A double summation
While going through the article given here (page 23), it seems that the following doublesummation is carried out
$sum_{i=1,j=1}^{N^2} B_{ij} = Big( sum_{i,j=1}^{N^2-1} + sum_{i=1,j=N^2}^{N^2} + sum_{i=N^2,j=1}^{N^2} - sum_{i=N^2,j=N^2}^{N^2} Big) B_{ij}$. Where $B_{ij} = c_{ij} F_i rho_S F_j^dagger$, is an operator.
Could anyone help me to understand how the double sum on the left breaks into the four sums on the right hand side?
sequences-and-series summation
asked Nov 29 at 18:56
Zilch
63
add a comment |
While going through the article given here (page 23), it seems that the following doublesummation is carried out
$sum_{i=1,j=1}^{N^2} B_{ij} = Big( sum_{i,j=1}^{N^2-1} + sum_{i=1,j=N^2}^{N^2} + sum_{i=N^2,j=1}^{N^2} - sum_{i=N^2,j=N^2}^{N^2} Big) B_{ij}$. Where $B_{ij} = c_{ij} F_i rho_S F_j^dagger$, is an operator.
Could anyone help me to understand how the double sum on the left breaks into the four sums on the right hand side?
sequences-and-series summation
asked Nov 29 at 18:56
Zilch
63
add a comment |
While going through the article given here (page 23), it seems that the following doublesummation is carried out
$sum_{i=1,j=1}^{N^2} B_{ij} = Big( sum_{i,j=1}^{N^2-1} + sum_{i=1,j=N^2}^{N^2} + sum_{i=N^2,j=1}^{N^2} - sum_{i=N^2,j=N^2}^{N^2} Big) B_{ij}$. Where $B_{ij} = c_{ij} F_i rho_S F_j^dagger$, is an operator.
Could anyone help me to understand how the double sum on the left breaks into the four sums on the right hand side?
sequences-and-series summation
asked Nov 29 at 18:56
Zilch
63
While going through the article given here (page 23), it seems that the following doublesummation is carried out
$sum_{i=1,j=1}^{N^2} B_{ij} = Big( sum_{i,j=1}^{N^2-1} + sum_{i=1,j=N^2}^{N^2} + sum_{i=N^2,j=1}^{N^2} - sum_{i=N^2,j=N^2}^{N^2} Big) B_{ij}$. Where $B_{ij} = c_{ij} F_i rho_S F_j^dagger$, is an operator.
Could anyone help me to understand how the double sum on the left breaks into the four sums on the right hand side?
sequences-and-series summation
sequences-and-series summation
asked Nov 29 at 18:56
Zilch
63
asked Nov 29 at 18:56
Zilch
63
asked Nov 29 at 18:56
Zilch
63
asked Nov 29 at 18:56
Zilch
63
asked Nov 29 at 18:56
Zilch
63
63
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
When you write $sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:
$i=1,ldots,N^2$; $j=N^2$
$i=N^2$; $j=1,ldots,N^2$
But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.
answered Nov 29 at 23:27
Martin Argerami
123k1176174
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%2f3019037%2fa-double-summation%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
When you write $sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:
$i=1,ldots,N^2$; $j=N^2$
$i=N^2$; $j=1,ldots,N^2$
But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.
answered Nov 29 at 23:27
Martin Argerami
123k1176174
add a comment |
When you write $sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:
$i=1,ldots,N^2$; $j=N^2$
$i=N^2$; $j=1,ldots,N^2$
But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.
answered Nov 29 at 23:27
Martin Argerami
123k1176174
add a comment |
When you write $sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:
$i=1,ldots,N^2$; $j=N^2$
$i=N^2$; $j=1,ldots,N^2$
But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.
answered Nov 29 at 23:27
Martin Argerami
123k1176174
When you write $sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:
$i=1,ldots,N^2$; $j=N^2$
$i=N^2$; $j=1,ldots,N^2$
But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.
answered Nov 29 at 23:27
Martin Argerami
123k1176174
answered Nov 29 at 23:27
Martin Argerami
123k1176174
answered Nov 29 at 23:27
Martin Argerami
123k1176174
answered Nov 29 at 23:27
Martin Argerami
123k1176174
123k1176174
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3019037%2fa-double-summation%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
