> In this talk we will introduce the ordinary power series generating function and consider its applications to binomial identities, recursion relations and combinatorial enumeration. Along the way we will encounter famous integer sequences such as the Fibonacci and Catalan Numbers. If time permits, we will discuss other kinds of generating functions.
Not yet, I am mostly busy on trying to get my chemistry DFT calculations to work, and after advice from my collegues, realised I forgot to substitute 32 electrons for one effective charge, thus end up using too much memory and failing the calculation
The only bit of generating function I have read so far is when DHMO and 2017 discussed about the method of snake oil
I remember some months ago someone ask about Absolute Zero. Now third law of thermodynamics gets its proof https://www.sciencealert.com/after-a-century-of-debate-cooling-to-absolute-zero-has-been-declared-mathematically-impossible so it looks tricky now the reach it :)
This is an example where if $\aleph_0$ physically exists, it will be interesting
In short: No you cannot mathematically reach absolute zero without an infinite reservior and infinite time
There's the additional complication with the third law that the ground state of a system may itself be degenerate at absolute zero, in which case the entropy needn't be zero.
(also an addenum: Negative heat capacity systems (like a clusters of population inverted sodium ions), while physical, are actually hotter because their entropy energy relationship (temperature) is of the opposite sign)
One thing to recognize is that, while this does matter in determining which component is which here, it doesn't matter in the magnitudes of those components.
Ojemba is sitting on a sled on the side of a hull inclined at 60 degrees. The combined weight of ojemba and the sled is 160 pounds. What is the magnitude of the force required for Mandisa to keep the sled from sliding down the hill
Don't question this beautiful book that uses racially diverse names of males and females instead of blindly asserting the white male patriarchal and racial dominance over society
(jk im sane, and you never said where the other person is applying their force)
Runge–Kutta methods are methods for the numerical solution of the ordinary differential equation
d
y
d
t
=
f
(
t
,
y
)
{\displaystyle {\frac {dy}{dt}}=f(t,y)\,}
which take the form
y
n
+
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=
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n
...
Right now, I am in the process of showing that a group $G$ of order $35$ must be cyclic, for the purpose of then showing that there is a unique group of order $35$, namely $\mathbb{Z}_{35}$.
To that effect, I was interested in applying an argument seen in the answer to the linked question given...
I have 1 question. So I am trying to prove the following thing The sequence $M_1 \rightarrow M_2 \rightarrow M_3$ is exact if for all modules $N$ $0 \rightarrow Hom(M_3,N) \rightarrow Hom(M_2,N) \rightarrow Hom(M_1,N)$.
so I am almost there
I proved everything but now I want to show $\ker(Hom(M_2,N) \rightarrow Hom(M_1,N))$ is equal to the $im(Hom(M_3,N) \rightarrow Hom(M_2,N))$
I showed one inclusion that is im is included in kernel I am almost there for the other side. But here is my issue
Let $\phi \in ker(Hom(M_2,N) \rightarrow Hom(M_1,N))$
call that map by $\bar{u}$ that is the map induced $M_1 \rightarrow M$
It is true yeah @arctictern More generally Given three modules $M_1,M_2, and M_3$ and two morphism $\phi_1 : M_1 \rightarrow M_3$ and $\phi_2 : M_1 \rightarrow M_2$ such that $ker(\phi_2) \subset \ker(\phi_1)$ then we have $\phi_3 : M_2 \rightarrow M_3$ making diagram commute in general it is defined as $\phi_3(m_2) = \phi_1(\phi_2^{-1}(\{m_2\})$ and this makes sense because $\phi_1$ fixes the fibers and we can prove this thing is A-module homomorphism
What you're really doing is just the FTC: $y(x)=\int\frac{dy}{dx}\,dx+C.$
(There are things that can be said of differential 1-forms, but for the purposes of ODEs it's enough to just take it as a formality.)
The main reason to do it is because it makes things easier to write stuff out. For instance, suppose $dy/dx=x/y$. One can rewrite this as $y(dy/dx)=x$ and then integrate both sides to get $\int y(x)\frac{dy}{dx}\,dx=\int y\,dy = \int x\,dx$.
I have always thought of it as really doing that first line; but I have been unsure why the other way doesn't work. I don't know much of anything about differential 1-forms; so the best I can do is to say that $dy$ on its own is not infinatesimal with respect to anything.
Well, if you're doing it by integration you have to stop for a moment and say "okay, I can use the substitution rule to trade one of the integrals for one over $y$."
With the differential notation, you take that for granted.
But, I'm someone who is used to the (probably abusive) differential form of it. So I don't really think about it much in practice.
Here linear algebra is in the first year as well as a bit of of ODEs in the analysis course, the actual diffeqs courses are a PDE one in the second year and a non mandatory ODE one in the third
@Julian Here it's linear algebra first/second year (it's weird), ODE & PDE are not mandatory, but has a prereq of analysis (so third year). Sometimes basic results on existence/uniqueness feature in analysis
Is there something like a "maximal normal neighborhood" of a point in a Riemannian manifold? I.e. a maximal open neighborhood $U$ of $p$ such that the exponential map $\exp_p$ is a diffeomorphism from $\exp_p^{-1}(U)$ onto $U$?
