hi, can someone help me with a hint: consider a point P in a triangle ABC. Let the feet of perpendiculars from P on AB, BC, CA be M, N, Q. If PM^2-AM.MB=PN^2-BN.NC=PQ^2-CQ.QA then prove/disprove that P is the orthocenter of ABC. I actually need it as a lemma for solving a math.se problem
Yeah, I had seen that exercise in the book but hadn’t solved it. Thanks.
Suppose that f has only roots p and q in the subset( one simple and the other being of order 3), then can I say that $ f(z)=(z-p)( z-q)^2 g(z)$ where g is non vanishing in the entire subset.
I was thinking of the similar thing. Thanks for confirming it. So we consider $f(z)/(z-p)$, z not equal to p. This quotient is holomorphic on the entire subset except at p. In particular, it is holomorphic near q. So near q, there exists a non vanishing g such that $f(z)/(z-p)= (z-q)^3 g(z)$
Define $\tilde g(z)= \frac{f(z)}{(z-p)(z-q)^3}$, then this $\tilde g$ is holomorphic except at p and at q.
and is actually an extension of g. So by the identity theorem, is the unique extension of g.
So we have: $f(z)= (z-p)(z-q)^3 \tilde g(z)$.
for all z in the subset.
@onepotatotwopotato I know the definition of removable singularities but is there any formula to check directly if a point is a removable singularity or not?
Thorgott: the first thing that comes to my mind while trying to prove the above H.E. is that CX is contractible. Am I in the right direction of proving this?
$V(f_1,\dotsc,f_n)=V(f_1^2+\dotsc+f_n^2)$ over $\mathbb{R}$, so it suffices to consider single polynomials, but I don't see a contradiction in sight anyway
Well, you can express it as an inequality easily enough. But how do you go from $x^2+y^2-1<0$ to $f(x,y)=0$? I don't feel like it's possible, but, as you say, I don't see any immediate reason why not
i dont know anything about any of this, if the disc is included in the zero set of a polynomial, doesn't taylor's formula or whatever imply that the polynomial is identically zero
a very surprising fact (to me, anyway, when I first heard it) is that over any field that isn't algebraically closed, the common zero locus of any set of polynomials can always be expressed as zero locus of a single polynomial
The original question is: if $\sin(A) = \frac{3}{5}, 0 \leq A \leq \frac{A}{2}$ and $\cos(B) = \frac{12}{13}, \frac{- \pi}{2} \leq B \leq 0$, find the value of $\sin(2(A-B))$
in general, there is a trick. pick $g\in k[x]$ without any zeros. then, $h(x,y)=y^{deg(g)}g(x/y)$ makes sense as a polynomial of two variables and $h(x,y)=0$ iff $(x,y)=(0,0)$, so the common zero locus of two polynomials $f_1,f_2\in k[x_1,\dotsc,x_n]$ is the zero locus of $h(f_1,f_2)\in k[x_1,\dotsc,x_n]$. the cases of more polynomials follow by an induction (and you only ever need finitely many by Hilbert's basis theorem).
if $k=\mathbb{R}$, you can pick $g=x^2+1$, yielding $h=x^2+y^2$ with the only zero $(0,0)$ and this recovers the previous trick
there's identities that reduce sin(2x) in terms of sin and cos of x. and then other identities, if x = A - B, that reduce that into other trig functions of A and B
and then there's crap about, if t is in some range, you can get cos(t) out of sin(t) and vice versa
is there more than that? i think we agree that the problem sucks
@XanderHenderson Sorry for late reply. I am not convinced it is surjection but I was thinking if I know how to prove it is injection then surjection will be not that hard to do so.
That is, essentially, the definition of "is an element of". I am asking what it means for $j$ to be an element of $A$, in terms of this function you have defined.
No, that is not what I am asking. Suppose that $f(x) = A$ (remember, $f$ is set valued). Now suppose that $j$ is an element of $A$. What does that say about $x$?
