Mathematics

12:26 AM

So $e_1,...,e_{n-1}$ is an orthonormal frame of $S^{n-1}$ with coframe $\omega_1,...,\omega_{n-1}$. We identify $T_xS^{n-1}\cong\{x\}^{\perp}\subseteq\mathbb{R}^n$ and $dx$ is the vector-valued $1$-form with entries $dx^1,...,dx^n$. So the claim reads as $\sum_{i=1}^ndx^ie_j^i=\omega_j$ ($e_j^i$ being the $i$-th entry of $e_j$), which we can check on a basis: $(\sum_{i=1}^ndx^ixe_j^i)(e_k)=\sum_{i=1}^ndx^i(e_k)e_j^i=\sum_{i=1}^ne_k^ie_j^i=\langle e_k,e_j\rangle=\delta_{ij}=\omega_j(e_k)$.

I don't see where the connection forms come into play

Binod

12:56 AM

Can anybody tell me how do i find the range of the expression:$x^3-6x^2+11x-6$ ?

feynhat

2:30 AM

@Binod differentiate

William Sun

2:54 AM

When we say “a $k[x]$-module is the same as a $k$-vector space $V$ with a linear map $V\to V$” what do we really mean by “the same as”?

I think it is a categorical equivalence between $k[x]$-modules and the category of “object with endomorphisms” $(V,\phi_V)$.

robjohn

@MaryStar Yes. It is reversed for $0\le a\le1$.

@Binod the range of a cubic polynomial is the whole real line (as long as the lead coefficient is not zero).

geocalc33

3:12 AM

-1

Q: Diffeomorphisms that permute angles?

I'm new to manifolds. I'm reading "An Intro to Manifolds" by Tu, when I can. A diffeomorphism is a conformal map if the pullback of the metric results in a conformally equivalent metric. Are there any references about diffeomorphisms that permute angles (not conformal)? (If this is impossible, p...

let me know what's unclear

a drew an amazing picture to illustrate my point

Binod

@robjohn is this true for any odd degree polynomial?

robjohn

@Binod yes

Mike Miller

@WilliamSun It's not just a categorical equivalence. It's an isomorphism of categories.

Binod

@robjohn I have one more question. If f(x) =$x^3+x^2+3x +|tan(x)|$ is it a bijective funtion?

robjohn

3:28 AM

@Binod No. $f(-\pi/2)=\infty$, $f(0)=0$, $f(\pi/2)=\infty$. Apply the intermediate value property.

geocalc33

When you add two vector field is this the same as thinking of them as fluids colliding and the net effect?

Binod

@robjohn I see. Thanx.

If there is a polynomial function say f(x) which is one-one . Will f(x) +sin(x) be also one -one ?

robjohn

@Binod $x+\sin(x)$ is one-one

Binod

Yes, but $f(x) \ne x$

robjohn

3:43 AM

@Binod oh, you're asking will it always be one-one? no. $\frac x2+\sin(x)$ is not one-one

Binod

@robjohn Can i ask one more ?

robjohn

sure

Binod

If $1+x^2=\sqrt 3x$ then $\Sigma_{n=1}^{24}(x^n-1/x^n)^2$=?

robjohn

Note that $\left(x^n-1/x^n\right)^2=x^{2n}-2+1/x^{2n}$

Binod

Yes

robjohn

3:57 AM

also, $x^2+2+1/x^2=3$

Binod

@robjohn I want to convert roots of this 1+x^2=\sqrt 3x$ into $\omega$ and then substitute this into this.

robjohn

@Binod That is a harder way.

Binod

Where $\omega$ is the cube root of unity.

@robjohn Yes.

@robjohn Then what do you suggest?

robjohn

You know that $x^2+1/x^2=1$. square and subtract 2 to get $x^4+1/x^4=-1$

square and subtract 2 to get $x^8+1/x^8=-1$

I'm working on $x^3+1/x^3$...

I guess we get $\left(x+1/x\right)^3=x^3+1/x^3+3\left(x+1/x\right)$

$3^{3/2}=x^3+1/x^3+3\sqrt3$, so $x^3+1/x^3=0$

Binod

@robjohn Isn't this becoming a little complicated?

robjohn

4:10 AM

well, then go ahead and solve for $x$ and plug it in.

Binod

@robjohn that will take me forever to solve.

robjohn

What, $1+x^2=\sqrt3x$? it's the quadratic equation.

Binod

@robjohn I am talking about calculating the summation.

robjohn

4:22 AM

Let $a_n=x^n+1/x^n$. Then $\sqrt3a_n=\left(x+1/x\right)\left(x^n+1/x^n\right)=x^{n+1}+1/x^{n+1}+x^{n-1}+1/x^{n-1}=a_{n+1}+a_{n-1}$ thus $a_{n+1}=\sqrt3a_n-a_{n-1}$

Or it might be better to let $b_n=x^{2n}+1/x^{2n}$ and derive a recursion for that

let me see....

