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Type | Label | Description |
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Statement | ||
Theorem | dcubic2 26901* | Reverse direction of dcubic 26903. Given a solution 𝑈 to the "substitution" quadratic equation 𝑋 = 𝑈 − 𝑀 / 𝑈, show that 𝑋 is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015.) |
⊢ (𝜑 → 𝑃 ∈ ℂ) & ⊢ (𝜑 → 𝑄 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑇 ∈ ℂ) & ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁)) & ⊢ (𝜑 → 𝐺 ∈ ℂ) & ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3))) & ⊢ (𝜑 → 𝑀 = (𝑃 / 3)) & ⊢ (𝜑 → 𝑁 = (𝑄 / 2)) & ⊢ (𝜑 → 𝑇 ≠ 0) & ⊢ (𝜑 → 𝑈 ∈ ℂ) & ⊢ (𝜑 → 𝑈 ≠ 0) & ⊢ (𝜑 → 𝑋 = (𝑈 − (𝑀 / 𝑈))) & ⊢ (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0) ⇒ ⊢ (𝜑 → ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇))))) | ||
Theorem | dcubic1 26902 | Forward direction of dcubic 26903: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.) |
⊢ (𝜑 → 𝑃 ∈ ℂ) & ⊢ (𝜑 → 𝑄 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑇 ∈ ℂ) & ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁)) & ⊢ (𝜑 → 𝐺 ∈ ℂ) & ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3))) & ⊢ (𝜑 → 𝑀 = (𝑃 / 3)) & ⊢ (𝜑 → 𝑁 = (𝑄 / 2)) & ⊢ (𝜑 → 𝑇 ≠ 0) & ⊢ (𝜑 → 𝑋 = (𝑇 − (𝑀 / 𝑇))) ⇒ ⊢ (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0) | ||
Theorem | dcubic 26903* | Solutions to the depressed cubic, a special case of cubic 26906. (The definitions of 𝑀, 𝑁, 𝐺, 𝑇 here differ from mcubic 26904 by scale factors of -9, 54, 54 and -27 respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.) |
⊢ (𝜑 → 𝑃 ∈ ℂ) & ⊢ (𝜑 → 𝑄 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑇 ∈ ℂ) & ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁)) & ⊢ (𝜑 → 𝐺 ∈ ℂ) & ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3))) & ⊢ (𝜑 → 𝑀 = (𝑃 / 3)) & ⊢ (𝜑 → 𝑁 = (𝑄 / 2)) & ⊢ (𝜑 → 𝑇 ≠ 0) ⇒ ⊢ (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇)))))) | ||
Theorem | mcubic 26904* | Solutions to a monic cubic equation, a special case of cubic 26906. (Contributed by Mario Carneiro, 24-Apr-2015.) |
⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑇 ∈ ℂ) & ⊢ (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2)) & ⊢ (𝜑 → 𝐺 ∈ ℂ) & ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3)))) & ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · 𝐶))) & ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − (9 · (𝐵 · 𝐶))) + (;27 · 𝐷))) & ⊢ (𝜑 → 𝑇 ≠ 0) ⇒ ⊢ (𝜑 → ((((𝑋↑3) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / 3)))) | ||
Theorem | cubic2 26905* | The solution to the general cubic equation, for arbitrary choices 𝐺 and 𝑇 of the square and cube roots. (Contributed by Mario Carneiro, 23-Apr-2015.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 0) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑇 ∈ ℂ) & ⊢ (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2)) & ⊢ (𝜑 → 𝐺 ∈ ℂ) & ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3)))) & ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶)))) & ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (;27 · ((𝐴↑2) · 𝐷)))) & ⊢ (𝜑 → 𝑇 ≠ 0) ⇒ ⊢ (𝜑 → ((((𝐴 · (𝑋↑3)) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / (3 · 𝐴))))) | ||
Theorem | cubic 26906* | The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4710 to convert the existential quantifier to a triple disjunction. This is Metamath 100 proof #37. (Contributed by Mario Carneiro, 26-Apr-2015.) |
