I'm trying to prove there are no simple groups of order 90. I've shown that, if there does exist such a group, then it is isomorphic to a subgroup of $S_6$ and that it's intersection with $A_6$ is either itself or the trivial group. I'm trying to rule out the possibility that the intersection is trivial and arrive at the fact that the group would be a subgroup of $A_6$. Then I can show that $A_6$ has no subgroup of order 90 and I'm good.
I think showing $H\rightarrow G/N$ will work, though, because that gives an injective map to the cyclic group of two elements.
I'm just trying to prove uncountable linear orders are not isomorphic
I think it is true since if a linear order is uncountable then there are many elements that may not be uniquely mapped. All it takes is either injection or surjection to fail
And its easier to do the injection fails part... But it seems that I can't write it in a more abstract tone :(
@YOUSEFY The Gödel sentence is pretty much built to be unprovable, the author is just wondering whether there are many naturally arising statements of mathematical interest which are also true but unprovable
I've got a question about Fourier Series (in complex space). If I want to approximate a periodic function, does that function need to contain its own center of mass?
I know that with real-valued Fourier Series the endpoints of one period need to be equal to the average value of the function along that period. So if I want to do a Fourier transform on f(x) where x is on [0,L], then f(0)=f(L)="average value of f on [0,L]"
So does this mean that a periodic function through complex space needs to intersect its own center of mass in order for it to be able to undergo a Fourier transform?
I have a question: Suppose given you sum and product of two integer positive numbers a and b greater than one. Now, can you know a and b? The answer is given here: https://math.stackexchange.com/questions/171407/finding-two-numbers-given-their-sum-and-their-product But, now suppose the a and b are positive real numbers, does this make difference?
Vector of differences. Suppose $x$ is an $n$ -vector. The associated vector of differences is the
$(n-1)$ -vector $d$ given by $d=\left(x_{2}-x_{1}, x_{3}-x_{2}, \ldots, x_{n}-x_{n-1}\right) .$ Express $d$ in terms of $x$
using vector operations $(e . g .,$ slicing notation, sum, difference, line...
Hi. Fix $n\in\mathbb{N}$ and let $A\subseteq \{1,\dots,n\}$ be a random set. Is it true that $E\left[\frac{1}{|A|}\sum_{i\in A}|X_i| \Bigg| |A| > 0\right] \leq \max_{i=1,\dots,n} E\left[|X_i| \Bigg| i\in A \right]$ if $A$ and $X_i$ are dependent? If $A$ were fixed, the answer would be yes, but now I can't see it.
No. see this example in this site "https://math.stackexchange.com/questions/2519390/if-the-graph-g-has-an-eulerian-circuit-prove-that-its-line-graph-has-a-hamilt" The idea is that if you have a counterexample that shows that a hamiltonian G doesn't correspond to Eulerian graph in line graph G', or NOT-hamiltonian G doesn't correspond to NOT-Eulerian graph in line graph G', then you just prove that the converse doesn't follow.
Given n and h, where n = a * b and h is the following equation: h is congruent to ca + rb mod n, where c and r are positive integer and a and b are PRIMES. Now, Is there any way to find a and b, given n and h?
suppose for example: n=6 and h is congruent to 10a + 20b mod 6, then how can you find a and b? I don't want to factor 6
There is a commutative ring $R$ with $1$ the elements of which correspond one-to-one with the context-free grammars of a single string $s \in \Sigma^*$. In $R$, $+$ is symmetric difference and $\cdot$ is intersection.
If you look at a CF grammar, say $g = \{S \to AAA, A \to aa\}$, for a string ...
How can I solve the functional equation to find all involutions whose reciprocals are non trivial involutions?
I think this functional equation can be expressed as:
$$ f(x)=\frac{1}{f(x)}, $$ with the conditions that $f(f(x))=x$ and $\frac{1}{f(f(x))}=\frac{1}{x}.$
$f(x)=\frac{1}{x}$ does not ...
So if you expect precisely an additive constant of integration as you wrote, then I would expect you're right, but I've never thought about this before.
I guess it's clear. If the $+C$ appears additively, then we know that $y'$ MUST depend just on $x$ (differentiate your expression).
Hmm, I'm not sure if I understand that argument of differentiating. If the solution was instead $y = Cg(x)$, it would differentiate to $y' = Cg'(x)$, but there are ODEs where $y$ appears explicitly with the solution $y = Cg(x)$, i.e., $y' = y$.
Like, if I tell you my bank account balance as a function of time, there are an infinite number of ways you could separate that between interest as a function of time and deposits as a function of time, correct?
You couldn't give me back the ODE accurately describing my interest (coefficient of the dependent variable) and deposits (forcing term) just from the balance as a function of time, I believe.
There is a commutative ring $R$ with $1$ the elements of which correspond one-to-one with the context-free grammars of a single string $s \in \Sigma^*$. In $R$, $+$ is symmetric difference and $\cdot$ is intersection.
If you look at a CF grammar, say $g = \{S \to AAA, A \to aa\}$, for a string ...
How can I solve the functional equation to find all involutions whose reciprocals are non trivial involutions?
I think this functional equation can be expressed as:
$$ f(x)=\frac{1}{f(x)}, $$ with the conditions that $f(f(x))=x$ and $\frac{1}{f(f(x))}=\frac{1}{x}.$
$f(x)=\frac{1}{x}$ does not ...