Then $b_n=b_{n+1}+b_{n-1}$, that is $b_{n+1}=b_n-b_{n-1}$

$b_0=2$ and $b_1=1$

$2,1,\overline{-1,-2,-1,1,2,1},$

Binod

$b_n=x^{2n}+1/x^{2n}$ from this how did you get the value of $b_0$ and$ b_1$?

robjohn

$b_1=1$

Binod

@robjohn Yes .

robjohn

$1\cdot b_n=\left(x^2+1/x^2\right)\left(x^{2n}+1/x^{2n}\right)=x^{2n+2}+1/x^{2n+2}+x^{2n-2}+1/x^{2n-2}=b_{n+1}+b_{n-1}$

we already computed that $b_1=x^2+1/x^2=\left(x+1/x\right)^2-2=3-2=1$

Therefore, $b_{n+1}=b_n-b_{n-1}$

$b_0=x^0+1/x^0=2$

so as computed above, $b_n$ repeats every $6$ terms

and the sum of those $6$ terms is $0$

So the sum is $48$

Binod

4:44 AM

@robjohn We started off with this $x^2+2+1/x^2=3$, right?

robjohn

@Binod that is squaring the original equation, yes

Binod

@robjohn after this , is this the next step?

robjohn

@Binod that says that $b_1=x^2+1/x^2=1$

Binod

@robjohn yes.

@robjohn After finding $b_1$ is this the next step?

robjohn

Then $1\cdot b_n=\left(x^2+1/x^2\right)\left(x^{2n}+1/x^{2n}\right)=\left(x^{2n+2}+1/x^{2n+1}\right)+\left(x^{2n-2}+1/x^{2n-2}\right)=b_{n+1}+b_{n-1}$

which gives $b_{n+1}=b_n-b_{n-1}$

Then we start out with $b_0,b_1,\dots=2,1,-1,-2,-1,1,2,1,\dots$

Note that when we have $2,1,\dots$, this repeats.

Binod

4:51 AM

@robjohn isn't this $b_{n-2} $ instead of $b_{n-1}$?

robjohn

No... $b_n=x^{2n}+1/x^{2n}$ so $b_{n+1}=x^{2n+2}+1/x^{2n+2}$ and $b_{n-1}=x^{2n-2}+1/x^{2n-2}$

Binod

@robjohn i see.

Mary Star

When $a\leq 0$ or $a\geq 1$ then $\left (1+\frac{1}{n}\right )^{\alpha}\geq 1+\frac{a}{n}$ and then $$ n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\geq \frac{a}{n^{1-\alpha}} $$
Taking the limit $n\rightarrow +\infty$ we get $$\lim_{n\rightarrow +\infty}n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\geq \lim_{n\rightarrow +\infty}\frac{a}{n^{1-\alpha}} \Rightarrow \lim_{n\rightarrow +\infty}n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\geq \begin{cases} 0 & \text{ if } 1-\alpha>0 \\ +\infty & \text{ if } 1-\alpha<0 \\ \alpha & \tex…

Binod

@robjohn what is this in this statement sir?

robjohn

@Binod the sequence of $b_n$. it repeats every $6$ terms

Binod

5:01 AM

@robjohn Sir if sum of each six terms is 0 then how did we arrive at that result that the sum is 48?

robjohn

@Binod Because $\left(x^n+1/x^n\right)^2=x^{2n}+2+1/x^{2n}$

the $x^{2n}+1/x^{2n}$ sum to $0$, so we are left with $24$ times $2$

Binod

Ah, because of that 2.

@robjohn Thank you so much sir.

robjohn

@MaryStar since you are computing $(n+1)^a-n^a$, it appears that the limit should be $\infty$ if $a\gt1$, $1$ if $a=1$, and $0$ if $a\lt1$.

Knight

6:54 AM

@robjohn I have to prove this inequality
$$
xyz \gt (y+z-x) ~(x+z -y) ~(x+y -z)
$$

I'm missing something, coz I know it's all about applying $ A.M. ~\gt G.M.$.

We can have something like this

$$
y+ z \gt 2 \sqrt{yz} \\
x+z \gt 2 \sqrt{xz} \\
x+y \gt 2 \sqrt{xy}
$$
But I need x, y and z all together.

Lelouch

7:40 AM

How do you get the Poincare Dual of a closed/compact $n$-form, and does it always exist ?

By poincare dual of an $m$-form $\eta$ on a $n$-manifold $M$, I mean a $n-m$-dimensional submanifold $N$ (possibly with boundary) such that for any $n-m$-form $\omega$, we have $\int_{N} i^* \omega = \int_{M} \omega \wedge \eta$

Also, is there a nice way to explicitly write down Poincare duals for $1$ dimensional submanifolds of surfaces ? And possibly unrelated, but it seems if you take a=an embedded one dimensional submanifold in a surface, take the dual of it, and integrate that along the submanifold, the result will be an integer ! I only checked it for toy cases, how do show this (or is this false ?) in general ? Can this be used to give explicit duals ?

Alessandro Codenotti

8:23 AM

@Edward Münster, why?

Edward Evans

8:43 AM

@Alessandro hahaha I thought so; I've also been looking at Münster for a PhD. There are two profs there who do things that I am potentially interested in

Alessandro Codenotti

9:13 AM

Nice! They have the less painful/bureaucratic application process of all the unis I've seen for PhDs so you should definitely apply haha @Edward

Edward Evans

Oh nice :D It looks quite streamlined lol, when does one have to apply? Like a year in advance or ? (E.g., if I wanted to start in October 2022 or smth)

Alessandro Codenotti

I applied in spring (applications where open from the end of Februry to the end of March iirc) for October of this year

Edward Evans

I see, thanks

Alessandro Codenotti

@EdwardEvans yeah you just fill in the online application form, get two professors to send a recommendation letter and you're done

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$Mathematics$ Mathematics