⊢ 𝑅 = {1, ((-1 + (i · (√‘3))) / 2), ((-1 − (i · (√‘3))) / 2)} & ⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 0) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑇 = (((𝑁 + (√‘𝐺)) / 2)↑𝑐(1 / 3))) & ⊢ (𝜑 → 𝐺 = ((𝑁↑2) − (4 · (𝑀↑3)))) & ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶)))) & ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (;27 · ((𝐴↑2) · 𝐷)))) & ⊢ (𝜑 → 𝑀 ≠ 0) ⇒ ⊢ (𝜑 → ((((𝐴 · (𝑋↑3)) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ 𝑅 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / (3 · 𝐴)))) | ||
Theorem | binom4 26907 | Work out a quartic binomial. (You would think that by this point it would be faster to use binom 15862, but it turns out to be just as much work to put it into this form after clearing all the sums and calculating binomial coefficients.) (Contributed by Mario Carneiro, 6-May-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + 𝐵)↑4) = (((𝐴↑4) + (4 · ((𝐴↑3) · 𝐵))) + ((6 · ((𝐴↑2) · (𝐵↑2))) + ((4 · (𝐴 · (𝐵↑3))) + (𝐵↑4))))) | ||
Theorem | dquartlem1 26908 | Lemma for dquart 26910. (Contributed by Mario Carneiro, 6-May-2015.) |
⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑆 ∈ ℂ) & ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2)) & ⊢ (𝜑 → 𝑀 ≠ 0) & ⊢ (𝜑 → 𝐼 ∈ ℂ) & ⊢ (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆))) ⇒ ⊢ (𝜑 → ((((𝑋↑2) + ((𝑀 + 𝐵) / 2)) + ((((𝑀 / 2) · 𝑋) − (𝐶 / 4)) / 𝑆)) = 0 ↔ (𝑋 = (-𝑆 + 𝐼) ∨ 𝑋 = (-𝑆 − 𝐼)))) | ||
Theorem | dquartlem2 26909 | Lemma for dquart 26910. (Contributed by Mario Carneiro, 6-May-2015.) |
⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑆 ∈ ℂ) & ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2)) & ⊢ (𝜑 → 𝑀 ≠ 0) & ⊢ (𝜑 → 𝐼 ∈ ℂ) & ⊢ (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆))) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → (((𝑀↑3) + ((2 · 𝐵) · (𝑀↑2))) + ((((𝐵↑2) − (4 · 𝐷)) · 𝑀) + -(𝐶↑2))) = 0) ⇒ ⊢ (𝜑 → ((((𝑀 + 𝐵) / 2)↑2) − (((𝐶↑2) / 4) / 𝑀)) = 𝐷) | ||
Theorem | dquart 26910 | Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 26906 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.) |
⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑆 ∈ ℂ) & ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2)) & ⊢ (𝜑 → 𝑀 ≠ 0) & ⊢ (𝜑 → 𝐼 ∈ ℂ) & ⊢ (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆))) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → (((𝑀↑3) + ((2 · 𝐵) · (𝑀↑2))) + ((((𝐵↑2) − (4 · 𝐷)) · 𝑀) + -(𝐶↑2))) = 0) & ⊢ (𝜑 → 𝐽 ∈ ℂ) & ⊢ (𝜑 → (𝐽↑2) = ((-(𝑆↑2) − (𝐵 / 2)) − ((𝐶 / 4) / 𝑆))) ⇒ ⊢ (𝜑 → ((((𝑋↑4) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ((𝑋 = (-𝑆 + 𝐼) ∨ 𝑋 = (-𝑆 − 𝐼)) ∨ (𝑋 = (𝑆 + 𝐽) ∨ 𝑋 = (𝑆 − 𝐽))))) | ||
Theorem | quart1cl 26911 | Closure lemmas for quart 26918. (Contributed by Mario Carneiro, 7-May-2015.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2)))) & ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8))) & ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4))))) ⇒ ⊢ (𝜑 → (𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ ∧ 𝑅 ∈ ℂ)) | ||
Theorem | quart1lem 26912 | Lemma for quart1 26913. (Contributed by Mario Carneiro, 6-May-2015.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2)))) & ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8))) & ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4))))) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑌 = (𝑋 + (𝐴 / 4))) ⇒ ⊢ (𝜑 → 𝐷 = ((((𝐴↑4) / ;;256) + (𝑃 · ((𝐴 / 4)↑2))) + ((𝑄 · (𝐴 / 4)) + 𝑅))) | ||
Theorem | quart1 26913 | Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2)))) & ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8))) & ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4))))) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑌 = (𝑋 + (𝐴 / 4))) ⇒ ⊢ (𝜑 → (((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = (((𝑌↑4) + (𝑃 · (𝑌↑2))) + ((𝑄 · 𝑌) + 𝑅))) | ||
