@ZachHauk Here's something interesting. Consider $ax^2+bx+c$. Assume $p>q$. Clearly, $p-q$ isn't symmetric, but $(p-q)^2$ is! Thus, we can find $p-q$ (in terms of the coefficients) by taking the square root.
Now, theoretically, once you have $s_1$ and $s_2$, you can use those to find $p$, $q$, and $r$, using an additional cube root or two. I don't quite remember how.
But this is the start of one way of solving the cubic.
Note that Abel's theorem still leaves open the possibility that every specific quintic has roots that can be written in terms of those operations; he only proved that there's no one equation that works for all of them.
It turns out it has to do with group theory. Every polynomial has what's called a "Galois group"; Galois proved that a polynomial has solvable roots if and only if the Galois group has a certain property
(Groups with that property are called "solvable groups")
Quick question: does someone know a good way to describe a line in the complex plane? (i.e. I have a problem statement that says "If f maps a domain D to a portion of a line," and I want to express that symbolically. I could say f'(z) = k, where k is a complex constant, but I want to make sure that makes sense.)
Define the reals $x,y$ as
$$ y = \max \Im ( W ( - exp(x)) exp(-x) ) $$
Where $W$ is the standard Lambert W function and $\Im$ means the imaginary part.
How to find $x$ and $y$ ?
Closed forms ( allowing integrals , sums etc ) , contour integrals , numerical methods ??
I know how to express th...
hello, I am an undergraduate Mathematics student with a weak foundation in all things math... and I would like some advice. Currently I study online because I am in the military and it is not possible for me to learn in a traditional environment. I am doing ok in Calculus 1, > 90% average, but that is more because of all the online tools available to help me solve problems. Does anyone have any advice I could utilize to shore up my weak areas?
@BalarkaSen @SteamyRoot I saw your discussion about curves disconnecting surfaces. If you allow intersection, two simple closed essential curves can cut your surfaces into simply connected pieces. (It is a cute exercise if you start from knowing that the curve graph associated to a genus $g>1$ surface is unbounded)
I just thought of an amazing way to play candyland.
Think of it like bullshit
But you have to lie about which card you get
If someone calls you out and they're wrong, you go back twice the amount you moved (2 reds forward? two reds back instead). If they're right, you go back twice the amount you moved.
Let K be a field and let V be a vector space over K. Define the product of two elements in $R = K \times V$ by $(a_1,v_1)(a_2,v_2) = (a_1 a_2, a_1v_2 + a_2v_1)$ where $a_1,a_2 \in K$ and $v_1,v_2$ in K. define sum component wise.
Prove R is a local ring. Prove that R has only one prime ideal : its unique maximal ideal.
So I proved that the ideals of R correspond to vector subspace.
vector subspaces.
that is it is of the form $\{0\}\times H$ for H vector subspace of V.
Elements in the complement of $\{0\} \times V$ are of the form $(a,v)$ where $a \neq 0$ and $v \in V$. They are units, since $(a,v) (-a^{-2}v) = (aa^{-1},-a^{-1}v + a^{-1}v) = (1,0)$. Thus $(a,v)$ is a unit. Then $\{0\} \times V$ is unique maximal ideal. Thus R is local.
@MikeMiller When you work in geometry/topology do you just get used to "tensor/forms" language, and how to translate geometric ideas into this language. As an example I have been reading a bit about Teichmuller theory, and measured foliations and it appears an important thing to associate with a measured foliation (you can go back an forth) is the quadratic differential.
Basically I have a hard time ever thinking I would come up with something like, and after spending some time with it, and translating and looking what is happening in local coordinates is it starting to make sense. Is there some sort of trick on how to interpret and come up with these sorts of things? (I am guessing you just have to get used to the language)
Suppose $I_p$ is a prime ideal. Let $(0,v)$ be arbitrarily element in $\{0\}\times V$. Then, $(0,v) (0,v) = (0,0) \in I_p$ Thus $(0,v) \in I_p$ which means $I_p = \{0\} \times V$. Thus has only one prime ideal its unique maximal ideal
essentially we are forming an associative algebra out of V by adjoining scalar units of an identity element and stipulating every product of vectors is zero
I have a function $f$ defined on $(-a,a)$ as $f(x) = 1$ when $x = 0$ and $0$ elsewhere. There is a sequence of smooth functions $\{f_n\}$ defined on $(-a,a)$, such that $f_n\to f$ pointwise. Also $V(f_n)\to V(f) = 2$ where $V(f)$ is the variation of function $f$ in the inetrval $(-a,a)$.
Lets de...
