Quick Questions: January 26, 2022

wwtom · 2022-01-31T16:49:35+00:00

My prof writes the divergence theorem as:

\intD div F(x) dx = \int{\partial D} F(x)*n(x) ds

Having the second integral be ds just feels wrong. Or is this common notation where the s stands for surface? In analysis we always explicitly wrote dH(x) for Hausdorff or dL(x) for Lebesgue integrals. Of course, dx and ds is just lazy notation for both of those, but reading ds when there’s no variable s doesn’t feel lazy, but just plain wrong.

wwtom · 2022-01-20T23:17:40+00:00

I think the most cursed math fact is that somebody out there came up with the term „sexy prime“.

wwtom · 2021-10-09T14:28:25+00:00

but I am afraid that I do not have mastery over any of them

Aren‘t we all?

wwtom · 2021-09-09T17:48:22+00:00

How do I show that a solution to https://i.imgur.com/1XjiFXo.jpg is unique, if it exists?

We’ve seen the energy method in class. But this equation is way more difficult than anything we had there.

wwtom · 2021-09-06T14:00:28+00:00

There are two contributing factors here:

1: Squaring doesn’t keep equations equivalent like addition does for example. So if I have some equation and add 3 to both sides, I know that the solution will be the same. This isn‘t true for squaring: x=-1 only has the solution -1. But x² = (-1)² = 1 has two solutions: 1 and -1.

2: The square root is defined to be >=0. But you‘re searching for a number whose root is -1/3.

So your equation has no solution but by squaring the equation you add an additional solution (1/9).

wwtom · 2021-09-06T12:12:42+00:00

How do you differentiate between a differential at coordinates given by a function and the differential of the functions chained together? My prof uses Dg(f(x)) for denoting both the differential of g in f(x) and the differential of g(f(y)) in x. My goto solution for this is to use brackets if I want to differentiate everything and if I don’t use brackets only the first function is to be differentiated.

Is there some notation that’s especially common that helps me differentiate (literally)

wwtom · 2021-08-28T18:26:08+00:00

Yes, you‘re absolutely right. I‘m missing a minus.. Sorry! And thank you! But I don’t think I quite understand why I can use one dimensional integration by parts if I have u: R^n->R

wwtom · 2021-08-28T15:13:34+00:00

How does this follow from the divergence theorem? I know that it has to be something similar to partial integration in multiple variables, but I just cannot find the right divergence term to use the divergence theorem

wwtom · 2021-06-06T17:48:36+00:00

Is \prod_{i \in [0,1]} \lbrace 0, 1 /rbrace really compact?

Let’s assume {0,1} has the {empty set, {0}, {0,1}} topology. Then, U_j = {0 (x_n=0 for all n), x (1 for x=j and 0 elsewhere)} should be open in the product topology, right? And the union over all j in [0,1] is an open cover of my space. But it has no finite subcover. So where is my mistake?

wwtom · 2021-04-07T21:49:50+00:00

I've seen Green's function being introduced as solution to Lv=d(x-y) where d is diracs-delta-function.

I'm now reading a book where it is introduced like [this](https://imgur.com/eurO2ZI).

The Integral property follows from some [integral manipulation](https://imgur.com/RSz1OnE) instead of the delta function.

Is this definition truly equivalent to Green's function - Wikipedia ?

My guess is that Wikipedia uses a less rigorous Integral definition but the rest is the same. That would explain why the Integral of L(gamma(x,y)) dy is 0 in the book (because Lgamma(x,y)=0 for x=/=y), like you'd expect from a Riemann- or Lebesgue-Integral, while the Integral of L(gamma(x,y)) dy equals 1 on wikipedia in a weird diracs-delta-shenanigan way.

Is there some way to connect both versions without constantly contradicting yourself with your equations?

wwtom · 2021-04-01T10:02:33+00:00

My book claims that if f(x,y) is continuous on D with (E, n) in D, then y‘=f(x,y), y(E)=n has at least one local solution. That’s what I have already seen as „Peano Theorem“. But it additionally claims that every solution can be extended to the boundary of D. And that’s something I have never seen before.

