This is the beginning of the Frobenius Theorem/criterion for a subbundle of the tangent bundle to be integrable (i.e., for there to be a $k$-dimensional foliation whose tangent spaces are given by the subbundle).
so far i've tried repeated l'hopital's. i tried l'hopital's and then try to fit the special trig limit of (1-cos x)/x, and something else which i don't remember.
I don't want to share the way I work it up because I can't prove that what I do will always work, though it does work here if you're interested in an infinitesimal argument.
Surely some will recoil in disgust, but I take $\infty = 0^{-1}$ and $f(x+0\cdot y) = f(x) + f'(x)\cdot 0\cdot y$ as axioms. This leads to: $${{\sqrt{2-2\,\cos \left(2\,\pi\,0\right)}}\over{0}}$$ $${{\sqrt{2-2\,\left(1-2\,\pi\,0^2\right)}}\over{0}}$$ $${{2\,\sqrt{\pi}\,0}\over{0}}$$
@Jeff The intuition comes from $f(x)$ and its tangent line at $x$ being equal over the set of numbers that are equal to $x$, lol. A more reasonable argument follows from a bastardization of the definition of the derivative
Hello everyone, how to find x through a? https://www.symbolab.com/solver/step-by-step/%5Clog_%7B6%7D%5Cleft(2%5Cright)%3Da%2C%20%5Clog_%7B3%7D%5Cleft(6%5Cright)%3Dx
Only difference being that the difference between the (squares of) the two sides can be represented without using angles as one would do in the R^2 case.
@0celo7 1) I am not going to write an essay on why your tone is not right. 2) Clearly more than one person has been annoyed by you(r tone?), so I thought it was appropriate to mention as you don't seem to realize why we are annoyed. 3) It is your job to figure out and fix what's wrong with your behavior.
Need to get x = 1/(1-a) from here: https://www.symbolab.com/solver/step-by-step/%5Clog_%7B6%7D%5Cleft(2%5Cright)%3Da%2C%20%5Clog_%7B3%7D%5Cleft(6%5Cright)%3Dx
but what i've tended to work on is stuff that can be broadly described as 'applications of asymptotic analysis'
I say 'applications of' because I don't really care about more foundational stuff. I'm more interested in "Hey, what kind of answers do I get if I apply these techniques?"
It's related, yeah, though steepest-descent itself is much simpler
Suppose you've got an integral like $\int_{-\infty}^{\infty} e^{-s f(x)}\,dx$
and for simplicity suppose that $f(x)$ is some smooth convex function with minimum $f(0)=A$ at 0
then that exponential function will be peaked at $x=0$ and decay away from it. moreover, the larger $s$ is, the more sharp that peak
in that case, one can approximate $f(x)=A+f''(0)x^2/2$ and get the estimate $e^{-s A}\int_{-\infty}^\infty e^{-s f''(0)x^2/2} = e^{-s A}\sqrt{s f''(0)/2}$ (I may be getting the Gaussian wrong)
that means we've managed to get an asymptotic estimate (i.e. the behavior as $s\to\infty$) of that integral in terms of just $f(0)$ and $f''(0)$
following up, when you do the same idea in the context of path integrals in physics, you get a leading 'classical' contribution plus a first quantum correction
and the requirement that one evaluate the contributions at the 'critical point' in this context leads to the classical equation of motion
Let $f:S^1\to S^1$ be smooth, then there is a smooth $g:\Bbb R\to\Bbb R$ such that $f(\cos t,\sin t)=(\cos g(t),\sin g(t))$ and $g(t+2\pi)=g(t)+2\pi q$, for some $q\in\Bbb Z$
