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Theorem List for Metamath Proof Explorer - 14401-14500   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremfprod1 14401* A finite product of only one term is the term itself. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝑘 = 𝑀𝐴 = 𝐵)       ((𝑀 ∈ ℤ ∧ 𝐵 ∈ ℂ) → ∏𝑘 ∈ (𝑀...𝑀)𝐴 = 𝐵)
 
Theoremprodsnf 14402* A product of a singleton is the term. A version of prodsn 14400 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝐵    &   (𝑘 = 𝑀𝐴 = 𝐵)       ((𝑀𝑉𝐵 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝐵)
 
Theoremclimprod1 14403 The limit of a product over one. (Contributed by Scott Fenton, 15-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)       (𝜑 → seq𝑀( · , (𝑍 × {1})) ⇝ 1)
 
Theoremfprodsplit 14404* Split a finite product into two parts. (Contributed by Scott Fenton, 16-Dec-2017.)
(𝜑 → (𝐴𝐵) = ∅)    &   (𝜑𝑈 = (𝐴𝐵))    &   (𝜑𝑈 ∈ Fin)    &   ((𝜑𝑘𝑈) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑘𝑈 𝐶 = (∏𝑘𝐴 𝐶 · ∏𝑘𝐵 𝐶))
 
Theoremfprodm1 14405* Separate out the last term in a finite product. (Contributed by Scott Fenton, 16-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑘 = 𝑁𝐴 = 𝐵)       (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (∏𝑘 ∈ (𝑀...(𝑁 − 1))𝐴 · 𝐵))
 
Theoremfprod1p 14406* Separate out the first term in a finite product. (Contributed by Scott Fenton, 24-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑘 = 𝑀𝐴 = 𝐵)       (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (𝐵 · ∏𝑘 ∈ ((𝑀 + 1)...𝑁)𝐴))
 
Theoremfprodp1 14407* Multiply in the last term in a finite product. (Contributed by Scott Fenton, 24-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...(𝑁 + 1))) → 𝐴 ∈ ℂ)    &   (𝑘 = (𝑁 + 1) → 𝐴 = 𝐵)       (𝜑 → ∏𝑘 ∈ (𝑀...(𝑁 + 1))𝐴 = (∏𝑘 ∈ (𝑀...𝑁)𝐴 · 𝐵))
 
Theoremfprodm1s 14408* Separate out the last term in a finite product. (Contributed by Scott Fenton, 27-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (∏𝑘 ∈ (𝑀...(𝑁 − 1))𝐴 · 𝑁 / 𝑘𝐴))
 
Theoremfprodp1s 14409* Multiply in the last term in a finite product. (Contributed by Scott Fenton, 27-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...(𝑁 + 1))) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑘 ∈ (𝑀...(𝑁 + 1))𝐴 = (∏𝑘 ∈ (𝑀...𝑁)𝐴 · (𝑁 + 1) / 𝑘𝐴))
 
Theoremprodsns 14410* A product of the singleton is the term. (Contributed by Scott Fenton, 25-Dec-2017.)
((𝑀𝑉𝑀 / 𝑘𝐴 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝑀 / 𝑘𝐴)
 
Theoremfprodfac 14411* Factorial using product notation. (Contributed by Scott Fenton, 15-Dec-2017.)
(𝐴 ∈ ℕ0 → (!‘𝐴) = ∏𝑘 ∈ (1...𝐴)𝑘)
 
Theoremfprodabs 14412* The absolute value of a finite product. (Contributed by Scott Fenton, 25-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑁𝑍)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)       (𝜑 → (abs‘∏𝑘 ∈ (𝑀...𝑁)𝐴) = ∏𝑘 ∈ (𝑀...𝑁)(abs‘𝐴))
 
Theoremfprodeq0 14413* Anything finite product containing a zero term is itself zero. (Contributed by Scott Fenton, 27-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑁𝑍)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)    &   ((𝜑𝑘 = 𝑁) → 𝐴 = 0)       ((𝜑𝐾 ∈ (ℤ𝑁)) → ∏𝑘 ∈ (𝑀...𝐾)𝐴 = 0)
 
