Mixing
Summation of series using definite integral
$begingroup$
I learned that definite integral gives the signed area under a curve by dividing the curve into small rectangular strips and "making" its width shrink to zero.
Using this knowledge summation of certain series can be found. I converted definite integral in this form
$$ int_{a}^{b} f(x) , mathrm{d}x = lim_{ntoinfty} frac{b-a}{n} sum_{r=1}^{n} fleft( a + left(frac{b-a}{n}right)r right) $$
However, another equivalent form is known to exist, provided below, and I find it much more convenient to use that form. I tried to manipulate the sum to arrive at the equivalent form but so far I've not able to succeed.
$$ int_{a}^{b} f(x) , mathrm{d}x = lim_{ntoinfty} frac{1}{n} sum_{r=g(n)}^{h(n)} fleft(frac{r}{n}right),
quad text{where } begin{cases}
lim_{ntoinfty} frac{g(n)}{n} = a, \
lim_{ntoinfty} frac{h(n)}{n} = b.
end{cases} $$
Can someone help (give me a hint perhaps) on how to convert to the second much more convenient expression.
I find second expression much more convenient because the limits of integration can be very easily found using it.
Any help will be appreciated.
EDIT: I've still not able to derive the equivalent relation mentioned above. Can someone hint me?
calculus integration limits definite-integrals summation
asked Feb 3 at 6:10
Shivansh JShivansh J
677
$endgroup$
|
show 2 more comments
$begingroup$
I learned that definite integral gives the signed area under a curve by dividing the curve into small rectangular strips and "making" its width shrink to zero.
Using this knowledge summation of certain series can be found. I converted definite integral in this form
$$ int_{a}^{b} f(x) , mathrm{d}x = lim_{ntoinfty} frac{b-a}{n} sum_{r=1}^{n} fleft( a + left(frac{b-a}{n}right)r right) $$
However, another equivalent form is known to exist, provided below, and I find it much more convenient to use that form. I tried to manipulate the sum to arrive at the equivalent form but so far I've not able to succeed.
$$ int_{a}^{b} f(x) , mathrm{d}x = lim_{ntoinfty} frac{1}{n} sum_{r=g(n)}^{h(n)} fleft(frac{r}{n}right),
quad text{where } begin{cases}
lim_{ntoinfty} frac{g(n)}{n} = a, \
lim_{ntoinfty} frac{h(n)}{n} = b.
end{cases} $$
Can someone help (give me a hint perhaps) on how to convert to the second much more convenient expression.
I find second expression much more convenient because the limits of integration can be very easily found using it.
Any help will be appreciated.
EDIT: I've still not able to derive the equivalent relation mentioned above. Can someone hint me?
calculus integration limits definite-integrals summation
asked Feb 3 at 6:10
Shivansh JShivansh J
677
$endgroup$
$begingroup$
Hmmm.. why isn't someone responding to my question? Sorry if it sounds a bit silly question.
$endgroup$
– Shivansh J
Feb 4 at 9:11
$begingroup$
The notation is a bit unclear to me. For the second sum to make sense, it should hold that $g(n)$ and $h(n)$ are always integers with $g(n) leq h(n)$ for any $n in mathbb N$.
$endgroup$
– Berni Waterman
Mar 5 at 21:00
$begingroup$
Could you give a practical example for the second expression?
$endgroup$
– user
Mar 5 at 21:07
$begingroup$
@user Consider when lower index of summation is 1 and upper index is 2n. Limits on integration will be given by
$endgroup$
– Shivansh J
Mar 5 at 21:10
$begingroup$
...a= 1/n (n tends to infinity) = 0 and b = 2n/n (n tends to infinity) = 2
$endgroup$
– Shivansh J
Mar 5 at 21:11
|
show 2 more comments
$begingroup$
I learned that definite integral gives the signed area under a curve by dividing the curve into small rectangular strips and "making" its width shrink to zero.
Using this knowledge summation of certain series can be found. I converted definite integral in this form
$$ int_{a}^{b} f(x) , mathrm{d}x = lim_{ntoinfty} frac{b-a}{n} sum_{r=1}^{n} fleft( a + left(frac{b-a}{n}right)r right) $$
However, another equivalent form is known to exist, provided below, and I find it much more convenient to use that form. I tried to manipulate the sum to arrive at the equivalent form but so far I've not able to succeed.
$$ int_{a}^{b} f(x) , mathrm{d}x = lim_{ntoinfty} frac{1}{n} sum_{r=g(n)}^{h(n)} fleft(frac{r}{n}right),
quad text{where } begin{cases}
lim_{ntoinfty} frac{g(n)}{n} = a, \
lim_{ntoinfty} frac{h(n)}{n} = b.
end{cases} $$
Can someone help (give me a hint perhaps) on how to convert to the second much more convenient expression.
I find second expression much more convenient because the limits of integration can be very easily found using it.
Any help will be appreciated.
EDIT: I've still not able to derive the equivalent relation mentioned above. Can someone hint me?
calculus integration limits definite-integrals summation
asked Feb 3 at 6:10
Shivansh JShivansh J
677
$endgroup$
I learned that definite integral gives the signed area under a curve by dividing the curve into small rectangular strips and "making" its width shrink to zero.
Using this knowledge summation of certain series can be found. I converted definite integral in this form
$$ int_{a}^{b} f(x) , mathrm{d}x = lim_{ntoinfty} frac{b-a}{n} sum_{r=1}^{n} fleft( a + left(frac{b-a}{n}right)r right) $$
However, another equivalent form is known to exist, provided below, and I find it much more convenient to use that form. I tried to manipulate the sum to arrive at the equivalent form but so far I've not able to succeed.
$$ int_{a}^{b} f(x) , mathrm{d}x = lim_{ntoinfty} frac{1}{n} sum_{r=g(n)}^{h(n)} fleft(frac{r}{n}right),
quad text{where } begin{cases}
lim_{ntoinfty} frac{g(n)}{n} = a, \
lim_{ntoinfty} frac{h(n)}{n} = b.
end{cases} $$
Can someone help (give me a hint perhaps) on how to convert to the second much more convenient expression.
I find second expression much more convenient because the limits of integration can be very easily found using it.
Any help will be appreciated.
EDIT: I've still not able to derive the equivalent relation mentioned above. Can someone hint me?
calculus integration limits definite-integrals summation
calculus integration limits definite-integrals summation
asked Feb 3 at 6:10
Shivansh JShivansh J
677
asked Feb 3 at 6:10
Shivansh JShivansh J
677
edited Mar 5 at 20:29
Shivansh J
asked Feb 3 at 6:10
Shivansh JShivansh J
677
asked Feb 3 at 6:10
Shivansh JShivansh J
677
asked Feb 3 at 6:10
Shivansh JShivansh J
677
677
$begingroup$
Hmmm.. why isn't someone responding to my question? Sorry if it sounds a bit silly question.
$endgroup$
– Shivansh J
Feb 4 at 9:11
$begingroup$
The notation is a bit unclear to me. For the second sum to make sense, it should hold that $g(n)$ and $h(n)$ are always integers with $g(n) leq h(n)$ for any $n in mathbb N$.
$endgroup$
– Berni Waterman
Mar 5 at 21:00
$begingroup$
Could you give a practical example for the second expression?
$endgroup$
– user
Mar 5 at 21:07
$begingroup$
@user Consider when lower index of summation is 1 and upper index is 2n. Limits on integration will be given by
$endgroup$
– Shivansh J
Mar 5 at 21:10
$begingroup$
...a= 1/n (n tends to infinity) = 0 and b = 2n/n (n tends to infinity) = 2
$endgroup$
– Shivansh J
Mar 5 at 21:11
|
show 2 more comments
$begingroup$
Hmmm.. why isn't someone responding to my question? Sorry if it sounds a bit silly question.
$endgroup$
– Shivansh J
Feb 4 at 9:11
$begingroup$
The notation is a bit unclear to me. For the second sum to make sense, it should hold that $g(n)$ and $h(n)$ are always integers with $g(n) leq h(n)$ for any $n in mathbb N$.
$endgroup$
– Berni Waterman
Mar 5 at 21:00
$begingroup$
Could you give a practical example for the second expression?
$endgroup$
– user
Mar 5 at 21:07
$begingroup$
@user Consider when lower index of summation is 1 and upper index is 2n. Limits on integration will be given by
$endgroup$
– Shivansh J
Mar 5 at 21:10
$begingroup$
...a= 1/n (n tends to infinity) = 0 and b = 2n/n (n tends to infinity) = 2
$endgroup$
– Shivansh J
Mar 5 at 21:11
$begingroup$
Hmmm.. why isn't someone responding to my question? Sorry if it sounds a bit silly question.
$endgroup$
– Shivansh J
Feb 4 at 9:11
$begingroup$
Hmmm.. why isn't someone responding to my question? Sorry if it sounds a bit silly question.
$endgroup$
– Shivansh J
Feb 4 at 9:11
$begingroup$
The notation is a bit unclear to me. For the second sum to make sense, it should hold that $g(n)$ and $h(n)$ are always integers with $g(n) leq h(n)$ for any $n in mathbb N$.
$endgroup$
– Berni Waterman
Mar 5 at 21:00
$begingroup$
The notation is a bit unclear to me. For the second sum to make sense, it should hold that $g(n)$ and $h(n)$ are always integers with $g(n) leq h(n)$ for any $n in mathbb N$.
$endgroup$
– Berni Waterman
Mar 5 at 21:00
$begingroup$
Could you give a practical example for the second expression?
