which means in radians the angle is worth \pi / 4.

Eureka! All we need now is to find an approximation for arctan(1) and we will be able to find an approximation for \pi / 4.

The approximation technique will be continued in part 2.

PS: For more Pi Day countdown articles, visit the fun corner

 

Pi Day Countdown #8 – Pi and Viete

In the 16th century a French lawyer named Francois Viete discovers a simple yet profound formula for computing Pi.

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Francois Viete, 1540 – 1603

Viete starts by using similar method to Archimedes. He starts by drawing a circle and then inscribe into it a square. By assuming that the circumference of the circle is equal to the perimeter of the square he estimates \pi to 4/ \sqrt{2}, approximately $latex 2.828… $. Viete then doubled the number of sides, the approximation then becomes:

2 * 2/ \sqrt{2} * 2/ \sqrt{2 + \sqrt{2}} \approx 3.061..

Each time Viete doubles the number of sides, he not only improves the approximation but he is multiplying into his approximation by a factor that is growing closer and closer to zero. For example the approximation of \pi for the octagon, is the approximation of \pi for the square times 2/ \sqrt{2 + \sqrt{2}}. From there Viete creates an infinite product that will give the exact solution for \pi:

\pi = 2 * 2/ \sqrt{2} * 2/ \sqrt{2 + \sqrt{2}} * 2/ \sqrt{2 + \sqrt{2 + \sqrt{2}}}* 2/ \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}} * ...

Notice that each term as a certain pattern to it. Each term will will have more nested square roots then the one previous to it. Unfortunately in order to produce an exact value of \pi one will need an infinite number of terms.

PS: For more Pi Day countdown articles, visit the fun corner

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Pi Day Countdown #7 – Pi and Infinity

Pi is the ratio of the circumference of the circle divided by its diameter. Simple, yet finding the value of Pi is complicated.

Great thinkers of the ancient world attempted to approximate Pi by using geometric techniques and theories known to them. Archimedes for example attempted to approximate Pi by estimating the length of the circumference. He did this by inscribing a regular polygon and using the circumference of the polygon as an approximation of the circumference. Liu Hui attempted to approximate Pi by inscribing a regular polygon and then using the area of that polygon as an approximation of Pi.

However in the late 17th century, another method of approximating Pi was discovered. Instead of using geometric shapes, mathematicians used a type of equation called an infinite series. An example of one of these used to approximate Pi is:

\pi / 4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...

Notice that here one can approximate \pi/4 simply by adding and subtracting all fractions where the denominator is an odd number. Of course in order to get the exact value of \pi one would need to add an infinite number of terms.

PS: For more Pi Day countdown articles, visit the fun corner

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