Percentage change of a quotient











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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)










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    up vote
    1
    down vote

    favorite












    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)










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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)










      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






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      edited Nov 26 at 2:21

























      asked Nov 26 at 1:51









      SolidSnake

      102




      102






















          1 Answer
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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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            1 Answer
            1






            active

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            active

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            up vote
            1
            down vote



            accepted










            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

























              up vote
              1
              down vote



              accepted










              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























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                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

                8,6421520




                8,6421520






























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