[VA] SRC#002- Almost integers and other beasties

[Valentin Albillo]

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Hi all, welcome to my SRC#002 - Almost integers and other beasties.

Here I'll show an assortment of the results I can get using my HP-71B IDENTIFY program version 2.0. The original version 1.0 was extensively discussed and demonstrated with examples galore in my article Boldly Going ... Identifying Constants, published more than 10 years ago. You can download it as a PDF document using this link:

Boldly Going ... Identifying Constants

Shortly after publishing it I expanded its already substantial capabilities with important additional features such as the ability to find Minimal Polynomials and other implicit expressions, which greatly increased the recognition of arbitrary constants, and further this version 2.0 can be used creatively to find interesting, uncanny expressions never before seen, like the following ones I found and which you might enjoy seeing and checking using your trusty HP calculator:

• if x = Ln(1+\(\pi\))               then  62x\(^2\)+123x =    300.000002
• if x = Sin e                 then  11x-9x\(^2\) =      2.9999227778868800
• if x = Sin 4                 then  5x(x-4)=      17.99979999
• if x = Sin 1 + Cos 2 + Tan 3 then  6\(^3\)x-x\(^2\) =       60.9999995
• if x = \(\frac{3}{7}\sqrt[3]{\frac{261}{\pi}}\)                 then  x\(^5\)-x =         21.0000000100
• if x = 1% of Acos\(\frac{-317}{664}\)         then  \(\sqrt{10}\)-x =       3.14159265358
• if x\(^X\)= \(\pi\)                     then  (x+6)(9x-x\(^3\)) = 80.999999999
• if x = Gamma \(\pi\)               then  38x\(^2\)-3x\(^3\) =     163.00000
  

• if x is the positive root of .7x\(^2\)-6x-236 = 0, then Ln x = 3.14159265
  

Go ahead, check them, and I'd love to see any and all comments you would have on the matter, as well as your own uncanny expressions of a similar nature (Gerson, I'm looking at you ), please post your very best, original ones discovered by you (no 3rd-party ones harvested on the Internet, please) as replies in this thread.

Regards.
V.
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[ttw]

The PDF had fonts the I couldn't read, but that doesn't stop me from commenting. (I is the internet.)

If interpreted your comments correctly, you compare by comparing fractions in lowest terms. This suggests (if you are not doing this already, converting decimals to continued fractions and do comparisons by generating a single partial quotient at each step. This could lead to a nice speedup as grossly different numbers could be eliminated quickly.

[Gerson W. Barbosa]

(12-13-2018 10:30 PM)Valentin Albillo Wrote:  Go ahead, check them, and I'd love to see any and all comments you would have on the matter, as well as your own uncanny expressions of a similar nature (Gerson, I'm looking at you ), please post your very best, original ones discovered by you (no 3rd-party ones harvested on the Internet, please) as replies in this thread.

Hello, Valentin,

More or less in the same vein,

• if x = \(\pi \sqrt{2}\)                   then  \(\left ( 12^{2}-5\times 10^{-5} \right )\)x+x\(^{-1}\) = 640.00000003

Here are a few more original near-integers and near-identities:

\[2\left ( \pi + e - \psi \right ) = 4.9999776\]

\[2\left ( e-\tan^{-1}\left ( e \right ) \right )=2.9999978\]

\[\ln \left ( \frac{16\ln 878}{\ln \left ( 16\ln 878 \right )}\right )=3.14159265377\]

\[\frac{e^{\frac{23}{4}}}{100+\frac{1}{100+\frac{1}{\sqrt{100\sqrt{5}}}}}=3.141592​65354\]

\[3.141593-\frac{\sqrt{3}}{5\times 10^{6}}=3.1415926535898\]

\[\frac{\ln \left ( \sqrt{8} \cdot 10^{8}\right )}{\ln \pi }=16.999994\]

\[\frac{\ln \left (2\cdot \varphi ^{39}\right )}{\ln \pi }=17.00000026\]

Best regards,

Gerson

[Thomas Klemm]

(12-13-2018 10:30 PM)Valentin Albillo Wrote:  no 3rd-party ones harvested on the Internet, please

[rprosperi]

(12-14-2018 03:14 PM)Gerson W. Barbosa Wrote:  \[\ln \left ( \frac{16\ln 878}{\ln \left ( 16\ln 878 \right )}\right )=3.14159265377\]

There is something beautiful and compelling about this one, at least for me!

