What are “complementary pair-wise comparable functions”?
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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
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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
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2 Answers
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$begingroup$
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
4,65511225
answered Apr 15 '18 at 12:35
SwastikSwastik
1
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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 |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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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
4,65511225
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
4,65511225
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
4,65511225
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
4,65511225
answered Apr 15 '18 at 12:35
SwastikSwastik
1
edited Apr 15 '18 at 12:53
Tyrone
4,65511225
edited Apr 15 '18 at 12:53
Tyrone
4,65511225
edited Apr 15 '18 at 12:53
Tyrone
4,65511225
4,65511225
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 |
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