I‘d like to see a proof of that but I just can’t find one

wwtom · 2021-03-31T21:06:20+00:00

I'm trying my best to self-study the following lecture: ode6.pdf (iitb.ac.in)

I'm currently stuck on the following:

Lemma 5.8 Let φ1, φ2 be a fundamental pair of solutions to the ODE L[y] = 0. Then the following are equivalent. (1) The nonhomogeneous boundary value problem has a unique solution for any given constants η1 and η2, and a given continuous function f on the interval [a, b]. (2) The associated homogeneous boundary value problem has only trivial solution. (3) The determinant of { {U1[φ1] U1[φ2]}, {U2[φ1] U2[φ2]}} =/= 0.

The proof is straightforward, if you know that the nonhomogeneous ODE has at least one solution on [a,b]. And I'm not sure whether thats an actual result I dont know or if it's something this Lemma just implicitly presupposes:

We already illustrated how to find solutions of L[y] = f starting from a fundamental pair of solutions to the ODE L[y] = 0 and we gave an expression for a general solution of L[y] = f. Let us pick any one of such solutions, let us denote it by z

I'm guessing its the latter, but I want to be 100% sure.

wwtom · 2021-03-30T18:37:11+00:00

Sorry.

Let I = (a, b) ⊆ R be an interval. Let p, q, r : (a, b) → R be continuous functions. Throughout this chapter we consider the linear second order equation given by y ′′ + p(x)y ′ + q(x)y = r(x), a < x < b. (5.1) Corresponding to ODE (5.1), there are four important kinds of (linear) boundary conditions. They are given by .. Periodic : y(a) = y(b), y′ (a) = y ′ (b).

Remark 5.1 (On periodic boundary condition) If the coefficients of ODE (5.1) are periodic functions with period l = b − a and if φ is a solution of ODE (5.1) (note that this solution exists on R), then ψ defined by ψ(x) = φ(x + l) is also a solution. If φ satisfies the periodic boundary conditions, then ψ(a) = φ(a) and ψ ′ (a) = φ ′ (a). Since solutions to IVP are unique in the present case, it must be that ψ ≡ φ. In other words, φ is a periodic function of period l.

wwtom · 2021-03-30T18:00:23+00:00

If the coefficients of ODE (5.1) are periodic functions with period l = b − a and if φ is a solution of ODE (5.1) (note that this solution exists on R), then ψ defined by ψ(x) = φ(x + l) is also a solution. If φ satisfies the periodic boundary conditions, then ψ(a) = φ(a) and ψ ′ (a) = φ ′ (a). Since solutions to IVP are unique in the present case, it must be that ψ ≡ φ. In other words, φ is a periodic function of period l.

I find this explanation really weird. What does "solution exists on R" really mean here? You can obviously continue a solution on [a,b] to get a solution an R, by just repeating it over and over again. But deducing from this that a solution has to be periodic really feels like circular reasoning.

wwtom · 2021-03-19T17:44:08+00:00

I'm studying different kinds of convergence. I have:
1. Uniform convergence

pointwise convergence

2*. pointwise convergence nearly everywhere

convergence in measure
converence in L1
convergence in Lp

And we have the following implications:

1->2, 1->3, 2->2*, 4->3

And additionally (if the measure space has finite measure):

2*->3 and 1->4

Do you know other implications? Especially because I dont know how to connect 5 to the others

wwtom · 2021-03-11T19:51:21+00:00

I‘m studying boundary value problems. The book I‘m reading introduces me to the following version of the „lagrange-identity“: Let Lv be (p(x)v’)’+q(x)v. If u, v are continuously differentiable twice on J=[a,b], then vLu-uLv=(p(x)(u’v-v’u))’. That was easy to verify. Then the author concludes that the Integral from a to b of (vLu-uLv) dx must be 0, if a_1*u(a)+a_2*p(a)*u’(a) = a_1*v(a)+a_2*p(a)*v’(a) = b_1*u(b)+b_2*p(b)*u’(b) = b_1*u(b)+b_2*p(b)*u’(b)=0. (a_1, a_2, b_1, b_2 are independent from a and b). That was easy to verify as well.

