Recent content by tuananh

Yeah, with the hint above, I have an idea: Suppose dimM < dimV and M has a basis {m1, m2, ..., mk}. Add to this basis some vectors called v1, v2,..., vn such that {m1, m2,...mk, v1, v2,..., vn} becomes a basis of V. Of course, v1, v2,..., vn are not in M. Define u1 = m1 + v1, u2 = m2 + v1...

Here is an counter-example: define f(x) = 1 if x in [0,1], f(x)=0 eslewhere Apparently, f is square-integrable on [0,infty) but f(x)/x is not.

By saying "H and K is homomorphic", I mean that there is an one-to-one mapping f from H to K such that both f and f^{-1} (the inverse of f) are continous. Perhaps, this concept should be called "isomorphic" and sorry if I made a mistake. If H=R and K=R^2, I think R and R^2 cannot be...

Hi all experts, I've just visitted another Maths forum and picked up two interesting questions: Question 1. Let H and K be homomorphic normed linear spaces. Is it necessary that H and K have the same dimension if both H and K are finite-dimension ? Is there possible a homomorphism between an...

Define M = sup{f(x) | a <= x < c}, we prove lim f(x) = M x->c- For every e>0, there exist an element a<= d < c such that M-e < f(d) <= M Thus, for all d < x < c, we have M-e < f(d) <= f(x) <= M < M+e This means lim f(x) = M x->c- The second equality can be proved...

"for every e>0 there exists x in A such that c-e<x<=c and there exists y in B such that c-e<y<=c so we have two elements that are both in A and B, but this is a contradiction." Wrong conclusion ! Ok, for every e>0, there will be x in A and y in B such that c-e<x<=c and c-e<y<=c, but from...

"for every e>0 there exists x in A such that c-e<x<=c and there exists y in B such that c-e<y<=c so we have two elements that are both in A and B, but this is a contradiction." Wrong conclusion ! Ok, for every e>0, there will be x in A and y in B such that c-e<x<=c and c-e<y<=c, but from...

Using this equality e^x >= x + 1 > x for all x, we have e^n > n for all n. Thus, ln(n) < n for all n. This implies 0 < ln(n)/n! < 1/(n-1)! for all n>1 The series (Sum 1/(n-1)!) is convergent, so is the series (Sum ln(n)/n!)

and here is the answer of the second question: Note that: 2^(2n) - 1 = 4^n -1 and use the equatity : a^n - b^n = (a-b)(a^(n-1)b - a^(n-2)b^2 + ...) with a=4, b=1 Or we can check all steps of the induction process as follow: With n=1: 4^1 -1 = 3 is divisible by 3 Suppose the...

I do not thinks the above equality is correct. Just take x = -1, we have arctan(x) = arctan(-1) = -Pi/4, and arctan(1/x) = arctan(1/-1) = arctan(-1) = -Pi/4. So, the sum of them is -Pi/2. However, let consider the function f(x) = arctan(x) + arctan(1/x), x>0 On this interval, f(x) is...