Hiya all, my two daily downvotes have come in. math.stackexchange.com/questions/480129/… both the question and answer here got downvoted - would love it if you guys took a look
Hi guys, quick question, if I subtract two numbers with tolerances how do I calculate the tolerance, is it the same as when adding numbers? (square, add, square root): x±0.04-y±0.04, would the tolerance be 0.0566?
Suppose I have the frequency $f$. According to just intonation, $\frac{3}{2} f$ is a perfect fifth.
Now compare the following:
$$ \left(\frac{3}{2}\right)^n \neq 2^m$$ for any integer $n,m$
But $$ \left(2^{\frac{7}{12}}\right)^{12} = 2^7 $$
Supposedly this difference explains why just inton...
Suppose I have the frequency $f$. According to just intonation, $\frac{3}{2} f$ is a perfect fifth.
Now compare the following:
$$ \left(\frac{3}{2}\right)^n \neq 2^m$$ for any integer $n,m$
But $$ \left(2^{\frac{7}{12}}\right)^{12} = 2^7 $$
Supposedly this difference explains why just inton...
I am told that a form of musical intonation called just intonation
Doesnt permit keys
For our purposes
I will call a key set a group of scales where each member of the group can be obtained by multiplying by a constant. For example, C# major is simply C major multiplied by $2^{1/12}$
So a musical key is a member of a key set
But I am confused because apprently one form of tuning ratios permits keys while another doesnt
@MikeMiller and i am fairly sure the explanation involves a simple mathematical property
Suppose I have the frequency $f$. According to just intonation, $\frac{3}{2} f$ is a perfect fifth.
Now compare the following:
$$ \left(\frac{3}{2}\right)^n \neq 2^m$$ for any integer $n,m$
But $$ \left(2^{\frac{7}{12}}\right)^{12} = 2^7 $$
Supposedly this difference explains why just inton...
@Silent Well, $n! = n(n-1)\cdots 3\cdot 2\cdot 1$ so if you divide by $q!$ (remembering that $q$ is fixed), you get something that is strictly larger than $(q+1)^{n-(q+1)}$
@DanielFischer Yesterday you said that "For any reasonable integral, you have $$\int_X h(x)\,d\mu \cdot b = \int_X h(x)\cdot b\,d\mu$$ whenever $h$ is a complex valued integrable function on $X$ and $b$ and element of a complex vector space."
Assume you have $a \in \mathcal{A}$, where $\mathcal{A}$ is Banach Algebra and a real valued function $f(x)$ on $[a,b] \subset \mathbb{R}$ then would you have $\int_{a}^{b}f(x)dxa = \int_{a}^{b}f(x)adx$?
I do some multivariable calc now. the idea being that I can tell SB that I now want to look at manifolds. I think that's a believable thing, and it's actually true.
Yes, it is definitely possible to learn differential topology without any strengthening of your foundations in analysis. In particular the content of "baby Rudin" is essentially irrelevant: rearrangement of series and gamma functions won't help you to understand, say, De Rham cohomology or Morse theory, nor even to compute in the tangent bundle! As an aside, I don't understand the American infatuation for that book, which I find boring, needlessly difficult and concise and without any show of enthusiasm for calculus, one of the most exciting and glorious achievements of mankind. — Georges ElencwajgDec 20 '13 at 10:49
@iwriteonbananas Hatcher's first exercise on ch. 3.3 is very cool. It says that if you drop the Hausdorff property for manifolds, then find an example of a 1-dimensional nonorientable manifolds :D
I have a question about the following lemma:
Assume that the characteristic of $F$ is $p$ and $p>2$.
Then $(t^m-1)/(t^n-1)$ is a square in $F[t, t^{-1}]$ if and only if $(\exists s \in \mathbb{Z}) m=np^s$.
($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the fi...
@DanielFischer But Riemann integration is defined for real valued functions, how can you just dump an $a \in \mathcal{A}$ in the integrand i.e. $\int_{a}^{b}f(x)adx$?
@DanielFischer Are you considering it as $\lim\limits_{i \to \infty }\sum a f(x_{i}) \Delta x $?
@Moses You need to extend the definition of the Riemann integral of course. The concept of a Riemann sum has an obvious extension. Then you define a function to be Riemann integrable if the limit of the Riemann sums [Caution, it's not a sequence, you need a net] exists, and the Riemann integral is that limit.
@DanielFischer It's not really a Riemann integral then it's some extension. When you have a chance could you please look at my last comments for the post. This is quite confusing...
I've just started reading Rudin's mathematical analysis and I'm a little confused by the proof in example 1.1 on pg 2 where we need to show that $p^2=2$ is not satisfied by any rational p. It's assumed that p can be written as $p=\frac{m}{n}$ where m and n are integers that are not even. Does he suppose this so that p remains a rational? Can someone please explain this?
ugh, @MikeMiller, seems like there's some subtle thing about continuity of multiplication on $G$ is involved. if I can get this right, I have a commutative diagram which ensures local consistency.
give me some more time, I am not very comfortable with Lie groups
yes, I completely get what you're trying to say. Maybe $X \vee Z$ is equivalent to some wacky adjunction space such that the attaching map was nullhomotopic. Who knows if that adjuction space isn't $Y \vee Z$?
So far I received no solution to $$\int _{0}^1\int _{0}^1\int _{0}^1\frac{1}{(x+y+z) (1+x y z)}\ dx \ dy \ dz$$ If I were a student at the famous universities in the world I would stress the professors every day with such questions.
