I'm encountering the above statement as part of a proof of this very theorem (roots vary continuously with the coefficients), which is leaving me quite confused
Having trouble formulating an accurate linear program. I keep getting a result that the solution space is unbounded, which is absurd because the task is "find a position for a middle hub such that the distances to three nearby hubs in minimized"
If $f$ is continuous function $f:\Bbb R\to\Bbb R$ then it has antiderivative. But, does continuous function $g:\Bbb C\to\Bbb C$ also has antiderivative?
OH!! Please verify this: a complex function that is continuous, but not differentiable, can't have antiderivative, because complex differentiable function infinitely differentiable.
@Silent yes that's correct; an irreducible polynomial $f$ over $\Bbb Q$ generates a maximal ideal of $\Bbb Q[X]$ and the quotient by this ideal is a field containing a root of $f$
Problem: Let $F$ be a field with $\text{char} F = p$ for some prime $p$. Show that if $X^p - X - a$ is reducible in $F[X]$, then it splits into distinct factors in $F[X]$
Solution in back of the book:
This solution confuses me slightly. Why is $f(X)$ splitting as two monic polynomials? ...
so... on facebook messenger there are well over a thousand groups that sell lsd, weed, you name it, where is SE on this you people have REALLY dropped the ball here
I mean just the traffic alone if you multiply that globally is a honeypot and it's all going to zucc
anyhoo numberphile have some great dvds on youtube, I gotta ask.. why.... why are his pupils so dilated? it's almost as if im peering into another parallel chronology of myself in another universe in the multiverse
@JennaSloan omg are you related to neal sLOan? him and I are like the best of pals he even replied to an email once albeit it ended in him threating to call the police but I feel as if our friendship can move past that
i don't mind either so long as get a ticket on on a space station to watch all the idiots die on live stream but it seems I'm not going to be getting much from Elon seeings i troll the #@$# out of him
The death of humanity is not enough for me, I want the permanent death of totality itself. Only by the erasure of God himself will this boring cycle stops repeating
(even death is too light a word, no words can describe that)
but real life hunger games.... even if i get left on the surface I'm totally going win guerrilla warfare with dark ages regulations i mean... yeah i think ive got a handle on that being on welfare for past few years
I'm not actually kidding here people will literally try to stab you for a $20 note
ah it wouldn't crumble if the control freaks accepted the that although fraught with potential problems, a free market is the optimal solution for humanity
you guys do a GFC scare every decade its not a fake news swindle that works ffs it's almost on the same level of lame as Africanised killer bees
anyhoo you're disturbing my deep state of focus on amazon
Does anyone have an example of a field of characteristic $p$ that isn't isomorphic to $\Bbb{F}_p$ or $\Bbb{Z}_p$ (these are basically the same fields)?
Suppose I need to design an experiment where I can test a set of people all at once, but the cost of each test increases as the number of people in the test increases. I also need to make sure that, for the total population of N people, every subset of this population of size k is in at least one of the test sets.
Is this a type of problem that's already been studied? Ideally, I'd like to find a rough bound for the maximum number of tests I need to run as a function of N and the size of each test set.
Given $$a + bx + cx^2 +dx^3 + x^4 = [(x-p)^2 + A(x-q)^2][B(x-p)^2 + (x-q)^2]$$, where $a,b,c$ and $d$ are integers. Find $p,q,A$ and $B$ in terms of $a,b,c$ and $d$. I have expanded both sides and compared which got me a messy system of equations. I wanted to know if there is any easy method to solve this or any key observation. Thanks.
Let A = 5 and B = 4, I can prove that they are exactly 5 and 4, and I did the summation of the two, but how can I prove that I did the summation and not some other operation
Let $F:\mathbb{R}^2 \rightarrow R$ be continuous and suppose there exists $C>0$ such that $|F(t,s_1)-F(t,s_2)|<C|s_1-s_2|$ for all $(t,s_1),(t,s_2) \in \mathbb{R}^2$.
For $f \in C([a.b])$, define the function $g$ on $[a,b]$ by $g(x)=\int_a^x F(t,f(t)) dt$. How do I show that $g \in C([a.b])$?
