Mixing
Prove that $int_a^b (f'(x))^2dx - hsum_{i=0}^{n-1} left(frac{f(x_{i+1})-f(x_i)}{h}right)^2 = O(h^2) $
$begingroup$
Prove that:
$int_a^b (f'(x))^2dx - hsum_{i=0}^{n-1} {(frac{f(x_{i+1})-f(x_i)}{h})}^2 = O(h^2) $
Where $h=frac{b-a}{n} , x_k=a+kh$ and $f in C^infty[a,b]$
First I've tried to estimate the following: $int_{x_i}^{x_{i+1}} (f'(x))^2dx - h{left(frac{f(x_{i+1})-f(x_i)}{h}right)}^2$
By Taylor series we know:
$f'(x)=frac{f(x+h)-f(x)}{h}-frac{h}{2}f''(alpha)$
(Where $ xleqalphaleq x+h$)
So $int_{x_i}^{x_{i+1}} (f'(x))^2dx = int_{x_i}^{x_{i+1}} left(frac{f(x+h)-f(x)}{h}-frac{h}{2}f''(alpha)right)^2dx$
Then I got stuck
integration numerical-methods
edited Jan 11 at 20:52
jameselmore
4,39432035
asked Jan 11 at 18:00
useruser
52
$endgroup$
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$begingroup$
Prove that:
$int_a^b (f'(x))^2dx - hsum_{i=0}^{n-1} {(frac{f(x_{i+1})-f(x_i)}{h})}^2 = O(h^2) $
Where $h=frac{b-a}{n} , x_k=a+kh$ and $f in C^infty[a,b]$
First I've tried to estimate the following: $int_{x_i}^{x_{i+1}} (f'(x))^2dx - h{left(frac{f(x_{i+1})-f(x_i)}{h}right)}^2$
By Taylor series we know:
$f'(x)=frac{f(x+h)-f(x)}{h}-frac{h}{2}f''(alpha)$
(Where $ xleqalphaleq x+h$)
So $int_{x_i}^{x_{i+1}} (f'(x))^2dx = int_{x_i}^{x_{i+1}} left(frac{f(x+h)-f(x)}{h}-frac{h}{2}f''(alpha)right)^2dx$
Then I got stuck
integration numerical-methods
edited Jan 11 at 20:52
jameselmore
4,39432035
asked Jan 11 at 18:00
useruser
52
$endgroup$
add a comment |
$begingroup$
Prove that:
$int_a^b (f'(x))^2dx - hsum_{i=0}^{n-1} {(frac{f(x_{i+1})-f(x_i)}{h})}^2 = O(h^2) $
Where $h=frac{b-a}{n} , x_k=a+kh$ and $f in C^infty[a,b]$
First I've tried to estimate the following: $int_{x_i}^{x_{i+1}} (f'(x))^2dx - h{left(frac{f(x_{i+1})-f(x_i)}{h}right)}^2$
By Taylor series we know:
$f'(x)=frac{f(x+h)-f(x)}{h}-frac{h}{2}f''(alpha)$
(Where $ xleqalphaleq x+h$)
So $int_{x_i}^{x_{i+1}} (f'(x))^2dx = int_{x_i}^{x_{i+1}} left(frac{f(x+h)-f(x)}{h}-frac{h}{2}f''(alpha)right)^2dx$
Then I got stuck
integration numerical-methods
edited Jan 11 at 20:52
jameselmore
4,39432035
asked Jan 11 at 18:00
useruser
52
$endgroup$
Prove that:
$int_a^b (f'(x))^2dx - hsum_{i=0}^{n-1} {(frac{f(x_{i+1})-f(x_i)}{h})}^2 = O(h^2) $
Where $h=frac{b-a}{n} , x_k=a+kh$ and $f in C^infty[a,b]$
First I've tried to estimate the following: $int_{x_i}^{x_{i+1}} (f'(x))^2dx - h{left(frac{f(x_{i+1})-f(x_i)}{h}right)}^2$
By Taylor series we know:
$f'(x)=frac{f(x+h)-f(x)}{h}-frac{h}{2}f''(alpha)$
