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Type | Label | Description |
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Statement | ||
Theorem | dquart 26201 | Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 26197 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 26202 | Closure lemmas for quart 26209. (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 26203 | Lemma for quart1 26204. (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 26204 | 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 26205 | Lemma for quart 26209. (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 26206 | Closure lemmas for quart 26209. (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 26207 | Closure lemmas for quart 26209. (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 26208 | Closure lemmas for quart 26209. (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 26209 | 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 33699) 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 26210 | The arcsine function. |
class arcsin | ||
Syntax | cacos 26211 | The arccosine function. |
class arccos | ||
Syntax | catan 26212 | The arctangent function. |
class arctan | ||
Definition | df-asin 26213 | 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 26214 | Define the arccosine function. See also remarks for df-asin 26213. 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 26215 | Define the arctangent function. See also remarks for df-asin 26213. 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 26216 | The argument to the logarithm in df-asin 26213 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → ((i · 𝐴) + (√‘(1 − (𝐴↑2)))) ≠ 0) | ||
Theorem | asinlem2 26217 | The argument to the logarithm in df-asin 26213 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 26218 | Lemma for asinlem3 26219. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) ≤ 0) → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))) | ||
Theorem | asinlem3 26219 | The argument to the logarithm in df-asin 26213 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))) | ||
Theorem | asinf 26220 | Domain and codomain of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arcsin:ℂ⟶ℂ | ||
Theorem | asincl 26221 | Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (arcsin‘𝐴) ∈ ℂ) | ||
Theorem | acosf 26222 | Domain and codoamin of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arccos:ℂ⟶ℂ | ||
Theorem | acoscl 26223 | Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) ∈ ℂ) | ||
Theorem | atandm 26224 | 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 26225 | This form of atandm 26224 is a bit more useful for showing that the logarithms in df-atan 26215 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 26226 | A compact form of atandm 26224. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (𝐴↑2) ≠ -1)) | ||
Theorem | atandm4 26227 | A compact form of atandm 26224. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ≠ 0)) | ||
Theorem | atanf 26228 | Domain and codoamin of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ arctan:(ℂ ∖ {-i, i})⟶ℂ | ||
Theorem | atancl 26229 | Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan → (arctan‘𝐴) ∈ ℂ) | ||
Theorem | asinval 26230 | 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 26231 | Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) = ((π / 2) − (arcsin‘𝐴))) | ||
Theorem | atanval 26232 | 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 26233 | 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 26234 | The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (arcsin‘-𝐴) = -(arcsin‘𝐴)) | ||
Theorem | acosneg 26235 | The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (arccos‘-𝐴) = (π − (arccos‘𝐴))) | ||
Theorem | efiasin 26236 | 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 26237 | 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 26240 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (sin‘(arcsin‘𝐴)) = 𝐴) | ||
Theorem | cosacos 26238 | The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 1-Apr-2015.) |
⊢ (𝐴 ∈ ℂ → (cos‘(arccos‘𝐴)) = 𝐴) | ||
Theorem | asinsinlem 26239 | Lemma for asinsin 26240. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → 0 < (ℜ‘(exp‘(i · 𝐴)))) | ||