A value varies from $x$ to $x+y$, assume $y=15$ and that the variation is $6%$. Evaluate $x$ and $x+y$. I tried this: the variation is $(x+y)-x=y$, and it is both $y=15$ and $(x+y)-y=6x/100$. Hence $15=6x/100$, so $x=250$. Finally, $x+y=265$. Is this a correct reasoning?
yeah, polynomials arise pretty naturally in the function context. if you know about a function f, and you know about adding and scaling and multiplying (and constant functions) as operations, you know about polynomials in f. if f is the identity, you know about polynomial polynomials.
yeah, you don't need that pointwise inequality, although you certainly have it if n is large enough.
@TedShifrin Is this your preferred surgeon? Is there a way to check out another that might be available sooner? I had to do that with urologists. It meant 2 months sooner for me.
Well, I saw one two years ago. This one comes extremely highly recommended by a close friend who's had other surgery and then this guy's. I'm hoping not to have surgery just yet, anyhow. I imagine yours was an urgent issue.
Assume $X$ is a locally compact, Hausdorff, locally connected space and $Y$ is locally connected metric space
Suppose we have a continuous family $g_y:U\rightarrow X$, of open embeddings where $U$ is open in $X$ for $y\in Y$. Now fix $y'\in Y$. Define $G:Y\times U\rightarrow Y\times X$ by $G(y,u)...
If I consider the map $\gamma:\Bbb{R}\rightarrow \Bbb{R}^3$ defined by $\gamma(t)=(t,e^{-1/t^2},0)$ if $t<0$, $\gamma(t)=(0,0,0)$ if $t=0$ and $\gamma(t)=(t,0,e^{-1/t^2})$ if $t>0$, then I got that $\gamma''(t)=\left(0,-\frac{(6t^2-4)e^{-\frac{1}{t^2}}}{t^6},0\right)$ for $t<0$, $\gamma''(t)=(0,0,0)$ for $t=0$ and $\gamma''(t)=\left(0,0,-\frac{(6t^2-4)e^{-\frac{1}{t^2}}}{t^6}\right)$ for $t>0$.
But now if $K$ is the curvature of $\alpha$ I get that $K(0)=0$ and $K(t)\neq 0$ for $t<0$ but in my opinion $K(t)\neq 0$ for $t>0$ is wrong since one can take for example $t=\frac{2}{\sqrt 6}$ and …
I looked at it, monoidal, and the notation was wrong and the question made no sense to me. I don't see how you can randomly pick $y'$ and expect anything to be reasonable.
I think I have given up on trying to use Safari on my (somewhat elderly) iMac. It just hung up the computer repeatedly. Everything seems copacetic using Firefox. I ran all the utilities to check the disk, etc., and it seemed fine.
Okay what about the following argument: Let $u\in U$. $(y',u)$ for any $u$. Assume we have an open set $V$ in $B\times X$ such that $(y',u)\in V$. Shrinking $V$ if necessary, we may assume $V=V_1\times V_2$, for open sets $V_1\owns y$ and $V_2\owns u$. Since $g_y$ is a continuous family, the map $F:Y\times U\rightarrow Y\times X$, given by $F(y,u)=(y,_g_y(u))$ is continuous. So by continuity, there exists open sets $B_1,V_1$ such that $(y',u)\in B_1\times V_1$ and such that
@TedShifrin I'm a bit confused with your notation since we used the same letters for different objects. I understood your idea as follows, if I take my $\alpha$, I know that there exists a unit speed reparametrization $\beta$ such that $\beta(s)=\alpha(\phi(s))$ for $\phi$ some diffeomorphism. But then $\beta'(s)=\alpha'(\phi(s))\cdot \alpha(\phi'(s))$
Write down the easiest example. Say $Y=X=\Bbb R$, and $g_y(x)=x+y^2$. Now choose $y'\ne 0$, say $y'=1$. Write down explicitly the mapping you're talking about and tell me it is continuous.
