Percentage change of a quotient












1














This was line in my Econ notes that was "by definition". If anyone can clarify this I would really appreciate.



We have a relationship:



$APL = frac{Y}{L}$



Now, define $Delta x$ as "percentage change in variable x". Then by definition:



$Delta APL = Delta Y - Delta L$



I am not sure how this is claimed "by definition". When I think about it, if I define $alpha$ and $beta$ as % changes, I get:



$Delta APL = frac{frac{Y times (1+alpha)}{Ltimes (1+beta)} -frac{Y}{L}}{frac{Y}{L}}$



$= frac{1+alpha}{1+beta} -1$



Which is not equal to $alpha - beta$ as claimed by my notes.



Where am I going wrong here? Is there some limit stuff I am forgetting?



(If anyone is curious the particular relationship is about the average product of labor. But this is not important. The claim being made is general)






calculus limits functions economics percentages







share|cite|improve this question





























    1














    This was line in my Econ notes that was "by definition". If anyone can clarify this I would really appreciate.



    We have a relationship:



    $APL = frac{Y}{L}$



    Now, define $Delta x$ as "percentage change in variable x". Then by definition:



    $Delta APL = Delta Y - Delta L$



    I am not sure how this is claimed "by definition". When I think about it, if I define $alpha$ and $beta$ as % changes, I get:



    $Delta APL = frac{frac{Y times (1+alpha)}{Ltimes (1+beta)} -frac{Y}{L}}{frac{Y}{L}}$



    $= frac{1+alpha}{1+beta} -1$



    Which is not equal to $alpha - beta$ as claimed by my notes.



    Where am I going wrong here? Is there some limit stuff I am forgetting?



    (If anyone is curious the particular relationship is about the average product of labor. But this is not important. The claim being made is general)






    calculus limits functions economics percentages







    share|cite|improve this question



























      1












      1








      1







      This was line in my Econ notes that was "by definition". If anyone can clarify this I would really appreciate.



      We have a relationship:



      $APL = frac{Y}{L}$



      Now, define $Delta x$ as "percentage change in variable x". Then by definition:



      $Delta APL = Delta Y - Delta L$



      I am not sure how this is claimed "by definition". When I think about it, if I define $alpha$ and $beta$ as % changes, I get:



      $Delta APL = frac{frac{Y times (1+alpha)}{Ltimes (1+beta)} -frac{Y}{L}}{frac{Y}{L}}$



      $= frac{1+alpha}{1+beta} -1$



      Which is not equal to $alpha - beta$ as claimed by my notes.



      Where am I going wrong here? Is there some limit stuff I am forgetting?



      (If anyone is curious the particular relationship is about the average product of labor. But this is not important. The claim being made is general)






      calculus limits functions economics percentages







      share|cite|improve this question















      This was line in my Econ notes that was "by definition". If anyone can clarify this I would really appreciate.



      We have a relationship:



      $APL = frac{Y}{L}$



      Now, define $Delta x$ as "percentage change in variable x". Then by definition:



      $Delta APL = Delta Y - Delta L$



      I am not sure how this is claimed "by definition". When I think about it, if I define $alpha$ and $beta$ as % changes, I get:



      $Delta APL = frac{frac{Y times (1+alpha)}{Ltimes (1+beta)} -frac{Y}{L}}{frac{Y}{L}}$



      $= frac{1+alpha}{1+beta} -1$



      Which is not equal to $alpha - beta$ as claimed by my notes.



      Where am I going wrong here? Is there some limit stuff I am forgetting?



      (If anyone is curious the particular relationship is about the average product of labor. But this is not important. The claim being made is general)






      calculus limits functions economics percentages




      calculus limits functions economics percentages






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 26 at 2:21

























      asked Nov 26 at 1:51









      SolidSnake

      102




      102






















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          What they do is the following:




          • They use the linear Taylor-approximation for the function $f(x,y) = frac{x}{y}$:


          $$f(x+Delta x,y+Delta y)- f(x,y) stackrel{Taylor}{approx} frac{partial f}{partial x}Delta x + frac{partial f}{partial y}Delta y$$



          So, you get
          $$f(x,y) = frac{x}{y} Rightarrow frac{partial f}{partial x} =color{blue}{frac{1}{y}} , : frac{partial f}{partial y} = color{blue}{-frac{x}{y^2}}$$



          For the percentage increase they divide by $f$:
          begin{eqnarray*} frac{Delta f}{f}
          & = & frac{color{blue}{frac{1}{y}}}{frac{x}{y}}Delta x + frac{color{blue}{-frac{x}{y^2}}}{frac{x}{y}}Delta y \
          & = & frac{Delta x}{x} - frac{Delta y}{y}
          end{eqnarray*}






          share|cite|improve this answer





















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            active

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            1














            What they do is the following:




            • They use the linear Taylor-approximation for the function $f(x,y) = frac{x}{y}$:


            $$f(x+Delta x,y+Delta y)- f(x,y) stackrel{Taylor}{approx} frac{partial f}{partial x}Delta x + frac{partial f}{partial y}Delta y$$



            So, you get
            $$f(x,y) = frac{x}{y} Rightarrow frac{partial f}{partial x} =color{blue}{frac{1}{y}} , : frac{partial f}{partial y} = color{blue}{-frac{x}{y^2}}$$



            For the percentage increase they divide by $f$:
            begin{eqnarray*} frac{Delta f}{f}
            & = & frac{color{blue}{frac{1}{y}}}{frac{x}{y}}Delta x + frac{color{blue}{-frac{x}{y^2}}}{frac{x}{y}}Delta y \
            & = & frac{Delta x}{x} - frac{Delta y}{y}
            end{eqnarray*}






            share|cite|improve this answer


























              1














              What they do is the following:




              • They use the linear Taylor-approximation for the function $f(x,y) = frac{x}{y}$:


              $$f(x+Delta x,y+Delta y)- f(x,y) stackrel{Taylor}{approx} frac{partial f}{partial x}Delta x + frac{partial f}{partial y}Delta y$$



              So, you get
              $$f(x,y) = frac{x}{y} Rightarrow frac{partial f}{partial x} =color{blue}{frac{1}{y}} , : frac{partial f}{partial y} = color{blue}{-frac{x}{y^2}}$$



              For the percentage increase they divide by $f$:
              begin{eqnarray*} frac{Delta f}{f}
              & = & frac{color{blue}{frac{1}{y}}}{frac{x}{y}}Delta x + frac{color{blue}{-frac{x}{y^2}}}{frac{x}{y}}Delta y \
              & = & frac{Delta x}{x} - frac{Delta y}{y}
              end{eqnarray*}






              share|cite|improve this answer
























                1












                1








                1






                What they do is the following:




                • They use the linear Taylor-approximation for the function $f(x,y) = frac{x}{y}$:


                $$f(x+Delta x,y+Delta y)- f(x,y) stackrel{Taylor}{approx} frac{partial f}{partial x}Delta x + frac{partial f}{partial y}Delta y$$



                So, you get
                $$f(x,y) = frac{x}{y} Rightarrow frac{partial f}{partial x} =color{blue}{frac{1}{y}} , : frac{partial f}{partial y} = color{blue}{-frac{x}{y^2}}$$



                For the percentage increase they divide by $f$:
                begin{eqnarray*} frac{Delta f}{f}
                & = & frac{color{blue}{frac{1}{y}}}{frac{x}{y}}Delta x + frac{color{blue}{-frac{x}{y^2}}}{frac{x}{y}}Delta y \
                & = & frac{Delta x}{x} - frac{Delta y}{y}
                end{eqnarray*}






                share|cite|improve this answer












                What they do is the following:




                • They use the linear Taylor-approximation for the function $f(x,y) = frac{x}{y}$:


                $$f(x+Delta x,y+Delta y)- f(x,y) stackrel{Taylor}{approx} frac{partial f}{partial x}Delta x + frac{partial f}{partial y}Delta y$$



                So, you get
                $$f(x,y) = frac{x}{y} Rightarrow frac{partial f}{partial x} =color{blue}{frac{1}{y}} , : frac{partial f}{partial y} = color{blue}{-frac{x}{y^2}}$$



                For the percentage increase they divide by $f$:
                begin{eqnarray*} frac{Delta f}{f}
                & = & frac{color{blue}{frac{1}{y}}}{frac{x}{y}}Delta x + frac{color{blue}{-frac{x}{y^2}}}{frac{x}{y}}Delta y \
                & = & frac{Delta x}{x} - frac{Delta y}{y}
                end{eqnarray*}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 3:02









                trancelocation

                9,0751521




                9,0751521






























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