I don’t know enough about chemistry to really answer your question, but if you’re interested in quantifying how a continuous phenomenon changes, you need calculus/differential equations
Is it possible to solve equations similar to
http://www.texify.com/img/\normalsize\!y%3Dt^3%20%2B%20t.gif
for t (or whatever variable t happens to be)?
I don’t think you can get an explicit formula for expressing t in terms of y in that case.
Does Finite Mathematics cover everything but Calculus?
Short answer, yes. Long answer, no. There’s a lot of discrete math that isn’t covered by finite mathematics.
I looked it up and found http://www.wolframalpha.com/. This site is awesome.
The top solution will graph a fairly accurate (slightly inaccurate according to my TI-84) inverse of
http://www.texify.com/img/\normalsize\!f(t)%20%3D%20t^3%20%2B%20t.gif
Is it wrong that I’m starting to hate Linear Algebra? So abstract as fuck. 
Yes. Subtract y from both sides. Then use numerical methods to approximate the roots.
Define “everything”
You’re frustrated. It’s normal. Post questions and I will answer them
Thanks but not now. For now I’m just reading my book and watching Khan Academy videos on concepts such as Linear Transformations and Isomorphisms. I feel that I have a basic idea of what’s going on but when the questions come, I can’t help but feel lost most of the time. I also feel that the questions aren’t as hard as I’m making them out to be but the way they word it makes it daunting.
Learning to translate what is being asked into what you know mathematically is an important skill to cultivate. It’s also important to notice the bigger picture, and there is a bigger picture. I’m sure you’ve already gone over that nice list of equivalent statements. Learning how to use that is useful. As I had mentioned before, understanding set theory is important since the idea of vector spaces uses that concept.
Ultimately, your success will be determined by how well you can deduce information from what you are given. If you’re finding something to be difficult, you need to go back and really understand what that concept is saying. I hope you aren’t just relying on the examples that are being given to get you through the course.
Everything. Well, everything that you will begin to see after algebra taken at an intermediate level. I was assuming this includes a combination of introductory stats, trigonometric problems, and the principles of mathematics to a degree.
Yea, something like that. Discrete/Finite Mathematics is a course that deals with everything else in the sense that calculus isn’t involved. There are no trig problems though. A good finite math course will cover some subset of the following:
Set theory
Propositional logic
Proofs
Counting techniques (permutations, combinations, pigeon-hole principle, inclusion-exclusion, etc)
Finite probability theory (sample spaces, conditional probability, expected value, Baye’s Theorem, Independence)
Recursion (a.k.a. how to find your place in a line without really counting)
Algorithms
Graph Theory (you work with dots called “vertices” and the lines that connect them)
If you’re lucky you might get to cover some number theory as well.
Graph Theory is amazing. I’m glad I worked for a professor deeply interested in it.
Okay so I’m a bit stumped here.
Find if p is a subspace of P2 and if so, find a basis for it.
{p(t):p’(1) = p(2)} p’ is the derivative
I checked that the three conditions on checking if a subset is a subspace pass (contains neutral element of 0, closed under addition, and closed under scalar multiplication) and they did pass but when finding the basis, the book says that the answer is ‘1 - t, 2 - t^2’ and now I’m trying to figure out how.
I did start however by finding the kernel on my own which is [1 1 2] (not sure if it’s supposed to be horizontal or vertical) by doing the following work shown below.
Any help is appreciated. 
You’re on the right track. You just use algebra to complete the problem.
Your set is all polynomials of the form at^2 + bt + c that satisfy p’(1) = p(2)
You correctly deduced that this implies that 2a + b + c = 0. But this is also a homogeneous equation of 1 equation and 3 unknowns, which has infinitely many solutions. So you have two free variables, and one variable c that you can solve for. So a = p is free, b = q is free, and c = -(2p + q)
If this confuses, you, note that c + 2a + b = 0 is the same system.
So the solution is
<a,b,c> = <p, q, -(2p+q) > = <p, 0, -2p> + <0, q, -q> = p<1, 0, -2> + q<0, 1, -1>
So the coefficients of the basis vectors are <1, 0, -2> and <0, 1, -1>
Let p = -1 and q = -1 for the free variables and plug in the corresponding values of a, b, and c in each basis vector for at^2 + bt + c and you have what your book wrote. Any real numbers p and q would work. They picked these so the solution would be increasing by degree for “style”.
I’m probably confusing myself even further by asking this but how do you decide what’s free and what’s not? Because I’m looking over your work and I can see what you’re doing, but now I’m asking myself why c is not free like the others. Also, I noticed how you changed the variables a bit from a + b + 2c to 2a + b + c, was mine wrong by any chance?
Also I’m going to be up all night since I have a midterm tomorrow at 4 pm and have a couple more questions to ask so you don’t mind do you? :lol:
It’s 1 equation in 3 unknowns in a homogeneous system. Therefore I can only solve for one unknown variable in the homogeneous system. c is not free because it’s what I solved for. If I had solved for a instead, then b and c would be free. Since addition is commutative, both systems are equivalent.
If n is the number of columns in a matrix, then n = [number of leading variables] + [number of free variables]
I changed the variables because I always saw the quadratic formula represented as ax^2 + bx + c. They’re just variables. You can have rw^2 + uw + e if you so desire.
You can post more questions if you’d like.
Okay I get it now thanks. Now for these next problems you’re supposed to find the transformation matrix and see if it’s an isomorphism, if not, then find the image, rank and basis of the kernel.
T(f(t) = f(-t) from P2 to P2 with regards to the basis [1 0] [0 1] [0 0] [0 0]
[0 0], [0 0], [1 0], [0 1]
and
T(M) = [1 1]M - M[1 1] from R^2x2 to R^2x2 with respect to the basis [1 1] [1 -1] [1 0] [0 1]
[1 1] [1 1] [-1 -1], [1 -1], [0 1], [1 0]
Again thanks for your help!
I’m a little confused about your notation, especially with regards to your basis vectors.
Are the bases defined in terms of matrices maybe? Could you clarify this?
Welp the forum blew my formatting for the basis vectors to hell, let me redo them.
for the first problem
[1 0] [0 1] [0 0] [0 0]
[0 0] [0 0] [1 0] [0 1]
for the second problem
[1 1] [1 1] [1 0] [0 1]
[-1 -1] [1 -1] [0 1] [1 0]
Did the best I could, sorry about that.