This is also rather natural if you think about how matrix multiplication works. Suppose you imagine multiplying out AX and you want to compute a certain column
what you do for that is work out what the electric potential of a single charge, then use that to express the electric potential of the entire charge configuration as an integral
Yeah, I'm supposed to be describing the one-point compactification. Though, if I were using this in an actual proof, I would just refer to it, instead of explaining what it is.
Let $\gamma\in \mathbb{R}$ and $A=\begin{pmatrix}1 & \gamma \\ 0 & 1\end{pmatrix}$. I want to calculate the norm $\|A\|_2:=\sup_{x\ne 0} \frac{\|Ax\|_2}{\|x\|_2}$.
We have that $$Ax=\begin{pmatrix}1 & \gamma \\ 0 & 1\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=\begin{pmatrix}x_1+\gamma \ x_2 \\ x_2\end{pmatrix}$$ Then $$\|Ax\|_2=\sqrt{\left (x_1+\gamma\right )^2+x_2^2} \ \text{ and } \ \||x\|_2=\sqrt{x_1^2+x_2^2}$$
So \begin{equation*}\|A\|_2:=\sup_{x\ne 0} \frac{\|Ax\|_2}{\|x\|_2}=\sup_{x\ne 0} \frac{\sqrt{\left (x_1+\gamma\right )^2+x_2^2}}{\sqrt{x_1^2+x_2^2}}\end{equati…
Regarding the cesàro sum, discussed also yesterday, I need some suggestions to rigorously write the proof. Let me reintroduce the problem. We have a sequence $(a_n)$ that converges to $a$. We want to prove that $\frac{1}{n}\sum_{k=1}^{n} a_n$ also converges to $a$. To prove it we state the implication of the supposition by definition. But when showing what we want to show, do we chose the same $N$'s of the first implication for each $\epsilon$?
I was thinking about that a little. The proof we attempted was to note that $$|s_n-a|=\left|\sum_{k=1}^n \frac{a_n-a}{n}\right|\leq \sum_{k=1}^n \frac{|a_n-a|}{n}$$
The trouble being that, for any $\epsilon>0$, the fact that $a_n\to a$ only implies that there's some $N$ such that $|a_n-a|<\epsilon$ for any $n\geq N$
well, what we want to do is bound them so much that we get $\sum_{k=1}^n \frac{|a_n-a|}{n}<\epsilon$ despite the fact that convergence only tells us about $|a_n-a|$ for $n\geq N$
Hello ! I'm looking for a 3rd year project, I've been searching for a long time and I can't find any exotic interesting questions, do you have any ideas in mind ? It can be fractals, chaos, graph, NP-Algorithms, Math applied to real life.. anything exotic I enjoy it!
I think I found it. Wolfram MathWorld says that if $a$ is a differential k-form of degree 1, then $a \wedge a = 0$, where the $\wedge$ symbol denotes the wedge product.
Let's try again. Pick some $\epsilon'>0$; there exists $N'$ such that $|a_n-a|<\epsilon'$ for $n\geq N$. Additionally, we have $|a_n-a|\leq \mu$ for $1\leq n<N$
So if $n\geq N'$, we have $$\sum_{k=1}^n \frac{|a_k-a|}{n}=\sum_{k=1}^{N'}\frac{|a_k-a|}{n}+\sum_{k=N'}^n \frac{|a_k-a|}{n}$$
From our prior statements, we can bound these sums as $$<\sum_{k=1}^{N'}\frac{\mu}{n}+\sum_{k=N'}^n \frac{\epsilon'}{n}$$
Hmm, I think I may see it. Take $\epsilon$ to be the larger of $\mu$ or $\epsilon'$
In that case, we have $\mu\leq \epsilon$ and $\epsilon'\leq \epsilon$, so we can bound this by $\sum_{k=1}^{N'}\frac{\epsilon}{n}+\sum_{k=N'}^n \frac{\epsilon}{n}=\sum_{k=1}^n \frac{\epsilon}{n}=\epsilon$
it doesn't work. if $|a_1-a|=1$ is the largest one in the entire sequence, then you've got $\mu=1$ and all we get is that the sum is bounded above by $1$---not useful
we need to be able to get $\epsilon$ arbitrarily small
is $\begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$ a isometry matrix? if so, is it rotation matrix or symmetry matrix?
@Semiclassical There are two things that worth noting.
First, I strongly think that I got the learning outcome that I should get, from this exercise. What I can write as a proof (or justification) is enough, I suppose.
The second is that it can be useful to remark that we have a proposition states that if we have a null sequence $(c_n)$ and a bounded sequence $(b_n)$, then the sequence $(c_nb_n)$ is also null.
