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 11 
 on: March 12, 2010, 04:16:46 AM 
Started by GODLOVEME - Last post by GODLOVEME
Show that |R2| = |R|. (Hint: replace R by {0, 1}N and construct a bijec-
tion by interleaving a pair of binary sequences to form a single sequence.)
please step by step

 12 
 on: February 24, 2010, 06:18:47 AM 
Started by KH_Global - Last post by KH_Global
Dear All,
              I have opened up a shop of storing fruit boxes. There is always fluctuating demand for the fruits and I am not able to keep the optimum level of stock. Somebody suggested me of applying the theory of probability to keep the appropriate leve of stock in my shop.

I welcome suggestion and recommendation from your side to solve this problem.


 13 
 on: February 21, 2010, 07:05:26 PM 
Started by John - Last post by webmaster
Check http://www.mathcs.org/analysis/reals/infinity/countble.html for sets and sequences that are countable yet contain intuitively "a lot" of numbers.

 14 
 on: February 21, 2010, 01:55:53 PM 
Started by John - Last post by John
How can a set be countable and dense, it would not have a one to one relation to the set of Natural Numbers?

I really really appreciate your help, friend.  Smiley

 15 
 on: February 21, 2010, 12:08:03 PM 
Started by John - Last post by webmaster
Perhaps it is not that bad. A sequence just means it must be a countable set so you are looking for a countable set that is 'dense', or in simpler terms kind of 'everywhere'. That might be a well known set, and since it is countable you can think of it as a sequence. If you find that set/sequence the rest might might not be too bad. Good luck!

 16 
 on: February 21, 2010, 10:45:26 AM 
Started by John - Last post by John
Who can think of a sequence X_n such that for every real number L there is a subsequence of X_n that converges to L?

So that sequence must have subsequences converging to every real number.... I stayed up all night trying to think of possible solutions, but this is out of my league I think now! Huh

 17 
 on: January 08, 2010, 12:53:17 AM 
Started by Saaqib - Last post by Saaqib
I am pursuing a PhD degree in Mathematics.Visually handicapped, I am unable to benefit much from the class lectures. I wish I could receive video lectures in English on all the core branches of Mathematics, notably Mathematical Analysis, Abstract Algebra, Differential Geometry, (Partial) Differential Equations, and more. Also, I would be extremely grateful if somebody could donate to me a selected range of books in good quality printing and in their latest editions on Analysis, Abstract Algebra, Differential Geometry, Differential Equations, and other major branches of Mathematics.
My postal address is as follows:
Saaqib Mahmood,
Kund Malyaar, Muhallah Musa Zai,
Nawan Shehr, Abbottabad,
PAKISTAN
Phone: +92-346-9527638

 18 
 on: December 22, 2009, 03:38:21 PM 
Started by deepnorth - Last post by ste
The question is very vague, but it sounds like f(x) is defined on the real number line where its range is the two point set {-1,1}, so f: A -> B where A = {the real line} and B = {-1,1}.

If the set A has the usual topology associated with the real line then open sets are of the form (a,b) where a < b and closed sets are of the form [a,b] and [a,a], where the latter denotes the singleton set {a}. If the set B is equipped with the discrete topology, then every singleton in B is an open set i.e. {-1} and {1} are both open in B, which implies that {-1} and {1} are also both closed in B (because the complement of an open set is closed and the complement of a closed set is open). So in this case, the complement of {-1} in B is {1} and the complement of {1} in B is {-1}.

Let U be a subset of the interval (0, infinity). The image set, f(U), regardless of whether U is open or closed, is {1} which is both open and closed.

This is not a complete solution, but look over the finer details of the question again and see if it helps.

Ste.

 19 
 on: December 22, 2009, 02:20:30 PM 
Started by ste - Last post by ste
I managed to work it out.

f(w) is entire because exp(w) is not equal to 0 for any w in the complex plane, however, (z-1)f(z-1) is not analytic at z = 0. Because the point z = 0 lies inside of the specified contour, we investigate the nature of this singularity by computing the Taylor series of f(z-1) about the point 0 and introducing a factor of z-1 to each term. Looking at the first term of the Taylor series, we see that res{z-1f(z-1);0} = the residue of z-1f(z-1) at 0 = -cos(0) = -1
 
By Cauchy's residue theorem the value of the integral is equal to -2(pi)i(res{z-1f(z-1);0}) = 2(pi)i.

 20 
 on: December 18, 2009, 06:42:27 PM 
Started by ste - Last post by ste
Does anyone know how to integrate f(z-1)/z with respect to z over the clockwise oriented circle of radius 2, centred at the origin (of the complex plane), where f(w) = (w - cos(w))/exp(w) and z & w are complex numbers?

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