What is the product of infinitely many infinitesimals?
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In general, taking the sequence for example, if $limlimits_{n to infty}a(n)=0$, we call the sequence $a(n)$ is an infinitesimal.
It's well known that, the product of a finite number of infinitesimals is still an infinitesimal, which can be proven by induction. Suppose that $limlimits_{n to infty}a_1(n)=limlimits_{n to infty}a_2(n)=0$. Then according to the rule of the limits product, $limlimits_{n to infty}[a_1(n)a_2(n)]=0$, which shows that the product of two infinitesimals is an infinitesimal. Thus, by induction, we can generalize the conclusion to the case when a finite number of infinitesimals multiply.
But what about the product of infinitely many infinitesimals? How to define such a product?
calculus infinitesimals
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,927618
$endgroup$
add a comment |
$begingroup$
In general, taking the sequence for example, if $limlimits_{n to infty}a(n)=0$, we call the sequence $a(n)$ is an infinitesimal.
It's well known that, the product of a finite number of infinitesimals is still an infinitesimal, which can be proven by induction. Suppose that $limlimits_{n to infty}a_1(n)=limlimits_{n to infty}a_2(n)=0$. Then according to the rule of the limits product, $limlimits_{n to infty}[a_1(n)a_2(n)]=0$, which shows that the product of two infinitesimals is an infinitesimal. Thus, by induction, we can generalize the conclusion to the case when a finite number of infinitesimals multiply.
But what about the product of infinitely many infinitesimals? How to define such a product?
calculus infinitesimals
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,927618
$endgroup$
3
$begingroup$
You can cook up an example where the limit of the infinite product is anything you want - have the decay of the first n be cancelled out by the next n. For example, set the first n to to 1/n, and the next n to n , and then the rest to one. Each one converges to zero (after maybe getting very large), and the product of all of them is 1.
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– Lorenzo
Dec 19 '18 at 6:05
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can you elaborate your example in details ?
$endgroup$
– mengdie1982
Dec 19 '18 at 6:28
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Define $a_i(n)$ ($n$ is the time variable) by setting it to be $1$ unless $i <= 2n$, at which time it is $1/n$ if $i <= n$ and otherwise it is set to $n$. Since eventually $n >= i$, each sequence $a_i$ looks eventually like the sequence $1/n$ so it converges to zero. The product at each time is 1 (if not adjust definition slightly), so the limit of the product is 1. At each time the infinite product is a finite product, since all but finitely many terms are 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 17:15
add a comment |
$begingroup$
In general, taking the sequence for example, if $limlimits_{n to infty}a(n)=0$, we call the sequence $a(n)$ is an infinitesimal.
It's well known that, the product of a finite number of infinitesimals is still an infinitesimal, which can be proven by induction. Suppose that $limlimits_{n to infty}a_1(n)=limlimits_{n to infty}a_2(n)=0$. Then according to the rule of the limits product, $limlimits_{n to infty}[a_1(n)a_2(n)]=0$, which shows that the product of two infinitesimals is an infinitesimal. Thus, by induction, we can generalize the conclusion to the case when a finite number of infinitesimals multiply.
But what about the product of infinitely many infinitesimals? How to define such a product?
calculus infinitesimals
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,927618
$endgroup$
In general, taking the sequence for example, if $limlimits_{n to infty}a(n)=0$, we call the sequence $a(n)$ is an infinitesimal.
It's well known that, the product of a finite number of infinitesimals is still an infinitesimal, which can be proven by induction. Suppose that $limlimits_{n to infty}a_1(n)=limlimits_{n to infty}a_2(n)=0$. Then according to the rule of the limits product, $limlimits_{n to infty}[a_1(n)a_2(n)]=0$, which shows that the product of two infinitesimals is an infinitesimal. Thus, by induction, we can generalize the conclusion to the case when a finite number of infinitesimals multiply.
But what about the product of infinitely many infinitesimals? How to define such a product?
calculus infinitesimals
calculus infinitesimals
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,927618
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,927618
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,927618
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,927618
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,927618
4,927618
3
$begingroup$
You can cook up an example where the limit of the infinite product is anything you want - have the decay of the first n be cancelled out by the next n. For example, set the first n to to 1/n, and the next n to n , and then the rest to one. Each one converges to zero (after maybe getting very large), and the product of all of them is 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 6:05
$begingroup$
can you elaborate your example in details ?
$endgroup$
– mengdie1982
Dec 19 '18 at 6:28
$begingroup$
Define $a_i(n)$ ($n$ is the time variable) by setting it to be $1$ unless $i <= 2n$, at which time it is $1/n$ if $i <= n$ and otherwise it is set to $n$. Since eventually $n >= i$, each sequence $a_i$ looks eventually like the sequence $1/n$ so it converges to zero. The product at each time is 1 (if not adjust definition slightly), so the limit of the product is 1. At each time the infinite product is a finite product, since all but finitely many terms are 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 17:15
add a comment |
3
$begingroup$
You can cook up an example where the limit of the infinite product is anything you want - have the decay of the first n be cancelled out by the next n. For example, set the first n to to 1/n, and the next n to n , and then the rest to one. Each one converges to zero (after maybe getting very large), and the product of all of them is 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 6:05
$begingroup$
can you elaborate your example in details ?
