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philmoore 2007-04-15 14:00

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.

mfgoode 2007-04-16 16:21

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:

davieddy 2007-04-17 06:21

Mozart's era:smile:

mfgoode 2007-04-17 08:59

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:

m_f_h 2007-04-27 17:32

[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.

jinydu 2007-04-28 20:37

[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]

mfgoode 2007-04-29 16:25

[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:

jinydu 2007-04-30 00:21

[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)

drew 2007-04-30 02:29

[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:

jinydu 2007-04-30 03:15

[QUOTE=drew;104895]Is that reduced? :wink:[/QUOTE]

I think so; Mathematica automatically reduces fractions.

mfgoode 2007-04-30 03:35

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:


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