is it common for korean textbook publishers to provide english translation of the title like that? i find that mildly odd. or, alternatively, odd that having done that, they don't also provide a romanization of the author's name, which an english reader could at least sometimes use to recognize the book in translation
basically every book on differentiable manifolds is titled something like "lectures on differentiable manifolds"
@DannyuNDos I read the name something else, a certain person from North Korea
Btw, clickbaity titles are so annoying. I saw a link that says "India's 52-year-old wait for an archery medal in the Olympics...". I clicked the link expecting it will continue with"is finally over". But it was "got even longer"
> Exercise: Independent repetitions of an experiment are performed. $A$ is an event that occurs with probability $p$, $0<p<1$. Let $T_k$ be the number of the performance at which $A$ occurs the $k$th time, $k=1,2,\ldots$. Compute (a) $E(T_3\mid T_1=5)$ (b) $E(T_1\mid T_3=5)$.
As I said, $T_1$ is simply $\text{Ge}(p)$ and $T_2$ I believe is negative binomial ($\text{NBin}(2,p)$) and $T_3$ is also negative binomial ($\text{NBin}(2,p)$), i.e. the number of trials until $2$ successes with success probability $p$. This exercise appears in a chapter on order statistics, and I notice that $T_1\leq T_2\leq T_3$, yet I am not sure how to obtain the joint pmf of $T_1$ and $T_3$. If I have that, then I think the problem is easily solved.
Not sure if this is well-formed question. Maybe someone can point me to correct literature, I didn't find about it on wiki. Consider Cayley Graph C of a virtually nilpotent group G. Suppose I take a finite part of C. How to find its diameter?
Why is it that if I have a homeomorphism of a compact Riemann surface with finitely many punctures, which is homotopic to the identity, then the extension of the homeomorphism must fix the punctures?
I tried to solve $y'' + y = \cos^2(x)$ so I first found the solution of the associated homogeneous equation by setting $y'' + y =0$ thus finding $y(x) = c_1 \cos (x) + c_2 \sin(x)$ , now I had the particular solution,
so I had seen that I could write $\cos^2(x) = \frac{1+\cos(2x)}{2} = \frac {1}{2} + \frac{\cos(2x)}{2}$ now I had to set $y'' + y = \frac{1}{2}$ and then try to use $y_p = A$ like this $A = \frac{1}{2}$ then $y'' + y = \frac{\cos(2x)}{2}$ and try using $y_p = B \cos(2x) + C \sin (2x)$ then differentiating up to $y''$. Then I continued until the solution.
My doubt goes back to finding the particular solution
Now I was trying to solve $y'' + y = \tan(x)$ and I don't know why I had to write that cosine in that way before (I admit that I saw it being done that way) now what should I do?
So I was thinking that the tangent should also be written some other way
oh, is it just because if the homeomorphism didn’t fix the punctures, and I draw a small loop around a puncture, that loop must move to a loop around another puncture via the Nomoto, but that means it crosses the puncture in finite time , which means one of the maps in the homotopy maps a point of the punctured Riemann surface to a puncture… ?
Or maybe put another way, the winding number of that loop around each puncture is continuous in the time parameter of the homotopy, which should prohibit the homeomorphism from not fixing the punctures if it is homotopic to the identity
because when you want to find a particular solution, when you have $=\sin ax$ or $=\cos bx$ you search $y_p$ as a linear combination of sine and cosine
If finitely generated group $G$ has polynomial growth rate $r^d$, then is it true that for arbitrary finite subgraph $\Gamma$ of $Cay(G)$ we have $diam(\Gamma)\leq r^d$?
Could anyone confirm for me that I can solve the integral$\int \frac{1}{\cos(x)} dx$ by setting $\cos(x) = \frac{1-t^2}{1+t^2 }$ and $dx = \frac{2}{1+t^2} dt$ so as to obtain $\int \frac{2}{1-t^2} dt$ , as a solution I obtain $-\ln\left| \frac{t-1}{t+1}\right|$ where $t = \tan \frac{x}{2}$ and therefore I obtain $-\ln\left|\frac{\tan \frac{x} {2}-1}{\tan \frac{x}{2}+1}\right|$
@Pizza Wrong verb. You don't "solve" integrals (you solve equations and inequalities). You "evaluate" integrals. You also don't "set" $cos(x)$ equal to something. Presumably, you are making a change of variables (i.e. a "substitution"). In any event, the approach you suggest looks fine to me (I am not going to take the time to work through the details).
