trying to think of how to word something. I've got two version of an inequality, one as $A^2\leq B^2$ and the other as $|A|\leq |B|$. Obviously these are equivalent, but how would I describe the latter in relation to the former?
how should I go about proving that a strictly increasing function is injective? I am not quite sure about it since the definition of a strictly increasing function negates the definition of an injective function.
Yes, so isn't it the negation of the definition of injectivity which suggests that there exists $x_1$ and $x_2$ such that if $f(x_1)$ = $f(x_2)$ ⟹ $x_1$ = $x_2$?
But the thing that confuses me is that since the function is strictly increasing, then how can two points be equal to each other and produce $x_1$ = $x_2$?
Yes, so isn't it the negation of the definition of injectivity which suggests that there exists $x_1$ and $x_2$ such that if $f(x_1)$ = $f(x_2)$ ⟹ $x_1$ = $x_2$?
no it doesn't say "there exists"
injectivity means that whenever we have $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$, then $x_1 = x_2$
I understand. I totally forgot about the contrapositive. So using the contrapositive we can basically show that since there doesn't exist $x_1$ and $x_2$ such that $x_1$ = $x_2$, therefore $f(x_1)$ won't be equal to $f(x_2)$
which is essentially the contrapositive of the definition.
But I'm still finding it hard to visualize how the function is injective. If the function is always increasing and doesn't have an instantaneous point where the slope is 0, then how can the definition of injectivity hold?
I understand that you can use the contrapositive, but I don't quite get the logic behind it..
Trichotomy just says: "Given two real numbers, one of three things has to be true: the first number is bigger than the second, the first number is equal to the second, or the first number is smaller than the second."
Ahh that sounds interesting. So far I have only touched negations, contrapositive and direct proof (Some statement, P that implies some other statement, Q). So, I still have much to cover :p
I see. That actually sounds really interesting and useful. I will definitely check it out at a deeper level.
So, just out of curiosity, would proving that an increasing function is injective using a direct proof (P implies Q) would be possible in this case? Are there any mathematical problems that can only be solved using a certain way due to their complexity?
I wanted to give an easy example of a non-constructive proof, or, more precisely, of a proof which states that an object exists, but gives no obvious recipe to create/find it.
Euclid's proof of the infinitude of primes came to mind, however there is an obvious way to "fix" it: just try all the n...
So for this example the complement is a disjoint union of two punctured tori and you just need to come up with absolutely any open set with non-diffeo complement
at which point you prove this by showing that any two discs oriented the same in a connected manifold are isotopic (this is a lemma of Palais I think but it's straightforward)
I am pretty sure that the connected sum of a manifold to itself is equivalent to the connected sum of that manifold to an $(n-1)$-sphere bundle over $S^1$
Language way too fancy. It's either the connected sum with $S^{n-1} \times S^1$ or with the $n$-dimensional Klein bottle, where you take $S^{n-1} \times I /(v, 0) \sim (r(v), 1)$, $r$ being some reflection
Depending on whether or not you do the gluing in an or-preserving or or-reversing fashion
References for the claim about discs being isotopic
Yes, you can perform that ambient isotopy: any oriented embedding $i: B^n \to M^n$ is isotopic to any other. (This is a lemma proven independently by Cerf and Palais1, but the idea is quite clear: shrink the image of $i$ until it's contained in the chart, then take the limit that defines the deri...
I once discussed this with a math people who told me that things could get fancy due to the whole spheres that could be homeomorphic but not diffeomorphic
Not sure if this is correct or relevent
Or if it doesn't matter if I assume everything smooth
Yeah that's indeed not a connected sum and indeed you can't say anything better about this than "connected sum with a sphere bundle" like you did above
The connected sum procedure is carried out as follows. Choose an oriented embedding $i_1, i_2$ of the disc into a manifold $M$, $M'$. Delete the interior of the images. Then the connected sum is what you get when you identify $i_1(x) = i_2(x)$ for points $x$ in the unit sphere.
I assume you mean they're poorly studied in physics which fine
Hello. Is the submersion $\mathbb R^2\to \mathbb R$ given by $(x,y)\mapsto xe^y$ a fiber bundle? It seems the gradient flow is complete and gives parallel transport. On the other hand, the Palais-Smale condition is not satisfied, so maybe I'm missing something...
@MikeMiller (I also think so. My question is about the Reeb foliation $(x^2-1)e^y$.) If you happen to have a couple of minutes later, I'd love to bug you. Thanks anyway!
