No, @masfenix. You just generalize spherical coords inductively, but you don't need an explicit formula as long as you realize it's of the form I gave you. What does Wikipedia offer?
@Ted it's more trouble than what it's worth. Have you ever had that happen to you?
Just let her calm down and try to talk to her again @MickLH
user4704
@MickLH You should probably call somebody you know back in whatever country she is in, and ask them to get her an ambulance. I don't think that SE chat is really the best place to get medical advice like this, beyond "call the police."
@TedShifrin before I go to bed do you recommend any books on Linear Algebra, I keep finding ones that are too basic, or too advanced (Like Roman's linear algebra book, or Serge Lang's) I feel like goldy-locks
@TedShifrin any more hints, im still stuck on this integral: $$\int_{\mathbb{R}^n} (1 + |\xi|^2)^{s/2} $$ I want to show that this converges when $s < -d/2$ through polar coordinates
Im having trouble write the gcd(x^3+x^2-x, x^5+x^4+2x^2-x-1)=1 as a linear combination. What I have so far is $ x^5+x^4+2x^2-x-1=(x^3+x^2-x)(x^2+1)+(x^2-1)$ $x^3+x^2-x=(x^2-1)(x+1)+1. I tried back substituting but it can't seem to work. The division is correct.
@PedroTamaroff okay so the improper integral is $$ \int_0^\infty (1 + r^2)^{s/2} r \, dr = \lim_{t\rightarrow \infty} \int_0^t(1 + r^2)^{s/2} r \, dr $$
@TedShifrin so I don't need really the explicit formula, I just need to look at $$\int_0^\infty (1 + r^2)^{s/2} r \, dr $$ and see how it behaves for large r$?
@PedroTamaroff so for $n=2$ I had $$\int_0^{2\pi} \int_0^{\infty} (1 + r^2)^s *r \, dr d\theta$$
the $1+r^2$ behaves like $r^2$ for large $r$, so we have the integrand which is $r^2s * r$ and it will only converge if $-(2s + 1) > ??$ what should i put in ?? is it 1 or 2?
in any case, it dosnt match what I need to show, that is $s < -n/2$
Can anyone help explain how to go about solving dr/dt = A/r^2 for t where A is a constant?
I've found a few questions on stack physics discussing it as it comes up when trying to find out long bodies take to collide due to gravity but I don't understand some of the intermediate steps
that was d^2r/dt^2 on the left sorry
acceleration equals gravitational constant and mass over distance squared
I found someone describing how to solve a problem like mine on a forum but I don't know how they're going between some of the steps physicsforums.com/showthread.php?t=71026
they start the same as I do but I basically lose track of what they're doing immediately
I'm not sure how they get from $\frac{d^2r}{dt^2} = \frac{A}{r^2}$ to $\frac{dr}{dt} = -2\sqrt{2A}\sqrt{\frac{1}{r} - \frac{1}{R}}$
@0x5f3759df Funny thing regarding that link. I run into the same problem on $r^2\frac{d^2r}{dt^2}=A$. I am prone to spending hours trying to find closed solutions to nonlinear ODE's that demand a numerical solution, but I am at a loss here. I suspect that this demands numerical integration, but would really love for someone to show the alternative. It meets no form I am completely familiar with.
if I integrate the right side with respect to t I get $\frac{-A}{r}$, and it looks like they're integrating from r to R or something which I don't understand.
it looked like there was a closed solution at the end of the first reply
Yeah the answer in that post is reasonable, he is effectively just solving the equivalent of $\frac{dr}{dt}=\frac{A}{r^2}$ and then introducing and applying an initial condition. I am at as much of a loss as you though as to how this would actually take your $r''$ and conclude with a solution $r$.
I took a master's degree in applied math, although out titles were probably different, so yes? Anyway, you may want to address the person who actually asked the question, I am just drinking beer and chatting. Not my question.
Any input would be thoroughly entertaining to me though, so fire away!
@usukidoll It is a total blast man! I miss it! I just graduated last May, and have been teaching at the junior college, but I am not quite getting the high that I got from school. I think I need to go back.
