Montreal Math ClubMathematics students in Montreal - Étudiants de mathématiques à Montréal

 Montreal Math Club :: Mathematics :: Calculus and Analysis Share |

# Exam question

AuthorMessage

Pythagoras

Posts : 17
Join date : 2009-11-09

 Subject: Exam question   Fri Nov 13, 2009 4:50 pm Hi. This is Jim. This is the problem that I showed Bruno last night from an old Concordia Ph. D. Comprehensive Examination. Question 1.(b)Prove: If f(x) -> + infinity and f'(x) -> 0 as x -> infinity then f is not a quotient of two polynomials.The solution I have seems too simple so I'd like to see if there is a proof using the logarithmic derivative of f that Bruno suggested.

Posts : 184
Join date : 2009-09-15
Age : 31
Location : the infinite, frictionless plane of uniform density

 Subject: Re: Exam question   Fri Nov 13, 2009 5:37 pm Hello Jim!The solution I suggested does not work after all. Indeed the log. derivative of a quotient of polynomial goes to 0 at infinity, so we can't use that!I would go like this : suppose f(x)=p(x)/q(x) is a quotient of polynomials. Then deg(p(x))>deg(q(x)), because of the first assumption. But f'(x)=(p'(x)q(x)-p(x)q'(x))/q(x)^2 is also a quotient of polynomials. The only way for this to converge to 0 at infinity is if we have : deg(p'(x)q(x)-p(x)q'(x))deg(q(x)). What's your solution?

Posts : 184
Join date : 2009-09-15
Age : 31
Location : the infinite, frictionless plane of uniform density

 Subject: Re: Exam question   Fri Nov 13, 2009 7:48 pm Here is my solution to #2a. Gave me some trouble!Let C be the set of conjugate subgroups of H, and N be the normalizer of H in G. Then we can rephrase the problem by saying that if G is the union of the elements of C, then H=G.If |C|>1 then we must have |C| > [G : H], because every element of C has order |H|, but since |C|>1 the union is not disjoint - the elements of C have the identity in common.A well known theorem states that |C| = [G : N]. Therefore [G : N] > [G : H], i.e. |G|/|N| > |G|/|H|, or |H|>|N|. This is absurd, because H is contained in N. Therefore |C|=1, H is normal in G and H=G.

Pythagoras

Posts : 17
Join date : 2009-11-09

 Subject: Re: Exam question   Sun Nov 15, 2009 2:05 pm Yes, that is exactly my solution. I haven't tried the group theory question yet though. So I will try it and see if I get the same answer.

Euclid

Posts : 95
Join date : 2009-09-15
Age : 56
Location : Alexandria

 Subject: Re: Exam question   Fri Feb 05, 2010 5:26 pm #2a is exercise 24, page 131 in Abstract Algebra Dummit & Foote.I'd say : if a subgroup contains a representative for each conjugacy class, it must be the entire group.

 Subject: Re: Exam question

 Exam question
 Page 1 of 1
 Similar topics
» Ibn Sina - Exam
» CS COURSE A GREAT CHANGE IN EXAM PATTERN: Introduction of Open Book Examination in Elective Subjects (Paper – 9) in Module-III of Professional Programme (New Syllabus)
» Question on Lambda symbol for light
» Question re generational curses/ healing prayer