In the aboe question, I can do parts (i), (ii) and (iii) without actually integrating. But is there a way to do part (iv) as well where there is no need to calculate f(x) explicitly?
I mean, if integrand was positive over [0,5] , then it was obvious that f assumes only one zero value at x = 0. But, since integrand dips below zero (twice), I am unable to solve (iv) without computing antiderivative.
silent: you'd certainly need to do something that accounted for the specific antiderivative you are working with there (i.e., it would not be enough to only be given that f'(x) = (x^2 - 5x + 4)(x^2 - 5x + 6), as the number of roots of f(x) on [0,5] could be anything between 0 and 5 consistent with that formula for f'(x). it really matters that f(0) = 0 here.
offhand it isn't immediately clear to me how you would even solve it with an express formula for f(x), since it's a fifth degree polynomial. certainly there are problems of this sort that could not be easily worked 'by hand,' i guess maybe this one turns out nicely (inferrable only from the fact that they are asking about it)
i guess you'd just plug in the local extrema and see if they correspond to crossing the axis or not. i don't see a "shortcut" to actually doing that
so that's a tractable calculation in this case, but its ease very much depends on f' having easily locatable integer roots. if the roots were instead quadratic expressions, even relatively simple ones, just figuring out the sign of f evaluated at each one of them could be considerably more work
@AlessandroCodenotti I'll start revising everything in the book to reinforce it, and then, like I said before, I thought about picking up Geometric aspects of general topology by Sakai
Definitely will continue along the topology route though
16 chapters is definitely a lot of knowledge so it does need to be reinforced I think
What is the precise reason why we only need to consider Cauchy sequences instead of Cauchy nets in a first countable topological space? Is it because given $x$ in the space and a Cauchy net converging to $x$ we can always find a Cauchy sequence converging to $x$ (and vice versa but that part is always true since sequences are Cauchy nets)?
@Jakobian Apologies---I misclicked a comment on the mobile version of the site, which deleted one of your comments, above. It was a mistake, but I cannot restore the missing comment. The mobile site sucks, and I've moved to the desktop site (where I have an actual keyboard). :/
@Jakobian Maybe this is helpful: I was asking in the context of completions. To construct the completion of a first countable topological space we can use Cauchy sequences instead of Cauchy nets / Cauchy filters.
Let $ f : [a, b] \rightarrow \mathbb{R} $ be a bounded function. Then for every $\epsilon > 0$, there exists $\delta > 0$such that for any partition $D$ of the interval $[a, b]$ with norm $\nu(D)$ smaller than $\delta$, the following holds:
$$
\left| U(f, D) - \int_a^b \overline{f}(x) \, dx \right|
Let $\varphi:\mathbb R\to\mathbb R_+$ be a convex function. Set $$\mathcal E_\varphi=\{(a,b)\in\mathbb R^2:\forall x\in\mathbb R,\varphi(x)\geq ax+b\}.$$ Then by convexity of $\varphi$, $$\varphi \left(x\right)=\underbrace{\sup_{\left(a{,}b\right)\in \mathcal E_{\varphi }}\left(ax+b\right)}_{g(x)}=\underbrace{\sup_{\left(a{,}b\right)\in \mathcal E_{\varphi }\cap \mathbb Q^2}\left(ax+b\right)}_{h(x)}.$$
I understand the first equality above, but not the second one. Clearly $g(x)\geq h(x)$, but why is $g(x)\leq h(x)$? Context; I am reading a proof of Jensen's inequality for conditional expectation, and it is key to consider only the supremum over rationals.
Upon thinking some more, this is actually clear now. Thanks for the help!
Of course, feel free to add some explanation if you've got one. The more the merrier.
Studiare la seguente forma differenziale: $\omega(x, y) = (ye^{xy} + 1) \, dx + xe^{xy} \, dy.$ Data la curva $\gamma(t) = (a \cos t, \sin t)$, $t \in [0, 2\pi]$, dire se esiste $a > 0$ tale che $\int_{\gamma} \omega = 1$.
$\gamma'(t) = (-a \sin t, \cos t)$
$\omega(\gamma(t)) = \left( \sin t \cdot e^{a \cos t \cdot \sin t} + 1 \right) d(a \cos t) + \left( a \cos t \cdot e^{a \cos t \cdot \sin t} \right) d(\sin t)$
$\int_0^{2\pi} \left[ \left( \sin t \cdot e^{a \cos t \cdot \sin t} + 1 \right)(-a \sin t) + a \cos t \cdot e^{a \cos t \cdot \sin t} \cos t \right] dt$
@Pizza In the rules, it does not explicitly state that you cannot use a language other than English in chat (SE network policy is that all questions and answers must be in English, except for a few non-English sites). So there is no rule which says you cannot use a non-English language in chat.
It is better to use English, when able, but no one is required to do so, and I'm not really into the idea of anyone policing which languages should be used.
Do note, however, that since I don't really speak anything other than English or Russian, I have a hair trigger when it comes to deleting flagged content in languages other than those, which means that you are more likely to get suspended if you are interacting in a language I don't understand.
I'd like to ask something if I have a function like $\sqrt{z^2 + 1}$, is it correct to place a branch cut like $[-i,\infty] \text{ and } [i,\infty]$ instead of along $\Im(z)\in [-i,+i]$
No ok I suppose it is correct, for one can always express $\sqrt{f(z)} = e^{0.5 \cdot \log{f(z)}}$, and then impose that $f(z) \ne -t, t \in [0,\infty[$
a field is conservative if, for example, the work done to move a particle from a point $P_0$ of the set $A$, to a point $P_1$ of $A$ does not depend on the curve $\gamma$ along which one moves moves, but only from the extremes $P_0$ and $P_1$ of it and from the direction of travel
@BinkyMcSquigglebottom another way to arrange this is to carry out the substitution $t=\ln(\tan x)$ first, rendering the integral as $$\int_{-\infty}^\infty \frac{1}{4t^2+\pi^2}\frac{dt}{1+e^{2t}}$$ at that point, using symmetry to simplify the integral is a bit more obvious
@ephe what matters is universal properties. You can complete uniform space uniquely up to a uniform isomorphism preserving it. It doesn't matter in which way you construct it then as long as it satisfies the definition of completion
You just need to show both constructions satisfy the definition of a completion. Then you invoke a theorem
I read about uniform spaces mostly in different setting, using pseudometrics. It was hard to construct a completion, but easy to show uniqueness. Perhaps your experience will differ but I suspect it won't.
Of course I'm assuming you are talking about uniform spaces - topological space by itself isn't endowed with structure that allows for Cauchy sequences
@Jakobian Well the problem is that I have no idea about what the category theoretic definition of completion is supposed to be. :/ I suppose its the one given by the inverse limit and the universal property you mention is the universal property of that, right?
It does seem like there is some weird linguistics issue that universal property can either mean something specific and well-defined in category theory, or just independence of construction
Isn't the former just the formalization of the latter? In the end we always seem to end up with our constructions being isomorphic up to unique isomorphism.
Yes, but to speak about the former you need to introduce what category you're talking about etc., which is useless when you just want to adress the latter
There is an issue of formalization induced baggage
So even though its a formalization, the latter is not a special case of the former
are those people serious mathematicians? and, if they are, are you certain you aren't missing an implicit understanding that they are only talking about their own and adjacent fields?
