You flip it when you multiply or divide by a negative.

Also remember that |X| < e is actually the inequality:

-e < X < e

Neither.

There is no 'right e' here. The point is that for *any* choice of e, there exists an N for which the inequality is satisfied by a_N, and all subsequent terms, if the sequence does indeed converge to l.

Perhaps a concrete example would help. Take a simple sequence a_n = 1/n, and we suspect that the limit is 0.

Suppose we choose e arbitrarily to start with, lets say we choose e = 1. Well then, the inequality |a_n-l| < e becomes |1/n| < 1, which is satisfied for n >= 2.

If we chose e = 1/10, then we need |1/n| < 1/10, which is satisfied for n>=11.

For *any* choice of positive e, we need |1/n| < e, which is satisfied for n >= 1/e. So all the terms after a_1/e satisfy the inequality regardless of the choice of e, and so a_n converges to zero.

Thanks, I think I might have a shot at this after all

what about the (infinite) direct sum of copies of Z_2?

That would just be (the limit of):

H_n : < a_1, a_2, ..., a_n | (a_1)^2 = (a_2)^2 = ... = (a_n)^2 = 1; for all i,j (a_i)(a_j) = (a_j)(a_i) >

Assume a_n converges to 1
(a) Show that there is N s.t. if n>=N then a_n<2
(b) Show that there is N_1 s.t. if n>=N_1 then a_n >1/2

This one is tough for me since a_n is not defined other than what it converges to.

Let N>e (Archimedean Property)

(a)

|a_n-1|<2

-1<a_n<3

(b)

|a_n-1|>1/2

|a_n|=|a_n-1+1|<=|a_n-1|+|1|<2+|1| for all n>=N??

Using triangle inequality

i.e. For all epsilon > 0, there exists an N_eps s.t. for all n >= N_eps, |a_n - 1| < epsilon.

Let epsilon = 1.

Let epsilon = 1/2.

So would this be the complete answer then

Assume a_n converges to 1
(a) Show that there is N s.t. if n>=N then a_n<2
(b) Show that there is N_1 s.t. if n>=N_1 then a_n >1/2

This one is tough for me since a_n is not defined other than what it converges to.

Let N>e (Archimedean Property)

(a)
For all e>0 There exists an N_e s.t. for all n>=N_e, |a_n-1|<e
|a_n-1|<1

0<a_n<2

(b)
For all e>0 There exists an N_e s.t. for all n>=N_e, |a_n-1|<e

|a_n-1|>1/2
1/2<a_n<3/2

The sequence a_1, a_2, .. converges to 1 if for all e > 0 there is some n such that 1-e < a_n < 1+e. Then consider the case when e = 1/2.

Complete sentences imo.

This is wrong. Should be: ...there is some n such that, for all m > n, 1-e < a_m < 1+e.

Edit: For example, to say that 0.99.. = 1 is really to say that the sequence 0.9, 0.99, .. converges to 1.

i like this idea and i'll probably be using it from time to time since i'm starting the meat of my ME courseload and i'm sure i'll have questions

just a thought on improvements, maybe have questions posted in another color, say green or red or whatever so that it's easier to distinguish between questions requiring help and possible solutions when skimming down the thread

Multi-variate Calculus

Consider the function



given by



Show that



is a 2-dimensional submanifold. Hint: It's image is a torus.

Now:



Next step: Find a non-singular submatrix.

The first entry of the jacobian is nonzero so long as , similarly the second and third entries are non-zero for , respectively.

Since the function is I expect there to be two parameters.

I've tried substituting the parametrically defined equations for a torus yet end up getting nowhere.

question: Let P(x,y) be a predicate defined on the universe of all ordered pairs x,y of natural numbers where 1 =< x, y =< 3. Exhibit finite disjunctions and conjunctions of propositions (without using quantifiers) which are logically equivalent to each of the following:

a) for all x P(x,0) or there exists x P(x,0)

b) For all x and For all y P(x,y)

i'm not even 100% sure what the question is asking for, if anyone could explain that it would be a big help, the section is called Predicate Forms and Quantifications. i don't know if it is asking to come up with propositions which fill these forms or if they just want equivalent statements that don't use quantifiers.

Hmmm...
Let me ponder this..
and return when I have solved the problem...
This may take awhile...
Rolling joint now..
brb

Solved it, just used the implicit function theorem. Realised f(x,y,z)=0 was the torus lol.

Classical Anlalysis

Consider the sequence a_n that takes on a value in the naturals

(a)give an example of such a sequence

I answered a_n = {2n|n is an element of the naturals}

(b) Show that a_n converges if and only if it is stationary. That is, if and onlu if there is a natural N such that if n>=N then a_n =a_N.

If you cold give me a clue on part b without answering. It is obvious this is true but I dont know how to prove it. So hint me in the right direction, then I will post my answer and see if it is correct.

Let L be the limit of {a_n}, and let epsilon = 1/2.

how would i set up the binomial distribution?

6 dice roll 1 6

(6 choose 1)(1/6)^1(1-(1/6))^5 doesnt look right
with this i get .401 the book says its .67

With 6 dice, to roll exactly one 6, the probability is:

6 * probability that the first die is a 6 and the other 5 are not

= 6 * (1/6) * (5/6)^5 = .402 (you should be more careful about rounding)

The probability that you roll at least one 6 is:

1 - probability of rolling all non-sixes

= 1 - (5/6)^6

= 1 - .3349

= .6651

disregard this, it is a really easy problem i was just reading 1=< x,y =<3 as 1=<x AND y=<3 when what it actually means is 1=< x =<3 AND 1=< y =<3

ok well i guess that was the gimmee,

if I wanna know the probabilty of rolling 2 or more 6's with 12 dice
then can i use the binomial distribution with n=12 k=2?

same thing three or more 6's with 18 dice n=18 k=3

whats my p?

2 or more 6's:

p = 1 - p(0 sixes) - p(1 six)

p(0 sixes) = (5/6)^12
p(1 six) = (12c1)*(1/6)*(5/6)^11

where NcR is the binomial coefficient N choose R.

I'll leave you to do the other one.

Assume a_n converges to l and is bounded below by m show that l>=m

This is obviously true just by definition of convergent. but what method should I use to show it?