Theorem | quartlem1 26914 | Lemma for quart 26918. (Contributed by Mario Carneiro, 6-May-2015.) |
⊢ (𝜑 → 𝑃 ∈ ℂ) & ⊢ (𝜑 → 𝑄 ∈ ℂ) & ⊢ (𝜑 → 𝑅 ∈ ℂ) & ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅))) & ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅)))) ⇒ ⊢ (𝜑 → (𝑈 = (((2 · 𝑃)↑2) − (3 · ((𝑃↑2) − (4 · 𝑅)))) ∧ 𝑉 = (((2 · ((2 · 𝑃)↑3)) − (9 · ((2 · 𝑃) · ((𝑃↑2) − (4 · 𝑅))))) + (;27 · -(𝑄↑2))))) | ||
Theorem | quartlem2 26915 | Closure lemmas for quart 26918. (Contributed by Mario Carneiro, 7-May-2015.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝐸 = -(𝐴 / 4)) & ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2)))) & ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8))) & ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4))))) & ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅))) & ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅)))) & ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3))))) ⇒ ⊢ (𝜑 → (𝑈 ∈ ℂ ∧ 𝑉 ∈ ℂ ∧ 𝑊 ∈ ℂ)) | ||
Theorem | quartlem3 26916 | Closure lemmas for quart 26918. (Contributed by Mario Carneiro, 7-May-2015.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝐸 = -(𝐴 / 4)) & ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2)))) & ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8))) & ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4))))) & ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅))) & ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅)))) & ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3))))) & ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2)) & ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3)) & ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3))) & ⊢ (𝜑 → 𝑇 ≠ 0) ⇒ ⊢ (𝜑 → (𝑆 ∈ ℂ ∧ 𝑀 ∈ ℂ ∧ 𝑇 ∈ ℂ)) | ||
Theorem | quartlem4 26917 | Closure lemmas for quart 26918. (Contributed by Mario Carneiro, 7-May-2015.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝐸 = -(𝐴 / 4)) & ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2)))) & ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8))) & ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4))))) & ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅))) & ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅)))) & ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3))))) & ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2)) & ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3)) & ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3))) & ⊢ (𝜑 → 𝑇 ≠ 0) & ⊢ (𝜑 → 𝑀 ≠ 0) & ⊢ (𝜑 → 𝐼 = (√‘((-(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆)))) & ⊢ (𝜑 → 𝐽 = (√‘((-(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆)))) ⇒ ⊢ (𝜑 → (𝑆 ≠ 0 ∧ 𝐼 ∈ ℂ ∧ 𝐽 ∈ ℂ)) | ||
Theorem | quart 26918 | The quartic equation, writing out all roots using square and cube root functions so that only direct substitutions remain, and we can actually claim to have a "quartic equation". Naturally, this theorem is ridiculously long (see quartfull 35149) if all the substitutions are performed. This is Metamath 100 proof #46. (Contributed by Mario Carneiro, 6-May-2015.