Mixing
Do these integrals have a simple expression in terms of local function properties?
$begingroup$
Recall the following identity, assuming continuity of $f$:
$$lim_{Delta xto0} frac{int_{x_0}^{x_0+Delta x}f(x)}{Delta x}=f(x_0)$$
Can similar expressions in terms of $f(x_0)$ or its derivatives be derived for the following generalization $I_n$ assuming continuity and infinite differentiability of $f$?
$$I_n = lim_{Delta xto0} frac{int_{x_0}^{x_0+Delta x}(x-x_0)^nf(x)}{(Delta x)^{n+1}}$$
COMMENT: I highly suspect that $I_n$ is proportional to $n$-th order derivatives of $f$ evaluated at $x_0$, but haven't been able to get this using typical integration by parts.
integration definite-integrals taylor-expansion
asked Jan 27 at 21:53
aghostinthefiguresaghostinthefigures
1,2641217
$endgroup$
|
show 2 more comments
$begingroup$
Recall the following identity, assuming continuity of $f$:
$$lim_{Delta xto0} frac{int_{x_0}^{x_0+Delta x}f(x)}{Delta x}=f(x_0)$$
Can similar expressions in terms of $f(x_0)$ or its derivatives be derived for the following generalization $I_n$ assuming continuity and infinite differentiability of $f$?
$$I_n = lim_{Delta xto0} frac{int_{x_0}^{x_0+Delta x}(x-x_0)^nf(x)}{(Delta x)^{n+1}}$$
COMMENT: I highly suspect that $I_n$ is proportional to $n$-th order derivatives of $f$ evaluated at $x_0$, but haven't been able to get this using typical integration by parts.
integration definite-integrals taylor-expansion
asked Jan 27 at 21:53
aghostinthefiguresaghostinthefigures
1,2641217
$endgroup$
$begingroup$
For $nge 1$, isn't $I_n$ zero by applying the identity you mentioned?
$endgroup$
– Eclipse Sun
Jan 27 at 22:00
$begingroup$
The first equality is wrong. There should be a $frac{1}{Delta x}$ in front of the integral (plus continuity of $f$ at $x_0$.)
$endgroup$
– Julián Aguirre
Jan 27 at 22:02
$begingroup$
@EclipseSun The identity doesn’t apply (at least not directly) because you always have at least an extra power of $x^n$ multiplying $f(x)$ in the integrand, no?
$endgroup$
– aghostinthefigures
Jan 27 at 22:05
$begingroup$
That’s correct @JuliánAguirre, will make the correction promptly.
$endgroup$
– aghostinthefigures
Jan 27 at 22:07
$begingroup$
Since $x_0,n$ are constant, why can't we read $(x-x_0)^nf(x)$ as a single $f$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:09
|
show 2 more comments
$begingroup$
Recall the following identity, assuming continuity of $f$:
$$lim_{Delta xto0} frac{int_{x_0}^{x_0+Delta x}f(x)}{Delta x}=f(x_0)$$
Can similar expressions in terms of $f(x_0)$ or its derivatives be derived for the following generalization $I_n$ assuming continuity and infinite differentiability of $f$?
$$I_n = lim_{Delta xto0} frac{int_{x_0}^{x_0+Delta x}(x-x_0)^nf(x)}{(Delta x)^{n+1}}$$
COMMENT: I highly suspect that $I_n$ is proportional to $n$-th order derivatives of $f$ evaluated at $x_0$, but haven't been able to get this using typical integration by parts.
integration definite-integrals taylor-expansion
asked Jan 27 at 21:53
aghostinthefiguresaghostinthefigures
1,2641217
$endgroup$
Recall the following identity, assuming continuity of $f$:
$$lim_{Delta xto0} frac{int_{x_0}^{x_0+Delta x}f(x)}{Delta x}=f(x_0)$$
Can similar expressions in terms of $f(x_0)$ or its derivatives be derived for the following generalization $I_n$ assuming continuity and infinite differentiability of $f$?
$$I_n = lim_{Delta xto0} frac{int_{x_0}^{x_0+Delta x}(x-x_0)^nf(x)}{(Delta x)^{n+1}}$$
COMMENT: I highly suspect that $I_n$ is proportional to $n$-th order derivatives of $f$ evaluated at $x_0$, but haven't been able to get this using typical integration by parts.
integration definite-integrals taylor-expansion
integration definite-integrals taylor-expansion
asked Jan 27 at 21:53
aghostinthefiguresaghostinthefigures
1,2641217
asked Jan 27 at 21:53
aghostinthefiguresaghostinthefigures
1,2641217
edited Jan 27 at 22:11
aghostinthefigures
asked Jan 27 at 21:53
aghostinthefiguresaghostinthefigures
1,2641217
asked Jan 27 at 21:53
aghostinthefiguresaghostinthefigures
1,2641217
asked Jan 27 at 21:53
aghostinthefiguresaghostinthefigures
1,2641217
1,2641217
$begingroup$
For $nge 1$, isn't $I_n$ zero by applying the identity you mentioned?
