So what is Hawking radiation if virtual particles are a purely made-up phenomena? Why did Stephan Hawking becomes so accredited for that radiation that it's named after him? What about the Casimir effect's "vacuum fluctuations"? Did physicsts lie when they said they reported the Casimir effect...
The carbon rings in the upper-middle of this page https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/react3.htm such as corannulene or coronene possess symmetries. But, they are not the typical dihedral arrangements of points, like a single hexagon or single pentagon or single equilateral...
Much of the theory of ordinary differential equations is based around continuous derivatives. A lot of nice theories came together with semi-group theory of linear systems and the Banach contraction theorem, but these are limited to continuous functions. Then you get into partial differential...
I don't think there's any implication of ##x = f^{-1}(x)## unless ##x## is identically the identity function. Both ##f## and ##f^{-1}## may have overlapping domains.
Instead of trying to "provoke" people which is clearly rude, you have a few options:
1 is, don't comment on a question you don't like, go do something else with your life
2 is be patient and understanding, work with the poster.
I think that's irrelevant to trying to manipulating someone with...
What I proposed is highly conventional and extends to non-function objects (operators), so I'm confused by what you're saying and still don't see a concise answer.
Consider an invertible linear transform ##T##, like say a square matrix. Then we may calculate ##T^{-1}\cdot T[x] = I x.## Notice...
Okay, after putting down the computer screen and having a chance to figure out what's going on in this problem, I figured out the crucial step that was left out and causing me confusion.
We start with ##\frac{d}{dx}y(x) = (ax + b)y(x)##. Then we make a substitution ##u = ax + b \implies...
The ability to learn is equally as important as the ability to communicate, and communicate respectfully at that.
The derivative of ##\frac{d}{dx} log(x)^2## is ##\frac{2log(x)}{x}##.
The derivative of ##e^{\cos(x)}## is ##-e^{\cos(x)} \cdot \sin(x)##.
My issue isn't that a don't understand the...
There's intermediate steps that you're leaving out. My substitution appears to be different than yours, so keep that in mind.
So, we start with the equation
##\frac{d}{dx}y(x) = (ax+b)y(x)##. Then I make the substitution ##u = ax+b## which implies ##x = \frac{u-b}{a}##.
our equation becomes...
##d/dx f(g(x)) = f'(g(x)) \cdot g'(x)##
That's the chain rule, that's what I'm familiar with, no problem there.
Going back to the original problem, we have
##\frac{d}{dx} y(x) = (ax+b)y(x)## a non-autonomous differential equation.
Now I say ## u = ax + b## which also implies ##\frac{u-b}{a} =...
Right, I'm aware of the chain rule, but it's ambiguous if I'm applying it correctly, and I also don't understand the intuition behind saying ##y(x) = z(ax+b)##. If ##y(x) = z(ax+b)##, what happens to the ##ax+b## coefficient of ##y## in the original equation?
Sure, I can brush up on multivariable calculus at some point.
But, this still leaves me wondering about the original method. Can the original substitution I used work? Or is it intrisically flawed? What went wrong in my original approach? I still don't see concise answers to those questions.