The span of S is defined as Span(S) = where for some field K. Span(S) is a subspace of V if it holds that:

- 0 is in span(S)
- for every v and w in span(S), v+w is in span(S)
- for every v in span(S) and a scalar c in K, cv is in span(S)

So all you need to do is check whether these conditions hold.

Suppose that some non-empty subset of S is *not* linearly independent and show that this leads to a contradiction.

This one of the most fundamental properties of linear independent sets and the proof is trivial when you consider that the only linear combination of any independent set that can be zero is 0v1 + .. + 0vi = 0.

Which textbook did you use? The chapters seem to be very similar in terms of content. We are using a book by Thomas Engel/Phillip Reid (2nd edition) called Physical Chemistry ldo.

I appreciate the help so far everyone, now I'm just stuck on a final set of questions that I would love some help with:

1. Determine, with proof, whether {(3,2,1) , (2,1,5) , (1,0,1)} is a basis for R3.

2. Invent a basis for each of the following subspaces of M(2,2)
a) the set of all 2x2 upper triangle matrices
b) the set of all 2x2 diagonal matrices

Lastly, how would I go about finding a basis, given a subspace?

Thank you again.

1) If f(x) = (x+1)/(1-2x), then f(-1/x) is equal to:
a) (x-1)/(x-2) b) (x-1)/(x+2) c) (x-1)/(2x+2)
d) (x+1)/(x+2) e) (x+1)/(x-2)


4) If 1/x - 1/y = 1/z, then y equals:
a) x/(x-z) b) xz/(x-z) c) z/(x+z) d) x/(z-x) e) xz/(z-x)


5) Let A= b(b-1)(b-2)....(1) where b is an integer. 1<b<60.
---------------------
b+1

For how many values of b is A an integer
a)41 b)40 c) 39 d)42 e)43



7) For the equation (a-b)^2 + a^2 = 169, a and b are integers and a is greater than or equal to 0. The number of different ordered pairs (a,b) that occur are:
a) 10 b) 11 c) 12 d) 6 e) 7

again, showing work is appreciated (i'm getting tested on this stuff later) but if your too lazy thats fine

Also question five if its not clear

A=

b(b-1)(b-2)....(1)
---------------------
b+1



b is an integer and 1<b<60

You should be able to figure this out. What different values can a take on?



And for each, what values can (a-b) take?

I am too lazy to work out 1 but the p-eigenbasis just means to solve for the wave equations in terms of p operators. Normally wave equations are found with respect to x, but by fourier transforming these wave equations you can switch to momentum space.

So after doing the Fourier transform your basis wave equations should be in terms of p. Thus your basis of wavefunctions depends on p; hence the name p-eigenbasis.

don't have a lot of time, so i can't elaborate a lot

1. f(-1/x) = (-1/x + 1)(1-2(-1/x))

make the bolded 1's equal to x/x and it's straight forward from there

4. 1/x -1/y = 1/z

1/x - 1/z = 1/y

get common denominator on the left hand side

(z-x)/xz = 1/y

cross multiply to isolate y / put y in the numerator

y = xz/(z-x)

the other 2 questions are trickier so i will get to the later if i have enough time.

For 1. Three vectors are a basis for R3 if they are linearly independent. LI here means that (0,0,0) = a(3,2,1)+b(2,1,5)+c(1,0,1) implies a,b and c are zero. Prove that this is the case (or find a counter-example) - it's a set of 3 simultaneous equations.

For 2. Pick the most obvious possible basis for M(2,2). Then remove some vectors from that basis to make a basis for the two subspace. If you can't think of an obvious basis for M(2,2), think what the standard basis for R^4 looks like, and it's similar to that.

Ya the 1-13 one aint hard, but I need to know how to work it out for larger numbers. Is there a shortcut?

I found the result I think you are looking for

http://sci.tech-archive.net/Archive/.../msg00251.html

So you can factorise your number and then see immediately how many ways it can be written as the sum of two squares.

For your 169 example, 169 = 13^2, so it can be written as the sum of two squares in 4*(2+1) = 12 ways. Of these, five are not possible by the requirement that a>=0; (-13,0), (-12,-7), (-12,-17), (-5,-17), (-5,7).

I think in general if n is not a perfect square, then exactly half of the solutions will not satisfy a>=0, if n is a perfect square then n/2-1 will not satisfy a>=0.

Question came up today which I couldn't do. Let a, b, c, d be positive integers such that abcd + abc + bcd + cda + dab + ab + bc + cd + da + ac + bd + a + b + c + d = 2009. What is a + b + c + d?

factorize something

maybe wait til next year

This had me laughing out loud.

I just checked this thread out for the first time and at this late hour don't have time to scope it out and see what's de riguer for the amount of assistance one is supposed to provide, but for the sake of deciphering thylacine's cryptic comment, add 1 to both sides. The LHS now factorizes quite nicely, and you can actually figure out what a, b, c and d are, up to permutation.

It wasn't cryptic, it was exact.

(a+1)(b+1)(c+1)(d+1) = 2010

Ha, for some reason I interpreted the problem as being a sum of base-10 numbers of 1,2,3, and 4 digits a,b,c and d which total 2009. I think there was a problem like that before with 2009. Mildly entertaining, and not too difficult. The answer to that one is a=1, b=1, c=2, d=3, so a+b+c+d = 7.

ok guys, i need help:

z^3=i

express z in the form z=a+bi

someone here able to help?

z= re^(i*theta)

So z^3 = r^3 e^(3*i*theta) = 1 * e^(i*pi/2, i*5pi/2, or i*9pi/2) = i

so r = 1, and theta = pi/6, 5pi/6, 9pi/6

e^(i*theta) = cos(theta) + i*sin(theta), so

z = cos(pi/6)+i*sin(pi/6)
z = cos(5pi/6)+i*sin(5pi/6)
z = cos(9pi/6)+i*sin(9pi/6)

are your 3 roots.

Hopefully this post is sufficiently general that you can solve other equations, like z^4 = 16i.

z = cos(x) + isin(x) for x = pi/6, 5pi/6, 9pi/6

ty guys, that was VERY helpfull and will for sure help a lot in the future. tyty

I spent a good 20 minutes figuring out how to use LaTeX so I see that your question has already been answered, but I'll post anyway (and hopefully in the future I'll be able to answer more quickly ):



Of course, you only need 3 values (i.e. substitute n=0,1,2 to get the values that Wyman/charlie gave), since cos and sin are periodic functions.

Tip: you can use the \text{} environment within mathmode to type normal text, so you get



edit: hmm, for some reason I can't put the first word in text mode.

edit 2: also, you can use \sin to get a decent lookin' "sin" i.e. instead of