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Happy Birthday, Leonhard Euler!
Born on this date in 1707, in addition to being a brilliant mathematician, his prolific articles and expositions were a major contribution to mathematics. By one estimate, Euler was personally responsible for about one-quarter (!) of all mathematics that was published in the 18th century. He worked most of his life in St. Petersburg and died in 1783.
Although he contributed to all areas of mathematics, from the point of view of the GIMPS project, he can be especially remembered for his computation that 2[sup]31[/sup]-1 is the eighth Mersenne prime. He did this by checking all possible prime factors of the form 62k + 1 less than the square root of 2[sup]31[/sup]-1 which were also equal to either 1 or 7 mod 8. This last condition followed from quadratic reciprocity, and essentially halved the required amount of work, which was then equivalent to dividing 2[sup]31[/sup]-1 by each of the 84 prime factors satisfying the conditions. He proved that all even perfect numbers were of Euclid's form, and therefore generated by the Mersenne primes. He also clarified statements made by Descartes and Frenicle about the possible form of any odd perfect numbers. In addition, he debunked Fermat's claim that 2[sup]m[/sup]+1 was always prime when m was a power of 2 by factoring 2[sup]32[/sup]+1 into its prime factors 641 times 6700417. Fermat had lived and worked a century before, and had clearly been a brilliant number theorist, as he had essentially invented the field of Diophantine equations and had also grasped the central role that prime numbers played in Diophantine problems. However, Fermat had published very little on his methods, leaving mere hints in his correspondence, and it was left to Euler to publish proofs of Fermat's claims and fill in the many gaps left in the exposition of this area. This Euler fulfilled admirably, extending Fermat's ideas and results in many directions. Euler also deduced the form of the law of quadratic reciprocity from particular cases, although a valid proof was not discovered until later by Gauss. The bulk of Euler's work was in analysis, and much of the notation now used in calculus classes became standard through Euler's books. His collected works now fill 79 volumes and are still in the process of being completed. His result that e raised to the power of pi times i is equal to -1 has been considered one of the most elegant formulas in all of mathematics. |
Euler !
:smile:
I'm a bit late on this one but Euler is immortal in mathematics and I can only add a footnote to Philmoore's brilliant summary of this mathematical giant. "Euler calculated without apparent effort, as men breathe, or as eagles sustain themselves in the wind" as Arago said, is not an exaggeration of the unequaled mathematical facility of Leonhard Euler, the most prolific mathematician in history and the man whom his contemporaries called 'Analysis incarnate' Even total blindness during the last seventeen years of his life did not retard his unparallelled productivity; indeed if anything, the loss of his eyesight sharpened Euler's perceptions in the inner world of his imagination. All hail Euler! Mally :coffee: |
Mozart's era:smile:
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Euler and infinite series.
[QUOTE=philmoore;103753]Born on this date in 1707, in addition to being a .....