Theoremfprodshft 14414* Shift the index of a finite product. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝜑𝐾 ∈ ℤ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   ((𝜑𝑗 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑗 = (𝑘𝐾) → 𝐴 = 𝐵)       (𝜑 → ∏𝑗 ∈ (𝑀...𝑁)𝐴 = ∏𝑘 ∈ ((𝑀 + 𝐾)...(𝑁 + 𝐾))𝐵)
 
Theoremfprodrev 14415* Reversal of a finite product. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝜑𝐾 ∈ ℤ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   ((𝜑𝑗 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑗 = (𝐾𝑘) → 𝐴 = 𝐵)       (𝜑 → ∏𝑗 ∈ (𝑀...𝑁)𝐴 = ∏𝑘 ∈ ((𝐾𝑁)...(𝐾𝑀))𝐵)
 
Theoremfprodconst 14416* The product of constant terms (𝑘 is not free in 𝐵.) (Contributed by Scott Fenton, 12-Jan-2018.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ ℂ) → ∏𝑘𝐴 𝐵 = (𝐵↑(#‘𝐴)))
 
Theoremfprodn0 14417* A finite product of nonzero terms is nonzero. (Contributed by Scott Fenton, 15-Jan-2018.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐵 ≠ 0)       (𝜑 → ∏𝑘𝐴 𝐵 ≠ 0)
 
Theoremfprod2dlem 14418* Lemma for fprod2d 14419- induction step. (Contributed by Scott Fenton, 30-Jan-2018.)
(𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑗𝐴) → 𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)    &   (𝜑 → ¬ 𝑦𝑥)    &   (𝜑 → (𝑥 ∪ {𝑦}) ⊆ 𝐴)    &   (𝜓 ↔ ∏𝑗𝑥𝑘𝐵 𝐶 = ∏𝑧 𝑗𝑥 ({𝑗} × 𝐵)𝐷)       ((𝜑𝜓) → ∏𝑗 ∈ (𝑥 ∪ {𝑦})∏𝑘𝐵 𝐶 = ∏𝑧 𝑗 ∈ (𝑥 ∪ {𝑦})({𝑗} × 𝐵)𝐷)
 
Theoremfprod2d 14419* Write a double product as a product over a two-dimensional region. Compare fsum2d 14213. (Contributed by Scott Fenton, 30-Jan-2018.)
(𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑗𝐴) → 𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐶 = ∏𝑧 𝑗𝐴 ({𝑗} × 𝐵)𝐷)
 
Theoremfprodxp 14420* Combine two products into a single product over the cartesian product. (Contributed by Scott Fenton, 1-Feb-2018.)
(𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   (𝜑𝐴 ∈ Fin)    &   (𝜑𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐶 = ∏𝑧 ∈ (𝐴 × 𝐵)𝐷)
 
Theoremfprodcnv 14421* Transform a product region using the converse operation. (Contributed by Scott Fenton, 1-Feb-2018.)
(𝑥 = ⟨𝑗, 𝑘⟩ → 𝐵 = 𝐷)    &   (𝑦 = ⟨𝑘, 𝑗⟩ → 𝐶 = 𝐷)    &   (𝜑𝐴 ∈ Fin)    &   (𝜑 → Rel 𝐴)    &   ((𝜑𝑥𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑥𝐴 𝐵 = ∏𝑦 𝐴𝐶)
 
Theoremfprodcom2 14422* Interchange order of multiplication. Note that 𝐵(𝑗) and 𝐷(𝑘) are not necessarily constant expressions. (Contributed by Scott Fenton, 1-Feb-2018.) (Proof shortened by JJ, 2-Aug-2021.)
(𝜑𝐴 ∈ Fin)    &   (𝜑𝐶 ∈ Fin)    &   ((𝜑𝑗𝐴) → 𝐵 ∈ Fin)    &   (𝜑 → ((𝑗𝐴𝑘𝐵) ↔ (𝑘𝐶𝑗𝐷)))    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐸 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐸 = ∏𝑘𝐶𝑗𝐷 𝐸)
 
