Why is there a $m-1$ when approximating a lower bound of a function through summation or integral?
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For a monotonically increasing f(x), why does the summation below, on the left hand side, always approximate the lower bound of the summation on the right hand side?
$int_{k=m-1}^n f(k) leq sum_{k=m}^n f(k)$
If $f(x)=k$, $m=2$, and for $ngeq 2$ then the left hand summation is no longer a lower bound, which contradicts the above.
This formula sheet from a class suggests the left hand side would be a lower bound: https://www.cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf
EDIT: I understand that I'm going wrong somewhere but I don't know where. The purpose of my comments is to demonstrate my reasoning for why I think there's a contradiction so that others can point out where my reasoning is incorrect. Thanks
EDIT 2: I've changed the summation symbol to an integral symbol to avoid confusion. It's worth nothing the results would be the same since the functions are called on integers (discrete intervals).
EDIT 3: The result is not the same with the change from a summation to integral because the $n$ on the left hand side is not included in the summation (so same as summation $m-1$ to $n-1$). Thanks Ross!
calculus functions summation upper-lower-bounds
asked Dec 30 '18 at 14:48
Omkar KonaraddiOmkar Konaraddi
42
$endgroup$
add a comment |
$begingroup$
For a monotonically increasing f(x), why does the summation below, on the left hand side, always approximate the lower bound of the summation on the right hand side?
$int_{k=m-1}^n f(k) leq sum_{k=m}^n f(k)$
If $f(x)=k$, $m=2$, and for $ngeq 2$ then the left hand summation is no longer a lower bound, which contradicts the above.
This formula sheet from a class suggests the left hand side would be a lower bound: https://www.cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf
EDIT: I understand that I'm going wrong somewhere but I don't know where. The purpose of my comments is to demonstrate my reasoning for why I think there's a contradiction so that others can point out where my reasoning is incorrect. Thanks
EDIT 2: I've changed the summation symbol to an integral symbol to avoid confusion. It's worth nothing the results would be the same since the functions are called on integers (discrete intervals).
EDIT 3: The result is not the same with the change from a summation to integral because the $n$ on the left hand side is not included in the summation (so same as summation $m-1$ to $n-1$). Thanks Ross!
calculus functions summation upper-lower-bounds
asked Dec 30 '18 at 14:48
Omkar KonaraddiOmkar Konaraddi
42
$endgroup$
$begingroup$
Wouldn't the left hand side with $f(x)=k$ be $(n-(m-1))k=(n-m+1)k$ and the right hand side be $(n-m)k$? EDIT: The comment I was responding to got deleted.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:03
$begingroup$
Your latest edit has made my answer inappropriate. The original version had two sums. Now there is not a left hand sum for my comments to apply to. The statement you now make is correct for $f$ increasing as the Riemann sum takes the right side of each interval.
$endgroup$
– Ross Millikan
Dec 30 '18 at 19:37
$begingroup$
I've understood my mistake, thank you!
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 20:28
add a comment |
$begingroup$
For a monotonically increasing f(x), why does the summation below, on the left hand side, always approximate the lower bound of the summation on the right hand side?
$int_{k=m-1}^n f(k) leq sum_{k=m}^n f(k)$
If $f(x)=k$, $m=2$, and for $ngeq 2$ then the left hand summation is no longer a lower bound, which contradicts the above.
This formula sheet from a class suggests the left hand side would be a lower bound: https://www.cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf
EDIT: I understand that I'm going wrong somewhere but I don't know where. The purpose of my comments is to demonstrate my reasoning for why I think there's a contradiction so that others can point out where my reasoning is incorrect. Thanks
EDIT 2: I've changed the summation symbol to an integral symbol to avoid confusion. It's worth nothing the results would be the same since the functions are called on integers (discrete intervals).
EDIT 3: The result is not the same with the change from a summation to integral because the $n$ on the left hand side is not included in the summation (so same as summation $m-1$ to $n-1$). Thanks Ross!
calculus functions summation upper-lower-bounds
asked Dec 30 '18 at 14:48
Omkar KonaraddiOmkar Konaraddi
42
$endgroup$
For a monotonically increasing f(x), why does the summation below, on the left hand side, always approximate the lower bound of the summation on the right hand side?
