@user21820 Thank you for the suggestion. I have not intent in writing sudoku solver by itself. I am learning propositional logic and there is sudoku solver as an example and I am trying to understand it now.
@LeakyNun Just show that for any finite fragment of ZFC you can find a transitive model M, such as V[k] for sufficiently large k, and then use a Henkin construction to construct a countable elementary submodel M' of M, and then Mostowski-collapse M' to a (countable) transitive model of that fragment. I've read that one can even obtain a countable transitive submodel of M, but not sure why, or maybe I read wrongly.
Because V[ω+ω] already has the powerset of powerset of N and hence has uncountable well-orderings, and so by Replacement on one of those you can construct ω[1], which has uncountable rank, appearing only after V[ω[1]].
The idea is that $V_\alpha$ already models most of the axioms (for $\alpha>\omega$ a limit), regular limit gives replacement and strongly inaccesible powerset
Hi any idea how can I solve $\cos(x)^2=\sin(4x)$? I derived it so much but couldn't end up with something clear. As a comment, it can be transformed into $\cos(2x)=2\sin(4x)+1$.
@user1732: If you're really interested in set theory, and you already know first-order logic, then you can take a look at the wikipedia page on ZFC. Replacement is one of the axiom schemas in ZFC. If what I just said makes no sense, then it's probably best for you to ignore ZFC. =D
separation: V[α] union: any transitive set extensionality: any transitive set foundation: any transitive set pairing: V[λ] for limit λ powerset: V[λ] for limit λ infinity: V[α] for α>ω replacement: V[α] for regular limit α
The idea is that $V_\alpha$ already models most of the axioms (for $\alpha>\omega$ a limit), regular limit gives replacement and strongly inaccesible powerset
then what did you mean by "strongly inaccessible [gives] powerset"?
Actually, if $\alpha$ is regular limit and $\lambda<\alpha\implies 2^{\lambda}<\alpha$ then $V_\alpha$ models replacement. If $\alpha$ is also bigger than $\omega$ it models the whole of ZFC
If you have 3 orthonormal vectors and from a3x3 matrix from them like $\begin{bmatrix} e_{1}& e_{2} & e_{3} \end{bmatrix}$
Then how can you show multiplication by it’s transpose form the left is the identity matrix? It’s easy with the right side but I can’t figure out the left
Is the $$\bigwedge_{i=1}^{9} \bigvee_{j=1}^{9} \bigwedge_{n=1}^{9}~p(i,j,n)$$ and $$\bigwedge_{i=1}^{9} \bigwedge_{n=1}^{9} \bigvee_{j=1}^{9}~p(i,j,n)$$ equal?
@JakeRose: That requires a theorem. You need to know that if $A$ is a square matrix and $BA = I$, then $A$ has to be invertible and $B=A^{-1}$, so $AB = I$ as well. It takes a bit of theory to get this. You can't get it by computation.
Let $A$ be a ring and let $A[[x]]$ be the ring of formal power series $f = \sum_{n=0}^{\infty}a_nx^n$ with coefficients in $A$. Show that $f$ is a unit in $A[[x]] \iff a_0$ is a unit in $A$.
I'm stuck on the reverse direction of this problem
Namely, given any matrix A, you can perform elementary operations to 'canonicalize' it to the form E R where E is a product of elementary matrices and R is a reduced row-echelon form.
From this alone you can observe a lot of things.
If A B = I then E R B = I and you can check that R must be identity.
So you would have shown that if a matrix has a right-inverse then it is a product of elementary matrices.
"Then I started to wonder if there's any "brutal" proof that does not visit the "higher" domain of algebraic structures and just uses the simple componentwise algebraic operations to prove that the dot product of the i-th row of B and the j-th column of A equals to the Kronecker delta from the given condition"
I suppose the OP's question can be restated as: "How does one prove that a matrix over a commutative ring is a stably finite ring with algebra alone?" — SemiclassicalAug 5 '14 at 13:27
is that supposed to be "ring of matrices over a commutative ring"?
@Perturbative let $g = \sum_{n=0}^\infty g_nx^n$ be an inverse, if it exists, for $f$. Try to compute $g_n$ in terms of $f_k$ and $g_{1},\ldots,g_{n-1}$.
@Semiclassical Well the proof I was sketching is literally elementary row operations alone, which I think is the easiest to explain. When you perform elementary operations E[1..n] on the left on A B = I, in that order, to reduce A to I, you will get A = E'[1] ... E'[n] and B = E[n] ... E[1], where E'[k] is the inverse of E[k]. It is then obvious that B A = I.
Let $f = \sum f_n x^n$. If an inverse $g$ exists, $fg = 1$, i.e. we must have $(\sum f_i x^i) (\sum g_j x^j) = 1$. Some computation gives $g_0 = 1/f_0$ and $$g_k = \frac1{f_0}\sum_{i=1}^k f_i g_{k-i}.$$ Hence if $g$ exists, $g_0 \in A$ so $f_0$ is invertible. Conversely, if $f_0$ is invertible, then with the given expression for the $g_k$, $g$ is an inverse for $f$.
reposting something I posted elsewhere: is there any sense in which $S^{-1}A$ "represents" the localisation functor $S^{-1}- : {\sf{Mod}}(A) \to {\sf{Mod}}(S^{-1}A)$? (motivated by $S^{-1}M \simeq M \otimes_A S^{-1}A$ and the tensor-hom adjunction)
I feel like my question might just be a ring theory problem, right? Isn't my question equivalent to showing that any pair of linear operators are associates?
@Semiclassical Well you are right that details would have to be provided, but that is how Gaussian elimination is proven to work. There are a couple of variants, but all of them essentially proceed by iteratively increasing the number of pivots at each round, which takes at most n elementary row operations. In total then we would need at most n^2 elementary row operations.