. His result that e raised to the power of pi times i is equal to -1 has been considered one of the most elegant formulas in all of mathematics.[/QUOTE] :rolleyes: Leonhard Euler Euler was working on the Basel Problem at the age of 24 in 1731 by calculating a numerical approximation. This is an arduous task by hand with a series which converges as slowly as this. In 1735 he arrived at the following exact result: 1+1/4+1/9+1/16 +1/k^2 ...... = (pi^2)/6 If that was not enough he went on to other similar series one of which I give below - the sum of the odd perfect squares 1+1/9+1/49 ....... =(pi^2)/8 These are truly remarkable results. No one expected the value p, the ratio of the circumference of a circle to the diameter, to appear in the formula for the sums. Where even Euler was mute and even after 200 years plus, the following problem remains unsolved. The sum of the reciprocals of the odd powers of the integers. 1+1/(2^3)+1/(3^3)+1/(4^3)+.......= ..? Any takers ? My conjecture is the sum will be (p/q) (pi^4) for some fraction p/q :smile: Mally :coffee: |
[quote=mfgoode;103869]
1+1/(2^3)+1/(3^3)+1/(4^3)+.......= ..? My conjecture is the sum will be (p/q) (pi^4) for some fraction p/q :smile: [/quote] [code]> sum(1/i^3,i=1..infinity); Zeta(3) > evalf(%/Pi^4,500); 0.0123402948369572013019496530957127837281768793763035283557246710406383171\ 472606606599884387353721352448829071859416507339043426742243816281494339481\ 025740237699802800592085775016081227010654573778747098505937998607393454487\ 093990910952235604718285983945262957094882983208280295207051978931846499288\ 712051686611058581604011047304010365010986773068219080259176456457241316727\ 988925876271690389762785626090195792712418631282529279771136022594284384039\ 72943907097645314685769007097543064750952546312171104[/code]Since digits don't seem to repeat up to here, either p,q are very large integers, or not integers at all. I think the second case is much more probable. PS: Disclaimer: I cheated - the first result is from maple on my laptop, but the second result was obtained from " Precision(500);ZetaFunction(3)/PI^4; " using a KASH online calculator (to ease copy-paste into this window "running" on another machine) ; however, my Maple gives me "7442134106" as the last digits... The results seem to agree up to the 7509 close to the end. |
[QUOTE=mfgoode;103869]
1+1/(2^3)+1/(3^3)+1/(4^3)+.......= ..? Any takers ? My conjecture is the sum will be (p/q) (pi^4) for some fraction p/q :smile: Mally :coffee:[/QUOTE] Why not [tex]\frac{p}{q}\pi^3[/tex]? I checked for that using Mathematica; if it were the case, p and q would have to have thousands of digits at least. My Fourier Analysis has a result that comes tantalizingly close: [tex]\Large{\sum_{k\ odd \geq 1}\frac{\sin(k\frac{\pi}{2})}{k^3}=1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+...=\frac{\pi^3}{32}}[/tex] If I could only get rid of all those minus signs, the problem would be solved; it is easy to show that the sum of all the odd terms is 7/8 of the final result. Oh well, at least it proves that: [tex]\Large{\zeta(3)=\frac{8}{7}(\frac{\pi^3}{32}+2(\frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\frac{1}{15^3}+...))}[/tex] [tex]\Large{\zeta(3)=\frac{\pi^3}{28}+\frac{16}{7}(\frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\frac{1}{15^3}+...)}[/tex] [tex]\Large{\zeta(3)=\frac{\pi^3}{28}+\frac{16}{7}(\frac{7}{8}\zeta(3)-(1+\frac{1}{5^3}+\frac{1}{9^3}+\frac{1}{13^3}+...))}[/tex] [tex]\Large{\zeta(3)=-\frac{\pi^3}{28}+\frac{16}{7}(1+\frac{1}{5^3}+\frac{1}{9^3}+\frac{1}{13^3}+...)}[/tex] |
[QUOTE=jinydu;104834]Why not [tex]\frac{p}{q}\pi^3[/tex]? I checked for that using Mathematica; if it were the case, p and q would have to have thousands of digits at least.