Theoremfprodcom2OLD 14423* Obsolete proof of fprodcom2 14422 as of 2-Aug-2021. (Contributed by Scott Fenton, 1-Feb-2018.) (Proof modification is discouraged.) (New usage is discouraged.)
(𝜑𝐴 ∈ Fin)    &   (𝜑𝐶 ∈ Fin)    &   ((𝜑𝑗𝐴) → 𝐵 ∈ Fin)    &   (𝜑 → ((𝑗𝐴𝑘𝐵) ↔ (𝑘𝐶𝑗𝐷)))    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐸 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐸 = ∏𝑘𝐶𝑗𝐷 𝐸)
 
Theoremfprodcom 14424* Interchange product order. (Contributed by Scott Fenton, 2-Feb-2018.)
(𝜑𝐴 ∈ Fin)    &   (𝜑𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐶 = ∏𝑘𝐵𝑗𝐴 𝐶)
 
Theoremfprod0diag 14425* Two ways to express "the product of 𝐴(𝑗, 𝑘) over the triangular region 𝑀𝑗, 𝑀𝑘, 𝑗 + 𝑘𝑁. Compare fsum0diag 14220. (Contributed by Scott Fenton, 2-Feb-2018.)
((𝜑 ∧ (𝑗 ∈ (0...𝑁) ∧ 𝑘 ∈ (0...(𝑁𝑗)))) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑗 ∈ (0...𝑁)∏𝑘 ∈ (0...(𝑁𝑗))𝐴 = ∏𝑘 ∈ (0...𝑁)∏𝑗 ∈ (0...(𝑁𝑘))𝐴)
 
Theoremfproddivf 14426* The quotient of two finite products. A version of fproddiv 14399 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐶 ≠ 0)       (𝜑 → ∏𝑘𝐴 (𝐵 / 𝐶) = (∏𝑘𝐴 𝐵 / ∏𝑘𝐴 𝐶))
 
Theoremfprodsplitf 14427* Split a finite product into two parts. A version of fprodsplit 14404 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑 → (𝐴𝐵) = ∅)    &   (𝜑𝑈 = (𝐴𝐵))    &   (𝜑𝑈 ∈ Fin)    &   ((𝜑𝑘𝑈) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑘𝑈 𝐶 = (∏𝑘𝐴 𝐶 · ∏𝑘𝐵 𝐶))
 
Theoremfprodsplitsn 14428* Separate out a term in a finite product. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   𝑘𝐷    &   (𝜑𝐴 ∈ Fin)    &   (𝜑𝐵𝑉)    &   (𝜑 → ¬ 𝐵𝐴)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   (𝑘 = 𝐵𝐶 = 𝐷)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → ∏𝑘 ∈ (𝐴 ∪ {𝐵})𝐶 = (∏𝑘𝐴 𝐶 · 𝐷))
 
Theoremfprodsplit1f 14429* Separate out a term in a finite product. A version of fprodsplit1 38559 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝑘𝐷)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   (𝜑𝐶𝐴)    &   ((𝜑𝑘 = 𝐶) → 𝐵 = 𝐷)       (𝜑 → ∏𝑘𝐴 𝐵 = (𝐷 · ∏𝑘 ∈ (𝐴 ∖ {𝐶})𝐵))
 
Theoremfprodn0f 14430* A finite product of nonzero terms is nonzero. A version of fprodn0 14417 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐵 ≠ 0)       (𝜑 → ∏𝑘𝐴 𝐵 ≠ 0)
 
Theoremfprodclf 14431* Closure of a finite product of complex numbers. A version of fprodcl 14390 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℂ)
 
Theoremfprodge0 14432* If all the terms of a finite product are nonnegative, so is the product. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)    &   ((𝜑𝑘𝐴) → 0 ≤ 𝐵)       (𝜑 → 0 ≤ ∏𝑘𝐴 𝐵)
 
