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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | isosctrlem2 26801 | Lemma for isosctr 26803. Corresponds to the case where one vertex is at 0, another at 1 and the third lies on the unit circle. (Contributed by Saveliy Skresanov, 31-Dec-2016.) |
⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) = (ℑ‘(log‘(-𝐴 / (1 − 𝐴))))) | ||
Theorem | isosctrlem3 26802* | Lemma for isosctr 26803. Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017.) |
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥)))) ⇒ ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0 ∧ 𝐴 ≠ 𝐵) ∧ (abs‘𝐴) = (abs‘𝐵)) → (-𝐴𝐹(𝐵 − 𝐴)) = ((𝐴 − 𝐵)𝐹-𝐵)) | ||
Theorem | isosctr 26803* | Isosceles triangle theorem. This is Metamath 100 proof #65. (Contributed by Saveliy Skresanov, 1-Jan-2017.) |
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥)))) ⇒ ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶 ∧ 𝐴 ≠ 𝐵) ∧ (abs‘(𝐴 − 𝐶)) = (abs‘(𝐵 − 𝐶))) → ((𝐶 − 𝐴)𝐹(𝐵 − 𝐴)) = ((𝐴 − 𝐵)𝐹(𝐶 − 𝐵))) | ||
Theorem | ssscongptld 26804* |
If two triangles have equal sides, one angle in one triangle has the
same cosine as the corresponding angle in the other triangle. This is a
partial form of the SSS congruence theorem.
This theorem is proven by using lawcos 26798 on both triangles to express one side in terms of the other two, and then equating these expressions and reducing this algebraically to get an equality of cosines of angles. (Contributed by David Moews, 28-Feb-2017.) |
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥)))) & ⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝐸 ∈ ℂ) & ⊢ (𝜑 → 𝐺 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 𝐵) & ⊢ (𝜑 → 𝐵 ≠ 𝐶) & ⊢ (𝜑 → 𝐷 ≠ 𝐸) & ⊢ (𝜑 → 𝐸 ≠ 𝐺) & ⊢ (𝜑 → (abs‘(𝐴 − 𝐵)) = (abs‘(𝐷 − 𝐸))) & ⊢ (𝜑 → (abs‘(𝐵 − 𝐶)) = (abs‘(𝐸 − 𝐺))) & ⊢ (𝜑 → (abs‘(𝐶 − 𝐴)) = (abs‘(𝐺 − 𝐷))) ⇒ ⊢ (𝜑 → (cos‘((𝐴 − 𝐵)𝐹(𝐶 − 𝐵))) = (cos‘((𝐷 − 𝐸)𝐹(𝐺 − 𝐸)))) | ||
Theorem | affineequiv 26805 | Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) ⇒ ⊢ (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐶 − 𝐵) = (𝐷 · (𝐶 − 𝐴)))) | ||
Theorem | affineequiv2 26806 | Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) ⇒ ⊢ (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐵 − 𝐴) = ((1 − 𝐷) · (𝐶 − 𝐴)))) | ||
Theorem | affineequiv3 26807 | Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶. (Contributed by AV, 22-Jan-2023.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) ⇒ ⊢ (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ (𝐴 − 𝐵) = (𝐷 · (𝐶 − 𝐵)))) | ||
Theorem | affineequiv4 26808 | Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶. (Contributed by AV, 22-Jan-2023.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) ⇒ ⊢ (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ 𝐴 = ((𝐷 · (𝐶 − 𝐵)) + 𝐵))) | ||
Theorem | affineequivne 26809 | Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶 if 𝐵 and 𝐶 are not equal. (Contributed by AV, 22-Jan-2023.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ≠ 𝐶) ⇒ ⊢ (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ 𝐷 = ((𝐴 − 𝐵) / (𝐶 − 𝐵)))) | ||
Theorem | angpieqvdlem 26810 | Equivalence used in the proof of angpieqvd 26813. (Contributed by David Moews, 28-Feb-2017.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 𝐵) & ⊢ (𝜑 → 𝐴 ≠ 𝐶) ⇒ ⊢ (𝜑 → (-((𝐶 − 𝐵) / (𝐴 − 𝐵)) ∈ ℝ+ ↔ ((𝐶 − 𝐵) / (𝐶 − 𝐴)) ∈ (0(,)1))) | ||
Theorem | angpieqvdlem2 26811* | Equivalence used in angpieqvd 26813. (Contributed by David Moews, 28-Feb-2017.) |
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥)))) & ⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 𝐵) & ⊢ (𝜑 → 𝐵 ≠ 𝐶) ⇒ ⊢ (𝜑 → (-((𝐶 − 𝐵) / (𝐴 − 𝐵)) ∈ ℝ+ ↔ ((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)) = π)) | ||
Theorem | angpined 26812* | If the angle at ABC is π, then 𝐴 is not equal to 𝐶. (Contributed by David Moews, 28-Feb-2017.) |
