@SchrodingersCat Well, as a replacement problem, you may be interested in finding a function such that every nth derivative evaluated at zero is (n!)^2
@SimplyBeautifulArt Well, we can obviously use Taylor expansion and plug that value... We get an awkward inf series.. you want something better as an answer...
@SimplyBeautifulArt: are you not proud that you're living in such a glorious time? you're freely discussing divergent series without any getting embroiled in any deeply heated controversy
Hello, how to prove that $\inf{2+\frac1n, n\in\mathbb{N}}=2$ without using limit, just the cararterisation of the if that is : $\forall \varepsilon>0, 2+\frac1n< \inf\{2+\frac1n\}+\varepsilon$
@LeakyNun I'm not confident that that statement is correct, Leaky. You ordinarily need a pair of complex conjugate roots to get a transposition in the Galois group.
@SimplyBeautifulArt the suggestion to write the reciprocal binomial coefficients as beta functions and then pass to the integral representation seems particularly smart
No, it's just that the problem maybe isn't really well-suited if you just learned about Galois theory. I was hoping that maybe you would come up with a more elegant solution than what I did
I have a clusterf*** of a last night dream: Imagine that, a Groundhog day style dream where it repeats more than 6 times and you try so hard to wake up!
When numerically evaluating $x=f(x)$, does $$a_{n+1}=\frac{f(a_n)(a_n-a_{n-1})+ a_n(f(a_n)-f(a_{n-1}))}{a_n-a_{n-1}+1}$$ generally converge quadratically, given $a_0\ne a_1$ are close enough to the root?