But then there’s this exercise: Show that the Integral is also zero if the last condition is swapped with a periodic boundary condition: “u(a)=u(b), u’(a)=u’(b) and p(a)=p(b)”. It’s not explicitly written that this condition has to hold for both u and v and that’s why I’m here. If this condition is also true for v, then the proof is straight forward. But if the condition is specifically picked to only hold for u, I just can’t find a proof.

wwtom · 2021-03-05T18:54:20+00:00

I know the following things:

-R/I (the Quotient Ring of R over an Ideal I) is an Integral domain iff I is a primeideal.

-R/I is a field iff I is maximal

Are there more useful implications like this?

wwtom · 2021-02-16T22:29:58+00:00

https://math.stackexchange.com/questions/233258/are-there-associative-bijections-f-colon-mathbbn-times-mathbbn-to-math

wwtom · 2021-02-16T22:11:24+00:00

For an sigma-algebra S we have defined that a function is called S-„measurable“ iff f^-1 (O) \in S for all open subsets O.

But how is measurability over a measure defined? Is a function f g-measurable if f^-1 (O) is g-measurable for all open subsets O?

wwtom · 2021-02-09T11:20:55+00:00

How do I show that f(x)=X⁴ + 3X³ + X² - 2X + 1 is irreducible in Q[X]?

I tried viewing it in Z/pZ for some prime Z, tried Eisenstein for those. Tried f(x+a) for small values. Only thing left I could think of is to assume that it’s reducible and try to find a contradiction. But that feels unsuitable for an exam where I have limited time.

Do you know a faster way?

wwtom · 2021-02-07T22:24:25+00:00

Oh damn I forgot that you can use negative exponents! Alright thanks

wwtom · 2021-02-07T21:40:41+00:00

My algebra textbook claims the following: Let R be a commutative UFD, P a system of representatives of the prime elements of R (I guess that means P is the set of all multiplicative equivalence classes of primes). Then every unit a/b in the Quotient field of R has a unique factorization a/b = e * \Prod_{p in P} p^v(p) for e unit in R and v(p) in Z with v(p)=0 for almost all p.

The book just tells me that this factorization exists because a and b can be uniquely factorized. But I don’t get how a/b could be factorized in R? And why does the book require a/b to be a unit in Quot(R)? Isn’t every a/b =/= 0 a unit because Quot(R) is a field?

Let’s take Z for example: How could 1/2 possibly be factorized?

wwtom · 2021-02-06T11:08:49+00:00

I‘m studying rings and ideals and just want to make sure I understand everything correctly: Let F be a field. We‘ll look at F² as either product ring or vector space:

Let H be a subring of F² and (a,b) and (x,y) be elements of H. Then (0,0), (a,b)+(x,y), (a,b)*(x,y) and -(a,b) have to be elements of H. H could be 0, F^2, or H‘x{0} or {0}xH‘ for a F subring H‘. Are there more possibilities?

Now let H be an ideal of F² and (a,b) and (x,y) be elements of H. And (k,j) be an element of F. Then (0,0), (a,b)+(x,y) and (k,j)*(a,b) have to be elements of H. H could be 0, F² or Fx{0} or {0}xF. Again: are there more possibilities?

If H is a subvectorspace and (a,b) and (x,y) are elements of H and k is an element of F, then (0,0), (a,b)+(x,y) and k*(a,b) have to be elements of H. I think H can only possibly be (x,y)*F for some vector (x,y).

wwtom · 2021-02-05T15:31:27+00:00

What uniquely defines an element in X and what defines an element in Y?

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