Another Ramanujan Trick

09292019, 06:45 AM
Post: #1




Another Ramanujan Trick
Just for fun, I came across with another Ramanujan trick:
Pi=Sqrt(Sqrt(2142/22)) It's better than 355/113 but at the cost of a couple of square roots (or a fourth root.) 

09292019, 09:17 AM
Post: #2




RE: Another Ramanujan Trick  
09292019, 09:55 AM
Post: #3




RE: Another Ramanujan Trick
(09292019 06:45 AM)ttw Wrote: Just for fun, I came across with another Ramanujan trick: Isn't it Pi=Sqrt(Sqrt(2143/22)) ? Greetings, Massimo +×÷ ↔ left is right and right is wrong 

09292019, 11:19 AM
(This post was last modified: 09292019 02:45 PM by Gerson W. Barbosa.)
Post: #4




RE: Another Ramanujan Trick
(09292019 09:55 AM)Massimo Gnerucci Wrote:(09292019 06:45 AM)ttw Wrote: Just for fun, I came across with another Ramanujan trick: Ciao, Massimo, Yes, you are right. Let us take our HP32S or 33S and do some “reverse engineering”: pi x^2 x^2 > 97.409091034 99 * > 96435.50001237 ENTER + IP > 19287 198 / FDISP > 97 9/22 1/x > 22/2143 => pi ~ sqrt(sqrt(2143/22)) = 3.14159265258 This fourth power oddity is common to both ln(pi) and e^x as well: (ln(pi))^4 = 1.7171652254 and (e^pi)^4 = 286751.313138, which lead to e^(sqrt(sqrt(170/99))) = 3.14159605245 and ln(sqrt(sqrt(28388380/99))) = 3.14159265359 These don’t make for efficient pi approximations, though. P.S.: You might want to check this out: \[\ln\left(\sqrt[4]{\frac{{305}^{3}+{25}^{3}+{5}^{3}+5+\frac{33}{{9}^{5}+{5}^{5}+{3}^{5}+3}}{99}} \right )\] P.P.S.: Or that one: \[\sqrt[4]{\frac{2143+{\left(6+\sqrt{\frac{6}{1+{6}^{6}}}\right)}^{6}}{22}}\] 

09292019, 01:28 PM
Post: #5




RE: Another Ramanujan Trick
(Sorry about the typo.)
The method works by developing the continued fraction for Pi^4. The continued fraction for Pi has a 292 real early and 355/113 comes from stopping just before the 292. The continued fraction for Pi^4 is (97; 2, 2, 3, 1, 16539...) so one stops before the 16539. Continued fractions (to anything) give a sequence of p(i))/q(i) of numerators and denominators. The error in the ith term is less than 1 part in q(i)*(q(i)+q(i1)). (Assuming I'm not off 1 in counting.) 

09292019, 03:23 PM
Post: #6




RE: Another Ramanujan Trick
(09292019 01:28 PM)ttw Wrote: Continued fractions (to anything) give a sequence of p(i))/q(i) of numerators and denominators. The statement is true, but you can tighten the bound with error < \(\Large{1 \over q_{i}q_{i+1} } \) (97; 2, 2, 3, 1) = 2143 / 22 (97; 2, 2, 3, 1, 16539) = 35444733 / 363875 Pi^4 must be between 2 convergents. gap = 1 / (22 * 363875) = 1 / 8005250 To get first fraction for Pi^4 with estimate better than 2143/22 Convergent before 2143/22 is 1656/17, solve for semiconvergent with less error XCas> fsolve(pi^4  (1656+2143k)/(17+22k) = 2143/22  pi^4, k) → k ≈ 8269.54 Semiconvergent error switched sign when k=8270. To confirm: XCas> float(pi^4 . [2143/22, 17724266/181957]) → (1.24912e7, 1.24898e7) 

09292019, 05:09 PM
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RE: Another Ramanujan Trick
OK, I was off by 1.


