complete the square, then it becomes obvious on what to set for x to minimise the value of y. With that, the relation of a,b with the minimum can be obtained
anybody could help with a math question I have not received yet a very satisfying answer it about finding the upper bound for the magnitude of the error in an approximation (with using integrals) math.stackexchange.com/questions/2366517/…
Hmm... it is easy to define exponentiation to base $e$ by using the fact that the exponential function is nondecreasing and the sandwich theorem above. For that we can establish that:
$e^{\omega}=\omega$
And thus $\alpha^{e^{\omega}}=\alpha^{\omega}$
and $\omega^{\alpha} = (e^{\omega})^{\alpha} = e^{\omega \alpha}$
what is less obvious though, is the value of $e^{\omega + n}$. Suppose $n < \omega$ then we know from the well ordering of ordinals $ \omega < \omega +n$. Now since exponentiation by any finite base is nondecreasing, we have $m < \omega,m^{\omega} \leq m^{\omega +n} \implies \omega \leq m^{\omega +n}$ so we only have a lower bound of the value of $e^{\omega + n}$
uh wait a sec..., maybe we can use $m^{\omega 2}$ as an upper bound...
so we have $\omega \leq m^{\omega +n} \leq \omega^2$
Hmm... $e^{\omega}n = e^{\omega}e^{\ln n}$, does not seemed very helpful
Actually, I think I need to revise how $a^{\beta\gamma} = (\alpha^{\beta})^{\gamma}$ is proved in the ordinals, that might give me some idea on how to handle the base $e$. Clearly the bounds $2 < e < 3$ is needed for some sandwich theorem
I have also an other question. We have the triangle ABC. We have that $X\in AC$, $Y\in BC$ and $Z=AY\cap BX$, where $X,Y\neq A,B,C$. I want to show that AB is parallel to XY iff $\frac{|CX|}{|CA|}=\frac{|ZB|}{|ZX|}=1$. Could you give me a hint? @MartinSleziak
When AB is parallel to XY then we have that then AZB and ZXY are congruent. What do we get from that? @MartinSleziak
We'll decide the details of how we want to organise this study group in the room I made for it. (tagging some people that I think were interested @Mike @Balarka @Dami @Astyx @Perturbative @Akiva)
It is easy to see that if AB // XY there is a pair of similar triangles, but it is not very obvious to me yet why the ratios between them must be 1:2 to obtain the required ratio of |CX|/|CA|=1
I mean, we can easily draw a pair of triangles sharing the same angle such that $\frac{|CX|}{|CA|}=\frac{|ZB|}{|ZX|}=1$ is false, so I felt like there is something else missing about the condition of Z to force the ratio to be 1:2
That will seemed so since $CX \subset CA$ based on what is give by the question, do you have to have the original question somewhere?
If $AB\parallel XY$, then since $BX \cap AY = Z$, $Z$ is colinear separately with $BX$ and $AY$. This means by opposite angles, $\angle BZA = \angle XZY$, by alternate angles for a pair of parallel line segments, $\angle BAZ = \angle XYZ$. Thus y interior angle of triangles the remaining pair of angles are equal. This means $\triangle ABZ \sim \triangle XYZ$. Now from the previous discussion, $\triangle ABC \sim \triangle XYZ$ and thus we have:
@LastIronStar Cheers, I will look into it. I am facinated on how big math really is. I always get amazed by all the previous discoveries that they present in this book for instance.
A finite dimensional real associative algebra with identity is a division algebra iff it has no zero divisors. One direction (divsion algebra$\implies$ no zero divisors) holds for infinite dimensional algebras as well, while the other doesn't, but I can't find a counterexample, any ideas?
the proof in the finite dimensional case uses the fact that if $A$ is the algebra without zero divisors then the map $A\to A$ that sends $a\mapsto ax$ for a fixed $x\in A\setminus\{0\}$ is injective, since its kernel is just $\{0\}$ (because $x$ isn't a zero divisor) and it must also be surjective since the dimension is finite
Hmm. But of course that last bit fails in infinite dimensions.
@AlessandroCodenotti Hm! Found a short article on JSTOR (jstor.org/stable/pdf/2321347.pdf) with the following sentence near the end:
"Also, there is an infinite dimensional algebra over the reals which has no zero divisors, yet is not a division algebra; it is the algebra of polynomials."
Let $\sigma_g$ be a reflection along a line and $\delta_Z$ a reflection around a point Z. The composition of reflection around a point is the composition of two reflections along perpendicular lines, a and b. So, we have that $\sigma_g\circ\delta_Z$ is equal to $\sigma_g\circ\sigma_a\circ\sigma_b$. This composition is a reflection if g is not parallel to a or b. Is this correct?
Quick question about function notation if I'm considering equations of form F(y/x) then does this follow: y/x * (-gamma + delta x)/(alpha - beta y) = F(y/x) since this is just some multiple of y/x and hence it's F(y/x) or am I misunderstanding the notation F(y/x)?
That's true for the first term clearly since it's just y/x but not the term with greek constants. So, no? To make sure I understand considering ky/x = F(y/x) then we have 2k for both pairs of coordinates. Whereas the term with greek constants differs => not F(y/x).
Suppose $\Bbb F$ is a field and $R$ is an integral domain which is also a finite dimensional $\Bbb F$-vector space, does it always hold that $R=\Bbb F[\alpha]$ for some $\alpha\in R$?
I'm 99% confident that this is true, so it's probably false
$R$ should actually be a field, pick a nonzero $r\in R$ and consider $\{1,r,r^2,\cdots,r^{n}\}$ (where $n=\text{dim}_{\Bbb F}(R)$), this set is linearly dependent so there exist $c_i\in\Bbb F$ such that $\sum\limits_{i=0}^nc_ir^i=0$, bring $c_0$ to the right side, multiply by $-c_0$ and you factor an $r$ on the left side
@Semiclassical Thanks for the input. Thought I had finally found canonical coordinates, but wasn't sure about F(y/x). This system is proving to be difficult.
well anyway putting those algebraic problems aside for a moment the only associative, unital, commutative, finite dimensional division algebras over a field are just the field extensions
(this was all motivated by trying to find a generalization of Frobenius theorem to fields different from $\Bbb R$)
Hence you can use the addition formula on those two angles and evidently obtain $\dfrac{\tan(\theta+\alpha)-\tan(\theta-\alpha)}{1+\tan(\theta-\alpha)\tan( \theta +\alpha)}$
the keys being 1) eliminating $\theta$ from $\theta+\alpha,\theta-\alpha$ means taking the difference of them, 2) being able to use the angle addition formula on that expression.
The Bernstein Polynominals B0,n-1 , .... , Bn-1,n-1 span the vector space of deg n-1 polynomials. I guess one proof this via induction, might be a bit tedious though. Does someone know/see another, preferable good approach?
One might be able to use the completeness relation somehow: $$\sum_{\nu=0}^n b_{\nu,n}(x)=\sum_{\nu=0}^n \binom{n}{\nu}x^\nu(1-x)^{n-\nu}=(x+(1-x))^n=1$$
(I'm using wikipedia's notation, so $b_{\nu,n}(x)$ instead of $B_{i,n}(x)$.)
I've decided to pop in maybe once a week, just to see what the dude is up to, and then do my own thing. Considering how he complained that he's 'sick of all his students', I figured one less student would improve his health a bit
Actually haven't been to any lectures after Day 1, I went up and picked his course outline from his office, and then was outta there, but I've only been hearing complaints from my friends also doing the course
i.e. "Though semiclassical methods are derived within quantum mechanics, the resulting formulae are usually expressed in terms of classical mechanics."
