Step
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1
1
1
1
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of Lemma
Taylor-theorem-case1
1. I : Interval
2. n : ℕ
3. F : ℕn + 2 ⟶ I ⟶ℝ
4. ∀b:{a:ℝ| a ∈ I}
((∀k:ℕn + 2. ∀x,y:{a:ℝ| a ∈ I} . ((x = y)
⇒ (F[k;x] = F[k;y])))
⇒ finite-deriv-seq(I;n + 1;i,x.F[i;x])
⇒ d(Taylor-approx(n;a;b;i,x.F[i;x]))/da = λx.b - x^n * (F[n + 1;x]/r((n)!)) on I)
5. a : {a:ℝ| a ∈ I}
6. b : {a:ℝ| a ∈ I}
7. ∀k:ℕn + 2. ∀x,y:{a:ℝ| a ∈ I} . ((x = y)
⇒ (F[k;x] = F[k;y]))
8. finite-deriv-seq(I;n + 1;i,x.F[i;x])
9. d(Taylor-approx(n;a;b;i,x.F[i;x]))/da = λx.b - x^n * (F[n + 1;x]/r((n)!)) on I
10. b - a ≠ r0
11. e : ℝ
12. r0 < e
13. R : ℝ
14. Taylor-remainder(I;n;b;a;k,x.F[k;x]) = R ∈ ℝ
15. [rmin(a;b), rmax(a;b)] ⊆ I
16. rmin(a;b) < rmax(a;b)
17. x : {t:ℝ| t ∈ [rmin(a;b), rmax(a;b)]}
18. y : {t:ℝ| t ∈ [rmin(a;b), rmax(a;b)]}
19. x = y
⊢ ((R/b - a) - b - y^n * (F[n + 1;x]/r((n)!))) = ((R/b - a) - b - y^n * (F[n + 1;y]/r((n)!)))
BY
{ ((Assert F[n + 1;x] = F[n + 1;y] BY Auto) THEN RWO "-1" 0 THEN Auto) }
Latex:
Latex:
1. I : Interval
2. n : \mBbbN{}
3. F : \mBbbN{}n + 2 {}\mrightarrow{} I {}\mrightarrow{}\mBbbR{}
4. \mforall{}b:\{a:\mBbbR{}| a \mmember{} I\}
((\mforall{}k:\mBbbN{}n + 2. \mforall{}x,y:\{a:\mBbbR{}| a \mmember{} I\} . ((x = y) {}\mRightarrow{} (F[k;x] = F[k;y])))
{}\mRightarrow{} finite-deriv-seq(I;n + 1;i,x.F[i;x])
{}\mRightarrow{} d(Taylor-approx(n;a;b;i,x.F[i;x]))/da = \mlambda{}x.b - x\^{}n * (F[n + 1;x]/r((n)!)) on I)
5. a : \{a:\mBbbR{}| a \mmember{} I\}
6. b : \{a:\mBbbR{}| a \mmember{} I\}
7. \mforall{}k:\mBbbN{}n + 2. \mforall{}x,y:\{a:\mBbbR{}| a \mmember{} I\} . ((x = y) {}\mRightarrow{} (F[k;x] = F[k;y]))
8. finite-deriv-seq(I;n + 1;i,x.F[i;x])
9. d(Taylor-approx(n;a;b;i,x.F[i;x]))/da = \mlambda{}x.b - x\^{}n * (F[n + 1;x]/r((n)!)) on I
10. b - a \mneq{} r0
11. e : \mBbbR{}
12. r0 < e
13. R : \mBbbR{}
14. Taylor-remainder(I;n;b;a;k,x.F[k;x]) = R
15. [rmin(a;b), rmax(a;b)] \msubseteq{} I
16. rmin(a;b) < rmax(a;b)
17. x : \{t:\mBbbR{}| t \mmember{} [rmin(a;b), rmax(a;b)]\}
18. y : \{t:\mBbbR{}| t \mmember{} [rmin(a;b), rmax(a;b)]\}
19. x = y
\mvdash{} ((R/b - a) - b - y\^{}n * (F[n + 1;x]/r((n)!))) = ((R/b - a) - b - y\^{}n * (F[n + 1;y]/r((n)!)))
By
Latex:
((Assert F[n + 1;x] = F[n + 1;y] BY Auto) THEN RWO "-1" 0 THEN Auto)
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