@abenthy Sure, but it would be highly non-unique. (Take the poset of open subsets of $T_pM$ for which the exponential map is a diffeomorphism onto its image and run Zorn's Lemma.)
Right..., too bad :| I would like to be able to say something like "the geodesic symmetry of a riemannian manifold is $s_p = \exp_p \circ (-\text{id}) \circ \exp_p^{-1}$, but it seems like we always have to put in the choice of a normal neighborhood $U$ of $p$. Is there a nice mathematical concept to put all these objects $(s_p,U)$ into "one object" so that I can rigorously speak of the geodesic symmetry at $p$?
@Mike Yeah, I think so too, one can probably formulate it elegantly in this language with something similar to sheafs/stalks. It's probably not worth the effort for me sadly, just curious :)
Didn't you have some kind of general theorem which says that if $|g'(p)|>1$ then the fixed point iterations do not converge?
You could use that instead of what you wrote in the image.
I'll have to leave, sorry. Maybe somebody else will pick this up. I have also mentioned this in calculus room - perhaps somebody will notice the problem there.
I want to teach a short course in probability and I'm looking for some counterintuitive examples for it. Results that seems to be obviously false but they true or vice versa...
I already found some things. For example these two videos:
Penney's game
How to Win a Guessing Game
In addition I'v...
@Dragneel Incidentally, that's why when you check a number for its prime factors, you only need to check up to the greatest prime lesser than the square root of the number.
The contrapositive and induction are a lot more effort than its worth. A contradiction is pretty straightforward but could also be easily rephrased as a direct proof
@Dragneel Hint- the inequality $k \le sqrt(n)$ is equivalent to $k \times k \le n$. Now remember that $n = kl$ and $1 \le k \le l <n$. How do you go from the given data to the equivalent conclusion?
So, in the case where $l=k$, then I square both sides of $k ≤ sqrt(n)$ which becomes $k^2 ≤ n$. Which is $k^2 ≤ kl$. Dividing both sides by $k$, the inequality becomes $k ≤ l$.
@MikeMiller It also answers the question, "Suppose I have a polynomial with nonnegative integer coefficients. How many values of the polynomial do you have to ask for in order to figure out what the polynomial is?"
[Method of snake oil thought] Let $$L_n(f(n,m))=\sum_{n\in I}f(n,m)=a_m$$ Then consider generating function: $$F(x)=L_m(a_mx^m)=\sum_{m \in J}a_mx^m=\sum_{m \in J}\sum_{n\in I}f(n,m)x^m$$ Now by linearity of $L$ $$F(x)=L_m(a_mx^m)=L_m(L_n(f(n,m))x^m)=L_m(L_n(f(n,m)x^m))$$ $$F(x)=L_n(L_m(f(n,m)x^m))$$ $$F(x)=L_m(a_mx^m)=L_n(L_m(f(n,m)x^m))$$
If $\displaystyle{L_m(f(n,m)x^m)=G(x)=L_n(b_nx^n)}$, then
$$F(x)=L_m(a_mx^m)=L_n(b_nx^n)$$ $$a_k=b_k$$
So method of snake oil works as long the inner sum that is resulted after interchange becomes a generating function, thus allow the coefficients to match up
If $G(x)$ does not exist, then method of snake oil fails
[Method of snake oil thought] Now what about: Let $$L_n(f(n,m))=\int_B f(n,m) dn= a(m)$$ Then consider some "general power series transform": $$F(x)=L_m(a(m)x^m)=\int_C a(m)x^m dm=\int_C\left(\int_B f(n,m) dn\right) x^m dm=\int_C\int_B f(n,m) x^m dndm$$ Suppose Fubini's theorem holds, then $$F(x)=\int_B \int_C f(n,m) x^m dmdn$$ Now if $\int_C f(n,m) x^m dm=G(x)=\int_B b(n)x^n dn$, then $$F(x)=\int_C a(m)x^m dm=\int_B b(n)x^n dn$$ Finally (condition to be found) $$a(k)=b(k)$$
I have an integral from - inf to + inf of 1/(x + i0). If I regularise it with an exponent like e^{- i 0 x} and use residue theorem, I get - 2 pi i. But if I just use Sokhotski formula, I get - pi i
Right now, I am in the process of showing that a group $G$ of order $35$ must be cyclic, for the purpose of then showing that there is a unique group of order $35$, namely $\mathbb{Z}_{35}$.
To that effect, I was interested in applying an argument seen in the answer to the linked question given...
I want to teach a short course in probability and I'm looking for some counterintuitive examples for it. Results that seems to be obviously false but they true or vice versa...
I already found some things. For example these two videos:
Penney's game
How to Win a Guessing Game
In addition I'v...
