Okay, but how do they show this then: $(a_1a_2\cdots a_k)\cdot(a_{k+1}\cdots a_n)=a_1a_2\cdots a_n$. What is their induction hypothesis? Do they start with $n=2$? Or $n=3$? And do they use the definition for a product of $n$ elements? Because I don't see why they'd do that
I am uncertain of what you call the definition below in English. Can someone confirm that the following definition is called "preimage of a set": $f^{-1}(H)=\{x ∈ X|f(x) ∈ H\} ⊆ H$
As it turns out, if you mess around with truncation the right way you get the result that I was not figuring out right, and which was really getting under my skin
While with $L^p$ spaces it seems like you have all these overpowered theorems that just make life work, and then just say "So $\ell^p = L^p(\mathbb{N},\mu)$ where $\mu$ is the counting measure, therefore special case"
I still don't know how to prove it. We assume that for some $n>2$, it holds that for all $k\leq n$, $$ a_1a_2\cdots a_n=(a_1\cdots a_{n-1})\cdot a_n. $$ Now consider a second product $a_1\cdots a_{n+1}$. We want to show that $ a_1\cdots a_{n+1}=(a_1\cdots a_n)\cdot a_{n+1}. $ We can write $a_1\cdots a_{n+1}=a_1\cdots a_na_{n+1}$. How can I argue that $a_1\cdots a_na_{n+1}=(a_1\cdots a_n)\cdot a_{n+1}$?
Say, could someone help me figure out this sequence's pattern? It was a question on a thing I had, so I assume theres a pattern, but I couldn't figure it out
@ShaVuklia The correct way to phrase this is that one defines the composition recursively using a specific bracketing and then proves using induction that this agrees with all other ways of putting brackets (by associativity)
Zach was talking about mathcamp the other day, seems he's not alone. A qualifiying question was posted over there. I answered without knowing, got upvotes and today it is deleted until March 20. If some questions looks like contest, be aware...
So we know that $a_1a_2a_3=(a_1a_2)a_3=a_1(a_2a_3)$. Now we want to give meaning to $a_1a_2a_3a_4$. Am I allowed to just split this? $a_1a_2a_3a_4=(a_1a_2a_3)a_4$?
@Sha Basically you're proving that you can do the bracketing however you like. The way proposed by your book is by induction. Define the product recursively with a certain bracketing, and prove by (strong) induction it's equal to any other bracketing of the same elements (in the same order).
@Astyx @TobiasKildetoft Ok.. maybe tomorrow I'll get it, when I look at it from a fresh point of view. Thanks a lot for you help - it's really appreciated
@Danu Spin structures are affine over real line bundles; if s otimes L = s', then the spin^c structures differ by L otimes C. This is clearly order 2 because L is.
Algebraically, tensoring with C is the bockstein homomorphism H^1(X;Z/2) -> H^2(X;Z) associated to the exact sequence Z -> Z -> Z/2.
Call that beta. Its image is exactly the elements of order 2.
math.stackexchange.com/questions/4632/… Aryabhata here has an answer that I have interest in clarification: why does he say the "order matters," when as I understand it 2+3+4 = 3+4+2? I am working on similar problem and want to find the combinations of given X dice of K sides, which combinations total exactly to N, being an integer
IANAM-->i am not a mathematician, a simple programmer indeed I am
@Semiclassical if I have an array in ruby like [[1,1],[1,2]....[6,6]] given two 6sided die, and I want to find the unique totals that equal 7, I'm going to flatten the array and map the totals to a hash, in order to find the totals, while stripping the values and checking against N to put them directly into the solutions (output) array, at least, this is my guess on how to approach the problem.
I understand this is elementary combinatorics -- my mathematics background stops at the stepping stones to calculus
