Percentage change of a quotient
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)
calculus limits functions economics percentages
add a comment |
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)
calculus limits functions economics percentages
add a comment |
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)
calculus limits functions economics percentages
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
edited Nov 26 at 2:21
asked Nov 26 at 1:51
SolidSnake
102
102
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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*}
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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*}
add a comment |
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*}
add a comment |
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*}
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
8,6421520
8,6421520
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013701%2fpercentage-change-of-a-quotient%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown