Percentage change of a quotient
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
asked Nov 26 at 1:51
SolidSnake
102
add a comment |
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
asked Nov 26 at 1:51
SolidSnake
102
add a comment |
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
asked Nov 26 at 1:51
SolidSnake
102
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
asked Nov 26 at 1:51
SolidSnake
102
asked Nov 26 at 1:51
SolidSnake
102
edited Nov 26 at 2:21
asked Nov 26 at 1:51
SolidSnake
102
asked Nov 26 at 1:51
SolidSnake
102
asked Nov 26 at 1:51
SolidSnake
102
102
add a comment |
add a comment |
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*}
answered Nov 26 at 3:02
trancelocation
9,0751521
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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votes
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*}
answered Nov 26 at 3:02
trancelocation
9,0751521
add a comment |
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*}
answered Nov 26 at 3:02
trancelocation
9,0751521
add a comment |
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*}
answered Nov 26 at 3:02
trancelocation
9,0751521
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*}
answered Nov 26 at 3:02
trancelocation
9,0751521
answered Nov 26 at 3:02
trancelocation
9,0751521
answered Nov 26 at 3:02
trancelocation
9,0751521
answered Nov 26 at 3:02
trancelocation
9,0751521
9,0751521
add a comment |
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