My Fourier Analysis has a result that comes tantalizingly close: [tex]\Large{\sum_{k\ odd \geq 1}\frac{\sin(k\frac{\pi}{2})}{k^3}=1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+...=\frac{\pi^3}{32}}[/tex] :smile: Well Jinydu that is good work on your part. I'm not into programming out these results but if you give me the sum of Zeta 3 to about 100 to 500 terms may be we could come up with something. I gave the fraction p/q (pi ^4) as Im sure there have been many trials on exponent 3 and they dont work out! Mally :coffee: |
[QUOTE=mfgoode;104872]
I'm not into programming out these results but if you give me the sum of Zeta 3 to about 100 to 500 terms may be we could come up with something. [/QUOTE] Do you mean a decimal expansion? That has already been done to billions of decimal places. I did a little bit of searching on this and according to [url]http://numbers.computation.free.fr/Constants/constants.html[/url], the current record is 2 billion decimal places: Cheng, H., Hanrot, G., Thomé, E., Zima, E., Zimmermann, P. "Time- and Space-Efficient Evaluation of Some Hypergeometric Constants." To appear in ISSAC 2007 ([url]http://www.cs.uwaterloo.ca/~issac07/)[/url]. The website of the first link actually has a downloadable program called PiFast that can easily calculate millions of decimal places. The first few are: 1.20205690315959428539973816151144999076498629234049... Or do you want an approximation in fractional form? Adding up the first 1000 terms gives: Numerator: 173505140632246515560188527476979249922446848506440759209193721322853315532916\ 968912652259452815699182592964065274265451062350793312831240024071661963287704\ 576671932036883405646938094448902393643239986268261600078001639566380412555865\ 874534557716497412687295647928815466299463138446146742162335968788021284726903\ 954636371513086653404682659836684901282308318584008866327082589930404211953252\ 122921087341883083059813476874072663142047634366271652514713654957009204531954\ 524558708142712746819053332107437637935776140598793340682492104904880708456332\ 945222255549865318656114021070241526304173921541453296293712613578747770044987\ 364405918897501781945104399983482862943447292229413187953181360782733383036289\ 895146986933600002923170154360703274898483654118959071986556214549368821032669\ 791591285416533720189461331332099822151800025293611841763274427437832701934116\ 109697402825151065106815896103541156285614670515624969131490494626985178524588\ 876156229177194142967265840085224021708821378427896807464374183580751890631672\ 992434177989640908370903172110187593511912471603295962272973642657229700104377\ 475745429187124036682560329294426524449121976034745442598253566255004312743317\ 091049776834009343040707205942313056296723928329916513186475437527351616318293\ 8191682313645607801513728707574638113894496689343 Denominator: 144340265651475079940803722695094530804808534539267637923458659622768945205199\ 273880003672286764646117301545381253857222969046353988513103483417849487617768\ 366484623836275659355076012125337901884144117710179905497238696337645421405601\ 928880246318860251577004495327305964839530205238599484867091080267593934108125\ 142518500638155027935402072335227478512573723248428379162749378512782364020590\ 366005399814947176382102152146685424903602132682510136735327306139298488865751\ 928806978809022487022697881371783820052656667899280821612085308067831104302361\ 915864665601706247324395423470100983887778374672499290418646045645343307597844\ 547774056805830861053967063925274245502529895649972311770546926339133414643074\ 928448454284001430032984184270932090933221006862934935472412386767903223538301\ 695313794450096871918945322625479131136002310407156748997841085545780590608636\ 062428902464566327026663437879137418546270958549836315759593623850005476278448\ 008894777391219744132972228072831647199227845855635704136849887435202158156543\ 964635117869089001644323227595795303969383645681137034412145910106758633987724\ 882130700527213382515660864259772551513413861081777768926743484733297478977299\ 445936244732632910716903192388035142200992367227859003046467599813368893711173\ 9438745765499868173004254052837982208000000000000 (Ignore the \'s) |
[QUOTE=jinydu;104891]Do you mean a decimal expansion? That has already been done to billions of decimal places. I did a little bit of searching on this and according to [url]http://numbers.computation.free.fr/Constants/constants.html[/url], the current record is 2 billion decimal places: Cheng, H., Hanrot, G., Thomé, E., Zima, E., Zimmermann, P. "Time- and Space-Efficient Evaluation of Some Hypergeometric Constants." To appear in ISSAC 2007 ([url]http://www.cs.uwaterloo.ca/~issac07/)[/url]. The website of the first link actually has a downloadable program called PiFast that can easily calculate millions of decimal places. The first few are:
1.20205690315959428539973816151144999076498629234049... Or do you want an approximation in fractional form? Adding up the first 1000 terms gives: Numerator: 173505140632246515560188527476979249922446848506440759209193721322853315532916\ 968912652259452815699182592964065274265451062350793312831240024071661963287704\ 576671932036883405646938094448902393643239986268261600078001639566380412555865\ 874534557716497412687295647928815466299463138446146742162335968788021284726903\ 954636371513086653404682659836684901282308318584008866327082589930404211953252\ 122921087341883083059813476874072663142047634366271652514713654957009204531954\ 524558708142712746819053332107437637935776140598793340682492104904880708456332\ 945222255549865318656114021070241526304173921541453296293712613578747770044987\ 364405918897501781945104399983482862943447292229413187953181360782733383036289\ 895146986933600002923170154360703274898483654118959071986556214549368821032669\ 791591285416533720189461331332099822151800025293611841763274427437832701934116\ 109697402825151065106815896103541156285614670515624969131490494626985178524588\ 876156229177194142967265840085224021708821378427896807464374183580751890631672\ 992434177989640908370903172110187593511912471603295962272973642657229700104377\ 475745429187124036682560329294426524449121976034745442598253566255004312743317\ 091049776834009343040707205942313056296723928329916513186475437527351616318293\ 8191682313645607801513728707574638113894496689343 Denominator: 144340265651475079940803722695094530804808534539267637923458659622768945205199\ 273880003672286764646117301545381253857222969046353988513103483417849487617768\ 366484623836275659355076012125337901884144117710179905497238696337645421405601\ 928880246318860251577004495327305964839530205238599484867091080267593934108125\ 142518500638155027935402072335227478512573723248428379162749378512782364020590\ 366005399814947176382102152146685424903602132682510136735327306139298488865751\ 928806978809022487022697881371783820052656667899280821612085308067831104302361\ 915864665601706247324395423470100983887778374672499290418646045645343307597844\ 547774056805830861053967063925274245502529895649972311770546926339133414643074\ 928448454284001430032984184270932090933221006862934935472412386767903223538301\ 695313794450096871918945322625479131136002310407156748997841085545780590608636\ 062428902464566327026663437879137418546270958549836315759593623850005476278448\ 008894777391219744132972228072831647199227845855635704136849887435202158156543\ 964635117869089001644323227595795303969383645681137034412145910106758633987724\ 882130700527213382515660864259772551513413861081777768926743484733297478977299\ 445936244732632910716903192388035142200992367227859003046467599813368893711173\ 9438745765499868173004254052837982208000000000000 (Ignore the \'s)[/QUOTE] Is that reduced? :wink: |
[QUOTE=drew;104895]Is that reduced? :wink:[/QUOTE]
I think so; Mathematica automatically reduces fractions. |
Decimal expansion.
:smile:
Hi Jinydu! I have not as yet looked or checked the websites you have given, so off hand, have you included pi^4 as a factor in your fraction to evaluate the fraction from the value of Zeta 3 ? The approximation was given by Albert H. Beiler author of 'Recreations in The theory of numbers' 1963 (pre-computer era) but he used pi^3 and so I guessed at pi^4. First evaluate the series: are the successive terms decreasing? in which case it will tend to a limit. If it is expanding I doubt if there is any point in pursuing the matter further. Mally :coffee: |
What I gave is a fractional approximation for [tex]\zeta(3)[/tex] itself.
Of course [tex]\sum_{n=1}^{\infty}\frac{1}{n^3}[/tex] converges. One can prove this by comparing it to [tex]\sum_{n=1}^{\infty}\frac{1}{n^2}[/tex], which is known to converge to [tex]\frac{\pi^2}{6}[/tex]; this argument also shows that [tex]\zeta(3)<\frac{\pi^2}{6}[/tex]. |
Irrational numbers!
:smile:
I’m sorry Jinydu, but the ‘hard’ work you put in that 1000 digit fraction was a waste of time! I was impressed at the interest you have taken. Well I dug further on my ‘scientific’. I recall that earlier I had tipped you off that the solution should be in irrationals or transcendentals. On further enquiry Zeta (3) is *‘known’* to be irrational. Hence it cannot be expressed as an integral fraction. I would have expected the normally informative Drew to have drawn your attention to this, inside of making wise cracks on your effort. Now with a bit of jugglery with my 10 digit Cal. I found an approximation (Appx) close enough for all practical purposes and in a fancy form. I got Zeta (3) to be near to the cube root of (pi/e) ^4 Which is equal to 1.212850266 …… Zeta (3) = 1.202056903… You may juggle around with other natural constants for a closer Appx. Regards, Mally :coffee: Kind Attention: XYZZY P.S. Kindly put it in Tex form and please Xzzzy put it as my formula above my pyramid avatar as its an original of mine. i.e Zeta (3) ~ cube root of (pi/e)^4 Thanks Mally :coffee: |
Yes, I knew that [tex]\zeta(3)[/tex] was irrational; that was (quite suddenly) proved by Apery in the 1970s. But I thought you were asking for a fractional approximation, so I provided one.