Both you guys truly amaze me... in a good way, just to be clear....

[Gerson W. Barbosa]

(12-14-2018 08:29 PM)Thomas Klemm Wrote:  

(12-13-2018 10:30 PM)Valentin Albillo Wrote:  no 3rd-party ones harvested on the Internet, please

I would humbly suggest an even more comprehensive test:

\(e^{\pi }-\pi + \left ( \frac{9}{3\times 10^{2}-\frac{2\sqrt{3}}{10^{3}-\ln \left ( 2+\sqrt{2} \right )}} \right )^{2}=20.000000000000000\)

[Gerson W. Barbosa]

Or this one, for 10-digit calculators:

\(\pi^{2}+\frac{e^{2}}{33\left(\ln\left(\pi\right)\right)^{4}}=10.00000000\)

[Valentin Albillo]

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Hi, ttw:

(12-14-2018 04:41 AM)ttw Wrote:  This suggests (if you are not doing this already, converting decimals to continued fractions and do comparisons by generating a single partial quotient at each step.

Thanks for your interest and comment. I do use a continued fraction algorithm to convert the arbitrary constant supplied by the user to a rational fraction, generating partial quotients one by one till the user-supplied accuracy is met.

Regards.
V.
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[Valentin Albillo]

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Hi, Gerson:

(12-14-2018 03:14 PM)Gerson W. Barbosa Wrote:  More or less in the same vein,

Thanks for your excellent findings, I was sure you'd never fail to contribute some amazing near-identities to this thread. As Bob Prosperi already pointed out, I too find this one particularly beautiful:

Quote:\[ \ln \left ( \frac{16\ln 878}{\ln \left ( 16\ln 878 \right )}\right )=3.14159265377 \]

Good finding indeed !

By the way, it's quite nice that the simple function x/Ln(x) sometimes gives almost-integer results for integer arguments (which means its graphic passes extremely close to integer-coordinates grid points), such as the following, in increasing order of "closeness":

             x               x/Ln(x)
          ---------------------------------
            17              6.0002541...
            163            31.9999987...
            53453        4910.0000012...
            110673       9529.0000006...
            715533      53078.0000004...

so that we have, for instance,

            53453/Ln(53453) = 4910.0000012...

In your case the argument x=16*Ln(878) results in x/Ln(x) being 23,1406926369... which is almost the famous Gelfond's constant = e^Pi (the easiest transcendental number to compute to high precision) so its natural logarithm is very nearly Pi itself.

Nice catch !

Have a fine weekend and best regards
V.
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[EdS2]

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(12-15-2018 09:56 PM)Valentin Albillo Wrote:  By the way, it's quite nice that the simple function x/Ln(x) sometimes gives almost-integer results for integer arguments...

Thanks, that leads to a rabbit hole of interesting links (OEIS and Mathoverflow.)

[ttw]

Exp(Pi*Sqrt(163)) is one of the classic examples. The explanation is rather complicated. Other expressions, for example:
({[Sqrt(5)+1]/2}^n)/Sqrt(5) is close to the Fibonacci numbers; in fact ({[Sqrt(5)+1]/2}^n-{[1-Sqrt(5)]/2}^n)/Sqrt(5) is the well-known Binet formula for Fibonacci numbers. This sequence works by successive approximation to an integer the 163 sequence just seems to happen.

[Valentin Albillo]

(12-16-2018 09:06 AM)EdS2 Wrote:  Thanks, that leads to a rabbit hole of interesting links (OEIS and Mathoverflow.)

Interesting. I did not consult the OEIS for the results I gave above for x/Ln(x), I simply obtained them myself by running this trivial HP-71B program I wrote in J-F Garnier's Emu71 to quickly find them:

        1  DESTROY ALL @ M=1 @ I=2
        2  X=I/LN(I) @ N=ABS(X-IROUND(X)) @ IF N<M THEN M=N @ DISP I;,X
        3  I=I+1 @ GOTO 2

        >RUN

        2   2.88539008178
        5   3.10667467281
        9   4.09607651981
        13   5.06832618827
        17   6.00025410569
        163   31.9999987385
        53453   4910.00000122
        110673   9529.00000068
        715533   53078.0000004
        1432276   101044
        ...       ...

