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## Hungry For MATH? Below Are Cases Where Some Tantalizing π Was Found

The ratio of circumference to diameter of a circle is denoted π. This irrational number is deeply imbedded in physics and mathematics.

Speaking rationally about irrational numbers, true math buffs always use the π symbol because π can never be expressed exactly using numbers. Using π=3.14 or to any decimal significance may provide grounds to get chastized in a math class.

Footnote: A legislature may have once decided to make π=3 to simplify computations. That surely would have made the fabric of the Universe shudder and groan.

• ### 1) AREA Under 2D Gaussian [TOP]

The above chart shows the Gaussian 'bell shaped' probability curve. Tha area bounded by the function y=e^(-(x)2)) and the X-axis (for all (x) from -infinity to +infinity) turns out to be the square root of π. This exponential function cannot be integrated in closed form by any known means (last time I checked) but can be easily evaluated by numerical integration in a spread sheet or simple looped computer program. The chart was created in a spread sheet and has added Paint annotations. The blue region under the red Gaussian function represents total area under the curve. Two violet integration elements are drawn (left and right of x=0) to point out that the exponential has symmetry which permits a correct result where 'under-summed area' error on left side is canceled by over-summed area on the right side of x=0. In actuality, the area under the left side is identical to the area under the right side of x=0.

In chosing this method, the total area must be summed from -x to +x. The exact value of π being an irrational number requires that the sum be tallied from -infinity to +infinity, but that would give the π result to an infinite number of places with infinite number of steps. Fortunately, it turns out that yx approaches zero so rapidly that π can be calculated to the accuracy of math in your computer without (x) ranging further than from -6 to +6.

### 2) VOLUME Under 3D-Gaussian [TOP] Created: 24-Jan-2011 07:04 PM

A 3D shape can be formed with area swept out by revolving the 2D gaussian function about the Z-axis. This forms a 3-Dimensional Gaussian solid; and what would the volume be in cubic units?, π of course?, yes. 1st order integration does not converge as well as the 2D case. Using cylindrical integration with 0.01 step size 1st order integration yields a value for π to an accuracy of about one part in 63000. The true exact volume under the 3D Gaussian is π cubic units. I find this fascinating and amazing.

### 3) AREA Under SIN(x)/x Integrated From minus to plus Infinity [TOP]

The function y(x)=SIN(x)/x looks somewhat like a damped COSINE wave whose amplitude=1 at x=0. Central portion of Chart is shown above. The amplitude declines as x in the denominator takes on larger plus or minus values further from x=0. If the sum of white areas below y=0 are subtracted from the sum of red areas above y=0, the total area converges to π when integrated for all x from -infinity to +infinity; chart shown for (x) from -50 to 50 radians.

### 4) SUM Of Series SIN(k)/k From k=1 to Infinity; How cool is this? [TOP]

Found the above 'piece of π' in American Scientist magazine in an article by David T. Kung on P-85; In the author's words; "the surprising sum of the series" I wrote piece of code and found the SUM converges to 1.070796327... I did not recognize this sum as 'surprising' so I emailed Mr. Kung who explained:

"If you double the number you got through numerical summation, you get 2.141592... I'm sure that looks more familiar! Thus the original sum converges to (π-1)/2." All these widely varied forms makes me think the number of summations converging to π or fractions of π may be unlimited. Mr. Kung's article also contained a mention of Fibonacci; see Fibonacci_Stuff.

### 5) SUM Of Alternating Sign Series for k From 1 to Infinity: [TOP]

This simplest looking series of odd reciprocal integer terms SUM(-1)k)/(1-2k) produces the sequence 1-1/3+1/5-1/7+1/9... The above chart shows the convergence SUM times 4 for the first 100 values of k. The series converges very slowly to π/4.

### 6) A Series Of Similar-Looking-SUMations Converging To π [TOP]

This section presents a series of similar looking SUMs that converge to π. The higher the power (n-exponent) the faster the convergence. I started tracking these π summations (some of which) found in class hand out notes in a Mathematical Methods course taught by Dr. E P Gray at JHU/APL ~1984.

Having seen a couple formulas of similar 'form': πn/C=SUM(1/kn) for k=1 to infinity, it did not take a rocket scientist to realize that a series of such even power exponent formulas must exist. I wrote looping algorithm to determine the (C) constant that made each summation converge to π by utilizing a binary convergence technique. Basically the algorithn contained the appropriate summing routine who's 'constant guess' incremented in binary jumps until the sum exceeded π, then the constant was decremented by ½ the difference in the two previous guesses until the SUM was less than π, then again incremented by ½ difference of last two, etc. Such a Binary search converges rapidly; tries = LOG(p)/LOG(2).

Side trip: Pick a number (p) from 0-1000, I make guesses, you tell me if the guess is high or low, Your number (p) will always be found in 10 or fewer tries by employing Binary Search. 1st guess is 500, next guess is 250 or 750 depending on previous guess being hi or lo. Good party trick?

But my (C) seeking algorithm was brute force, beating the bushes, so to speak. So I went to Dr. Gray's office to inquire. I still have his scratch sheet 'off the top of his head' on how to determine the constant given the power for this Similar SUMmation Series. An aside, he worked on other neat problems like spherical harmonics to determine geoids of planetary bodies. Dr. Gray was very well versed in 'π' stuff, trust me. To n=10:

• π2/6=SUM(1/n2) For n=1 to infinity, THEREFORE: π=(6*SUM(1/n2))½
• π4/90=SUM(1/n4) For n=1 to infinity, THEREFORE: π=(90*SUM(1/n4))¼
• π6/940=SUM(1/n6) For n=1 to infinity, THEREFORE: π=(940*SUM(1/n6))(1/6)
• π8/9450=SUM(1/n8) For n=1 to infinity, THEREFORE: π=(9450*SUM(1/n8))(1/8)
• π10/93555=SUM(1/n10) For n=1 to infinity, THEREFORE: π=(93555*SUM(1/n10))(1/10)

### The Code Below Contains 6 BASIC Subroutines For Those Interested In Algorithms [TOP]

Be aware that a computer can only take numerical solution out to the number of places near the significant number of places for math in that computer, usually 12-16. Noting that limitation, I started to develop an algorithm using 12 place math (and strings) to compute functions of invert (1/x) and multiply good to any significant places chosen. I did manage to finish an invert routine, may publish that some time. Inputs were: number of significant places and the number to invert, it works well. To do quotents, just invert the denominator then multiply that by numerator. So a high-significant-figure math invert and multiply routine should be all one needs to compute PI or any summation to a desired fixed place accuracy.

Maybe things like integer summations lead to a quotation attributed to mathematician Leopold Kronecker who once wrote that "God made the integers; all else is the work of man".

• PI Computation Basic Program
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