In any case, I don't think it is surprising that [tex]\zeta(3)[/tex] is irrational. In a way, this is what you would expect because [tex]\zeta(2),\zeta(4),\zeta(6),\zeta(8),...\zeta(2n),...[/tex] are all rational multiples of [tex]\pi^{2n}[/tex] and are hence irrational (in fact transcendental). In my opinion, a more interesting question is whether [tex]\frac{\zeta(3)}{\pi^3}[/tex] is rational; the pattern for the even natural numbers suggests that it should be. As far as I know, this question is still unsolved, although it is easy to use a computer to prove that if [tex]\frac{\zeta(3)}{\pi^3}[/tex] is rational, then the numerator and denominator must be very large (at least thousands of digits). |
[QUOTE=jinydu;105073]In my opinion, a more interesting question is whether [tex]\frac{\zeta(3)}{\pi^3}[/tex] is rational; the pattern for the even natural numbers suggests that it should be.[/QUOTE]
But isn't that a specious extrapolation? Why would you expect that because Zeta(n) behaves a certain way for all the evens, that it should behave at all similarly for the odds, especially given the fact that no closed-form expressions have been found for the odds? What you say above strikes me as analogous to reasoning that "the pattern for the even natural numbers suggests that the odds should all also be divisible by 2." As far as I can tell, there is *no* reason to expect a similar pattern to hold for Zeta(odd) as for Zeta(even). |
[QUOTE=ewmayer;105099]But isn't that a specious extrapolation?
~ ~ ~ As far as I can tell, there is *no* reason to expect a similar pattern to hold for Zeta(odd) as for Zeta(even).[/QUOTE] :rolleyes: Exactly so Ernst. I couldnt have said it better! In any case one could not determine it exactly and it is like the rest of the irrationals. Its like claiming that root 2 can be expressed as a fraction though it can be expressed as a length Its the same argument. Also the necessity of it involving the exponent 3 for pi holds no water. I have shown this in my approximation which is very close to the true value. Mally :coffee: |
[QUOTE=mfgoode;105103]Exactly so Ernst. I couldnt have said it better![/QUOTE]
Except that you're the one who suggested there should be similar pattern for Zeta(3) as for Zeta(2) to begin with! Hello? Post #4, scroll up a bit? Oh here, I'll even save you the scrolling: [quote=mfgoode;103869]Where even Euler was mute and even after 200 years plus, the following problem remains unsolved. The sum of the reciprocals of the odd powers of the integers. 1+1/(2^3)+1/(3^3)+1/(4^3)+.......= ..? Any takers ? My conjecture is the sum will be (p/q) (pi^4) for some fraction p/q[/quote] |
[quote=mfgoode;104898]off hand, have you included pi^4 as a factor in your fraction to evaluate the fraction from the value of Zeta 3 ?
