Herstein defines integral domain as: a commutative ring with no zero divisors, whereas Dummit and Foote defines it to as a commutative ring with an identity (unit element) that has no zero divisors.
i have seen some variation there but my impression is that it is not like whether 0 is a natural number, where there is an even split. most people these days want rings to have 1.
i don't generally accept wikipedia as an authority on this kind of thing. i direct you to "multiplicative identity and the term 'ring,'" indicating a shift in textbooks that began in the 1960s. so yes, there's different definitions in use. but herstein is on the wrong side of history.
in my humble opinion. either way, it isn't a crisis, it's just something to pay attention to if you look in some random reference.
which side of the fence they're on. i guess in that respect it is like whether 0 is a natural number.
something i haven't seen in a while is in group theory, a few books use cycle notation but adopt the convention that permutations are applied on the left. so (13)(12) sends 1 to 3, not to 2. that can be confusing. it seems to be pretty rare though.
i had a harmonic analysis class once from a prof who also struggled with it. his homework assignments would contain stuff like "Problem 1. In the formula in on page 50 of my notes, check whether it should be xy^{-1} or y^{-1}x or some other thing"
has something to do with punnett squares, which is why i brought it up.
one time a mentor asked me how my research was going and i said, ehh, not great, and he said 'wander over to the biology department. maybe you can be like hardy in hardy-weinberg and be famous for something easy.'
he was kidding but to be honest it's a good idea.
hardy is of course math famous but i heard about hardy-weinberg hardy in high school.
in math he's often paired with his frequent collaborator littlewood. he seemed destined to have a hyphen after his name.
good on him for having a name nearer to the beginning of the alphabet.
I wrote this question for my schools math olympiad, give it a try: Given that $a^{n} = a \times a \times a....$ for $a,n \in \mathbb{R}$ for $n$ number of times and that $a^{-n} = \frac{1}{a^n}$. And given that $e^{\infty} = \infty$ what is the value of $e^{-\infty}$? \\ $A) \infty$ $B)$ Undefined $C) 0$ $D)1$ $E)-\infty$
youtube.com/watch?v=HDiaEYl-39s refers to a Math.SE post about why using the small-angle approximation to solve $\lim_{x\to0}\frac{1}{sin^2x}-\frac{1}{x^2}$ gives the wrong answer, but the link to that post is missing. Does anyone know which post this is?
approximations are not additive in general. Let me state one definition so that the earlier statement makes sense. Let's say that $f\sim g$ as $x\to a$, where $a\in \{-\infty\}\cup \{\infty\}\cup \mathbb R$ iff $\lim_{x\to a}\frac{f(x)}{g(x)}=1$ It is not in general true that $f\sim g$ and $h\sim k\implies f+h\sim g+k$, where f,g,h,k are real valued functions defined on R.
The limit $\lim_{x\to 0}\frac 1{\sin^2x}-\frac 1{x^2}$ is one such example. Note that even though $\sin^2x\sim x^2$ and based on this, if the approximations were additive, we would get the limit as $\lim_{x\to 0}\frac 1{x^2}-\frac 1{x^2}=0$. But that's not true.
The expression $\frac 1{\sin^2x}-\frac 1{x^2}$ is in indeterminate form ($\infty -\infty$) as $x\to 0$ so nothing can be said from the expression in that form. It requires simplification as showed in the link shared by Leslie.
you should try anki for that. that's what I use to learn little bits and not forget them in the long run. so you can keep adding bits here and there over a long time and not forget any of it
and cool, french grammar is difficult hehe
but i mean, compared to maths most things are easier lol
like, in all languages there are a ton of things that don't make sense, and have just been inherited from the past so you wouldn't know if someone didn't tell you
Let $E$ be a (analytic) rank $1$ elliptic curve over $\bQ$ with torsion group $\bZ/2\bZ$.\\ Let $E'$ be $\bQ$-isogenous to E with torsion group $\bZ/2\bZ\oplus\bZ/2\bZ$.\\\\ (1) Assume $\Sha_{E' /\bQ}$ is nontrivial \\\\ What conditions would have to hold for (1) to imply that $\Sha_{E/\bQ}$ is also nontrivial?
what is an example of two periodic functions $f$ and $g$ such that every period of f is incommensurate with every period of g and that f+g is still periodic?
