Is a “curve” considered “linear”?
$begingroup$
In linear regression, we are fitting a polynomial to a set of data points. In Bishop's book of Pattern Recognition & Machine Learning, there are a few examples where the fit is a curve or a straight line. I am a bit confused if a curve is linear or not. The term linear means that the fit should be a linear function or a polynomial of degree 1 i.e., a straight line. But in many resources, examples are shown where the fit can be a polynomial of degree 3,9 etc. So, are these higher order polynomials linear?
linear-regression terminology
edited 19 hours ago
Wes
33011
asked Feb 16 at 21:43
Srishti MSrishti M
2326
$endgroup$
add a comment |
$begingroup$
In linear regression, we are fitting a polynomial to a set of data points. In Bishop's book of Pattern Recognition & Machine Learning, there are a few examples where the fit is a curve or a straight line. I am a bit confused if a curve is linear or not. The term linear means that the fit should be a linear function or a polynomial of degree 1 i.e., a straight line. But in many resources, examples are shown where the fit can be a polynomial of degree 3,9 etc. So, are these higher order polynomials linear?
linear-regression terminology
edited 19 hours ago
Wes
33011
asked Feb 16 at 21:43
Srishti MSrishti M
2326
$endgroup$
add a comment |
$begingroup$
In linear regression, we are fitting a polynomial to a set of data points. In Bishop's book of Pattern Recognition & Machine Learning, there are a few examples where the fit is a curve or a straight line. I am a bit confused if a curve is linear or not. The term linear means that the fit should be a linear function or a polynomial of degree 1 i.e., a straight line. But in many resources, examples are shown where the fit can be a polynomial of degree 3,9 etc. So, are these higher order polynomials linear?
linear-regression terminology
edited 19 hours ago
Wes
33011
asked Feb 16 at 21:43
Srishti MSrishti M
2326
$endgroup$
In linear regression, we are fitting a polynomial to a set of data points. In Bishop's book of Pattern Recognition & Machine Learning, there are a few examples where the fit is a curve or a straight line. I am a bit confused if a curve is linear or not. The term linear means that the fit should be a linear function or a polynomial of degree 1 i.e., a straight line. But in many resources, examples are shown where the fit can be a polynomial of degree 3,9 etc. So, are these higher order polynomials linear?
linear-regression terminology
linear-regression terminology
edited 19 hours ago
Wes
33011
asked Feb 16 at 21:43
Srishti MSrishti M
2326
edited 19 hours ago
Wes
33011
asked Feb 16 at 21:43
Srishti MSrishti M
2326
edited 19 hours ago
Wes
33011
edited 19 hours ago
Wes
33011
edited 19 hours ago
Wes
33011
33011
asked Feb 16 at 21:43
Srishti MSrishti M
2326
asked Feb 16 at 21:43
Srishti MSrishti M
2326
asked Feb 16 at 21:43
Srishti MSrishti M
2326
2326
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Lets give an example:
1. y = w*x
This is linear in terms of both weight (w) and data (x).
2. y = w*(x^2) OR y = w*z ; where z = x^2
This is still linear in terms of weight (w) and still treated as a linear regression for the transformed data (z)
answered 2 days ago
Mankind_008Mankind_008
41117
$endgroup$
add a comment |
$begingroup$
If instead of using feature x, you use its square, you get a curve.
It is a linear function of its variables, but you may enter the square or a cube of a variable, therefore making the graph appear as a curve.
In this sense it is still linear while in essence it is a polynomial curve.
answered 2 days ago
useruser
62251027
$endgroup$
add a comment |
$begingroup$
Typically the word "curve" is not used to describe a linear model since at that point it is usually just described as a "line", although mathematically a line is just a specific kind of curve. I think this is similar to what you stated - a linear model is a polynomial of degree 1, but usually in common usage people will not say "polynomial" when referring to a "linear" model.
However, higher-order polynomials are indeed not linear by definition.
answered 2 days ago
WesWes
33011
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Would love to have a comment on why this was downvoted.
$endgroup$
– Wes
8 hours ago
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: "557"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
},
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%2fdatascience.stackexchange.com%2fquestions%2f45703%2fis-a-curve-considered-linear%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Lets give an example:
1. y = w*x
This is linear in terms of both weight (w) and data (x).
2. y = w*(x^2) OR y = w*z ; where z = x^2
This is still linear in terms of weight (w) and still treated as a linear regression for the transformed data (z)
answered 2 days ago
Mankind_008Mankind_008
41117
$endgroup$
add a comment |
$begingroup$
Lets give an example:
1. y = w*x
This is linear in terms of both weight (w) and data (x).
2. y = w*(x^2) OR y = w*z ; where z = x^2
This is still linear in terms of weight (w) and still treated as a linear regression for the transformed data (z)
answered 2 days ago
Mankind_008Mankind_008
41117
$endgroup$
add a comment |
$begingroup$
Lets give an example:
1. y = w*x
This is linear in terms of both weight (w) and data (x).
