just checking the last line and

Thank you sir

Quick pre-image question, where A and B are subsets of the reals and f(x) = x^2.

f(A) = [0,4]
f(B) = [-1,1]

f^(-1)(A) = [-2,2]
f^(-1)(B) = [-1,1]

The second one is sort of strange because [-1,0) isn't hit by the pre-image I give and that's the best we can do with reals. What're we supposed to do here...?

Thanks for the help,
Mariogs

technically, f^-1(Y) = {x in X : f(x) in Y}

So if f:R-->R, x|-->x^2, then f^-1({1, apple, cumquat, 0}) = {-1, 1, 0}

but it's a stupid question

edit: and although [-1,0) is not in the range (or image) of f, if we define f:R-->R, at least it's in the codomain. Still stupid though.

it's in the codomain?

I just meant that it's asking us what maps to the reals included in [-1,1]...and obviously [0,1] as an interval works, and we just ignore the fact that we can't get to the [-1,0) portion of [-1,1]

Thanks for the help, Wyman. Algebra not offered in the summer but real analysis is and Abbott's book has been great so far.

Hey guys. Can anybody help me with the following:

I have a bunch of data point (say 10k) with different values and I want a program that builds the type of distribution that best fits them. You can imagine it like representing the data with many small bars and the curve passes through the top of each one. Any ideas how this can be done?

I think the easiest and cheapest way is creating a histogram in Excel (assuming you already have Excel). You'll have to enable the Analysis Toolpak. Here's a little tutorial I found using Google for creating histograms in Excel.

Given choices of distributions, it's easy to measure which one is best. But there are infinitely many choices of plausible distributions, so you have to make a choice first.

If you're a little more explicit about what you're doing, maybe we can suggest something.

I know what you meant. I'm saying that when I say

F: X --> Y

this doesn't mean that all of Y is in the image of F. Take the map F:RxR-->R, F(x,y) = 0. This is a map from RxR to R, but the range or image of F is {0}. But R is the codomain (some set that includes the range) we specified. Semantics.

I'm saying it's silly for us to ask for the preimage (under F) of a set in the codomain that doesn't lie entirely in the image of F. But whatever. The preimage of a set in codomain \ image is just { } I guess.

You can use the statistical programming software R and use QQ plots.

Also, it sounds like you are describing a histogram, which R can give quite easily.

Feeling dumb but trying to show sq rt(3) is irrational.

Assume it's rational, so (a/b)^2 = 3 for some integers, a, b.

We have a^2 = 3b^2, and we know that a and b can't both be even (since then we could reduce (a/b) further.

If a is odd, a^2 is odd and b must be odd. So a = 2m+1, b = 2n+1 for some integers, m, n.

Now we have (2m+1)^2 = 3(2n+1)^2.
Simplifying, we get:

4m^2+4m+1 = 3(4n^2+4n+1)

We know the left term is odd (even + even + 1) and right term is odd * odd = odd...so what's the issue here?

When I looked online they did some weird algebra to show that one side had to be odd and the other even, so our assumption that sq rt(3) is rational must have been wrong, aka sq rt(3) is rational.

What am I missing here, guys?

Thanks for the help,
Mariogs

Left hand side is congruent to 1 mod 4
Right hand side is congruent to 3 mod 4

(assuming you did the algebra correctly)

Yeah, and the way to think about mods without saying "mod" is to pull everything divisible by 4 onto one side:

4m^2+4m+1 = 3(4n^2+4n+1)
...

4(m^2 + m - 3n^2 - 3n) = 2

m^2 + m - 3n^2 - 3n is an integer of course, but 4x = 2 has no solutions in the integers.

if sqrt(16) = ± 4

then, if I am asked to evaluate the expression:
16^(3/2)

thats the same as sqrt(16)^3 right?
which is the same as
4^3 and (-4)^3 right?

so the answer is 64 or -64 correct?

I'm reviewing some stuff using an answer key and it says the answer is 64.

Although there are two solutions to x^2 = 16, when someone writes sqrt(16) they are referring only to the positive solution. This is why, when you have y = x^2, you have to write x = {plus or minus} sqrt(y) -- because sqrt(y) by itself refers only to the positive solution.

