What are “complementary pair-wise comparable functions”?
$begingroup$
I got this term while studying periodic functions; my book writes:
If $f_1(x)$ & $f_2(x)$ are periodic functions with periods $T_1$ & $T_2$ respectively, then we have $h(x) = f_1(x) + f_2(x)$ has period, as $dfrac{1}{2} text{L.C.M. of }; {T_1 ,T_2}$, if $f_1(x)$ & $f_2(x)$ are complementary pair-wise comparable functions.
Don't know how the author deduced the rule. But my main question is what do this "complementary, pair-wise, comparable function" mean? I've googled this but it was in vain.
functions
$endgroup$
add a comment |
$begingroup$
I got this term while studying periodic functions; my book writes:
If $f_1(x)$ & $f_2(x)$ are periodic functions with periods $T_1$ & $T_2$ respectively, then we have $h(x) = f_1(x) + f_2(x)$ has period, as $dfrac{1}{2} text{L.C.M. of }; {T_1 ,T_2}$, if $f_1(x)$ & $f_2(x)$ are complementary pair-wise comparable functions.
Don't know how the author deduced the rule. But my main question is what do this "complementary, pair-wise, comparable function" mean? I've googled this but it was in vain.
functions
$endgroup$
add a comment |
$begingroup$
I got this term while studying periodic functions; my book writes:
If $f_1(x)$ & $f_2(x)$ are periodic functions with periods $T_1$ & $T_2$ respectively, then we have $h(x) = f_1(x) + f_2(x)$ has period, as $dfrac{1}{2} text{L.C.M. of }; {T_1 ,T_2}$, if $f_1(x)$ & $f_2(x)$ are complementary pair-wise comparable functions.
Don't know how the author deduced the rule. But my main question is what do this "complementary, pair-wise, comparable function" mean? I've googled this but it was in vain.
functions
$endgroup$
I got this term while studying periodic functions; my book writes:
If $f_1(x)$ & $f_2(x)$ are periodic functions with periods $T_1$ & $T_2$ respectively, then we have $h(x) = f_1(x) + f_2(x)$ has period, as $dfrac{1}{2} text{L.C.M. of }; {T_1 ,T_2}$, if $f_1(x)$ & $f_2(x)$ are complementary pair-wise comparable functions.
Don't know how the author deduced the rule. But my main question is what do this "complementary, pair-wise, comparable function" mean? I've googled this but it was in vain.
functions
functions
asked Sep 6 '15 at 18:07
user142971
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think it means if you put $(π/2-x)$ in place of $x$ in one of the functions and get the second function, then they are called complementary. Like $sin(x)$ and $cos(x)$.
edited Apr 15 '18 at 12:53
Tyrone
4,72011225
answered Apr 15 '18 at 12:35
SwastikSwastik
1
$endgroup$
add a comment |
$begingroup$
Consider a set of functions $F$.
The set is pairwise comparable if, for any two functions $f(x)$ and $g(x)$ within the set $F$, we know that either:
$f(x) ge g(x)$ for any value of $x$, or
$f(x) le g(x)$ for any value of $x$.
answered Mar 26 '16 at 18:52
NavneetNavneet
1
$endgroup$
$begingroup$
and what does it mean when you put "complementary" in there?
$endgroup$
– GEdgar
Mar 26 '16 at 18:55
$begingroup$
This interpretation doesn't seem to make much sense in the context of the question.
$endgroup$
– Eric Wofsey
Mar 26 '16 at 19:27
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%2f1424456%2fwhat-are-complementary-pair-wise-comparable-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think it means if you put $(π/2-x)$ in place of $x$ in one of the functions and get the second function, then they are called complementary. Like $sin(x)$ and $cos(x)$.
edited Apr 15 '18 at 12:53
Tyrone
4,72011225
answered Apr 15 '18 at 12:35
SwastikSwastik
1
$endgroup$
add a comment |
$begingroup$
I think it means if you put $(π/2-x)$ in place of $x$ in one of the functions and get the second function, then they are called complementary. Like $sin(x)$ and $cos(x)$.
edited Apr 15 '18 at 12:53
Tyrone
4,72011225
answered Apr 15 '18 at 12:35
SwastikSwastik
1
$endgroup$
add a comment |
$begingroup$
I think it means if you put $(π/2-x)$ in place of $x$ in one of the functions and get the second function, then they are called complementary. Like $sin(x)$ and $cos(x)$.