Substituting  X=I/LN(I) at line 2 by some other function and rerunning the program will result in a new set of almost-integer values, for instance:

        2 X=I/TAN(I) ...

        >RUN

         2   -.915315108721
         3   -21.0457576543
         7   8.03260795684
         37   -44.0072133321
         48   39.9957590124
         128   -123.004197859
         170   460.0010337
         1489   -12899.9995967
         2106   986.000155144
         11923   15493.9999873

i.e.: 1489/Tan(1489) = -12899.9995967 ~ -12900

and so on and so forth. Trivial variations of this trivial program will produce an infinitude of almost-integer valued expressions of all kinds.

Thanks for your interest and links.

V.
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[Valentin Albillo]

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Hi, all:

(12-16-2018 10:43 PM)I Wrote:  Trivial variations of this trivial program will produce an infinitude of almost-integer valued expressions of all kinds.

A few additional, nice almost-integer results obtained that way:

      5e^Acos(178/181)  = 6.0000000066

      9e^Acos(538/541)  = 10.00000000023

      8e^Acos(430/433)  = 9.00000000048

      Ln 146 + Sin 614  = 4.00000800014

      Ln 455 + Cos 188  = 7.00000034

      Ln 231 + Tan 87   = 4.00000023

      Gamma(314/709)    = 2.00000047

All trig functions, in radians.

V.
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[Gerson W. Barbosa]

640000*x^5-768000*φ^2*x^4+3000+ln(2)=0

[Valentin Albillo]

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Hi, Gerson:

(01-02-2019 11:45 PM)Gerson W. Barbosa Wrote:  640000*x^5-768000*φ^2*x^4+3000+ln(2)=0

I only have an iPad at hand right now so running this extremely quick'n'dirty Newton on it produces the intended root of your polynomial, namely:

10 def fnf(x)=640000*x^5-768000*p^2*x^4+3000+log(2)
20 def fnd(x)=(fnf(x+0.0001)-fnf(x-0.0001))/0.0002
30 p=(1+sqr(5))/2:input x0:home
35 for i=1 to 15
40 x1=x0-fnf(x0)/fnd(x0)
50 print x1;" ";fnf(x1)
60 x0=x1
70 next i

Run

?10
8.167851990851785 14317006965.841368
6.715802401810718 4653139559.305097
5.573461555087618 1501801189.590903
4.687642795302511 477761112.5232644
4.021032838991736 147136983.11165047
3.551962197641171 41803117.36241603
3.2712995361031516 9506051.502593396
3.1593486482128053 1132110.8530623894
3.141983056899174 24349.060661761047
3.141592847539331 12.090443311217141
3.1415926535896097 0.0000030975368739971643
3.14159265358956 -3.170697671084355e-8
3.1415926535895604 -1.904654323148236e-9
3.1415926535895604 -1.904654323148236e-9
3.1415926535895604 -1.904654323148236e-9

which is a nice approximation to Pi, congrats and thanks for sharing. Perhaps it's even more accurate than what the iPad produces but right now I can't tell ...

Regards.
V
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[Paul Dale]

The underline finishes one digit too far:

(01-03-2019 06:29 AM)Valentin Albillo Wrote:  3.1415926535895604 -1.904654323148236e-9

3.1415926535897932

Learning some leading digits of \( \pi \) is useful (for trolling).

Pauli

[Valentin Albillo]

(01-03-2019 06:48 AM)Paul Dale Wrote:  The underline finishes one digit too far:

Nope, it ends exactly where I intended it to end.

And yes, you are trolling.

V.
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[Gerson W. Barbosa]

(01-03-2019 06:48 AM)Paul Dale Wrote:  3.1415926535897932

Learning some leading digits of \( \pi \) is useful (for trolling).

Pauli

Or for spotting mistakes like this one (YouTube video).
I used to remember the first 16 digits when I used my HP-200LX on a regular basis (its screen has gone dark and I have no spare part to replace it again). Anyway, I don't remember any "747" sequence occurring so early in \( \pi \).

Gerson.

[Paul Dale]

The first "747" occurs at position 740 (thanks to the \(\pi\) searcher).

In the fullness of \(\pi\), this is right near the start (of course).


Pauli

[Valentin Albillo]

(12-19-2018 10:23 PM)I Wrote:  A few additional, nice almost-integer results obtained that way:

For completeness, I forgot to include this remarkable one which I discovered and posted here last March:

            Sin(9*(Sin 1 + Cos 40)) = 0.999999999999999830826985368...

which differs from the integer 1 by about 1e-16.

V.
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