Taylor polynomials on a compact interval
$begingroup$
Let $I ⊂ mathbb{R}$ be a compact interval. Show that for every $f ∈ C^2(I , mathbb{R})$ it exists a $C > 0$ such that:
$|f(a)+f(b)-2f(frac{a+b}{2})|≤C(b-a)^2 $
$∀a, b ∈ I .$
Let $a,b ∈I$ and $a<b$, it exists by definition a $ξ_1∈(a, (a+b)/2)$ with
$f(a)=f(frac{a+b}{2})+f'(frac{a+b}{2})(frac{a-b}{2})+frac{1}{2}f''(ξ_1)frac{(a-b)^2}{4}$
and it exists a $ξ_2∈((a+b)/2, b)$ with
$f(b)=f(frac{a+b}{2})+f'(frac{a+b}{2})(frac{b-a}{2})+frac{1}{2}f''(ξ_2)frac{(b-a)^2}{4}$
Summing the two identities we get
$f(a)+f(b)-2f(frac{a+b}{2})=frac{f''(ξ_1)+f''(ξ_2)}{8}(b-a)^2$
How can I say that the first part of the right side is equal to $C$ and introduce the absolute value with ≤?
real-analysis
asked Dec 6 '18 at 16:18
DadaDada
7510
$endgroup$
add a comment |
$begingroup$
Let $I ⊂ mathbb{R}$ be a compact interval. Show that for every $f ∈ C^2(I , mathbb{R})$ it exists a $C > 0$ such that:
$|f(a)+f(b)-2f(frac{a+b}{2})|≤C(b-a)^2 $
$∀a, b ∈ I .$
Let $a,b ∈I$ and $a<b$, it exists by definition a $ξ_1∈(a, (a+b)/2)$ with
$f(a)=f(frac{a+b}{2})+f'(frac{a+b}{2})(frac{a-b}{2})+frac{1}{2}f''(ξ_1)frac{(a-b)^2}{4}$
and it exists a $ξ_2∈((a+b)/2, b)$ with
$f(b)=f(frac{a+b}{2})+f'(frac{a+b}{2})(frac{b-a}{2})+frac{1}{2}f''(ξ_2)frac{(b-a)^2}{4}$
Summing the two identities we get
$f(a)+f(b)-2f(frac{a+b}{2})=frac{f''(ξ_1)+f''(ξ_2)}{8}(b-a)^2$
How can I say that the first part of the right side is equal to $C$ and introduce the absolute value with ≤?
real-analysis
asked Dec 6 '18 at 16:18
DadaDada
7510
$endgroup$
add a comment |
$begingroup$
Let $I ⊂ mathbb{R}$ be a compact interval. Show that for every $f ∈ C^2(I , mathbb{R})$ it exists a $C > 0$ such that:
$|f(a)+f(b)-2f(frac{a+b}{2})|≤C(b-a)^2 $
$∀a, b ∈ I .$
Let $a,b ∈I$ and $a<b$, it exists by definition a $ξ_1∈(a, (a+b)/2)$ with
$f(a)=f(frac{a+b}{2})+f'(frac{a+b}{2})(frac{a-b}{2})+frac{1}{2}f''(ξ_1)frac{(a-b)^2}{4}$
and it exists a $ξ_2∈((a+b)/2, b)$ with
$f(b)=f(frac{a+b}{2})+f'(frac{a+b}{2})(frac{b-a}{2})+frac{1}{2}f''(ξ_2)frac{(b-a)^2}{4}$
Summing the two identities we get
$f(a)+f(b)-2f(frac{a+b}{2})=frac{f''(ξ_1)+f''(ξ_2)}{8}(b-a)^2$
How can I say that the first part of the right side is equal to $C$ and introduce the absolute value with ≤?
real-analysis
asked Dec 6 '18 at 16:18
DadaDada
7510
$endgroup$
Let $I ⊂ mathbb{R}$ be a compact interval. Show that for every $f ∈ C^2(I , mathbb{R})$ it exists a $C > 0$ such that:
$|f(a)+f(b)-2f(frac{a+b}{2})|≤C(b-a)^2 $
$∀a, b ∈ I .$
Let $a,b ∈I$ and $a<b$, it exists by definition a $ξ_1∈(a, (a+b)/2)$ with
$f(a)=f(frac{a+b}{2})+f'(frac{a+b}{2})(frac{a-b}{2})+frac{1}{2}f''(ξ_1)frac{(a-b)^2}{4}$
and it exists a $ξ_2∈((a+b)/2, b)$ with
$f(b)=f(frac{a+b}{2})+f'(frac{a+b}{2})(frac{b-a}{2})+frac{1}{2}f''(ξ_2)frac{(b-a)^2}{4}$
Summing the two identities we get
$f(a)+f(b)-2f(frac{a+b}{2})=frac{f''(ξ_1)+f''(ξ_2)}{8}(b-a)^2$
How can I say that the first part of the right side is equal to $C$ and introduce the absolute value with ≤?
real-analysis
real-analysis
asked Dec 6 '18 at 16:18
DadaDada
7510
asked Dec 6 '18 at 16:18
DadaDada
7510
asked Dec 6 '18 at 16:18
DadaDada
7510
asked Dec 6 '18 at 16:18
DadaDada
7510
asked Dec 6 '18 at 16:18
DadaDada
7510
7510
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint : Any continuous function $f $ from a compact space to $Bbb R $ is bounded and attains its bounds.
answered Dec 6 '18 at 16:29
Thomas ShelbyThomas Shelby
2,8871421
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add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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$begingroup$
Hint : Any continuous function $f $ from a compact space to $Bbb R $ is bounded and attains its bounds.
answered Dec 6 '18 at 16:29
Thomas ShelbyThomas Shelby
2,8871421
$endgroup$
add a comment |
$begingroup$
Hint : Any continuous function $f $ from a compact space to $Bbb R $ is bounded and attains its bounds.
answered Dec 6 '18 at 16:29
Thomas ShelbyThomas Shelby
2,8871421
$endgroup$
add a comment |
$begingroup$
Hint : Any continuous function $f $ from a compact space to $Bbb R $ is bounded and attains its bounds.
answered Dec 6 '18 at 16:29
Thomas ShelbyThomas Shelby
2,8871421
$endgroup$
Hint : Any continuous function $f $ from a compact space to $Bbb R $ is bounded and attains its bounds.
answered Dec 6 '18 at 16:29
Thomas ShelbyThomas Shelby
2,8871421
answered Dec 6 '18 at 16:29
Thomas ShelbyThomas Shelby
2,8871421
answered Dec 6 '18 at 16:29
Thomas ShelbyThomas Shelby
2,8871421
answered Dec 6 '18 at 16:29
Thomas ShelbyThomas Shelby
2,8871421
2,8871421
add a comment |
add a comment |
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