2. y = w*(x^2) OR y = w*z ; where z = x^2
This is still linear in terms of weight (w) and still treated as a linear regression for the transformed data (z)
answered 2 days ago
Mankind_008Mankind_008
41117
$endgroup$
Lets give an example:
1. y = w*x
This is linear in terms of both weight (w) and data (x).
2. y = w*(x^2) OR y = w*z ; where z = x^2
This is still linear in terms of weight (w) and still treated as a linear regression for the transformed data (z)
answered 2 days ago
Mankind_008Mankind_008
41117
edited 2 days ago
answered 2 days ago
Mankind_008Mankind_008
41117
answered 2 days ago
Mankind_008Mankind_008
41117
answered 2 days ago
Mankind_008Mankind_008
41117
41117
add a comment |
add a comment |
$begingroup$
If instead of using feature x, you use its square, you get a curve.
It is a linear function of its variables, but you may enter the square or a cube of a variable, therefore making the graph appear as a curve.
In this sense it is still linear while in essence it is a polynomial curve.
answered 2 days ago
useruser
62251027
$endgroup$
add a comment |
$begingroup$
If instead of using feature x, you use its square, you get a curve.
It is a linear function of its variables, but you may enter the square or a cube of a variable, therefore making the graph appear as a curve.
In this sense it is still linear while in essence it is a polynomial curve.
answered 2 days ago
useruser
62251027
$endgroup$
add a comment |
$begingroup$
If instead of using feature x, you use its square, you get a curve.
It is a linear function of its variables, but you may enter the square or a cube of a variable, therefore making the graph appear as a curve.
In this sense it is still linear while in essence it is a polynomial curve.
answered 2 days ago
useruser
62251027
$endgroup$
If instead of using feature x, you use its square, you get a curve.
It is a linear function of its variables, but you may enter the square or a cube of a variable, therefore making the graph appear as a curve.
In this sense it is still linear while in essence it is a polynomial curve.
answered 2 days ago
useruser
62251027
answered 2 days ago
useruser
62251027
answered 2 days ago
useruser
62251027
answered 2 days ago
useruser
62251027
62251027
add a comment |
add a comment |
$begingroup$
Typically the word "curve" is not used to describe a linear model since at that point it is usually just described as a "line", although mathematically a line is just a specific kind of curve. I think this is similar to what you stated - a linear model is a polynomial of degree 1, but usually in common usage people will not say "polynomial" when referring to a "linear" model.
However, higher-order polynomials are indeed not linear by definition.
answered 2 days ago
WesWes
33011
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Would love to have a comment on why this was downvoted.
$endgroup$
– Wes
8 hours ago
add a comment |
$begingroup$
Typically the word "curve" is not used to describe a linear model since at that point it is usually just described as a "line", although mathematically a line is just a specific kind of curve. I think this is similar to what you stated - a linear model is a polynomial of degree 1, but usually in common usage people will not say "polynomial" when referring to a "linear" model.
However, higher-order polynomials are indeed not linear by definition.
answered 2 days ago
WesWes
33011
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Would love to have a comment on why this was downvoted.
$endgroup$
– Wes
8 hours ago
add a comment |
$begingroup$
Typically the word "curve" is not used to describe a linear model since at that point it is usually just described as a "line", although mathematically a line is just a specific kind of curve. I think this is similar to what you stated - a linear model is a polynomial of degree 1, but usually in common usage people will not say "polynomial" when referring to a "linear" model.
However, higher-order polynomials are indeed not linear by definition.
answered 2 days ago
WesWes
33011
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Typically the word "curve" is not used to describe a linear model since at that point it is usually just described as a "line", although mathematically a line is just a specific kind of curve. I think this is similar to what you stated - a linear model is a polynomial of degree 1, but usually in common usage people will not say "polynomial" when referring to a "linear" model.
However, higher-order polynomials are indeed not linear by definition.
answered 2 days ago
WesWes
33011
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
WesWes
33011
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
WesWes
33011
answered 2 days ago
WesWes
33011
33011
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Would love to have a comment on why this was downvoted.
$endgroup$
– Wes
8 hours ago
add a comment |
$begingroup$
Would love to have a comment on why this was downvoted.
$endgroup$
– Wes
8 hours ago
$begingroup$
Would love to have a comment on why this was downvoted.
$endgroup$
– Wes
8 hours ago
$begingroup$
Would love to have a comment on why this was downvoted.
$endgroup$
– Wes
8 hours ago
add a comment |
Thanks for contributing an answer to Data Science 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%2fdatascience.stackexchange.com%2fquestions%2f45703%2fis-a-curve-considered-linear%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
