31,769 signifigant digits. DAMN. 😁
Platonic Solids
55 posts
Pi and e to 100000 decimal places, courtesy of Mathematica. It took about 3 seconds to calculate it. I would have done them to 1000000 but that would have probably taken about half an hour.
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I could have sworn that I attached it...
I could have sworn that I attached it...
The only reason that huge amount of significant digits is useful is for statistical analysis for randomness and such. For most practical purposes, 3.1415 is enough precision 🙂 (whenever I say this to a mathematician, they usually attempt to come up with some excuse why a larger precision would really be better!).
Don't usually find myself over here, but my eye was caught by the title.
Has anyone tried any concave polyhedra yet.
May I suggest the Great Dodecahedron - it is formed from just 12 intersecting pentagons and looks easy ( but what do I know ).
It also breaks the Euler formula that links vertices, edges and faces.
Sorry to intrude, please carry on 🙂
UJ
Has anyone tried any concave polyhedra yet.
May I suggest the Great Dodecahedron - it is formed from just 12 intersecting pentagons and looks easy ( but what do I know ).
It also breaks the Euler formula that links vertices, edges and faces.
Sorry to intrude, please carry on 🙂
UJ
The only problem is that you need a good way of depth-culling each face against the others. This isn't easy in AVS. You could simulate a z-buffer as color value (using maximum blend), but then you'd need to tesselate each face into short line segments to get any sort of accuracy.
Or you could split it up in non-intersecting triangles, guaranteed to be fun :P.
Or you could split it up in non-intersecting triangles, guaranteed to be fun :P.
A relatively simple way to not draw hidden faces is to make sure that the co-ordinates are entered in a certain order, clockwise or anti-clockwise, then to see if from the perspective of the viewport the co-ordinates are being place clockwise or anticlockwise, if they are in the wrong order then you don't draw the face. The only problem with that method is that it only works 100% for convex objects.
As for pi I can remember being told something like 'if the universe were a perfect sphere and its radius was the distance from us to the furthest known object, then you could calculate it volume to within a cubic meter of accuracy by using pi to 10dp'. I can't remeber the exact number, it could have been 15 or 20dp but it was definately a lot less than 30000.
EDIT: The other thing is that if you are doing a calculation where you need to multiply pi several times then the error will increase more and more.
As for pi I can remember being told something like 'if the universe were a perfect sphere and its radius was the distance from us to the furthest known object, then you could calculate it volume to within a cubic meter of accuracy by using pi to 10dp'. I can't remeber the exact number, it could have been 15 or 20dp but it was definately a lot less than 30000.
EDIT: The other thing is that if you are doing a calculation where you need to multiply pi several times then the error will increase more and more.
Jher: I guess I should've used a different term. If you look at that great dodecahedron you can see every face intersects practically every other face. Traditional culling isn't going to work here, you need z-buffer-ish techniques to handle that. And getting a smoothly shaded pentagon isn't easy either.
By the way, I haven't actually encountered powers of pi in any formula. I know about pi^2/6 as the limit of some expression, but that's about it.
By the way, I haven't actually encountered powers of pi in any formula. I know about pi^2/6 as the limit of some expression, but that's about it.
Yeah, but you could split it into triangles like you said. I'm sure that the normal back face culling method would work fine then, although i remember reading somewhere that it fails for certain convex objects.
I'll let the powers of pi thing slip though, and just pretend that you didn't say that (any trig function working in degrees/graduations). I'll admit that having pi to any excessively large accuracy serves no practical purpose yet.
I'll let the powers of pi thing slip though, and just pretend that you didn't say that (any trig function working in degrees/graduations). I'll admit that having pi to any excessively large accuracy serves no practical purpose yet.
Number of parsecs (Smallest possible unit, WITHOUT involving quantum physics) between 2 edges of universe = Digits required.
Thats a huge number of digits... several millions of billions.
I'd say in the order of a googleplex or more.
I think it's googolplex though. The search engine Google is derived from that, but it's misspelled on purpose.