$f(x)=\sin x, g(x)=\sin (\sqrt 2 x)$ give an example where $f+g$ is not periodic.
If you have a space of bounded killing flow lines say spanning points $p$ and $p'$ s.t. the flow vanishes at these points and the boundary, maps the boundary to itself, and you take the space of flow lines at a given radius from a center point, you'd get a flow on some surface of revolution. Is there a name for a surface like this?
'' Algebra is the offer made by the devil to the mathematician. The devil says: `I will give you this powerful machine, it will answer any question you like. All you need to do is give me your soul: give up geometry and you will have this marvellous machine.''
- Michael Atiyah (2004). Collected works. Vol. 6. Oxford Science Publications. The Clarendon Press Oxford University.
A condition for the intersection of two smooth manifolds to be a smooth manifold is that they intersect transversally. Is this only an obstruction because of the smooth structure?
Question: Is the intersection of two topological manifolds always a topological manifold? If not, are there conditio...
does anyone know why in this en.wikipedia.org/wiki/… , $\omega$ in local coordinates is supposed to be real? Wouldn't that mean that $h_{\alpha \overline{\beta}}$ would have to be wholy imaginary ?
basically i dont see how they are getting that form, when I computed it myself I got the coefficient of $dz_{\alpha} \wedge \overline{dz_{\beta}}$ to be minus the imaginary part of $h_{\alpha \overline{\beta}}$
@geocalc33 If you understand what goes wrong when transversality fails, you have plenty of examples where the intersection is not even a topological manifold. Just intersect planes with a torus (the usual picture stands it up and takes horizontal planes).
then ofcourse if it has period T then $h(x)=f(x)+g(x)=h(x+T)\implies f(x+T)-f(x)=g(x)-g(x+T)$. Also, $h(x+kT_1)=f(x)+g(x+kT_1)$ and $h(x+mT_2)=f(x+mT_2)+g(x)$. From here, I didn't know what to do.
there's not much to do from here, I thought and then looked at the solution in the back of the book. The solution used a result from another exercise and I find that non-intuitive.
nevertheless, the solution works. I am now open to the idea that sometimes there may not be a way that involves their ideas like the earlier question I asked today. I was expecting some elementary function (involving log, exponential, trigonometric, inverse trigonometric etc.) but apparently there isn't such function!!
@anak I had done that. That took me only upto: $h(x+T_1)-h(x)$ is $T_2-$periodic.
By interchanging roles of $T_1, T_2,$, one gets: $h(x+T_2)-h(x)$ is $T_1$- periodic.
This is quite intuitive. Think about $m\alpha + n$ as $m,n$ vary over integers. What is the difference between $\alpha$ rational and $\alpha$ irrational?
Sounds like anakhro and I are saying the same thing.
This is related to very famous and very important exercises — like the points $e^{ik\theta}$ on the unit circle when $\theta$ is a rational/irrational multiple of $\pi$.
suppose for i=1. Then $h(x+T_1)-h(x)$ has two incommensurate periods $T$ and $T_2$. Since $h(x+T_1)-h(x)$ is continuous at x, $n_kT_2+m_kT\to x\implies h(n_kT_2+m_kT+T_1)-h(x)\to h(x+T_1)-h(x)\implies h(T_1)=h(x)$
Let $f:[0,1] \to \mathbb{R}$ be a continuous function. Prove that
$$\lim_{n \to \infty} \frac{1}{2^n}\sum_{k=0}^n(-1)^k {n\choose k}f\left(\frac{k}{n} \right)=0$$
I know that $f$ is uniformly continuous and I tried to get some inequalities for the terms $f\left(\frac{k}{n} \right)$. For al...
i don't think that any statement of that sort could be taken literally, or 100% seriously, but it probably expresses something that he did think was true.