$endgroup$
– Eclipse Sun
Jan 27 at 22:00
$begingroup$
The first equality is wrong. There should be a $frac{1}{Delta x}$ in front of the integral (plus continuity of $f$ at $x_0$.)
$endgroup$
– Julián Aguirre
Jan 27 at 22:02
$begingroup$
@EclipseSun The identity doesn’t apply (at least not directly) because you always have at least an extra power of $x^n$ multiplying $f(x)$ in the integrand, no?
$endgroup$
– aghostinthefigures
Jan 27 at 22:05
$begingroup$
That’s correct @JuliánAguirre, will make the correction promptly.
$endgroup$
– aghostinthefigures
Jan 27 at 22:07
$begingroup$
Since $x_0,n$ are constant, why can't we read $(x-x_0)^nf(x)$ as a single $f$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:09
|
show 2 more comments
$begingroup$
For $nge 1$, isn't $I_n$ zero by applying the identity you mentioned?
$endgroup$
– Eclipse Sun
Jan 27 at 22:00
$begingroup$
The first equality is wrong. There should be a $frac{1}{Delta x}$ in front of the integral (plus continuity of $f$ at $x_0$.)
$endgroup$
– Julián Aguirre
Jan 27 at 22:02
$begingroup$
@EclipseSun The identity doesn’t apply (at least not directly) because you always have at least an extra power of $x^n$ multiplying $f(x)$ in the integrand, no?
$endgroup$
– aghostinthefigures
Jan 27 at 22:05
$begingroup$
That’s correct @JuliánAguirre, will make the correction promptly.
$endgroup$
– aghostinthefigures
Jan 27 at 22:07
$begingroup$
Since $x_0,n$ are constant, why can't we read $(x-x_0)^nf(x)$ as a single $f$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:09
$begingroup$
For $nge 1$, isn't $I_n$ zero by applying the identity you mentioned?
$endgroup$
– Eclipse Sun
Jan 27 at 22:00
$begingroup$
For $nge 1$, isn't $I_n$ zero by applying the identity you mentioned?
$endgroup$
– Eclipse Sun
Jan 27 at 22:00
$begingroup$
The first equality is wrong. There should be a $frac{1}{Delta x}$ in front of the integral (plus continuity of $f$ at $x_0$.)
$endgroup$
– Julián Aguirre
Jan 27 at 22:02
$begingroup$
The first equality is wrong. There should be a $frac{1}{Delta x}$ in front of the integral (plus continuity of $f$ at $x_0$.)
$endgroup$
– Julián Aguirre
Jan 27 at 22:02
$begingroup$
@EclipseSun The identity doesn’t apply (at least not directly) because you always have at least an extra power of $x^n$ multiplying $f(x)$ in the integrand, no?
$endgroup$
– aghostinthefigures
Jan 27 at 22:05
$begingroup$
@EclipseSun The identity doesn’t apply (at least not directly) because you always have at least an extra power of $x^n$ multiplying $f(x)$ in the integrand, no?
$endgroup$
– aghostinthefigures
Jan 27 at 22:05
$begingroup$
That’s correct @JuliánAguirre, will make the correction promptly.
$endgroup$
– aghostinthefigures
Jan 27 at 22:07
$begingroup$
That’s correct @JuliánAguirre, will make the correction promptly.
$endgroup$
– aghostinthefigures
Jan 27 at 22:07
$begingroup$
Since $x_0,n$ are constant, why can't we read $(x-x_0)^nf(x)$ as a single $f$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:09
$begingroup$
Since $x_0,n$ are constant, why can't we read $(x-x_0)^nf(x)$ as a single $f$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:09
|
show 2 more comments
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$begingroup$
For $nge 1$, isn't $I_n$ zero by applying the identity you mentioned?
$endgroup$
– Eclipse Sun
Jan 27 at 22:00
$begingroup$
The first equality is wrong. There should be a $frac{1}{Delta x}$ in front of the integral (plus continuity of $f$ at $x_0$.)
$endgroup$
– Julián Aguirre
Jan 27 at 22:02
$begingroup$
@EclipseSun The identity doesn’t apply (at least not directly) because you always have at least an extra power of $x^n$ multiplying $f(x)$ in the integrand, no?
$endgroup$
– aghostinthefigures
Jan 27 at 22:05
$begingroup$
That’s correct @JuliánAguirre, will make the correction promptly.
$endgroup$
– aghostinthefigures
Jan 27 at 22:07
$begingroup$
Since $x_0,n$ are constant, why can't we read $(x-x_0)^nf(x)$ as a single $f$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:09