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝐸 = -(𝐴 / 4)) & ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2)))) & ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8))) & ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4))))) & ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅))) & ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅)))) & ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3))))) & ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2)) & ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3)) & ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3))) & ⊢ (𝜑 → 𝑇 ≠ 0) & ⊢ (𝜑 → 𝑀 ≠ 0) & ⊢ (𝜑 → 𝐼 = (√‘((-(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆)))) & ⊢ (𝜑 → 𝐽 = (√‘((-(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆)))) ⇒ ⊢ (𝜑 → ((((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = 0 ↔ ((𝑋 = ((𝐸 − 𝑆) + 𝐼) ∨ 𝑋 = ((𝐸 − 𝑆) − 𝐼)) ∨ (𝑋 = ((𝐸 + 𝑆) + 𝐽) ∨ 𝑋 = ((𝐸 + 𝑆) − 𝐽))))) | ||
Syntax | casin 26919 | The arcsine function. |
class arcsin | ||
Syntax | cacos 26920 | The arccosine function. |
class arccos | ||
Syntax | catan 26921 | The arctangent function. |
class arctan | ||
Definition | df-asin 26922 | Define the arcsine function. Because sin is not a one-to-one function, the literal inverse ◡sin is not a function. Rather than attempt to find the right domain on which to restrict sin in order to get a total function, we just define it in terms of log, which we already know is total (except at 0). There are branch points at -1 and 1 (at which the function is defined), and branch cuts along the real line not between -1 and 1, which is to say (-∞, -1) ∪ (1, +∞). (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arcsin = (𝑥 ∈ ℂ ↦ (-i · (log‘((i · 𝑥) + (√‘(1 − (𝑥↑2))))))) | ||
Definition | df-acos 26923 | Define the arccosine function. See also remarks for df-asin 26922. Since we define arccos in terms of arcsin, it shares the same branch points and cuts, namely (-∞, -1) ∪ (1, +∞). (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arccos = (𝑥 ∈ ℂ ↦ ((π / 2) − (arcsin‘𝑥))) | ||
Definition | df-atan 26924 | Define the arctangent function. See also remarks for df-asin 26922. Unlike arcsin and arccos, this function is not defined everywhere, because tan(𝑧) ≠ ±i for all 𝑧 ∈ ℂ. For all other 𝑧, there is a formula for arctan(𝑧) in terms of log, and we take that as the definition. Branch points are at ±i; branch cuts are on the pure imaginary axis not between -i and i, which is to say {𝑧 ∈ ℂ ∣ (i · 𝑧) ∈ (-∞, -1) ∪ (1, +∞)}. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arctan = (𝑥 ∈ (ℂ ∖ {-i, i}) ↦ ((i / 2) · ((log‘(1 − (i · 𝑥))) − (log‘(1 + (i · 𝑥)))))) | ||
Theorem | asinlem 26925 | The argument to the logarithm in df-asin 26922 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → ((i · 𝐴) + (√‘(1 − (𝐴↑2)))) ≠ 0) | ||
Theorem | asinlem2 26926 | The argument to the logarithm in df-asin 26922 has the property that replacing 𝐴 with -𝐴 in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (((i · 𝐴) + (√‘(1 − (𝐴↑2)))) · ((i · -𝐴) + (√‘(1 − (-𝐴↑2))))) = 1) | ||
Theorem | asinlem3a 26927 | Lemma for asinlem3 26928. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) ≤ 0) → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))) | ||
Theorem | asinlem3 26928 | The argument to the logarithm in df-asin 26922 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))) | ||
Theorem | asinf 26929 | Domain and codomain of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arcsin:ℂ⟶ℂ | ||
Theorem | asincl 26930 | Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (arcsin‘𝐴) ∈ ℂ) | ||
Theorem | acosf 26931 | Domain and codoamin of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arccos:ℂ⟶ℂ | ||
Theorem | acoscl 26932 | Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) ∈ ℂ) | ||
Theorem | atandm 26933 | Since the property is a little lengthy, we abbreviate 𝐴 ∈ ℂ ∧ 𝐴 ≠ -i ∧ 𝐴 ≠ i as 𝐴 ∈ dom arctan. This is the necessary precondition for the definition of arctan to make sense. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ 𝐴 ≠ -i ∧ 𝐴 ≠ i)) | ||
