What are “complementary pair-wise comparable functions”?
up vote
1
down vote
favorite
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
add a comment |
up vote
1
down vote
favorite
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
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
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
up vote
0
down vote
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 at 12:53
Tyrone
4,06011125
answered Apr 15 at 12:35
Swastik
1
add a comment |
up vote
-2
down vote
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
Navneet
1
and what does it mean when you put "complementary" in there?
– GEdgar
Mar 26 '16 at 18:55
This interpretation doesn't seem to make much sense in the context of the question.
– Eric Wofsey
Mar 26 '16 at 19:27
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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 at 12:53
Tyrone
4,06011125
answered Apr 15 at 12:35
Swastik
1
add a comment |
up vote
0
down vote
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 at 12:53
Tyrone
4,06011125
answered Apr 15 at 12:35
Swastik
1
add a comment |
up vote
0
down vote
up vote
0
down vote
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 at 12:53
Tyrone
4,06011125
answered Apr 15 at 12:35
Swastik
1
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 at 12:53
Tyrone
4,06011125
answered Apr 15 at 12:35
Swastik
1
edited Apr 15 at 12:53
Tyrone
4,06011125
edited Apr 15 at 12:53
Tyrone
4,06011125
edited Apr 15 at 12:53
Tyrone
4,06011125
4,06011125
answered Apr 15 at 12:35
Swastik
1
answered Apr 15 at 12:35
Swastik
1
answered Apr 15 at 12:35
Swastik
1
1
add a comment |
add a comment |
up vote
-2
down vote
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
Navneet
1
and what does it mean when you put "complementary" in there?
– GEdgar
Mar 26 '16 at 18:55
This interpretation doesn't seem to make much sense in the context of the question.
– Eric Wofsey
Mar 26 '16 at 19:27
add a comment |
up vote
-2
down vote
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
Navneet
1
and what does it mean when you put "complementary" in there?
– GEdgar
Mar 26 '16 at 18:55
This interpretation doesn't seem to make much sense in the context of the question.
– Eric Wofsey
Mar 26 '16 at 19:27
add a comment |
up vote
-2
down vote
up vote
-2
down vote
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
Navneet
1
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
Navneet
1
edited Mar 26 '16 at 18:56
user249332
answered Mar 26 '16 at 18:52
Navneet
1
answered Mar 26 '16 at 18:52
Navneet
1
answered Mar 26 '16 at 18:52
Navneet
1
1
and what does it mean when you put "complementary" in there?
– GEdgar
Mar 26 '16 at 18:55
This interpretation doesn't seem to make much sense in the context of the question.
– Eric Wofsey
Mar 26 '16 at 19:27
add a comment |
and what does it mean when you put "complementary" in there?
– GEdgar
Mar 26 '16 at 18:55
This interpretation doesn't seem to make much sense in the context of the question.
– Eric Wofsey
Mar 26 '16 at 19:27
and what does it mean when you put "complementary" in there?
– GEdgar
Mar 26 '16 at 18:55
and what does it mean when you put "complementary" in there?
– GEdgar
Mar 26 '16 at 18:55
This interpretation doesn't seem to make much sense in the context of the question.
– Eric Wofsey
Mar 26 '16 at 19:27
This interpretation doesn't seem to make much sense in the context of the question.
– Eric Wofsey
Mar 26 '16 at 19:27
add a comment |
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
