Antonio Vargas

oh neat

it's a tough decision to make

@Semiclassical come join me in oregon, we'll do consulting :D

tbh I wish I had learned more stats earlier. Bayesian stuff is actually awesome.

@Semiclassical permanent. I'm trying to do stats now, outside of academia.

yeah I "died" as Erdos would say

:)

hey semiclassical

^

nasa is streaming a launch on twitch

@Semiclassical greetings from the netherside

Is it just me, or are these kinds of questions almost as bad as "guess the next number" questions?

hey

@Daminark I'd avoid humor that might anger future readers trying to fill in the details

@Semiclassical Yo dude

I've definitely seen it a few times @Daminark

We thus expect that we can approximate the sum by approximating the slowly-varying factors of the summand near this peak at $k \approx x$: $$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!} L(k) \approx \sum_{k=0}^{\infty} \frac{x^k}{k!} L(x) = L(x) e^x.$$ In particular, this heuristic holds when $L(x) = \log x$ and $L(x) = 1/\sqrt{x}$, as in here and here.

If there is a peak term, it would occur approximately when $$\frac{x^k}{k!} L(k) \approx \frac{x^{k+1}}{(k+1)!} L(k+1),$$ which is the same as $$\frac{L(k)}{L(k+1)} \approx \frac{x}{k+1}.$$ But if $k$ is large (this will be true for large $x$), then $L(k)/L(k+1) \approx 1$, so we have $$k \approx x - 1 \approx x.$$

@GabrielRomon Suppose we have a power series $$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!} L(k),$$ where $L$ is a function satisfying $$\lim_{x \to \infty} \frac{L(x)}{L(x+1)} = 1.$$ If we want to determine the behavior of $f(x)$ as $x \to \infty$, one way to proceed would be to determine which terms of the series contribute most to its size.

Superpower

Savage!

But I like you.

You are ignored, Kasmir

@GFauxPas Yes, I think something went wrong somewhere.

@KasmirKhaan, not only are we handsome, we are literate too

@GFauxPas The problem itself. I don't see how $f(z^*) = f(z)^*$ is relevant to the problem at all, since it's not true in general for the functions the problem asks about.

@GFauxPas Try to see what happens in a simple case, like $f(z) = c_0 + c_1 z$.

@GFauxPas $f(z^*) = f(z)^*$, assuming $*$ means conjugate.

@GFauxPas If $f$ is defined by a power series, then it's true iff all the series coefficients are real.

@GabrielRomon The asymptotics chapter in Graham, Knuth, and Patashnik's Concrete Mathematics is a pretty good intro too, but covers less. You could also take a look at Green and Knuth's Mathematics for the Analysis of Algorithms.

I think de Bruijn's would be better as an introduction though.

@GabrielRomon Miller's Applied Asymptotic Analysis is pretty great. It contains lots of applications to differential equations.

I'm not sure what an image of a topological space would look like, but perhaps TikZ could do it? @gian

hello. quiet in here.

good luck!

I ask questions to help students

hope I wasn't too rude

sure thing!

pointwise, right

(less than 1 in absolute value)

Yes, since the other factor is |...| <= 1

Glad to help.${}$

Correct. But you have been given two different approaches where you can eventually use the DCT. The first was in my second comment, and the second was in Umberto's answer.

First, $g$ must not have an $n$ in it. Second, the point I'm trying to make is that there is no such $g$. It seems like it would be a good exercise for you to convince yourself of this. You can't prove that $\int f_n \to 0$ in this case because it's not true!

Indeed, you have not said what your $g$ would be! That's the dominating function.

Pointwise, sure. But what's the key part of the dominated convergence theorem? What's your domination function? The dominated convergence theorem is not just "$f_n \to 0$ pointwise implies $\int f_n \to 0$". There's a reason it has "dominated" in the name.

A hint: Substitute $y = nx$.

First, your justification for integrability is right, but it's pretty superficial. Do you know how to fill in the details? So what if $ne^{-nx} \frac{x^2+1}{x^2+x+1} \sim ne^{-nx}$? If $f(x) \sim g(x)$ and $g$ is integrable, what lets you conclude that $f$ is integrable? Second, just saying "apply the dominated convergence theorem" isn't enough. You actually have to apply it. If you did, you would find that the limit you guessed is incorrect.

this question is misguided for the reasons I outlined in the comments.

I posted this in the main chat but maybe it should have gone here instead: Is it just me, or are these kinds of questions almost as bad as "guess the next number" questions?