@user1294729 Your chain rule is not correct.
@monoidal Your first instinct should always be to try simple examples to understand things.
You don't want to compute that, but its derivative is not bad. It's just the speed.
That's the whole point (if you actually read my text). You need arclength parametrization for theory, but in practice you avoid it and apply the chain rule immediately.
Does the following make sense: Suppose $g_y:U\rightarrow X$ for $U$ open in $X$ and $y\in Y$ are open embeddings such that $G(y,u)=(y,g_{u}(y))$ is continuous. Fix $y'$ in $Y$ and identify $g_{y'}$ with the inclusion $U\hookrightarrow X$. Then for $y$ sufficiently close to $y'$, $g_y$ lies in any open set containing the inclusion.
@user1294729 Why don't you take 5 minutes and actually read the pages I suggested? You actually want to compute everything with the $t$ derivatives, and never put in $\phi$ explicitly.
@monoidal What does "identify $g_{y'}$ with the inclusion" actually mean?
@user1294729 I suspect your course that gave you this exercise taught you something rather than giving this exercise out of the blue.
So if you're assuming $g_{y'}$ is the inclusion map, then the exercise is mere tautology. What do you mean by "$g_y$ lies in any open set containing the inclusion"? Are you doing the topology on the function space?
So I have seen the definition for curvature for a unit speed curve and also this representation with $\phi$ I wrote above. But then I need to derive the representation twice and then I don't see how further
Because in you second example you apply the formula from above but I don't see how you found formula.
Yes, that is correct @TedShifrin . But the only thing I don't understand is how does one 'identify' $g_{y'}$ with inclusion for it to make sense to say for $y$ close enough to $y'$, $g_y$ is close enough to inclusion
Okay so I think my question is: If $g_y$ is continuous and for none of the $y$, $g_y$ is inclusion. Is there a way to construct a continuous family $h_y$,which contains each $g_y$ and the inclusion?
because if $g_y$ is already continuous family and for some $y_0$, $g_{y_0}$ is inclusion then the result is clear
@user1294729 I am not going to do your work for you. If you do not want to understand the way I suggest, then just do your chain rule stuff carefully and follow your formula. I would recommend you try a simpler example first.
@monoidal No. What I suspect they want you to do is compose with the inverse of $g_{y'}$ in the appropriate way so that you then get the inclusion. But you really need to ask the person who wrote the question, not us.
@TedShifrin no I don't want you to do the work for me. But I want to understand if I have the formula for $\beta'(t)$ as I wrote above what does it helps me because I don't see this formula in your script and therefore I'm confused
Yeah, the person that wrote the exam is no longer at the uni and I asked the one teaching the course and he said he'll get back to me, but it's been 3 weeks now
That formula appears, as I said, on p. 14, @user1294729, except I differentiated $\alpha$ rather than $\beta$ because that is physically more natural. $\phi(s(t))=t$ is the connection and so you can differentiate by the chain rule to get $\phi'(s(t))s'(t)=1$. $s'(t)$ is the speed and is the meaningful thing to think about physically, rather than the inverse.
and where do I see this connection? I really have no pre knowledge, we only had one lecture.
I am really lost, I read your pages but I don't see any connection to my knowledge because you use this freemen formulas also on page 14 and we also have not seen those yet @TedShifrin
Are scalar triple products necessary? Surely there's other ways to compute the volume of a parallelepiped right? Why did I learn it? This math ain't mathing
@user1294729 I defined $\alpha(t)=\beta(s(t))$ and you defined $\beta(s) = \alpha(\phi(s))$. Compare those equations carefully.
Hades: Nothing is necessary. If you know about determinants in 3D, then feel free to compute the volume that way. It turns out it's identical to using the scalar triple product.
Physics uses scalar triple products all over the place.
