@PaulPlummer Nah, contest math is too boring and point-set topology is too analysis-ish. I'm just studying algebra for now. I've finished the first three sections of the rings chapter, but I've now doubled back and started working through the rest of the problems in the groups chapter (as well as from Artin, because the prof said to).
I actually got interested in algebraic geometry. dodges kicks from Balarka
Well, considering my early fascination with -- don't wrinkle your noses -- how you could say stuff about the shape of a conic just by looking at the equation, it makes sense.
@columbus8myhw An isolated point does not need either successors, or predecessors. For example $(\mathbb{R} \setminus [-1,1] ) \cup \{0\}$, has $0$ as an isolated point, but it has no successor or predecessor. But isolated points can have both, think of the integers. I don't think your last statement about only be homeomorphic to either of those is true. For example $\mathbb{Q} \setminus [-1,1]$, is not homeomorphic to wither of those. $\mathbb{Q}$ is the unique dense, countable, linear order
For instance, how do you create a function that takes in two vectors of the same length? You could, of course, take in two vectors and blow the world up if they differed in length. But that's inelegant.
In Idris, there's a vector type parametrized by the length.
A theory is a bunch of axioms inside an ambient space of a deductive system, so it has two "layers". From what I understand of type theory, it's something that puts the two layers on an equal footing.
Yes. In programming, usually the type system (which says what functions are allowed to take and what they're allowed to spit out) is subordinate to the rest of the guts. In a proper functional programming language, they're equal.
For example, it's known that a intuitionistic propositional calculus is the same as a cartesian closed category. now, an intuitionistic prop. calculus can be completely described in terms of dependent type theory. and dependent type theory in turn is an internal language for locally cartesian closed categories.
so it's something the lies "in the middle" of the two theories : categorical logic, and mathematical logic. seems like exactly the thing that can unify a lot of mathematical ideas.
I had a professor that worked on New Foundations, and NFU (a type theory). It sounds sort of interesting, for example (unless I am misremembering) it does not really have problems with talking about the whole universe, and powersets of the universe.
seems nice. but right at this moment, I'd like to read more on the categorical perspectives of type theory, which that article doesn't talk about. maybe I'll read this one later.
[Added explanation]
I found this exercise as follows in Hungerford : Abstract algebra (3rd edition) page 236, exercise number 40. Stated as follows :
C.40. Prove that every element of $A_{n}$ is a product of $n$-cycles.
$n$-cycle is explained in the book like the permutation : $(1 2 3 4 .....
Five people did not agree. I have no opinion either way on your question. If you want it to be reopened, I suggest properly capitalizing and punctuating your question, as well as moving some version of that comment into the question itself.
I do not think most people read comments to questions if the question itself doesn't grip them.
$\int{\frac{12x+24}{2x^{2}+8x+9}dx}$ becomes $\int{\frac{12x+24}{u}*\frac{du}{4(x+2)}}$, which I can then simplify to $\int{\frac{12(x+2)}{u * 4(x+2)}*du}$. This makes it $3\int{\frac{1}{u}*du}$, which is $3ln|2x^{2}+8x+9|$
Exactly. Instead of going through the whole substitution procedure, it's easier to pull out certain factors to get the derivative multiplied by the function to whichever power it's raised to.
I thought of that too, but I'm worried people might interpret that as a matrix product, where x would refer to the n*n matrix with x_i as its ith column.
That's true, but I wouldn't want people interpreting x^TAx as the product of three n*n matrices. I could, of course, write \sum x_i^T A_{ij} x_j, but somehow I'd like to suppress the indices altogether.
I don't think the expression you've got there is exactly what I mean, either. What I mean is the sum of the (i,j)th entry of A times the dot product of x_i and x_j, summed over i = 1 to n and j = 1 to n.
I don't believe this is the same thing as the sum of x_i^T A x_j over i = 1 to n and j = 1 to n.
Yeah, they're definitely not the same. If you take a simple example with n = 2, where A is the 2*2 identity matrix, x_1 = (1, 2), and x_2 = (3, 4), the expression you have is 52 while the expression I want equal to 30.
*is equal to
Hmm, I guess $\sum_{i,j=1}^n A_{ij} x_i \cdot x_j$ is probably the clearest way to write this, then.
Perhaps it's not amongst the toughest integrals, but it's interesting to try to find an elegant
approach for the integral
$$I_1=\int_0^1 \frac{\log (x)}{\sqrt{x (x+1)}} \, dx$$
$$=4 \text{Li}_2\left(-\sqrt{2}\right)-4 \text{Li}_2\left(-1-\sqrt{2}\right)+2 \log ^2\left(1+\sqrt{2}\right)-4 \log \...
@Khallil: usually, when i run into convoluted notation like that, i find it useful to try a few small $n$. for instance, $n=2$ (not one!) should give back the usual integration by parts.
another, equivalent way to say it: if you move the integrals to the same side, it becomes $$\int\sum_{k=1}^n \left(\prod_{i=1}^n u_i\right)\frac{1}{u_k}du_k=\prod_{i=1}^n u_i$$
and so it requires that the LHS be just $\int d(\prod_{i=1}^n u_i)$, which amounts to the differential identity i alluded to earlier
[Added explanation]
I found this exercise as follows in Hungerford : Abstract algebra (3rd edition) page 236, exercise number 40. Stated as follows :
C.40. Prove that every element of $A_{n}$ is a product of $n$-cycles.
$n$-cycle is explained in the book like the permutation : $(1 2 3 4 .....
Perhaps it's not amongst the toughest integrals, but it's interesting to try to find an elegant
approach for the integral
$$I_1=\int_0^1 \frac{\log (x)}{\sqrt{x (x+1)}} \, dx$$
$$=4 \text{Li}_2\left(-\sqrt{2}\right)-4 \text{Li}_2\left(-1-\sqrt{2}\right)+2 \log ^2\left(1+\sqrt{2}\right)-4 \log \...