[/quote] I already had posted zeta(3)/pi^4 to 500 digits (first post after your conjecture). The best way to check if a number is rational is (IMHO) to compute the continued fraction expansion: [code]gp > default(realprecision,500); contfrac(zeta(3)/Pi^4) time = 0 ms. %287 = [0, 81, 28, 3, 2, 1, 1, 1, 1, 1, 10, 1, 2, 5, 1, 3, 1, 5, 1, 1, 3, 2, 14, 2, 1, 1, 3, 1, 6, 2, 1, 3, 153, 1, 47, 1, 1, 2, 1, 6, 1, 1, 8, 1, 1, 13, 23, 1, 3, 1, 1, 1, 3, 2, 1, 2, 1, 1, 2, 1, 2, 16, 3, 1, 5, 1, 1, 9, 1, 1, 3, 1, 1, 25, 2, 1, 8, 8, 1, 1, 1, 14, 2, 1, 82, 2, 1, 42, 2, 1, 1, 18, 29, 1, 8, 3, 1, 7, 1, 2, 1, 2, 2, 1, 1, 8, 3, 1, 34, 2, 6034, 4, 1, 18, 1, 42, 1, 1, 12, 14, 2, 5, 2, 5, 1, 1, 4, 1, 10, 1, 7, 7, 13, 27, 25, 2, 3, 1, 6, 5, 1, 81, 17, 6, 3, 3, 1, 15, 1, 2, 6, 5, 2, 2, 1, 17, 1, 1, 4, 10, 1, 24, 1, 3, 1, 1, 117, 1, 4, 1, 2, 2, 2, 24, 1, 1, 1, 1, 1, 4, 1, 6, 1, 2, 2, 11, 1, 7, 1, 1, 3, 1, 4, 2, 38, 1, 1, 1, 1, 10, 1, 12, 1, 1, 1, 3, 1, 2, 1, 2, 1, 3, 2, 11, 1, 2, 9, 1, 6, 7, 1, 4, 1, 1, 2, 1, 1, 3, 1, 11, 1, 2, 2, 1, 1, 7, 13, 8, 1, 6, 1, 7, 24, 2, 5, 1, 5, 1, 2, 42, 1, 2, 1, 2, 1, 4, 3, 231, 224, 1, 3, 5, 3, 1, 1, 1, 4, 1, 1, 1, 5, 1, 5, 1, 11, 3, 2, 74, 1, 7, 2, 29, 1, 1, 1, 1, 4, 1, 1, 1, 3, 2, 2, 1, 4, 1, 2, 2, 2, 1, 3, 1, 237, 9, 13, 4, 2, 1, 2, 30, 1, 1, 1, 6, 2, 4, 1, 4, 1, 1, 1, 22, 1, 1, 10, 5, 1, 2, 6, 32, 7, 2, 23, 1, 6, 3, 2, 1, 1, 1, 1, 1, 1, 29, 2, 1, 10, 1, 2, 297, 1, 2, 12, 5, 1, 1, 7, 4, 1, 1, 3, 3, 1, 3, 1, 2, 1, 1, 7, 2, 1, 1, 2, 1, 3, 1, 1, 1, 2, 6, 1, 24, 1, 2, 1, 1, 2, 14, 1, 1, 4, 4, 4, 14, 2, 1, 1, 7, 3, 1, 1, 14, 3, 1, 1, 1, 1, 2, 2, 1, 26, 4, 1, 3, 2, 3, 1, 3, 1, 3, 2, 1, 29, 15, 1, 15, 29, 3, 1, 2, 2, 1, 1, 1, 1, 305, 2, 3, 20, 1, 4, 1, 1, 14, 1, 1, 1, 3, 1, 2, 35, 1, 6, 2, 62, 1, 1, 1, 5, 4, 1, 1, 3, 17, 1, 1, 1, 13, 1, 6, 1, 1, 1, 1, 2, 2, 4, 2, 1, 1, 15, 1, 2, 3, 1, 8, 1, 1, 1, 1, 7, 2] [/code](The 6034 would be a good point to stop in order to have a "simple" fraction. The presence of 1's indicates that the corresponding rest is "the most irrational it can be".) I also think that one cannot expect a similar relation for even powers than for odd powers. However, question for amateurs: why is the minus sign a problem here, while sum(1/i^2) = pi^2/6 sum((-1)^i/i^2) = -pi^2/12 sum(1/i^3) = zeta(3) sum((-1)^i/i^3) = -zeta(3) *3/4 sum(1/i^4) = pi^4/90 sum((-1)^i/i^4) = -pi^4 * 7/720 etc.? |
[QUOTE=ewmayer;105099]But isn't that a specious extrapolation? Why would you expect that because Zeta(n) behaves a certain way for all the evens, that it should behave at all similarly for the odds, especially given the fact that no closed-form expressions have been found for the odds? What you say above strikes me as analogous to reasoning that
"the pattern for the even natural numbers suggests that the odds should all also be divisible by 2." As far as I can tell, there is *no* reason to expect a similar pattern to hold for Zeta(odd) as for Zeta(even).[/QUOTE] I agree that it is not at all rigorous; I just think it makes it grounds for curiosity. m_f_h, look more carefully at my first post in this thread. If I could find the alternating sum for the entire [tex]\zeta(3)[/tex] series, I would have solved the problem. To see this, let X be the sum of the odd terms in [tex]\zeta(3)[/tex] and Y be the sum of the even terms. It is easy to show that X = 7Y. So I knew the value of X-Y, I would be able to solve for both and hence find X + Y = [tex]\zeta(3)[/tex]. However, the alternating series I originally only alternate among the [I]odd[/I] terms; the even terms don't show up at all. By the way, I seems likely that the same trick cannot be used to split the series even further into (1 mod 8 terms + 5 mod 8 terms) or (3 mod 8 terms + 7 mod 8 terms). In both cases, the differences between the two parts do not seem to be a rational multiple of [tex]\pi^3[/tex] for any rational number with a numerator or denominator less than 1000 digits, according to Mathematica. |
Rational vs irrational.
