On Jun 12, 7:31 am, WM <mueck...@rz.fh-augsburg.de> wrote: > On 12 Jun., 09:17, Rainer Rosenthal <r.rosent...@web.de> wrote: > > > > > > > WM schrieb: > > > > On 11 Jun., 14:13, Rainer Rosenthal <r.rosent...@web.de> wrote: > > >> WM wrote: > > >>> My proof rests upon the fact that after the set of all finite paths of > > >>> the form 0.1, 0.11, 0.111, ... has been constructed, there is no > > >>> chance to construct the path 0.111... in addition. > > >> Why not start with path 0.111... and then add those other paths > > >> 0.1, 0.11, 0.111, ...? > > > > [Many lines snipped, because I felt they didn't get to the point I > > > was trying to ask] > > > However, your contribution to this problem is of high value and I > > > appreciate it very much. > > > Thank you, but it seems as if I wasn't clear in my question. > > Didn't you say: "If I am going to enumerate the real numbers > > in such and such a way then there will be some of them not > > in my list"? > > It is not enumerating all real numbers. > What I do is to use actually infinite paths (at the end we will see > that they do not exist, but I assume their existence) for constructing > the complete binary tree. As a simple example I use only such paths > that have tails of the form 000..., i.e., "terminating" paths. > > The result is: They exhaust the complete infinite binary tree. There > is no node that remains uncovered. And there remains no node and no > set of nodes that could be used to construct a path that is not yet in > the tree. The tree is complete at the end. With the tip of your finger > you can follow every path, even 1/pi or 1/3. > > > > > But then, that doesn't prove all too much, because it's obvious > > that there are many ways of not counting the reals. > > The point is that there is no way of counting the reals. > > As I said, I do not count the reals. I cover all possible infinite > sequences of bits by means of a countable set of paths = a countable > set of binary sequences. > > And your idea to start with some paths like 0.111... or 1/pi that do > not belong to the countable set (of "terminating" paths, i.e., such > with tails 000...) that I use for construction, shows us, that the > result is in no way different. But how can we explain that the result > is not influenced by those paths? In my opinion by recognizing that > such paths do not exist. > > Regards, WM- Hide quoted text - > > - Show quoted text -
Welcome to the "real" mathematics and not the fake "Matrix" movie sorry excuse for chess you've been playing for the last 20 years.
(Founder, Inverse 19 Mathematics) Not the only, but one without it would not exist.
Efforts by mathematicians to determine the value of pi since have led to the conclusion that it ... Unlike rational numbers, pi cannot be represented by a common fraction, ....
But what about an "uncommon fraction"?
Will you look at my drawing and in particular how the twelve points are the outside points of the "sphere" are all "rationally" reached by "decimal" numbers?
6.2035884*2 and doubled an infinite number of times should yield a constant decimal binary expansion (what I mean by this is the numbers to the right of the decimal point should always end in "4, then 8, then 6, then 2, then 4, then 8, then 6, then 2...repeating for infinity).
So in binary, by starting with this number we predict for n>2 the infinite binary tree will end with decimal progression: 2, 4, 8, 6
For n<2 the infinite binary tree will begin and end (depends on if you are increasing or decreasing the value) 1, 5, 5, 5 (repeating 5 infinitely) as we decrease to infinitely small.
If you would like to do the honors Mr. Plouffe, I endear you to calculate the angles and lengths of the number of ways, it is quite obviously possible to "square a circle" based on the geometry and numerical progression.
As is self-evident to anyone with even a modest understanding of mathematics and geometry after visual inspection of the drawn proof (http://www.flickr.com/photos/24869933@N07/3619373444/sizes/l/) it is quite easily accomplished to connect and triangulate the outer points of the rectangles to the outer points of the square and circle and even by dimension sphere.
Yes, the drawing clearly by definition of inverse symmetry (take the drawing cut it in half at the "equator" and turn the paper over and rotate it and compare the top and bottom "mirrored symmetry" and you will see it is an exact inverse of symmetry itself) the drawing explains how to properly and mathematically soundly represent a fourth dimension on a two dimensional plane.
In plane language if you take the picture and separate the top half of the sphere by the bottom half of the sphere they are identical but only if the sphere existed on a complete three dimensional plane through expansion.
Since the paper is flat, this is an impossibility, nonetheless it is completely expressed and retractable and countable infinitely.
By the definition of our standard we have established any k-digit approximation to pi is rational.
