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## #1 Re: Help Me ! » probability question (don't know if I did this right) » 2014-02-12 16:59:28

thank you for the help : )

## #2 Re: Help Me ! » probability question (don't know if I did this right) » 2014-02-12 16:39:39

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

## #3 Re: Help Me ! » probability question (don't know if I did this right) » 2014-02-12 16:28:55

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 : )

## #4 Help Me ! » probability question (don't know if I did this right) » 2014-02-12 15:41:05

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

Am I doing it right?

## #5 Re: Help Me ! » continuity » 2011-12-19 21:03:14

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

## #6 Help Me ! » linear algebra » 2011-11-11 19:40:31

Replies: 1

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

Replies: 0

## #8 Help Me ! » Dense and separable » 2011-07-04 19:51:41

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

## #9 Help Me ! » taylor's theorem? » 2011-06-21 15:51:04

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?

## #10 Re: Help Me ! » Powers of 3 and integers » 2011-06-20 15:12:34

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.

## #11 Re: Help Me ! » Joint Density » 2011-06-17 17:22:17

For 3,

so you are right ,gj

## #12 Re: Help Me ! » linear algebra proof » 2011-04-11 11:20:50

If you use Sylvesters rank inequality, you will get the answer right away.
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

## #13 Re: Help Me ! » convergence of a sequence » 2011-03-31 20:44:31

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 well
Hence
converges too, since its increasing and non negative.

## #14 Re: Help Me ! » Help!!!on calculus prove questions » 2011-03-31 19:34:43

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

## #15 Re: Help Me ! » Algebra problem! » 2011-03-31 18:51:35

never mind. I got it wrong : (

## #16 Re: Help Me ! » Intersections » 2011-01-22 17:14:50

a full fledged quartic is just nightmare.

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

## #17 Re: Help Me ! » Intersections » 2011-01-22 17:04:03

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.

## #18 Re: Help Me ! » Intersections » 2011-01-22 16:53:56

I got the equation for intersection is  (x-a)^2 +( 1/x -b)^2= r^2 , for circle centered at (a,b). And its of degree 4th. So it should have 4 roots. so At most 4 real roots. but the tricky part is that, y=1/x has two symmetric parts. x>0 and x<0. So the 4 intersections should be 2 for each one. I wish I just could argue from the symmetry but I dont know the mathematics for that.

## #19 Help Me ! » Intersections » 2011-01-22 16:28:58

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 : (

## #20 Help Me ! » Metric space topology » 2011-01-21 11:43:07

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.

## #21 Re: Help Me ! » A silly question about Mandelbrot set » 2011-01-06 19:44:51

hmm, so the generated graph is not the complete picture of it. So my question has not been a well defined one.
thanks

## #22 Help Me ! » A silly question about Mandelbrot set » 2011-01-05 13:57:18

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)

## #23 Re: Help Me ! » Linear algebra » 2010-11-23 20:36:03

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

## #24 Help Me ! » Linear algebra » 2010-11-23 18:15:13

Replies: 1

Let

be a subspace of
and its orthogonal complement respectively

defined projection operator

. And
has the property

Then my book says, the range of

is
, this is obvious

but I never get how it could conclude that  the null space of

is

I try assuming ,

and can't reach any contradiction.