(Where $ xleqalphaleq x+h$)
So $int_{x_i}^{x_{i+1}} (f'(x))^2dx = int_{x_i}^{x_{i+1}} left(frac{f(x+h)-f(x)}{h}-frac{h}{2}f''(alpha)right)^2dx$
Then I got stuck
integration numerical-methods
integration numerical-methods
edited Jan 11 at 20:52
jameselmore
4,39432035
asked Jan 11 at 18:00
useruser
52
edited Jan 11 at 20:52
jameselmore
4,39432035
asked Jan 11 at 18:00
useruser
52
edited Jan 11 at 20:52
jameselmore
4,39432035
edited Jan 11 at 20:52
jameselmore
4,39432035
edited Jan 11 at 20:52
jameselmore
4,39432035
4,39432035
asked Jan 11 at 18:00
useruser
52
asked Jan 11 at 18:00
useruser
52
asked Jan 11 at 18:00
useruser
52
52
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We have
begin{align*}
A(x) &overset{text{def}}{=} int_a^b [f'(x)]^2dx - sum_{k=0}^{n-1}left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h \
&= sum_{k=0}^{n-1}left{color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h}right}
end{align*}
Now, for $x in [x_k, x_{k+1}]$,
begin{align*}
[f'(x)]^2 = [f'(x_k)]^2 + 2f'(x_k)f''(x_k)(x - x_k) + {[f''(x_k)]^2 + f'(x_k)f'''(x_k)}(x - x_k)^2 + O(h^3)
end{align*}
and so
begin{align*}
color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} &= color{green}{[f'(x_k)]^2h} + color{orange}{2f'(x_k)f''(x_k)cdotfrac{1}{2}h^2} + {[f''(x_k)]^2 + f'(x_k)f'''(x_k)}cdotfrac{1}{3}h^3 + O(h^4) \
&=color{green}{[f'(x_k)]^2h} + color{orange}{2f'(x_k)f''(x_k)cdotfrac{1}{2}h^2} + O(h^3)
end{align*}
Next,
begin{align*}
f(x_{k+1}) = f(x_k) + hf'(x_k) + frac{h^2}{2}f''(x_k) + O(h^3)
end{align*}
and so
begin{align*}
color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h} &= left(f'(x_k) + frac{h}{2}f''(x_k) + O(h^2)right)^2 h \
&= color{green}{[f'(x_k)]^2h} + color{orange}{f'(x_k)f''(x_k)h^2}+O(h^3)
end{align*}
Therefore,
begin{align*}
color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h} = O(h^3)
end{align*}
Finally,
begin{align*}
A(x) = sum_{k=0}^{n-1}O(h^3) = O(h^2)
end{align*}
answered Jan 11 at 22:20
Tom ChenTom Chen
913513
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$begingroup$
I was only able to get $O(h)$. Perhaps someone cleverer than I can point out where I was loose.
By the mean value theorem,
$$frac{f(x_{i+1}) - f(x_i)}{h} = f'(x_0)$$
for some $x_0 in [x_i, x_{i+1}]$.
Because $f in C^2[a,b]$, there exists $M$ such that $max{|f'(x)|,|f''(x)|} le M$ for all $x in [a,b]$.
This implies
$$|(f'(x))^2 - (f'(x_0))^2|
le 2 M^2 |f'(x) - f'(x_0)|
le 2 M^3 h$$
for all $x in [x_i, x_{i+1}]$.
