I meant f(x) is that expression and not (f(x))^-1.
But well...I just got to know that the problem can be solved not using ordinary generating functions but will require "something" called exponential generating functions.
Can Someone here please give me a motivation for the concept of...
We are about to begin this topic soon in class...
I'd like to know all about generating functions and their application. I tried reading it up...One thing I'd like to know is, once you have a generating function...then what? You get some information...but what is it?
I tried to find no. of...
have been thinking about prob 3....i guess you take A6 (the set of even permutations of 6 elements), possibly find a subgroup H of order 4 and then look at the 3 distinct cosets of H.
suppose G is abelian and has 2 elements of order 2, say a and b...then {e,a,b,ab} becomes a subgroup of G. but 4 does not divide 2p unless p=2. if p=2 then anyway {e,a} is a subgroup of G of order 2( p in this case).
i guess i must( or rather i am supposed to) add my thought process...
since 3 does not divide n (n,3)=1 so we have un +3v=1...now for any x in G, x = x^(un+3v)= x^3v.
so G ={y^3 / y is in G}...using this i think we're supposed to show that ab=ba for all a in G. here's where i'm stuck.
seeing lots of group theory here after a really long time...
let G be a finite group of order n, where n is not divisible by 3. suppose
(ab)^3 = a^3 b^3 ,for a, b in G . prove that G is abelian.
Here’s an interesting question…
Let R be a commutative ring and ‘a’ an element in R. If the principal ideal Ra is a maximal ideal of R then show that ‘a’ is an irreducible element.
If a is prime, this is pretty obvious…if a is not prime, then we say a= bc for some b,c in R. Now we need to show...
this seems to be a very fundamental problem...but i need help...
prove or disprove : let D be a euclidean domain with size function d, then for a,b in D, b != 0, there exist unique q,r in D such that a= qb+r where r=0 or d(r) < d(b).
first of all, what is size function? next...do we only...
ok...so am + bn= 1 implies......1-bn = am
that implies... bn(1-bn) = abmn = 0...
so bn is an idempotent element ....
i looked at " am +bn =1" a hundred times before posting this question....but it flashed just now! thanks a ton!
here's a real tough one ( at least for me) ....show that the ring Z/mnZ where m ,n are relatively prime has an idempotent element other than 0 and 1.
i looked at examples and it works....
do we look for solutions of the equation a^2 -a = kmn , for some k in Z( that is, other than 0 and 1)?
help!
thanks...that makes it clear...
but here's another question :redface: does this mean i could have done this problem without considering separate cases???
let F= { 0=a1, a2,a3,a4.....an} be a finite field. show that
(1+a2)(1+a3).........(1+an) = 0.
when n is odd, it's simple since 1 belongs to F. then odd number of elements are left( they're distinct from 0 and 1). at least one of them, say x,must have itself as its inverse. x^2 = 1...
hi!
would like to know what a homeomorphism means ( how do you geometrically visualize it?)
AND is the symbol 8 homeomorphic to the symbol X? Why or why not?
( from whatever little i know intuitively about homeomorphisms, i think it is not....)
Yeah, you use cayley-hamilton theorem
so, you have p(A)=0...
That implies a*A^n + b*A^(n-1) + c*A^(n-2)........+ I= 0
(i've used a,b,c as coefficients)...then take the identity matrix to the other side. Multiply both sides by inverse of A. Then RHS becomes -A^(-1) the LHS shows that the...