Theoremfprodeq0g 14433* Any finite product containing a zero term is itself zero. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   (𝜑𝐶𝐴)    &   ((𝜑𝑘 = 𝐶) → 𝐵 = 0)       (𝜑 → ∏𝑘𝐴 𝐵 = 0)
 
Theoremfprodge1 14434* If all of the terms of a finite product are larger or equal to 1, so is the product. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)    &   ((𝜑𝑘𝐴) → 1 ≤ 𝐵)       (𝜑 → 1 ≤ ∏𝑘𝐴 𝐵)
 
Theoremfprodle 14435* If all the terms of two finite products are nonnegative and compare, so do the two products. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)    &   ((𝜑𝑘𝐴) → 0 ≤ 𝐵)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℝ)    &   ((𝜑𝑘𝐴) → 𝐵𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 ≤ ∏𝑘𝐴 𝐶)
 
Theoremfprodmodd 14436* If all factors of two finite products are equal modulo 𝑀, the products are equal modulo 𝑀. (Contributed by AV, 7-Jul-2021.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℤ)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   ((𝜑𝑘𝐴) → (𝐵 mod 𝑀) = (𝐶 mod 𝑀))       (𝜑 → (∏𝑘𝐴 𝐵 mod 𝑀) = (∏𝑘𝐴 𝐶 mod 𝑀))
 
5.10.12.5  Infinite products
 
Theoremiprodclim 14437* An infinite product equals the value its sequence converges to. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐴)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)    &   (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝐵)       (𝜑 → ∏𝑘𝑍 𝐴 = 𝐵)
 
Theoremiprodclim2 14438* A converging product converges to its infinite product. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐴)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)       (𝜑 → seq𝑀( · , 𝐹) ⇝ ∏𝑘𝑍 𝐴)
 
Theoremiprodclim3 14439* The sequence of partial finite product of a converging infinite product converge to the infinite product of the series. Note that 𝑗 must not occur in 𝐴. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘𝑍𝐴)) ⇝ 𝑦))    &   (𝜑𝐹 ∈ dom ⇝ )    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)    &   ((𝜑𝑗𝑍) → (𝐹𝑗) = ∏𝑘 ∈ (𝑀...𝑗)𝐴)       (𝜑𝐹 ⇝ ∏𝑘𝑍 𝐴)
 
Theoremiprodcl 14440* The product of a non-trivially converging infinite sequence is a complex number. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐴)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑘𝑍 𝐴 ∈ ℂ)
 
Theoremiprodrecl 14441* The product of a non-trivially converging infinite real sequence is a real number. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐴)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℝ)       (𝜑 → ∏𝑘𝑍 𝐴 ∈ ℝ)
 
Theoremiprodmul 14442* Multiplication of infinite sums. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐴)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)    &   (𝜑 → ∃𝑚𝑍𝑧(𝑧 ≠ 0 ∧ seq𝑚( · , 𝐺) ⇝ 𝑧))    &   ((𝜑𝑘𝑍) → (𝐺𝑘) = 𝐵)    &   ((𝜑𝑘𝑍) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝑍 (𝐴 · 𝐵) = (∏𝑘𝑍 𝐴 · ∏𝑘𝑍 𝐵))
 
5.10.13  Falling and Rising Factorial
 
Syntaxcfallfac 14443 Declare the syntax for the falling factorial.
class FallFac
 
Syntaxcrisefac 14444 Declare the syntax for the rising factorial.
class RiseFac
 
Definitiondf-risefac 14445* Define the rising factorial function. This is the function (𝐴 · (𝐴 + 1) · ...(𝐴 + 𝑁)) for complex 𝐴 and nonnegative integers 𝑁. (Contributed by Scott Fenton, 5-Jan-2018.)
RiseFac = (𝑥 ∈ ℂ, 𝑛 ∈ ℕ0 ↦ ∏𝑘 ∈ (0...(𝑛 − 1))(𝑥 + 𝑘))
 