09292019, 09:36 PM
Post: #8




RE: Another Ramanujan Trick
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Hi, Gerson: (09292019 11:19 AM)Gerson W. Barbosa Wrote: You might want to check this out:\[\ln\left(\sqrt[4]{\frac{{305}^{3}+{25}^{3}+{5}^{3}+5+\frac{33}{{9}^{5}+{5}^{5}+{3}^{5}+3}}{99}} \right )\]Or that one:\[\sqrt[4]{\frac{2143+{\left(6+\sqrt{\frac{6}{1+{6}^{6}}}\right)}^{6}}{22}}\] Nice, they're indeed aesthetically pleasing but as approximations to Pi they use far too many digits, functions and operations for the accuracy they achieve. On the other hand, the following theoreticallybased approximation to Pi is much better in that regard: just ten individual digits, two functions and two arithmetic operations to achieve almost 17 digits (save for 2 units in the last place):\[ \frac{3 Ln 640320}{\sqrt{163}} =3,1415926535897930... \] A small modification, essentially adding just three digits and one sum gets you almost 32 digits (again save for 2 units in the last place).\[ \frac{Ln{({ 640320}^{3}+744})}{\sqrt{163}} =3,1415926535897932384626433832797... \]Regards. V. . All My Articles & other Materials here: Valentin Albillo's HP Collection 

09302019, 03:10 AM
(This post was last modified: 09302019 03:11 AM by Gerson W. Barbosa.)
Post: #9




RE: Another Ramanujan Trick
(09292019 09:36 PM)Valentin Albillo Wrote:(09292019 11:19 AM)Gerson W. Barbosa Wrote: You might want to check this out:\[\ln\left(\sqrt[4]{\frac{{305}^{3}+{25}^{3}+{5}^{3}+5+\frac{33}{{9}^{5}+{5}^{5}+{3}^{5}+3}}{99}} \right )\]Or that one:\[\sqrt[4]{\frac{2143+{\left(6+\sqrt{\frac{6}{1+{6}^{6}}}\right)}^{6}}{22}}\] Hello, Valentin, My second approximation uses 10 digits and produces 17 significant digits, but it requires four functions, twice as much compared to the first approximation you mention. By no means I intend to compete with Ramanujan, Dedekind and others. That was just an idle Sunday morning play. Anyway, thanks for bringing these really nice approximations to my attention again. They have prompted me to search for an 82digit one belonging to the same group. I’ve finally found it in Tito Pieza’s blog here. Best regards, Gerson. 

09302019, 10:24 PM
Post: #10




RE: Another Ramanujan Trick
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Hi again, Gerson: (09302019 03:10 AM)Gerson W. Barbosa Wrote: My second approximation uses 10 digits and produces 17 significant digits, but it requires four functions, twice as much compared to the first approximation you mention. Funny, it seems that my counting methods and yours do differ. In your second approximation I count 13 digits (4,2,1,4,3,6,6,1,6,6,6,2,2) and 9 operations (3 additions, 2 divisions, 2 roots and 2 powers). Quote:They have prompted me to search for an 82digit one belonging to the same group. I’ve finally found it in Tito Pieza’s blog here I did follow the link you mention but I was less than impressed. It may produce 82 correct digits but it's far too complex to be regarded as a remarkable approximation, the number of digits and operations needed is on the same ballpark as the number of correct digits produced. Compare it with the two truly remarkable approximations I gave, or even the simple 355/133, where 6 digits and 1 division gets you almost 8 digits (save for 2 or 3 units in the last place, depending on rounding). Best regards. V. . All My Articles & other Materials here: Valentin Albillo's HP Collection 

10022019, 04:03 AM
(This post was last modified: 10022019 04:12 AM by Gerson W. Barbosa.)
Post: #11




RE: Another Ramanujan Trick
(09302019 10:24 PM)Valentin Albillo Wrote: . I would consider one 4 and the two instances of 6 as functions (fourth root and the reciprocals of sixth roots). But I have yet another trick up my sleeve: 10 DESTROY ALL 20 OPTION BASE 1 @ DIM A(35) @ COMPLEX B(34) 30 A(1)=3 @ A(34)=1.E+12 @ A(35)=2.4E+17 40 MAT B=PROOT(A) 50 DISP REPT(B(2)) The four significant digits in line 30 of the HP71B program above yield 10 correct digits of pi. Best regards, Gerson. 

