What is the Exact Number to Pi?

[Anything]

Pi

does anyone know the exact number to pi (yes the one over 1000 characters long:P) i need it a.s.a.p

 

[amcavoy]

That is only an approximation. You can find certain amounts of digits on Google, but they aren't exact.

 

[VietDao29]

I don't know what's that for... But anyway, here it is:
Pi to 1,000,000 places.

 

[Zurtex]

does anyone know the exact number to pi (yes the one over 1000 characters long:P) i need it a.s.a.p

You can't write it fully in decimal form as it is irrational.

 

[hypermorphism]

does anyone know the exact number to pi (yes the one over 1000 characters long:P) i need it a.s.a.p

You can calculate pi's decimal expansion using the algorithms here.

 

[finchie_88]

Why do you need pi to that accuracy, and why have people got computers working out pi to that number of decimal places? What is the use, I don't get it...

 

[shaner-baner]

pi approximations

There is a mathematician named Borwien who has come up with some neat algorithms. The best one in terms of speed of convergence vs. complexity
converge quartically, i.e. the number of correct digit roughly quadruples every iteration, furthermore the algorithms only involve square roots and fourth roots, nothing as complicated as Ramujan's stuff. Using this algorithm a mere 15 iterations gives something like a billion digits. His derivation of it is an interesting mix of elliptic intergrals and the arithmetic-geometric mean of a number. What takes computers so long these days is just the mechanics of doing arithmatic on numbers with a billion digits. Look up Borwein if you are interested,

 

[matt grime]

Why do you need pi to that accuracy, and why have people got computers working out pi to that number of decimal places? What is the use, I don't get it...

Conjecture: pi is normal. Normal roughly means that the digits of pi behave randomly in some sense. We can't prove this, but knowing pi to many places has provided strong supporting evidence for it to be true.

Further, having a known (transcendetal) constant is useful for benchmarking alogrithms (ie how fast they converge)

 

[GregA]

I remember trying to figure out a formula for calculating pi a few months ago and came up with: 2x(sin(90/x)) where a very big value of x gets a close approximation for pi.
My calculator could only evaluate it to 10 digits though...the other millions of digits where given in standard form

 

[Manchot]

I remember trying to figure out a formula for calculating pi a few months ago and came up with: 2x(sin(90/x)) where a very big value of x gets a close approximation for pi.
My calculator could only evaluate it to 10 digits though...the other millions of digits where given in standard form

Taylor expand sin(90°/x) = sin(\frac{\pi}{2x}) around x=0, and see what you get. :)

 

[Aditya89]

Can anybody tell me how to plot pi on anumber line?

 

[GregA]

Taylor expand sin(90°/x) = sin(\frac{\pi}{2x}) around x=0, and see what you get. :)

Gonna be a while before I can do that

 

[uart]

Can anybody tell me how to plot pi on anumber line?

Sure. Just place a dot at about 3.14 and then label it \pi.

 

[HallsofIvy]

Can anybody tell me how to plot pi on anumber line?

What do you mean by "plot pi"?

If you mean just mark a point on a line, pi is approximately 3.14. Marking that point should be sufficient. If you mean "construct a line segment, using compass and straight edge, having length pi, given a line segment of length 1", it is impossible. pi is transcendental and so not constructible.

 

[bomba923]

Indeed, as discussed here:
https://www.physicsforums.com/showthread.php?t=99233

 

[Manchot]

Gonna be a while before I can do that

Ok, well basically, using calculus, you can show that

\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...

where x is in radians. (You can easily verify this for yourself on a graphing calculator, if you have one.) This infinite series converges for all x, but it is centered around zero, so it converges for low numbers "before" it converges for high ones. In fact, for numbers that are infinitely close to zero, you can just drop the higher order terms and say that \sin(x) = x. You were taking x to be infinitely high, so \frac{\pi}{2x} would be very low, and you can say that 2x\sin(\frac{\pi}{2x}) = 2x(\frac{\pi}{2x}) = 2(\frac{\pi}{2}) = \pi. Nice, huh?

(Incidentally, I arrived at a formula very similar to yours when I was a sophomore in high school, and knew geometry but not calculus. Would I be correct in guessing that you arrived at it by calculating the circumference of a regular polygon with an infinite number of sides?)

 

[Robokapp]

does anyone know the exact number to pi (yes the one over 1000 characters long:P) i need it a.s.a.p

There is no exact number. It is Irrational. It's sum{Pi^2/2}=1/n^2 If I'm not mistaking...and imagine how many natural numbers there are...

 

[Curious3141]

There is no exact number. It is Irrational. It's sum{Pi^2/2}=1/n^2 If I'm not mistaking...and imagine how many natural numbers there are...

I think you're driving at zeta (2).
\zeta(2) = \Sigma_{i=1}^{\infty}\frac{1}{i^2} = \frac{\pi^2}{6}
The proof of the irrationality (or indeed transcendence) of pi does not depend on that equality, though.

 

[Robokapp]

I don't know what's that for... But anyway, here it is:
Pi to 1,000,000 places.

I wonder...why do you need it? are you trying to crack it? If you succeed let me know!

 

[Aditya89]

What do you mean by "plot pi"?
If you mean just mark a point on a line, pi is approximately 3.14. Marking that point should be sufficient. If you mean "construct a line segment, using compass and straight edge, having length pi, given a line segment of length 1", it is impossible. pi is transcendental and so not constructible.

By that, I meant that can u plot it by any other method. And I have also got the answer now. It is constructible by some other apparatus.

 

[GregA]

Ok, well basically, using calculus, you can show that
\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...
where x is in radians. (You can easily verify this for yourself on a graphing calculator, if you have one.) This infinite series converges for all x, but it is centered around zero, so it converges for low numbers "before" it converges for high ones. In fact, for numbers that are infinitely close to zero, you can just drop the higher order terms and say that \sin(x) = x. You were taking x to be infinitely high, so \frac{\pi}{2x} would be very low, and you can say that 2x\sin(\frac{\pi}{2x}) = 2x(\frac{\pi}{2x}) = 2(\frac{\pi}{2}) = \pi. Nice, huh?
(Incidentally, I arrived at a formula very similar to yours when I was a sophomore in high school, and knew geometry but not calculus. Would I be correct in guessing that you arrived at it by calculating the circumference of a regular polygon with an infinite number of sides?)

very nice!
with regards to how I arrived at the formula, it's a similar method to yours but instead I chose right-angled triangles as opposed to n-sided polygons... on a cartesian graph (my easiest way to describe it) make a circle with it's centre the origin and work on just one quadrant. Make a chord from the point where the circle crosses the x-axis and where it crosses the y-axis. This is now an isocolese triangle...bisect this chord with a line from the origin and make two chords from where this line hits the circumfrence..you now have two isocolese triangles...you can repeat the process indefinitely but if you cut these triangles in half you have x number of right-angled triangles (x also being equal to the number of times 90 degrees has been subdivided into smaller angles) using sin (90/x) and then multiplying this by x you have the approximate ratio of half a semicircle's circumfrence to its radius and you just multiply this by two.