Let $(t_n)_n$ be a real sequence such that $t_n\rightarrow t$. Consider $|F(t_n,f(t_n))-F(t,f(t))|\le|F(t_n,f(t_n))-F(t_n,f(t))|+|F(t_n,f(t))-F(t,f(t))|\le C|f(t_n)-f(t)|+|F(t_n,f(t))-F(t,f(t))|$.
Actually, we have $(t_n,f(t_n))\rightarrow (t,f(t))$ regardless due to the continuity of $f$, so this should already imply $F(t_n,(f_n))\rightarrow F(t,f(t))$, hence the continuity, meaning that the assumption on the existence of $C$ could be dropped entirely. Not sure whether I'm overlooking some subtle detail or the assumption is actually unnecessary.
@Thorgott I think this is required in the second part. (b) Show that $T:C([a,b]) \to C([a,b])$ defined by $Tf(x)=\int_x^a F(t,f(t))$ is a continuous map.
So am I. The thing that made me skeptical is that not using a hypothesis is usually a very good indicator of doing something wrong, but since it gets used in part $(b)$ and I believe the argument above is sound, I'm convinced now.
If $f(x)=\frac{\sin[x]}{[x]}$ if $[x]\neq 0$ and $f(x)=0$ if $[x]=0$, where $[x]$ denotes the greatest integer less than or equal to $[x]$, then $\lim_{x\to 0} f(x)$ is: (A) 0 (B) 1 (C) $\infty$ (D) does not exist
I got the left hand limit to be equal to $\frac{\sin(-1)}{-1}$ and the right hand limit to be equal to be to $\frac{\sin(0)}{0}$.
Well, $0/0$ is not defined and strongly indicates you made a mistake somewhere. Look at the definition of $f(x)$ again. It makes a special exception so that this case actually does not occur.
@Thorgott There is a third part as well. It says that show that $T(B)$ is an equicontinuous family of functions in $C([a,b])$ provided $B \subset C([a,b])$ is a bounded set. I am having trouble estimating $\int_x^y |F(t,f(t))| dt$ provided $|f(t)|<M$.
@MrAp As you correctly noted, when $x$ approaches $0$ from the right, $[x]$ is eventually $0$, not only approaches it. How is $f(x)$ defined when $[x]=0$?
@Thorgott, I wonder what would be the case if this was not a piecewise function. I mean $f(x)=\frac{sin[x]}{[x]}$ for all $x \in R$. What do you think will be the R.H.L in that case.
The definition does not say $\frac{\sin[x]}{[x]}=0$ whenever $[x]=0$; it says $f(x)=0$ whenever $[x]=0$. This is an important distinction as that fraction is definitely undefined.
That $f(x)$ would not be defined in the interval $[0,1)$ so it makes no sense to ask what its limit is (at least the RHS limit; the LSH limit remains $\sin(1)$).
@Thorgott when we are calculating the right hand limit of $f(x)$ then it means that $x$ is a positive number close to $0$ , which means that $[x] = 0$, and if $[x] == 0$ then $f(x)$ is also $0$. So don't you think the right hand limit will be $0$ ?
@Thorgott Does this result follow from part $(a)$ as we showed that $F(t,f(t))$ is continuous over $[a,b]$? Then, $\int_x^y |F(t,f(t))| dt \leq int_x^y |F(t_0,f(t_0))| dt=N|x-y|$. But then, where did we use the fact that $|f(t)|<M$.
Undefined, because both the nominator and denominator would be constant $0$ in the interval $[0,1)$. Indeterminate is a term only used to describe limits and an indeterminate form like $0/0$ means that we are looking at an expression $f(x)/g(x)$ as $x$ approaches something and we have that both $f(x)$ and $g(x)$ individually go to $0$. The expression $f(x)/g(x)$ in those cases is usually not defined at the point $x$ approaches, but only in a neighborhood thereof.
@Apptica Yes, in the original question, the right-hand limit is $0$.
@user330477 The issue is that we need an upper bounded that works for all $f\in B$ simultaneously. Give me a minute to think about it.
If $G \simeq S_3$ and $g_1,g_2 \in G$ are of order $2$2 and $3$, respectively, can I say that $g_1 \mapsto (1,2)$ and $g_2 \mapsto (1,2,3)$ defines an isomorphism?