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥)))) & ⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 𝐵) & ⊢ (𝜑 → 𝐵 ≠ 𝐶) ⇒ ⊢ (𝜑 → (((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)) = π → 𝐴 ≠ 𝐶)) | ||
Theorem | angpieqvd 26813* | The angle ABC is π iff 𝐵 is a nontrivial convex combination of 𝐴 and 𝐶, i.e., iff 𝐵 is in the interior of the segment AC. (Contributed by David Moews, 28-Feb-2017.) |
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥)))) & ⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 𝐵) & ⊢ (𝜑 → 𝐵 ≠ 𝐶) ⇒ ⊢ (𝜑 → (((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)) = π ↔ ∃𝑤 ∈ (0(,)1)𝐵 = ((𝑤 · 𝐴) + ((1 − 𝑤) · 𝐶)))) | ||
Theorem | chordthmlem 26814* | If 𝑀 is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 26804 to observe that, since AMQ and BMQ have equal sides, the angles QMB and QMA must be equal. Since they are supplementary, both must be right angles. (Contributed by David Moews, 28-Feb-2017.) |
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥)))) & ⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝑄 ∈ ℂ) & ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2)) & ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄))) & ⊢ (𝜑 → 𝐴 ≠ 𝐵) & ⊢ (𝜑 → 𝑄 ≠ 𝑀) ⇒ ⊢ (𝜑 → ((𝑄 − 𝑀)𝐹(𝐵 − 𝑀)) ∈ {(π / 2), -(π / 2)}) | ||
Theorem | chordthmlem2 26815* | If M is the midpoint of AB, AQ = BQ, and P is on the line AB, then QMP is a right angle. This is proven by reduction to the special case chordthmlem 26814, where P = B, and using angrtmuld 26790 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.) |
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥)))) & ⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝑄 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℝ) & ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2)) & ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵))) & ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄))) & ⊢ (𝜑 → 𝑃 ≠ 𝑀) & ⊢ (𝜑 → 𝑄 ≠ 𝑀) ⇒ ⊢ (𝜑 → ((𝑄 − 𝑀)𝐹(𝑃 − 𝑀)) ∈ {(π / 2), -(π / 2)}) | ||
Theorem | chordthmlem3 26816 | If M is the midpoint of AB, AQ = BQ, and P is on the line AB, then PQ 2 = QM 2 + PM 2 . This follows from chordthmlem2 26815 and the Pythagorean theorem (pythag 26799) in the case where P and Q are unequal to M. If either P or Q equals M, the result is trivial. (Contributed by David Moews, 28-Feb-2017.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝑄 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℝ) & ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2)) & ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵))) & ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄))) ⇒ ⊢ (𝜑 → ((abs‘(𝑃 − 𝑄))↑2) = (((abs‘(𝑄 − 𝑀))↑2) + ((abs‘(𝑃 − 𝑀))↑2))) | ||
Theorem | chordthmlem4 26817 | If P is on the segment AB and M is the midpoint of AB, then PA · PB = BM 2 − PM 2 . If all lengths are reexpressed as fractions of AB, this reduces to the identity 𝑋 · (1 − 𝑋) = (1 / 2) 2 − ((1 / 2) − 𝑋) 2 . (Contributed by David Moews, 28-Feb-2017.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ (0[,]1)) & ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2)) & ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵))) ⇒ ⊢ (𝜑 → ((abs‘(𝑃 − 𝐴)) · (abs‘(𝑃 − 𝐵))) = (((abs‘(𝐵 − 𝑀))↑2) − ((abs‘(𝑃 − 𝑀))↑2))) | ||
Theorem | chordthmlem5 26818 | If P is on the segment AB and AQ = BQ, then PA · PB = BQ 2 − PQ 2 . This follows from two uses of chordthmlem3 26816 to show that PQ 2 = QM 2 + PM 2 and BQ 2 = QM 2 + BM 2 , so BQ 2 − PQ 2 = (QM 2 + BM 2 ) − (QM 2 + PM 2 ) = BM 2 − PM 2 , which equals PA · PB by chordthmlem4 26817. (Contributed by David Moews, 28-Feb-2017.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝑄 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ (0[,]1)) & ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵))) & ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄))) ⇒ ⊢ (𝜑 → ((abs‘(𝑃 − 𝐴)) · (abs‘(𝑃 − 𝐵))) = (((abs‘(𝐵 − 𝑄))↑2) − ((abs‘(𝑃 − 𝑄))↑2))) | ||
