A Taylor theorem for Hölder continuous function?
$begingroup$
Let $ C^{m,s}_{b} $ be the space of bounded function $ u: mathbb{R} rightarrow mathbb{R} $ which satisfies
begin{alignat*}{2}
bigg| u^{(m)}(x) - u^{(m)}(y) bigg| leq C | x - y |^{s}.
end{alignat*}
I'm looking for a Taylor - type theorem which could bound the remainder of the following polynomial
$$ u(x) = sum_{k geq N} frac{(x-y)^{k}}{k!} u^{(k)}(y) + R_{N}(x) $$
by $$| R_{N}(x) | leq |x-y|^{m+s}.$$
In the simple case that $ N=1 $ I know that
$$ u(x) = u(y) + u'(y)(x-y) + int_{0}^{1} Big( u'big(y + t(x-y) big) - u'(y) Big)dt (x-y) .$$
But I have yet been successful in finding the right expression for the arbitrary case. Could anyone please shed some light on the problem? Thanks in advance!
real-analysis analysis holder-spaces
asked Dec 23 '18 at 11:37
MeagainMeagain
18910
$endgroup$
add a comment |
$begingroup$
Let $ C^{m,s}_{b} $ be the space of bounded function $ u: mathbb{R} rightarrow mathbb{R} $ which satisfies
begin{alignat*}{2}
bigg| u^{(m)}(x) - u^{(m)}(y) bigg| leq C | x - y |^{s}.
end{alignat*}
I'm looking for a Taylor - type theorem which could bound the remainder of the following polynomial
$$ u(x) = sum_{k geq N} frac{(x-y)^{k}}{k!} u^{(k)}(y) + R_{N}(x) $$
by $$| R_{N}(x) | leq |x-y|^{m+s}.$$
In the simple case that $ N=1 $ I know that
$$ u(x) = u(y) + u'(y)(x-y) + int_{0}^{1} Big( u'big(y + t(x-y) big) - u'(y) Big)dt (x-y) .$$
But I have yet been successful in finding the right expression for the arbitrary case. Could anyone please shed some light on the problem? Thanks in advance!
real-analysis analysis holder-spaces
asked Dec 23 '18 at 11:37
MeagainMeagain
18910
$endgroup$
add a comment |
$begingroup$
Let $ C^{m,s}_{b} $ be the space of bounded function $ u: mathbb{R} rightarrow mathbb{R} $ which satisfies
begin{alignat*}{2}
bigg| u^{(m)}(x) - u^{(m)}(y) bigg| leq C | x - y |^{s}.
end{alignat*}
I'm looking for a Taylor - type theorem which could bound the remainder of the following polynomial
$$ u(x) = sum_{k geq N} frac{(x-y)^{k}}{k!} u^{(k)}(y) + R_{N}(x) $$
by $$| R_{N}(x) | leq |x-y|^{m+s}.$$
In the simple case that $ N=1 $ I know that
$$ u(x) = u(y) + u'(y)(x-y) + int_{0}^{1} Big( u'big(y + t(x-y) big) - u'(y) Big)dt (x-y) .$$
But I have yet been successful in finding the right expression for the arbitrary case. Could anyone please shed some light on the problem? Thanks in advance!
real-analysis analysis holder-spaces
asked Dec 23 '18 at 11:37
MeagainMeagain
18910
$endgroup$
Let $ C^{m,s}_{b} $ be the space of bounded function $ u: mathbb{R} rightarrow mathbb{R} $ which satisfies
begin{alignat*}{2}
bigg| u^{(m)}(x) - u^{(m)}(y) bigg| leq C | x - y |^{s}.
end{alignat*}
I'm looking for a Taylor - type theorem which could bound the remainder of the following polynomial
$$ u(x) = sum_{k geq N} frac{(x-y)^{k}}{k!} u^{(k)}(y) + R_{N}(x) $$
by $$| R_{N}(x) | leq |x-y|^{m+s}.$$
In the simple case that $ N=1 $ I know that
$$ u(x) = u(y) + u'(y)(x-y) + int_{0}^{1} Big( u'big(y + t(x-y) big) - u'(y) Big)dt (x-y) .$$
But I have yet been successful in finding the right expression for the arbitrary case. Could anyone please shed some light on the problem? Thanks in advance!
real-analysis analysis holder-spaces
real-analysis analysis holder-spaces
asked Dec 23 '18 at 11:37
MeagainMeagain
18910
asked Dec 23 '18 at 11:37
MeagainMeagain
18910
edited Dec 23 '18 at 12:11
Meagain
asked Dec 23 '18 at 11:37
MeagainMeagain
18910
asked Dec 23 '18 at 11:37
MeagainMeagain
18910
asked Dec 23 '18 at 11:37
MeagainMeagain
18910
18910
add a comment |
add a comment |
1 Answer
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$begingroup$
Take the Taylor formula of order $m-1$ with the Lagrange remainder
$$
u(x) = sum_{k=0}^{m-1} frac{(x-y)^{k}}{k!} u^{(k)}(y) + frac{(x-y)^{m}}{m!} u^{(m)}(xi),
$$
where $xi$ is between $y$ and $x$, and rewrite it as
$$
u(x) = sum_{k=0}^{m} frac{(x-y)^{k}}{k!} u^{(k)}(y) + frac{(x-y)^{m}(u^{(m)}(xi)-u^{(m)}(y))}{m!}.
$$
Using the Holder condition for the remainder one has
$$
|R_m(x)|=left|frac{(x-y)^{m}(u^{(m)}(xi)-u^{(m)}(y))}{m!} right|le Cfrac{|x-y|^{m+s}}{m!}.
$$
answered Dec 23 '18 at 13:54
AndrewAndrew
9,58211946
$endgroup$
$begingroup$
Thanks! Happy Xmas!