Definitiondf-fallfac 14446* Define the falling factorial function. This is the function (𝐴 · (𝐴 − 1) · ...(𝐴𝑁)) for complex 𝐴 and nonnegative integers 𝑁. (Contributed by Scott Fenton, 5-Jan-2018.)
FallFac = (𝑥 ∈ ℂ, 𝑛 ∈ ℕ0 ↦ ∏𝑘 ∈ (0...(𝑛 − 1))(𝑥𝑘))
 
Theoremrisefacval 14447* The value of the rising factorial function. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) = ∏𝑘 ∈ (0...(𝑁 − 1))(𝐴 + 𝑘))
 
Theoremfallfacval 14448* The value of the falling factorial function. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) = ∏𝑘 ∈ (0...(𝑁 − 1))(𝐴𝑘))
 
Theoremrisefacval2 14449* One-based value of rising factorial. (Contributed by Scott Fenton, 15-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) = ∏𝑘 ∈ (1...𝑁)(𝐴 + (𝑘 − 1)))
 
Theoremfallfacval2 14450* One-based value of falling factorial. (Contributed by Scott Fenton, 15-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) = ∏𝑘 ∈ (1...𝑁)(𝐴 − (𝑘 − 1)))
 
Theoremfallfacval3 14451* A product representation of falling factorial when 𝐴 is a nonnegative integer. (Contributed by Scott Fenton, 20-Mar-2018.)
(𝑁 ∈ (0...𝐴) → (𝐴 FallFac 𝑁) = ∏𝑘 ∈ ((𝐴 − (𝑁 − 1))...𝐴)𝑘)
 
Theoremrisefaccllem 14452* Lemma for rising factorial closure laws. (Contributed by Scott Fenton, 5-Jan-2018.)
𝑆 ⊆ ℂ    &   1 ∈ 𝑆    &   ((𝑥𝑆𝑦𝑆) → (𝑥 · 𝑦) ∈ 𝑆)    &   ((𝐴𝑆𝑘 ∈ ℕ0) → (𝐴 + 𝑘) ∈ 𝑆)       ((𝐴𝑆𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ 𝑆)
 
Theoremfallfaccllem 14453* Lemma for falling factorial closure laws. (Contributed by Scott Fenton, 5-Jan-2018.)
𝑆 ⊆ ℂ    &   1 ∈ 𝑆    &   ((𝑥𝑆𝑦𝑆) → (𝑥 · 𝑦) ∈ 𝑆)    &   ((𝐴𝑆𝑘 ∈ ℕ0) → (𝐴𝑘) ∈ 𝑆)       ((𝐴𝑆𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ 𝑆)
 
Theoremrisefaccl 14454 Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℂ)
 
Theoremfallfaccl 14455 Closure law for falling factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ ℂ)
 
Theoremrerisefaccl 14456 Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℝ)
 
Theoremrefallfaccl 14457 Closure law for falling factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ ℝ)
 
Theoremnnrisefaccl 14458 Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℕ)
 
Theoremzrisefaccl 14459 Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℤ)
 
Theoremzfallfaccl 14460 Closure law for falling factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ ℤ)
 
Theoremnn0risefaccl 14461 Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℕ0𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℕ0)
 
Theoremrprisefaccl 14462 Closure law for rising factorial. (Contributed by Scott Fenton, 9-Jan-2018.)
((𝐴 ∈ ℝ+𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℝ+)
 
Theoremrisefallfac 14463 A relationship between rising and falling factorials. (Contributed by Scott Fenton, 15-Jan-2018.)
((𝑋 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝑋 RiseFac 𝑁) = ((-1↑𝑁) · (-𝑋 FallFac 𝑁)))
 
Theoremfallrisefac 14464 A relationship between falling and rising factorials. (Contributed by Scott Fenton, 17-Jan-2018.)
((𝑋 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝑋 FallFac 𝑁) = ((-1↑𝑁) · (-𝑋 RiseFac 𝑁)))
 
Theoremrisefall0lem 14465 Lemma for risefac0 14466 and fallfac0 14467. Show a particular set of finite integers is empty. (Contributed by Scott Fenton, 5-Jan-2018.)
(0...(0 − 1)) = ∅
 