$endgroup$
– user
Mar 5 at 21:07
$begingroup$
Could you give a practical example for the second expression?
$endgroup$
– user
Mar 5 at 21:07
$begingroup$
@user Consider when lower index of summation is 1 and upper index is 2n. Limits on integration will be given by
$endgroup$
– Shivansh J
Mar 5 at 21:10
$begingroup$
@user Consider when lower index of summation is 1 and upper index is 2n. Limits on integration will be given by
$endgroup$
– Shivansh J
Mar 5 at 21:10
$begingroup$
...a= 1/n (n tends to infinity) = 0 and b = 2n/n (n tends to infinity) = 2
$endgroup$
– Shivansh J
Mar 5 at 21:11
$begingroup$
...a= 1/n (n tends to infinity) = 0 and b = 2n/n (n tends to infinity) = 2
$endgroup$
– Shivansh J
Mar 5 at 21:11
|
show 2 more comments
1 Answer
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It more make sense that we focus on the behavior of $$frac1nsum_{r=h(n)}^{g(n)}f(r/n)$$
as $nto infty$. Think of the $1/n$ as the width of each subinterval of $[a,b]$, and think of $r/n$ as $x$-values in each subinterval. As $ntoinfty$, each subinterval approaches a single point, namely $lim_{ntoinfty}r/n$. So it's just a sum of $text{height}timestext{width}$.
answered Mar 19 at 18:15
clathratusclathratus
5,1441439
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1 Answer
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$begingroup$
It more make sense that we focus on the behavior of $$frac1nsum_{r=h(n)}^{g(n)}f(r/n)$$
as $nto infty$. Think of the $1/n$ as the width of each subinterval of $[a,b]$, and think of $r/n$ as $x$-values in each subinterval. As $ntoinfty$, each subinterval approaches a single point, namely $lim_{ntoinfty}r/n$. So it's just a sum of $text{height}timestext{width}$.
answered Mar 19 at 18:15
clathratusclathratus
5,1441439
$endgroup$
add a comment |
$begingroup$
It more make sense that we focus on the behavior of $$frac1nsum_{r=h(n)}^{g(n)}f(r/n)$$
as $nto infty$. Think of the $1/n$ as the width of each subinterval of $[a,b]$, and think of $r/n$ as $x$-values in each subinterval. As $ntoinfty$, each subinterval approaches a single point, namely $lim_{ntoinfty}r/n$. So it's just a sum of $text{height}timestext{width}$.
answered Mar 19 at 18:15
clathratusclathratus
5,1441439
$endgroup$
add a comment |
$begingroup$
It more make sense that we focus on the behavior of $$frac1nsum_{r=h(n)}^{g(n)}f(r/n)$$
as $nto infty$. Think of the $1/n$ as the width of each subinterval of $[a,b]$, and think of $r/n$ as $x$-values in each subinterval. As $ntoinfty$, each subinterval approaches a single point, namely $lim_{ntoinfty}r/n$. So it's just a sum of $text{height}timestext{width}$.
answered Mar 19 at 18:15
clathratusclathratus
5,1441439
$endgroup$
It more make sense that we focus on the behavior of $$frac1nsum_{r=h(n)}^{g(n)}f(r/n)$$
as $nto infty$. Think of the $1/n$ as the width of each subinterval of $[a,b]$, and think of $r/n$ as $x$-values in each subinterval. As $ntoinfty$, each subinterval approaches a single point, namely $lim_{ntoinfty}r/n$. So it's just a sum of $text{height}timestext{width}$.
answered Mar 19 at 18:15
clathratusclathratus
5,1441439
answered Mar 19 at 18:15
clathratusclathratus
5,1441439
answered Mar 19 at 18:15
clathratusclathratus
5,1441439
answered Mar 19 at 18:15
clathratusclathratus
5,1441439
5,1441439
add a comment |
add a comment |
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$begingroup$
Hmmm.. why isn't someone responding to my question? Sorry if it sounds a bit silly question.
$endgroup$
– Shivansh J
Feb 4 at 9:11
$begingroup$
The notation is a bit unclear to me. For the second sum to make sense, it should hold that $g(n)$ and $h(n)$ are always integers with $g(n) leq h(n)$ for any $n in mathbb N$.
$endgroup$
– Berni Waterman
Mar 5 at 21:00
$begingroup$
Could you give a practical example for the second expression?
$endgroup$
– user
Mar 5 at 21:07
$begingroup$
@user Consider when lower index of summation is 1 and upper index is 2n. Limits on integration will be given by
$endgroup$
– Shivansh J
Mar 5 at 21:10
$begingroup$
...a= 1/n (n tends to infinity) = 0 and b = 2n/n (n tends to infinity) = 2
$endgroup$
– Shivansh J
Mar 5 at 21:11