Thus
$$left|int_{x_i}^{x_{i+1}} (f'(x))^2, dx - h left(frac{f(x_{i+1}) - f(x_i)}{h}right)^2right|
le int_{x_i}^{x_{i+1}} |(f'(x))^2 - (f'(x_0))^2| , dx le 2 M^3 h^2.$$
Summing over $i = 0,1,ldots, n-1$ yields $2M^3 h$.
answered Jan 11 at 19:54
angryavianangryavian
40.8k23380
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$begingroup$
Let $g$ be any Lipschitz function. Then the atomar expression in question is the minimum of a quadratic function
$$
int_x^{x+h}g(s)^2ds-frac1hleft(int_x^{x+h}g(s)dsright)^2 = min_{minBbb R}int_x^{x+h}(g(s)-m)^2ds
$$
By the mean value theorem, $m=frac1hint_x^{x+h}g(s)ds$ is equal to one of the function values of $g$.
begin{align}
min_{cin[x,x+h]}int_x^{x+h}(g(s)-g(c))^2ds
&leint_x^{x+h}|g(s)-g(x+h/2)|^2ds\
&leint_x^{x+h}L^2|s-(x+h/2)|^2ds=frac{L^2}{12}h^3
end{align}
In the given case $g=f'$ the Lipschitz constant can be chosen as $L=|f''|_infty$,
$$
int_{x_i}^{x_{i+1}}f'(s)^2ds-hleft(frac{f(x_{i+1})-f(x_i)}hright)^2
lefrac{|f''|_infty^2}{12}h^3
$$
and in the sum
$$
int_{x_0}^{x_n}f'(s)^2ds-hsum_{i=0}^{n-1}left(frac{f(x_{i+1})-f(x_i)}hright)^2
lefrac{|f''|_infty^2}{12}(x_n-x_0)h^2
$$
answered Jan 12 at 8:27
LutzLLutzL
58.1k42054
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
We have
begin{align*}
A(x) &overset{text{def}}{=} int_a^b [f'(x)]^2dx - sum_{k=0}^{n-1}left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h \
&= sum_{k=0}^{n-1}left{color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h}right}
end{align*}
Now, for $x in [x_k, x_{k+1}]$,
begin{align*}
[f'(x)]^2 = [f'(x_k)]^2 + 2f'(x_k)f''(x_k)(x - x_k) + {[f''(x_k)]^2 + f'(x_k)f'''(x_k)}(x - x_k)^2 + O(h^3)
end{align*}
and so
begin{align*}
color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} &= color{green}{[f'(x_k)]^2h} + color{orange}{2f'(x_k)f''(x_k)cdotfrac{1}{2}h^2} + {[f''(x_k)]^2 + f'(x_k)f'''(x_k)}cdotfrac{1}{3}h^3 + O(h^4) \
&=color{green}{[f'(x_k)]^2h} + color{orange}{2f'(x_k)f''(x_k)cdotfrac{1}{2}h^2} + O(h^3)
end{align*}
Next,
begin{align*}
f(x_{k+1}) = f(x_k) + hf'(x_k) + frac{h^2}{2}f''(x_k) + O(h^3)
end{align*}
and so
begin{align*}
color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h} &= left(f'(x_k) + frac{h}{2}f''(x_k) + O(h^2)right)^2 h \
&= color{green}{[f'(x_k)]^2h} + color{orange}{f'(x_k)f''(x_k)h^2}+O(h^3)
end{align*}
Therefore,
begin{align*}
color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h} = O(h^3)
end{align*}
Finally,
begin{align*}
A(x) = sum_{k=0}^{n-1}O(h^3) = O(h^2)
end{align*}
answered Jan 11 at 22:20
Tom ChenTom Chen
913513
$endgroup$
add a comment |
$begingroup$
We have
begin{align*}
A(x) &overset{text{def}}{=} int_a^b [f'(x)]^2dx - sum_{k=0}^{n-1}left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h \
&= sum_{k=0}^{n-1}left{color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h}right}
end{align*}
Now, for $x in [x_k, x_{k+1}]$,