Theoremrisefac0 14466 The value of the rising factorial when 𝑁 = 0. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝐴 ∈ ℂ → (𝐴 RiseFac 0) = 1)
 
Theoremfallfac0 14467 The value of the falling factorial when 𝑁 = 0. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝐴 ∈ ℂ → (𝐴 FallFac 0) = 1)
 
Theoremrisefacp1 14468 The value of the rising factorial at a successor. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac (𝑁 + 1)) = ((𝐴 RiseFac 𝑁) · (𝐴 + 𝑁)))
 
Theoremfallfacp1 14469 The value of the falling factorial at a successor. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac (𝑁 + 1)) = ((𝐴 FallFac 𝑁) · (𝐴𝑁)))
 
Theoremrisefacp1d 14470 The value of the rising factorial at a successor. (Contributed by Scott Fenton, 19-Mar-2018.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝐴 RiseFac (𝑁 + 1)) = ((𝐴 RiseFac 𝑁) · (𝐴 + 𝑁)))
 
Theoremfallfacp1d 14471 The value of the falling factorial at a successor. (Contributed by Scott Fenton, 19-Mar-2018.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝐴 FallFac (𝑁 + 1)) = ((𝐴 FallFac 𝑁) · (𝐴𝑁)))
 
Theoremrisefac1 14472 The value of rising factorial at one. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝐴 ∈ ℂ → (𝐴 RiseFac 1) = 𝐴)
 
Theoremfallfac1 14473 The value of falling factorial at one. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝐴 ∈ ℂ → (𝐴 FallFac 1) = 𝐴)
 
Theoremrisefacfac 14474 Relate rising factorial to factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝑁 ∈ ℕ0 → (1 RiseFac 𝑁) = (!‘𝑁))
 
Theoremfallfacfwd 14475 The forward difference of a falling factorial. (Contributed by Scott Fenton, 21-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ) → (((𝐴 + 1) FallFac 𝑁) − (𝐴 FallFac 𝑁)) = (𝑁 · (𝐴 FallFac (𝑁 − 1))))
 
Theorem0fallfac 14476 The value of the zero falling factorial at natural 𝑁. (Contributed by Scott Fenton, 17-Feb-2018.)
(𝑁 ∈ ℕ → (0 FallFac 𝑁) = 0)
 
Theorem0risefac 14477 The value of the zero rising factorial at natural 𝑁. (Contributed by Scott Fenton, 17-Feb-2018.)
(𝑁 ∈ ℕ → (0 RiseFac 𝑁) = 0)
 
Theorembinomfallfaclem1 14478 Lemma for binomfallfac 14480. Closure law. (Contributed by Scott Fenton, 13-Mar-2018.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ0)       ((𝜑𝐾 ∈ (0...𝑁)) → ((𝑁C𝐾) · ((𝐴 FallFac (𝑁𝐾)) · (𝐵 FallFac (𝐾 + 1)))) ∈ ℂ)
 
Theorembinomfallfaclem2 14479* Lemma for binomfallfac 14480. Inductive step. (Contributed by Scott Fenton, 13-Mar-2018.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ0)    &   (𝜓 → ((𝐴 + 𝐵) FallFac 𝑁) = Σ𝑘 ∈ (0...𝑁)((𝑁C𝑘) · ((𝐴 FallFac (𝑁𝑘)) · (𝐵 FallFac 𝑘))))       ((𝜑𝜓) → ((𝐴 + 𝐵) FallFac (𝑁 + 1)) = Σ𝑘 ∈ (0...(𝑁 + 1))(((𝑁 + 1)C𝑘) · ((𝐴 FallFac ((𝑁 + 1) − 𝑘)) · (𝐵 FallFac 𝑘))))
 
Theorembinomfallfac 14480* A version of the binomial theorem using falling factorials instead of exponentials. (Contributed by Scott Fenton, 13-Mar-2018.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → ((𝐴 + 𝐵) FallFac 𝑁) = Σ𝑘 ∈ (0...𝑁)((𝑁C𝑘) · ((𝐴 FallFac (𝑁𝑘)) · (𝐵 FallFac 𝑘))))
 