Theorem | asinsin 26240 | The arcsine function composed with sin is equal to the identity. This plus sinasin 26237 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 26244). (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → (arcsin‘(sin‘𝐴)) = 𝐴) | ||
Theorem | acoscos 26241 | 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 26242 | The arcsine of 1 is π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (arcsin‘1) = (π / 2) | ||
Theorem | acos1 26243 | The arccosine of 1 is 0. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (arccos‘1) = 0 | ||
Theorem | reasinsin 26244 | 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 26245 | 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 26246 | 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 26247 | 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 26248 | 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 26249 | Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ (-(π / 2)[,](π / 2))) | ||
Theorem | asinrecl 26250 | The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ ℝ) | ||
Theorem | acosrecl 26251 | The arccosine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ (-1[,]1) → (arccos‘𝐴) ∈ ℝ) | ||
Theorem | cosasin 26252 | 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 26253 | 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 26254 | 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 26255 | The arctangent function is odd. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (arctan‘-𝐴) = -(arctan‘𝐴)) | ||
Theorem | atan0 26256 | The arctangent of zero is zero. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (arctan‘0) = 0 | ||
Theorem | atandmcj 26257 | The arctangent function distributes under conjugation. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ dom arctan → (∗‘𝐴) ∈ dom arctan) | ||
Theorem | atancj 26258 | 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 26255 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 26259 | The arctangent function is real for all real inputs. (Contributed by Mario Carneiro, 31-Mar-2015.) |
⊢ (𝐴 ∈ ℝ → (arctan‘𝐴) ∈ ℝ) | ||
Theorem | efiatan 26260 | 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 26261 | Lemma for atanlogadd 26262. (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 26262 | 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 26263 | Lemma for atanlogsub 26264. (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 26264 | A variation on atanlogadd 26262, 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 26265 | 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 26266 | 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 26267 | The arctangent function is an inverse to tan. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (tan‘(arctan‘𝐴)) = 𝐴) | ||
Theorem | atandmtan 26268 | 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 26269 | The cosine of an arctangent. (Contributed by Mario Carneiro, 3-Apr-2015.) |
⊢ (𝐴 ∈ dom arctan → (cos‘(arctan‘𝐴)) = (1 / (√‘(1 + (𝐴↑2))))) | ||
Theorem | cosatanne0 26270 | 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 26271 | 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 26272 | 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 26273 | Lemma for atanbnd 26274. (Contributed by Mario Carneiro, 5-Apr-2015.) |
⊢ (𝐴 ∈ ℝ+ → (arctan‘𝐴) ∈ (-(π / 2)(,)(π / 2))) | ||
Theorem | atanbnd 26274 | 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 26275 | 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 26276 | The arctangent of 1 is π / 4. (Contributed by Mario Carneiro, 2-Apr-2015.) |
⊢ (arctan‘1) = (π / 4) | ||