10022019, 11:14 PM
Post: #12




RE: Another Ramanujan Trick
(10022019 04:03 AM)Gerson W. Barbosa Wrote: I would consider one 4 and the two instances of 6 as functions (fourth root and the reciprocals of sixth roots). The fourth root would be the function nthroot(4,x) so the nthroot would be one function and the 4 would be one digit. Same with the various 6, they would be power(6,x), i.e: one function, power, and onedigit argument, 6. And that 's being generous and not counting the "" as one unary operation. Else, you could have 2^1,651496129472 = 3,141592653589+ and claim that the underlined power is just one function. Nope. Quote:But I have yet another trick up my sleeve: You forgot to include the <justjoking> ... </justjoking> tags. If something as immensely complicated as a function capable of finding the real part of one root of a 34th degree polynomial with 32 zero coefficients and 3 realvalued coefficients is to be counted as just one function applied to 3 parameters then you can go the whole hog and simply use: 4*arctan(1) = 3,1415926535897932384626433832795+ which uses just 2 digits and one function, which is many orders of magnitude simpler than your function which gives the root of the 34thdegree polynomial, and further agrees with infinitely many correct digits of Pi. Anyway, quite ingenious on your part, but hopelessly useless as a simple approximate formula. Best regards. V. . All My Articles & other Materials here: Valentin Albillo's HP Collection 

10032019, 07:21 AM
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RE: Another Ramanujan Trick
(10022019 11:14 PM)Valentin Albillo Wrote: If something as immensely complicated as a function capable of finding the real part of one root of a 34th degree polynomial with 32 zero coefficients and 3 realvalued coefficients is to be counted as just one function applied to 3 parameters then you can go the whole hog and simply use: Surely going the whole hog would be arccos(1)? — Ian Abbott 

10032019, 05:47 PM
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RE: Another Ramanujan Trick
(10032019 07:21 AM)ijabbott Wrote: Surely going the whole hog would be arccos(1)? Surely ! Did you visit my new HP site ? If yes, did some contents gather your interest ? Any comments welcome, especially constructive ones. Regards, V. . All My Articles & other Materials here: Valentin Albillo's HP Collection 

10032019, 07:58 PM
(This post was last modified: 10032019 08:04 PM by Gerson W. Barbosa.)
Post: #15




RE: Another Ramanujan Trick
(10022019 11:14 PM)Valentin Albillo Wrote:(10022019 04:03 AM)Gerson W. Barbosa Wrote: I would consider one 4 and the two instances of 6 as functions (fourth root and the reciprocals of sixth roots). Two nested square roots woudn't help, so I'll leave it as is. (10022019 11:14 PM)Valentin Albillo Wrote: Same with the various 6, they would be power(6,x), i.e: one function, power, and onedigit argument, 6. And that 's being generous and not counting the "" as one unary operation. You do have a point here. (10022019 11:14 PM)Valentin Albillo Wrote:Quote:But I have yet another trick up my sleeve: No need to. I knew I could count on your sense of humor. (10022019 11:14 PM)Valentin Albillo Wrote: Anyway, quite ingenious on your part, but hopelessly useless as a simple approximate formula. Yes, I am aware of its uselessness as an approximation to pi, but I thought the associated polynomial is interesting just the same. It arises from noticing that the 34th power of pi is 8.00010471505E16, whose underlined digits match the first digits of pi/3. Best regards, Gerson. P.S.: I forgot to mention that it take more than 3 minutes for the HP71B to show the result. The HP50g is faster, still it takes 38 seconds which certainly most of you will find too much time. « { 1 35 } 0 CON 1 3 PUT 34 1.E12 PUT 35 2.4E17 PUT PROOT 6 GET RE » EVAL > 3.14159265364 