$endgroup$
– Meagain
Dec 24 '18 at 13:33
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take the Taylor formula of order $m-1$ with the Lagrange remainder
$$
u(x) = sum_{k=0}^{m-1} frac{(x-y)^{k}}{k!} u^{(k)}(y) + frac{(x-y)^{m}}{m!} u^{(m)}(xi),
$$
where $xi$ is between $y$ and $x$, and rewrite it as
$$
u(x) = sum_{k=0}^{m} frac{(x-y)^{k}}{k!} u^{(k)}(y) + frac{(x-y)^{m}(u^{(m)}(xi)-u^{(m)}(y))}{m!}.
$$
Using the Holder condition for the remainder one has
$$
|R_m(x)|=left|frac{(x-y)^{m}(u^{(m)}(xi)-u^{(m)}(y))}{m!} right|le Cfrac{|x-y|^{m+s}}{m!}.
$$
answered Dec 23 '18 at 13:54
AndrewAndrew
9,58211946
$endgroup$
$begingroup$
Thanks! Happy Xmas!
$endgroup$
– Meagain
Dec 24 '18 at 13:33
add a comment |
$begingroup$
Take the Taylor formula of order $m-1$ with the Lagrange remainder
$$
u(x) = sum_{k=0}^{m-1} frac{(x-y)^{k}}{k!} u^{(k)}(y) + frac{(x-y)^{m}}{m!} u^{(m)}(xi),
$$
where $xi$ is between $y$ and $x$, and rewrite it as
$$
u(x) = sum_{k=0}^{m} frac{(x-y)^{k}}{k!} u^{(k)}(y) + frac{(x-y)^{m}(u^{(m)}(xi)-u^{(m)}(y))}{m!}.
$$
Using the Holder condition for the remainder one has
$$
|R_m(x)|=left|frac{(x-y)^{m}(u^{(m)}(xi)-u^{(m)}(y))}{m!} right|le Cfrac{|x-y|^{m+s}}{m!}.
$$
answered Dec 23 '18 at 13:54
AndrewAndrew
9,58211946
$endgroup$
$begingroup$
Thanks! Happy Xmas!
$endgroup$
– Meagain
Dec 24 '18 at 13:33
add a comment |
$begingroup$
Take the Taylor formula of order $m-1$ with the Lagrange remainder
$$
u(x) = sum_{k=0}^{m-1} frac{(x-y)^{k}}{k!} u^{(k)}(y) + frac{(x-y)^{m}}{m!} u^{(m)}(xi),
$$
where $xi$ is between $y$ and $x$, and rewrite it as
$$
u(x) = sum_{k=0}^{m} frac{(x-y)^{k}}{k!} u^{(k)}(y) + frac{(x-y)^{m}(u^{(m)}(xi)-u^{(m)}(y))}{m!}.
$$
Using the Holder condition for the remainder one has
$$
|R_m(x)|=left|frac{(x-y)^{m}(u^{(m)}(xi)-u^{(m)}(y))}{m!} right|le Cfrac{|x-y|^{m+s}}{m!}.
$$
answered Dec 23 '18 at 13:54
AndrewAndrew
9,58211946
$endgroup$
Take the Taylor formula of order $m-1$ with the Lagrange remainder
$$
u(x) = sum_{k=0}^{m-1} frac{(x-y)^{k}}{k!} u^{(k)}(y) + frac{(x-y)^{m}}{m!} u^{(m)}(xi),
$$
where $xi$ is between $y$ and $x$, and rewrite it as
$$
u(x) = sum_{k=0}^{m} frac{(x-y)^{k}}{k!} u^{(k)}(y) + frac{(x-y)^{m}(u^{(m)}(xi)-u^{(m)}(y))}{m!}.
$$
Using the Holder condition for the remainder one has
$$
|R_m(x)|=left|frac{(x-y)^{m}(u^{(m)}(xi)-u^{(m)}(y))}{m!} right|le Cfrac{|x-y|^{m+s}}{m!}.
$$
answered Dec 23 '18 at 13:54
AndrewAndrew
9,58211946
answered Dec 23 '18 at 13:54
AndrewAndrew
9,58211946
answered Dec 23 '18 at 13:54
AndrewAndrew
9,58211946
answered Dec 23 '18 at 13:54
AndrewAndrew
9,58211946
9,58211946
$begingroup$
Thanks! Happy Xmas!
$endgroup$
– Meagain
Dec 24 '18 at 13:33
add a comment |
$begingroup$
Thanks! Happy Xmas!
$endgroup$
– Meagain
Dec 24 '18 at 13:33
$begingroup$
Thanks! Happy Xmas!
$endgroup$
– Meagain
Dec 24 '18 at 13:33
$begingroup$
Thanks! Happy Xmas!
$endgroup$
– Meagain
Dec 24 '18 at 13:33
add a comment |
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