So sqrt(16) = 16^(1/2) = 4. Thus 16^(3/2) = sqrt(16)^3 = 4^3 = 64.

Let set A be bounded below and B = {b in B s.t. b is a lower bound of A}.

Aim to show: sup B = inf A.

Proof:
Assume not: Either sup B > inf A, or sup B < inf A

If sup B > inf A, there is a b in B that is not a lower bound of A (that is, there is an b in B greater than an a in A). But B is the set of all lower bounds of A and we've just discovered a b in B that is not a lower bound of A. Contradiction.

If sup B < inf A, there is a lower bound of A greater than any b in B. But B is the set of all lower bounds of A. Contradiction.

So it must be the case that sup B = inf A.

Look right?

your definition for B seems off.

B = {b in B such that...} is self-referential

Why? What b is that?

This is wordy

Why?

Grading this, I want to see that you understand the definition of inf and sup, so you should invoke them.

1) Sup B = least upper bound of B, which is the set of all lower bounds of A. So if Sup B > inf A (the greatest lower Bound of A), there must be a b in B which is greater than at least one element in A (inf A). But if this is true, this b is not a lower bound for A, which is a problem since B is the set of all lower bounds of A.

2) If Sup B < inf A, there must be at least one element, x, between Sup B and inf A. The existence of X means that our original Sup B isn't the least upper bound of B (the lower bounds of A), since Sup B < x < inf A. In short, x is also a lower bound of A, which is problematic since x > Sup B, and there should not be any elements in B > sup B.

Ish?

Thx for the help, Wyman. This stuff's tricky heh.

Still not happy.

What is B, by the way? Your definition was off. Is it "the set of lower bounds for A"?

B is the set of all lower bounds of A. What still needs work?

Hey, I have a homework question I'm having trouble with.

Let A be an open set in Rn.
f: A -->Rn is differentiable and has a non-singular derivative.

Prove that f(A) is open.

My first thought was to use the Inverse Function Theorem, but we're only given that f is differentiable, not continuously differentiable. Does anybody have any clue how to solve this?

A is a set with a lower bound, and B is the set of lower bounds of A. Prove sup B = inf A.

1) sup B <= inf A.
If x is a lower bound for A, then x <= inf(A) [def'n of inf]
Every element of B is a lower bound for A, so for each b in B, b <= inf(A).
Hence inf(A) is an upper bound for B.
Hence inf(A) >= sup(B) [by the def'n of sup].

2) sup B >= inf A.
inf A is (by definition) a lower bound for A. Hence it is in B. Hence it is <= sup B, since sup B is (by definition) an upper bound for B.

Hence sup(B)=inf(A)

My guess is that the intent was that you assume f is continuously differentiable.

There's a more general thing going on here, called Brouwer's Invariance of Domain Theorem, which says that f need only be continuous and injective. But of more relevance to you is:
http://terrytao.wordpress.com/2011/0...entiable-maps/
which describes an inverse function theorem that relaxes the "continuously" portion of the continuously differentiable hypothesis.

It's cool that this theorem fails if A is not in R^n. Take A = (a,b) (open interval in R^1) and f:R-->R^2, f(x) = (x,x). Certainly the image of A is not open in R^2.

GL.

Simple Trig question I'm sure, but I can't find the answer.

The problem I'm trying to solve:

note, let y = sin(x)

Find all solutions of the following equation in the interval [0,2pi)
3y^2 - 7y + 2 = 0

...
(3y-1)(y-2) = 0
sin(x) = 1/3 or 2 (and 2 is invalid, so only 1/3)
sin(x) = 1/3

The answer is x = ???

I'm familiar with the unit circle and reference angles, but basically I know only about 2 right triangles:

One with sides 1, 1, and sqrt(2) and one with sides 1, sqrt(3), 2.

So for example, if sin(x) = sqrt(3)/2, then x = pi/6, but I only know that because I'm familiar with that triangle.

How do I find x where sin(x) = a/b (or sin(x) = 1/3 for example)?

Thanks.