  Members

7

## Community Reputation

0 Neutral

• Rank
Lepton
1. Okay, but is your name a hybrid of student-idiot, or stud-idiot, or stud-iota, or student-iota, or did you misspell studious, or something else entirely?
2. I don't know what "mathml" is, I use latex. It says there is a latex editor here but it didn't work the first time, so...
3. I'm wondering what the general process is for this. In general being well defined means there is only exactly one output for each input or that if x=y, f(x) = f(y) which looks like the reverse of proving injectivity, I don't know if that is a coincidence or not. Is a Fourier transform of a real function is still always real? I suppose the idea is that the imaginary component decays to 0 as you take the integral from -infinity to infinity so that it evaluates to a single finite real number, or actually, does the output of a the Fourier transform of a real valued function need to be real? W
4. Again there's no edit button for some reason so I have to waste another post saying f(g(t)) should have been f(h(t)).
5. That isn't the equation I meant unfortunately, unless possibly you are proposing that as a substitution. In the original equation, f(g(t)) is a composite function, and I represented its derivative under the chain rule as f'(h(t))*h'(t). Although alpha is a constant, it seems like you are missing the extra function on the outside of h+alpha to make f(h(t)+alpha).
6. Not quite because then you have an extra factor of h'(h-inverse(w)).
7. Suppose you have an equation (I don't see a latex editor or know how to use math tags here) f'(h(t))*h'(t) = f(h(t)+ \alpha) where f' is differentiated with respect to t, following from the chain rule on f(h(t)). Is there a substitution that will transform this differential equation into the form of f'(w) = f(w+\alpha) ? It seems reasonable but I am not finding an easy way to do it.
×