Which could have many practical scientific applications. For instance, search:
ON THE THEOREM OF IVA$EV-MUSATOV IIIx e F(l) + F(2) + ... + F(p). This remark is the basis of our proof of Theorem .... -y(M + 1), x' + y(M + 1)] = 0 it is easy to check that Conditions (i), ...... Now set \i = pi x. *jj, v. Since p, x is positive,. (i)" fi is positive, ... plms.oxfordjournals.org/cgi/reprint/s3-53/1/143.pdf by TW Korner - 1986 - Cited by 7 - Related articles
REPRESENTATION OF FUNCTIONSBY SERIES AND CLASSES ?(1)with measure \E\> 2? -- ? and such that the trigonometric series for g(x) converges uniformly on [0, 2?]. .... trigonometric series are due to Ivashev-Musatov  who has shown that there are null series (in the sense of ..... function ?(?) e C(0, 1) with ?(0) = ?(1) = 0 such that ...... <*i<Pi(0= ? ?;(0-4-a ... www.iop.org/EJ/article/0036-0279/27/2/R01/RMS_27_2_R01.pdf by PL Ul'yanov - 1972 - Cited by 24 - Related articles
Full text of "Bulletin of the British Museum (Natural History)"X 20. FIG. 5. Well preserved large fragment showing the different aspect and ...... 0-1-0-5 P i n height) so giving the shell surface the appearance of ...... 94-95, i pi. Maseru. BRINK, A. S. 1963. Two cynodonts from the Ntawere ...... MUSATOV, D. L, NEMIROVSKAYA, V. N., SHIROKOVA, E. V. & ZHURAVLEVA, I. T. 1961. ... www.archive.org/stream/bulletinofbritis17geollond/bulletinofbritis17geollond_djvu.txt
Trigonometric series with rapidly decreasing coefficients a i0, 5 ...Nt,+i - Pi = n,. N tl+j. -pi =N tl+j. -i +pi +1. ..... number 8 6 (0,1) for the product (32) such that for every point x 6 E there are ... www.iop.org/EJ/article/1064-5616/187/11/A01/MSB_187_11_A01.pdf by SS Galstyan - 1996 - Related articles - All 7 versions
Plouffe's formula for pi - sci.math | Google Groups3 posts - 2 authors - Last post: 41 minutes ago x=pi e^(ipi)+1=0. Martin Musatov. Reply Forward. Martin Musatov to Robert show details May 3 Reply. Thank you, Sir. No problem. ... groups.google.com/group/sci.math/browse_thread/thread/65ba40fd9e5e8f46/ bf6d1ee13d87270b?lnk=raot - 41 minutes ago
[PDF] ???? ?-?File Format: PDF/Adobe Acrobat Gli storici ci dicono che, nel X secolo, il principe Vladi- ...... i pi`u anziani e carismatici si evidenziavano i nomi di ...... tin Somov, Viktor Borisov-Musatov, Aleksandr Benois e Lev Bakst ... 5/0; Judovin 5/0; Korovin 2/1; Tukacov 1/0; Goli- cyn 1/1; Zacharov 1/1; Jurkunas 5/0; Krasauskas 2/0;. Bulaka 3/0. ... www.esamizdat.it/eSamizdat_2005_(III)_2-3.pdf
I am testing these identities:
(1/2) * ( - k ) * ( k - 1/2 ) * ( - 3 ) / k!
This equation produces
M = k! ( - 3 ) * ( k - 1/2) * ( - k ) * (1/2)
Thank you for reading and writing the great comments back!
---------- Forwarded message ---------- From: inverse 19 <hope9...@verizon.net> Date: Apr 11, 7:41 pm Subject: CHALLENGE TO MATHEMATICIANS OF THE WORLD To: sci.math
On Apr 11, 11:40 am, "Harris Moran" <inva...@example.com> wrote:
> L@@K! A caps-lock google-poster!
> "inverse 19" <hope9...@verizon.net> wrote in message
NOTE THE PROGRESSION OF NUMBERS PELLS PROPORTION X^2=3 or 4 = 3 as the numbers progression, WOW if this does not proof this what will
1. Every odd number , number as Y value till infinity is exactly solvable with a proportionate result using the base Pells equation i.e 3,5,7,9,etc etc till infinty and the answer is as in 2(2^2)+1=3^2 and so on till infinity with a whole number integer.
2.Every even number , without exception, using Pells equation, the integer always is a .75 till infinity consistant
as 63.75(2^4)=1=4^4 till infinity every integer will end with .75. is that the same proportion as 3 over 4, Y=3 for the least with numbers proportion
There is a definite pattern by pells to numbers, and I note that in that pattern the number 19 proportion is strikingly proprtionate, and I clearly surmise that that circluar progression is the same as numbers for vector 19 progression.