@user330477 I believe the following should work. Since $B$ is uniformly bounded, we have an $M$ s.t. $|f(t)|<M$ for all $t\in[a,b]$ and $f\in B$. So $f(t)$ lies in the compact interval $[-M,M]$ for all $t\in[a,b]$ and $f\in B$. Then $(t,f(t))$ lies in the compact set $[a,b]\times[-M,M]$ for all $t\in[a,b]$ and $f\in B$. Thus, we can bound $|F(t,f(t))|$ above for all $t\in[a,b]$ and all $f\in B$ simultaneously. This should suffice after choosing a sufficiently small $\delta$.
@Thorgott I meant to say this is probably not the expected answer. I know it is rigorous and convincing. But it requires knowledge of topology, which we have not talked to much about in my course.
That the continuous image of a compact set is compact (in particular, bounded) is a standard result. I'm not sure what other approach there even is to bound $F(t,f(t))$ as all we have to work with are continuity assumptions.
As James noted in his comment, generating sets are not unique, since if $A$ is a set that generates the group, then any set containing $A$ will also be a generating set. However, I assume you are trying to find a smallest set of generators.
If you allow an element $g$ to be a generator, then eve...
@Thorgott And bounding $|F(t,f(t))|$ is the only way of showing that $T(B)$ is an equicontinuous family of functions in $C([a,b])$ provided $B \subset C([a,b])$ is a bounded set.
Since the graph is complete, for any pair of vertices u and v, they are connected, and their permutes s(u) and s(v) must also be connected, so the biconditional is true (since both sides are always true).
However, the fact I'm using can be proven rather succinctly (if you've seen a proof of the extreme value theorem, this is the same idea). Let $F:[a,b]\times[c,d]\rightarrow\mathbb{R}$ be a continuous function. Assume $F([a,b]\times[c,d])$ is unbounded. Then pick a sequence $(x_n,y_n)$ such that $|F(x_n,y_n)|\rightarrow\infty$. By Bolzano-Weierstrass, there is a subsequence s.t. $(x_{n_k},y_{n_k})\rightarrow(x,y)$. But by continuity, $F(x_{n_k},y_{n_k})\rightarrow F(x,y)$. Contradiction.
@user76284 I don't think so. Here's what I have. Let $\sigma \in S_n$. Define $\tilde{\sigma} : K_n \to K_n$ to be $\tilde{\sigma}(v_i) = v_{\sigma (i)}$ and $\tilde{\sigma}(e_i) = e_{\sigma (i)}$, where $v_i$ is a vertex, $e_i$ an edge. it is easy to show that $\tilde{\sigma}$ is a bijection, but showing that it preserves adjacency is a little tougher. I need to show that $Ends(e) = \{v_i,v_j\}$ implies $Ends(\tilde{\sigma}(e)) = \{\tilde{\sigma}(v_i),\tilde{\sigma}(v_j)\}$.
“an automorphism of a graph G = (V,E) is a permutation σ of the vertex set V, such that the pair of vertices (u,v) form an edge if and only if the pair (σ(u),σ(v)) also form an edge”
@Mathphile $\sin$ and $\cos$ are defined without any reference to a number base whatsoever, so I'm afraid the question does not make sense. Same applies to the logarithm unless you mean the base of the logarithm, in which case they are different functions.
Are the edges labeled in any way? I find it easier to think purely in terms of vertices and then ask whether a particular pair is in the “edge set” or not.
Ok, in that case I do not understand the question. The way we define the functions $\sin,\cos$ and $\log$ is completely independent of any numeral system.
I do know that in $K_n$ that there is a unique edge between any two vertices, so perhaps uniqueness will guarantee that $Ends(\tilde{\sigma}(e)) = \{\tilde{\sigma}(v_i),\tilde{\sigma}(v_j)\}$
I think edges matter according to my book. The book I am using is a book on Geometry Group Theory and their conventions are different from those of graph theorists. E.g., to a geometric group theorist, a graph is a 1 dimensional CW complex.
Looks like it’s odd with respect to the middle of the interval?