begin{align*}
[f'(x)]^2 = [f'(x_k)]^2 + 2f'(x_k)f''(x_k)(x - x_k) + {[f''(x_k)]^2 + f'(x_k)f'''(x_k)}(x - x_k)^2 + O(h^3)
end{align*}
and so
begin{align*}
color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} &= color{green}{[f'(x_k)]^2h} + color{orange}{2f'(x_k)f''(x_k)cdotfrac{1}{2}h^2} + {[f''(x_k)]^2 + f'(x_k)f'''(x_k)}cdotfrac{1}{3}h^3 + O(h^4) \
&=color{green}{[f'(x_k)]^2h} + color{orange}{2f'(x_k)f''(x_k)cdotfrac{1}{2}h^2} + O(h^3)
end{align*}
Next,
begin{align*}
f(x_{k+1}) = f(x_k) + hf'(x_k) + frac{h^2}{2}f''(x_k) + O(h^3)
end{align*}
and so
begin{align*}
color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h} &= left(f'(x_k) + frac{h}{2}f''(x_k) + O(h^2)right)^2 h \
&= color{green}{[f'(x_k)]^2h} + color{orange}{f'(x_k)f''(x_k)h^2}+O(h^3)
end{align*}
Therefore,
begin{align*}
color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h} = O(h^3)
end{align*}
Finally,
begin{align*}
A(x) = sum_{k=0}^{n-1}O(h^3) = O(h^2)
end{align*}
answered Jan 11 at 22:20
Tom ChenTom Chen
913513
$endgroup$
add a comment |
$begingroup$
We have
begin{align*}
A(x) &overset{text{def}}{=} int_a^b [f'(x)]^2dx - sum_{k=0}^{n-1}left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h \
&= sum_{k=0}^{n-1}left{color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h}right}
end{align*}
Now, for $x in [x_k, x_{k+1}]$,
begin{align*}
[f'(x)]^2 = [f'(x_k)]^2 + 2f'(x_k)f''(x_k)(x - x_k) + {[f''(x_k)]^2 + f'(x_k)f'''(x_k)}(x - x_k)^2 + O(h^3)
end{align*}
and so
begin{align*}
color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} &= color{green}{[f'(x_k)]^2h} + color{orange}{2f'(x_k)f''(x_k)cdotfrac{1}{2}h^2} + {[f''(x_k)]^2 + f'(x_k)f'''(x_k)}cdotfrac{1}{3}h^3 + O(h^4) \
&=color{green}{[f'(x_k)]^2h} + color{orange}{2f'(x_k)f''(x_k)cdotfrac{1}{2}h^2} + O(h^3)
end{align*}
Next,
begin{align*}
f(x_{k+1}) = f(x_k) + hf'(x_k) + frac{h^2}{2}f''(x_k) + O(h^3)
end{align*}
and so
begin{align*}
color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h} &= left(f'(x_k) + frac{h}{2}f''(x_k) + O(h^2)right)^2 h \
&= color{green}{[f'(x_k)]^2h} + color{orange}{f'(x_k)f''(x_k)h^2}+O(h^3)
end{align*}
Therefore,
begin{align*}
color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h} = O(h^3)
end{align*}
Finally,
begin{align*}
A(x) = sum_{k=0}^{n-1}O(h^3) = O(h^2)
end{align*}
answered Jan 11 at 22:20
Tom ChenTom Chen
913513
$endgroup$
We have
begin{align*}
A(x) &overset{text{def}}{=} int_a^b [f'(x)]^2dx - sum_{k=0}^{n-1}left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h \
&= sum_{k=0}^{n-1}left{color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h}right}
end{align*}
Now, for $x in [x_k, x_{k+1}]$,
begin{align*}
[f'(x)]^2 = [f'(x_k)]^2 + 2f'(x_k)f''(x_k)(x - x_k) + {[f''(x_k)]^2 + f'(x_k)f'''(x_k)}(x - x_k)^2 + O(h^3)
end{align*}
and so
begin{align*}