Theorem | bndatandm 26277 | 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 26278* | 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 26279* | 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 26280* | 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 26281* | 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 26282* | 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 26283* | 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 26284* | 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 26285* | 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 26286* | 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 26287* | 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 26288 | Lemma for leibpi 26290. (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 26289* | 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 26290 | 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 15568). (2) Using leibpilem2 26289 to rewrite the series as a power series, it is the 𝑥 = 1 special case of the Taylor series for arctan (atantayl2 26286). (3) Although we cannot directly plug 𝑥 = 1 into atantayl2 26286, Abel's theorem (abelth2 25799) 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 26284) 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 26291 | The Leibniz formula for π. This version of leibpi 26290 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) | ||
Theorem | log2cnv 26292 | Using the Taylor series for arctan(i / 3), produce a rapidly convergent series for log2. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ 𝐹 = (𝑛 ∈ ℕ0 ↦ (2 / ((3 · ((2 · 𝑛) + 1)) · (9↑𝑛)))) ⇒ ⊢ seq0( + , 𝐹) ⇝ (log‘2) | ||
Theorem | log2tlbnd 26293* | Bound the error term in the series of log2cnv 26292. (Contributed by Mario Carneiro, 7-Apr-2015.) |
⊢ (𝑁 ∈ ℕ0 → ((log‘2) − Σ𝑛 ∈ (0...(𝑁 − 1))(2 / ((3 · ((2 · 𝑛) + 1)) · (9↑𝑛)))) ∈ (0[,](3 / ((4 · ((2 · 𝑁) + 1)) · (9↑𝑁))))) | ||
Theorem | log2ublem1 26294 | Lemma for log2ub 26297. The proof of log2ub 26297, which is simply the evaluation of log2tlbnd 26293 for 𝑁 = 4, takes the form of the addition of five fractions and showing this is less than another fraction. We could just perform exact arithmetic on these fractions, get a large rational number, and just multiply everything to verify the claim, but as anyone who uses decimal numbers for this task knows, it is often better to pick a common denominator 𝑑 (usually a large power of 10) and work with the closest approximations of the form 𝑛 / 𝑑 for some integer 𝑛 instead. It turns out that for our purposes it is sufficient to take 𝑑 = (3↑7) · 5 · 7, which is also nice because it shares many factors in common with the fractions in question. (Contributed by Mario Carneiro, 17-Apr-2015.) |
⊢ (((3↑7) · (5 · 7)) · 𝐴) ≤ 𝐵 & ⊢ 𝐴 ∈ ℝ & ⊢ 𝐷 ∈ ℕ0 & ⊢ 𝐸 ∈ ℕ & ⊢ 𝐵 ∈ ℕ0 & ⊢ 𝐹 ∈ ℕ0 & ⊢ 𝐶 = (𝐴 + (𝐷 / 𝐸)) & ⊢ (𝐵 + 𝐹) = 𝐺 & ⊢ (((3↑7) · (5 · 7)) · 𝐷) ≤ (𝐸 · 𝐹) ⇒ ⊢ (((3↑7) · (5 · 7)) · 𝐶) ≤ 𝐺 | ||
Theorem | log2ublem2 26295* | Lemma for log2ub 26297. (Contributed by Mario Carneiro, 17-Apr-2015.) |
⊢ (((3↑7) · (5 · 7)) · Σ𝑛 ∈ (0...𝐾)(2 / ((3 · ((2 · 𝑛) + 1)) · (9↑𝑛)))) ≤ (2 · 𝐵) & ⊢ 𝐵 ∈ ℕ0 & ⊢ 𝐹 ∈ ℕ0 & ⊢ 𝑁 ∈ ℕ0 & ⊢ (𝑁 − 1) = 𝐾 & ⊢ (𝐵 + 𝐹) = 𝐺 & ⊢ 𝑀 ∈ ℕ0 & ⊢ (𝑀 + 𝑁) = 3 & ⊢ ((5 · 7) · (9↑𝑀)) = (((2 · 𝑁) + 1) · 𝐹) ⇒ ⊢ (((3↑7) · (5 · 7)) · Σ𝑛 ∈ (0...𝑁)(2 / ((3 · ((2 · 𝑛) + 1)) · (9↑𝑛)))) ≤ (2 · 𝐺) | ||
Theorem | log2ublem3 26296 | Lemma for log2ub 26297. In decimal, this is a proof that the first four terms of the series for log2 is less than 53056 / 76545. (Contributed by Mario Carneiro, 17-Apr-2015.) (Proof shortened by AV, 15-Sep-2021.) |
⊢ (((3↑7) · (5 · 7)) · Σ𝑛 ∈ (0...3)(2 / ((3 · ((2 · 𝑛) + 1)) · (9↑𝑛)))) ≤ ;;;;53056 | ||
Theorem | log2ub 26297 | log2 is less than 253 / 365. If written in decimal, this is because log2 = 0.693147... is less than 253/365 = 0.693151... , so this is a very tight bound, at five decimal places. (Contributed by Mario Carneiro, 7-Apr-2015.) (Proof shortened by AV, 16-Sep-2021.) |
⊢ (log‘2) < (;;253 / ;;365) | ||
Theorem | log2le1 26298 | log2 is less than 1. This is just a weaker form of log2ub 26297 when no tight upper bound is required. (Contributed by Thierry Arnoux, 27-Sep-2017.) |
⊢ (log‘2) < 1 | ||
Theorem | birthdaylem1 26299* | Lemma for birthday 26302. (Contributed by Mario Carneiro, 17-Apr-2015.) |
⊢ 𝑆 = {𝑓 ∣ 𝑓:(1...𝐾)⟶(1...𝑁)} & ⊢ 𝑇 = {𝑓 ∣ 𝑓:(1...𝐾)–1-1→(1...𝑁)} ⇒ ⊢ (𝑇 ⊆ 𝑆 ∧ 𝑆 ∈ Fin ∧ (𝑁 ∈ ℕ → 𝑆 ≠ ∅)) | ||
Theorem | birthdaylem2 26300* | For general 𝑁 and 𝐾, count the fraction of injective functions from 1...𝐾 to 1...𝑁. (Contributed by Mario Carneiro, 7-May-2015.) |
⊢ 𝑆 = {𝑓 ∣ 𝑓:(1...𝐾)⟶(1...𝑁)} & ⊢ 𝑇 = {𝑓 ∣ 𝑓:(1...𝐾)–1-1→(1...𝑁)} ⇒ ⊢ ((𝑁 ∈ ℕ ∧ 𝐾 ∈ (0...𝑁)) → ((♯‘𝑇) / (♯‘𝑆)) = (exp‘Σ𝑘 ∈ (0...(𝐾 − 1))(log‘(1 − (𝑘 / 𝑁))))) |
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