Does the function $fcirc g(x) = (fcirc g)(x)$ have same properties?












2












$begingroup$


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.










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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
















2












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 at 9:12









Fabio Lucchini

9,34111426




9,34111426










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


















  • $begingroup$
    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










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
$$(fcirc g (x))(y)=f (g (x)(y)) $$






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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
    $$(fcirc g (x))(y)=f (g (x)(y)) $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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
      $$(fcirc g (x))(y)=f (g (x)(y)) $$






      share|cite|improve this 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
        $$(fcirc g (x))(y)=f (g (x)(y)) $$






        share|cite|improve this answer









        $endgroup$



        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
        $$(fcirc g (x))(y)=f (g (x)(y)) $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 26 at 9:19









        Fabio LucchiniFabio Lucchini

        9,34111426




        9,34111426






























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