color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} &= color{green}{[f'(x_k)]^2h} + color{orange}{2f'(x_k)f''(x_k)cdotfrac{1}{2}h^2} + {[f''(x_k)]^2 + f'(x_k)f'''(x_k)}cdotfrac{1}{3}h^3 + O(h^4) \
&=color{green}{[f'(x_k)]^2h} + color{orange}{2f'(x_k)f''(x_k)cdotfrac{1}{2}h^2} + O(h^3)
end{align*}
Next,
begin{align*}
f(x_{k+1}) = f(x_k) + hf'(x_k) + frac{h^2}{2}f''(x_k) + O(h^3)
end{align*}
and so
begin{align*}
color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h} &= left(f'(x_k) + frac{h}{2}f''(x_k) + O(h^2)right)^2 h \
&= color{green}{[f'(x_k)]^2h} + color{orange}{f'(x_k)f''(x_k)h^2}+O(h^3)
end{align*}
Therefore,
begin{align*}
color{red}{int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - color{blue}{left(frac{f(x_{k+1}) - f(x_k)}{h}right)^2h} = O(h^3)
end{align*}
Finally,
begin{align*}
A(x) = sum_{k=0}^{n-1}O(h^3) = O(h^2)
end{align*}
answered Jan 11 at 22:20
Tom ChenTom Chen
913513
edited Jan 11 at 22:25
answered Jan 11 at 22:20
Tom ChenTom Chen
913513
answered Jan 11 at 22:20
Tom ChenTom Chen
913513
answered Jan 11 at 22:20
Tom ChenTom Chen
913513
913513
add a comment |
add a comment |
$begingroup$
I was only able to get $O(h)$. Perhaps someone cleverer than I can point out where I was loose.
By the mean value theorem,
$$frac{f(x_{i+1}) - f(x_i)}{h} = f'(x_0)$$
for some $x_0 in [x_i, x_{i+1}]$.
Because $f in C^2[a,b]$, there exists $M$ such that $max{|f'(x)|,|f''(x)|} le M$ for all $x in [a,b]$.
This implies
$$|(f'(x))^2 - (f'(x_0))^2|
le 2 M^2 |f'(x) - f'(x_0)|
le 2 M^3 h$$
for all $x in [x_i, x_{i+1}]$.
Thus
$$left|int_{x_i}^{x_{i+1}} (f'(x))^2, dx - h left(frac{f(x_{i+1}) - f(x_i)}{h}right)^2right|
le int_{x_i}^{x_{i+1}} |(f'(x))^2 - (f'(x_0))^2| , dx le 2 M^3 h^2.$$
Summing over $i = 0,1,ldots, n-1$ yields $2M^3 h$.
answered Jan 11 at 19:54
angryavianangryavian
40.8k23380
$endgroup$
add a comment |
$begingroup$
I was only able to get $O(h)$. Perhaps someone cleverer than I can point out where I was loose.
By the mean value theorem,
$$frac{f(x_{i+1}) - f(x_i)}{h} = f'(x_0)$$
for some $x_0 in [x_i, x_{i+1}]$.
Because $f in C^2[a,b]$, there exists $M$ such that $max{|f'(x)|,|f''(x)|} le M$ for all $x in [a,b]$.
This implies
$$|(f'(x))^2 - (f'(x_0))^2|
le 2 M^2 |f'(x) - f'(x_0)|
le 2 M^3 h$$
for all $x in [x_i, x_{i+1}]$.
Thus
$$left|int_{x_i}^{x_{i+1}} (f'(x))^2, dx - h left(frac{f(x_{i+1}) - f(x_i)}{h}right)^2right|
le int_{x_i}^{x_{i+1}} |(f'(x))^2 - (f'(x_0))^2| , dx le 2 M^3 h^2.$$
Summing over $i = 0,1,ldots, n-1$ yields $2M^3 h$.
answered Jan 11 at 19:54
angryavianangryavian
40.8k23380
$endgroup$
add a comment |
$begingroup$
I was only able to get $O(h)$. Perhaps someone cleverer than I can point out where I was loose.
By the mean value theorem,
$$frac{f(x_{i+1}) - f(x_i)}{h} = f'(x_0)$$
for some $x_0 in [x_i, x_{i+1}]$.
Because $f in C^2[a,b]$, there exists $M$ such that $max{|f'(x)|,|f''(x)|} le M$ for all $x in [a,b]$.
