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location Buenos Aires, Argentina
age 21
visits member for 3 years, 10 months
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(my about me is currently blank)


Sep
10
comment Is it true that $ \sum \limits_{i=1}^{\infty} f(i) = \lim_{n \to \infty} \sum \limits_{i=1}^{n} f(i) $?
By the way, it doesn't make sense to ask if this holds for all $n$, since $n$ isn't a free variable.
Sep
8
comment $f(x)=f(x^2+ 1/4)$ , $f$ is continuous from $\mathbb{R}$ to $\mathbb{R}$
$f$ can't be one-to-one, since for example $f(0) = f(\frac14)$.
Sep
7
comment Problem with differentiation as a concept.
Small correction: The definition is either $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ or $\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$.
Sep
4
comment Integration of function
The binomial theorem proper only works when the power is a positive integer. You could get a series expansion, but otherwise I doubt this has a closed form. If it was a definite integral, you may be able to relate to the beta function using the substitution $u=\frac{a}{x}$.
Sep
4
comment Integration of function
Please make sure I did the LaTeX right. Is it $1+\frac{a}{x}$ or $\frac{1+a}{x}$?
Sep
2
comment Easy way to find the streamlines
If you're not supposed to do it rigorously, then sketching $\mathbf{F}$ at a few select points can give you an idea.
Sep
2
comment What's the Period of This Function?
Try graphing it, for particular values of $k$.
Aug
30
comment The Inequalities
I'm not sure I understand the question, and I don't have the book. Would you mind rephrasing it a bit?
Aug
24
comment why $x^2 = y^3$ is not smooth?
Usually you require that the derivative is never zero, which in this case happens at $t = 0$.
Aug
23
comment How to indicate keeping the sign when squaring a number.
While $x \cdot |x|$ certainly works, there's nothing wrong with the cases definition, and it's actually easier to understand.
Aug
16
comment Can $0\le\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau\le a$ imply $\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau$ exists?
The issue here is that you ask if from $0 \le \lim \int_0^t f \le a$ we can deduce the limit in question ($\lim \int_0^t f$) exists. But if it turns out it doesn't exist, then the question didn't make sense in the first place. I'm not sure what the best and rigorous way to describe the same meaning is, because the way I see it, there is no meaning. I hope this doesn't sound rude; I'm just trying to use the mathematical terms. The question as stated has no meaning because you are using something that you don't know exists in a way that can only be used if it in fact exists.
Aug
16
comment Can $0\le\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau\le a$ imply $\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau$ exists?
Since you accepted Clayton's answer, it would seem that $\int_0^t f(\tau)\ d\tau$ is indeed what you mean. The problem isn't the limit symbol; by definition, $\int_0^\infty f = \lim_{t\to \infty} \int_0^t f$. The problem is that by saying that this integral is between $0$ and $a$, you are implicitly assuming that it exists, otherwise you couldn't compare because you can't compare a number with something that doesn't exist. (cont.)
Aug
15
comment Can $0\le\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau\le a$ imply $\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau$ exists?
No, what you can say is that $f(t)$ is bounded; then, if for example you know that $f(t)$ is increasing, then you know that its limit exists and that $\lim\ f(t) \le \lim g(t)$. You can't talk about limits without first knowing if they exist. If (again) I'm understanding what you're saying correctly, your inequality should be $0 \le \int_0^t f(\tau)\ d\tau \le a$.
Aug
15
comment Can $0\le\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau\le a$ imply $\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau$ exists?
Am I not understanding something? To compare the limit with other numbers, first it has to exist, or else the comparison makes no sense.
Aug
15
comment Why does $\frac{\partial^2f}{\partial x \partial y} = \frac{\partial^2f}{\partial y \partial x}$
You need $f$ to be $C^2$, I believe.
Aug
13
comment Are electrodynamic problems in the complex plane relevant to real life?
Maybe you should ask this in Physics.SE?
Aug
10
comment Is my proof correct about limit of $\sin\left(\frac{1}{x}\right)$?
It's not a big deal, but you should mention what happens when $|A| > 1$.
Aug
7
comment Differentiate $\ln(\cos2x)$ With respect to $x$.
@Amzoti: Please, there's no need to be like that. I understand what you wrote, and the asker of the question does too. But if you read my comment, you'll see that not only did I not even begin to imply that something was offensive, but also that I think that writing things like $\ln u =\frac1{u}$ only helps further the confusion of students who write that when asked to differentiate $\ln u$.
Aug
7
comment Differentiate $\ln(\cos2x)$ With respect to $x$.
Yes, please don't write things like $\ln u = \frac1{u}$. Students are confused enough already. Either say $(\ln u)'$, or $\frac{d}{du}\ln u$, or else "the derivative of $\ln u$ is $\frac1{u}$".
Aug
6
comment Are parallel vectors always scalar multiple of each others?
@AmitTomar: So really any lines that don't intersect? Because that's not the definition of parallel.