Theorem | chordthm 26819* | The intersecting chords theorem. If points A, B, C, and D lie on a circle (with center Q, say), and the point P is on the interior of the segments AB and CD, then the two products of lengths PA · PB and PC · PD are equal. The Euclidean plane is identified with the complex plane, and the fact that P is on AB and on CD is expressed by the hypothesis that the angles APB and CPD are equal to π. The result is proven by using chordthmlem5 26818 twice to show that PA · PB and PC · PD both equal BQ 2 − PQ 2 . This is similar to the proof of the theorem given in Euclid's Elements, where it is Proposition III.35. This is Metamath 100 proof #55. (Contributed by David Moews, 28-Feb-2017.) |
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥)))) & ⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑃 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 𝑃) & ⊢ (𝜑 → 𝐵 ≠ 𝑃) & ⊢ (𝜑 → 𝐶 ≠ 𝑃) & ⊢ (𝜑 → 𝐷 ≠ 𝑃) & ⊢ (𝜑 → ((𝐴 − 𝑃)𝐹(𝐵 − 𝑃)) = π) & ⊢ (𝜑 → ((𝐶 − 𝑃)𝐹(𝐷 − 𝑃)) = π) & ⊢ (𝜑 → 𝑄 ∈ ℂ) & ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄))) & ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐶 − 𝑄))) & ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐷 − 𝑄))) ⇒ ⊢ (𝜑 → ((abs‘(𝑃 − 𝐴)) · (abs‘(𝑃 − 𝐵))) = ((abs‘(𝑃 − 𝐶)) · (abs‘(𝑃 − 𝐷)))) | ||
Theorem | heron 26820* | Heron's formula gives the area of a triangle given only the side lengths. If points A, B, C form a triangle, then the area of the triangle, represented here as (1 / 2) · 𝑋 · 𝑌 · abs(sin𝑂), is equal to the square root of 𝑆 · (𝑆 − 𝑋) · (𝑆 − 𝑌) · (𝑆 − 𝑍), where 𝑆 = (𝑋 + 𝑌 + 𝑍) / 2 is half the perimeter of the triangle. Based on work by Jon Pennant. This is Metamath 100 proof #57. (Contributed by Mario Carneiro, 10-Mar-2019.) |
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥)))) & ⊢ 𝑋 = (abs‘(𝐵 − 𝐶)) & ⊢ 𝑌 = (abs‘(𝐴 − 𝐶)) & ⊢ 𝑍 = (abs‘(𝐴 − 𝐵)) & ⊢ 𝑂 = ((𝐵 − 𝐶)𝐹(𝐴 − 𝐶)) & ⊢ 𝑆 = (((𝑋 + 𝑌) + 𝑍) / 2) & ⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 𝐶) & ⊢ (𝜑 → 𝐵 ≠ 𝐶) ⇒ ⊢ (𝜑 → (((1 / 2) · (𝑋 · 𝑌)) · (abs‘(sin‘𝑂))) = (√‘((𝑆 · (𝑆 − 𝑋)) · ((𝑆 − 𝑌) · (𝑆 − 𝑍))))) | ||
Theorem | quad2 26821 | The quadratic equation, without specifying the particular branch 𝐷 to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 0) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → (𝐷↑2) = ((𝐵↑2) − (4 · (𝐴 · 𝐶)))) ⇒ ⊢ (𝜑 → (((𝐴 · (𝑋↑2)) + ((𝐵 · 𝑋) + 𝐶)) = 0 ↔ (𝑋 = ((-𝐵 + 𝐷) / (2 · 𝐴)) ∨ 𝑋 = ((-𝐵 − 𝐷) / (2 · 𝐴))))) | ||
Theorem | quad 26822 | The quadratic equation. (Contributed by Mario Carneiro, 23-Apr-2015.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐴 ≠ 0) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝐷 = ((𝐵↑2) − (4 · (𝐴 · 𝐶)))) ⇒ ⊢ (𝜑 → (((𝐴 · (𝑋↑2)) + ((𝐵 · 𝑋) + 𝐶)) = 0 ↔ (𝑋 = ((-𝐵 + (√‘𝐷)) / (2 · 𝐴)) ∨ 𝑋 = ((-𝐵 − (√‘𝐷)) / (2 · 𝐴))))) | ||
Theorem | 1cubrlem 26823 | The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.) |
⊢ ((-1↑𝑐(2 / 3)) = ((-1 + (i · (√‘3))) / 2) ∧ ((-1↑𝑐(2 / 3))↑2) = ((-1 − (i · (√‘3))) / 2)) | ||
Theorem | 1cubr 26824 | The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.) |
⊢ 𝑅 = {1, ((-1 + (i · (√‘3))) / 2), ((-1 − (i · (√‘3))) / 2)} ⇒ ⊢ (𝐴 ∈ 𝑅 ↔ (𝐴 ∈ ℂ ∧ (𝐴↑3) = 1)) | ||
Theorem | dcubic1lem 26825 | Lemma for dcubic1 26827 and dcubic2 26826: simplify the cubic equation under the substitution 𝑋 = 𝑈 − 𝑀 / 𝑈. (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 → 𝑈 ∈ ℂ) & ⊢ (𝜑 → 𝑈 ≠ 0) & ⊢ (𝜑 → 𝑋 = (𝑈 − (𝑀 / 𝑈))) ⇒ ⊢ (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ (((𝑈↑3)↑2) + ((𝑄 · (𝑈↑3)) − (𝑀↑3))) = 0)) | ||
