Solution to $x_i=a(1+sum_{ine j}frac{b_ix_j}{1+b_ix_j})$
I'm dealing with a set of variables $x_i$ and following equations:
$forall i$ $x_i=a(1+sum_{ine j}frac{b_ix_j}{1+b_ix_j})$
Wolfram mathematica's NSolve could help, but it's not very accurate and is very limited in capacity for the number of $x_i$s, like 8.
I'm aiming for 20-100 $x_i$s so, therefore, looking for other ways like an algorithm to find $x_i$s. Any suggestion?
algorithms
add a comment |
I'm dealing with a set of variables $x_i$ and following equations:
$forall i$ $x_i=a(1+sum_{ine j}frac{b_ix_j}{1+b_ix_j})$
Wolfram mathematica's NSolve could help, but it's not very accurate and is very limited in capacity for the number of $x_i$s, like 8.
I'm aiming for 20-100 $x_i$s so, therefore, looking for other ways like an algorithm to find $x_i$s. Any suggestion?
algorithms
add a comment |
I'm dealing with a set of variables $x_i$ and following equations:
$forall i$ $x_i=a(1+sum_{ine j}frac{b_ix_j}{1+b_ix_j})$
Wolfram mathematica's NSolve could help, but it's not very accurate and is very limited in capacity for the number of $x_i$s, like 8.
I'm aiming for 20-100 $x_i$s so, therefore, looking for other ways like an algorithm to find $x_i$s. Any suggestion?
algorithms
I'm dealing with a set of variables $x_i$ and following equations:
$forall i$ $x_i=a(1+sum_{ine j}frac{b_ix_j}{1+b_ix_j})$
Wolfram mathematica's NSolve could help, but it's not very accurate and is very limited in capacity for the number of $x_i$s, like 8.
I'm aiming for 20-100 $x_i$s so, therefore, looking for other ways like an algorithm to find $x_i$s. Any suggestion?
algorithms
algorithms
edited Dec 1 at 11:09
asked Dec 1 at 4:22
Anais
286
286
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Adding and subtracting $dfrac{x_i}{1+x_i}$ inside the parenthesis gives
$$x_i = a(1 + sum_{j}frac{x_j}{1+x_j} - frac{x_i}{1+x_i})$$
The sum does not depend on $i$; it is a constant. Can you continue?
1
Sum is not a constant and depends on all $x_i$s.
– Anais
Dec 1 at 5:34
@Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
– user58697
Dec 1 at 5:39
My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
– Anais
Dec 1 at 5:47
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%2f3020984%2fsolution-to-x-i-a1-sum-i-ne-j-fracb-ix-j1b-ix-j%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
Adding and subtracting $dfrac{x_i}{1+x_i}$ inside the parenthesis gives
$$x_i = a(1 + sum_{j}frac{x_j}{1+x_j} - frac{x_i}{1+x_i})$$
The sum does not depend on $i$; it is a constant. Can you continue?
1
Sum is not a constant and depends on all $x_i$s.
– Anais
Dec 1 at 5:34
@Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
– user58697
Dec 1 at 5:39
My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
– Anais
Dec 1 at 5:47
add a comment |
Adding and subtracting $dfrac{x_i}{1+x_i}$ inside the parenthesis gives
$$x_i = a(1 + sum_{j}frac{x_j}{1+x_j} - frac{x_i}{1+x_i})$$
The sum does not depend on $i$; it is a constant. Can you continue?
1
Sum is not a constant and depends on all $x_i$s.
– Anais
Dec 1 at 5:34
@Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
– user58697
Dec 1 at 5:39
My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
– Anais
Dec 1 at 5:47
add a comment |
Adding and subtracting $dfrac{x_i}{1+x_i}$ inside the parenthesis gives
$$x_i = a(1 + sum_{j}frac{x_j}{1+x_j} - frac{x_i}{1+x_i})$$
The sum does not depend on $i$; it is a constant. Can you continue?
Adding and subtracting $dfrac{x_i}{1+x_i}$ inside the parenthesis gives
$$x_i = a(1 + sum_{j}frac{x_j}{1+x_j} - frac{x_i}{1+x_i})$$
The sum does not depend on $i$; it is a constant. Can you continue?
answered Dec 1 at 5:22
user58697
1,829512
1,829512
1
Sum is not a constant and depends on all $x_i$s.
– Anais
Dec 1 at 5:34
@Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
– user58697
Dec 1 at 5:39
My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
– Anais
Dec 1 at 5:47
add a comment |
1
Sum is not a constant and depends on all $x_i$s.
– Anais
Dec 1 at 5:34
@Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
– user58697
Dec 1 at 5:39
My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
– Anais
Dec 1 at 5:47
1
1
Sum is not a constant and depends on all $x_i$s.
– Anais
Dec 1 at 5:34
Sum is not a constant and depends on all $x_i$s.
– Anais
Dec 1 at 5:34
@Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
– user58697
Dec 1 at 5:39
@Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
– user58697
Dec 1 at 5:39
My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
– Anais
Dec 1 at 5:47
My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
– Anais
Dec 1 at 5:47
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%2f3020984%2fsolution-to-x-i-a1-sum-i-ne-j-fracb-ix-j1b-ix-j%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