Who can point me to good material on Matsubara summation?
Actually no, let's break it down further
So I have a summation that i'm representing as a sum of residues of a function
How does the contour deformation work precisely?
There is no pole at infinity, it's a zero instead
So if I have a sum of poles on the real axis
And two poles in the lower half plane
(say)
I'm saying my summation is over the real axis poles, can I deform the clockwise contours around these real axis poles to a contour enclosing the complete lower half plane only ?
Or option 2: do I need to also have a contour in the upper half plane?
The imaginary part of the weighting function is positive in the upper half plane
The imaginary part of the two lower half plane poles is negative after the transformation x -> x+iepsilon
well, the problem here is the "codomain" is the proper class, V
I don't know how functions that maps to proper class works
The powerset operator can be described as mapping between one set $S$ to another of cardinality $2^{|S|}$ along with the subset properties this new set has to obey
Let $X$ be a finite, commutative, boolean monoid with a zero element $0$. Define the set $C$ of subsets of $X$ to be $C = \{ x \subset X : 0 \in x, $ and $ \forall a,b \in x, \ ab \in \{a, b, 0\}\}$.
Define for $x, y \in C$:
$$
x + y = \{0\}\cup (x \Delta y) \setminus \{u \in x \Delta y: \exis...
@JakeRose If you're trying to figure out what a manifold (in Euclidean space) is, you might want to take a look at my YouTube lectures (linked in my profile) 3500 day 40 and 3510 day 23. The parametric definition you copied above appears as one of three equivalent ways of understanding manifolds.
@MatheinBoulomenos can you classify all the subspaces $V$ of $\Bbb R^\Bbb N$ closed under $\delta$ satisfying the property that there is a linear map $f: V \to \Bbb R$ such that $f(v) = v(0) + f(\delta(v))$ for all $v$?
I'm not sure whether for my presentation tomorrow I should cite the Albert--Brauer--Hasse--Noether--theorem without proof or define $\Bbb C \otimes_\Bbb R V$ as $V + iV$
@LeakyNun sorry, i am a bit confused - so is L'Hospital the right way to go (as i can interprete enumator and denominator as functions in $t$) , or not?
Limes je jedan od osnovnih pojmova u matematičkoj analizi.
== Limes niza ==
Neka je
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n
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{\displaystyle (a_{n})}
niz realnih ili kompleksnih brojeva. Reći ćemo da niz
(
a
n
)
{\displaystyle (a_{n})}
konvergira broju L (realan ili kompleksan broj) ako vrijedi
(
∀
ϵ
>
0
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∃
n…
Limes je jedan od osnovnih pojmova u matematičkoj analizi.
== Limes niza ==
Neka je
(
a
n
)
{\displaystyle (a_{n})}
niz realnih ili kompleksnih brojeva. Reći ćemo da niz
(
a
n
)
{\displaystyle (a_{n})}
konvergira broju L (realan ili kompleksan broj) ako vrijedi
(
∀
ϵ
>
0
(
∃
n...
anyway @T_01 taylor series gives you sin(tv2) = tv2 + t^3 v2^3/6 + ...
Limes, in mathematica, est quantitas ad quam alia quantitas adpropinquit. Definitio haec est:
lim
x
→
a
f
(
x
)
=
b
{\displaystyle \lim _{x\rightarrow a}f(x)=b}
significat
∀
ϵ
>
0
∃
δ
>
0
ut
|
x
−
a
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δ
⇒...
the version for finite Galois extension is: if $L/K$ is a finite Galois extension with Galois group $G$, then if we consider the category $L^*[G]$-Mod which consists of $L$-vector spaces together with a semilinear $G$-action, then $L^*[G]$-Mod is equivalent to k-Mod via thefunctors $V \mapsto V^G$ and $W \mapsto L\otimes_K W$
that's actually an equivalence of symmetric monoidal categories if we equip $L^*[G]$-Mod with the tensor product over $L$ and $K$-Mod with the tensor product over $K$
thus by abstract nonsense, you also get an equivalence of the corresponding categories of monoid objects and commutative monoid objects
which are $K$-algebras and $L$-algebras with a semilinear $G$-action by ring automorphisms (respectively the commutative ones for commutative monoid objects)
and then continuing by abstract nonsense, you get a category equivalence of the corresponding cogroup objects in commutative $K$-algebras which are dual to affine algebraic groups
there's also some Galois descent for projective varieties, but that's harder