Discrete math still makes me cringe lol
i really hate to ask but i’ve been working on this trig substitution for about 2 hours now and watching youtube lectures to help me figure it out.
it’s integral of x/sqrt( (x^2)+6x+12 dx
when i complete the square i get x/ (x+3)^2+3
then i draw my triangle with sqrt ((x+3)^2+3 as the hypotenuse
i make the following substitutions
sqrt(3)secΘ = (x+3)^2+3
x = sqrt(3) tanΘ-3
dx = sqrt(3)sec^2Θ
then end up with integral
(sqrt(3) tanΘ-3)(sqrt(3)sec^2Θ)/sqrt(3)secΘ
when i simplify i get integral
sqrt(3) tanΘsecΘ-3secΘ
am i correct so far?
I think that GPA is a flawed ass system. The thing is though, having to know how to do 90% or more of the material (like Calculus) only to have a slight error causing you to get 1 question wrong on a **4 QUESTION EXAM WORTH 35% OF YOUR GRADE **really sucks ass because then you end up getting like a C and in turn, makes your GPA look like crap because GPA only is calculated using only letter grades…
eh, i know lots of people with wholly mediocre GPAs that got great jobs after college. most of their interviews involved impromptu tests and scenarios, so as long as you can bang out the work, you’re good.
as long as you can wrap your head around induction, discrete math shouldn’t be too bad.
another thing about gpa is that it depends on major specifically, a high gpa in a science major looks a lot better than one in criminal justice or english. Here’s not knocking any of those lol.
yeah my gpa is high enough (not 4.0!) where Cs bring me down, worst shit ever when you are in a class you can’t really stand but you have to take.
I don’t think there is anything wrong with GPA, just the way people use it and how it’s interpreted. To me, GPA is just that, the sum of the grades times hours / sum of hours, iirc. Problem is, people go off on a tangent and start saying stuff like 3.2-ish “are really hands on and are more suited for the field while 3.9 are really analytical”. Ooookkkaayyy.
I’m pretty sure I’ve met a 3.9 from a name brand university that was one of the most moronic idiots I’ve met while I’ve also met a 3.9 who was really good at what he was doing but taking outside of his comfort zone, he was completely derailed.
O and by the way, artistic and linguistic and rhetoric are equally as important science and math. You have to be creative, just plugging and chugging numbers into an equation doesn’t jack crack. That’s trig sub right? Do you know where it comes from? Do you know the logic behind? Are there assumptions made? Do you know when it’s appropriate and when it doesn’t work?
http://www.thehcmr.org/issue1_1/gaitsgory.pdf
Hahaha. I had to think for a minute whether it’s obvious that the determinants are zero without any calculations. I came to the conclusion that it’s not “immediately obvious”, but it is “after a minute obvious”. And that last sentence killed me.
How the heck do you get to teach math at Harvard and think for a generic matrix the determinant is nonzero?
I haven’t taken linear algebra in forever and even I know that’s completely bogus
I’m sure he didn’t really think that. He meant that for a random matrix, the probability that Det is zero is zero, but it’s not like his students would be able to figure this out. But then to get zero for a clearly not random matrix and not try to figure out the column dependence is pretty bad. There are only three columns and it takes about 30 seconds to find it…
Here’s something cool (geogebra or some type of software will help):
Graph any parabola with two distinct real zeros and mark them. Now reflect the parabola across the horizontal line that passes through its vertex (so you have mirror images touching each other). Plot the complex zeros of this parabola in the x-y plane and you should notice something.
question
@HoneyBBQGrundle @Warrior’s Dreams @Tekno Virus
I have to prove that S = {a set of all real numbers} is a supspace of complex numbers.
I’m lost as to how to set this up…
If i let X, and Y be any element of any real numbers, and p,q,r,s be an element of any real numbers, where q=0 and s=0
is x and y an element of the complex plane where
C^n = {a + bi : blank} and x = p/q and y = r/s
is this even remotely close, or am I really off base.
Is this the question stated verbatum? The reason I ask is that there is an n there. At first I thought that based on the form of an element of a set a + bi, that n = 2, but then I thought that if n > 2, then C^2 would be isomorphic to C^n anyways. So I’m a little confused by the notation there.
But you said to prove that R is a subspace of C, so that is what I will go with.
In any event, I suppose this will depend on how much you are allowed to assume.
Of course, real numbers are just complex numbers with b = 0 using your nomenclature.
The first thing that jumped into my mind is to use the fact that C and R are both vector spaces. Then, there is a theorem from linear algebra where all you have to do is show that complex addition and scalar multiplication to a complex number still holds.
The longer approach would be to show that all 10 vector space axioms hold, meaning that you’d have to show 8 more things.
But basically, the proof goes like this
x+y = (a + bi) + (c+di) = (a + c + bi + di) = (a+c) + (b+d)i = (a + c) + (0+0)i = (a+c) + (0)i= a + c
kx = k(a+bi) = (ka+kbi) = (ka) + (kb)i, = ka + (k*0)i = ka + 0i = ka, where k is a real number
That’s pretty much it. Notice that you started out with two vectors u and v, and ended up showing that the vector u+v is also an element by using the definition. That is the general way to solve this kind of problem.