This implies
$$|(f'(x))^2 - (f'(x_0))^2|
le 2 M^2 |f'(x) - f'(x_0)|
le 2 M^3 h$$
for all $x in [x_i, x_{i+1}]$.
Thus
$$left|int_{x_i}^{x_{i+1}} (f'(x))^2, dx - h left(frac{f(x_{i+1}) - f(x_i)}{h}right)^2right|
le int_{x_i}^{x_{i+1}} |(f'(x))^2 - (f'(x_0))^2| , dx le 2 M^3 h^2.$$
Summing over $i = 0,1,ldots, n-1$ yields $2M^3 h$.
answered Jan 11 at 19:54
angryavianangryavian
40.8k23380
$endgroup$
I was only able to get $O(h)$. Perhaps someone cleverer than I can point out where I was loose.
By the mean value theorem,
$$frac{f(x_{i+1}) - f(x_i)}{h} = f'(x_0)$$
for some $x_0 in [x_i, x_{i+1}]$.
Because $f in C^2[a,b]$, there exists $M$ such that $max{|f'(x)|,|f''(x)|} le M$ for all $x in [a,b]$.
This implies
$$|(f'(x))^2 - (f'(x_0))^2|
le 2 M^2 |f'(x) - f'(x_0)|
le 2 M^3 h$$
for all $x in [x_i, x_{i+1}]$.
Thus
$$left|int_{x_i}^{x_{i+1}} (f'(x))^2, dx - h left(frac{f(x_{i+1}) - f(x_i)}{h}right)^2right|
le int_{x_i}^{x_{i+1}} |(f'(x))^2 - (f'(x_0))^2| , dx le 2 M^3 h^2.$$
Summing over $i = 0,1,ldots, n-1$ yields $2M^3 h$.
answered Jan 11 at 19:54
angryavianangryavian
40.8k23380
answered Jan 11 at 19:54
angryavianangryavian
40.8k23380
answered Jan 11 at 19:54
angryavianangryavian
40.8k23380
answered Jan 11 at 19:54
angryavianangryavian
40.8k23380
40.8k23380
add a comment |
add a comment |
$begingroup$
Let $g$ be any Lipschitz function. Then the atomar expression in question is the minimum of a quadratic function
$$
int_x^{x+h}g(s)^2ds-frac1hleft(int_x^{x+h}g(s)dsright)^2 = min_{minBbb R}int_x^{x+h}(g(s)-m)^2ds
$$
By the mean value theorem, $m=frac1hint_x^{x+h}g(s)ds$ is equal to one of the function values of $g$.
begin{align}
min_{cin[x,x+h]}int_x^{x+h}(g(s)-g(c))^2ds
&leint_x^{x+h}|g(s)-g(x+h/2)|^2ds\
&leint_x^{x+h}L^2|s-(x+h/2)|^2ds=frac{L^2}{12}h^3
end{align}
In the given case $g=f'$ the Lipschitz constant can be chosen as $L=|f''|_infty$,
$$
int_{x_i}^{x_{i+1}}f'(s)^2ds-hleft(frac{f(x_{i+1})-f(x_i)}hright)^2
lefrac{|f''|_infty^2}{12}h^3
$$
and in the sum
$$
int_{x_0}^{x_n}f'(s)^2ds-hsum_{i=0}^{n-1}left(frac{f(x_{i+1})-f(x_i)}hright)^2
lefrac{|f''|_infty^2}{12}(x_n-x_0)h^2
$$
answered Jan 12 at 8:27
LutzLLutzL
58.1k42054
$endgroup$
add a comment |
$begingroup$
Let $g$ be any Lipschitz function. Then the atomar expression in question is the minimum of a quadratic function
$$
int_x^{x+h}g(s)^2ds-frac1hleft(int_x^{x+h}g(s)dsright)^2 = min_{minBbb R}int_x^{x+h}(g(s)-m)^2ds
$$
By the mean value theorem, $m=frac1hint_x^{x+h}g(s)ds$ is equal to one of the function values of $g$.