Theorembinomrisefac 14481* A version of the binomial theorem using rising factorials instead of exponentials. (Contributed by Scott Fenton, 16-Mar-2018.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → ((𝐴 + 𝐵) RiseFac 𝑁) = Σ𝑘 ∈ (0...𝑁)((𝑁C𝑘) · ((𝐴 RiseFac (𝑁𝑘)) · (𝐵 RiseFac 𝑘))))
 
Theoremfallfacval4 14482 Represent the falling factorial via factorials when the first argument is a natural. (Contributed by Scott Fenton, 20-Mar-2018.)
(𝑁 ∈ (0...𝐴) → (𝐴 FallFac 𝑁) = ((!‘𝐴) / (!‘(𝐴𝑁))))
 
Theorembcfallfac 14483 Binomial coefficient in terms of falling factorials. (Contributed by Scott Fenton, 20-Mar-2018.)
(𝐾 ∈ (0...𝑁) → (𝑁C𝐾) = ((𝑁 FallFac 𝐾) / (!‘𝐾)))
 
Theoremfallfacfac 14484 Relate falling factorial to factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝑁 ∈ ℕ0 → (𝑁 FallFac 𝑁) = (!‘𝑁))
 
5.10.14  Bernoulli polynomials and sums of k-th powers
 
Syntaxcbp 14485 Declare the constant for the Bernoulli polynomial operator.
class BernPoly
 
Definitiondf-bpoly 14486* Define the Bernoulli polynomials. Here we use well-founded recursion to define the Bernoulli polynomials. This agrees with most textbook definitions, although explicit formulae do exist. (Contributed by Scott Fenton, 22-May-2014.)
BernPoly = (𝑚 ∈ ℕ0, 𝑥 ∈ ℂ ↦ (wrecs( < , ℕ0, (𝑔 ∈ V ↦ (#‘dom 𝑔) / 𝑛((𝑥𝑛) − Σ𝑘 ∈ dom 𝑔((𝑛C𝑘) · ((𝑔𝑘) / ((𝑛𝑘) + 1))))))‘𝑚))
 
Theorembpolylem 14487* Lemma for bpolyval 14488. (Contributed by Scott Fenton, 22-May-2014.) (Revised by Mario Carneiro, 23-Aug-2014.)
𝐺 = (𝑔 ∈ V ↦ (#‘dom 𝑔) / 𝑛((𝑋𝑛) − Σ𝑘 ∈ dom 𝑔((𝑛C𝑘) · ((𝑔𝑘) / ((𝑛𝑘) + 1)))))    &   𝐹 = wrecs( < , ℕ0, 𝐺)       ((𝑁 ∈ ℕ0𝑋 ∈ ℂ) → (𝑁 BernPoly 𝑋) = ((𝑋𝑁) − Σ𝑘 ∈ (0...(𝑁 − 1))((𝑁C𝑘) · ((𝑘 BernPoly 𝑋) / ((𝑁𝑘) + 1)))))
 
Theorembpolyval 14488* The value of the Bernoulli polynomials. (Contributed by Scott Fenton, 16-May-2014.)
((𝑁 ∈ ℕ0𝑋 ∈ ℂ) → (𝑁 BernPoly 𝑋) = ((𝑋𝑁) − Σ𝑘 ∈ (0...(𝑁 − 1))((𝑁C𝑘) · ((𝑘 BernPoly 𝑋) / ((𝑁𝑘) + 1)))))
 
Theorembpoly0 14489 The value of the Bernoulli polynomials at zero. (Contributed by Scott Fenton, 16-May-2014.)
(𝑋 ∈ ℂ → (0 BernPoly 𝑋) = 1)
 
Theorembpoly1 14490 The value of the Bernoulli polynomials at one. (Contributed by Scott Fenton, 16-May-2014.)
(𝑋 ∈ ℂ → (1 BernPoly 𝑋) = (𝑋 − (1 / 2)))
 
