Mixing
Does the function $fcirc g(x) = (fcirc g)(x)$ have same properties?
$begingroup$
Is $fcirc g(x) = (fcirc g)(x)$? Do they have similar properties on how they work as a functions.
The function $(fcirc g)(x)$ is equal to $f(g(x))$. But I am unaware on how $fcirc g(x)$ works.
My guess is that $fcirc g(x)$ is the same as $(fcirc g)(x)$ since you compose the $x$ in $(fcirc g)(x)$ together to achieve $f circ g(x)$. But a problem that i noticed that if we compose $(fcirc g)(x)$ it together as I stated to get $fcirc g(x)$, it would also imply that you can achieve $f(x)circ g$
Due to that problem, I am unaware what is the appropriate answer. Furthermore, I am unable to find anything online.
functions proof-verification
edited Jan 26 at 9:12
Fabio Lucchini
9,34111426
asked Jan 26 at 3:02
naru sinnaru sin
203112
$endgroup$
|
show 1 more comment
$begingroup$
Is $fcirc g(x) = (fcirc g)(x)$? Do they have similar properties on how they work as a functions.
The function $(fcirc g)(x)$ is equal to $f(g(x))$. But I am unaware on how $fcirc g(x)$ works.
My guess is that $fcirc g(x)$ is the same as $(fcirc g)(x)$ since you compose the $x$ in $(fcirc g)(x)$ together to achieve $f circ g(x)$. But a problem that i noticed that if we compose $(fcirc g)(x)$ it together as I stated to get $fcirc g(x)$, it would also imply that you can achieve $f(x)circ g$
Due to that problem, I am unaware what is the appropriate answer. Furthermore, I am unable to find anything online.
functions proof-verification
edited Jan 26 at 9:12
Fabio Lucchini
9,34111426
asked Jan 26 at 3:02
naru sinnaru sin
203112
$endgroup$
$begingroup$
Well... what do you think?
$endgroup$
– LordVader007
Jan 26 at 3:07
$begingroup$
@LordVader007 I updated it
$endgroup$
– naru sin
Jan 26 at 3:08
2
$begingroup$
I believe it is notation for the same thing. For example, you have $f(x) = sin(x)$ and $g(x) = log(x)$, so $f(g(x))= sin(log(x))$. Writing $(sin circ log)(x)$ is same as saying $sin circ log(x)$.
$endgroup$
– LordVader007
Jan 26 at 3:13
$begingroup$
The o in fog(x) is the sign of composition and x is not multiplied. Functions are not multiplied by the argument but the whole composition is operated on x.
$endgroup$
– SNEHIL SANYAL
Jan 26 at 3:14
$begingroup$
@SNEHILSANYAL i am aware that is a composition and I will fix that immediately
$endgroup$
– naru sin
Jan 26 at 3:15
|
show 1 more comment
$begingroup$
Is $fcirc g(x) = (fcirc g)(x)$? Do they have similar properties on how they work as a functions.
The function $(fcirc g)(x)$ is equal to $f(g(x))$. But I am unaware on how $fcirc g(x)$ works.
My guess is that $fcirc g(x)$ is the same as $(fcirc g)(x)$ since you compose the $x$ in $(fcirc g)(x)$ together to achieve $f circ g(x)$. But a problem that i noticed that if we compose $(fcirc g)(x)$ it together as I stated to get $fcirc g(x)$, it would also imply that you can achieve $f(x)circ g$
Due to that problem, I am unaware what is the appropriate answer. Furthermore, I am unable to find anything online.
functions proof-verification
edited Jan 26 at 9:12
Fabio Lucchini
9,34111426
asked Jan 26 at 3:02
naru sinnaru sin
203112
$endgroup$
Is $fcirc g(x) = (fcirc g)(x)$? Do they have similar properties on how they work as a functions.
The function $(fcirc g)(x)$ is equal to $f(g(x))$. But I am unaware on how $fcirc g(x)$ works.
