Freedom — speech, religion, press
Rifles — bear arms
Enforced — quartering soldiers
Entry — search & seizure
Silence — self-incrimination/due process
Trial — speedy/public trial
Associates — civil jury
Torture — cruel/unusual punishment
Extras — unenumerated rights
States — reserved powers

Associates is the weakest I think.

Or as a story:
Free riflemen evicted entrants silently; trials assembled, tortured extra states.

https://preview.redd.it/tw9f5um06zgh1.png?width=743&format=png&auto=webp&s=cb6797209e265064ada9414b63b58b790ce221db

Nana
Short for nanofyourbusiness.

Used as a one-word reply when someone asks a nosy question.

Example:
“Why were you there?”
“Nana.”

I've been trying to pin down, as cleanly as possible, why/how the two standard characterizations of the exponential are equivalent/related:

• Algebraic: f(x+y) = f(x)·f(y) (homomorphism from an additive structure to a multiplicative one).
• Analytic: f′ = λ·f (eigenfunction of the derivative).

The cleanest unification I know is the Lie-theoretic one: exp : 𝔤 → G is simultaneously the analytic object (flow of a left-invariant vector field) and the algebraic object (intertwines + on the algebra with × on the group on commuting elements).

But I tried to find the minimal set of abstract properties on a derivation-like operator K such that any eigenfunction of K (normalized to 1 at 0) automatically satisfies the additive-to-multiplicative functional equation.

Setup

Let (A, +, 0) be an additive monoid, (B, +, ·, 0, 1) a unital commutative ring, and Func(A, B) the ring of functions A → B with pointwise operations. Define the shift (T_y g)(x) := g(x+y).

Suppose K : Func(A, B) → Func(A, B) satisfies:

• (A) Additivity: K(g + h) = K(g) + K(h)
• (L) Leibniz: K(g·h) = K(g)·h + g·K(h)
• (C) Kills constants: K(c_b) = 0 for any constant function c_b
• (T) Translation invariance: K ∘ T_y = T_y ∘ K

And suppose f and λ ∈ B satisfy:

• (E) Eigenfunction: K(f) = λ·f
• (N) Normalization: f(0) = 1
• (U) Uniqueness: evaluation at 0 is injective on ker(K − λI).

Claim

f(x + y) = f(x)·f(y).

Proof sketch

Fix y and let g_y := T_y f. By (T) and (E), K(g_y) = λ·g_y. Define

g(x) := f(x+y) − f(x)·f(y)  =  g_y − f·c_{f(y)}.

Then g(0) = f(y) − f(0)·f(y) = 0 by (N), and using (A), (L), (C), (E):

K(g) = K(g_y) − [K(f)·c_{f(y)} + f·K(c_{f(y)})]
     = λ·g_y − λ·f·c_{f(y)} − 0
     = λ·g.

So g ∈ ker(K − λI) with g(0) = 0, hence g ≡ 0 by (U). ∎

The additive-exponential property is forced by (A), (L), (C), (T), (U). Among these, (L) and (T) feel like the real reason: Leibniz is what allows you to split the product, and the translation invariance lets you treat T_y f as another eigenfunction.

Questions

1. Is this minimal?
2. Is there a slicker/more standard formulation?
3. What's the right reference for the equivalence as an explanatory matter, not just as a theorem?
4. Am I missing a hypothesis?

A couple of notes on the proof itself:

• Commutativity of B is only used in the last line (f(x)·f(y) vs f(y)·f(x)); everything before works in a noncommutative ring with care about left/right multiplication. This foreshadows the Lie-group case where exp(X+Y) ≠ exp(X)·exp(Y) unless [X,Y] = 0 (otherwise BCH)
• (C) follows from (L) + (A) in many (not all) settings: K(1) = K(1·1) = 2·K(1) so K(1) = 0, then extend by (A).