Mathematics

8:01 PM

@Alizter Err, I'm not entirely sure how you would propose to. That's pretty much the only way I know how to show two things are isomorphic.

I guess you could do it by invoking the classification of 1-dimensional Lie groups. But that makes you a jerk.

Daniel Fischer

@MikeMiller And doesn't that also use properties of the exponential function?

Mike Miller

@DanielFischer I don't think so. It uses the exponential map, which is different.

First prove that simply connected Lie groups <-> Lie algebras; then because there's only one 1-dimensional Lie algebra, there's only one simply connected 1-dimensional Lie group, which is $\Bbb R$. Then all 1-dimensional groups have $\Bbb R$ has their universal covering group, and by covering space theory they're all given by $\Bbb R/\Gamma$, where $\Gamma$ is a discrete subgroup, hence $\{0\}$ or $\Bbb Z$.

Link

Hey, studying/reviewing for an exam and had a question: tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx Under the double roots, they get a second solution to be $y_2(x) = x^y ln(x)$ where do they get this from?

Mike Miller

Since $S^1$ isn't $\Bbb R$ it must be isomorphic to $\Bbb R/\Bbb Z$.

Chris's sis

I just posted on main a very nice integral!

0

Q: Evaluating $\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx$

One of the ways to compute the integral $$\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx$$ is to make use of the series of $\log(\sin(x))$, but the result I got after doing that wasn't that friendly. Is it possible to find a neat way of evaluating the integral?

UserX

8:15 PM

@Link why bother learning all these particular solutions when you can solve the most general case $ax^2y''+bxy'+cy=0$? Just assume $y=x^r$ and everything flows naturally

Link

@UserX Thank you! That saves me so much time.

@UserX Though, can you explain how I would get the second portion for the double roots example just using the general case?

The $ln$ part.

UserX

I didn't read it, I just read the tittle :P

Link

In essence he gets a quadratic which $r = 4$, and says we need a second solution, so he adds on a $x^y ln(x)$ somehow.

@UserX

UserX

I got you to approaches to get it more intuitively. $x^y \ln x$ is a trial solution like $y=x^r$

Link

@UserX Ah... I see.

UserX

8:28 PM

The first one is to not assume $y=x^r$ but $y=e^u$ to reduce the equation to a linear DE with constant coefficients

misspeled two...

The second is to apply the following change of variables $t\colon =\ln x$ and $y(x)=g(\ln x)=g(t)$

Link

So, essentially if it doesn't work the first way, try the second way using the two approaches?

UserX

I think the more intuitive will be the second one to understand how the double roots cause problems

No, not at all. These are not practical. These are to gain intuition on why the double roots require a different assumption. If you meet an Euler equation go for $y=x^r$ and $y=x^r\ln x$ if it has double roots

Link

Oh, alrighty. Thanks x.x

UserX

I was actually expanding the " In this case IT CAN BE SHOWN that the second solution will be"

Alizter

@MikeMiller For ismorphism theorems can the case for groups and rings be proved under a single umbrella? e.g. ideas$\sim$normal subgroups etc. Is it something to do with Category theory?

Link

8:35 PM

Oh, I see. Exam is on 2nd order and higher order linear ODE's. So this is coming up in it.

Alizter

Oh wait I found the answer dw @MikeMiller

UserX

That's how you show it. Honestly, intuition is great but if you cram for exams, learn those formulas and stress on their proofs(those are not exactly proofs) later

Link

@UserX Yea, I kinda tend to get a bit cramming for them. Where can I find how to get the general solution from characteristic roots and a characteristic equation? I can't seem to find any good simple readings.

UserX

Paul's online notes I guess

daOnlyBG

Hello everyone

UserX

8:42 PM

My best advice would be to try examples out of your head in W|A and see the patern

That's what gave me intuition about them. Some years ago I had an eureka moment when I realised this without studying it

In fact I think I got you a better way to gain intuition on them. Assume $y=e^{\lambda x}$ and insert it in the DE. You'll see how characteristic equations show up

Remember that $e^{\lambda x}$ can't be $0$ for finite $\lambda$ and you have a finite $\lambda$

Mike Miller

OK @Alizter

Zach Saucier

I don't understand this homework :(

Mike Miller

I understand very little in life

Zach Saucier

too many hours invested, so little are solved, lol

UserX

@ZachSaucier The definition of my life.

Mike Miller

8:58 PM

Such is the way of things.

UserX

I wonder how things would be if everything was solvable with minimum work

Would we still be trying?

Zach Saucier

Can someone be so kind as to walk me through this problem?