My guess is that $fcirc g(x)$ is the same as $(fcirc g)(x)$ since you compose the $x$ in $(fcirc g)(x)$ together to achieve $f circ g(x)$. But a problem that i noticed that if we compose $(fcirc g)(x)$ it together as I stated to get $fcirc g(x)$, it would also imply that you can achieve $f(x)circ g$
Due to that problem, I am unaware what is the appropriate answer. Furthermore, I am unable to find anything online.
functions proof-verification
functions proof-verification
edited Jan 26 at 9:12
Fabio Lucchini
9,34111426
asked Jan 26 at 3:02
naru sinnaru sin
203112
edited Jan 26 at 9:12
Fabio Lucchini
9,34111426
asked Jan 26 at 3:02
naru sinnaru sin
203112
edited Jan 26 at 9:12
Fabio Lucchini
9,34111426
edited Jan 26 at 9:12
Fabio Lucchini
9,34111426
edited Jan 26 at 9:12
Fabio Lucchini
9,34111426
9,34111426
asked Jan 26 at 3:02
naru sinnaru sin
203112
asked Jan 26 at 3:02
naru sinnaru sin
203112
asked Jan 26 at 3:02
naru sinnaru sin
203112
203112
$begingroup$
Well... what do you think?
$endgroup$
– LordVader007
Jan 26 at 3:07
$begingroup$
@LordVader007 I updated it
$endgroup$
– naru sin
Jan 26 at 3:08
2
$begingroup$
I believe it is notation for the same thing. For example, you have $f(x) = sin(x)$ and $g(x) = log(x)$, so $f(g(x))= sin(log(x))$. Writing $(sin circ log)(x)$ is same as saying $sin circ log(x)$.
$endgroup$
– LordVader007
Jan 26 at 3:13
$begingroup$
The o in fog(x) is the sign of composition and x is not multiplied. Functions are not multiplied by the argument but the whole composition is operated on x.
$endgroup$
– SNEHIL SANYAL
Jan 26 at 3:14
$begingroup$
@SNEHILSANYAL i am aware that is a composition and I will fix that immediately
$endgroup$
– naru sin
Jan 26 at 3:15
|
show 1 more comment
$begingroup$
Well... what do you think?
$endgroup$
– LordVader007
Jan 26 at 3:07
$begingroup$
@LordVader007 I updated it
$endgroup$
– naru sin
Jan 26 at 3:08
2
$begingroup$
I believe it is notation for the same thing. For example, you have $f(x) = sin(x)$ and $g(x) = log(x)$, so $f(g(x))= sin(log(x))$. Writing $(sin circ log)(x)$ is same as saying $sin circ log(x)$.
$endgroup$
– LordVader007
Jan 26 at 3:13
$begingroup$
The o in fog(x) is the sign of composition and x is not multiplied. Functions are not multiplied by the argument but the whole composition is operated on x.
$endgroup$
– SNEHIL SANYAL
Jan 26 at 3:14
$begingroup$
@SNEHILSANYAL i am aware that is a composition and I will fix that immediately
$endgroup$
– naru sin
Jan 26 at 3:15
$begingroup$
Well... what do you think?
$endgroup$
– LordVader007
Jan 26 at 3:07
$begingroup$
Well... what do you think?
$endgroup$
– LordVader007
Jan 26 at 3:07
$begingroup$
@LordVader007 I updated it
$endgroup$
– naru sin
Jan 26 at 3:08
$begingroup$
@LordVader007 I updated it
$endgroup$
– naru sin
Jan 26 at 3:08
2
2
$begingroup$
I believe it is notation for the same thing. For example, you have $f(x) = sin(x)$ and $g(x) = log(x)$, so $f(g(x))= sin(log(x))$. Writing $(sin circ log)(x)$ is same as saying $sin circ log(x)$.
$endgroup$
– LordVader007
Jan 26 at 3:13
$begingroup$
I believe it is notation for the same thing. For example, you have $f(x) = sin(x)$ and $g(x) = log(x)$, so $f(g(x))= sin(log(x))$. Writing $(sin circ log)(x)$ is same as saying $sin circ log(x)$.
$endgroup$
– LordVader007
Jan 26 at 3:13
$begingroup$
The o in fog(x) is the sign of composition and x is not multiplied. Functions are not multiplied by the argument but the whole composition is operated on x.
$endgroup$
– SNEHIL SANYAL
Jan 26 at 3:14
$begingroup$
The o in fog(x) is the sign of composition and x is not multiplied. Functions are not multiplied by the argument but the whole composition is operated on x.
$endgroup$
– SNEHIL SANYAL
Jan 26 at 3:14
$begingroup$
@SNEHILSANYAL i am aware that is a composition and I will fix that immediately
$endgroup$
– naru sin
Jan 26 at 3:15
$begingroup$
@SNEHILSANYAL i am aware that is a composition and I will fix that immediately
$endgroup$
– naru sin
Jan 26 at 3:15
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
First note that $fcirc g (x) $ means $fcirc (g (x)) $ and this is, generally, different from $(fcirc g)(x) $.