@Balarka: Here is a strategy to obtain a counterexample. Find groups $G,H,H'$ that admit finite CW models for $K(-,1)$ and such that $G * H \cong G *H'$. (There are finitely presented examples of groups that don't cancel like that.)
I would challenge the whole board of professors, no matter it's about MIT, Princeton, Harvard, every day I would challenge them (with my creations, of course).
More precisely, once $1 \in G$ is given a local orientation, I want to get the continuity of orientation work for an open nbhd $U$ of $1$, and then push that neighborhood to a neighborhood $U_g$ of $g$ by multiplication by $g$.
Then some crazy commutative diagram work does the trick.
That is, if my orientation is continuous around $1$, then it's continuous everywhere. I am not sure how to make it work around $1$
Oh, I see. As usual I came to nag you about diff. geo, Mike, albeit probably a bit easier this time: the curve $\gamma \colon I \to \mathbb{R^2}$ defined by $t \mapsto (t^2, t^2)$ is not extendible, right?
You have a map $(0,\infty) \to \Bbb R^2$ and you want to know if you can extend it to, say, $(0-\varepsilon,\infty)$ in a $C^1$ manner that always has nonzero derivative?
A vector field along a curve, $V \colon I \to TM$ s.t $V(t) \in T_{\gamma(t)} M$ is extendible if there is a vector field $V' \colon M \to TM$ s.t $V(t) = V'_{\gamma(t)}$
I am going to be road tripping and living out of my SUV for a while. I need a way to power a tiny portable cooler/fridge 24/7 and would like to be able to power my cell phone, laptop, and maybe a few other things when the vehicle is off.
I have been looking into the possibility of mounting a sol...
Ok, no, this makes perfect sense. $\dot{\gamma}(-)$ is a vector field along $\gamma$. Suppose there is a vector field $V' \colon M \to TM$ s.t $\dot{\gamma}(-) = V'_{\gamma(-)}$.
Since $\dot{\gamma}(-1) \ne \dot{\gamma}(1)$, we do not want these mapping to the same tangent space, right? Which is exactly the case under the assumption.
Sorry for interrupting the conversation, but it seems like a charming idea to write vector fields as paths on the tangent bundle. Is there any interpretation of $\pi_1(TM)$ in that way?
@BalarkaSen: That's not what a vector field is. A vector field is defined over the whole manifold.
Not over a curve.
@AndrewThompson: Yes, I agree with the above, and it trivializes the problem. But the thing should still not be extendible even if you remove silly obstructions like this.
Your vector field, along $x = y$, is $(x,y) \mapsto (\sqrt{x},\sqrt{y})$.
The claim is that you cannot extend this in a smooth way. But to do this it would suffice to show that there is no smooth extension of $\sqrt{x}$ to all of $\Bbb R$.
Mathematical operations on binary trees represented by binary numbers, to transform them? For example if x = 1.001011 = (()()) then x + (2^7)x = (()())(()()), a couple more calculations and ((()())(()())) which is a valid tree again, although this is a very trivial adjustment that in this case is probably better done with nodes connected by references/pointers, something like square root just scrambles it and makes it invalid, I'm thinking of making a SO question perhaps to look for useful ones.
Ohh, maybe I could ask here instead, what operations (factorial, powers, reciprocal, n chooses k, etc...) on all, or just fractional binary numbers between 0 and 1 where the number of 1s in the number is equal to the number of 0s, preserves this equal ratio of 0's and 1's.
@TedShifrin If I leave now, it may take an hour and a half to get to UCLA, and that is just a bit before your 12-2. If I get there at 2, I need to leave by 3. It looks a bit tight today.
Hello all! Anyone here know a bit of a history? I just came across something that makes no sense to me as to how things progressed in this way (if what I've read is correct; I've not been able to find supportive evidence nor evidence to refute the claim...)
Euler proved (formally) his product formula for the harmonic series in $1737$. Riemann wrote his revolutionary paper in $1859$. Then Leopold Kronecker proved $\sum n^{-s}=\prod_p (1-p^{-s})^{-1}$ in $1876$? Is this really true? If so, how did it get that way?
@ted Alright I see what you mean now... The easiest way I know to find where its coordonates $(\alpha, \beta)$ given by $\alpha = \frac{-b}{2a}$ and $\beta=f(\alpha)$
@Ramanewbie I don't get your question at all . you have a quadratic equation so it should have two roots (complex or real(by fundamental theorem of algebra)). There is an added condition according to which ac<0 implies discriminant will be positive and hence distinct real roots which are $\alpha$ and $\beta$
@AlecTeal That may be tolerated in chat (depends on who's present), but when a comment containing those is flagged, the comment goes. And if you do it too often, you may be sent on a vacation from the site.
When someone's being really pedantic or weasel-like @TobiasKildetoft I could call them "a rimmer" but I think that might go down badly :P (Flaggers sid off, Rimmer is a character)
@DanielFischer we really need to sort this though. I need some approved terms to use.
Suppose I have the frequency $f$. According to just intonation, $\frac{3}{2} f$ is a perfect fifth.
Now compare the following:
$$ \left(\frac{3}{2}\right)^n \neq 2^m$$ for any integer $n,m$
But $$ \left(2^{\frac{7}{12}}\right)^{12} = 2^7 $$
Supposedly this difference explains why just inton...
I have a question about the following lemma:
Assume that the characteristic of $F$ is $p$ and $p>2$.
Then $(t^m-1)/(t^n-1)$ is a square in $F[t, t^{-1}]$ if and only if $(\exists s \in \mathbb{Z}) m=np^s$.
($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the fi...
