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bobbym contacted me in my email (something im truly grateful for since i havent visited this forum in a while) and let me know.

Taylor, how are you man ??? Its been so many YEARS i thought i might never come in contact with you. bobbym gave me your email so I will contact you through there.

Thanks bobbym.

Thanks so much. I had worked up to the part:

And i didnt know how to continue. However, you said the above implied the following:

And I still dont know how? Yes, I see that the r is becoming r^2 and the angle of sin and cos is becoming double, but is there a way to keep working algebraicly until the last statement appears?

Thanks.

**LuisRodg**- Replies: 4

Hello all,

Question is:

Show that (x,y) --> (x^2 - y^2, 2xy) maps the first quadrant {x>0, y>0}into the upper half plane {y>0}.

Now if i were to "brute force" this then I can easily do it. By "brute force" i mean I consider any arbitrary (x', y') in the upper half plane and show this point can be expressed as (x^2 - y^2, 2xy) where x>0, y>0 (first quadrant). This shows that every point in the upper half plane can be obtained. Then show that if (x,y) is chosen in the first quadrant, then the transformation cannot give a point in the lower half half plane.

So done. However, its not beautiful.

In the book they hint to use polar coordinates and then in the answers in the back they say:

(x,y) --> (x^2 - y^2, 2xy) becomes:

(r, 0) --> (r^2, 2*theta)

and I just dont see how?

Thanks for any help.

**LuisRodg**- Replies: 3

Im just brainstorming here.

Lets say i have the following two numbers:

2.31

2.39

How many digits are they accurate to? 1 or 2? So basically, for accuracy do you go to the decimal digits to check how many digits match? (so in this case it will be accurate to 1 digit?)

Thanks.

Yes I took algebraic topology and algebraic geometry as well.

Wow! My post from 4 years ago!

Well, don't know what to say, other than I ended up graduating with a math major and a minor in computer science :-)

I'm now a masters student in math.

Time passes incredibly fast, and you cannot let obstacles get in the way of your dreams. That is what I have learned...

**LuisRodg**- Replies: 0

Let A be a set. Denote by dA the set of boundary points of A. Denote by intA the interior of A. Denote A' the closure of A.

It was asked of me to prove the following:

(i) dA' is a subset of dA

(ii) d(intA) is a subset of dA

and I have already done so. However, now I'm asked to find a example of a set A such that dA, dA', and d(intA) are all different. This set can be as weird as you like and you can consider any topology on any space.

I been trying to find such a set for some time but haven't been able. Such a set A cannot be open because then A = intA and obviously their boundary points will be the same. Similarly if A is closed A = A'

**LuisRodg**- Replies: 0

The question is as follows:

Is it possible to find a metric in R^2 such that the ball of radius 1 centered at 0 looks like:

[-10, 10] x [-1/10, 1/10] U [-1/10, 1/10] x [-10, 10]

At first I thought it was possible and scratched my head many hours and after writing down a lot of equations and verifying I came up with the following "metric":

d((x1,y1), (x2,y2)) = min { max { 10|x1-x2|, |y1-y2|/10 }, max { |x1-x2|/10, 10|y1-y2| } }

Surprisingly this DOES give the above picture as the ball, however, triangular inequality fails, and hence its not a metric.

Turns out that it is not possible to find a metric. Professor hinted "you can use the vector space structure of R^2", but i dont know what to make of it.

Also, now that I think about it. Probably the reason no metric can give such a ball is because the figure is not convex? Although as far as I know, a norm will always give convex balls, while metrics can give both convex and nonconvex?

Any help....? I have suffered enough.

**LuisRodg**- Replies: 1

Definition: an Abelian group G is p-primary for any prime p, if for every g in G there is n>0 such that g*p^n = 0

So then given any arbitrary Abelian group, not necessary p-primary, we can define it's p-primary components:

Gp = {g in G : there exist n>0 such that g*p^n = 0} for any prime p.

So that means that for any prime p, an Abelian group has a p-primary component Gp.

So then there is a theorem that says that G is the direct sum (finite!) of it's p-primary components. My confusion is, since for any prime p there is a p-primary component, that means that there are infinitely many p-primary components. In the decomposition of G into the direct sum of it's p-primary components, which of the infinitely many components do you choose for the decomposition into a finite direct sum of components?

I don't know if I'm explaining myself. A group G has infinite components, but there is a finite decomposition into direct sum of those components, how do you choose those finite components from the infinite list?

Thanks.

**LuisRodg**- Replies: 0

Here is one I was able to do:

1) Given the following curve:

Find the coordinate ring

: = because the division by Y eliminates all Y and leaves only remainders of X.So now I need to do another:

2) Given the following curve:

Find the coordinate ring

: and now I dont know what happens...Any help? If anyone needs any clarification on the notation, please tell me.

Thanks.

I didnt get either of them?

**LuisRodg**- Replies: 7

I do not understand the last part, how they get the iff statement. Where did 3pi/4 come from?

Could someone please explain? Thanks.

**LuisRodg**- Replies: 1

Any idea how to prove this corollary?

Thanks.

**LuisRodg**- Replies: 4

The derivative of a matrix is the matrix containing the derivative of each entry. I thought this was by definition but it is not.

From the above definition of the derivative of a matrix, how do I derive that:

**LuisRodg**- Replies: 2

Im reading a booklet on Methods of Applied Analysis and they mention this theorem and say the proof can be looked at in any linear algebra book. I looked in my linear algebra book and could not find such theorem.

Could anyone care to provide a link to a proof or state it here if its readily available? Thanks.

**LuisRodg**- Replies: 2

How do you prove this? Any ideas?

**LuisRodg**- Replies: 1

Now, in the case that n is prime, the contradiction easily follows that n divides both p and q. What about the case where n is non-prime?

**LuisRodg**- Replies: 1

Any hints? Thanks.

Im sorry but how do you arrive at that hint?

^ (in response to Ricky's hint which is now deleted?)

Would help a lot if you put it in latex.

It converges to zero I think.