Below is a problem I did from the book Differential Equations by K.A. Stroud and Dexter Booth. I got the right answer but I am not sure I did it right especially when I took the inverse Laplace Transform. Therefore, I am hoping that somebody here can check it for me.
Thanks,
Bob
Problem:
Solve...
In graph theory, a simple path is a path that contains no repeated vertices. But, in a directed graph, the directions of the arrows must be respected, right? That is A -> B <- C is not a path? However, I have an example which states that would also be a simple path
I want to calculate the solution $p\in \mathbb{P}^4$ of the interpolation exercise $$p(0)=2 , \ p'(0)=3, \ p''(0)=1 \\ p(1)=2, \ p'(1)=0$$
For that we have to write a general polynomial of degree 4, $p(x)=ax^4+bx^3+cx^2+dx+e$ and then with the given conditions we have to solve a linear system.
Is that correct or am I supposed to do something else?
@MaryStar That seems like a reasonable approach to me. Of course, you already know some coefficients, since $p(x) = p(0) + p'(0) x + \frac{p''(0)}2x^2 + \frac{p'''(0)}{3!} x^3 + \dots$. But you get exactly the same values from the system of linear equations - in three of the equations there is only one non-zero coefficient.
How to calculate $|Stab_{S_7}((1 \ 2 \ 3)(4 \ 5 \ 6))|=\frac{|G|}{|G((123)(456))|}$? I do understand that $|G|=7!$ because it has 7 elements but how to calculate $|G((123)(456))|$?
(There's also an holomorphic bijection between the disk and the universal cover of $\Bbb C\setminus\{0,1\}$, where the fundamental polygon looks like that. Pretty sure)
@nbro section 11.3 "Directed Paths and Connectedness" defines directed walks and calls a walk on the underlying graph a "semiwalk", with the special case of a "semipath"
a semipath is a path in a digraph that ignores the directions, a simple path is a path in which no vertex is repeated, so a simple semipath would be a semipath with no repeated vertices
In my experience, this is what causes more problems: inconsistent notation and terminology. The concepts, at the end, are learnable, but if you have inconsistent notation, and you are not familiar with the inconsistent notation, that is, e.g., that you can call a simple semipath just a simple path, then you will get confused, because a simple path is usually not a simple semi-path. How do you know for sure that they meant a simple semi-path, as opposed to a simple path?
It seems to me that the definition of limit is not precise.. but in Math we alwys say precise things, we define everything precisely.. so where I am going wrong?
It's an amalgamated product of Z_4 and Z_6. Even better, it's a Z_2-extension of PSL_2(Z) which is Z_2 * Z_3 which is hyperbolic. So it in turn is hyperbolic.
I'm reading this paper, it's a gentle introduction to the Grigorchuk group (the first group that was shown to have intermediate growth), you'd probably like it too
What I said was that Aff(R^2) has no F_2 subgroup. Alessandro gave the argument that since it's R semidirect R^x, and those two groups are abelian, it's amenable and amenable groups don't contain F_2.
About stratified spaces on and off. I have taken up symplectic geometry as a side reading.
I realized in my probability-II course that the probability density function is precisely the Radon-Nikodym derivative of the pullback measure by the random variable wrt the Lebesgue measure
So I want to read the proof of R-N now
I have the group cohomology talk next weekend, so I also want to rethink about that for a bit
One of my friend is studying differential forms, and another is studying linear algebra (various normal forms, the ideal-subspace correspondence, ...) so I am forced to think about those often :P
Let $H$ be a subgroup of $G$. I want to show that $gHg^{-1} = \{ ghg^{-1} : g \in G, h \in H \}$ is also a subgroup of $G$. But I am having some trouble showing that $gHg^{-1}$ is closed under composition. Any hints?
Can someone please example? (link: https://math.stackexchange.com/questions/3080545/calculating-the-orbit-of-a-group) I can't seem to understand how to calculate.
