I will share my work (if I am posting too much, then just delete, I guess)
I've just been playing around with graphing software, and it looks like my bounding functions are bounded by $y = 1$, although it's not clear to me why this is necessary, or even if this is what my professor is asking me for.
I understand if nobody wants to read three pages. I'll send a message to my professor. I guess it's obvious that my bounding functions are bounded by $y = 1$, since they're all $x/(1+x)$ times e^[negative number].
Does any body know how to use ChatJax for rendering LaTeX equation in Chat? I tried to follow the instructions in the link to $\LaTeX$ in chat but to no avail.
@OliverDíaz Sorry, I have it working on Windows 10 and Mac OS. I just do what I said. Not sure what other advice to offer.
@TedShifrin It's not integrable, but I don't see where this is required. I am using the definition of a uniformly bounded family of functions from Wikipedia. Maybe he means that the integrals need to be uniformly bounded, in which case I'm probably toast, because the integral of my $g_m$ is $m + 1$.
Yes, well, I thought I needed to make the theorem apply over $(0, \infty)$, which is why I considered all those intervals and functions. There must be something very obvious I'm missing, because I have no idea what's going on.
@Novice: with regards to your problem (differtiability of $t\mapsto \int^\infty_0\frac{e^{-tx}}{1+x}\,dx$, choose an interval $[a,b]\subset(0,\infty)$ for $t$ and the result in your notes. That way you get differentiability in the interval $(a,b)$.
@Novice: $\frac{|e^{-tx}-e^{-t_0x}|}{|t-t_0|}\leq |x|e^{-\xi x}\leq|x|e^{-cx}$ for all $t$ in some interval $[a,b]\subset(0,\infty)$ that contains say, $t_0$.
@OliverDíaz Thanks, but I have to think more about this. I've been going back and forth with my professor over email, and chatting here, and I don't understand anything anybody is trying to tell me.
I may be mistaken, but my professor seems to think that I can use a single bounding function $g$, which doesn't seem possible, unless I've overlooked something terribly obvious.
@Novice: I don't think there is anything more to it. since you are dominating the expression $\frac{|e^{-tx}-e^{-t_0x}|}{|t-t_0|}$ by a function $g(x)=xe^{-cx}\in L_1(dx/(1+x))$. Recall that differntiability is a local thing, so to check differntiability at any point $t_0$, you only need to control what happens near $t_0$. Of course, there may not be a "uniform" dominating function $g$ that works for all $t$'s, but that is not what one needs for differntiation here.
@TedShifrin That is what I was taking about. A bound for each bounded closed interval. That will do what @Novice needs... (but of course you se that)...
@robjohn: I just got it to work with Safari on my new M1-processor computer. I just have to drag it to the favorites bar (maybe bar is to the name but I think you know what I mean).
Let $X\sim_\alpha Y$ when $\alpha X \cong \alpha Y$ and $X\sim_\beta Y$ when $\beta X \cong \beta Y$ where $\alpha$ is one-point compactification and $\beta$ is Stone-Čech compactification
Can we say something about those equivalence relations
Is there a special name for First-Order Logic with all terms restricted to variables (i.e no function terms)? Is it simpler to decide satisfiability in this case?
in the puzzling chat we have identified a five-tuple of words that solves over 95% of wordles, i.e. you can play the same five words every day and be guaranteed to win on the sixth guess, with that high probability. if your dignity can take the hit of averaging 6 guesses.
my own average has dipped just below 4.
it's SHALE/POINT/CURVY/GAMED/BEEFY.
olivia came into my office a minute ago meowing. she exercised her claws on my chair and rubbed my leg and then left for my daughter's room. i don't know what she wants in there.
I use the same first word every time. But I change up the second word according to the letters I have right in the first. My average is almost exactly 3.75.
I play in "hard mode" almost exclusively, so I have to reuse correct letters. I think I've broken that rule only once.
why is complex analysis considered the good twin and real analysis considered the evil twin? [even though my intuition tells me that jumping from ℝ to ℂ was supposed to make things harder?]
many of the hypotheses that appear in the beginnings of complex analysis books have far-reaching consequences that make the world of stuff satisfying those hypotheses tractable.
many of the hypotheses that appear in the beginnings of real analysis books do not.
The complex derivative puts such a structure on the functions that good things follow. The real derivative does not impose this same structure and as a result there are many real differentiable functions that can behave badly.
Of course, essential singularities are the black holes of complex analysis.
for any result in finite dimensional linear algebra, there are a zillion proofs, and two of them push the symbols in just the right way so that they generalize. and you just go and find those proofs.
if you can draw a picture of what the argument is trying to do, that's a sign it won't work in infinite dimensions.
there's a piece of those arguments that doesn't work. when you try to find something nonzero and orthogonal to a set that you only know isn't everything. that set could be dense, and still not be everything.
other than that, yes.
i didn't phrase that well. but sometimes your hypotheses that give you enough room in finite dimensions don't give you enough room in infinite dimensions.
it's a good thing. it gave me a job for a few years.
it didn't pay very well, which is why i went into suing people, but, that's another story.
ted: my daughter reacted to a touchy moment this morning by accusing her mother of being a bad mom. "you're a BAD MOM." she said this several times. it was in response to not being allowed to watch the garbage truck pick up our garbage. how would you handle?
the problem is that by my own existence i have established a precedent of talking s--t basically all the time.
My professor released my final grade, so I suppose he's not going to reply to me anymore. I suppose that's the end of my saga. Still don't see what was wrong with the answer I submitted.