[QUOTE=jinydu;105073]Yes, I knew that [tex]\zeta(3)[/tex] was irrational;
In my opinion, a more interesting question is whether [tex]\frac{\zeta(3)}{\pi^3}[/tex] is rational; the pattern for the even natural numbers suggests that it should be. As far as I know, this question is still unsolved, although it is easy to use a computer to prove that if [tex]\frac{\zeta(3)}{\pi^3}[/tex] is rational, then the numerator and denominator must be very large (at least thousands of digits).[/QUOTE] :smile: Thank you for your reply and clarification on the subject. To me its a voyage of discovery all along. My question is :can an irrational fraction of numerator and denominator produce a rational number? For instance can root 2/root 3 be rational as it is stands? Well if you square both numbers it becomes rational but not as presented. To test your methods can you give me the fraction when Zeta (26) is divided by pi^2 ? I confirm it is rational and not irrational because Zeta (3) is itself rational. This question is also directed for m_f_h with his continued fraction algorithm. Awaiting your reply, Mally :coffee: |
[QUOTE=mfgoode;105368]
My question is :can an irrational fraction of numerator and denominator produce a rational number? [/QUOTE] Sure it can. For instance, take [tex]\frac{\sqrt{2}}{\sqrt{2}}[/tex]. [QUOTE=mfgoode;105368] To test your methods can you give me the fraction when Zeta (26) is divided by pi^2 ? I confirm it is rational and not irrational because Zeta (3) is itself rational. [/QUOTE] Actually, [tex]\frac{\zeta(26)}{\pi^2}[/tex] is a rational multiple of [tex]\pi^{24}[/tex] and is hence irrational, while [tex]\zeta(3)[/tex] itself is irrational. After computing the first 10 terms of the continued fraction expansion for [tex]\frac{\zeta(26)}{\pi^2}[/tex] and combining it into a simple fraction, I get: [tex]\frac{158157}{1560947}[/tex] |
similar pattern!
[QUOTE=ewmayer;105249]Except that you're the one who suggested there should be similar pattern for Zeta(3) as for Zeta(2) to begin with! Hello? Post #4, scroll up a bit? Oh here, I'll even save you the scrolling:[/QUOTE]
:smile: Well I agreed to your excellent interjection that what's good for the goose is not good for the gander. You have quoted from post no. 4 which was my first or (second?) Subsequently I clarified that p/q will not be rational but irrational. [QUOTE=Mally]On further enquiry Zeta (3) is *‘known’* to be irrational. Hence it cannot be expressed as an integral fraction.[/QUOTE] Evidently you have missed out on this post and hence lost the flow and development of the thread! Now I disdain posts "you said", "he said", and "I said" posts as I affirm it is an attack against the poster and not the post it self. How come you did not tap me on the back for an excellent appx I gave? Surprisingly no one commented on this and neither have they bettered it. Please give us helpful hints like the goose and gander or any thing more mathematical rather than obliquely criticising the work of others (esp mine) and their endeavours to get nearer to the truth! Mally :coffee: |
Typo error!