Theorem | dcubic2 26826* | Reverse direction of dcubic 26828. 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 26827 | Forward direction of dcubic 26828: 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 26828* | Solutions to the depressed cubic, a special case of cubic 26831. (The definitions of 𝑀, 𝑁, 𝐺, 𝑇 here differ from mcubic 26829 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 26829* | Solutions to a monic cubic equation, a special case of cubic 26831. (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 26830* | 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 26831* | The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4712 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 26832 | Work out a quartic binomial. (You would think that by this point it would be faster to use binom 15817, 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 26833 | Lemma for dquart 26835. (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 26834 | Lemma for dquart 26835. (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 26835 | Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 26831 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 26836 | Closure lemmas for quart 26843. (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 26837 | Lemma for quart1 26838. (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 26838 | 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 26839 | Lemma for quart 26843. (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 26840 | Closure lemmas for quart 26843. (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 26841 | Closure lemmas for quart 26843. (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 26842 | Closure lemmas for quart 26843. (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 26843 | 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 34908) 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 26844 | The arcsine function. |
class arcsin | ||
Syntax | cacos 26845 | The arccosine function. |
class arccos | ||
Syntax | catan 26846 | The arctangent function. |
class arctan | ||
Definition | df-asin 26847 | 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 26848 | Define the arccosine function. See also remarks for df-asin 26847. 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 26849 | Define the arctangent function. See also remarks for df-asin 26847. 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 26850 | The argument to the logarithm in df-asin 26847 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → ((i · 𝐴) + (√‘(1 − (𝐴↑2)))) ≠ 0) | ||
Theorem | asinlem2 26851 | The argument to the logarithm in df-asin 26847 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 26852 | Lemma for asinlem3 26853. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) ≤ 0) → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))) | ||
Theorem | asinlem3 26853 | The argument to the logarithm in df-asin 26847 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))) | ||
Theorem | asinf 26854 | Domain and codomain of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arcsin:ℂ⟶ℂ | ||
Theorem | asincl 26855 | Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (arcsin‘𝐴) ∈ ℂ) | ||
Theorem | acosf 26856 | Domain and codoamin of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arccos:ℂ⟶ℂ | ||
Theorem | acoscl 26857 | Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) ∈ ℂ) | ||
Theorem | atandm 26858 | 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 26859 | This form of atandm 26858 is a bit more useful for showing that the logarithms in df-atan 26849 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 26860 | A compact form of atandm 26858. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (𝐴↑2) ≠ -1)) | ||
Theorem | atandm4 26861 | A compact form of atandm 26858. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ≠ 0)) | ||
Theorem | atanf 26862 | Domain and codoamin of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arctan:(ℂ ∖ {-i, i})⟶ℂ | ||
