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#1 Help Me ! » Discrete » 2010-08-12 10:22:46

samuel12
Replies: 2

Hi guys just some quick questions (need answered asap please smile)

(i) Prove that in any digraph with p vertices

p-1 >= RC(u) >= 1-p        Where RC(u) = outreach(u) - inreach(u)     and u is some vertex.

(ii) Prove that in any acyclic digraph, if (u,v) is an arc then RC(u)>RC(v)

(iii) Prove that if u is any source in an acyclic digraph and v is a sink, then
                          RC(u)>=RC(v)

under what circumstances can equality occur.

Thanks guys=)

#3 Re: Help Me ! » Big O notation » 2010-07-31 13:25:48

Hi mathsyperson, I'm just as confused as you are sad
An example of one we did in class is :

Show that

is

for

and find

  since log is increasing

So if someone could maybe go through this one, it would be much appreciated, cheers=)

Find

And maybe even (if you're feeling lucky big_smile)

Let f(n) = n^2 and g(n) = 2n^2. Find O(g(n)-f(n))

Thanks for your time smile

#4 Help Me ! » linear algebra » 2010-07-30 12:30:34

samuel12
Replies: 1

For the question:
Give a basis for the vector space V and find the dimension of V

where
means polynomial of the form

I could only find x in which this would work.

So would the basis be {x}??

I would appreciate some guidance rather than a direct answer(if this is possible=)) since this is an assignment question.

Thanks all

#5 Help Me ! » Big O notation » 2010-07-28 09:12:03

samuel12
Replies: 2

Hey guys just need some help with this one, thanks.

(a) If

and

show that

#6 Re: Help Me ! » linear algebra » 2010-07-25 11:27:25

Well yeah different notation in different countries i guess haha

#7 Re: Help Me ! » linear algebra » 2010-07-23 15:28:26

Hi;



now finally R2 - R1

Do I store R2-R1 back into R2?

Umm yeah so like R2---> R2 - R1

if that makes sense=)

#8 Re: Help Me ! » vector spaces and subspaces » 2010-07-22 08:04:34

Solution:

is a nonempty set because it contins the zero vector 0.

(i.e. Let)



Let u  and  v  be in 

Say,

and

Then

So u+v is also in

  (because it has the right form)

Siilarly if k is a scalar, then

So ku is in

Thus,

is a nonempty set of
that is closed under addition and scalar multiplication. Therefore,
is a subspace of
, by Theorem 5.2

#9 Re: Help Me ! » vector spaces and subspaces » 2010-07-21 21:55:07

Nevermind i think i've got it, I'll post my solution tomorrow.

#10 Help Me ! » vector spaces and subspaces » 2010-07-21 21:18:16

samuel12
Replies: 2

This is Theorem 5.2

Let V be a vector space and let W be a nonempty subset of V. Then w is a subspace of V if and only if the following conditions hold:

a. If u and v are in W, then u + v is in W.
b. If u is in W and c is a scalar, then cu is in W.

Now for the question:

In exercise 25, use Theorem 5.2 to determine whether W is a subspace of V.

Can someone please explain to me what is going on here, thanks smile  (if it makes a difference the column vector above is also ment to have squigly brackets around it, but i couldnt do that AND the parenthesis.)

#11 Re: Help Me ! » more linear algebra » 2010-07-21 09:11:53

Ok after a little thought power i managed to solve it....

From MATLAB B*A is:

Now we simply set up the equations

and solve each one for c1 c2 c3 by augmenting each system of equations and using rref or back substitution...

for the first equation:

will do.

for the second equation:

will do.

and lastly, for the third equation:

will do.

So we have written each column of BA as a linear combination of the columns of B.

#12 Help Me ! » more linear algebra » 2010-07-20 17:58:53

samuel12
Replies: 2

Hi can anyone explain a method to solve this problem

let A,B be two matrices where

A=[1,0,-2;-3,1,1;2,0,-1]  (where ; seperates the rows of the matrix)
B=[2,3,0;1,-1,1;-1,6,4]

Use the matrix-column representation of the product to write each column of BA as a linear combination of the columns of B.


cheers guys

#13 Re: Help Me ! » linear algebra » 2010-07-20 16:32:06

the first row operation will be...

minus R1 from each of  R2 through to Rn

then

divide all rows from R2 through to Rn by n

then

minus (n-1)R2 from each R3 to Rn (where n is the row number i.e. a constant)

now finally R2 - R1

to get....

[1 2 3 ... n]
[0 -1 -2 ... 1-n]
[0 0 0 ... 0]
[0 0 0 ... 0]
etc etc

which is the given nxn matrix in row echelon form (i think haha)

#14 Help Me ! » logic » 2010-07-20 08:59:22

samuel12
Replies: 1

I'm having a few troubles with 'validly deduce'

Here is the question:

Q: For any set S of sentences, if T is a tautology, then you can validly deduce T from S.

Can anyone explain to me what it is to 'validly deduce' something, pretty confused here, cheers smile

#15 Re: Help Me ! » linear algebra » 2010-07-20 08:56:08

Umm i'm not going to bother posting what i got unless anyone actually wants the answer smile if so speak up.

#16 Re: Help Me ! » linear algebra » 2010-07-18 13:04:52

yeah i think i got a bit confused=/ let me get back to you haha

#17 Help Me ! » linear algebra » 2010-07-18 12:20:42

samuel12
Replies: 11

just a quick question and sorry for being messy =/

[    1                2              3        ...          n   ]                                     
[  n+1            n+2          n+3      ...         2n  ]                           
[ 2n+1          2n+2         2n+3     ...         3n  ]                       
[   .                                                         .   ]                                         
[   .                                                         .   ]                                         
[   .                                                         .   ]
[(n^2 -n +1)(n^2 -n +2) (n^2 -n +3) ... n^2  ]

I am asked to find the row echelon form of the nxn matrix.
What i did is minused n from row 2 then minused 2n from row 3 and so on... untill the last row where i minused (n-1)n

this gave me an nxn matrix with identical rows containing 1 2 3 ... n     

Would this be sufficent as the row echelon form of the matrix or simplify it further?

Thanks in advance for the help smile

#18 Re: Help Me ! » probability » 2010-07-17 18:53:25

ok thanks so much bobbym that helped lots smile

#19 Re: Help Me ! » probability » 2010-07-17 18:33:21

Umm yes please that would be great if you dont mind

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