Theorembpolycl 14491 Closure law for Bernoulli polynomials. (Contributed by Scott Fenton, 16-May-2014.) (Proof shortened by Mario Carneiro, 22-May-2014.)
((𝑁 ∈ ℕ0𝑋 ∈ ℂ) → (𝑁 BernPoly 𝑋) ∈ ℂ)
 
Theorembpolysum 14492* A sum for Bernoulli polynomials. (Contributed by Scott Fenton, 16-May-2014.) (Proof shortened by Mario Carneiro, 22-May-2014.)
((𝑁 ∈ ℕ0𝑋 ∈ ℂ) → Σ𝑘 ∈ (0...𝑁)((𝑁C𝑘) · ((𝑘 BernPoly 𝑋) / ((𝑁𝑘) + 1))) = (𝑋𝑁))
 
Theorembpolydiflem 14493* Lemma for bpolydif 14494. (Contributed by Scott Fenton, 12-Jun-2014.)
(𝜑𝑁 ∈ ℕ)    &   (𝜑𝑋 ∈ ℂ)    &   ((𝜑𝑘 ∈ (1...(𝑁 − 1))) → ((𝑘 BernPoly (𝑋 + 1)) − (𝑘 BernPoly 𝑋)) = (𝑘 · (𝑋↑(𝑘 − 1))))       (𝜑 → ((𝑁 BernPoly (𝑋 + 1)) − (𝑁 BernPoly 𝑋)) = (𝑁 · (𝑋↑(𝑁 − 1))))
 
Theorembpolydif 14494 Calculate the difference between successive values of the Bernoulli polynomials. (Contributed by Scott Fenton, 16-May-2014.) (Proof shortened by Mario Carneiro, 26-May-2014.)
((𝑁 ∈ ℕ ∧ 𝑋 ∈ ℂ) → ((𝑁 BernPoly (𝑋 + 1)) − (𝑁 BernPoly 𝑋)) = (𝑁 · (𝑋↑(𝑁 − 1))))
 
Theoremfsumkthpow 14495* A closed-form expression for the sum of 𝐾-th powers. (Contributed by Scott Fenton, 16-May-2014.) This is Metamath 100 proof #77. (Revised by Mario Carneiro, 16-Jun-2014.)
((𝐾 ∈ ℕ0𝑀 ∈ ℕ0) → Σ𝑛 ∈ (0...𝑀)(𝑛𝐾) = ((((𝐾 + 1) BernPoly (𝑀 + 1)) − ((𝐾 + 1) BernPoly 0)) / (𝐾 + 1)))
 
Theorembpoly2 14496 The Bernoulli polynomials at two. (Contributed by Scott Fenton, 8-Jul-2015.)
(𝑋 ∈ ℂ → (2 BernPoly 𝑋) = (((𝑋↑2) − 𝑋) + (1 / 6)))
 
Theorembpoly3 14497 The Bernoulli polynomials at three. (Contributed by Scott Fenton, 8-Jul-2015.)
(𝑋 ∈ ℂ → (3 BernPoly 𝑋) = (((𝑋↑3) − ((3 / 2) · (𝑋↑2))) + ((1 / 2) · 𝑋)))
 
Theorembpoly4 14498 The Bernoulli polynomials at four. (Contributed by Scott Fenton, 8-Jul-2015.)
(𝑋 ∈ ℂ → (4 BernPoly 𝑋) = ((((𝑋↑4) − (2 · (𝑋↑3))) + (𝑋↑2)) − (1 / 30)))
 
Theoremfsumcube 14499* Express the sum of cubes in closed terms. (Contributed by Scott Fenton, 16-Jun-2015.)
(𝑇 ∈ ℕ0 → Σ𝑘 ∈ (0...𝑇)(𝑘↑3) = (((𝑇↑2) · ((𝑇 + 1)↑2)) / 4))
 
5.11  Elementary trigonometry
 
5.11.1  The exponential, sine, and cosine functions
 
Syntaxce 14500 Extend class notation to include the exponential function.
class exp
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