Theorem | atandm2 26934 | This form of atandm 26933 is a bit more useful for showing that the logarithms in df-atan 26924 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 − (i · 𝐴)) ≠ 0 ∧ (1 + (i · 𝐴)) ≠ 0)) | ||
Theorem | atandm3 26935 | A compact form of atandm 26933. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (𝐴↑2) ≠ -1)) | ||
Theorem | atandm4 26936 | A compact form of atandm 26933. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ≠ 0)) | ||
Theorem | atanf 26937 | Domain and codoamin of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arctan:(ℂ ∖ {-i, i})⟶ℂ | ||
Theorem | atancl 26938 | Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan → (arctan‘𝐴) ∈ ℂ) | ||
Theorem | asinval 26939 | Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (arcsin‘𝐴) = (-i · (log‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))) | ||
Theorem | acosval 26940 | Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) = ((π / 2) − (arcsin‘𝐴))) | ||
Theorem | atanval 26941 | Value of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan → (arctan‘𝐴) = ((i / 2) · ((log‘(1 − (i · 𝐴))) − (log‘(1 + (i · 𝐴)))))) | ||
Theorem | atanre 26942 | A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℝ → 𝐴 ∈ dom arctan) | ||
Theorem | asinneg 26943 | The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (arcsin‘-𝐴) = -(arcsin‘𝐴)) | ||
Theorem | acosneg 26944 | The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (arccos‘-𝐴) = (π − (arccos‘𝐴))) | ||
Theorem | efiasin 26945 | The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (exp‘(i · (arcsin‘𝐴))) = ((i · 𝐴) + (√‘(1 − (𝐴↑2))))) | ||
Theorem | sinasin 26946 | The arcsine function is an inverse to sin. This is the main property that justifies the notation arcsin or sin↑-1. Because sin is not an injection, the other converse identity asinsin 26949 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (sin‘(arcsin‘𝐴)) = 𝐴) | ||
Theorem | cosacos 26947 | The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (cos‘(arccos‘𝐴)) = 𝐴) | ||
Theorem | asinsinlem 26948 | Lemma for asinsin 26949. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → 0 < (ℜ‘(exp‘(i · 𝐴)))) | ||
Theorem | asinsin 26949 | The arcsine function composed with sin is equal to the identity. This plus sinasin 26946 allow to view sin and arcsin as inverse operations to each other. For ease of use, we have not defined precisely the correct domain of correctness of this identity; in addition to the main region described here it is also true for some points on the branch cuts, namely when 𝐴 = (π / 2) − i𝑦 for nonnegative real 𝑦 and also symmetrically at 𝐴 = i𝑦 − (π / 2). In particular, when restricted to reals this identity extends to the closed interval [-(π / 2), (π / 2)], not just the open interval (see reasinsin 26953). (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → (arcsin‘(sin‘𝐴)) = 𝐴) | ||
Theorem | acoscos 26950 | The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (0(,)π)) → (arccos‘(cos‘𝐴)) = 𝐴) | ||
Theorem | asin1 26951 | The arcsine of 1 is π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (arcsin‘1) = (π / 2) | ||
Theorem | acos1 26952 | The arccosine of 1 is 0. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (arccos‘1) = 0 | ||
Theorem | reasinsin 26953 | The arcsine function composed with sin is equal to the identity. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ (-(π / 2)[,](π / 2)) → (arcsin‘(sin‘𝐴)) = 𝐴) | ||
