Mixing
Approximation of a sum with an integral…
Let $G$ a continuous function in $C([0,1], mathbb R)$. I think that $$frac{1}{N}sum_{x ;text{odd}in {1,ldots, N}}GBig (frac{x}{N}Big )xrightarrow{Nto +infty}frac{1}{2}int_0^1G(r)dr,$$
and I would like to prove that. I think to do a change of variables as follows:
$$frac{1}{N}sum_{x ;text{odd}in {1,ldots, N}}GBig (frac{x}{N}Big )=frac{1}{N}sum_{k=0}^{frac{N-1}{2}}GBig(frac{2k+1}{N}Big)=frac{1}{2}frac{1}{N}sum_{z=1}^NGBig(zBig),$$
where in the last step I defined $z=2k+1$ and I applied a kind of change of variables for the series that usually holds for integrals ($dk=frac{1}{2}dz$). Is that correct?
real-analysis integration definite-integrals approximation-theory
asked Nov 20 '18 at 15:32
user495333
916
add a comment |
Let $G$ a continuous function in $C([0,1], mathbb R)$. I think that $$frac{1}{N}sum_{x ;text{odd}in {1,ldots, N}}GBig (frac{x}{N}Big )xrightarrow{Nto +infty}frac{1}{2}int_0^1G(r)dr,$$
and I would like to prove that. I think to do a change of variables as follows:
$$frac{1}{N}sum_{x ;text{odd}in {1,ldots, N}}GBig (frac{x}{N}Big )=frac{1}{N}sum_{k=0}^{frac{N-1}{2}}GBig(frac{2k+1}{N}Big)=frac{1}{2}frac{1}{N}sum_{z=1}^NGBig(zBig),$$
where in the last step I defined $z=2k+1$ and I applied a kind of change of variables for the series that usually holds for integrals ($dk=frac{1}{2}dz$). Is that correct?
real-analysis integration definite-integrals approximation-theory
asked Nov 20 '18 at 15:32
user495333
916
What you wrote is currently not true since $G$ is not defined on $x=2$.
– Keen-ameteur
Nov 20 '18 at 15:59
The first substitution was already $x=2k+1$ so introducing $z$ will just take you back to the first formula renaming $x$ as $z$. There’s no way to double the number of summands with just a change of variables.
– Ben
Nov 20 '18 at 16:16
I think you meant “$G(z/N)$” at the end. Still the formula is not true. The question is equivalent to whether sampling the function over odd numerators is equal to summing over even numerators. It’s easy to picture counterexamples to this - a sine function with peaks at even numerators and troughs at odd numerators.
– Ben
Nov 20 '18 at 16:20
Seeing as the function has to stay constant with $N$, I don't think you can find such a sine function, as it will have to have infinite amounts of peaks.
– Keen-ameteur
Nov 20 '18 at 16:47
@Keen-amateur The comment above is addressing the case of fixed $N$, as in the question the OP asked about using an analog of the formula for differential forms in the discrete case.
– Ben
Nov 20 '18 at 16:55
add a comment |
Let $G$ a continuous function in $C([0,1], mathbb R)$. I think that $$frac{1}{N}sum_{x ;text{odd}in {1,ldots, N}}GBig (frac{x}{N}Big )xrightarrow{Nto +infty}frac{1}{2}int_0^1G(r)dr,$$
and I would like to prove that. I think to do a change of variables as follows:
$$frac{1}{N}sum_{x ;text{odd}in {1,ldots, N}}GBig (frac{x}{N}Big )=frac{1}{N}sum_{k=0}^{frac{N-1}{2}}GBig(frac{2k+1}{N}Big)=frac{1}{2}frac{1}{N}sum_{z=1}^NGBig(zBig),$$
where in the last step I defined $z=2k+1$ and I applied a kind of change of variables for the series that usually holds for integrals ($dk=frac{1}{2}dz$). Is that correct?
real-analysis integration definite-integrals approximation-theory
asked Nov 20 '18 at 15:32
user495333
916
Let $G$ a continuous function in $C([0,1], mathbb R)$. I think that $$frac{1}{N}sum_{x ;text{odd}in {1,ldots, N}}GBig (frac{x}{N}Big )xrightarrow{Nto +infty}frac{1}{2}int_0^1G(r)dr,$$
and I would like to prove that. I think to do a change of variables as follows:
$$frac{1}{N}sum_{x ;text{odd}in {1,ldots, N}}GBig (frac{x}{N}Big )=frac{1}{N}sum_{k=0}^{frac{N-1}{2}}GBig(frac{2k+1}{N}Big)=frac{1}{2}frac{1}{N}sum_{z=1}^NGBig(zBig),$$
where in the last step I defined $z=2k+1$ and I applied a kind of change of variables for the series that usually holds for integrals ($dk=frac{1}{2}dz$). Is that correct?
real-analysis integration definite-integrals approximation-theory
real-analysis integration definite-integrals approximation-theory
asked Nov 20 '18 at 15:32
user495333
916
asked Nov 20 '18 at 15:32
user495333
916
asked Nov 20 '18 at 15:32
user495333
916
asked Nov 20 '18 at 15:32
user495333
916
asked Nov 20 '18 at 15:32
user495333
916
916
What you wrote is currently not true since $G$ is not defined on $x=2$.
– Keen-ameteur
Nov 20 '18 at 15:59
The first substitution was already $x=2k+1$ so introducing $z$ will just take you back to the first formula renaming $x$ as $z$. There’s no way to double the number of summands with just a change of variables.
– Ben
Nov 20 '18 at 16:16
I think you meant “$G(z/N)$” at the end. Still the formula is not true. The question is equivalent to whether sampling the function over odd numerators is equal to summing over even numerators. It’s easy to picture counterexamples to this - a sine function with peaks at even numerators and troughs at odd numerators.
– Ben
Nov 20 '18 at 16:20
Seeing as the function has to stay constant with $N$, I don't think you can find such a sine function, as it will have to have infinite amounts of peaks.
– Keen-ameteur
Nov 20 '18 at 16:47
@Keen-amateur The comment above is addressing the case of fixed $N$, as in the question the OP asked about using an analog of the formula for differential forms in the discrete case.
– Ben
Nov 20 '18 at 16:55
add a comment |
What you wrote is currently not true since $G$ is not defined on $x=2$.
– Keen-ameteur
Nov 20 '18 at 15:59
The first substitution was already $x=2k+1$ so introducing $z$ will just take you back to the first formula renaming $x$ as $z$. There’s no way to double the number of summands with just a change of variables.
– Ben
Nov 20 '18 at 16:16
I think you meant “$G(z/N)$” at the end. Still the formula is not true. The question is equivalent to whether sampling the function over odd numerators is equal to summing over even numerators. It’s easy to picture counterexamples to this - a sine function with peaks at even numerators and troughs at odd numerators.
– Ben
Nov 20 '18 at 16:20
Seeing as the function has to stay constant with $N$, I don't think you can find such a sine function, as it will have to have infinite amounts of peaks.
– Keen-ameteur
Nov 20 '18 at 16:47
@Keen-amateur The comment above is addressing the case of fixed $N$, as in the question the OP asked about using an analog of the formula for differential forms in the discrete case.
– Ben
Nov 20 '18 at 16:55
What you wrote is currently not true since $G$ is not defined on $x=2$.
– Keen-ameteur
Nov 20 '18 at 15:59
What you wrote is currently not true since $G$ is not defined on $x=2$.
– Keen-ameteur
Nov 20 '18 at 15:59
The first substitution was already $x=2k+1$ so introducing $z$ will just take you back to the first formula renaming $x$ as $z$. There’s no way to double the number of summands with just a change of variables.
– Ben
Nov 20 '18 at 16:16
The first substitution was already $x=2k+1$ so introducing $z$ will just take you back to the first formula renaming $x$ as $z$. There’s no way to double the number of summands with just a change of variables.
– Ben
Nov 20 '18 at 16:16
I think you meant “$G(z/N)$” at the end. Still the formula is not true. The question is equivalent to whether sampling the function over odd numerators is equal to summing over even numerators. It’s easy to picture counterexamples to this - a sine function with peaks at even numerators and troughs at odd numerators.
– Ben
Nov 20 '18 at 16:20
I think you meant “$G(z/N)$” at the end. Still the formula is not true. The question is equivalent to whether sampling the function over odd numerators is equal to summing over even numerators. It’s easy to picture counterexamples to this - a sine function with peaks at even numerators and troughs at odd numerators.
– Ben
Nov 20 '18 at 16:20
Seeing as the function has to stay constant with $N$, I don't think you can find such a sine function, as it will have to have infinite amounts of peaks.
– Keen-ameteur
Nov 20 '18 at 16:47
Seeing as the function has to stay constant with $N$, I don't think you can find such a sine function, as it will have to have infinite amounts of peaks.
– Keen-ameteur
Nov 20 '18 at 16:47
@Keen-amateur The comment above is addressing the case of fixed $N$, as in the question the OP asked about using an analog of the formula for differential forms in the discrete case.
– Ben
Nov 20 '18 at 16:55
@Keen-amateur The comment above is addressing the case of fixed $N$, as in the question the OP asked about using an analog of the formula for differential forms in the discrete case.
– Ben
Nov 20 '18 at 16:55
add a comment |
3 Answers
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It would perhaps be worth recalling that for a sequence of partitions $ Big { 0=y_0^{(m)}< y_1^{(m)}<....< y_{n^{(m)}}^{(m)} =1 Big}_{m=1}^infty $ of $[0,1]$ such that $underset{kin n^{(m)}}{min} vert y_k^{(m)}-y_{k-1}^{(m)} vert overset {mrightarrow infty}{rightarrow}0$, that:
$underset{k=1}{ overset{n^{(m)} }{sum} } G(zeta_k^{(m)}) cdot (y_k^{(m)}-y_{k+1}^{(m)}) overset{mrightarrow infty}{rightarrow} int_0^1 G(t) dt $
where $zeta_k^{(m)}in [y_k^{(m)}, y_{k+1}^{(m)}]$ for all $kin {1,...,n^{(m)} }$.
answered Nov 20 '18 at 16:17
Keen-ameteur
1,292316
add a comment |
As stated in the comments above, the second formula in the original post is not correct. However, the statement about the limit is still true.
In the second formula, the use of the change-of-variables formula for differential forms is incorrect. The correct discrete analog of such a formula is as follows. Suppose you have values $z_0,ldots,z_n$ in the domain and you want to calculate the right-hand Riemann sum:
$$S_n = sum_{i=1}^n G(z_i) Delta z_i$$
Where $Delta z_i := z_i - z_{i-1}$. Then you may make a change of variable $z_i = 2k_i + 1$. The discrete difference then satisfies $Delta k_i = frac12 Delta z_i$ so that $S_n = 2cdotsum_{i=1}^n G(2k_i+1)Delta k_i$.
As for the limit formula, you may think of it as a midpoint Riemann sum ($1/N$ is midpoint from $0$ to $2/N$, $3/N$ is the midpoint from $2/N$ to $4/N$, etc.) except $1/N$ has been used when the intervals really have size $2/N$. (Technically the midpoint sum doesn’t work out for $N$ even but this missing term dies in the limit.)
answered Nov 20 '18 at 16:49
Ben
2,563616
add a comment |
The points $x/N$ (add $0,1$ if needed) for odd $x$ form a partition of $[0,1]$ with norm $2/N$ and hence the Riemann sum $$frac{2}{N}sum_{xtext{ odd}} G(x/N) $$ tends to $int_{0}^{1}G(r),dr$ as $Ntoinfty $ and the proof of the result in question in complete. Note that the result holds for all Riemann integrable functions $G$ and not just for continuous functions.
answered Nov 20 '18 at 16:49
Paramanand Singh
48.9k555159
The downvote indicates some issue with the post. Let me know if it can be improved in some manner.
– Paramanand Singh
Nov 21 '18 at 5:41
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
It would perhaps be worth recalling that for a sequence of partitions $ Big { 0=y_0^{(m)}< y_1^{(m)}<....< y_{n^{(m)}}^{(m)} =1 Big}_{m=1}^infty $ of $[0,1]$ such that $underset{kin n^{(m)}}{min} vert y_k^{(m)}-y_{k-1}^{(m)} vert overset {mrightarrow infty}{rightarrow}0$, that:
$underset{k=1}{ overset{n^{(m)} }{sum} } G(zeta_k^{(m)}) cdot (y_k^{(m)}-y_{k+1}^{(m)}) overset{mrightarrow infty}{rightarrow} int_0^1 G(t) dt $
where $zeta_k^{(m)}in [y_k^{(m)}, y_{k+1}^{(m)}]$ for all $kin {1,...,n^{(m)} }$.
answered Nov 20 '18 at 16:17
Keen-ameteur
1,292316
add a comment |
It would perhaps be worth recalling that for a sequence of partitions $ Big { 0=y_0^{(m)}< y_1^{(m)}<....< y_{n^{(m)}}^{(m)} =1 Big}_{m=1}^infty $ of $[0,1]$ such that $underset{kin n^{(m)}}{min} vert y_k^{(m)}-y_{k-1}^{(m)} vert overset {mrightarrow infty}{rightarrow}0$, that:
$underset{k=1}{ overset{n^{(m)} }{sum} } G(zeta_k^{(m)}) cdot (y_k^{(m)}-y_{k+1}^{(m)}) overset{mrightarrow infty}{rightarrow} int_0^1 G(t) dt $
where $zeta_k^{(m)}in [y_k^{(m)}, y_{k+1}^{(m)}]$ for all $kin {1,...,n^{(m)} }$.
answered Nov 20 '18 at 16:17
Keen-ameteur
1,292316
add a comment |
It would perhaps be worth recalling that for a sequence of partitions $ Big { 0=y_0^{(m)}< y_1^{(m)}<....< y_{n^{(m)}}^{(m)} =1 Big}_{m=1}^infty $ of $[0,1]$ such that $underset{kin n^{(m)}}{min} vert y_k^{(m)}-y_{k-1}^{(m)} vert overset {mrightarrow infty}{rightarrow}0$, that:
$underset{k=1}{ overset{n^{(m)} }{sum} } G(zeta_k^{(m)}) cdot (y_k^{(m)}-y_{k+1}^{(m)}) overset{mrightarrow infty}{rightarrow} int_0^1 G(t) dt $
where $zeta_k^{(m)}in [y_k^{(m)}, y_{k+1}^{(m)}]$ for all $kin {1,...,n^{(m)} }$.
answered Nov 20 '18 at 16:17
Keen-ameteur
1,292316
It would perhaps be worth recalling that for a sequence of partitions $ Big { 0=y_0^{(m)}< y_1^{(m)}<....< y_{n^{(m)}}^{(m)} =1 Big}_{m=1}^infty $ of $[0,1]$ such that $underset{kin n^{(m)}}{min} vert y_k^{(m)}-y_{k-1}^{(m)} vert overset {mrightarrow infty}{rightarrow}0$, that:
$underset{k=1}{ overset{n^{(m)} }{sum} } G(zeta_k^{(m)}) cdot (y_k^{(m)}-y_{k+1}^{(m)}) overset{mrightarrow infty}{rightarrow} int_0^1 G(t) dt $
where $zeta_k^{(m)}in [y_k^{(m)}, y_{k+1}^{(m)}]$ for all $kin {1,...,n^{(m)} }$.
answered Nov 20 '18 at 16:17
Keen-ameteur
1,292316
answered Nov 20 '18 at 16:17
Keen-ameteur
1,292316
answered Nov 20 '18 at 16:17
Keen-ameteur
1,292316
answered Nov 20 '18 at 16:17
Keen-ameteur
1,292316
1,292316
add a comment |
add a comment |
As stated in the comments above, the second formula in the original post is not correct. However, the statement about the limit is still true.
In the second formula, the use of the change-of-variables formula for differential forms is incorrect. The correct discrete analog of such a formula is as follows. Suppose you have values $z_0,ldots,z_n$ in the domain and you want to calculate the right-hand Riemann sum:
$$S_n = sum_{i=1}^n G(z_i) Delta z_i$$
Where $Delta z_i := z_i - z_{i-1}$. Then you may make a change of variable $z_i = 2k_i + 1$. The discrete difference then satisfies $Delta k_i = frac12 Delta z_i$ so that $S_n = 2cdotsum_{i=1}^n G(2k_i+1)Delta k_i$.
As for the limit formula, you may think of it as a midpoint Riemann sum ($1/N$ is midpoint from $0$ to $2/N$, $3/N$ is the midpoint from $2/N$ to $4/N$, etc.) except $1/N$ has been used when the intervals really have size $2/N$. (Technically the midpoint sum doesn’t work out for $N$ even but this missing term dies in the limit.)
answered Nov 20 '18 at 16:49
Ben
2,563616
add a comment |
As stated in the comments above, the second formula in the original post is not correct. However, the statement about the limit is still true.
In the second formula, the use of the change-of-variables formula for differential forms is incorrect. The correct discrete analog of such a formula is as follows. Suppose you have values $z_0,ldots,z_n$ in the domain and you want to calculate the right-hand Riemann sum:
$$S_n = sum_{i=1}^n G(z_i) Delta z_i$$
Where $Delta z_i := z_i - z_{i-1}$. Then you may make a change of variable $z_i = 2k_i + 1$. The discrete difference then satisfies $Delta k_i = frac12 Delta z_i$ so that $S_n = 2cdotsum_{i=1}^n G(2k_i+1)Delta k_i$.
As for the limit formula, you may think of it as a midpoint Riemann sum ($1/N$ is midpoint from $0$ to $2/N$, $3/N$ is the midpoint from $2/N$ to $4/N$, etc.) except $1/N$ has been used when the intervals really have size $2/N$. (Technically the midpoint sum doesn’t work out for $N$ even but this missing term dies in the limit.)
answered Nov 20 '18 at 16:49
Ben
2,563616
add a comment |
As stated in the comments above, the second formula in the original post is not correct. However, the statement about the limit is still true.
In the second formula, the use of the change-of-variables formula for differential forms is incorrect. The correct discrete analog of such a formula is as follows. Suppose you have values $z_0,ldots,z_n$ in the domain and you want to calculate the right-hand Riemann sum:
$$S_n = sum_{i=1}^n G(z_i) Delta z_i$$
Where $Delta z_i := z_i - z_{i-1}$. Then you may make a change of variable $z_i = 2k_i + 1$. The discrete difference then satisfies $Delta k_i = frac12 Delta z_i$ so that $S_n = 2cdotsum_{i=1}^n G(2k_i+1)Delta k_i$.
As for the limit formula, you may think of it as a midpoint Riemann sum ($1/N$ is midpoint from $0$ to $2/N$, $3/N$ is the midpoint from $2/N$ to $4/N$, etc.) except $1/N$ has been used when the intervals really have size $2/N$. (Technically the midpoint sum doesn’t work out for $N$ even but this missing term dies in the limit.)
answered Nov 20 '18 at 16:49
Ben
2,563616
As stated in the comments above, the second formula in the original post is not correct. However, the statement about the limit is still true.
In the second formula, the use of the change-of-variables formula for differential forms is incorrect. The correct discrete analog of such a formula is as follows. Suppose you have values $z_0,ldots,z_n$ in the domain and you want to calculate the right-hand Riemann sum:
$$S_n = sum_{i=1}^n G(z_i) Delta z_i$$
Where $Delta z_i := z_i - z_{i-1}$. Then you may make a change of variable $z_i = 2k_i + 1$. The discrete difference then satisfies $Delta k_i = frac12 Delta z_i$ so that $S_n = 2cdotsum_{i=1}^n G(2k_i+1)Delta k_i$.
As for the limit formula, you may think of it as a midpoint Riemann sum ($1/N$ is midpoint from $0$ to $2/N$, $3/N$ is the midpoint from $2/N$ to $4/N$, etc.) except $1/N$ has been used when the intervals really have size $2/N$. (Technically the midpoint sum doesn’t work out for $N$ even but this missing term dies in the limit.)
answered Nov 20 '18 at 16:49
Ben
2,563616
edited Nov 22 '18 at 6:31
answered Nov 20 '18 at 16:49
Ben
2,563616
answered Nov 20 '18 at 16:49
Ben
2,563616
answered Nov 20 '18 at 16:49
Ben
2,563616
2,563616
add a comment |
add a comment |
The points $x/N$ (add $0,1$ if needed) for odd $x$ form a partition of $[0,1]$ with norm $2/N$ and hence the Riemann sum $$frac{2}{N}sum_{xtext{ odd}} G(x/N) $$ tends to $int_{0}^{1}G(r),dr$ as $Ntoinfty $ and the proof of the result in question in complete. Note that the result holds for all Riemann integrable functions $G$ and not just for continuous functions.
answered Nov 20 '18 at 16:49
Paramanand Singh
48.9k555159
The downvote indicates some issue with the post. Let me know if it can be improved in some manner.
