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typos fixes and removal of unneed hypothesis - 10.3 10.4 #503
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@@ -57,7 +57,6 @@ theorem inverse_function_theorem {X Y: Set ℝ} {f: ℝ → ℝ} {g:ℝ → ℝ} | |
| (hfXY: ∀ x ∈ X, f x ∈ Y) (hgYX: ∀ y ∈ Y, g y ∈ X) | ||
| (hgf: ∀ x ∈ X, g (f x) = x) (hfg: ∀ y ∈ Y, f (g y) = y) | ||
| {x₀ y₀ f'x₀: ℝ} (hx₀: x₀ ∈ X) (hfx₀: f x₀ = y₀) (hne : f'x₀ ≠ 0) | ||
| (hcluster: ClusterPt x₀ (.principal (X \ {x₀}))) | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. This assumption is not needed. Full proof without it - https://github.com/rkirov/analysis/blob/main/Analysis/Section_10_4.lean#L81-L113 |
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| (hf: HasDerivWithinAt f f'x₀ X x₀) (hg: ContinuousWithinAt g Y y₀) : | ||
| HasDerivWithinAt g (1/f'x₀) Y y₀ := by | ||
| -- This proof is written to follow the structure of the original text. | ||
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@@ -74,27 +73,27 @@ theorem inverse_function_theorem {X Y: Set ℝ} {f: ℝ → ℝ} {g:ℝ → ℝ} | |
| convert (hf _ hx _).inv₀ _ using 2 with n <;> grind | ||
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| /-- Exercise 10.4.1(a) -/ | ||
| example {n:ℕ} (hn: n > 0) : ContinuousOn (fun x:ℝ ↦ x^(1/n:ℝ)) (.Ici 0) := by sorry | ||
| example {n:ℕ} : ContinuousOn (fun x:ℝ ↦ x^(1/n:ℝ)) (.Ioi 0) := by sorry | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more.
The |
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| /-- Exercise 10.4.1(b) -/ | ||
| example {n:ℕ} (hn: n > 0) {x:ℝ} (hx: x ∈ Set.Ici 0) : HasDerivWithinAt (fun x:ℝ ↦ x^(1/n:ℝ)) | ||
| ((n:ℝ)⁻¹ * x^((n:ℝ)⁻¹-1)) (.Ici 0) x := by sorry | ||
| example {n:ℕ} {x:ℝ} (hx: x ∈ Set.Ioi 0) : HasDerivWithinAt (fun x:ℝ ↦ x^(1/n:ℝ)) | ||
| ((n:ℝ)⁻¹ * x^((n:ℝ)⁻¹-1)) (.Ioi 0) x := by sorry | ||
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| /-- Exercise 10.4.2(a) -/ | ||
| example (q:ℚ) {x:ℝ} (hx: x ∈ Set.Ici 0) : | ||
| HasDerivWithinAt (fun x:ℝ ↦ x^(q:ℝ)) (q * x^(q-1:ℝ)) (.Ici 0) x := by | ||
| example (q:ℚ) {x:ℝ} (hx: x ∈ Set.Ioi 0) : | ||
| HasDerivWithinAt (fun x:ℝ ↦ x^(q:ℝ)) (q * x^(q-1:ℝ)) (.Ioi 0) x := by | ||
| sorry | ||
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| /-- Exercise 10.4.2(b) -/ | ||
| example (q:ℚ) : (nhdsWithin 1 (.Ici 0 \ {1})).Tendsto (fun x:ℝ ↦ (x^(q:ℝ)-1)/(x-1)) (nhds q) := by | ||
| example (q:ℚ) : (nhdsWithin 1 (.Ioi 0 \ {1})).Tendsto (fun x:ℝ ↦ (x^(q:ℝ)-1)/(x-1)) (nhds q) := by | ||
| sorry | ||
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| /-- Exercise 10.4.3(a) -/ | ||
| example (α:ℝ) : (nhdsWithin 1 (.Ici 0 \ {1})).Tendsto (fun x:ℝ ↦ (x^α-1^α)/(x-1)) (nhds α) := by | ||
| example (α:ℝ) : (nhdsWithin 1 (.Ioi 0 \ {1})).Tendsto (fun x:ℝ ↦ (x^α-1^α)/(x-1)) (nhds α) := by | ||
| sorry | ||
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| /-- Exercise 10.4.3(b) -/ | ||
| example (α:ℝ) {x:ℝ} (hx: x ∈ Set.Ici 0) : HasDerivWithinAt (fun x:ℝ ↦ x^α) (α * x^(α-1)) (.Ici 0) x := by | ||
| example (α:ℝ) {x:ℝ} (hx: x ∈ Set.Ioi 0) : HasDerivWithinAt (fun x:ℝ ↦ x^α) (α * x^(α-1)) (.Ioi 0) x := by | ||
| sorry | ||
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| end Chapter10 | ||
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vacuously true even if
a < bis not true.