Closed-form representation of area below a function












1












$begingroup$


Let $f:[0,1]rightarrow[0,1]$. $f$ is strictly increasing in $x$ if $xin[0,x^*]$ and strictly decreases otherwise.



Suppose that I'm interested in finding the area of the domain where $f(x)leq t$. That is, if
$$t=f(x)=f(x+y)$$ for some $t$, $xleq x^*$ and $x+ygeq x^*$, I want to find $y$.



Is there any closed form representation of $y$ in terms of $f$?



Further, is there any way that I can find $frac{dy}{dt}$? Any suggestions?






area







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $f:[0,1]rightarrow[0,1]$. $f$ is strictly increasing in $x$ if $xin[0,x^*]$ and strictly decreases otherwise.



    Suppose that I'm interested in finding the area of the domain where $f(x)leq t$. That is, if
    $$t=f(x)=f(x+y)$$ for some $t$, $xleq x^*$ and $x+ygeq x^*$, I want to find $y$.



    Is there any closed form representation of $y$ in terms of $f$?



    Further, is there any way that I can find $frac{dy}{dt}$? Any suggestions?






    area







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $f:[0,1]rightarrow[0,1]$. $f$ is strictly increasing in $x$ if $xin[0,x^*]$ and strictly decreases otherwise.



      Suppose that I'm interested in finding the area of the domain where $f(x)leq t$. That is, if
      $$t=f(x)=f(x+y)$$ for some $t$, $xleq x^*$ and $x+ygeq x^*$, I want to find $y$.



      Is there any closed form representation of $y$ in terms of $f$?



      Further, is there any way that I can find $frac{dy}{dt}$? Any suggestions?






      area







      share|cite|improve this question









      $endgroup$




      Let $f:[0,1]rightarrow[0,1]$. $f$ is strictly increasing in $x$ if $xin[0,x^*]$ and strictly decreases otherwise.



      Suppose that I'm interested in finding the area of the domain where $f(x)leq t$. That is, if
      $$t=f(x)=f(x+y)$$ for some $t$, $xleq x^*$ and $x+ygeq x^*$, I want to find $y$.



      Is there any closed form representation of $y$ in terms of $f$?



      Further, is there any way that I can find $frac{dy}{dt}$? Any suggestions?






      area




      area






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 11 '18 at 15:54









      GreenteamaniacGreenteamaniac

      345




      345






















          1 Answer
          1






          active

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          1












          $begingroup$

          Let's write $$f(x)=begin{cases}f_1(x),xin[0,x^*]\f_2(x),xin(x^*,1]end{cases}$$
          We know that both these branches are invertible (one is strictly increasing, one of them is strictly decreasing), so we can calculate $f_{1,2}^{-1}(y)$. If you sketch some graph of $f$, draw the line parallel to $x$ axis through $t$. You notice that it intersects $f_1$ at $(f_1^{-1}(t),t)$ and $f_2$ at $(f_2^{-1}(t),t)$. From these $$y=f_2^{-1}(t)-f_1^{-1}(t)$$
          You can now take the derivative to get $frac{dy}{dt}$.



          In the same sketch of $f$, you can write the area under the curve in terms of integral along $y$ axis as well: $$A=int_0^t(f_2^{-1}(y)-f_1^{-1}(y))dy-yt$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks @Andrei. Can we define $frac{dy}{dt}$ at $t^*=f(x^*)$? If not, can we define a limit that is approached from above, say $lim_{trightarrow t^*}y'(t)$?
            $endgroup$
            – Greenteamaniac
            Dec 12 '18 at 1:55










          • $begingroup$
            Note that $t^*$ is a maximum, so you can only approach it from below. You can define the limit, but there is no guarantee that it exists
            $endgroup$
            – Andrei
            Dec 12 '18 at 2:08










          • $begingroup$
            Alright, thanks!
            $endgroup$
            – Greenteamaniac
            Dec 12 '18 at 2:48











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Let's write $$f(x)=begin{cases}f_1(x),xin[0,x^*]\f_2(x),xin(x^*,1]end{cases}$$
          We know that both these branches are invertible (one is strictly increasing, one of them is strictly decreasing), so we can calculate $f_{1,2}^{-1}(y)$. If you sketch some graph of $f$, draw the line parallel to $x$ axis through $t$. You notice that it intersects $f_1$ at $(f_1^{-1}(t),t)$ and $f_2$ at $(f_2^{-1}(t),t)$. From these $$y=f_2^{-1}(t)-f_1^{-1}(t)$$
          You can now take the derivative to get $frac{dy}{dt}$.



          In the same sketch of $f$, you can write the area under the curve in terms of integral along $y$ axis as well: $$A=int_0^t(f_2^{-1}(y)-f_1^{-1}(y))dy-yt$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks @Andrei. Can we define $frac{dy}{dt}$ at $t^*=f(x^*)$? If not, can we define a limit that is approached from above, say $lim_{trightarrow t^*}y'(t)$?
            $endgroup$
            – Greenteamaniac
            Dec 12 '18 at 1:55










          • $begingroup$
            Note that $t^*$ is a maximum, so you can only approach it from below. You can define the limit, but there is no guarantee that it exists
            $endgroup$
            – Andrei
            Dec 12 '18 at 2:08










          • $begingroup$
            Alright, thanks!
            $endgroup$
            – Greenteamaniac
            Dec 12 '18 at 2:48
















          1












          $begingroup$

          Let's write $$f(x)=begin{cases}f_1(x),xin[0,x^*]\f_2(x),xin(x^*,1]end{cases}$$
          We know that both these branches are invertible (one is strictly increasing, one of them is strictly decreasing), so we can calculate $f_{1,2}^{-1}(y)$. If you sketch some graph of $f$, draw the line parallel to $x$ axis through $t$. You notice that it intersects $f_1$ at $(f_1^{-1}(t),t)$ and $f_2$ at $(f_2^{-1}(t),t)$. From these $$y=f_2^{-1}(t)-f_1^{-1}(t)$$
          You can now take the derivative to get $frac{dy}{dt}$.



          In the same sketch of $f$, you can write the area under the curve in terms of integral along $y$ axis as well: $$A=int_0^t(f_2^{-1}(y)-f_1^{-1}(y))dy-yt$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks @Andrei. Can we define $frac{dy}{dt}$ at $t^*=f(x^*)$? If not, can we define a limit that is approached from above, say $lim_{trightarrow t^*}y'(t)$?
            $endgroup$
            – Greenteamaniac
            Dec 12 '18 at 1:55










          • $begingroup$
            Note that $t^*$ is a maximum, so you can only approach it from below. You can define the limit, but there is no guarantee that it exists
            $endgroup$
            – Andrei
            Dec 12 '18 at 2:08










          • $begingroup$
            Alright, thanks!
            $endgroup$
            – Greenteamaniac
            Dec 12 '18 at 2:48














          1












          1








          1





          $begingroup$

          Let's write $$f(x)=begin{cases}f_1(x),xin[0,x^*]\f_2(x),xin(x^*,1]end{cases}$$
          We know that both these branches are invertible (one is strictly increasing, one of them is strictly decreasing), so we can calculate $f_{1,2}^{-1}(y)$. If you sketch some graph of $f$, draw the line parallel to $x$ axis through $t$. You notice that it intersects $f_1$ at $(f_1^{-1}(t),t)$ and $f_2$ at $(f_2^{-1}(t),t)$. From these $$y=f_2^{-1}(t)-f_1^{-1}(t)$$
          You can now take the derivative to get $frac{dy}{dt}$.



          In the same sketch of $f$, you can write the area under the curve in terms of integral along $y$ axis as well: $$A=int_0^t(f_2^{-1}(y)-f_1^{-1}(y))dy-yt$$






          share|cite|improve this answer









          $endgroup$



          Let's write $$f(x)=begin{cases}f_1(x),xin[0,x^*]\f_2(x),xin(x^*,1]end{cases}$$
          We know that both these branches are invertible (one is strictly increasing, one of them is strictly decreasing), so we can calculate $f_{1,2}^{-1}(y)$. If you sketch some graph of $f$, draw the line parallel to $x$ axis through $t$. You notice that it intersects $f_1$ at $(f_1^{-1}(t),t)$ and $f_2$ at $(f_2^{-1}(t),t)$. From these $$y=f_2^{-1}(t)-f_1^{-1}(t)$$
          You can now take the derivative to get $frac{dy}{dt}$.



          In the same sketch of $f$, you can write the area under the curve in terms of integral along $y$ axis as well: $$A=int_0^t(f_2^{-1}(y)-f_1^{-1}(y))dy-yt$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 18:39









          AndreiAndrei

          11.8k21126




          11.8k21126












          • $begingroup$
            Thanks @Andrei. Can we define $frac{dy}{dt}$ at $t^*=f(x^*)$? If not, can we define a limit that is approached from above, say $lim_{trightarrow t^*}y'(t)$?
            $endgroup$
            – Greenteamaniac
            Dec 12 '18 at 1:55










          • $begingroup$
            Note that $t^*$ is a maximum, so you can only approach it from below. You can define the limit, but there is no guarantee that it exists
            $endgroup$
            – Andrei
            Dec 12 '18 at 2:08










          • $begingroup$
            Alright, thanks!
            $endgroup$
            – Greenteamaniac
            Dec 12 '18 at 2:48


















          • $begingroup$
            Thanks @Andrei. Can we define $frac{dy}{dt}$ at $t^*=f(x^*)$? If not, can we define a limit that is approached from above, say $lim_{trightarrow t^*}y'(t)$?
            $endgroup$
            – Greenteamaniac
            Dec 12 '18 at 1:55










          • $begingroup$
            Note that $t^*$ is a maximum, so you can only approach it from below. You can define the limit, but there is no guarantee that it exists
            $endgroup$
            – Andrei
            Dec 12 '18 at 2:08










          • $begingroup$
            Alright, thanks!
            $endgroup$
            – Greenteamaniac
            Dec 12 '18 at 2:48
















          $begingroup$
          Thanks @Andrei. Can we define $frac{dy}{dt}$ at $t^*=f(x^*)$? If not, can we define a limit that is approached from above, say $lim_{trightarrow t^*}y'(t)$?
          $endgroup$
          – Greenteamaniac
          Dec 12 '18 at 1:55




          $begingroup$
          Thanks @Andrei. Can we define $frac{dy}{dt}$ at $t^*=f(x^*)$? If not, can we define a limit that is approached from above, say $lim_{trightarrow t^*}y'(t)$?
          $endgroup$
          – Greenteamaniac
          Dec 12 '18 at 1:55












          $begingroup$
          Note that $t^*$ is a maximum, so you can only approach it from below. You can define the limit, but there is no guarantee that it exists
          $endgroup$
          – Andrei
          Dec 12 '18 at 2:08




          $begingroup$
          Note that $t^*$ is a maximum, so you can only approach it from below. You can define the limit, but there is no guarantee that it exists
          $endgroup$
          – Andrei
          Dec 12 '18 at 2:08












          $begingroup$
          Alright, thanks!
          $endgroup$
          – Greenteamaniac
          Dec 12 '18 at 2:48




          $begingroup$
          Alright, thanks!
          $endgroup$
          – Greenteamaniac
          Dec 12 '18 at 2:48


















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