Looking R as a subset of R^2, we see that it has no interior but it is connected. Same with the circle in R^2, connected with no interior. these two exist in the the "intuitively speaking", higher dimension space? like an interval in R would be open relative to R, but not R^2, similar considerations for the circle aswell; are there any other examples of connected sets with non connected interiors?
From Linear Algebra I always knew that the standard inner product on the vector space $F^n$(, where $F=\Bbb C$ or $\Bbb R$ ) is given by $<a,b>$ and $a,b\in F^n$ such that $<a,b>=\sum_ia_i\bar{b_i}.$ This is often referred as the dot product in elementary courses.
Now, say $F=\Bbb C$ and $n=1$ th...
For posts looking for feedback or verification of a proposed solution. "Is this proof correct?" is too broad or missing context. Instead, the question must identify precisely which step in the proof is in doubt, and why so.
Its not the job of people on this site to find your mistakes
@Jakobian that's really confusing me badly. Is this a convention or something? Like when studying complex analysis, we must think of C as R^2 and when studying linear algebra we must think C to be a vector space over itself (,unless otherwise specified) ?
If $\Bbb C=\Bbb R^2$ then I get what you're saying. But the question is, is it?
Consider a smooth (nice) function $\gamma(t)$. Can $\gamma(t)$ be a solution to two distinct initial value differential problems $(D_1, \gamma(t_0))$ and $(D_2, \gamma(t_0))$ where $D_i$ is a differential equation ?
Hi everyone. How can I check that the analytic function $h(u) = \frac{iR + u + i \sqrt{R^2 − u^2}}{2}$, with $u = x_1 +ix_2$, defines a conformal map from the region outside a semicircle, $\mathbb{C} \setminus S_R = \left\{(x_1, x_2):|x_1|^2 + |x_2|^2 = R^2, x_2 \leq 0 \right\}$ to the region outside a disk $D$ of radius $\frac{R}{\sqrt 2}$ centred at the origin, $\mathbb{C} \setminus D$?
I struggle with this concept. Let $\mathcal B_1\lor\mathcal B_2$ denote the smallest sigma algebra that contains $\mathcal B_1\cup\mathcal B_2$. Is a generating set for $\mathcal B_1\lor\mathcal B_2$ simply $$\{B_1\cap B_2:B_1\in\mathcal B_1,B_2\in\mathcal B_2\}?$$Why?
I am looking for innocent looking cases of positive integers "surpringly" divisible by the square of a large prime. "Surprisingly" means that no construction is involved or hidden algebraic factors. An example is $87641^2\mid 5^{31}+24^{10}$ , but I look for more striking examples.
@Jakobian it matters wrt this situation, you see. In the book, I am studying from, the dot product is defined by regarding $\Bbb C$ as a vector space over $\Bbb R.$ But the point is(, as you mention as well) we can do the dot product in different ways for eg: by considering $\Bbb C$ as a vector space over itself. It all depends on what context are we are talking.
@Jakobian and why do you think that? I like your explanations and on several occassions if I recall correctly, I expressed by admiration for the explanations you provide.
And I again, mention, I remain grateful to anyone who tries helping me out.
So that counts you as well. You help me often when I get confused. You try to use some of your time in helping me out, just for the reason that I have a better understanding and this really means a lot to me.
I hate to sound like a broken record, but reread your text, with a pencil in hand. If the authors define an inner product / dot product for you, then that is the inner product. Period. Unless the text gives you some reason to care, don't worry about whether $\mathbb{C}$ is isomorphic (in whatever sense you are imagining) to $\mathbb{R}^2$ or not. If the authors of the text you are reading don't care enough to make this point, then it likely doesn't matter for the purposes of their exposition. — Xander Henderson ♦5 mins ago
@XanderHenderson Thanks! I get it. That makes it clear.
How I wish that I was a student under your guidance! Not only you're a good teacher but also a kind-hearted person. Your students are so lucky to have you as their mentor.
It's always a great privilege to chat with you. Thanks, again!
Hi everyone. How can I check that the analytic function $h(u) = \frac{iR + u + i \sqrt{R^2 − u^2}}{2}$, with $u = x_1 +ix_2$, defines a conformal map from the region outside a semicircle, $\mathbb{C} \setminus S_R = \left\{(x_1, x_2):|x_1|^2 + |x_2|^2 = R^2, x_2 \leq 0 \right\}$ to the region outside a disk $D$ of radius $\frac{R}{\sqrt 2}$ centred at the origin, $\mathbb{C} \setminus D$?
Hmm... I don't really know whats the definition of a conformal map other than its "angle-preserving" map, whatever that means. So I won't help
Oh okay. Its just a holomorphic map with non-zero derivative
I don't really understand how we understand the square root here
Can I really get a conformal map from the region outside a semicircle, $\mathbb{C} \setminus S_R = \left\{(x_1, x_2):|x_1|^2 + |x_2|^2 = R^2, x_2 \leq 0 \right\}$ to the region outside a disk $D$ of radius $\frac{R}{\sqrt 2}$ centred at the origin, $\mathbb{C} \setminus D$? I should end up with $...
The falling factorial is defined to be the number of permutations of length $k$ from an $n$ element set, $$n^{(k)}=n(n-1)\cdots(n-k+1).$$ Is this simply $0$ when $k>n$ (just like $\binom{n}{k}$ is zero when $n<k$)?
The solution to $y' = a(t)y$ is $y = Ce^{\int a(t)dt}$, so if I start with a general solution and assume it's from a linear first order ODE with no forcing term, I can solve for $a(t)$ to recover the ODE. I'm having trouble doing this with recurrences. The only solution I can find to $z[n + 1] = b[n]z[n]$ is $z[n] = D\Pi_{i = 1}^{n - 1} b[i]$. Is there a way I can solve this for $b[n]$?