Let $f:[0,1] \to \mathbb{R}$ be a continuous function. Prove that
$$\lim_{n \to \infty} \frac{1}{2^n}\sum_{k=0}^n(-1)^k {n\choose k}f\left(\frac{k}{n} \right)=0$$
I know that $f$ is uniformly continuous and I tried to get some inequalities for the terms $f\left(\frac{k}{n} \right)$. For al...
user, one thing to keep in mind is that some people who work in realms of geometry/topology where algebraic invariants are computed sometimes use "linear algebra" to mean something significantly broader than what it means in a normal university curriculum.
i've heard people use it to embrace basically anything that happens after you apply a functor into the world of algebra.
I want to comment to that answer: -1. very badly written answer with no explanation at all. What is $\binom{n-1}{k-1}$ when k=0? and why are the sums equal?
But that will be rude :(.
The answer posted is same as given in the solution manual.
@Koro it would be $0$, they are using the fact that the number of ways of choosing $k$ things from $n$ is the same as the number of ways of doing so so that a particular fixed item is always chosen, and such that this particular item is neve chosen
then its just a matter of expanding the sum, and rewriting the summation indices
@porridgemathematics: I expanded the sum as that seemed natural to me and I ended up with a sum written above and there is this one extra term that doesn't cancel.
Had they put limit before the sums, then that would make more sense.
The equality of the two sums doesn't look trivial to me at all.
Consider the metric $ds^2=(1-\frac{2m}{r})dt^2+(1-\frac{2m}{r})^{-1}dr^2 + r^2d\theta^2 + r^2sin^2\theta d\phi^2$.
Suppose a particle very large starts at the initial radius $R$ and then radially infalls.
My text then states: It can be shown that the parameter defining the particle's trajectory (...
@porridgemathematics I was wrong earlier. It's not left out. We have $(-1)^n(n,n) f(n/n)=(-1)^n (n-1,n-1) f(n/n)$ and this combines with preceding term.
'' Algebra is the offer made by the devil to the mathematician. The devil says: `I will give you this powerful machine, it will answer any question you like. All you need to do is give me your soul: give up geometry and you will have this marvellous machine.''
- Michael Atiyah (2004). Collected works. Vol. 6. Oxford Science Publications. The Clarendon Press Oxford University.
Hello!! In which cases is the probability equal to the expected value ? In general do we not have that the expected value is equal to the probability multiplied by the amount of times the event happens, i.e. μ = P(x) * n ?
but I also don't like that formula for binomial coefficients. It often gives values that do not work in many useful binomial formulas. The bottom argument should always be an integer and any bottom argument less than $0$ should give a $0$ coefficient
You see, I don't know why I didn't see it sooner, but you can get a truncated approximation to the natural logarithm through floored logarithms by increasing the order of the domain.
Suppose I want $\log_2(3)$ to 64 bits. $2^{-64}\lfloor\log_2(3^{2^{64}})\rfloor \approx \log_2(3)$. In other words, more generally: $$\lim_{n\to\infty^{+}} \lfloor\log_2(x^n)\rfloor = \log_2(x)$$ for real $x$ and positive integer $n$.
All due to the fact that $\ln(x^n) = n \ln(x)$
Unless you're insane, no one computes $x^{2^{64}}$ and counts the number of bits in the result. If you can find the largest power of two just greater than $3^{2^{64}}$ without computing the product, then you can just compute the quotient of those two exponents. All it requires is comparison.
I'm trying to recall something I saw once about a surprisingly simple statement known to be undecidable. I think it was undecidable by Peano Arithmetic, but maybe some other well-known system, or a slight modification of one. I think it used just ordinary algebra operations, and there were probably low-degree polynomials in one or two variables. Any guesses?
@RandomVariable No, I think I posted that his command was not what my mother thought. He took command of VAW-114, which I think was with the Kitty Hawk at the time.
Here He was promoted to Captain at the change of command we attended.
the carrier was there, he was being promoted to captain, he was leaving with the carrier, so I can see why there might have been confusion.
Let $f:[0,1] \to \mathbb{R}$ be a continuous function. Prove that
$$\lim_{n \to \infty} \frac{1}{2^n}\sum_{k=0}^n(-1)^k {n\choose k}f\left(\frac{k}{n} \right)=0$$
I know that $f$ is uniformly continuous and I tried to get some inequalities for the terms $f\left(\frac{k}{n} \right)$. For al...