I'll let the powers of pi thing slip though, and just pretend that you didn't say that (any trig function working in degrees/graduations).No, I'm serious. The degrees-to-radians conversion just uses pi or 1/pi, but no powers with absolute value above 1. What do you have in mind then?
cos x= 1 + (x^2)/2 + (x^4)/4! + (x^6)/6!...
You plug anything involving pi into that and you get powers of pi. You can't convert the value of cos x afterwards, not without using another infinite series (as far as I know).
You plug anything involving pi into that and you get powers of pi. You can't convert the value of cos x afterwards, not without using another infinite series (as far as I know).
I believe you're mistaken, Jheriko. The series is closer to:
inf
E (x^(4n))/((4n)!)-(x^(4n+2))/((4n+2)!) = cos x
n=0
(E being the closest thing to epsilon I could think of and inf being infinity)
inf
E (x^(4n))/((4n)!)-(x^(4n+2))/((4n+2)!) = cos x
n=0
(E being the closest thing to epsilon I could think of and inf being infinity)
Oh crap.. i missed some more minuses. There is a better way to express that series.
BTW: E should be a capital sigma to represent summation, epsilon is used to represent arbitary small values that tend to zero. You can use the symbol font for those greek letters.
Anyway, what you want is:
n->inf
S(-1^(n) * x^(2n)/(2n)!) = cos(x)
n=0
and replacing 2n with 2n+1 for sin(x).
You can also express them as a product of a series or in terms of hypergeometric functions, either way you still end up with a power series of some kind.
BTW: E should be a capital sigma to represent summation, epsilon is used to represent arbitary small values that tend to zero. You can use the symbol font for those greek letters.
Anyway, what you want is:
n->inf
S(-1^(n) * x^(2n)/(2n)!) = cos(x)
n=0
and replacing 2n with 2n+1 for sin(x).
You can also express them as a product of a series or in terms of hypergeometric functions, either way you still end up with a power series of some kind.
The taylor series expansion of cos x can hardly be seen as a formula 🙂. Besides, it still has nothing to do with the accuracy of pi itself. You'll never fill in pi in that series, because cos and sin of pi are well known values 😛.
Lets say that you want to find cos of any number in degrees then
cos(x) becomes cos(x*pi/180). or you might want to do cos(pi/7) or something.
cos(x) becomes cos(x*pi/180). or you might want to do cos(pi/7) or something.
Jheriko I think that the value of cos(pi*n) whith n being an element of Q can be calculated using formulas for sin(x+y),cos(x+y),sin(x/2),cos(x/3) and such.
Ugh...consider the pi issue closed :P.
sorry to have started it... 🙁
Aww crap...I'm taking Greek, I should know that shit :P
And one of the series for pi is something like:
inf
E 1/(2n)-1/(2n+2)
n=0
I believe that converges on pi/4
NOW it's closed 😉
*mumbles, 'alpha, vita, ghamma, thelta, epsilon, zita, ita, thita, yota...'*
And one of the series for pi is something like:
inf
E 1/(2n)-1/(2n+2)
n=0
I believe that converges on pi/4
NOW it's closed 😉
*mumbles, 'alpha, vita, ghamma, thelta, epsilon, zita, ita, thita, yota...'*
Bah, f(x) = x/((1-x^2)^.5) is the inverse function of a sigmoid. It's also the derivative of a semi circle. Circles use pi. I'm sure there's some way of calculating pi by using this relationship.
Completely off topic, if you want to find the derivative of a whole circle you can use a parametric equation like x=cos t, y=sin t and find the derivative by using:
dy/dx=(dy/dt)/(dx/dt)
This will give an expression for dy/dx in terms of t (for a circle its -cot t), which you can replace with x by doing some substitution. You can find higher derivatives by using the chain rule and solving some simple differential equations.
dy/dx=(dy/dt)/(dx/dt)
This will give an expression for dy/dx in terms of t (for a circle its -cot t), which you can replace with x by doing some substitution. You can find higher derivatives by using the chain rule and solving some simple differential equations.
😔 hmm... only an exercise: