Convergence of sum to integral
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I would like to estimate the absolute value of the following difference
$$ Delta(L) = sum_{alpha=-L+1}^L frac{1}{1+2 L} e^{i t sec^2left(pifrac{alpha - 1/2}{2 L+1}right)} - int_{-frac{1}{2}}^frac{1}{2} e^{i t sec^2(pi x)}dx $$
where $L$ is an integer, for $L to infty$. I am assuming $t$ has a positive imaginary part.
Numerically, I have evidence that the error decays like $|Delta(L)| approx c L^{-d} e^{- a L^b}$, with $b$ approximately $frac{1}{2}$ or $frac{2}{3}$. I would like to find $b$ and $a$ (I'm not so interested in $d$ and $c$).
I tried to use the Euler-Maclaurin formula to estimate the error, but all derivatives of the boundary function $e^{i t sec^2left(pifrac{x - 1/2}{2 L+1}right)}$ vanish in the strict $L to infty$ limit, and I'm not sure how to resum them for large $L$.
sequences-and-series riemann-sum euler-maclaurin
asked Nov 16 at 20:42
chubecca
112
add a comment |
up vote
2
down vote
favorite
I would like to estimate the absolute value of the following difference
$$ Delta(L) = sum_{alpha=-L+1}^L frac{1}{1+2 L} e^{i t sec^2left(pifrac{alpha - 1/2}{2 L+1}right)} - int_{-frac{1}{2}}^frac{1}{2} e^{i t sec^2(pi x)}dx $$
where $L$ is an integer, for $L to infty$. I am assuming $t$ has a positive imaginary part.
Numerically, I have evidence that the error decays like $|Delta(L)| approx c L^{-d} e^{- a L^b}$, with $b$ approximately $frac{1}{2}$ or $frac{2}{3}$. I would like to find $b$ and $a$ (I'm not so interested in $d$ and $c$).
I tried to use the Euler-Maclaurin formula to estimate the error, but all derivatives of the boundary function $e^{i t sec^2left(pifrac{x - 1/2}{2 L+1}right)}$ vanish in the strict $L to infty$ limit, and I'm not sure how to resum them for large $L$.
sequences-and-series riemann-sum euler-maclaurin
asked Nov 16 at 20:42
chubecca
112
Since you have the means to investigate the error numerically, maybe you could try linear regression on $$log(- log | Delta|)$$ Then you can estimate $a$ and $b$ if the dependence indeed has the form you suggest
– Yuriy S
Nov 16 at 21:05
@YuriyS that's a good point, I should try that more, however since the coefficient $a$ depends on $t$ that's not too easy. By the way why you study the double log?
– chubecca
Nov 16 at 21:06
@chubecca, from your claim we have $-log |Delta|=a L^b$. Then to find both $a$ and $b$ we can plot $log(a L^b)=log a + b log L$ vs $log L$, which should look like a straight line. With least squares you can find the parameters
– Yuriy S
Nov 16 at 21:11
@YuriyS sorry, I edited the post to make my statement more accurate
– chubecca
Nov 16 at 21:14
The exponential behavior will still be more prominent for large $L$, so you could try my suggestion.
– Yuriy S
Nov 16 at 21:22
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I would like to estimate the absolute value of the following difference
$$ Delta(L) = sum_{alpha=-L+1}^L frac{1}{1+2 L} e^{i t sec^2left(pifrac{alpha - 1/2}{2 L+1}right)} - int_{-frac{1}{2}}^frac{1}{2} e^{i t sec^2(pi x)}dx $$
where $L$ is an integer, for $L to infty$. I am assuming $t$ has a positive imaginary part.
Numerically, I have evidence that the error decays like $|Delta(L)| approx c L^{-d} e^{- a L^b}$, with $b$ approximately $frac{1}{2}$ or $frac{2}{3}$. I would like to find $b$ and $a$ (I'm not so interested in $d$ and $c$).
I tried to use the Euler-Maclaurin formula to estimate the error, but all derivatives of the boundary function $e^{i t sec^2left(pifrac{x - 1/2}{2 L+1}right)}$ vanish in the strict $L to infty$ limit, and I'm not sure how to resum them for large $L$.
sequences-and-series riemann-sum euler-maclaurin
asked Nov 16 at 20:42
chubecca
112
I would like to estimate the absolute value of the following difference
$$ Delta(L) = sum_{alpha=-L+1}^L frac{1}{1+2 L} e^{i t sec^2left(pifrac{alpha - 1/2}{2 L+1}right)} - int_{-frac{1}{2}}^frac{1}{2} e^{i t sec^2(pi x)}dx $$
where $L$ is an integer, for $L to infty$. I am assuming $t$ has a positive imaginary part.
Numerically, I have evidence that the error decays like $|Delta(L)| approx c L^{-d} e^{- a L^b}$, with $b$ approximately $frac{1}{2}$ or $frac{2}{3}$. I would like to find $b$ and $a$ (I'm not so interested in $d$ and $c$).
I tried to use the Euler-Maclaurin formula to estimate the error, but all derivatives of the boundary function $e^{i t sec^2left(pifrac{x - 1/2}{2 L+1}right)}$ vanish in the strict $L to infty$ limit, and I'm not sure how to resum them for large $L$.
sequences-and-series riemann-sum euler-maclaurin
sequences-and-series riemann-sum euler-maclaurin
asked Nov 16 at 20:42
chubecca
112
asked Nov 16 at 20:42
chubecca
112
edited Nov 16 at 21:09
asked Nov 16 at 20:42
chubecca
112
asked Nov 16 at 20:42
chubecca
112
asked Nov 16 at 20:42
chubecca
112
112
Since you have the means to investigate the error numerically, maybe you could try linear regression on $$log(- log | Delta|)$$ Then you can estimate $a$ and $b$ if the dependence indeed has the form you suggest
– Yuriy S
Nov 16 at 21:05
@YuriyS that's a good point, I should try that more, however since the coefficient $a$ depends on $t$ that's not too easy. By the way why you study the double log?
– chubecca
Nov 16 at 21:06
@chubecca, from your claim we have $-log |Delta|=a L^b$. Then to find both $a$ and $b$ we can plot $log(a L^b)=log a + b log L$ vs $log L$, which should look like a straight line. With least squares you can find the parameters
– Yuriy S
Nov 16 at 21:11
@YuriyS sorry, I edited the post to make my statement more accurate
– chubecca
Nov 16 at 21:14
The exponential behavior will still be more prominent for large $L$, so you could try my suggestion.
– Yuriy S
Nov 16 at 21:22
add a comment |
Since you have the means to investigate the error numerically, maybe you could try linear regression on $$log(- log | Delta|)$$ Then you can estimate $a$ and $b$ if the dependence indeed has the form you suggest
– Yuriy S
Nov 16 at 21:05
@YuriyS that's a good point, I should try that more, however since the coefficient $a$ depends on $t$ that's not too easy. By the way why you study the double log?
– chubecca
Nov 16 at 21:06
@chubecca, from your claim we have $-log |Delta|=a L^b$. Then to find both $a$ and $b$ we can plot $log(a L^b)=log a + b log L$ vs $log L$, which should look like a straight line. With least squares you can find the parameters
– Yuriy S
Nov 16 at 21:11
@YuriyS sorry, I edited the post to make my statement more accurate
– chubecca
Nov 16 at 21:14
The exponential behavior will still be more prominent for large $L$, so you could try my suggestion.
– Yuriy S
Nov 16 at 21:22
Since you have the means to investigate the error numerically, maybe you could try linear regression on $$log(- log | Delta|)$$ Then you can estimate $a$ and $b$ if the dependence indeed has the form you suggest
– Yuriy S
Nov 16 at 21:05
Since you have the means to investigate the error numerically, maybe you could try linear regression on $$log(- log | Delta|)$$ Then you can estimate $a$ and $b$ if the dependence indeed has the form you suggest
– Yuriy S
Nov 16 at 21:05
@YuriyS that's a good point, I should try that more, however since the coefficient $a$ depends on $t$ that's not too easy. By the way why you study the double log?
– chubecca
Nov 16 at 21:06
@YuriyS that's a good point, I should try that more, however since the coefficient $a$ depends on $t$ that's not too easy. By the way why you study the double log?
– chubecca
Nov 16 at 21:06
@chubecca, from your claim we have $-log |Delta|=a L^b$. Then to find both $a$ and $b$ we can plot $log(a L^b)=log a + b log L$ vs $log L$, which should look like a straight line. With least squares you can find the parameters
– Yuriy S
Nov 16 at 21:11
@chubecca, from your claim we have $-log |Delta|=a L^b$. Then to find both $a$ and $b$ we can plot $log(a L^b)=log a + b log L$ vs $log L$, which should look like a straight line. With least squares you can find the parameters
– Yuriy S
Nov 16 at 21:11
@YuriyS sorry, I edited the post to make my statement more accurate
– chubecca
Nov 16 at 21:14
@YuriyS sorry, I edited the post to make my statement more accurate
– chubecca
Nov 16 at 21:14
The exponential behavior will still be more prominent for large $L$, so you could try my suggestion.
– Yuriy S
Nov 16 at 21:22
The exponential behavior will still be more prominent for large $L$, so you could try my suggestion.
– Yuriy S
Nov 16 at 21:22
add a comment |
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Since you have the means to investigate the error numerically, maybe you could try linear regression on $$log(- log | Delta|)$$ Then you can estimate $a$ and $b$ if the dependence indeed has the form you suggest
– Yuriy S
Nov 16 at 21:05
@YuriyS that's a good point, I should try that more, however since the coefficient $a$ depends on $t$ that's not too easy. By the way why you study the double log?
– chubecca
Nov 16 at 21:06
@chubecca, from your claim we have $-log |Delta|=a L^b$. Then to find both $a$ and $b$ we can plot $log(a L^b)=log a + b log L$ vs $log L$, which should look like a straight line. With least squares you can find the parameters
– Yuriy S
Nov 16 at 21:11
@YuriyS sorry, I edited the post to make my statement more accurate
– chubecca
Nov 16 at 21:14
The exponential behavior will still be more prominent for large $L$, so you could try my suggestion.
– Yuriy S
Nov 16 at 21:22