Theorem | asinsinb 26954 | Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (-(π / 2)(,)(π / 2))) → ((arcsin‘𝐴) = 𝐵 ↔ (sin‘𝐵) = 𝐴)) | ||
Theorem | acoscosb 26955 | Relationship between cosine and arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (0(,)π)) → ((arccos‘𝐴) = 𝐵 ↔ (cos‘𝐵) = 𝐴)) | ||
Theorem | asinbnd 26956 | The arcsine function has range within a vertical strip of the complex plane with real part between -π / 2 and π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (ℜ‘(arcsin‘𝐴)) ∈ (-(π / 2)[,](π / 2))) | ||
Theorem | acosbnd 26957 | The arccosine function has range within a vertical strip of the complex plane with real part between 0 and π. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (ℜ‘(arccos‘𝐴)) ∈ (0[,]π)) | ||
Theorem | asinrebnd 26958 | Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ (-(π / 2)[,](π / 2))) | ||
Theorem | asinrecl 26959 | The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ ℝ) | ||
Theorem | acosrecl 26960 | The arccosine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ (-1[,]1) → (arccos‘𝐴) ∈ ℝ) | ||
Theorem | cosasin 26961 | The cosine of the arcsine of 𝐴 is √(1 − 𝐴↑2). (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (cos‘(arcsin‘𝐴)) = (√‘(1 − (𝐴↑2)))) | ||
Theorem | sinacos 26962 | The sine of the arccosine of 𝐴 is √(1 − 𝐴↑2). (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (sin‘(arccos‘𝐴)) = (√‘(1 − (𝐴↑2)))) | ||
Theorem | atandmneg 26963 | The domain of the arctangent function is closed under negatives. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → -𝐴 ∈ dom arctan) | ||
Theorem | atanneg 26964 | The arctangent function is odd. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (arctan‘-𝐴) = -(arctan‘𝐴)) | ||
Theorem | atan0 26965 | The arctangent of zero is zero. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (arctan‘0) = 0 | ||
Theorem | atandmcj 26966 | The arctangent function distributes under conjugation. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan → (∗‘𝐴) ∈ dom arctan) | ||
Theorem | atancj 26967 | The arctangent function distributes under conjugation. (The condition that ℜ(𝐴) ≠ 0 is necessary because the branch cuts are chosen so that the negative imaginary line "agrees with" neighboring values with negative real part, while the positive imaginary line agrees with values with positive real part. This makes atanneg 26964 true unconditionally but messes up conjugation symmetry, and it is impossible to have both in a single-valued function. The claim is true on the imaginary line between -1 and 1, though.) (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ≠ 0) → (𝐴 ∈ dom arctan ∧ (∗‘(arctan‘𝐴)) = (arctan‘(∗‘𝐴)))) | ||
Theorem | atanrecl 26968 | The arctangent function is real for all real inputs. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℝ → (arctan‘𝐴) ∈ ℝ) | ||
Theorem | efiatan 26969 | Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (exp‘(i · (arctan‘𝐴))) = ((√‘(1 + (i · 𝐴))) / (√‘(1 − (i · 𝐴))))) | ||
Theorem | atanlogaddlem 26970 | Lemma for atanlogadd 26971. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ ((𝐴 ∈ dom arctan ∧ 0 ≤ (ℜ‘𝐴)) → ((log‘(1 + (i · 𝐴))) + (log‘(1 − (i · 𝐴)))) ∈ ran log) | ||
Theorem | atanlogadd 26971 | The rule √(𝑧𝑤) = (√𝑧)(√𝑤) is not always true on the complex numbers, but it is true when the arguments of 𝑧 and 𝑤 sum to within the interval (-π, π], so there are some cases such as this one with 𝑧 = 1 + i𝐴 and 𝑤 = 1 − i𝐴 which are true unconditionally. This result can also be stated as "√(1 + 𝑧) + √(1 − 𝑧) is analytic". (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → ((log‘(1 + (i · 𝐴))) + (log‘(1 − (i · 𝐴)))) ∈ ran log) | ||