But I'm curious if such group action can be defined also in the case of a principle bundle? @Jakobian — Bastam Tajik14 mins ago
@Thorgott someone asked me a question, but I don't even know what a principle bundle is. Moreover I forgot what I was talking about in the question of mine.
@Jakobian ok maybe not super clear, but if you think of e.g. $E(X\mid Y=y)$, then this is simply the integral or sum over $y$ times the conditional density or pmf, which would be the same in both cases, or?
assuming $Y = y$ is a event of positive probability
then $E(X|Y=y) = \frac{E[X\cdot 1_{Y = y}]}{P(Y = y)}$
and then you can use that since $X\cdot 1_{\{y\}}(Y)$ has the same distribution as $Z\cdot 1_{\{y\}}(W)$ and $Y$ has the same distribution as $W$, $P(X|Y = y) = P(Z| W = y)$
from definition we could say that $E[E[X_i|X_1+...+X_n]\cdot 1_H] = E[X_i\cdot 1_H] = E[X_j\cdot 1_H]$ for any $H\in\sigma(X_1+...+X_n)$ and $E[X_i|X_1+...+X_n]$ is $\sigma(X_1+...+X_n)$-measurable
since it satisfies the definition of what it means to be $E[X_j|X_1+...+X_n]$ we see that the two are equal in the sense that one can be chosen to represent the other or vice versa
@SineoftheTime I'm not really interested in analysis
Well it's been a long while since I had my nose in PDE or Calculus of Variations but at a first look, it looks correct to me but take this with a grain of salt. It's really sad that this question has received four upvotes but zero helpful comments. The site is becomming exactly like I feared. Less and less people are interacting on graduate or higher level topics and most of the attention is focussed on middle school, high school or basic college level mathematics. Sorry that I could not be of more help but it should follow from Riesz Representation theorem and the fundamental lemma as you did — Mr.Gandalf Sauron1 hour ago
Honestly, I should probably make that number higher. Attorneys, who spend less time in school than I have, typically charge more than \$300 per hour. And that is at the low end.
where does the $I$ in $T - I\lambda$ come from (context is condition of scalar to be an eigenvalue for a linear operator $T$ i.e., $Tv = \lambda v$ for some $T \in \mathcal{L}(V,V), v \in V, \lambda \in \textbf{F}$)
You could have defined an operator $T_\lambda(x) = \lambda x$. But then $T_\lambda = \lambda I$, so why not just write $\lambda I$? So people don't write $T-T_\lambda$, they write $T-\lambda I$. And that's why $I$ is there
@SineoftheTime If instead of the sin(x) (the starting one) there was the cos(x), by chance I have to do it from scratch or I can solve the one with the sin(x) and then modify the result for that of the cos( x)?
I don't see how you can use the integral with $\sin $
maybe integration by parts
find the general formulas for $\int e^{ax} \sin bx $ and $\int e^{ax}\cos bx$ and then plug in $a$ and $b$ instead of redoing the same integrals everytime
@jakobian The meaning of $T - \lambda I$ being not injective means for any two vectors $x,y \in V$, $(T-\lambda I)x = (T-\lambda I)y \implies x = y$ right
So $Tx - \lambda x = Ty - \lambda y \implies x = y$
i think this question about what it means for a linear map to be injective could be improved by adding in even more notation. what if T is a sum of linear maps S_1, T_3, and H_alpha
obliv injectivity of a linear map is often most easy to think about/check in terms of the kernel or nullspace (different words for same thing) being zero or not. because when a map S is linear and x and y are vectors, Sx = Sy is equivalent to S(x-y) = 0, while x = y is equivalent to x-y being zero
in case you want to reduce the number of symbols under consideration even further
e.g. linear map S "not" being injective (what you seemed to be initially asking about, although you then quoted a definition of injectivity) is equivalent to the existence of one nonzero vector v with Sv = 0
and T - lambda I "not" being injective is the equivalent of there being a nonzero v with (T - lambda I) v = 0, or equivalently, with Tv being lambda v
you can prove that an operator S isn't injective by finding exactly one nonzero vector v and showing that Sv is 0. so in principle you can do it by calculating Sv for a single vector v. but finding such a vector v is often some work