$endgroup$
– mengdie1982
Dec 19 '18 at 6:28
$begingroup$
Define $a_i(n)$ ($n$ is the time variable) by setting it to be $1$ unless $i <= 2n$, at which time it is $1/n$ if $i <= n$ and otherwise it is set to $n$. Since eventually $n >= i$, each sequence $a_i$ looks eventually like the sequence $1/n$ so it converges to zero. The product at each time is 1 (if not adjust definition slightly), so the limit of the product is 1. At each time the infinite product is a finite product, since all but finitely many terms are 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 17:15
3
3
$begingroup$
You can cook up an example where the limit of the infinite product is anything you want - have the decay of the first n be cancelled out by the next n. For example, set the first n to to 1/n, and the next n to n , and then the rest to one. Each one converges to zero (after maybe getting very large), and the product of all of them is 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 6:05
$begingroup$
You can cook up an example where the limit of the infinite product is anything you want - have the decay of the first n be cancelled out by the next n. For example, set the first n to to 1/n, and the next n to n , and then the rest to one. Each one converges to zero (after maybe getting very large), and the product of all of them is 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 6:05
$begingroup$
can you elaborate your example in details ?
$endgroup$
– mengdie1982
Dec 19 '18 at 6:28
$begingroup$
can you elaborate your example in details ?
$endgroup$
– mengdie1982
Dec 19 '18 at 6:28
$begingroup$
Define $a_i(n)$ ($n$ is the time variable) by setting it to be $1$ unless $i <= 2n$, at which time it is $1/n$ if $i <= n$ and otherwise it is set to $n$. Since eventually $n >= i$, each sequence $a_i$ looks eventually like the sequence $1/n$ so it converges to zero. The product at each time is 1 (if not adjust definition slightly), so the limit of the product is 1. At each time the infinite product is a finite product, since all but finitely many terms are 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 17:15
$begingroup$
Define $a_i(n)$ ($n$ is the time variable) by setting it to be $1$ unless $i <= 2n$, at which time it is $1/n$ if $i <= n$ and otherwise it is set to $n$. Since eventually $n >= i$, each sequence $a_i$ looks eventually like the sequence $1/n$ so it converges to zero. The product at each time is 1 (if not adjust definition slightly), so the limit of the product is 1. At each time the infinite product is a finite product, since all but finitely many terms are 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 17:15
add a comment |
1 Answer
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Dear Sir @Lorenzo, if what you mean is the one above, it would be fair to say that the example is not valid.
Notice that there are $k$ terms greater than $dfrac{1}{2}$ in ${x_n^{(k)}}_{n=1}^{infty}$. Then let $k to infty$, there are infinitely many terms greater than $dfrac{1}{2}$. As result, $prodlimits_{k=1}^{infty} x_n^{(k)}$ is not the product of infinitely many infinitesimals.
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,927618
$endgroup$
$begingroup$
This is not what I mean. The first term should be (1/2, 2, 1,1,1 ...) Then (1/3,1/3,3,3,1,1,1...)
$endgroup$
– Lorenzo
Dec 20 '18 at 15:01
$begingroup$
Although maybe I'm misreading your example. If those are the sequences, the example you wrote works. The product at any time is 1, but each sequence goes to zero. There are never infinitely many terms greater than 1/2.
$endgroup$
– Lorenzo
Dec 20 '18 at 15:04
add a comment |
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$begingroup$
Dear Sir @Lorenzo, if what you mean is the one above, it would be fair to say that the example is not valid.
Notice that there are $k$ terms greater than $dfrac{1}{2}$ in ${x_n^{(k)}}_{n=1}^{infty}$. Then let $k to infty$, there are infinitely many terms greater than $dfrac{1}{2}$. As result, $prodlimits_{k=1}^{infty} x_n^{(k)}$ is not the product of infinitely many infinitesimals.
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,927618
$endgroup$
$begingroup$
This is not what I mean. The first term should be (1/2, 2, 1,1,1 ...) Then (1/3,1/3,3,3,1,1,1...)
$endgroup$
– Lorenzo
Dec 20 '18 at 15:01
$begingroup$
Although maybe I'm misreading your example. If those are the sequences, the example you wrote works. The product at any time is 1, but each sequence goes to zero. There are never infinitely many terms greater than 1/2.
$endgroup$
– Lorenzo
Dec 20 '18 at 15:04
add a comment |
$begingroup$
Dear Sir @Lorenzo, if what you mean is the one above, it would be fair to say that the example is not valid.
