Enjoy!Can the Sum of Reciprocal Powers be Represented by an Infinite Product?

[quddusaliquddus]

Know any 'nice' proofs in maths? Or know an alternative and simpler/nicer proof to common method employed? Post here ==>

 

[Zorodius]

I always liked http://www.usna.edu/MathDept/mdm/pyth.html

 

[quddusaliquddus]

Hmmm...seem to have a hard time going to the link.

 

[Zorodius]

You can try copying and pasting this into your address bar if the link isn't working for you:
http://www.usna.edu/MathDept/mdm/pyth.html

 

[quddusaliquddus]

Thanks ... it might jus be my slow connection ...

 

[Gokul43201]

More links :

http://math.colgate.edu/faculty/dlantz/Pythpfs/Bhaskapf.html

http://www.math.ntnu.no/~hanche/pythagoras/

 

[fourier jr]

More links :

http://math.colgate.edu/faculty/dlantz/Pythpfs/Bhaskapf.html

http://www.math.ntnu.no/~hanche/pythagoras/

/\ =

I like Euclid's proof better:

http://math.colgate.edu/faculty/dlantz/Pythpfs/Euclidpf.html

 

[quddusaliquddus]

/\ =

I like Euclid's proof better:

http://math.colgate.edu/faculty/dlantz/Pythpfs/Euclidpf.html

lol ... you got to admit ... his ones look prettier

 

[Gokul43201]

Euclid's proof was what I was taught in school.

Thought it was much nicer to draw a square inside a square and write "Behold !".

 

[quddusaliquddus]

lol...know any others?

 

[cragwolf]

This book contains many beautiful proofs.

 

[fourier jr]

This book contains many beautiful proofs.

definitely. I showed my topology prof the topological proof that there are infinitely-many primes & he liked it. There are lots of other good proofs in there too.

 

[1+1=1]

the infinitely many primes proof is cool. it seems funny how just saying "suppose not" wrecks the whole proof. (at least that is the way i proved it in HW)

 

[uart]

I've always liked the proof that,
Sum{ k=1,2,3..., 1/k^a } = Product{ p=2,3,5,7,11,13,17..., 1/(1-1/p^a) }.

I mostly like it because those two things, one a sum over all positive integers and the other a product over all primes, just seem so impossibly unlikely to be equal yet are fairly easily shown to be so.

 

[styler]

There is a very nice book re-issuedlast year which collects togther a bunch of extremely elegant proofs of classical theorems. In honor of legendary mathematician Paul Erdos, this tome, titled, Proofs From the Book, is actually a text aimed at advanced undergraduate math students.
Erdos was famous for, among other things, his semi-serious notion of proofs that were 'From the Book" This phrase, "From the book" references the idea that given any theorem there is a proof of it which is most pure and most fundamental so that if god were to write a collection of proofs of statements which were true and could be proved mathematically that this book of god's would include these elegant proofs.
The book is easy to read and eay to obtain. I know a few courses have been taught with it already.
But (I) don't take the from the book idea too seriously. It has already be proved that most things can't be proved (even if they are true) and some metamathemticians now support the idea that mathematics is really just as much a collection of unrelated, disconnected ideas as any other discipline expressed as a formal language. There is some motion among semioticians/mathematicians to try to contruct better arguments for the idea that mathematics is deply deficient (much more so than Godel showed) as a result of fundamental limits of constructs of the mind. Some would contend that one day pure math might seem as absurd as indrect proof by contradiction.
But its a cool book. And it does contain soem proofs of familiar theorems which are terse and others which are really pretty nice and help to elucidate why somehting is true (or better, HOW something is true).

 

[futb0l]

I am bookmarking this thread. I need good resources to learn number theory.

 

[futb0l]

hmm .. I don't get this one ...
Sum{ k=1,2,3..., 1/k^a } = Product{ p=2,3,5,7,11,13,17..., 1/(1-1/p^a) }.
can anybody explain it clearer?

 

[uart]

hmm .. I don't get this one ...
Sum{ k=1,2,3..., 1/k^a } = Product{ p=2,3,5,7,11,13,17..., 1/(1-1/p^a) }.
can anybody explain it clearer?

If you apply the binomial expansion "1/(1-x) = 1 + x + x^2 + x^3 + ..." to the right hand side (RHS) of the equation (with x =1/p^a) then you get an infinite product of infinite series,

RHS = ( 1 + (1/p1)^a + (1/p1^2)^a + (1/p1^3)^a + ... ) ( 1 + (1/p2)^a + (1/p2^2)^a + (1/p2^3)^a + ... ) ...

At first sight the above expansion looks like it's made things more complicated instead of simpler. Note however that if you start expanding out the products then you see that each product of reciprocal prime powers actually corresponds to a unique reciprocal integer power.

For example, say you have an integer n that has it's unique prime factorization as n=p1 p2^3 p3 (just as a simple concrete example, n=270 in this case), then note that (1/p1)^a (1/p2^3)^a (1/p3) = 1/(p1 p2^3 p3)^a = 1/n^a.

From the above example you should be able the see that the RHS expansion contains "1/n^a" for every possible positive integer n. Further, due to the uniqueness of prime factorization, there are no repeated terms, thus the RHS is none other an the infinite sum of 1/n^a terms exactly as per the LHS.

If you like you can test the consistency of this theorem with a simple numerical example. Try for example doing the few dozen terms with a=2 and you should see the LHS sum and the RHS product both converging to Pi^2 / 6.

That is, both these should converge to Pi^/6,
LHS = 1/1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + 1/49 + ...
RHS = 1/(1-1/4) 1/(1-1/9) 1/(1-1/25) 1/(1-1/49) 1/(1-1/121) ...