Happy Birthday, Leonhard Euler! - Page 2

jinydu · 2007-05-01, 01:16

What I gave is a fractional approximation for $\zeta(3)$ itself.

Of course $\sum_{n=1}^{\infty}\frac{1}{n^3}$ converges. One can prove this by comparing it to $\sum_{n=1}^{\infty}\frac{1}{n^2}$ , which is known to converge to $\frac{\pi^2}{6}$ ; this argument also shows that $\zeta(3)<\frac{\pi^2}{6}$ .

mfgoode · 2007-05-02, 06:00

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

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

jinydu · 2007-05-02, 08:28

Yes, I knew that $\zeta(3)$ 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 $\zeta(3)$ is irrational. In a way, this is what you would expect because $\zeta(2),\zeta(4),\zeta(6),\zeta(8),...\zeta(2n),...$ are all rational multiples of $\pi^{2n}$ and are hence irrational (in fact transcendental).

In my opinion, a more interesting question is whether $\frac{\zeta(3)}{\pi^3}$ 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 $\frac{\zeta(3)}{\pi^3}$ is rational, then the numerator and denominator must be very large (at least thousands of digits).

ewmayer · 2007-05-02, 15:53

Quote:

Originally Posted by jinydu

In my opinion, a more interesting question is whether $\frac{\zeta(3)}{\pi^3}$ is rational; the pattern for the even natural numbers suggests that it should be.

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).

mfgoode · 2007-05-02, 16:10

Quote:

Originally Posted by ewmayer

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).

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

ewmayer · 2007-05-04, 17:51

Quote:

Originally Posted by mfgoode

Exactly so Ernst. I couldnt have said it better!

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:

Originally Posted by mfgoode

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

m_f_h · 2007-05-04, 21:59

Quote:

Originally Posted by mfgoode

off hand, have you included pi^4 as a factor in your fraction to evaluate the fraction from the value of Zeta 3 ?

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]

(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.?

jinydu · 2007-05-04, 22:58

Quote:

Originally Posted by ewmayer

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).

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 $\zeta(3)$ series, I would have solved the problem. To see this, let X be the sum of the odd terms in $\zeta(3)$ 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 = $\zeta(3)$ . However, the alternating series I originally only alternate among the odd 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 $\pi^3$ for any rational number with a numerator or denominator less than 1000 digits, according to Mathematica.

mfgoode · 2007-05-06, 07:22

Quote:

Originally Posted by jinydu

Yes, I knew that $\zeta(3)$ was irrational;

In my opinion, a more interesting question is whether $\frac{\zeta(3)}{\pi^3}$ 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 $\frac{\zeta(3)}{\pi^3}$ is rational, then the numerator and denominator must be very large (at least thousands of digits).

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

jinydu · 2007-05-06, 07:33

Quote:

Originally Posted by mfgoode

My question is :can an irrational fraction of numerator and denominator produce a rational number?

Sure it can. For instance, take $\frac{\sqrt{2}}{\sqrt{2}}$ .

Quote:

Originally Posted by mfgoode

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.

Actually, $\frac{\zeta(26)}{\pi^2}$ is a rational multiple of $\pi^{24}$ and is hence irrational, while $\zeta(3)$ itself is irrational.

After computing the first 10 terms of the continued fraction expansion for $\frac{\zeta(26)}{\pi^2}$ and combining it into a simple fraction, I get:

$\frac{158157}{1560947}$

mfgoode · 2007-05-06, 07:46

Quote:

Originally Posted by ewmayer

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:

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:

Originally Posted by Mally

On further enquiry Zeta (3) is *‘known’* to be irrational.
Hence it cannot be expressed as an integral fraction.

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

2007-05-02, 06:00	#13
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2²×3³×19 Posts	Irrational numbers! 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 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 Last fiddled with by mfgoode on 2007-05-02 at 06:09 Reason: Adding P.S.

2007-05-02, 08:28	#14
jinydu Dec 2003 Hopefully Near M48 1758₁₀ Posts	Yes, I knew that $\zeta(3)$ 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 $\zeta(3)$ is irrational. In a way, this is what you would expect because $\zeta(2),\zeta(4),\zeta(6),\zeta(8),...\zeta(2n),...$ are all rational multiples of $\pi^{2n}$ and are hence irrational (in fact transcendental). In my opinion, a more interesting question is whether $\frac{\zeta(3)}{\pi^3}$ 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 $\frac{\zeta(3)}{\pi^3}$ is rational, then the numerator and denominator must be very large (at least thousands of digits). Last fiddled with by jinydu on 2007-05-02 at 08:29

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2007-05-01, 01:16	#12
jinydu Dec 2003 Hopefully Near M48 2×3×293 Posts	What I gave is a fractional approximation for $\zeta(3)$ itself. Of course $\sum_{n=1}^{\infty}\frac{1}{n^3}$ converges. One can prove this by comparing it to $\sum_{n=1}^{\infty}\frac{1}{n^2}$ , which is known to converge to $\frac{\pi^2}{6}$ ; this argument also shows that $\zeta(3)<\frac{\pi^2}{6}$ .