Recent content by alberto7

Sorry, I meant ##f(0)=0##.

I think you are in the good path. Just apply Liouville's theorem to the auxiliary function $$g(z)=\frac{f(z)}{z}\text,$$ which is entire because ##f(z)=0## and we can continue ##g## to the origin as $$g(0)=\lim_{z\to 0}\frac{f(z)}{z}\text.$$

As an example of ##\mathcal C^\infty## but not analytic? As a way to glue ##\mathcal C^\infty##-ly? Such a function is usually explained in Differential Geometry books in a section title Partitions of Unity.

I think we use ##\arcsin## for the functional inverse of ##\sin## and ##\csc## for its multiplicative inverse, instead of ##\sin^{-1}##, in order to avoid this confusion. As ##\sin^n## is multiplicative for ##n>0##, I would say that ##\sin^{-n}=\csc^n##. If one needed to write many times the...

Do Ruffini again: try with the divisors of -15 of both signs.

Redo the quotient (z^4-4z^3+6z^2-4z-15)/(z+1), since the leading term must be z^3, not -z^3.

Proceed with Ruffini. You'll find another root (because the problem is easy) and the remaining factor is quadratic, whose solutions you get with the formula.

Divergent series are solutions as formal series, and sometimes they define functions, but in sectors, of course not in discs. See Borel summability and multisummablity.

I haven't seen limits as |x|→∞, but I have seen limits as x→±∞ or limits as x→∞ with this meaning. In the latter case they distinguished limits as x→+∞ and as x→-∞. I think they pictured ∞ as a single point outside the line, making it a circle, its Alexandroff compactification, considering...

If ##f## is a polynomial, there are results for bounding the zeros of ##f'## know the zeros of ##f##. The Grace-Heawood theorem says that if ##z_1## and ##z_2## are distinct zeros of ##f##, ##f'## has a zero in the disk with center ##\frac12(z_1+z_2)## and radius ##\frac12|z_1-z_2|\cot(\pi/n)##...