It looks like you're using an Ad Blocker.
Please white-list or disable AboveTopSecret.com in your ad-blocking tool.
Thank you.
Some features of ATS will be disabled while you continue to use an ad-blocker.
Originally posted by amantine
Originally posted by Nans DESMICHELS
4atan(1)-2atan(142857142857)
0.000000000014000000000014
No, it approximates that. The real answer is: 1.400000000001399999999978533333333 26613333333386981333333646746666665 28794666665250090666669509998222284 00325902221823327094691342655094489 78243061285172058104162119194660237 75810799130537397842460345463572786 42768243070097724234732764952218828 60815089318196497016898223287070808 00499675285137518904359280141809953 69450977776856994023529684191017152 64480512615163104521082110292790539 77859809823095001437644765545856567 51930523580865859801162359386628497 07024483383867440604191331571072722 57703436365728995411837*10^-11
Originally posted by amantineI quote my post in another thread about the irrationality proof of pi we learn at school. A similar proof can be found here.
One proof of irrationality of pi:
Look at the following integral
That iintegral has as values
I(2) = -2^2 + 24
I(3) = -24^2 + 240
I(4) = 2^4 - 360^2 + 3360
I(5) = 60^4 - 6720^2 + 60480
I(6) = -2^6 + 1680^4 - 151200^2 + 1330560
All sums of powers of . We will determine a maximum for the integral:
x(-x) 0.5(-0.5 )
0.5(-0.5 ) = 0.25^2
0.25^2 3
x(-x) 3
Now the other part:
0 sin(x) 1
sin(x) 1
Filling this in and calculating the integral gives you an upperlimit for I(n) of *3^n/n!
Now let's assume is rational (and repeats): =p/q, where p and q are two whole numbers and the expression has been simplified as much as possible.
That means for example with I(4):
(2p^4 - 360p^2q^2 + 3360q^4)/q^4
Because this is an positive numbers that has been simplified as much as possible we can say:
1/q^4 (2p^4 - 360p^2q^2 + 3360q^4)/q^4
This is true for every I(n), so you can say:
1/q^n I(n)
We determined a upper limit for I(n), so 1/q^n is always smaller than that:
1/q^n *3^n/n!
1 *(3q)^n/n!
But lim(x-> ) x^n/n! goes to 0. That would mean:
lim(x-> ) 1 *(3q)^n/n!
gives
1 0
This is ofcourse not true. That problem is our assumption that is rational. Q.E.D.
[edit on 18-6-2004 by amantine]
Originally posted by mOjOm
How about this one:
If you add the first '144' Pi decimals it adds up to 666 too.
Numbers are fun, but I still don't get what you're trying to do here exactly.
Originally posted by Nans DESMICHELS
What I try to show is that when I withdraw 2*atan(142857) (the cycle number) with 4*atan(1), I don't obtain a random pattern, like if I'd tryed it with an random number (like 123456) but a pattern based on 14 (2*7).
arctan(142857)*2=3.14157865............................. = PI
I'm trying to understand why when I do :
2*atan(142857142857142857)
I'm close to PI
and when I do 2*atan(142857142857142857142857)
I'm still closer to PI.
Originally posted by foxtrot_uniform
im lost????????????????????????????????????????????????? isnt pi an infinite number? like no matter how high of a number you can think of you just add one and its bigger?