Man @0x5f3759df why not sign in to math.SE, link your accounts and post this as a really great question. I will be the first to upvote, and may even give response. It sounds like @usukidoll might have more insight into solving this second order nonlinear ODE. He will probably post a response after he eats his dinner.
one of the physics answers came to the same (incorrect) answer that I got
some of the other others invoked conservation of energy and other physics things I don't really want mixed in with my math and any of them that were actually successful I just don't understand
I haven't taken a formal differential equations course (only calculus) but I had been hoping this would be an easy equation because it seems like $f'' = f^n$ would be the simplest case of a second order nonlinear ODE
Anyone can post, just show all your research, and give references. Post it as a purely mathematical question with reference to the physics.
I like the question of solving $\frac{d^2r}{dt^2}=\frac{A}{r^2}$ as a general question with all of your research. A post would give this a broader forum.
I wager two pennies that there is no known closed solution, but sometimes there are more important things than a closed solution. For example a complete stability analysis. Just knowing how it behaves over parameters and initial conditions. This we can do.
Again, I am really aching for someone to show I am wrong here. :))
Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values however)
Plane inside cylinder. The portion of the plane $z = -x$ inside the cylinder $x^2$+$y^2$=4
getting my stuff hold on
so we need to convert to cylindrical coordinates/polar coordinates
$x=rcos(\theta)$ $y=rsin(\theta)$ and z = z
we know that $x^2+y^2$ is $r^2$ in polar coordinates
so $r^2$=$4$ which makes $r=2$
I have to use the area of the surface formula which involves paramentrization and cross product
the $r(r, \theta) = rcos(\theta)i+rsin(\theta)j-rcos(\theta)k$
because $z=-x$ the k is $-rcos(\theta)$
Now I have to take partial derivatives in terms of $r$ and $\theta$
@0x5f3759df you know that in that post there are solutions for say $b=1$, or with $b=2$ the Weierstrass P function could come into play. I am especially curious about what people have to say for integers $b<0$, and in your particular case $b=-2$
@usukidoll I'm sorry, you're not being ignored. I know all about what you are writing, but I am sort of fixated on solutions to $\frac{d^2y}{dx^2} = ay^b$ in particular for negative integers $b$
@0x5f3759df you might want to tighten the scope of your question to ask the question of $b=-2$. It is sort of broad the way it is stated. Nothing wrong with the broad generalization of all exponents $b$, but the post might not wind up giving a definitive answer to your original question.
I already numerically integrated this thing, and I know we can all find exact solutions for say $b=1$, but the question of $b=-2$ steps into the territory of the unknown for me. This interests me.
@usukidoll I am sorry I was fixated on this differential equation. Are you asking if it solves that?
Well, no, but if you got the right answer I would probably name a child after you. I don't think there is a closed solution to that post for $\frac{d^2y}{dx^2} = ay^{-2}$.
k.. so the problem I'm doing is similar to this https://docs.google.com/file/d/0B54O0sv0MQILYTdqQlU3WHRtUU0/edit Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values however)
Plane inside cylinder. The portion of the plane $z=−x$ inside the cylinder $x^2+y^2=4$
@0x5f3759df The values of constants have a superficial effect except in the case of $b$. In that case, it has a profound effect. This is the nonlinearity. It is profound.
@0x5f3759df The extent to which this thing is even solvable revolves around that. That is why I think you might tighten your scope to the original physics quandry that generated your question.
I was expecting that the choice of b (aside from b=1 being simpler) would only change what value roots and exponents would take throughout the solution.
this is the problem again... Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values however)
For the complex version of the geometric series, $\sum_{i = 1}^\infty r^n e^{in\theta}$, isn't this equivalent to $\frac{1}{1-re^{i\theta}}$ given $0 < r < 1$?
I am reading a solution to a problem stating $\frac{re^{i\theta}}{1-re^{i\theta}}$ so I am a bit confused
I have a site question. Suppose a new user with small rep posts a question. I know they can always comment on their own question, but is it the case that they cannot comment on answers to their own question with insufficient rep?
if $x\in\bar{S}$ then: $\quad x\not\in{\rm int}\,S~\Leftrightarrow~$ no nbhd of $x$ is $\subset S~\Leftrightarrow~$ every nbhd of $x$ intersects $S^c~\Leftrightarrow~ x\in\overline{S^c}$
@N3buchadnezzar or you can easily see the solution of $f(x+1)=f(x)/p$, combine it with $\int_0^1 f(x)\mathrm{d}x= 1$ and then integrate over the required interval.