Theorem | atanlogsublem 26972 | Lemma for atanlogsub 26973. (Contributed by Mario Carneiro, 4-Apr-2015.) |
⊢ ((𝐴 ∈ dom arctan ∧ 0 < (ℜ‘𝐴)) → (ℑ‘((log‘(1 + (i · 𝐴))) − (log‘(1 − (i · 𝐴))))) ∈ (-π(,)π)) | ||
Theorem | atanlogsub 26973 | A variation on atanlogadd 26971, to show that √(1 + i𝑧) / √(1 − i𝑧) = √((1 + i𝑧) / (1 − i𝑧)) under more limited conditions. (Contributed by Mario Carneiro, 4-Apr-2015.) |
⊢ ((𝐴 ∈ dom arctan ∧ (ℜ‘𝐴) ≠ 0) → ((log‘(1 + (i · 𝐴))) − (log‘(1 − (i · 𝐴)))) ∈ ran log) | ||
Theorem | efiatan2 26974 | Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (exp‘(i · (arctan‘𝐴))) = ((1 + (i · 𝐴)) / (√‘(1 + (𝐴↑2))))) | ||
Theorem | 2efiatan 26975 | Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (exp‘(2 · (i · (arctan‘𝐴)))) = (((2 · i) / (𝐴 + i)) − 1)) | ||
Theorem | tanatan 26976 | The arctangent function is an inverse to tan. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (tan‘(arctan‘𝐴)) = 𝐴) | ||
Theorem | atandmtan 26977 | The tangent function has range contained in the domain of the arctangent. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (cos‘𝐴) ≠ 0) → (tan‘𝐴) ∈ dom arctan) | ||
Theorem | cosatan 26978 | The cosine of an arctangent. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (cos‘(arctan‘𝐴)) = (1 / (√‘(1 + (𝐴↑2))))) | ||
Theorem | cosatanne0 26979 | The arctangent function has range contained in the domain of the tangent. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (cos‘(arctan‘𝐴)) ≠ 0) | ||
Theorem | atantan 26980 | The arctangent function is an inverse to tan. (Contributed by Mario Carneiro, 5-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → (arctan‘(tan‘𝐴)) = 𝐴) | ||
Theorem | atantanb 26981 | Relationship between tangent and arctangent. (Contributed by Mario Carneiro, 5-Apr-2015.) |
⊢ ((𝐴 ∈ dom arctan ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (-(π / 2)(,)(π / 2))) → ((arctan‘𝐴) = 𝐵 ↔ (tan‘𝐵) = 𝐴)) | ||
Theorem | atanbndlem 26982 | Lemma for atanbnd 26983. (Contributed by Mario Carneiro, 5-Apr-2015.) |
⊢ (𝐴 ∈ ℝ+ → (arctan‘𝐴) ∈ (-(π / 2)(,)(π / 2))) | ||
Theorem | atanbnd 26983 | The arctangent function is bounded by π / 2 on the reals. (Contributed by Mario Carneiro, 5-Apr-2015.) |
⊢ (𝐴 ∈ ℝ → (arctan‘𝐴) ∈ (-(π / 2)(,)(π / 2))) | ||
Theorem | atanord 26984 | The arctangent function is strictly increasing. (Contributed by Mario Carneiro, 5-Apr-2015.) |
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (𝐴 < 𝐵 ↔ (arctan‘𝐴) < (arctan‘𝐵))) | ||
Theorem | atan1 26985 | The arctangent of 1 is π / 4. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (arctan‘1) = (π / 4) | ||
Theorem | bndatandm 26986 | A point in the open unit disk is in the domain of the arctangent. (Contributed by Mario Carneiro, 5-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → 𝐴 ∈ dom arctan) | ||
Theorem | atans 26987* | The "domain of continuity" of the arctangent. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐷 = (ℂ ∖ (-∞(,]0)) & ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷} ⇒ ⊢ (𝐴 ∈ 𝑆 ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ∈ 𝐷)) | ||
Theorem | atans2 26988* | It suffices to show that 1 − i𝐴 and 1 + i𝐴 are in the continuity domain of log to show that 𝐴 is in the continuity domain of arctangent. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐷 = (ℂ ∖ (-∞(,]0)) & ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷} ⇒ ⊢ (𝐴 ∈ 𝑆 ↔ (𝐴 ∈ ℂ ∧ (1 − (i · 𝐴)) ∈ 𝐷 ∧ (1 + (i · 𝐴)) ∈ 𝐷)) | ||