begin{align}
min_{cin[x,x+h]}int_x^{x+h}(g(s)-g(c))^2ds
&leint_x^{x+h}|g(s)-g(x+h/2)|^2ds\
&leint_x^{x+h}L^2|s-(x+h/2)|^2ds=frac{L^2}{12}h^3
end{align}
In the given case $g=f'$ the Lipschitz constant can be chosen as $L=|f''|_infty$,
$$
int_{x_i}^{x_{i+1}}f'(s)^2ds-hleft(frac{f(x_{i+1})-f(x_i)}hright)^2
lefrac{|f''|_infty^2}{12}h^3
$$
and in the sum
$$
int_{x_0}^{x_n}f'(s)^2ds-hsum_{i=0}^{n-1}left(frac{f(x_{i+1})-f(x_i)}hright)^2
lefrac{|f''|_infty^2}{12}(x_n-x_0)h^2
$$
answered Jan 12 at 8:27
LutzLLutzL
58.1k42054
$endgroup$
add a comment |
$begingroup$
Let $g$ be any Lipschitz function. Then the atomar expression in question is the minimum of a quadratic function
$$
int_x^{x+h}g(s)^2ds-frac1hleft(int_x^{x+h}g(s)dsright)^2 = min_{minBbb R}int_x^{x+h}(g(s)-m)^2ds
$$
By the mean value theorem, $m=frac1hint_x^{x+h}g(s)ds$ is equal to one of the function values of $g$.
begin{align}
min_{cin[x,x+h]}int_x^{x+h}(g(s)-g(c))^2ds
&leint_x^{x+h}|g(s)-g(x+h/2)|^2ds\
&leint_x^{x+h}L^2|s-(x+h/2)|^2ds=frac{L^2}{12}h^3
end{align}
In the given case $g=f'$ the Lipschitz constant can be chosen as $L=|f''|_infty$,
$$
int_{x_i}^{x_{i+1}}f'(s)^2ds-hleft(frac{f(x_{i+1})-f(x_i)}hright)^2
lefrac{|f''|_infty^2}{12}h^3
$$
and in the sum
$$
int_{x_0}^{x_n}f'(s)^2ds-hsum_{i=0}^{n-1}left(frac{f(x_{i+1})-f(x_i)}hright)^2
lefrac{|f''|_infty^2}{12}(x_n-x_0)h^2
$$
answered Jan 12 at 8:27
LutzLLutzL
58.1k42054
$endgroup$
Let $g$ be any Lipschitz function. Then the atomar expression in question is the minimum of a quadratic function
$$
int_x^{x+h}g(s)^2ds-frac1hleft(int_x^{x+h}g(s)dsright)^2 = min_{minBbb R}int_x^{x+h}(g(s)-m)^2ds
$$
By the mean value theorem, $m=frac1hint_x^{x+h}g(s)ds$ is equal to one of the function values of $g$.
begin{align}
min_{cin[x,x+h]}int_x^{x+h}(g(s)-g(c))^2ds
&leint_x^{x+h}|g(s)-g(x+h/2)|^2ds\
&leint_x^{x+h}L^2|s-(x+h/2)|^2ds=frac{L^2}{12}h^3
end{align}
In the given case $g=f'$ the Lipschitz constant can be chosen as $L=|f''|_infty$,
$$
int_{x_i}^{x_{i+1}}f'(s)^2ds-hleft(frac{f(x_{i+1})-f(x_i)}hright)^2
lefrac{|f''|_infty^2}{12}h^3
$$
and in the sum
$$
int_{x_0}^{x_n}f'(s)^2ds-hsum_{i=0}^{n-1}left(frac{f(x_{i+1})-f(x_i)}hright)^2
lefrac{|f''|_infty^2}{12}(x_n-x_0)h^2
$$
answered Jan 12 at 8:27
LutzLLutzL
58.1k42054
answered Jan 12 at 8:27
LutzLLutzL
58.1k42054
answered Jan 12 at 8:27
LutzLLutzL
58.1k42054
answered Jan 12 at 8:27
LutzLLutzL
58.1k42054
58.1k42054
add a comment |
add a comment |
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