$int_{k=m-1}^n f(k) leq sum_{k=m}^n f(k)$
If $f(x)=k$, $m=2$, and for $ngeq 2$ then the left hand summation is no longer a lower bound, which contradicts the above.
This formula sheet from a class suggests the left hand side would be a lower bound: https://www.cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf
EDIT: I understand that I'm going wrong somewhere but I don't know where. The purpose of my comments is to demonstrate my reasoning for why I think there's a contradiction so that others can point out where my reasoning is incorrect. Thanks
EDIT 2: I've changed the summation symbol to an integral symbol to avoid confusion. It's worth nothing the results would be the same since the functions are called on integers (discrete intervals).
EDIT 3: The result is not the same with the change from a summation to integral because the $n$ on the left hand side is not included in the summation (so same as summation $m-1$ to $n-1$). Thanks Ross!
calculus functions summation upper-lower-bounds
calculus functions summation upper-lower-bounds
asked Dec 30 '18 at 14:48
Omkar KonaraddiOmkar Konaraddi
42
asked Dec 30 '18 at 14:48
Omkar KonaraddiOmkar Konaraddi
42
edited Dec 30 '18 at 20:30
Omkar Konaraddi
asked Dec 30 '18 at 14:48
Omkar KonaraddiOmkar Konaraddi
42
asked Dec 30 '18 at 14:48
Omkar KonaraddiOmkar Konaraddi
42
asked Dec 30 '18 at 14:48
Omkar KonaraddiOmkar Konaraddi
42
42
$begingroup$
Wouldn't the left hand side with $f(x)=k$ be $(n-(m-1))k=(n-m+1)k$ and the right hand side be $(n-m)k$? EDIT: The comment I was responding to got deleted.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:03
$begingroup$
Your latest edit has made my answer inappropriate. The original version had two sums. Now there is not a left hand sum for my comments to apply to. The statement you now make is correct for $f$ increasing as the Riemann sum takes the right side of each interval.
$endgroup$
– Ross Millikan
Dec 30 '18 at 19:37
$begingroup$
I've understood my mistake, thank you!
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 20:28
add a comment |
$begingroup$
Wouldn't the left hand side with $f(x)=k$ be $(n-(m-1))k=(n-m+1)k$ and the right hand side be $(n-m)k$? EDIT: The comment I was responding to got deleted.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:03
$begingroup$
Your latest edit has made my answer inappropriate. The original version had two sums. Now there is not a left hand sum for my comments to apply to. The statement you now make is correct for $f$ increasing as the Riemann sum takes the right side of each interval.
$endgroup$
– Ross Millikan
Dec 30 '18 at 19:37
$begingroup$
I've understood my mistake, thank you!
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 20:28
$begingroup$
Wouldn't the left hand side with $f(x)=k$ be $(n-(m-1))k=(n-m+1)k$ and the right hand side be $(n-m)k$? EDIT: The comment I was responding to got deleted.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:03
$begingroup$
Wouldn't the left hand side with $f(x)=k$ be $(n-(m-1))k=(n-m+1)k$ and the right hand side be $(n-m)k$? EDIT: The comment I was responding to got deleted.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:03
$begingroup$
Your latest edit has made my answer inappropriate. The original version had two sums. Now there is not a left hand sum for my comments to apply to. The statement you now make is correct for $f$ increasing as the Riemann sum takes the right side of each interval.
$endgroup$
– Ross Millikan
Dec 30 '18 at 19:37
$begingroup$
Your latest edit has made my answer inappropriate. The original version had two sums. Now there is not a left hand sum for my comments to apply to. The statement you now make is correct for $f$ increasing as the Riemann sum takes the right side of each interval.
$endgroup$
– Ross Millikan
Dec 30 '18 at 19:37
$begingroup$
I've understood my mistake, thank you!
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 20:28
$begingroup$
I've understood my mistake, thank you!
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 20:28
add a comment |
1 Answer
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The upper limit on the left should be $n-1$. The integral you are approximating is from $m-1$ to $n$. You have ends and break points at $m-1,m,m+1,m+2ldots n-1,n$. Using the left side takes the value of $f(x)$ from the left hand end of each interval while the right side takes the value of $f(x)$ from the right hand side of each interval. As $f(x)$ is increasing, the right side will be larger. The true integral will be between the two values.
Your formula sheet gives for $f(x)$ increasing
$$int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)le int_{m}^{n+1}f(x)dx$$
When you try to bound the left integral by a sum you need to make the limits on the integral match. If you shift the limits on the right integral down by $1$ you get the left integral. The lower bound sum then has to have both its limits decreased by $1$, giving the correct formula
$$sum_{k=m-1}^{n-1}f(k) le int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)$$
This shows the upper limit on the left sum should be $n-1$. The easiest way to see the problem is to use $f(x)=1$. Your left sum is then $n-m+1$ and your right sum is $n-m$. The integral is also $n-m$. You are adding one too many terms on the left.
answered Dec 30 '18 at 14:54
Ross MillikanRoss Millikan
302k24200375
$endgroup$
$begingroup$
This contradicts the formula sheet here: cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf and here: staff.ustc.edu.cn/~csli/graduate/algorithms/book6/chap03.htm (CTRL + F "approximation by integrals". Both use $m-1$ for the lower bound and do not use $n-1$ for the upper bound.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:00
$begingroup$
You want to sum the same number of terms on both sides. I have usually seen the right starting at $m$ going to $n-1$ and the left starting at $m+1$ going to $n$. The logic is the same. Just take $f(x)=1$ to see that you need the number of terms to match.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:07
$begingroup$
The formula sheet you linked in the question has the sum bounded by integrals, but the idea is the same. The range of the integrals is $n-m+1$ and the number of terms in the sum is $n-m+1$
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:10
$begingroup$
How is the number of terms the same if one side is going from $m-1$ to $n$ and the other side is going from $m$ to $n$? The left hand side has 1 more term in addition to all the terms of the right hand side. Thus, the left hand side must be larger whenever $m geq 2$ and $n geq 2$.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:26
$begingroup$
For example, if $f(x)=x$, $m=2$ and $n=3$ then $sum_{x=1}^{3} x leq sum_{x=2}^{3} x $ does not hold true (because 1 + 2 + 3 is greater than 2 + 3)
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:29
|
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1 Answer
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$begingroup$
The upper limit on the left should be $n-1$. The integral you are approximating is from $m-1$ to $n$. You have ends and break points at $m-1,m,m+1,m+2ldots n-1,n$. Using the left side takes the value of $f(x)$ from the left hand end of each interval while the right side takes the value of $f(x)$ from the right hand side of each interval. As $f(x)$ is increasing, the right side will be larger. The true integral will be between the two values.
Your formula sheet gives for $f(x)$ increasing
$$int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)le int_{m}^{n+1}f(x)dx$$
When you try to bound the left integral by a sum you need to make the limits on the integral match. If you shift the limits on the right integral down by $1$ you get the left integral. The lower bound sum then has to have both its limits decreased by $1$, giving the correct formula
$$sum_{k=m-1}^{n-1}f(k) le int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)$$
This shows the upper limit on the left sum should be $n-1$. The easiest way to see the problem is to use $f(x)=1$. Your left sum is then $n-m+1$ and your right sum is $n-m$. The integral is also $n-m$. You are adding one too many terms on the left.
answered Dec 30 '18 at 14:54
Ross MillikanRoss Millikan
302k24200375
$endgroup$
$begingroup$
This contradicts the formula sheet here: cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf and here: staff.ustc.edu.cn/~csli/graduate/algorithms/book6/chap03.htm (CTRL + F "approximation by integrals". Both use $m-1$ for the lower bound and do not use $n-1$ for the upper bound.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:00
$begingroup$
You want to sum the same number of terms on both sides. I have usually seen the right starting at $m$ going to $n-1$ and the left starting at $m+1$ going to $n$. The logic is the same. Just take $f(x)=1$ to see that you need the number of terms to match.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:07
$begingroup$
The formula sheet you linked in the question has the sum bounded by integrals, but the idea is the same. The range of the integrals is $n-m+1$ and the number of terms in the sum is $n-m+1$
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:10
$begingroup$
How is the number of terms the same if one side is going from $m-1$ to $n$ and the other side is going from $m$ to $n$? The left hand side has 1 more term in addition to all the terms of the right hand side. Thus, the left hand side must be larger whenever $m geq 2$ and $n geq 2$.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:26
$begingroup$
For example, if $f(x)=x$, $m=2$ and $n=3$ then $sum_{x=1}^{3} x leq sum_{x=2}^{3} x $ does not hold true (because 1 + 2 + 3 is greater than 2 + 3)
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:29
|
show 5 more comments
$begingroup$
The upper limit on the left should be $n-1$. The integral you are approximating is from $m-1$ to $n$. You have ends and break points at $m-1,m,m+1,m+2ldots n-1,n$. Using the left side takes the value of $f(x)$ from the left hand end of each interval while the right side takes the value of $f(x)$ from the right hand side of each interval. As $f(x)$ is increasing, the right side will be larger. The true integral will be between the two values.
Your formula sheet gives for $f(x)$ increasing
$$int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)le int_{m}^{n+1}f(x)dx$$
When you try to bound the left integral by a sum you need to make the limits on the integral match. If you shift the limits on the right integral down by $1$ you get the left integral. The lower bound sum then has to have both its limits decreased by $1$, giving the correct formula
$$sum_{k=m-1}^{n-1}f(k) le int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)$$
This shows the upper limit on the left sum should be $n-1$. The easiest way to see the problem is to use $f(x)=1$. Your left sum is then $n-m+1$ and your right sum is $n-m$. The integral is also $n-m$. You are adding one too many terms on the left.
answered Dec 30 '18 at 14:54
Ross MillikanRoss Millikan
302k24200375
$endgroup$
$begingroup$
This contradicts the formula sheet here: cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf and here: staff.ustc.edu.cn/~csli/graduate/algorithms/book6/chap03.htm (CTRL + F "approximation by integrals". Both use $m-1$ for the lower bound and do not use $n-1$ for the upper bound.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:00
$begingroup$
You want to sum the same number of terms on both sides. I have usually seen the right starting at $m$ going to $n-1$ and the left starting at $m+1$ going to $n$. The logic is the same. Just take $f(x)=1$ to see that you need the number of terms to match.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:07
$begingroup$
The formula sheet you linked in the question has the sum bounded by integrals, but the idea is the same. The range of the integrals is $n-m+1$ and the number of terms in the sum is $n-m+1$
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:10
$begingroup$
How is the number of terms the same if one side is going from $m-1$ to $n$ and the other side is going from $m$ to $n$? The left hand side has 1 more term in addition to all the terms of the right hand side. Thus, the left hand side must be larger whenever $m geq 2$ and $n geq 2$.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:26
$begingroup$
For example, if $f(x)=x$, $m=2$ and $n=3$ then $sum_{x=1}^{3} x leq sum_{x=2}^{3} x $ does not hold true (because 1 + 2 + 3 is greater than 2 + 3)
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:29
|
show 5 more comments
$begingroup$
The upper limit on the left should be $n-1$. The integral you are approximating is from $m-1$ to $n$. You have ends and break points at $m-1,m,m+1,m+2ldots n-1,n$. Using the left side takes the value of $f(x)$ from the left hand end of each interval while the right side takes the value of $f(x)$ from the right hand side of each interval. As $f(x)$ is increasing, the right side will be larger. The true integral will be between the two values.