Theorem | atansopn 26989* | The domain of continuity of the arctangent is an open set. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐷 = (ℂ ∖ (-∞(,]0)) & ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷} ⇒ ⊢ 𝑆 ∈ (TopOpen‘ℂfld) | ||
Theorem | atansssdm 26990* | The domain of continuity of the arctangent is a subset of the actual domain of the arctangent. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐷 = (ℂ ∖ (-∞(,]0)) & ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷} ⇒ ⊢ 𝑆 ⊆ dom arctan | ||
Theorem | ressatans 26991* | The real number line is a subset of the domain of continuity of the arctangent. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐷 = (ℂ ∖ (-∞(,]0)) & ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷} ⇒ ⊢ ℝ ⊆ 𝑆 | ||
Theorem | dvatan 26992* | The derivative of the arctangent. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐷 = (ℂ ∖ (-∞(,]0)) & ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷} ⇒ ⊢ (ℂ D (arctan ↾ 𝑆)) = (𝑥 ∈ 𝑆 ↦ (1 / (1 + (𝑥↑2)))) | ||
Theorem | atancn 26993* | The arctangent is a continuous function. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐷 = (ℂ ∖ (-∞(,]0)) & ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷} ⇒ ⊢ (arctan ↾ 𝑆) ∈ (𝑆–cn→ℂ) | ||
Theorem | atantayl 26994* | The Taylor series for arctan(𝐴). (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ (((i · ((-i↑𝑛) − (i↑𝑛))) / 2) · ((𝐴↑𝑛) / 𝑛))) ⇒ ⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → seq1( + , 𝐹) ⇝ (arctan‘𝐴)) | ||
Theorem | atantayl2 26995* | The Taylor series for arctan(𝐴). (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(2 ∥ 𝑛, 0, ((-1↑((𝑛 − 1) / 2)) · ((𝐴↑𝑛) / 𝑛)))) ⇒ ⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → seq1( + , 𝐹) ⇝ (arctan‘𝐴)) | ||
Theorem | atantayl3 26996* | The Taylor series for arctan(𝐴). (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐹 = (𝑛 ∈ ℕ0 ↦ ((-1↑𝑛) · ((𝐴↑((2 · 𝑛) + 1)) / ((2 · 𝑛) + 1)))) ⇒ ⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → seq0( + , 𝐹) ⇝ (arctan‘𝐴)) | ||
Theorem | leibpilem1 26997 | Lemma for leibpi 26999. (Contributed by Mario Carneiro, 7-Apr-2015.) (Proof shortened by Steven Nguyen, 23-Mar-2023.) |
⊢ ((𝑁 ∈ ℕ0 ∧ (¬ 𝑁 = 0 ∧ ¬ 2 ∥ 𝑁)) → (𝑁 ∈ ℕ ∧ ((𝑁 − 1) / 2) ∈ ℕ0)) | ||
Theorem | leibpilem2 26998* | The Leibniz formula for π. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐹 = (𝑛 ∈ ℕ0 ↦ ((-1↑𝑛) / ((2 · 𝑛) + 1))) & ⊢ 𝐺 = (𝑘 ∈ ℕ0 ↦ if((𝑘 = 0 ∨ 2 ∥ 𝑘), 0, ((-1↑((𝑘 − 1) / 2)) / 𝑘))) & ⊢ 𝐴 ∈ V ⇒ ⊢ (seq0( + , 𝐹) ⇝ 𝐴 ↔ seq0( + , 𝐺) ⇝ 𝐴) | ||
Theorem | leibpi 26999 | The Leibniz formula for π. This proof depends on three main facts: (1) the series 𝐹 is convergent, because it is an alternating series (iseralt 15717). (2) Using leibpilem2 26998 to rewrite the series as a power series, it is the 𝑥 = 1 special case of the Taylor series for arctan (atantayl2 26995). (3) Although we cannot directly plug 𝑥 = 1 into atantayl2 26995, Abel's theorem (abelth2 26500) says that the limit along any sequence converging to 1, such as 1 − 1 / 𝑛, of the power series converges to the power series extended to 1, and then since arctan is continuous at 1 (atancn 26993) we get the desired result. This is Metamath 100 proof #26. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐹 = (𝑛 ∈ ℕ0 ↦ ((-1↑𝑛) / ((2 · 𝑛) + 1))) ⇒ ⊢ seq0( + , 𝐹) ⇝ (π / 4) | ||
Theorem | leibpisum 27000 | The Leibniz formula for π. This version of leibpi 26999 looks nicer but does not assert that the series is convergent so is not as practically useful. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ Σ𝑛 ∈ ℕ0 ((-1↑𝑛) / ((2 · 𝑛) + 1)) = (π / 4) |
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