[QUOTE=jinydu;105369]Sure it can. For instance, take [tex]\frac{\sqrt{2}}{\sqrt{2}}[/tex].
Actually, [tex]\frac{\zeta(26)}{\pi^2}[/tex] is a rational multiple of [tex]\pi^{24}[/tex] and is hence irrational, while [tex]\zeta(3)[/tex] itself is irrational. After computing the first 10 terms of the continued fraction expansion for [tex]\frac{\zeta(26)}{\pi^2}[/tex] and combining it into a simple fraction, I get: [tex]\frac{158157}{1560947}[/tex][/QUOTE] :smile: [QUOTE=jinydu]Sure it can. For instance, take [tex]\frac{\sqrt{2}}{\sqrt{2}}[/tex]. [/QUOTE] Can you give another example not coming to 1 but any other integer? I'm extremely sorry jinydu. That was a typographical error in my post. I meant [tex]\frac{\zeta(26}{\pi^26}[\tex] i.e. Zeta 26/pi^26 in case the tex doesnt show up Now what is the fraction? Mally :coffee: P.S. BTW what went wrong with my tex language? |
[QUOTE=mfgoode;105393]
Can you give another example not coming to 1 but any other integer? [/quote] How about [tex]\frac{n\sqrt{2}}{\sqrt{2}}[/tex] for any integer [tex]n[/tex]. [QUOTE=mfgoode;105393] I meant [tex]\frac{\zeta(26}{\pi^26}[\tex] i.e. Zeta 26/pi^26 in case the tex doesnt show up Now what is the fraction? [/quote] [tex]\frac{1315862}{11094481976030578125}[/tex] [QUOTE=mfgoode;105393]P.S. BTW what went wrong with my tex language?[/QUOTE] Use /tex not \tex |
Fraction
[QUOTE=jinydu;105395]How about [tex]\frac{n\sqrt{2}}{\sqrt{2}}[/tex] for any integer [tex]n[/tex].
[tex]\frac{1315862}{11094481976030578125}[/tex] Use /tex not \tex[/QUOTE] :smile: Your fraction is correct. Thanks Jinydu. Mally :coffee: |
There's a useful trick for special-values-of-functions:
go to [url]http://integrals.wolfram.com/index.jsp[/url] and ask to integrate Zeta[26] and the output comes out as 1315862 pi^26 x / 11094481976030578125 I don't think this is the intended use of integrals.wolfram.com, but I don't know a full-fat Web Mathematica. |
Summation!
[QUOTE=fivemack;105602]There's a useful trick for special-values-of-functions:
go to [url]http://integrals.wolfram.com/index.jsp[/url] and ask to integrate Zeta[26] and the output comes out as 1315862 pi^26 x / 11094481976030578125 I don't think this is the intended use of integrals.wolfram.com, but I don't know a full-fat Web Mathematica.[/QUOTE] :rolleyes: Thank you fivemack for the 'useful trick' for summing up of series. I have noted it down At first glance I wondered how can one integrate a series without a variable, so I tried the URL you gave disbelievingly. Seeing an 'x' in the answer I caught on as strictly speaking it is actually the summing up of Zeta[26]x^0 hence on integration one must get x. This is a fine example for summation of series. Well we all live and learn which is an unending process!. The next logical step is to plug in Zeta (3) and lets see what comes up. I haven't done it right now and will let you know what comes up. Great Guns, fivemack! Mally :coffee: P.S. I got xZeta[3] Not even a decimal answer which jinydu got from his site. |
[QUOTE=fivemack;105602]There's a useful trick for special-values-of-functions:
go to [url]http://integrals.wolfram.com/index.jsp[/url] and ask to integrate Zeta[26] and the output comes out as 1315862 pi^26 x / 11094481976030578125 I don't think this is the intended use of integrals.wolfram.com, but I don't know a full-fat Web Mathematica.[/QUOTE] That shouldn't be too much of a surprise. The integral of a constant is the variable times that constant. |
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