> Integrate $\int_C (x^{3}+y^{3}) dy$ where C is the curve consisting of two edges of the rectangle in the first quadrant having opposite vertices at the origin and the point (6, 5), where the first edge traversed starts at the origin and is horizontal.

UserX

@ZachSaucier I guess you should start by finding your C.

Zach Saucier

my attempt was x = t, y = 0 where 0 <= t <=6 and x = 0, y = t where 0 <= t <=5 but that doesn't seem right

robjohn

9:13 PM

@ZachSaucier $\mathrm{d}y$ is $0$ except on the vertical pieces of $C$, the first being $\int_0^5(6^3+y^3)\,\mathrm{d}y$ the other being $\int_5^0(0^3+y^3)\,\mathrm{d}y$

The answer is therefore $\int_0^5(6^3+y^3)\,\mathrm{d}y-\int_0^5(0^3+y^3)\,\mathrm{d}y =\int_0^56^3\,\mathrm{d}y=1080$

Alizter

@MikeMiller What is the name for a module M such that (M, +) does not form an abelian group?

Not a module any more

robjohn

@ZachSaucier Unless I have $C$ wrong... Could you describe $C$ a bit more? (my integral is over the entire rectangle)

Zach Saucier

that's the description I have. Evidently It's the line y = 0 from x = 0 to x = 6, then the line x = 6 from y = 0 to y = 5

so it's just the two lines, not the whole rectangle I believe

robjohn

@ZachSaucier Ah, then it is just the first of the integrals that I gave...

$\int_0^5(6^3+y^3)\,\mathrm{d}y$

because $\mathrm{d}y$ is $0$ on the horizontal piece

Zach Saucier

so the dy just removes the need of the first line? since the change in y is 0?

Américo Tavares

9:21 PM

@robjohn Do you know if there has been recently any modification on how the posts are shown to logged in users. I've answered math.stackexchange.com/q/1015411/752 but it didn't appear in the first page. However when I visited the same page without being logged in I saw it.

robjohn

@ZachSaucier well, the first line can be parametrized as $(6t,0)$

Zach Saucier

that makes sense

robjohn

@AméricoTavares It's at the top of the second page for me.

Américo Tavares

@robjohn Thanks!

robjohn

@ZachSaucier so when integrating, $\mathrm{d}x=6\mathrm{d}t$ and $\mathrm{d}y=0$

Mike Miller

9:24 PM

I've never heard a name for those, @Alizter. Probably because nobody cares.

robjohn

@AméricoTavares did you find it?

@ZachSaucier The second contour is $(6,5t)$

So $\mathrm{d}x=0$ and $\mathrm{d}y=5\mathrm{d}t$

So you could do the integral as $\int_0^1(6^3+125t^3)5\,\mathrm{d}t$

Or you could parametrize the contour as $(6,t)$ where $t$ goes from $0$ to $5$

Then the integral would be $\int_0^5(6^3+t^3)\,\mathrm{d}t$

Américo Tavares

@robjohn No, because I see only one page with 50 posts. I don't know if it has something to do with my ignored tags, although none is one of them. This strange behauvior is recent. Maybe the second time it happed.

robjohn

@AméricoTavares There isn't a bar along the bottom to see pages 1, 2, 3, etc?

Américo Tavares

@robjohn NO! And I think there has never been.

robjohn

@AméricoTavares That is weird...

skullpatrol

9:31 PM

very weird...

Mike Miller

Click "Questions" at the top. It should show you the bar.

robjohn

@AméricoTavares try this link

Américo Tavares

Ah, there's a link to "Browse the complete list of questions".

Alizter

@MikeMiller What is the most straight forward way to prove this about the R-module $\Bbb R[X]/\Bbb R\cong \Bbb R[X]$

robjohn

@AméricoTavares perhaps than makes the buttons I showed above appear

Alizter

9:34 PM

I came up with a proof but I may be overdoing it.

Zach Saucier

@robjohn Thanks a lot for the help!

robjohn

@ZachSaucier np

Alizter

Consider the surjective endomorphism $D:x\to x'$ so by iso theorem we have $\Bbb R[X]/\ker(D)\cong \Bbb R[X]$ and we have $\ker(D)\cong \Bbb R$ so it holds.

UserX

@robjohn I cannot see more than the 1st page on SE networks that I'm not signed up.

Mike Miller

Seems reasonable @Alizter

Wait, no. You're taking derivatives?

UserX

9:37 PM

Ah it was solved nvm

Alizter

@MikeMiller ... Yes?

robjohn

@UserX Okay, but Américo is signed up on math.se