The trace of $\mathrm M$ is the directional derivative of the determinant in the direction of $\mathrm M$ at $\mathrm I_n$, i.e.,
$$\det (\mathrm I_n + h \mathrm M) = 1 + h \, \mbox{tr} (\mathrm M) + O (h^2)$$
In Tao's words, "near the identity, the determinant behaves like the trace" [0]. More...
Otherwise there are a buckload of examples; symmetric matrices are diagonalizable so just pick eigenvalues d_1, ..., d_n large enough, so (d_1 ... d_n) - (d_1 + ... + d_n) will easily be bigger than 1
The difference of the determinant and the trace is 2. The determinant is 8 and the trace is 6.
Anyway, counterexample to the question. [2, 0, 0;1, 2, 0;0, 0, 2] - the determinant once again is 8 and the trace is once again 6. The matrix is clearly not symmetric.
Just adding a nontrivial element to a single slot doesn't change the story yet makes things asymmetric.
Define the reals $x,y$ as
$$ y = \max \Im ( W ( - exp(x)) exp(-x) ) $$
Where $W$ is the standard Lambert W function and $\Im$ means the imaginary part.
How to find $x$ and $y$ ?
Closed forms ( allowing integrals , sums etc ) , contour integrals , numerical methods ??
I know how to express th...
@DanielFischer any opinion on Alles Mathematik: Von Pythagoras zu Big Data from Aigner ? I want to buy it as a gift to a German teenager with olympic background
So for dehydration of ethanol, we use concentrated sulfuric acid, like this:
but for hydration of ethylene, we use dilute sulfuric acid, like this:
so why is one concentrated, and one dilute? For that matter, how does the sulfuric acid even help in the first place? Thanks.
@BalarkaSen It is related, I am not sure of all the benefits and downsides, but I guess the 1-forms induces local orientations which sometimes you don't want
@BalarkaSen The idea is pretty simple, it decomposing you surface into lines, and singular points (which you need for surfaces of genus >1) and you use these curves as a ruler. You can give a transverse measure in terms of a 1-form, basically it measures how far you have gone in horizontal or vertical direction
For quadratic differentials is that they give a vertical and horizontal foliation, and there isn't an orientation on anything
If you look locally, you see a bit of plane with a bunch of horizontal lines, and if you had a curve, transverse to the horizontal lines, the way you would measure the length is measure how far the horizontal lines are from the endpoints of the curve. That is is is basically $|dy|$ or $|dx|$ @BalarkaSen
Anyway I should get going, need to do some stuff. It is sort of helpful for me to attempt to explain this stuff @BalarkaSen
My physics professor defines a 1-form $\Theta=\sum\Theta_\alpha dx^\alpha$ and then calculates $d\Theta=\frac12(\frac{\partial \theta\alpha}{\partial x^\beta}-\frac{\partial\theta_\beta}{\partial x^\alpha})(dx^\beta\otimes dx^\alpha)$ (summations over alpha and beta are implied where necessary), I don't understand why he gets a tensor product there rather than an exterior one though
@DHMO If $$\phi(x)=\cos(x)-\int_0^x(x-t)\phi(t)dt$$ then what is the value of $\phi(x)+\phi''(x)$ ? OPTIONS: (a) sin(x) (b) -sin(x) (c) cos(x) (d) -cos(x)
@Alessandro Very strange; that should indeed be an exterior product. How did he define the exterior derivative? It's noteworthy that tensor product, antisymmetrized, is the exterior product
I suppose there is some weird physics convention going on there
@BalarkaSen that's the problem, he didn't define it, he was extremely handwavy, even for a physicist. He did talk about antisymmetrizing things though, can you elaborate on that?
@Alessandro For sure. The thing is that instead of looking at antisymmetric/alternating multilinear maps $V \times V \times \cdots \times V \to \Bbb R$, you could just look at multilinear maps (with no other convention). The space of such objects form a vector space, denoted as $\bigotimes^k V^*$, or $T^k V^*$, known as the "tensor algebra".
Define the reals $x,y$ as
$$ y = \max \Im ( W ( - exp(x)) exp(-x) ) $$
Where $W$ is the standard Lambert W function and $\Im$ means the imaginary part.
How to find $x$ and $y$ ?
Closed forms ( allowing integrals , sums etc ) , contour integrals , numerical methods ??
I know how to express th...
So for dehydration of ethanol, we use concentrated sulfuric acid, like this:
but for hydration of ethylene, we use dilute sulfuric acid, like this:
so why is one concentrated, and one dilute? For that matter, how does the sulfuric acid even help in the first place? Thanks.
