What is the product of infinitely many infinitesimals?
$begingroup$
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
4,932618
$endgroup$
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
4,932618
$endgroup$
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.
$endgroup$
– 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 |
$begingroup$
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
4,932618
$endgroup$
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
4,932618
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,932618
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,932618
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,932618
asked Dec 19 '18 at 5:50
mengdie1982mengdie1982
4,932618
4,932618
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
1
active
oldest
votes
$begingroup$
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
4,932618
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046045%2fwhat-is-the-product-of-infinitely-many-infinitesimals%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
4,932618
$endgroup$
$begingroup$
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
4,932618
$endgroup$
$begingroup$
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
4,932618
$endgroup$
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
4,932618
edited Dec 20 '18 at 3:10
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,932618
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,932618
answered Dec 20 '18 at 3:05
mengdie1982mengdie1982
4,932618
4,932618
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046045%2fwhat-is-the-product-of-infinitely-many-infinitesimals%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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