@DHMO, if you are trying to find $\arctan$ or $\arcsin$ or whatever, it is just the inverse of $\tan$ or $\sin$, right? So if you have 25 for $\sin$, than $\arcsin$ would be $1/25$, right?
@Mahmoud Suppose that it's true for some $k\in\Bbb N$, i.e. $5^{3^k}+1=m(3^k+1)$. Then, $5^{3^{k+1}}+1=5^{3^k\times3}+1=(5^{3^k})^3+1=(m(3^k+1)-1)^3+1$
it means, that if $\sin\left(\dfrac\pi3\right) = \dfrac{\sqrt{3}}2$, then $\arcsin\left(\dfrac{\sqrt3}2\right)=\dfrac\pi3$
inverse function
that means, if we look like functions as machines that takes an input and gives an output, then the inverse function is the anti-machine that takes the output and gives the input
@Mahmoud Suppose that it's true for some $k\in\Bbb N$, i.e. $5^{3^k}+1=m(3^{k+1})$. Then, $5^{3^{k+1}}+1=5^{3^k\times3}+1=(5^{3^k})^3+1=(m(3^{k+1})-1)^3+1$
@DHMO, oh, okay. But then why is this used in trigonometry? I looked it up on Khan Academy and it is calculus. Doesn't trig come before calculus? Because the problem came up on Khan Academy's trig course, and I tried to look at the hints that explained the problem, but they were rather confusing.
@Mussulini If you can see that it does, what kind of extra explanation are you looking for?
It can also be seen from the functional equation $\Gamma(z+1) = z\Gamma(z)$. Divide by $z$: $\Gamma(z) = \Gamma(z+1)/z$, then think about the limit $z \to 0$, then the limit $z \to -1$, then $z \to -2$, etc.
I don't get why it diverges for negative integers but not for all negative reals, the graphs of $\frac{x^a}{e^x}$ look similar if a is negative whether or not it is integer
Since I am trying to get the bounty room going, perhaps a reasonable thing could be to adverties it here. So here is link to the relevant meta post and here is the room.
A quick query: If I were to distribute 4 vectors (of equal magnitude) around a point in space (so they're co-initial) how'd I go about doing it? (Would I have to position each vector at an angle of 109.5 degrees from each other? )
and I wondered if anyone here had a spare moment to help me?
The problem is that I have an image with lines drawn on it. I am using a mapping tool much like google maps which tiles images of the current map on the screen. Each square is drawn individually so there may be times that a line intersects the current square. There may also be occasions where the line could be contained in a square. Lines can be in any direction.
I have the start and end points for each line
for the whole drawing
But I need to calculate the start and end points of the line for the given tile.
I've been thinking about this problem so much I can't sleep at night and when I do finally get some sleep I dream about it
So if boundaries of rectangles are line of the form x=x0 (vertical) and y=y0 (horizontal) then finding intersection between this boundary line and arbitrary line should be rather easy.
You simply need to find the point on the line with x-coordinate equal to x0. Similarly for horizontal boundary lines.
Ok, what I am trying to say (and you can probably transform it to the way you need), that if you have a square given by corners (x0,y0) and (x1,y1) and you also have line given by two endpoints (a,b) and (c,d), then it is relatively easy to find the intersection between the line and the square.
but then because the lines (which is determined by the content of the floor plan) are random, there is no analytic equation to check whether they intersect the grid. Thus the only thing I knew is that you need to somehow get an array of data for that underlying image before these can be checked with the gridlines to see if there is an intersection?
BTW @PrimeByDesign if you want to get somebody's attention, you can use pings - in the form @username.
Given two points (a,b) and (c,d) you can get easily the equation of the line between them. For example, you can write it as (y-b)/(x-a)=(d-b)/(c-a).
You can also easily check whether the endpoints are in the give tile.
If not you simply want intersections of the given line with the lines x=x0, y=y0, x=x1 and y=y1, which for the boundary of the square.
However, this is probably oversimplification of your original problem, since I am talking about straight lines and you probably draw Bezier curves or something even more complicated.
BTW I am not sure what are you using, but probably any decent library for drawing lines and curves should already contain something like this.
BTW @PrimeByDesign you can probably find some stuff about this online. For example, this SO post line-rectangle collision detection and also the other posts linked there seem related to your problem. (Or at least to my simplification of your problem.)
Hey guys I have a elementary calculus problem: Suppose I want to show that the arbitrary constants of an integration does not matter: $$\int uv' dx=u(v + C_1)-\int u'(v+C_2) dx$$ $$=uv + uC_1-\int u'v+ u'C_2 dx$$ $$=uv + uC_1-\int u'v dx- (u + C_3)C_2$$ $$=uv-\int u'v dx + u(C_1-C_2) - C_3C_2$$ But I got an extra arbitrary constant that is multiplied to u. Is there a reason why $C_1=C_2$?
in particular, there is a unique one that is both compatible with the holomorphic structure in the sense that its $(0,1)$-part is $\bar\partial$ and compatible with the Hermitian structure in the sense analogous to metric connections
Hey! (Sorry to interrupt.) A short question. Is there consensus on the definition of a directional derivative of a vector field? Say $$\vec{a}(x_1,\ldots,x_n)=\sum_{i=1}^{k}f_i(x_1,\ldots,x_n)\vec{e_i} $$. Is the directional derivative (a.) the Jacobian matrix of $\vec{a}$ multiplied with a vector $\vec{u}$ of any length; (b) or must $\vec{u}$ be a unit vector?
@LinearChristmas I've seen it both ways, too. It seems to make a bit more quantitative sense when considering unit vectors but there may be good reasons for both and that's a bit above my pay-grade. Seems like something where you'd just defer to the author's convention
@DHMO, okay. I looked at your comment and didn't quite understand what was meant by $(90, -90)$ (or the interval bit, for that matter). So what exactly does that mean?
@Fargle: that is what I thought. Thanks! In the mean time I also read Wiki's talk page on the issue where there is the quote: '[---]if you say "derivative along a vector", that vector does not need to be of length one. If you say "derivative along a direction", then yes, a direction by convention is normalized to length 1.'
oh, sure, sometimes doing problems without reading is unreasonable. but it's a great great feeling that will actually stick to you when you try a problem for some time and then go back, stumble upon something in the text that solves it
maybe if you just kept reading, you'd shrug that something in the text off as irrelevant and obvious
So eh.. Consider $\Bbb R$-coefficients, then by universal coefficients cohomology is dual of homology. From the SES for $(M,\partial M)$ we get natural homomorphisms $H^{2n}(M)\to H^{2n}(\partial M)$ and the other way around on homology
Using the LES on cohomology and homology you have this commutative ladder
@Secret So is it equal to $w_2 \cdot 2$ where $w_2=2 \pi$ (the length of the unit circle ) and $2$ is the radius of the ball over which we are integrating?