Theorem | atancl 26863 | Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan → (arctan‘𝐴) ∈ ℂ) | ||
Theorem | asinval 26864 | 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 26865 | Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) = ((π / 2) − (arcsin‘𝐴))) | ||
Theorem | atanval 26866 | 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 26867 | 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 26868 | The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (arcsin‘-𝐴) = -(arcsin‘𝐴)) | ||
Theorem | acosneg 26869 | The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (arccos‘-𝐴) = (π − (arccos‘𝐴))) | ||
Theorem | efiasin 26870 | 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 26871 | 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 26874 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (sin‘(arcsin‘𝐴)) = 𝐴) | ||
Theorem | cosacos 26872 | The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (cos‘(arccos‘𝐴)) = 𝐴) | ||
Theorem | asinsinlem 26873 | Lemma for asinsin 26874. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → 0 < (ℜ‘(exp‘(i · 𝐴)))) | ||
Theorem | asinsin 26874 | The arcsine function composed with sin is equal to the identity. This plus sinasin 26871 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 26878). (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → (arcsin‘(sin‘𝐴)) = 𝐴) | ||
Theorem | acoscos 26875 | 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 26876 | The arcsine of 1 is π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (arcsin‘1) = (π / 2) | ||
Theorem | acos1 26877 | The arccosine of 1 is 0. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (arccos‘1) = 0 | ||
Theorem | reasinsin 26878 | 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 26879 | 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 26880 | 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 26881 | 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 26882 | 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 26883 | Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ (-(π / 2)[,](π / 2))) | ||
Theorem | asinrecl 26884 | The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ ℝ) | ||
Theorem | acosrecl 26885 | The arccosine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ (-1[,]1) → (arccos‘𝐴) ∈ ℝ) | ||
Theorem | cosasin 26886 | 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 26887 | 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 26888 | 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 26889 | The arctangent function is odd. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (arctan‘-𝐴) = -(arctan‘𝐴)) | ||
Theorem | atan0 26890 | The arctangent of zero is zero. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (arctan‘0) = 0 | ||
Theorem | atandmcj 26891 | The arctangent function distributes under conjugation. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan → (∗‘𝐴) ∈ dom arctan) | ||
Theorem | atancj 26892 | 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 26889 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 26893 | The arctangent function is real for all real inputs. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℝ → (arctan‘𝐴) ∈ ℝ) | ||
Theorem | efiatan 26894 | 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 26895 | Lemma for atanlogadd 26896. (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 26896 | 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 26897 | Lemma for atanlogsub 26898. (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 26898 | A variation on atanlogadd 26896, 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 26899 | 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 26900 | 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)) |
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