Recent content by NoDoubts

guys, need help. this should be easy. pls let me know if anything is unclear. the statement about different field homomorphisms being linearly independent is called Dedekind lemma. My question is why being linearly independent in this case implies existence of n elements a1, ..., an \in K...

I mean, one direction is clear i.e. if there exist n elements with the det != 0 then homomorphisms are clearly not linearly independent... I don't see the other direction i.e. given n linearly independent homomorphisms why there exists n elements with the above property?

Should be simple, but can't figure out :) Why is that , for a field K, the linear independence of field homomorphisms g1, ..., gn: K -> K equivalent to the existence of elements a1, ..., an \in K such that the determinant det| gi(aj)| != 0 (...so, just like in a case of linear...

cheers. now I can move on relieved :)

yep. I think I nailed it by Rouche's theorem http://en.wikipedia.org/wiki/Rouch%C3%A9's_theorem you can majorize the |3zn - ez | by the |7*z^n| on the unit circle |z|=1, then the result follows.

do you know how else I can solve this problem?

yes. this is obvious, but how does it answer the question? what we need to do is, according to argument principle, to calculate an integral on closed countour which is unit circle. http://en.wikipedia.org/wiki/Argument_principle

anyone?

well, f'(z)/f(z) = (1863*z^620 - e^z}) / (3*z^621 - e^z). Now, we need to integrate that along the contour C. I've tried to integrate it directly by setting z = e^{theta*i) where 0 < theta <= 2pi , but got stuck.

sure. and how does it help?

or use any of the high precision c++ libraries...

this is problem 7 from here http://www.math.umass.edu/~hajir/m621/m621hw6.pdf anyone?

How many zeros inside the unit disk does the following function have? f(z) = 3 z^621 - e^z. (in words, three times z to the power 621 minus e^z :) ...argument principle answers this question, but I have a problem evaluating the integral of f'(z)/f(z).

hmmm...OK. looks like Herr Gauss has solved it long time ago :))) http://books.google.com.hk/books?id=JFX8IBiAoaoC&pg=PA8&lpg=PA8&dq=gauss+irreducible+polynomial+rational+functions&source=bl&ots=W12oFhmrIh&sig=l7RSmPQiDos4ZxswQ25DUkwVse4&hl=en&sa=X&oi=book_result&resnum=4&ct=result#PPA8,M1

yes, later on it says that "if f (polynomial from k[x,y]) is irreducible then it remains irreducible as an element of k(y)[x]" can someone explain to me why it is so? ...I guess, if we assume f = gh where g,h are from k(y)[x] then we can multiply both sides by the product of common...