Your formula sheet gives for $f(x)$ increasing
$$int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)le int_{m}^{n+1}f(x)dx$$
When you try to bound the left integral by a sum you need to make the limits on the integral match. If you shift the limits on the right integral down by $1$ you get the left integral. The lower bound sum then has to have both its limits decreased by $1$, giving the correct formula
$$sum_{k=m-1}^{n-1}f(k) le int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)$$
This shows the upper limit on the left sum should be $n-1$. The easiest way to see the problem is to use $f(x)=1$. Your left sum is then $n-m+1$ and your right sum is $n-m$. The integral is also $n-m$. You are adding one too many terms on the left.
answered Dec 30 '18 at 14:54
Ross MillikanRoss Millikan
302k24200375
$endgroup$
The upper limit on the left should be $n-1$. The integral you are approximating is from $m-1$ to $n$. You have ends and break points at $m-1,m,m+1,m+2ldots n-1,n$. Using the left side takes the value of $f(x)$ from the left hand end of each interval while the right side takes the value of $f(x)$ from the right hand side of each interval. As $f(x)$ is increasing, the right side will be larger. The true integral will be between the two values.
Your formula sheet gives for $f(x)$ increasing
$$int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)le int_{m}^{n+1}f(x)dx$$
When you try to bound the left integral by a sum you need to make the limits on the integral match. If you shift the limits on the right integral down by $1$ you get the left integral. The lower bound sum then has to have both its limits decreased by $1$, giving the correct formula
$$sum_{k=m-1}^{n-1}f(k) le int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)$$
This shows the upper limit on the left sum should be $n-1$. The easiest way to see the problem is to use $f(x)=1$. Your left sum is then $n-m+1$ and your right sum is $n-m$. The integral is also $n-m$. You are adding one too many terms on the left.
answered Dec 30 '18 at 14:54
Ross MillikanRoss Millikan
302k24200375
edited Dec 30 '18 at 19:34
answered Dec 30 '18 at 14:54
Ross MillikanRoss Millikan
302k24200375
answered Dec 30 '18 at 14:54
Ross MillikanRoss Millikan
302k24200375
answered Dec 30 '18 at 14:54
Ross MillikanRoss Millikan
302k24200375
302k24200375
$begingroup$
This contradicts the formula sheet here: cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf and here: staff.ustc.edu.cn/~csli/graduate/algorithms/book6/chap03.htm (CTRL + F "approximation by integrals". Both use $m-1$ for the lower bound and do not use $n-1$ for the upper bound.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:00
$begingroup$
You want to sum the same number of terms on both sides. I have usually seen the right starting at $m$ going to $n-1$ and the left starting at $m+1$ going to $n$. The logic is the same. Just take $f(x)=1$ to see that you need the number of terms to match.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:07
$begingroup$
The formula sheet you linked in the question has the sum bounded by integrals, but the idea is the same. The range of the integrals is $n-m+1$ and the number of terms in the sum is $n-m+1$
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:10
$begingroup$
How is the number of terms the same if one side is going from $m-1$ to $n$ and the other side is going from $m$ to $n$? The left hand side has 1 more term in addition to all the terms of the right hand side. Thus, the left hand side must be larger whenever $m geq 2$ and $n geq 2$.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:26
$begingroup$
For example, if $f(x)=x$, $m=2$ and $n=3$ then $sum_{x=1}^{3} x leq sum_{x=2}^{3} x $ does not hold true (because 1 + 2 + 3 is greater than 2 + 3)
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:29
|
show 5 more comments
$begingroup$
This contradicts the formula sheet here: cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf and here: staff.ustc.edu.cn/~csli/graduate/algorithms/book6/chap03.htm (CTRL + F "approximation by integrals". Both use $m-1$ for the lower bound and do not use $n-1$ for the upper bound.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:00
$begingroup$
You want to sum the same number of terms on both sides. I have usually seen the right starting at $m$ going to $n-1$ and the left starting at $m+1$ going to $n$. The logic is the same. Just take $f(x)=1$ to see that you need the number of terms to match.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:07
$begingroup$
The formula sheet you linked in the question has the sum bounded by integrals, but the idea is the same. The range of the integrals is $n-m+1$ and the number of terms in the sum is $n-m+1$
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:10
$begingroup$
How is the number of terms the same if one side is going from $m-1$ to $n$ and the other side is going from $m$ to $n$? The left hand side has 1 more term in addition to all the terms of the right hand side. Thus, the left hand side must be larger whenever $m geq 2$ and $n geq 2$.
