@MartianCactus It's harder for other numbers, but it's still technically possible for every quadratic equation as far as I know. The quadratic formula is just an easier method in the general case.
The splitting method is equivalent to the quadratic formula. If you write out what you do for the splitting method for a general ax^2 + bx + c you just get the quadratic formula.
Like, $x^2 + 1$ can be factored by doing $x^2 + ix - ix + 1$ which you can then factor as $x(x+i) - i(x+i) = (x-i)(x+i)$, but that would be hard to see if the quadratic formula didn't already give you $\pm i$ as an answer.
Wait, maybe I misunderstand what the splitting formula is. For me, it is not splitting the whole quadratic into a square. It is writing the quadratic as a square plus a constant term
@MartianCactus, you need to prove that every polynomial has a root over C. That is non-trivial (high school mathematics won't do). Once you have that, you get the following: if P(x) is a polynomial, and c is a root, then P(x)/(x - c) is a polynomial again. And you continue with that polynomial until you have all n roots.
@MartianCactus, a double root really looks different from a single root. (Note how a quadratic with a double root just touches the x-axis, whereas a single root of polynomial with two roots really crosses it.)
Yes. The notation is a little informal. Really, what I mean by $P(x)/(x-c)$ is "the polynomial which is the same as $P(x)/(x-c)$ for all values $x \neq c$".
Maybe an example is illuminating: suppose that we have the polynomial $x^2 - 5x + 6$. Do you know a root of this polynomial?
Uh, I used wolfram-alpha in this case, because I'm lazy. The proper way to do it is long division of polynomials, but if you're in high school that is a little bit beyond you, probably.
Now, this is what we mean when we say that $-2$ is a single root of $x^3 + 8 = 0$. Namely, it is a root, but once you divide that factor, you are left with something of which -2 is not a root.
> As a less trivial example, consider the space Q of all rational numbers with their ordinary topology, and the set A of all positive rational numbers whose square is bigger than 2. Using the fact that $\sqrt {2}$ is not in Q, one can show quite easily that A is a clopen subset of Q. (Note also that A is not a clopen subset of the real line R; it is neither open nor closed in R.)
@BalarkaSen Alright. That's the last type I learned about (of course in physics you usually work with the curvature tensor immediately and then derive other curvature invariants from there)
That's just being informal with language. The correct statement is "For a quadratic polynomial, the sum of the multiplicities of the roots of the polynomial is 2."
But that's a mouthful. So we refer to a root of multiplicity 2 as if it is two different roots that occur at the same point (and this makes sense, actually), to save us that awkward phrasing.
@BalarkaSen Sure, whatever :D So consider some connection on a bundle $E$. It's a map that sends sections of $E$ to sections of $T^*M\otimes E$, right.
Hello @s.harp Let $u(x) \geq 0$ be harmonic in $\Omega$ and $u|_{\partial{\Omega'}}=0$ where the space $\Omega'$ is a proper subset of $\Omega$. I want to prove that $u(x) \equiv 0$ in $\Omega$.
If we suppose that $u \in C^0(\overline{\Omega'})$ we can deduce that $u(x)=0$ in $\Omega'$.
But how can we prove that $u(x) \equiv 0$ in $\Omega$ ?
@BalarkaSen So you have something from 0-forms to 1-forms, and it's probably a natural idea to extend this to an operator from $k$-forms to $k+1$-forms
@DHMO Yes. Clopen implies open, and the other way around, if every subset is open, then for an open subset $A$, its complement $A^c$ is open, hence $A$ is closed.
Sort of, yes. If you say, "a root of $P(x)$ is a number $c$ such that $P(x) = c$", then you have to define what the multiplicity of a root is (which I sort of did above), and then your theorem becomes as I did above.
Maybe, another way of looking at it will help you. Suppose you have numbers $c_1,c_2$. Then $(x-c_1)(x-c_2)$ is a polynomial, right?
Because if a quadratic $Q(x)$ has a root $c_1$, then $Q(x)/(x - c_1)$ will again be a polynomial, and it must be linear, so you can write $Q(x) = (x - c_1)(ax + b)$, which you can write into the form above.
