[VA] SRC #009 - Pi Day 2021 Special
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03-23-2021, 10:20 PM
Post: #38
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RE: [VA] SRC #009 - Pi Day 2021 Special
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:
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) = \(\pi\) ee/e! = 11.1735566407 and this simple HP-71B command-line expression will evaluate I(X) and \(\pi\) XX/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 { \(\pi\) ee / e! } ?PI 15.9359953238 15.9359953238 { \(\pi\) \(\pi\)\(\pi\) / \(\pi\)! } Additional Notes:
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, W0(X) eW0(X) = X Particularizing its value for X = 1 and exchanging W0(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:
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:
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:
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:
That's all for now. Again, thanks for your interest and participation in this SRC, glad you enjoyed it ! Best regards. V. All My Articles & other Materials here: Valentin Albillo's HP Collection |
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