edited Apr 15 '18 at 12:53
Tyrone
4,72011225
answered Apr 15 '18 at 12:35
SwastikSwastik
1
$endgroup$
I think it means if you put $(π/2-x)$ in place of $x$ in one of the functions and get the second function, then they are called complementary. Like $sin(x)$ and $cos(x)$.
edited Apr 15 '18 at 12:53
Tyrone
4,72011225
answered Apr 15 '18 at 12:35
SwastikSwastik
1
edited Apr 15 '18 at 12:53
Tyrone
4,72011225
edited Apr 15 '18 at 12:53
Tyrone
4,72011225
edited Apr 15 '18 at 12:53
Tyrone
4,72011225
4,72011225
answered Apr 15 '18 at 12:35
SwastikSwastik
1
answered Apr 15 '18 at 12:35
SwastikSwastik
1
answered Apr 15 '18 at 12:35
SwastikSwastik
1
1
add a comment |
add a comment |
$begingroup$
Consider a set of functions $F$.
The set is pairwise comparable if, for any two functions $f(x)$ and $g(x)$ within the set $F$, we know that either:
$f(x) ge g(x)$ for any value of $x$, or
$f(x) le g(x)$ for any value of $x$.
answered Mar 26 '16 at 18:52
NavneetNavneet
1
$endgroup$
$begingroup$
and what does it mean when you put "complementary" in there?
$endgroup$
– GEdgar
Mar 26 '16 at 18:55
$begingroup$
This interpretation doesn't seem to make much sense in the context of the question.
$endgroup$
– Eric Wofsey
Mar 26 '16 at 19:27
add a comment |
$begingroup$
Consider a set of functions $F$.
The set is pairwise comparable if, for any two functions $f(x)$ and $g(x)$ within the set $F$, we know that either:
$f(x) ge g(x)$ for any value of $x$, or
$f(x) le g(x)$ for any value of $x$.
answered Mar 26 '16 at 18:52
NavneetNavneet
1
$endgroup$
$begingroup$
and what does it mean when you put "complementary" in there?
$endgroup$
– GEdgar
Mar 26 '16 at 18:55
$begingroup$
This interpretation doesn't seem to make much sense in the context of the question.
$endgroup$
– Eric Wofsey
Mar 26 '16 at 19:27
add a comment |
$begingroup$
Consider a set of functions $F$.
The set is pairwise comparable if, for any two functions $f(x)$ and $g(x)$ within the set $F$, we know that either:
$f(x) ge g(x)$ for any value of $x$, or
$f(x) le g(x)$ for any value of $x$.
answered Mar 26 '16 at 18:52
NavneetNavneet
1
$endgroup$
Consider a set of functions $F$.
The set is pairwise comparable if, for any two functions $f(x)$ and $g(x)$ within the set $F$, we know that either:
$f(x) ge g(x)$ for any value of $x$, or
$f(x) le g(x)$ for any value of $x$.
answered Mar 26 '16 at 18:52
NavneetNavneet
1
edited Mar 26 '16 at 18:56
user249332
answered Mar 26 '16 at 18:52
NavneetNavneet
1
answered Mar 26 '16 at 18:52
NavneetNavneet
1
answered Mar 26 '16 at 18:52
NavneetNavneet
1
1
$begingroup$
and what does it mean when you put "complementary" in there?
$endgroup$
– GEdgar
Mar 26 '16 at 18:55
$begingroup$
This interpretation doesn't seem to make much sense in the context of the question.
$endgroup$
– Eric Wofsey
Mar 26 '16 at 19:27
add a comment |
$begingroup$
and what does it mean when you put "complementary" in there?
$endgroup$
– GEdgar
Mar 26 '16 at 18:55
$begingroup$
This interpretation doesn't seem to make much sense in the context of the question.
$endgroup$
– Eric Wofsey
Mar 26 '16 at 19:27
$begingroup$
and what does it mean when you put "complementary" in there?
$endgroup$
– GEdgar
Mar 26 '16 at 18:55
$begingroup$
and what does it mean when you put "complementary" in there?
$endgroup$
– GEdgar
Mar 26 '16 at 18:55
$begingroup$
This interpretation doesn't seem to make much sense in the context of the question.
$endgroup$
– Eric Wofsey
Mar 26 '16 at 19:27
$begingroup$
This interpretation doesn't seem to make much sense in the context of the question.
$endgroup$
– Eric Wofsey
Mar 26 '16 at 19:27
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%2f1424456%2fwhat-are-complementary-pair-wise-comparable-functions%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