Equivalently: if you take that figure and rotate it by 180 degrees, the purple part looks to be unchanged
Easiest way to check that, if you’re plotting a list of y-values, is to reverse that list and add it to itself. If it really is symmetric in that manner, then the resulting sum of lists should be identically zero
Given $$a + bx + cx^2 +dx^3 + x^4 = [(x-p)^2 + A(x-q)^2][B(x-p)^2 + (x-q)^2]$$, where $a,b,c$ and $d$ are integers. Find $p,q,A$ and $B$ in terms of $a,b,c$ and $d$. I have expanded both sides and compared which got me a messy system of equations. I wanted to know if there is any easy method to solve this or any key observation. Thanks.
@user193319 saying $\tilde{\sigma}(e_i)=e_{\sigma(i)}$ doesn't make sense. $S_n$ acts on $\{1, \dots, n\}$ and there are $\binom{n}{2}$ edges in $K_n$, so you can't index them by elements from the set on which $S_n$ acts
for a graph with no multiple edges, the notions for graph homomorphisms coincide, any map between vertices that preserves adjacency induces a unique compatible map on the edges
Given $$a + bx + cx^2 +dx^3 + x^4 = [(x-p)^2 + A(x-q)^2][B(x-p)^2 + (x-q)^2]$$, where $a,b,c$ and $d$ are integers. Find $p,q,A$ and $B$ in terms of $a,b,c$ and $d$. I have expanded both sides and compared which got me a messy system of equations. I wanted to know if there is any easy method to solve this or any key observation. Thanks.
If you look just 10 messages above, i have wrote what i got
Suppose $R$ is a ring with $R^\times$ cyclic of order $5$. Then $\mathrm{char}(R)=2$, because else $-1$ would have order $2$ in $R^\times$ contadicting Lagrange's theorem
so $R$ contains $\Bbb F_2$. Let $u \in R^\times$ be a generator and consider the subring generated by $u$, this a quotient of $\Bbb F_2[x]$ (via sending $x$ to $u$) and the kernel contains $x^5-1$, since $u$ has order $5$, so it is in fact a quotient of $\Bbb F_2[x]/(x^5-1)$
the factorization of $x^5-1$ over $\Bbb F_2$ into irreducibles is $(x-1)(x^4+x^3+x^2+x+1)$, so there are three possibilities for this subring: - it is isomorphic to $\Bbb F_2[x]/(x-1) \cong \Bbb F_2$ - it is isomorphic to $\Bbb F_2[x]/(x^4+x^3+x^2+x+1) \cong \Bbb F_{16}$ - it is isomorphic to $\Bbb F_2[x]/(x^5-1) \cong \Bbb F_2 \times \Bbb F_{16}$
in the first case, the unit group of $\Bbb F_2$ is trivial, so it doesn't contain an element of order 5, in the other two cases, the unit group has order 15
Alright, a chunk of that went over my head, I won't deny. Though, what I'm most curious about is why it is that $\text{char}(R)\neq 2\implies (-1)^2=1$
The complete graph on $n$ vertices has exactly one edge joining each pair of distinct vertices, and is denoted by $K_n$
A symmetry of a graph $\Gamma$ is a bijection $\alpha$ taking vertices to vertices and edges to edges such that if $Ends(e) = \{v,w\}$, then $Ends(\alpha(e)) = \{\alp...
Perhaps someone could clarify what exactly a symmetry of a graph. The definition my book gives isn't terribly clear (e.g., what exactly is a edge? Is a symmetry a function on the vertices, edges, or both?)
@user193319 I'd say it is a function on both vertices and edges. But if there are no multiple edges (as graph theorists like to assume), then the information what the symmetry does on the edges is redundant, as it is determined by the action on the vertices, so in this case authors usually omit the action on the edges
Okay, so in the case of multiple edges (such as in this math.stackexchange.com/questions/3197468/… question), how does one deal with edges? How does one label them? Without multiple edges, to denote an edge formed by vertices $v$ and $w$ I would just write $\{v,w\}$ (as does the author of the book I'm using); but with multiple edges this is ambiguous...
Meh. I took courses all the way through, and I encouraged students to do it, too. Even doing research, you should continue to learn some stuff out of your area ... or more deeply in your area.
Problem: Let $F$ be a field with $\text{char} F = p$ for some prime $p$. Show that if $X^p - X - a$ is reducible in $F[X]$, then it splits into distinct factors in $F[X]$
Solution in back of the book:
This solution confuses me slightly. Why is $f(X)$ splitting as two monic polynomials? ...