C^n was to denote a complex plane in any dimension. not raised. lol,
But thank you for that. the hardest part of this is setting it up, doing the axioms is the easy part, but that’s to be expected since that’s essentially algebra…hahah
@Warrior’s Dreams @HoneyBBQGrundle
A⊕B = -(A+B) and kA = -kA
So I have to prove axioms 5 and 6, which prove something is a Vector space. Now Axiom 5 states that the addition of a zero vector, results in the same thing
axiom 5 0⊕u = u
and axiom 6 states that a additie inverse must exist. u⊕(-u) = -u⊕u = 0.
Coincedentially, axiom 6 fails automatically if axiom 5 fails. You must have a zero vector for axiom 6 to be satisfied. So, my question is, why does axiom 5 fail?
A⊕B = -(A+B), where u⊕0 = u, therefore, there must be an element that when added, A⊕B ⇒ A⊕0 = -(A) correct?
If I assume B = 0 then
-(A+B) = -(A+0) = -A.
Now I’m not understanding what’s going on. And the solution that I’m supposed to get is
-(A+B) = A ⇒B = -2A
Why do I have -2A, is it because when I add the zero vector, this is what it looks like
-(A+B) + -(A-B) = -2A
If that’s the case, where am I getting this additional -(A-B). Thanks
OK, so more ideas. Algebraically, we want A + (-A) =0 so that A+ 0 =A, hence why you can’t have axiom 6 without axiom 5. the reason I’m thinking of it this way, is because essentially we are doing algebra, except we define what addition means, and what multiplication by a scalar means, which is why it’s called abstract addition and abstract subtraction. So everything on the right hand side can be freely manipulated algebraically provided, you follow the rules and start ith the given definition. right?
If I assume B = 0
A⊕B = -(A+B) = -(A+0) = -A, so -A = A (the supposed zero vector)?
therefore, if u⊕0 = (A⊕B)⊕A
(A⊕B)⊕A = -(A+B)+(-A) = -(A+0)+(-A) = -2A = 2A =! A⊕B
or am I reaching hard here.
[s]You are confusing the symbol for the additive inverse with the element in that vector space that has all zeros that when using the usual rule for addition just happens to also use the same symbol.
With the usual rule, B = 0 is the additive inverse because it satisfies A⊕B = A
However, with the new rule you have for addition,
A⊕B = -(A+B),
So 0 = B does not satisfy it because -(A+0) = -A, which is not equal to A.
So you don’t get back A, you get -A instead.
-2A works as the additive inverse because it gives you back the original element when you use the new rule
-(A + (-2A) ) = -(-A) = A
So with this addition rule, it is not 0, which is in the set, that is the additive inverse, it is -2A.
You are most likely confused because in your text it says there must exist some element 0 that is the additive inverse
The 0 there, is just a symbol that represents the additive inverse of a vector space
With the usual addition, 0 the element is also the additive inverse, so it makes sense there
The notion of a vector space generalizes that concept of how 0 works in that space though.
Unfortunately, that is the symbol they decided to use.
So with the new addition rule, your ‘0’ (that is, your additive inverse in this vector space) is -2A.
Said another way, ‘0’ = -2A
This is not the same as saying the element 0 = -2A
This is a good lesson. It shows you that the element that you would normally consider to be the additive inverse is not always going to be the additive inverse in a different vector space. It depends on how you define addition and scalar multiplication.
Said another way, the vector that has all zeros as components is not always the vector called the additive inverse (that just happens to use the bold symbol ‘0’ ) that when added to any vector (call it D) in the set will give you back that same vector D, It is with the usual addition and scalar multiplication, but not with the addition and multiplication you have here.
As a bonus to yourself, try showing that this additive inverse -2A when added to itself in this new vector space, returns -2A (which it better if it is to actually be the additive inverse).
[/s]
Edit: I guess that’s what I get for not taking a closer look
Wait up,
the solution provided by the professor and solutions manual says, the additive inverse does not exist because there is no zero vector, that when added you get back A+B.
And that’s what’s confusing me. If you try and find the zero vector by setting B = 0, naturally you should get to the zero vector. But in this case, the zero vector does not exist, and, my book says that without a zero vector, the additive inverse does not exist since essentially
A+0 = A
A-A = 0
And that’s why I rationlized it the way I did…Maybe i’m being dense here.
No, they’re right. You had said that the answer you’re supposed to get is -2A for the additive inverse, so I didn’t check to see if the additive identity even existed.
Now that I think about it, I don’t even know what set A and B belong to… R? C? R^2?
I’ll assume they belong to R
So if A⊕B = -(A+B), and you want to find ‘0’ such that A⊕’0’ = A, then B can’t be nonzero because if it was, you can’t get back A by adding B because (i.e. -(A + B) < A if B > 0 and -(A + B) > A if B < 0).
If B = 0, then you get back -A. This shows that none of the elements in R can be the zero element. Consequently, no additive inverse can exist for all elements in R.
is this linear algebra or discrete math?
I have a question of my own, along that line.
Theoretical only in here, or applied math as well?
Linear Algebra
Applied is fine, though admittedly I have more experience with the former than the latter