The complex exponential and subset sums #
Properties of t ↦ exp(-(t · x)) and the expansion of ∏ (1 + exp x) over a
multiset of roots into zero- and nonzero-sum subset contributions, the analytic
heart of Niven's proof of the transcendence of π.
For x ∈ ℂ, one has exp(x) * exp(-x) = 1.
For x ∈ ℂ, the map t ↦ exp(-(t * x)) is continuous on [0, 1] at t.
If P(π i) = 0 and P ≠ 0, then ∏_{x ∈ roots(P)} (1 + e^x) = 0.
The map t ↦ e^(∑_{x ∈ t} x) on multisets of complex numbers.
Equations
Instances For
The sum ∑_{t ⊆ s} e^(∑_{x ∈ t} x) splits into the zero-sum and nonzero-sum parts.
The product ∏_{x ∈ s} (1 + e^x) equals to
∑_{t ⊆ s, ∑_{y ∈ t} y = 0} e^(∑_{y ∈ t} y) + ∑_{t ⊆ s, ∑_{y ∈ t} y ≠ 0} e^(∑_{y ∈ t} y).
The multiset of all nonzero subset sums of s, i.e. { ∑_{x ∈ t} x ≠ 0 | t ⊆ s }.
Equations
- nonzeroSubsetSums s = Multiset.filter (fun (x : α) => x ≠ 0) (subsetSums s)
Instances For
Every element of { ∑_{x ∈ t} x ≠ 0 | t ⊆ s } is nonzero.
If t ⊆ s and ∑_{x ∈ t} x = 0, then e^(∑_{x ∈ t} x) = 1.
Applying t ↦ e^(∑_{x ∈ t} x) to all t ⊆ s with ∑_{x ∈ t} x = 0 gives only 1s.
The sum ∑_{t ⊆ s, ∑_{x ∈ t} x = 0} e^(∑_{x ∈ t} x) equals
#{t ⊆ s | ∑_{x ∈ t} x = 0}.
If B(π i) = 0 and roots(B) is the multiset of roots of B, then
∑_{t ⊆ roots(B), ∑_{x ∈ t} x ≠ 0} e^(∑_{x ∈ t} x) = -#{t ⊆ roots(B) | ∑_{x ∈ t} x = 0}.
If f⁽ᵏ⁾ is differentiable at t * x for every k ≤ n and every t ∈ [0, 1], then
t ↦ exp(-t * x) * ( f⁽ⁿ⁾(t * x) - f(t * x) ) is continuous on [0, 1].
If f⁽ᵏ⁾ is differentiable at t x for every k ≤ n, then
t ↦ exp(-t * x) * ∑ₖ₌₀ⁿ⁺¹ f⁽ᵏ⁾(t * x) is differentiable at t.
If f⁽ᵏ⁾ is differentiable at t x for every k ≤ n, then with hasDeriv we have
d/dt (exp(-t * x) * ∑ₖ₌₀ⁿ⁺¹ f⁽ᵏ⁾(t * x)) = x * exp(- t * x) * (f⁽ⁿ⁺¹⁾(t * x) - f(t * x)).
If f⁽ᵏ⁾ is differentiable at t x for every k ≤ n, then with deriv we have
d/dt (exp(-t * x) * ∑ₖ₌₀ⁿ⁺¹ f⁽ᵏ⁾(t * x)) = x * exp(- t * x) * (f⁽ⁿ⁺¹⁾(t * x) - f(t * x)).
If f⁽ᵏ⁾ is differentiable at t x for every k ≤ n + 1 and every t ∈ [0, 1], then
∫₀¹ x * exp(-t x) * (f⁽ⁿ⁺¹⁾(t x) - f(t x)) dt = exp(- x) * ∑ₖ₌₀ⁿ f⁽ᵏ⁾(x) - ∑ₖ₌₀ⁿ f⁽ᵏ⁾(0).
If a₀, …, aₙ₋₁ enumerate the nonzero subset sums of the roots of B
and bᵢ = 1 for all i, then ∑ᵢ bᵢ exp(aᵢ) = -k.