How to evaluate $$\sum_{\substack{k_1,\ \cdots,\ k_n \in \mathbb{N}_0 \\ k_1\ne\ \cdots\ \ne k_n}}\frac1{m^{k_1+\cdots k_n}}$$ I have that this is a geometric series with weights $a_k$. $a_k$ should be the number of distinct weak compositions of length n of k. So it should be $a_k=\binom{k+n-1}{n-1}$ right? However if I put this back into the geo series and use the generating function I get an answer but it is inconsistent with simple cases of $m$ and $n$.
Hello @alizter, lol. Could you explain the current A Level system to me? There are two subjects, Math and Further Math, and you can get A or A* for each, is that right? Is there another math subject? Is there something called an S paper? I don't get it.
So if two dice are rolled, the probability of the sum being 10+ is $\frac{7}{36}$. And if you roll two dice three times, the probability of at least one roll having a sum of 10+ is still $\frac{7}{36}$, right?
@Alizter It sounds complicated. Here in local bookstores, I see books covering Pure Math 1,2,3, Mechanics 1,2 and Probability and Statistics 1,2. Is that the old syllabus? Is that only 1 math subject?
Is any closed proper subset if S_1 simply connected? I beleive it is since if given any closed curve on this subset, we can contract it to a point. However, a friend of mine said that it isn't simply connected because we can't contract the whole space to an entire point. Who's correct?
So if two dice are rolled, the probability of the sum being 10+ is 7/36 . And if you roll two dice three times, the probability of at least one roll having a sum of 10+ is still 7/36 , right?
Yes, I was constructing an example of a nested sequence of closed sets, each of which is simply connected and whose union is not. The union in my example happens to be S_1
I feel so unhappy living where I am now. So many of the things I consider important in life are not here, because people choose to remove them or ban them on purpose. Sick, sick, sick.
@user130018 OK, I am just trying to compare the different education systems around the world, especially with regard to pre-university math, which I don't know about.
Question: I'm working through the details of showing $\widehat{f'}(\xi) = 2\pi i \xi \widehat f(\xi)$ and I use integration by parts to get $\lim_{N \to \infty} \bigg[ f(x)e^{-2\pi i \xi x} \bigg]_{-N}^N + 2\pi i \xi \widehat f(\xi)$. I'm trying to show that the middle term goes to zero but I'm not seeing it?
@TedShifrin I feel bad for you..even though you tell them explicitly the questions on the exam, they still don't get it? Maybe they thought it was reverse psychology trying to trick them lol
@Studentmath: I told them a computer needs to be rebooted on average once a day, then asked for a reasonable random variable expressing the number of reboots in a day. About 1/4 got it ...
@TedShifrin Yes, that's part of it. He can't expect to pass just by attending and looking at the homework questions, he needs to sit down and do some exercises
@TedShifrin It's the same to me. In $M[a^{-1}]$, you have polinomials elements of the form $\sum r_i a^{-i}m_i$. I don't think it is a sin to use that notation.
@alizter I want to ask you. Are there any other books other than the 18 you showed me that are also in the same series? It seems there are none, but I want to be sure before I get a complete set.
@Alizter Too bad the Bostock and Chandler books are no longer in print. But I think Pure Math 1 and 2 is going to be published by OUP, since OUP has taken over Nelson and Thornes. However, they don't list Applied Math 1 and 2 on their website. The strange thing is that even the Edexcel website does not contain all 18 books.
@Alizter First, when I navigate to X, it does not list all 18 books you listed. Second, when I navigate to Y, it starts to list those that are published by OUP instead. So, all different books, and all incomplete, lol.
@Alizter Physics. He did undergrad physics in Princeton, US. He is my best friend, and one of the wisest people on earth. I will ask him for help to solve my life problems when he returns.
@Alizter since the numbers are distinct, you can arrange a solution in increasing order .. $k_1 < k_2 \cdots < k_n$, and successive elements in this list differ by atleast one, so make the change of variable $k_{i+1} - k_i = x_i + 1$, for $n-1 \ge i \ge 1$, then you are counting number of non-negative integer solutions .. does that sound correct ? :O
@alizter Sorry to bother you again. I was looking at the Cambridge A level Math books on CUP and OUP websites. How come there seems to be only Math and no Further Math? I don't get it.