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The function field model came up in GH from MO's answer, and then also in user54038's. I just want to add some detail to explain how good of an analogy the function field model is. The Riemann zeta …

As Fedor Petrov expects in the comments, this is true for $x=a/b$ if and only if the order of $2$ modulo $b$ is greater than the order of $2$ modulo the product of prime factors of $b$. To prove this, …

Residues satisfy $$\operatorname{res}_x(fdg) + \operatorname{res}_x(gdf) = \operatorname{res}_x(d (fg)) =0$$ which tells you that they are antisymmetric, and many antisymmetric pairings in math arise …

Consider the map $f\colon z \mapsto z^3-z$. Then there is a path component $M$ of $\mathcal G$ consisting of germs of the inverse map to $f$. More precisely $M$ is isomorphic to $\mathbb C \setminus \ …

The $k$-Hodge conjecture is false for any $k$ containing an irrational real $\alpha$. Indeed we may clearly assume $\alpha$ is negative. Consider the elliptic curve $E = \mathbb C / \langle 1, \sqrt{ …

We have $$ \frac{d}{dx}\left ( f(x) e^{h(x)}\right) = \left( \frac{ f'(x)}{f(x)} + h'(x) \right) \left( f(x) e^{h(x)}\right).$$ Thus $$ \sum_n a_n n x^{n-1} = \left( \frac{ f'(x)}{f(x)} + h'(x) \rig …

$$\int_x^\infty \{ u\} u^{-s-1} du = \int_x^\infty \left( \{ u\}-\frac{1}{2}\right) u^{-s-1} du + \frac{1}{2} \frac{x^{-s}}{s}$$ and by integration by parts, $$\int_x^\infty \left( \{ u\}-\frac{1}{2} …

If $f(z)$ is a holomorphic function in a neighborhood of $z=0$ then $f( p^{-s})$ is a Dirichlet series for any prime (or really any natural number) $p$: we take $a_n=0$ if $n$ is not a power of $p$ an …

Bounds for the value of zeta take the form $$ |\zeta(\sigma +it) | \ll t^{ f(\sigma)+\epsilon} $$ where $|t|>2$ and $f$ is some function of $\sigma$. We can take $f(1)=0$ and $f(0)=1/2$ and we can al …

No (for one interpretation of the question). There does not exist a continuous function $b $ from the set of systems of polynomial equations to $\mathbb R^{>0} \cup \{\infty\}$ that takes a finite val …

If you know them as algebras over $M(SL_2(\mathbb Z))$, almost exactly the procedure you describe works. Since the modular forms with weight a multiple of $12$ are sections of powers of an ample line …

1 Sort of. It's not totally obvious what "the eigenvalues of the curvature" are because the curvature is not a matrix, but a matrix valued in 2-forms. Also, there's the $1/2\pi i$ thing you mention. O …

Suppose $\alpha = a/b$ is rational. All but finitely many primes are relatively prime to $b$. For these primes, we have $$ e^{ 2\pi i ap/b} = \sum_{\chi: (\mathbb Z/b)^\times \to \mathbb C^\times} \c …

No. Take a convex set and use an exponential function $e^{2\pi i z/c}$, where $c$ is the difference between two points in the set. The image of the set is no longer simply connected, and thus cannot b …

Let $\Gamma$ be a non-algebraic analytic curve. Let $z_1,\dots, z_n$ be points of $\Gamma$. Let $X_{n,d}$ be the set of algebraic curves of degree $d$ that contain $z_1,\dots, z_n$. Now consider a ran …