– Paramanand Singh
Nov 21 '18 at 5:41
add a comment |
The points $x/N$ (add $0,1$ if needed) for odd $x$ form a partition of $[0,1]$ with norm $2/N$ and hence the Riemann sum $$frac{2}{N}sum_{xtext{ odd}} G(x/N) $$ tends to $int_{0}^{1}G(r),dr$ as $Ntoinfty $ and the proof of the result in question in complete. Note that the result holds for all Riemann integrable functions $G$ and not just for continuous functions.
answered Nov 20 '18 at 16:49
Paramanand Singh
48.9k555159
The downvote indicates some issue with the post. Let me know if it can be improved in some manner.
– Paramanand Singh
Nov 21 '18 at 5:41
add a comment |
The points $x/N$ (add $0,1$ if needed) for odd $x$ form a partition of $[0,1]$ with norm $2/N$ and hence the Riemann sum $$frac{2}{N}sum_{xtext{ odd}} G(x/N) $$ tends to $int_{0}^{1}G(r),dr$ as $Ntoinfty $ and the proof of the result in question in complete. Note that the result holds for all Riemann integrable functions $G$ and not just for continuous functions.
answered Nov 20 '18 at 16:49
Paramanand Singh
48.9k555159
The points $x/N$ (add $0,1$ if needed) for odd $x$ form a partition of $[0,1]$ with norm $2/N$ and hence the Riemann sum $$frac{2}{N}sum_{xtext{ odd}} G(x/N) $$ tends to $int_{0}^{1}G(r),dr$ as $Ntoinfty $ and the proof of the result in question in complete. Note that the result holds for all Riemann integrable functions $G$ and not just for continuous functions.
answered Nov 20 '18 at 16:49
Paramanand Singh
48.9k555159
answered Nov 20 '18 at 16:49
Paramanand Singh
48.9k555159
answered Nov 20 '18 at 16:49
Paramanand Singh
48.9k555159
answered Nov 20 '18 at 16:49
Paramanand Singh
48.9k555159
48.9k555159
The downvote indicates some issue with the post. Let me know if it can be improved in some manner.
– Paramanand Singh
Nov 21 '18 at 5:41
add a comment |
The downvote indicates some issue with the post. Let me know if it can be improved in some manner.
– Paramanand Singh
Nov 21 '18 at 5:41
The downvote indicates some issue with the post. Let me know if it can be improved in some manner.
– Paramanand Singh
Nov 21 '18 at 5:41
The downvote indicates some issue with the post. Let me know if it can be improved in some manner.
– Paramanand Singh
Nov 21 '18 at 5:41
add a comment |
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What you wrote is currently not true since $G$ is not defined on $x=2$.
– Keen-ameteur
Nov 20 '18 at 15:59
The first substitution was already $x=2k+1$ so introducing $z$ will just take you back to the first formula renaming $x$ as $z$. There’s no way to double the number of summands with just a change of variables.
– Ben
Nov 20 '18 at 16:16
I think you meant “$G(z/N)$” at the end. Still the formula is not true. The question is equivalent to whether sampling the function over odd numerators is equal to summing over even numerators. It’s easy to picture counterexamples to this - a sine function with peaks at even numerators and troughs at odd numerators.
– Ben
Nov 20 '18 at 16:20
Seeing as the function has to stay constant with $N$, I don't think you can find such a sine function, as it will have to have infinite amounts of peaks.
– Keen-ameteur
Nov 20 '18 at 16:47
@Keen-amateur The comment above is addressing the case of fixed $N$, as in the question the OP asked about using an analog of the formula for differential forms in the discrete case.
– Ben
Nov 20 '18 at 16:55