If in a theorem I use and explain notation (i. e. something like "where X is Y"). Can I then re-use that notation in another theorem without having to explain it again?
@Vrouvrou At this point I agree with @JasperLoy, you may need to spend some time internalizing it, working out the minor details, and try to make it make sense to you but it is all here.
@teadawg1337 I see your point, but that is arranged such that I could use a brilliant way. In the main proof I only needed $m \in \mathbb{N}$, so I only needed to treat this kind of value for $m$.
I'm fully recovered after that fail, and I have more energy, creativity and inspiration than ever. Every single day and every single experience are meant to make me more more powerful than ever.
I just discovered a new very nice kind of series to investigate.
How would one describe an integral curve using a commutative diagram?
Or alternatively is there a nicer categorical way to describe the following construction: Let $c: (-\epsilon,\epsilon)=I \to M$ be a curve in a manifold. Define $c':I \to TM$ as the curve $t_0 \mapsto [c(t+t_0)] \in T_{c(t_0)}M$
Nominating math.stackexchange.com/questions/1204369/… for reopening. The OP included his attempt at a solution in answer form below, which seems to have been missed by the last one or two close-voters.
uhm., I wouldn't think so. As far as I'm aware, it is only a markup/display language. You might be able to create or find a program which dynamically displays text that auto-updates using input-fields... but that would be a separate entity.
Wave equation: $u_{tt}=au_{xx}, a>0$
We are looking for solutions of the wave equation of the form of a wave function.
We suppose that $u(x,t)=A \cos(kx- \omega t)$ is a solution of $u_{tt}=au_{xx}, a>0$.
We have:
$$u_x(x,t)=-Ak \sin(kx- \omega t)\\u_{xx}(x,t)=-Ak^2 \cos(kx-\omega t)\\u_t(x,t...
In my notes there is the following example about the energy method.
$$u_{tt}(x, t)-u_{xxtt}(x, t)-u_{xx}(x, t)=0, 0<x<1, t>0 \\ u(x, 0)=0 \\ u_t(x, 0)=0 \\ u_x(0, t)=0 \\ u_x(1, t)=0$$
$$\int_0^1(u_tu_{tt}-u_tu_{xxtt}-u_tu_{xx})dx=0 \tag 1$$
$$\int_0^1 u_tu_{tt}dx=\int_0^1\frac{1}{2}(u_t^2)...
@Vrouvrou You only have a nice equality like that for the Hilbert space ($p=2$). For the Banach space, you will have the Minkowski inequality, but nothing as nice as you get when you have an inner product.
A ok... The answer at the post is fine!! I have edited my post and added what I have tried but I didn't get an answer to that yet. That's why I asked it here... @robjohn
Hello @robjohn @quid :) If we have a series $\sum_n a_n x^n$ and the limit $\frac{a_{n+1}}{a_n}$ is equal to $0$, do we deduce that the radius of convergence is $+\infty$?
@quid Nice :) So if we are looking for a solution of the form $\sum_{n=0}^{\infty} a_n x^n$ and we know the general formula for $a_{3k}$ and $a_{3k+1}$ and also that $a_{3k+2}=0$ and that the radii of convergence of $\sum_{k} a_{3k} x^{3k}$ and $\sum_{k} a_{3k+1} x^{3k+1}$ is $+\infty$, then we deduce that the solution is $\sum_k a_{3k} x^{3k}+ \sum_k a_{3k+1} x^{3k+1}$ , that it converges everywhere on $\mathbb{R}$ since the radius of convergence is $+\infty$, right?
This seems correct @evinda The radius of convergence of the sum of powerseries is at least the smaller of the two. In thsi case both are infinite so it is infinite;
If you have a powerseries $\sum b_n x^n$ and you know that $b_n=0$ for all $n$ than you can say the radiuus of convergence is infinity. As argument you can either give "clearly it converges for all x" or "$\sqrt[n]{b_n}$" is $0$ and thus the radius is 1/0 that is infinity. @evinda
Yah, I thought that was weird too, I am pretty sure I have seen some comments along the lines that he refuses to use chat and does not like it., maybe it was someone else idk
@robjohn, how to prove that there do not exist three distinct positive integers a, b and c such that each integer divides the difference of the other two?
Wave equation: $u_{tt}=au_{xx}, a>0$
We are looking for solutions of the wave equation of the form of a wave function.
We suppose that $u(x,t)=A \cos(kx- \omega t)$ is a solution of $u_{tt}=au_{xx}, a>0$.
We have:
$$u_x(x,t)=-Ak \sin(kx- \omega t)\\u_{xx}(x,t)=-Ak^2 \cos(kx-\omega t)\\u_t(x,t...
In my notes there is the following example about the energy method.
$$u_{tt}(x, t)-u_{xxtt}(x, t)-u_{xx}(x, t)=0, 0<x<1, t>0 \\ u(x, 0)=0 \\ u_t(x, 0)=0 \\ u_x(0, t)=0 \\ u_x(1, t)=0$$
$$\int_0^1(u_tu_{tt}-u_tu_{xxtt}-u_tu_{xx})dx=0 \tag 1$$
$$\int_0^1 u_tu_{tt}dx=\int_0^1\frac{1}{2}(u_t^2)...