Aaa... I have constructed a dfa that makes the addition in the binary system but I don't know how to prove that the dfa that I have constructed is the right one @AntonioVargas
How should I interpret this? The input consists of a sequence of pairs, should I take the $i$th input as the $i$th digit of each binary number, starting with least significant?
@MeesdeVries Can we not have as input a pair of numbers? I thought that the first number of the first pair for example is the last digit of the first sequence and the second number of this pair is the last digit of the second sequence . Yes, the digit after the / is the output
Also, you can have pairs of digits as input, that is fine. You could encode it as single numbers if you prefer, let $(0,0) ~ 0$, $(0,1) ~ 1$, $(1,0) ~ 1$ and $(1,1) ~ 3$. Either way is fine.
Alright. Well, by any reasonable interpretation, this DFA seems correct to me.
@MeesdeVries, what prerequisites apart from a solid understanding of General Topology and Analysis/Abstract Algebra do you think is needed to tackle it?
@MeesdeVries, I'm about to go through Spivak's Calculus on Manifolds and Munkre's book on Topology and I was wondering if I could go straight to Lee's book
I read the first chapter, and I loved the way he treated the contents of it
But most other books in 'Differential Geometry' take a different style, Spivak's own Volumes are a very different read compared to Lee's book
I haven't read Spivak or Munkres, but I am pretty sure you should be fine. Honestly, if you're just doing this for yourself anyway, just start in Lee, and go on until you get stuck. And then see if there's anything you need to read up on.
The only thing I could hold against that book is perhaps being a bit long-winded but I guess that the high level of precision is actually a good thing when just starting out in the topic.
It is my understanding that an identity relation is defined even when all possible ordered pairs $(x,x)$ for $x$ belonging to $A$ are not included. This is true for symmetric and transitive relations too. The only relation in which it is absolutely necessary to have all possible ordered pairs $(x,y)$ is the reflexive relation.
The smooth manifolds one is also in some sense elementary, but there is just so much structure on smooth manifolds that it's okay---you can fill a big book about simple stuff on manifolds.
@SteamyRoot One writes $g_n(x)=n[f(x+1/n)-f(x)]$, for then $g_n\to f'$ pointwise. But a pointwise limit of continuous functions is continuous on a $G_\delta$ dense set, so any derivative is continuous on a $G_\delta$ dense set.
So, there is no function whose derivative is continuous on just $\Bbb Q$.
But one can have a function with derivative continuous on $\Bbb R-\Bbb Q$.
If $k$ is a multiple of $n$ such that $\mathrm{digsum}(k) = n$, then appending $k$ to itself $10^m$ times and then appending $m$ zeroes gives a multiple of $10^mn$ with digit sum $10^mn$, I think.
@DanielCortild Let $n=2^a5^bc$ where $c$ is co-prime with $10$. Let the maximum of $a$ and $b$ be $m$. The required number is $\displaystyle 10^m \sum_{i\mathop=0}^{n-1}10^{(c-1)i}$.
We have that $f=x^3-3x-1\in \mathbb{Q}[x]$ is irreducible in $\mathbb{Q}[x]$. Let $a\in \mathbb{C}$ be a root of $f$. We have that $2-a^2$ is a root of $f$ and the extension $\mathbb{Q}(a)/\mathbb{Q}$ is normal. Let $n$ be a positive integer and $c_0, c_1, c_2\in \mathbb{Z}$ such that $(3+a-a^2)^n=c_0+c_1a+c_2a^2$. For each $n$ there exist such $c_0, c_1, c_2$.
How could we show that then also the relation $(1-a)^n=(c_0+2c_1+4c_2)+c_2a-(c_1+c_2)a^2$ holds? Could you give me some hints?
Hello @DanielFischer !! Do you maybe have an idea?
This is basically copied from my answer on this question, which I've now updated some.
Let's let $\|n\|$ denote the smallest number of 1's needed to write n using an arbitrary combination of addition and multiplication. For instance, ||11||=8, because $11=(1+1)(1+1+1+1+1)+1$, and there's no sho...
