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1d
comment Riemannian metrics and how spaces look
(Should have said distances and angles.) The metric is something you get to decide yourself, but only insofar as you get to decide the shape of the space you're imagining yourself. But keep in mind that given a space from the outset (say, a smooth surface embedded in $\Bbb R^3$ that makes for easy visualization) with all the distances/angles already defined, the metric is determined by this figure - you don't have any say in the matter. And conversely, this figure is determined by the metric (at least "intrinsically" anyway - angles and distances on it).
1d
comment Riemannian metrics and how spaces look
Think of all the possible ways you could take the open unit disk and bend it to make whatever (smooth, contractible) surface your heart desires. They can look very different. And every one of them can be described by putting a metric on the open unit disk. The whole point of the metric is to be able to describe distances, and if you mess with all of the distances between points then you warp the space! I recommend reading my answer here to understand the motivation of metrics.
1d
comment How do I interpret “K2 mod K1”?
Note further that (a) saying the vector is "in $K_2$ mod $K_1$" (somewhat abusively) and (b) the fact the vector is chosen to be in $K_2$ but not in $K_1$ are unrelated comments.
1d
comment How do I interpret “K2 mod K1”?
It absolutely does not mean set-theoretic difference. With $\Bbb C/\Lambda$, it is both a quotient group and a quotient space (which ends up being a complex manifold / Riemann surface). You should really look into quotients if you are having trouble understanding what they are. They are important! Anyway, for what it's worth, saying that one is choosing something in $A$ mod $B$ in this context likely means one is choosing something in $A$, and additionally is parenthetically saying what will happen next will only depend on which coset of $B$ you chose it from.
1d
comment How do I interpret “K2 mod K1”?
Your question mentions you are vaguely aware we might be speaking of quotient spaces. Have you looked up the term? Anything in, say, the Wikipedia article that you're having trouble following? Anyway, the time it would take to discuss what quotient spaces are and how this comment might be interpreted in light of them makes their word choice unjustified in my opinion. They should just say "pick a vector $v\in K_2\setminus K_1$."
1d
comment Necessity of being well-defined in Group Homomorphism?
You mean homomorphism, not homeomorphism. They are different things. The latter is from topology. You are putting way too much stock into the phrase "well-defined." There are no ill-defined maps, ever, period - only bona fide functions vs. failed attempts to define a function that didn't work out.
2d
comment Polynomial of 11th degree
Hint: The RHS is divisible by $x+1\iff $ RHS $=0$ when $x=-1$. :-)
2d
comment Polynomial of 11th degree
Solving for $f(x)$ from your equation will in general yield a rational function - there is a unique choice of $A$ for which it simplifies to a polynomial. Can you find that value for $A$?
Jul
22
comment Find $\lim\limits_{x \rightarrow \infty} x e^{x^2} \int_x^{\infty}e^{-u^2}du $
Then I'll fix your exam for you - edited. Do you think $\lim\limits_{x\to\infty}x\cdot x^{-1}$ is $\infty$ because $x\to\infty$ but $x^{-1}$ is bounded for $x>1$? Notice how $x^{-1}\to0$ though!
Jul
22
revised Find $\lim\limits_{x \rightarrow \infty} x e^{x^2} \int_x^{\infty}e^{-u^2}du $
added 4 characters in body; edited title
Jul
22
comment Find $\lim\limits_{x \rightarrow \infty} x e^{x^2} \int_x^{\infty}e^{-u^2}du $
Do you mean $\int_x^\infty e^{-u^2}du$? You can't have $x$ be the integral's dummy variable and have $x$ appear outside of the integral. You just can't.
Jul
21
comment How to describe the degree of symmetry of an object?
It's possible for two things to have the same number of symmetries, but different types of symmetry. Meaning, different symmetry groups.
Jul
20
comment A question about product of transpositions
No, that doesn't make any sense at all. The notation $(12)$ describes the function $f$ with $f(1)=2$ and $f(2)=1$. Composing twice, we get $f(f(1))=f(2)=1$ and $f(f(2))=f(1)=2$. So $f^2$ is the identity map. Swapping two things twice is the same as having not swapped them at all.
Jul
18
comment Does this suggest a unique additive factorization on the rational numbers?
No - you can have any number of primes in a multiplicative factorization, but the coefficients of your $\frac{1}{n!}$s are restricted. You can write any positive integer in base $10$ uniquely using digits from $0,\cdots,9$ and we don't call this a "factorization" either.
Jul
17
comment How to show that $\int\limits_1^{\infty} \frac{1}{x}dx$ diverges(without using the harmonic series)?
No - you can use the harmonic series' divergence to prove the integral's divergence. But the fact you're asking how to prove it a different way implicitly suggests this is how it is done in practice ... and that's not my experience.
Jul
16
comment How to show that $\int\limits_1^{\infty} \frac{1}{x}dx$ diverges(without using the harmonic series)?
Given $N$, we know $\log x>N$ for all $x>e^N$. So yes, $\log x$ diverges. Your logic concerning the harmonic series is backwards, BTW. We do not use the harmonic series' divergence to prove $\int_1^\infty\frac{1}{x}dx$ or $\log x$ diverges - we use the latter to prove the former!
Jul
14
awarded  Necromancer
Jul
14
comment Why is the average of a sum equal the sum of the averages?
$\sum_y P_{XY}(x,y)=P_X(x)$ and $\sum_x P_{XY}(x,y)=P_Y(y)$
Jul
12
comment Why is there a $q_i$ such that $q_j|q_i$?
(1) Monic polynomial. (2) Apparently you already know what it means for an element of a ring to be prime. Since $F[X]$ is a UFD, every irreducible element is prime, so every irreducible element is an example. Most students are familiarized with factoring polynomials over $\Bbb R$ or $\Bbb C$ before abstract algebra... (3) By induction, if a prime divides any product of things, it divides at least one of those things.
Jul
12
comment Prove $\{a(x,y,z)=(ax,y,z)\}$ is a vector space
Yes, that's true as well - your example utilizes the distributive property (and incidentally so does proof of $0{\bf x}={\bf 0}$ does). I suppose I am responding to OP's comment on the other answer, technically..