@usukidoll One thing which you can do to make the integral less annoying is the substitution $u=y/\theta$
that eliminates most of the $\theta$'s from the problem, and all you really end up needing to compute is the definite integral $\int_0^1 u(1-u)^{n-1}\,du$
the link I saw spoiled the appearance of Russell's paradox, alas
i.e. if it's a finite game, then choosing to play hypergame is a legal first move. but then you can play infinitely by just offering to play hypergame again
if it's not infinite, then choosing to play isn't a legal first move...in which case it's finite.
I don't think you run into the problem if, for instance, you limit yourself to games which finish in less than 100 million turns (picking an arbitrary cutoff)
since then you could only choose the hypergame finitely many times
there's probably some more principled way of doing it
but frankly idc
I would be curious if there's a game-theoretic version of the Yablo paradox
what happens if I need to find $E[Y^{2}]$ ? could I do the same u sub but then there will be an extra y in it? $u = \frac{y}{\theta}$ followed by $uy(1-u)^{n-1}$
This may be borderline off-topic but this is the only place to ask.
I've always seen images of Leonhard Euler with a "hat" or head covering that is unfamiliar to me.
Is there a name for the head covering shown in Emanuel Handmann 1753 painting of Euler? Does it cary any specific significance o...
This is problem 11 (b) from the first chapter of "Basic Topology" by M.A. Armstrong. The author hasn't had time to develop many theorems or mathematical machinery, so this problem should be able to be solved by just picturing a series of intermediate steps. It goes
Imagine all the spaces sh...
The terms in $\sum_{n=1}^{\infty}\frac {(-n)^{n-1}}{n!}t^n$ tends to zero, but they do not do so monotonically, meaning that alternative series test cannot be used to check its convergence
no I mean you cannot interchange $\int$ and $\sum$ in the step $$- \int \frac {W(t)}{te^t}dt= -\int \frac{1}{te^t}\sum_{n=1}^{\infty}\frac {(-n)^{n-1}}{n!}t^n dt = -\sum_{n=1}^{\infty} \int\frac {(-n)^{n-1}}{(n!)te^t}t^n dt$$
@RithikKapoor Well, by just looking at that integrand, the chance for it to have a closed form is highly unlikely. In fact, that tetration only converges for some values
I came across infinite tetrations on wikipedia (https://en.wikipedia.org/wiki/Tetration) it says that the infinite tetration converges if and only if $\ e^{-e} \leq x \leq e^{1/e}$. I was wondering if there is a proof of this.
Hi all, my professor mentioned in class that one cannot use L'Hôpital's rule to solve the well known lim x-> 0 (sin x)/x = 1. I was wondering what condition of L'Hôpital's rule it failed.
Magma is one of those beautiful words of Greek origin (μάγμα) that arouses the child and the wild in me, making me think of volcanoes. I just found out, though, that it is also used in mathematics to mean a type of algebraic structure (a set paired with a binary operation on it)! I am very curiou...
Definition: An algebraic structure $(V,\cdot, E(\cdot))$ is called a volcano if its elements $M \in V$ are magmas such that there exists a unary operator $E$ such that $E(M) \in V$ are groups
(where $E$ follows the joke as it represents eruption, and thus cooled down magmas E(M) are groups)
The group algebra of a topological group G is the collection of all continuous functions $f : G \to \Bbb{C}$ with compact support, denoted by $C_c(G)$? And when $G$ is discrete, compact support implies finite support, so we identify $f : G \to \Bbb{C}$ with the sum $\sum_{g \in G} f(g) g$, which shows that the group ring and group algebra are isomorphic for discrete groups. Does this sound right?
If a person wanted to learn some "basics" in non-associative algebra, what would be a good place to start? Anyone know that field well enough to offer some advice?
Grading linear algebra homeworks at the moment. Also, looking over the results of a search algorithm I put together yesterday. So, pretty good. How about you?
Kind of tired, just reading up on the history of abstract algebra because it didn't look too taxing mentally, and sometimes it's nice to get a historical perspective on these things.
as by looking at the desmos graph desmos.com/calculator/8qufo3tuco , you can see that both the graphs match in the range of $e^{-e} \lt x \le e^{1/e}$. This is the same range at which the integral of $x^{x^{x^{.^{.......}}}}$ is defined.
Your estimate of $e^{-e}$ for the lower bound of the approximation seems off. It's closer to 0.68. (If you're on mobile, I don't know, but on desktop you should have the link in your bookmarks and you just select it from there.)
Though, your manipulation of the functions seem solid enough.
Graphs are useful, but they aren't proof. Don't take them for granted.
Like, without a proof, for all we know, as $a$ increases, one function could steadily closer approximate a second function, until $a\geq 10^{10^{100}}$, at which point it begins to decrease in how good the approximation is.
But, I do believe you've provided a proof, so long as I've not missed anything.
I was actually skeptical because my HS math teacher is a PHD and she said the the integral of x^x^x… needed its own function any couldn't be defined by any other function for example in this case the incomplete gamma function.