Indeed, $fcirc g (x) $ is a function when $g (x) $ is a function.
More precisely, let $g:Xto Z^Y $ and $f:Zto W $.
Then for all $xin X$ we have $g (x):Yto Z $ hence $fcirc g (x):Yto W $ and
$$(fcirc g (x))(y)=f (g (x)(y)) $$
answered Jan 26 at 9:19
Fabio LucchiniFabio Lucchini
9,34111426
$endgroup$
add a comment |
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1 Answer
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$begingroup$
First note that $fcirc g (x) $ means $fcirc (g (x)) $ and this is, generally, different from $(fcirc g)(x) $.
Indeed, $fcirc g (x) $ is a function when $g (x) $ is a function.
More precisely, let $g:Xto Z^Y $ and $f:Zto W $.
Then for all $xin X$ we have $g (x):Yto Z $ hence $fcirc g (x):Yto W $ and
$$(fcirc g (x))(y)=f (g (x)(y)) $$
answered Jan 26 at 9:19
Fabio LucchiniFabio Lucchini
9,34111426
$endgroup$
add a comment |
$begingroup$
First note that $fcirc g (x) $ means $fcirc (g (x)) $ and this is, generally, different from $(fcirc g)(x) $.
Indeed, $fcirc g (x) $ is a function when $g (x) $ is a function.
More precisely, let $g:Xto Z^Y $ and $f:Zto W $.
Then for all $xin X$ we have $g (x):Yto Z $ hence $fcirc g (x):Yto W $ and
$$(fcirc g (x))(y)=f (g (x)(y)) $$
answered Jan 26 at 9:19
Fabio LucchiniFabio Lucchini
9,34111426
$endgroup$
add a comment |
$begingroup$
First note that $fcirc g (x) $ means $fcirc (g (x)) $ and this is, generally, different from $(fcirc g)(x) $.
Indeed, $fcirc g (x) $ is a function when $g (x) $ is a function.
More precisely, let $g:Xto Z^Y $ and $f:Zto W $.
Then for all $xin X$ we have $g (x):Yto Z $ hence $fcirc g (x):Yto W $ and
$$(fcirc g (x))(y)=f (g (x)(y)) $$
answered Jan 26 at 9:19
Fabio LucchiniFabio Lucchini
9,34111426
$endgroup$
First note that $fcirc g (x) $ means $fcirc (g (x)) $ and this is, generally, different from $(fcirc g)(x) $.
Indeed, $fcirc g (x) $ is a function when $g (x) $ is a function.
More precisely, let $g:Xto Z^Y $ and $f:Zto W $.
Then for all $xin X$ we have $g (x):Yto Z $ hence $fcirc g (x):Yto W $ and
$$(fcirc g (x))(y)=f (g (x)(y)) $$
answered Jan 26 at 9:19
Fabio LucchiniFabio Lucchini
9,34111426
answered Jan 26 at 9:19
Fabio LucchiniFabio Lucchini
9,34111426
answered Jan 26 at 9:19
Fabio LucchiniFabio Lucchini
9,34111426
answered Jan 26 at 9:19
Fabio LucchiniFabio Lucchini
9,34111426
9,34111426
add a comment |
add a comment |
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$begingroup$
Well... what do you think?
$endgroup$
– LordVader007
Jan 26 at 3:07
$begingroup$
@LordVader007 I updated it
$endgroup$
– naru sin
Jan 26 at 3:08
2
$begingroup$
I believe it is notation for the same thing. For example, you have $f(x) = sin(x)$ and $g(x) = log(x)$, so $f(g(x))= sin(log(x))$. Writing $(sin circ log)(x)$ is same as saying $sin circ log(x)$.
$endgroup$
– LordVader007
Jan 26 at 3:13
$begingroup$
The o in fog(x) is the sign of composition and x is not multiplied. Functions are not multiplied by the argument but the whole composition is operated on x.
$endgroup$
– SNEHIL SANYAL
Jan 26 at 3:14
$begingroup$
@SNEHILSANYAL i am aware that is a composition and I will fix that immediately
$endgroup$
– naru sin
Jan 26 at 3:15