Notice that there are $k$ terms greater than $dfrac{1}{2}$ in ${x_n^{(k)}}_{n=1}^{infty}$. Then let $k to infty$, there are infinitely many terms greater than $dfrac{1}{2}$. As result, $prodlimits_{k=1}^{infty} x_n^{(k)}$ is not the product of infinitely many infinitesimals.
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,927618
$endgroup$
$begingroup$
This is not what I mean. The first term should be (1/2, 2, 1,1,1 ...) Then (1/3,1/3,3,3,1,1,1...)
$endgroup$
– Lorenzo
Dec 20 '18 at 15:01
$begingroup$
Although maybe I'm misreading your example. If those are the sequences, the example you wrote works. The product at any time is 1, but each sequence goes to zero. There are never infinitely many terms greater than 1/2.
$endgroup$
– Lorenzo
Dec 20 '18 at 15:04
add a comment |
$begingroup$
Dear Sir @Lorenzo, if what you mean is the one above, it would be fair to say that the example is not valid.
Notice that there are $k$ terms greater than $dfrac{1}{2}$ in ${x_n^{(k)}}_{n=1}^{infty}$. Then let $k to infty$, there are infinitely many terms greater than $dfrac{1}{2}$. As result, $prodlimits_{k=1}^{infty} x_n^{(k)}$ is not the product of infinitely many infinitesimals.
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,927618
$endgroup$
Dear Sir @Lorenzo, if what you mean is the one above, it would be fair to say that the example is not valid.
Notice that there are $k$ terms greater than $dfrac{1}{2}$ in ${x_n^{(k)}}_{n=1}^{infty}$. Then let $k to infty$, there are infinitely many terms greater than $dfrac{1}{2}$. As result, $prodlimits_{k=1}^{infty} x_n^{(k)}$ is not the product of infinitely many infinitesimals.
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,927618
edited Dec 20 '18 at 3:10
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,927618
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,927618
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,927618
4,927618
$begingroup$
This is not what I mean. The first term should be (1/2, 2, 1,1,1 ...) Then (1/3,1/3,3,3,1,1,1...)
$endgroup$
– Lorenzo
Dec 20 '18 at 15:01
$begingroup$
Although maybe I'm misreading your example. If those are the sequences, the example you wrote works. The product at any time is 1, but each sequence goes to zero. There are never infinitely many terms greater than 1/2.
$endgroup$
– Lorenzo
Dec 20 '18 at 15:04
add a comment |
$begingroup$
This is not what I mean. The first term should be (1/2, 2, 1,1,1 ...) Then (1/3,1/3,3,3,1,1,1...)
$endgroup$
– Lorenzo
Dec 20 '18 at 15:01
$begingroup$
Although maybe I'm misreading your example. If those are the sequences, the example you wrote works. The product at any time is 1, but each sequence goes to zero. There are never infinitely many terms greater than 1/2.
$endgroup$
– Lorenzo
Dec 20 '18 at 15:04
$begingroup$
This is not what I mean. The first term should be (1/2, 2, 1,1,1 ...) Then (1/3,1/3,3,3,1,1,1...)
$endgroup$
– Lorenzo
Dec 20 '18 at 15:01
$begingroup$
This is not what I mean. The first term should be (1/2, 2, 1,1,1 ...) Then (1/3,1/3,3,3,1,1,1...)
$endgroup$
– Lorenzo
Dec 20 '18 at 15:01
$begingroup$
Although maybe I'm misreading your example. If those are the sequences, the example you wrote works. The product at any time is 1, but each sequence goes to zero. There are never infinitely many terms greater than 1/2.
$endgroup$
– Lorenzo
Dec 20 '18 at 15:04
$begingroup$
Although maybe I'm misreading your example. If those are the sequences, the example you wrote works. The product at any time is 1, but each sequence goes to zero. There are never infinitely many terms greater than 1/2.
$endgroup$
– Lorenzo
Dec 20 '18 at 15:04
add a comment |
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3
$begingroup$
You can cook up an example where the limit of the infinite product is anything you want - have the decay of the first n be cancelled out by the next n. For example, set the first n to to 1/n, and the next n to n , and then the rest to one. Each one converges to zero (after maybe getting very large), and the product of all of them is 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 6:05
$begingroup$
can you elaborate your example in details ?
$endgroup$
– mengdie1982
Dec 19 '18 at 6:28
$begingroup$
Define $a_i(n)$ ($n$ is the time variable) by setting it to be $1$ unless $i <= 2n$, at which time it is $1/n$ if $i <= n$ and otherwise it is set to $n$. Since eventually $n >= i$, each sequence $a_i$ looks eventually like the sequence $1/n$ so it converges to zero. The product at each time is 1 (if not adjust definition slightly), so the limit of the product is 1. At each time the infinite product is a finite product, since all but finitely many terms are 1.
$endgroup$
– Lorenzo
Dec 19 '18 at 17:15