Let $\mathcal{H}$ be some Hilbert space, let $B(\mathcal{H})$ denote the bounded linear operators acting on $\mathcal{H}$, and let $M_n(B(\mathcal{H}))$ denote the $n \times n$ matrices with operator-valued entries. Let $A = [a_{ij}]$ be one such matrix. My question is,
If $\sum_{i,j=1}^n u_i...
Hm, in the commutative case (which I just realised is almost certainly not the case!) I think DVRs satisfy this. All nonzero elements are associate means there's a unique ideal, so it's a ... local ring of dimension 1?
A plane is missing and it is presumed that it was equally likely to have gone down in any of three possible regions. Let $1 − \alpha_{i}$ denote the probability the plane will be found upon a search of the $i^{th}$ region when the plane is, in fact, in that region, $i = 1, 2, 3$. What is the probability that the plane is in the $i^{th}$ region, given that a search of region $1$ is unsuccessful, $i = 1, 2, 3$? My try : Given that search of $1$st region is unsuccesfull implies that the plane is in either region $2$nd or $3$rd with probability $0.5$, therefore probability that it will be found…
@user23571113 I don't think my maths knowledge is solid enough to run for moderator. I expect a maths moderator to be relatively well versed in many fields of maths
@user21820 @Semiclassic: I wasn't suggesting the theoretical arguments were anything deep or difficult. I just meant to say that you can't in the case of @JakeS's explicit question see it in a direct computational fashion.
The starting point for this justification is the Dirichlet series associated with the formula for the von Mangoldt function:
$$\Lambda(m)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|m} \frac{\mu(d)}{d^{(s-1)}} \tag{1}$$
which is a infinite symmetric matrix where the $n$-th row has peri...
Latex is so bad. It is like art. You never really know if it is true or not. Whereas Mathematica code never lies. Or at least it gives you an indication if you are on the right track. I like Mathematica. I wish everybody did.
@Tanuj: The easiest thing is to write down the Taylor polynomial for $\ln(1-x)$ and substitute $x^2$ in it. I have no idea what you know (rigorously or not).
The function $\ln(1-x^2)$ is approximated about $x=0$ by an nth degree Taylor's Polynomial. Find n such that
$|Error|<0.1$ on $0\leq x\leq0.5$.
My Try: So I know to evaluate the remainder term of the Taylor polynomial, I can use the following expression
$$R_n (x)=\frac{x^{n+1}}{(n+1)!}\cdo...
yup @EricSilva. I had MRIs regularly before/after my cancer surgery on my arm. Being jammed in that noisy constricted space for an hour or two ain't fun.
First of all you need to find a group $G$ that has a proper subgroup isomorphic to $G$, obviously this can't happen for finite $G$ but it'd be even better if $G$ weren't finitely generated, just to be sure
Hmm, what about this easier problem: Can you find a vector space $V$ and a linear map $L\colon V\to V$ such that $V/\ker(L)\simeq V$ but $\ker(L)\neq\{0\}$? If you can then by saying "$\Bbb Z$-module" instead of vector space you'll have the counterexample you're looking for
What are your thoughts on the effectivity on passive learning such as lectures, @TedShifrin? Research shows that it's not the most effective way to do it, but everyone everywhere still does it.
I still like non-passive lectures, @Oskar. I tried to engage students as much as I could. Plus then working with them individually or in groups. If I hadn't retired, I might have tried having students watch my videos of the lectures and then having class be a bit more just problem-solving. I'm not sure.
No, @Tanuj. You replace $x$ with $x^2$. But the $\xi$ term doesn't change ... except that now $\xi$ lives where?
Okay. I've noticed that I learn much more by doing projects than in class. I guess that everyone's different. Maybe that's just the best way to do it for me.
I think that working on a problem and talking with someone who's giving you clues is a really good way of doing it. I've read a few research articles, and it's definitely an acquired taste.
@Oskar: Yes, and I was fortunate enough to be able to do that a lot with students, but it is hard to work that way effectively with 50 or more students (or even 25).
Yeah! Here at Uppsala University I heard that for every class, which is two hours here, the professor gets eight hours of preparation time. That's a lot of time to sit around in small groups and talk about questions that have popped up.
Hey guys, I'm not sure what I'm doing wrong with my calculations here, I'd appreciate help.
We know that given some number $x$ in the range $a$ to $b$ we can work out what percentage of $x$ that is in between $a$ and $b$ with $\text{Percentage}(x,a,b)=\frac{x-a}{b-a}$ as explained in this post https://math.stackexchange.com/q/754156.
Now, given the following function $f(x) = x^2 *10$. The values for the functions are as such https://gyazo.com/197f5e950772a5844ae509868f0c5ce8
Let us arbitrarily choose $a=10$, $b=40$, and $x=25$ which is $50%$. We can confirm $25$ is $50%$ using the equation
Let $\mathcal{H}$ be some Hilbert space, let $B(\mathcal{H})$ denote the bounded linear operators acting on $\mathcal{H}$, and let $M_n(B(\mathcal{H}))$ denote the $n \times n$ matrices with operator-valued entries. Let $A = [a_{ij}]$ be one such matrix. My question is,
If $\sum_{i,j=1}^n u_i...
The function $\ln(1-x^2)$ is approximated about $x=0$ by an nth degree Taylor's Polynomial. Find n such that
$|Error|<0.1$ on $0\leq x\leq0.5$.
My Try: So I know to evaluate the remainder term of the Taylor polynomial, I can use the following expression
$$R_n (x)=\frac{x^{n+1}}{(n+1)!}\cdo...