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:26
$begingroup$
For example, if $f(x)=x$, $m=2$ and $n=3$ then $sum_{x=1}^{3} x leq sum_{x=2}^{3} x $ does not hold true (because 1 + 2 + 3 is greater than 2 + 3)
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– Omkar Konaraddi
Dec 30 '18 at 15:29
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This contradicts the formula sheet here: cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf and here: staff.ustc.edu.cn/~csli/graduate/algorithms/book6/chap03.htm (CTRL + F "approximation by integrals". Both use $m-1$ for the lower bound and do not use $n-1$ for the upper bound.
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– Omkar Konaraddi
Dec 30 '18 at 15:00
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This contradicts the formula sheet here: cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf and here: staff.ustc.edu.cn/~csli/graduate/algorithms/book6/chap03.htm (CTRL + F "approximation by integrals". Both use $m-1$ for the lower bound and do not use $n-1$ for the upper bound.
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– Omkar Konaraddi
Dec 30 '18 at 15:00
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You want to sum the same number of terms on both sides. I have usually seen the right starting at $m$ going to $n-1$ and the left starting at $m+1$ going to $n$. The logic is the same. Just take $f(x)=1$ to see that you need the number of terms to match.
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– Ross Millikan
Dec 30 '18 at 15:07
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You want to sum the same number of terms on both sides. I have usually seen the right starting at $m$ going to $n-1$ and the left starting at $m+1$ going to $n$. The logic is the same. Just take $f(x)=1$ to see that you need the number of terms to match.
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– Ross Millikan
Dec 30 '18 at 15:07
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The formula sheet you linked in the question has the sum bounded by integrals, but the idea is the same. The range of the integrals is $n-m+1$ and the number of terms in the sum is $n-m+1$
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– Ross Millikan
Dec 30 '18 at 15:10
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The formula sheet you linked in the question has the sum bounded by integrals, but the idea is the same. The range of the integrals is $n-m+1$ and the number of terms in the sum is $n-m+1$
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– Ross Millikan
Dec 30 '18 at 15:10
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How is the number of terms the same if one side is going from $m-1$ to $n$ and the other side is going from $m$ to $n$? The left hand side has 1 more term in addition to all the terms of the right hand side. Thus, the left hand side must be larger whenever $m geq 2$ and $n geq 2$.
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– Omkar Konaraddi
Dec 30 '18 at 15:26
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How is the number of terms the same if one side is going from $m-1$ to $n$ and the other side is going from $m$ to $n$? The left hand side has 1 more term in addition to all the terms of the right hand side. Thus, the left hand side must be larger whenever $m geq 2$ and $n geq 2$.
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– Omkar Konaraddi
Dec 30 '18 at 15:26
$begingroup$
For example, if $f(x)=x$, $m=2$ and $n=3$ then $sum_{x=1}^{3} x leq sum_{x=2}^{3} x $ does not hold true (because 1 + 2 + 3 is greater than 2 + 3)
$endgroup$
– Omkar Konaraddi
Dec 30 '18 at 15:29
$begingroup$
For example, if $f(x)=x$, $m=2$ and $n=3$ then $sum_{x=1}^{3} x leq sum_{x=2}^{3} x $ does not hold true (because 1 + 2 + 3 is greater than 2 + 3)
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– Omkar Konaraddi
Dec 30 '18 at 15:29
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Wouldn't the left hand side with $f(x)=k$ be $(n-(m-1))k=(n-m+1)k$ and the right hand side be $(n-m)k$? EDIT: The comment I was responding to got deleted.
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– Omkar Konaraddi
Dec 30 '18 at 15:03
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Your latest edit has made my answer inappropriate. The original version had two sums. Now there is not a left hand sum for my comments to apply to. The statement you now make is correct for $f$ increasing as the Riemann sum takes the right side of each interval.
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– Ross Millikan
Dec 30 '18 at 19:37
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I've understood my mistake, thank you!
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– Omkar Konaraddi
Dec 30 '18 at 20:28