Yes I see, but what i said was, to prove R/P has no zero divisors, we need to take elements that are not in P, and their product is 0 , and this is a contradiction
because P is prime, two elements not in P , do not get in P when we multiply them
if I get an element in P , ie their product is zero , but P is prime it cant happen
[a] =0 iff a is an element of P , [a] [b] = [ab] = 0 iff ab is an element of P [a] is a zero divisor iff a is not in P, and there is an element in R minus P, st ab is in P.
but this is saying R/P has zero divisiors iff P is not a prime ideal
Hi everyone, I was trying to verify the claims at the end of the page 125 of Hatcher's AT, namely that "(delta_1)^n - (delta_2)^n in the first group corresponds to the cycle (delta_1)^n" part, but I had (delta_1)^n - const. Now, I am trying to show that const is in the boundary and showed it for odd n (namely it's the image of the "upper" const map) but couldn't show it for even n. Can someone help with that sentence from the book and also give me a hint to show that const is always a boundary?
Consider $s\in\{0,..,n\}$. Let $H$ be a set of all the $s$-subsets of $[n]$. Also consider the action of $S_n$ on $X$ by $\sigma\cdot \{x_1,...,x_s\}=\{\sigma(x_1),...,\sigma(x_s)\}$. let $x\in X$.
Is the quotient of a locally path connected space locally path connected? I'm thinking that $[0,1]/ \sim$ with $x \sim y$ if and only if $x,y \in [0,1)$ is a counterexample, but I am not sure how to show this.
The neighborhoods are so big in that quotient that local path connectedness is just the same thing as path connectedness, and any quotient of the interval is clearly path connected.
Now as it turns out quotients of locally path connected spaces indeed are locally path connected, so you will have trouble coming up with any counterexample. I don't remember the proof but it was an exercise in my first topology class, so it can't be that nasty.
IIRC I relied on the following characterization. X is locally path connected iff for any open set U, the path components of U are open in U.
Interesting, since the continuous image of a locally path connected space is not necessarily locally path connected I thought it sounded suspicious for quotients as well
The book was Kinsey's "topology of surfaces". It is ok, I like the discussion of the classification of surfaces. But a lot of what I learned was through supplementary exercises so I don't really have anything to share anymore.
@TedShifrin If I paramaterise a function x= f(t) and y=g(t) like x=t and y=t^2 why is it that the intersection in the x,y,t plane, I only get a point? not y=x^2? I understand the method of eliminating the variable just not why it works, may you please explain?
Consider an integral domain $A$ with a multiplicative norm $N$.
$$N(a) \space N(b) = N(ab) $$
For all elements $a,b$ in $A$.
Since we are talking about an integral domain there are no zero-divisors.
Thus $N(x) = 0 $ iff $x=0$.
Let $k$ be the number of units.
Let $n$ be an integer and let $y...
Yeah i'm reasonably interested in it, I'm more interested in mathematics itself but I enjoy teaching and try to do it in an interesting/non-formulaic way
Oh, I actually like the idea of software that provides access to tutorial and supplementary problems based on established weaknesses or gaps. I think that's ideal use of the computer in teaching.
It turns out that many mathematics education majors (in the US) actually do not like mathematics and are not good teachers. Some are wonderful, of course.
Let $ n > 1 $ be an integer.
Consider The prime factorization
$$ n = x_1 x_2 x_3 ... $$
Now define
$$ t(n) = t( x_1 x_2 x_3 ...) = t(x_1) + t(x_2) + t(x_3) + ... + t( x_1 + x_2 + x_3 + ... ) $$
Clearly this function is completely determined by its values at primes. This gives us multiple so...
(Truth be told, I'm mainly doing it as a Plan B because my masters application got rejected. If I don't get it I'll probably send a quick application to Frankfurt university)
@mick There is only one possible choice of $t$, the $0$ function. Let $p$ be prime. Then, $t(p)=t(p)+t(p)$ the first term comes from applying $t$ to each prime factor in turn and summing. The second term comes from applying $t$ to the sum of the prime factors. Thus, $t(p)=0$. Then, extend to get $t\equiv 0$.
Below is a problem I did from the book Differential Equations by K.A. Stroud and Dexter Booth. I got the right answer but I am not sure I did it right especially when I took the inverse Laplace Transform. Therefore, I am hoping that somebody here can check it for me.
Thanks,
Bob
Problem:
Solve...
Consider an integral domain $A$ with a multiplicative norm $N$.
$$N(a) \space N(b) = N(ab) $$
For all elements $a,b$ in $A$.
Since we are talking about an integral domain there are no zero-divisors.
Thus $N(x) = 0 $ iff $x=0$.
Let $k$ be the number of units.
Let $n$ be an integer and let $y...
