You are not logged in.

- Topics: Active | Unanswered

thank you for the help : )

my number is exactly the same as yours. So I think I got it right

Oh cool! thanks, didn't know there was this type of problem. Looks like I got the right answer

The reason I got this problem is , one of my english teacher, she groups students in 8 and ask them to make one comment on other group member other than themselves. So in this way, only 37% chance everyone gets a comment which is a massive fail lol. I am going to email her about it : )

**Dragonshade**- Replies: 7

It is probably really simple, but I don't have an easy solution ( not even sure if I did it right)

so 8 people have their assigned seats, If we shuffle them around, what is the probably of all 8 not sitting on their assigned seats

Here is what I did

I solve this recursion with computer program, it gives

is about 37%Am I doing it right?

I'd say only one, that is at x=1

**Dragonshade**- Replies: 1

Is it true that every vector space (infinite dimension) contains a nontrivial (not just 0) finite dimension subspace?

**Dragonshade**- Replies: 0

problem solved. Delete this thread please

**Dragonshade**- Replies: 1

How to prove the space of bounded function on a closed interval B[a,b] is non-separable (does not have a countable dense subset) ?

the metric is the sup |g(t)-f(t)| .

I think setting up an uncountable and disjoint collection of subset, and then if a dense set exists, it would contains uncountably many elements.

but I dont know how to do that

or does anyone have a better way?

thanks

**Dragonshade**- Replies: 0

J is a functional

the book I am reading says the following

I don't see how taylor's theorem would lead to the last step. What version of taylor theorem is this?

Here is another algorithm I work out

1.Write the number in base 3

2.Replace the 2 , if theres any, before each term with (3-1)

3.simplify

4.if there are still 2 within the terms, repeat the process

for example

I think this process would make all the 2s vanish eventually.

For 3,

so you are right ,gj

If A is a m-by-n matrix and B n-by-k, then

Rank(A)+Rank(B)-n <= Rank(AB). Which is just 2Rank(A)-m<= Rank(A^2) =Rank(0)

2Rank(A) <=m/2

try this

consider two vectors

By Schwarz's inequality,

converges

We know C_n is convergent, C_n is bounded by some number M

for any epsilon > 0, there exist N, such that |C_n - b| < epsilon/ M for n>N

then

So

converges as wellHence converges too, since its increasing and non negative.

For part a of the first problem

Suppose such a y exists

then

Now we could use induction on it

Suppose n=k holds

when n=k+1

never mind. I got it wrong : (

a full fledged quartic is just nightmare.

is there any theorem about the intersection of curves based on curvature

Ah, sorry, I didnt see that cuz the page didnt load fully. I am looking at it

But I mean any arbitrary circle, not necessary the one center at the origin.

**Dragonshade**- Replies: 8

how to prove that an arbitrary circle only intersects y=1/x (x>0) at most 2 points. I tried brute force, polynomial equation with degree 4 is not fun at all. Curvature might be a way, but I have no ideas : (

**Dragonshade**- Replies: 0

Show that the set S={(x,y) | xy=1, x>0 } is closed in R^2

I came up with a way to show that S is homeomorphic to x axis by some projections, therefore closed. but my class has only talked about open, closed set in a metric space so far. I don't know how to prove it by showing that the complement is open. Since constructing an open ball for every point outside the curve has been mathematically daunting. ( always bumped into some almost uncalculatable polynomial expressions). Help.

thanks

**Dragonshade**- Replies: 2

I noticed that one would eventually be looking at the center of the mandelbrot shape by picking any point on the graph and zooming deep enough. If the center of that shape is colored black. Why wouldn't the entire graph look black? (assuming there is no limit to the computing speed to the computer)

Ah, nvm, it follows right from the orthogonal decomposition of a vector.

**Dragonshade**- Replies: 1

Let

be a subspace of and its orthogonal complement respectivelydefined projection operator

. And has the propertyThen my book says, the range of

is , this is obviousbut I never get how it could conclude that the null space of

isI try assuming ,

and can't reach any contradiction.thanks in advance.

That explains it very clearly, Thanks guys