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Understanding Pi


violetendncy

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Hello friends, it has occured to me that I know depressingly little about mathematics in general. I mean, I can do all the stuff in class they want you to do, but I really am not comfortable with any of the stuff. In saying this, I mean that I feel as though there are certain things I haven't learned....probably because they were attempted to be taught to me by southern women entertaining too many children....

 

Anywise, I was fiddling around with a circle the other day, and I wrote a TI-BASIC program that computes the Pi constant from scratch. If any of you are interested, here is the code (Ti-83 plus):

 

:PROGRAM:CIRCLE

:ClrHome

:Input "Radius:",R

:0-->F

:R-->A

:A-->B

:3-->K

:sqrt( (AA) + (BB))-->C

:2AB-->E

:F+E-->F

:While K<200

:0.5©-->A

:(R-sqrt( (RR) - (AA) ))-->B

:sqrt( (AA) + (BB) )-->C

:(2^(K-1))(AB)-->E

:(1+K)-->K

:F+E-->F

:End

:Disp "Area=",F

:Disp "Pi=",(F/(RR))

 

Now, you can follow the program flow for a bit and realise how close it gets. Actually, I think I've worked it so that it calculates EXACTLY pi. But I am just unsure. As I said, I'm probably making it all too simple....but basically, It divides a circle into polygons and then solves for the right triangles that emerge from continuously halving your angles, beginning with 360 degrees. I wish I could post a picture to illustrate this more properly, but alas that's not the point.

 

What I want to know is if any of you have better ways of explaining pi (and/or calculating it).

 

Also, I'd like to know if any of you can show some manner of dissection for the volume of a sphere, a priori.

 

Thanks!

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I'm pretty sure that discription is accurate. Isn't pi calculated using integration, the area under the curve using an infinite number of rectangles? I know that's how the formula's for rectangles and triangles are derived. Can anyone expand?

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Isn't pi calculated using integration, the area under the curve using an infinite number of rectangles?

 

I think that's what I did, the program finds triangles to double their area and then multiply by some power of two.

 

Another thing that bothers me about this method is that I'm relying on the Pythagorean Theorem so heavily, but I really don't know how or why it was originally developed....I mean, did that school of thought or whatever just arbitrarily stumble upon the fact?

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I think that's what I did' date=' the program finds triangles to double their area and then multiply by some power of two.

 

Another thing that bothers me about this method is that I'm relying on the Pythagorean Theorem so heavily, but I really don't know how or why it was originally developed....I mean, did that school of thought or whatever just arbitrarily stumble upon the fact?[/quote']Maybe. But it is certainly proven in Euclid's Elements. What should really bother you is how you legitimise the limiting process.

 

Being that it's Christmas, I'm a little too drunk to figure out what you have done, but it looks similar to the first half of Archimedes' method, which is perhaps the oldest evaluation of pi.

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Hi there,

i see pi as being the area of a circle with radius 1unit.

considering a circle is a equilateral polygon with infinite sides:

ergo area of polygon= (∞-2).1/2.sin(∞-2).pi/∞

 

As has been stated only the boundaries can be worked out

see if it helps

**Happy Holidays Guys!!

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infinity -2.. or pi/infinity makes no sense, as those binary operations involving "infinity" is undefined.

 

you could however circumsribe the circle with a polygon outside and inside the circle... then take the infimum of all the polygons outside the circle and supremum of all the polygons inside the circle and if those two value correspond, then thats the value of the area of the circle...

 

thats just the basic principle of reimann integration.. where u basically take step functions above and below the function

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if you think of a circle as a polygon with infinite sides, it will have infinite triangles-2. so if you use the formula to calculate the are of a triangle n

n=number of sides

A=1/2abSIN© you can substitute C for (n-2)pi/n

if you then multiply the area of 1 triangle by infinity-2 you will have pi or close enough.

try it out with 99999 sides

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