Hi,

all:

Thanks to all of you who posted messages and/or comments to this

SRC #009 - \(\pi\) Day 2021 Special (namely

J-F Garnier,

Gerson W. Barbosa,

Albert Chan,

PeterP,

robve,

Maximilian Hohmann,

Ángel Martín and

Massimo Gnerucci), for your interest and valuable contributions, much appreciated.

Now, these are my original results and comments for the various parts of this

SRC:

- Caveat: Don't expect anything ground-breaking here, this is a simple SRC, not a full-fledged challenge or even a challenge proper, and some of you already gave the correct results and explained them thoroughly as well, so there's not much to add, this will be brief.

a. The root of this equation in

[3,4] is indeed

X = \(\pi\) ~ 3.14159265359. The base integral is:

so that we have:

I(\(\pi\)) = \(\pi\) \(\pi\)^{\(\pi\)}/\(\pi\)! = 15.9359953238 ,

I(e) = \(\pi\) e^{e}/e! = 11.1735566407
and this simple

HP-71B command-line expression will evaluate

I(X) and

\(\pi\) X^{X}/X! for any given

X ≥ 0:

>INPUT X @ INTEGRAL(0,PI,0,(SIN(IVAR)*EXP(IVAR/TAN(IVAR))/IVAR)^X);PI*X^X/GAMMA(X+1)

?1 3.14159265359 3.14159265359 { \(\pi\) }

?2 6.28318530717 6.2831853072 { 2 \(\pi\) }

?EXP(1) 11.1735566407 11.1735566407 { \(\pi\) e^{e} / e! }

?PI 15.9359953238 15.9359953238 { \(\pi\) \(\pi\)^{\(\pi\) }/ \(\pi\)! }

Additional Notes:
- robve asked which quadrature procedures have I written that are better than Romberg. Well, actually several for various machines, and using them I've been able to compute definite integrals (even difficult ones) with high precision (say 100 digits and more) very fast. For the HP-71B's implementation I've achieved speeds faster than the ones achieved by the Math ROM's Romberg-based quadrature, despite the latter having the tremendous advantage of being implemented in assembly language (vs. my BASIC code) and using 15-digit/50,000-exponent precision for extended accuracy (vs. the 12-digit/499-exponent available to my BASIC code).

I'll say no more about it as my implementation is the subject matter of my article "VA036 - Boldly Going - Outsmarting INTEGRAL", to be published soon.

b. The root of this equation in

[3,4] is also

X = \(\pi\) ~ 3.14159265359. This time the base integral is:

which is a representation of the ubiquitous

Lambert W function as a parametric definite integral, which though not new (there are other various integral representations) seems to me rather awesome nevertheless: a

definite integral solves a

transcendental equation,

W_{0}(X) e^{W0(X) } = X
Particularizing its value for

X = 1 and exchanging

W_{0}(1) and

\(\pi\) we get:

from where the equation is then obtained. This

HP-71B program lets us try different values of

X and returns the difference between the

LHS and the

RHS, to see how it approaches

0 when

X approaches

\(\pi\):

1 DEF FNF(T)=LN(1+SIN(T)/T*EXP(T/TAN(T)))

2 DEF FNI(X)=INTEGRAL(0,X,0,FNF(IVAR))

3 DESTROY ALL @ W=FNROOT(0,1,FVAR*EXP(FVAR)-1)

4 INPUT X @ DISP FNI(X)/W-X @ GOTO 4

[RUN]

? .1 .131342478087

? .5 .63098466783

? 1 1.1030252775

? 1.5 1.27459847697

? 2 1.08227897742

? 2.5 .6404117652

? 3 .14159265358

? 3.14 .00159265358

? 3.1415 .00009265358

? PI -.00000000001

For

X >

\(\pi\), it gives an error:

ERR L1:LOG(neg). As expected, for

X =

\(\pi\) it returns

~ 0 but notice that for

X ≥ 3 it returns ~

\(\pi\) -

X. Using a graphing calculator to plot the above values into a continuous graph will show why.

c. Though this seems to be a striking

finite evaluation which gives

\(\pi\) in terms of

e in a much simpler and direct way than

Euler's formula and without involving

complex values, the magic is tarnished somewhat by the fact that

e isn't really needed here, as the basic

identity is:

\(\pi\) = 4 * ( Arctan X - Arctan \(\frac{X - 1}{X + 1}\) )
which is valid for all

X > -1, as some of you explained and proved. That said, there are many interesting particular cases which you can use to trick your colleagues into believing you've discovered a brand-new, amazing evaluation for

\(\pi\). For instance:

- Using \(\pi\) itself (!) to get \(\pi\): \(\pi\) = 4 * (Arctan \(\pi\) - Arctan \(\frac{\pi - 1}{\pi + 1}\))

>4*(ATN(PI)-ATN((PI-1)/(PI+1))) -> 3.14159265359

- Using the Golden Ratio \(\phi\) to get \(\pi\): \(\pi\) = 4 * (Arctan \(\phi\) - Arctan \(\frac{\phi - 1}{\phi + 1}\))

>P=(1+SQR(5))/2 @ 4*(ATN(P)-ATN((P-1)/(P+1))) -> 3.14159265360

- Using the current year (2021) to get \(\pi\): \(\pi\) = 4 * (Arctan 2021 - Arctan \(\frac{2020}{2022}\))

>4*(ATN(2021)-ATN(2020/2022)) -> 3.14159265358
You can also trick them by using your phone number, your birthday or any number personally related to you or the person being tricked !

Additional Notes:
- J-F Garnier commented that he doesn't think there's a relation which can be used to get \(\pi\) from e but added he was thinking about finite formulae. Indeed, there are various relatively simple infinite series involving e and returning \(\pi\) that would do fine.

He also insisted that deriving \(\pi\) from Euler's formula as a logarithm base e (namely \(\pi\) = log_{e}(-1) / i ) doesn't mean that e is involved because he can use arctan instead and this is the method used to compute log(z) in various HP calculators. To this I say that the computing methods used are irrelevant, we're talking here about theoretical math, not practical implementation details, and theoretically the derivation of \(\pi\) from Euler's formula holds and fundamentally involves e as the log base. I could give a pertinent example related to cubic equations to make it clearer but it would make this too long.

Gerson suggested replacing e by c (the speed of light in whatever units) in the formula and correctly gave the range of constants and resulting values, as well as a link to an interesting formula which, well, links e and \(\pi\) but I don't consider that a bona-fide way to derive \(\pi\) from e or vice versa, and the other example he gave is a definite integral, also quite unsatisfactory to me for the purpose: all sorts of integrals have \(\pi\) as a result, thus demonstrating that \(\pi\) is ubiquitous but nothing else. Just change \(\pi\) to 5 and you'll see what I mean: a definite integral is no way to "derive" 5 from anything.

Last, Albert Chan and robve also explained several times and in various ways why the identity works.

d. The summation for even dimensions from

0 to infinity of the volumes enclosed by the respective

n-dimensional unit spheres

(R = 1) is

e^{\(\pi\)} ~

23.1406926328 (

aka Gelfond's constant) and thus its

\(\pi\)-th root is

e, or conversely we could say that

\(\pi\) is its natural logarithm. The

\(\pi\)-th root of the summation is readily obtained with the

HP-71B by executing this from the command line:

>V=0 @ FOR D=0 TO 50 STEP 2 @ V=V+PI^(D/2)/FACT(D/2) @ NEXT D @ V^(1/PI)

2.71828182846

which agrees with

e ~

2.71828182846. For the 12-digit

HP-71B we stop at dimension

50 because

\(\pi\)^{25}/25! ~

1.73.10^{-13}, so adding further terms won't affect the result.

Also, as I read for the first time in some Martin Gardner's book many decades ago, the volume enclosed by the

n-dimensional unit sphere tends to

0 with growing

n, and reaches a

maximum for a (fractional) dimension between

5 (Vol = 8

\(\pi\)^{2}/15 ~

5.264) and

6 (Vol =

\(\pi\)^{3}/6 ~

5.168). See if you can find this

unique dimension and the corresponding

maximum volume.

Additional Notes:
- Albert Chan commented that the formula gives 1 for the volume of a 0-dimensional sphere (whatever its "radius"), which seems weird to him. Well, that's by definition. Come to that, the formula also gives 2 for the volume of the 1-dimensional unit "sphere", which is but a line segment that obviously has no 3D "volume", and it also gives \(\pi\) for the volume of the 2-dimensional unit "sphere", which is but a 2D circle with no 3D "volume" either. We tend to think of volume in terms of dimensions 3 or greater but mathematically that's not necessarily so.

e. The song "

\(\pi\)" by Kate Bush is indeed awesome, as is most of her music (

"Cloudbusting", mentioned in this thread, certainly is, as is the video for it, almost a whole tragic movie told in a few minutes), and part of it appears in

The Simpsons' 26th-season finale,

"Mathlete's Feat", featuring about one minute of the song or so.

\(\pi\) also appears in at least two other episodes which I watched at the time: in one of them, Prof. Frink unexpectedly says aloud that

\(\pi\) is exactly equal to

3 as a way to get some much needed attention (he sure gets it !), and in another Homer and Marge are visiting some school for

"Snotty Girls and Mama's Boys" (i.e., gifted children) and two of them are singing a hand-clapping song with lyrics they've concocted to help them remember a few digits of

\(\pi\). There may be many more ... (episodes featuring

\(\pi\), that is).

Additional Notes:
- Ángel Martín briefly visited the thread to express his love for the song (and Massimo Gnerucci did likewise for the artist herself), as did Maximilian Hohmann, who also mentioned "Cloudbusting" and its pseudo-scientific background, which I knew about from reading Martin Gardner's and James Randi's books in the distant past.

f. First, those

31,415,926,535,897 decimal places were reported by Emma Haruka Iwao on 2019's

\(\pi\) Day after 121 days of computation. Then, on January 2020 Tim Mullican computed

50 trillion digits in some 300 days, which if I'm not wrong it's the current world record as of 2021.

The resulting string of digits passes all

normality tests, where we consider a real number

R to be

normal in some base

B if any string of

N base-

B digits appears in the base-

B representation of

R with frequency

B^{-N}, e.g: in base

10 every digit

0-9 appears

1/10-th of the time, every 2-digit sequence

00-99 appears

1/100-th of the time, etc.

It can be proved that

almost all real numbers are normal in any and all bases

B, but rigurously proving that a

"naturally-occurring" real

R (i.e., not artificially defined), say

\(\pi\), is normal for even just

one base

B (say base

2 or

10) is

excruciatingly difficult and not a single result is known so far, though the expectations are that all irrational numbers actually are, e.g. if you compute a trillion bits of \(\sqrt{2}\), you'll find about the same number of

0's and

1's and about the same number of

00's,

01's, 10's and

11's, etc.

Not all is lost, however. If we can't (yet) prove than

\(\pi\) is normal in base 2 or 10, say, we can try to

estimate the credibility of the decision

"\(\pi\) is not normal", which somewhat resembles

probabilistic primality testing, where we can't rigurously prove that a certain number is prime in reasonable time but we may certify it as a

probable prime if we can quickly prove that the probability of it being composite is

extremely small. In the case of

\(\pi\)'s normality, this has been done by analyzing the first

16 trillion bits of

\(\pi\), and the result is that the decision

"\(\pi\) is not normal" has credibility

4.3497.

10^{-3064}, which in my dictionary is akin to

"impossible".

Second, the bilingual joke about

"Fear of number \(\pi\)" being called

"Trescatorcephobia" is a pun on

3.14 being usually said as

"tres catorce" in Spanish.

Finally, re the

"peer reviewed" papers demonstrating that

\(\pi\)'s value is some

algebraic number, I'm astonished that anyone would give them credit (let alone publish them) except for non-academic reasons. Next category: papers succinctly proving

Riemann's Hypothesis or

Fermat's Last Theorem in a few pages, almost casually.

Additional Notes:
- Peter P commented "where was my simple proof of Fermat's Last Theorem again? Gotta get that published lest someone steals my brilliant idea...". LOL !

FLT is almost false, in the sense that there are infinitely many almost-counterexamples where the arbitrarily large LHS and RHS differ by just 1 (two other almost-counterexamples appeared in The Simpsons as well though not so close, namely 1782^{12} + 1841^{12} = 1922^{12} and 3987^{12} + 4365^{12} = 4472^{12}). Also, a nice "flipped" variant of FLT can actually be proved using a truly "brilliant" idea.

robve gave very interesting comments and links re so-called predatory journals, as well as a related personal experience while attending an IEEE conference.

That's all for now. Again,

thanks for your interest and participation in this

SRC, glad you enjoyed it !

Best regards.

V.