10032019, 08:23 PM
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RE: Another Ramanujan Trick
(10032019 05:47 PM)Valentin Albillo Wrote:(10032019 07:21 AM)ijabbott Wrote: Surely going the whole hog would be arccos(1)? I've visited it, but haven't delved into the articles yet, apart from the "Time Voyager" story that someone linked to earlier (although I'm pretty sure I'd already read your story some time ago, and definitely wasn't confusing it with Paul Nahin's "Newton's Gift" story from Omni magazine). — Ian Abbott 

10032019, 08:26 PM
Post: #17




RE: Another Ramanujan Trick
(10032019 05:47 PM)Valentin Albillo Wrote:(10032019 07:21 AM)ijabbott Wrote: Surely going the whole hog would be arccos(1)? Not quite! If you are interested only in the exact representation of the constant, than there is nothing more simple than \(\pi\) . It looks like somehow I've failed to make myself clear this is all about original approximations to \(\pi\), not about exact formulas or other well known approximations. My bad. The real thing Just an approximation Best regards, Gerson. 

10032019, 09:44 PM
Post: #18




RE: Another Ramanujan Trick
(10032019 07:58 PM)Gerson W. Barbosa Wrote:(10022019 11:14 PM)Valentin Albillo Wrote: You forgot to include the <justjoking> ... </justjoking> tags. Of course you can count on it. Quote:I thought the associated polynomial is interesting just the same. It arises from noticing that the 34th power of pi is 8.00010471505E16, whose underlined digits match the first digits of pi/3. Nice find, indeed. Did you mention it in the past ? I seem to remember having read about it in one of your old posts but I may just be mistaken. Quote:I forgot to mention that it take more than 3 minutes for the HP71B to show the result. To be fair, you must consider that it's not simply computing just the one root you're interested in for your approximation to Pi, but rather it's simultaneously computing all 34 complex roots to full accuracy (using 15 digits internally). By the time the 3 minutes have elapsed you have the 68 components of the 34 complex roots readily stored in complex array B(), though you then go and use just one of them. Considering this fact, 3 minutes doesn't seem that long, it comes to less than 5.3 seconds per 12digit root. And, perhaps, using FNROOT instead to compute just the one root you're interested in would be much faster than using PROOT to compute all 34 roots. Did you try ? Regards. V. All My Articles & other Materials here: Valentin Albillo's HP Collection 

10042019, 04:14 PM
Post: #19




RE: Another Ramanujan Trick
(10032019 09:44 PM)Valentin Albillo Wrote:(10032019 07:58 PM)Gerson W. Barbosa Wrote: I thought the associated polynomial is interesting just the same. It arises from noticing that the 34th power of pi is 8.00010471505E16, whose underlined digits match the first digits of pi/3. I think I have used that in a nearinteger expression, but I can't find it right now. (10032019 09:44 PM)Valentin Albillo Wrote:Quote:I forgot to mention that it take more than 3 minutes for the HP71B to show the result. No, I didn't. But that surely would have been much faster. At first, I tried Wolfram Alpha, which does it instantaneously, but I thought the HP71B code would fit better here. Anyway, here it is, just for the record: Solve x^34  (1/3)*10^12*x  8*10^16 = 0 Best regards, Gerson. 

10062019, 09:07 PM
Post: #20




RE: Another Ramanujan Trick
(09302019 10:24 PM)Valentin Albillo Wrote:(09302019 03:10 AM)Gerson W. Barbosa Wrote: They have prompted me to search for an 82digit one belonging to the same group. I’ve finally found it in Tito Pieza’s blog here Ok, then the following might almost double that. Well, sort of cheating, but at least no use of any inverse trig function that would just kill the fun. ln(640320^3 + 744)/√163 + sinh(sin(ln(640320^3 + 744)/√163)) Best regards, Gerson. 

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