It's been awhile since I've done the kind of math a macro problem set is asking for, so I was wondering if someone at Hatrack could help me out a bit. It's homework, but the homework doesn't count for any sort of grade...
One doesn't need to know the background on the macro model for the math - just know that a, s, e, n, and x are parameters.
Given: s * f(k) = (e+n+x)k y = f(k) a = k * f ' (k)/y (that's f prime, if it isn't clear)
Find (dy/dn)(n/y) - i.e. the derivative of y with respect to n times (n/y).
I also need to find (dy/ds)(s/y), but if I have an example to go on, I think I'll be okay... I have the feeling that this is really (relatively)simple and I'm just missing something.
[ September 18, 2007, 12:05 AM: Message edited by: Jhai ]
Posted by Shigosei (Member # 3831) on :
Can I assume that d, x, k, and s are not functions of n? Are they just numbers you plug in without regard to what n is? In that case, I'd just substitute y = f(k) into the first equation.
s*y = (d+n+x)*k
y = dk/s + nk/s + xk/s
Then I'd take the derivative. As long as d,k,s, and x are constant with respect to n, then I can just treat them as constants, while n is the variable.
So dy/dn = k/s. I think.
Posted by Jhai (Member # 5633) on :
Sorry - there's a problem with my notation. One d is a meant as a derivative. I've corrected it in the original problem.
Posted by HollowEarth (Member # 2586) on :
I'm getting (dy/dn)(n/y) = k^2 * n *(e+n+x)/s^2 ?
I'm thinking that something important might still be missing since, you don't need a = ... for anything. Also is f' another f or is that df/dk?
Posted by Shigosei (Member # 3831) on :
Okay. I'm not entirely certain I'm doing this right, but I'll give it a shot.
y = f(k) You can use a = k*f'(k)/y to solve for f(k) in terms of k and a. You use separation of variables for this.
a= (dy/dk)*(k/y)
(a/k)dk = (1/y)dy
Integrate to get
a*ln(k) = ln(y) + P, where P is a constant dependent on boundary conditions. It'll cancel eventually, so it's not all that important.
e^(a*ln(k)) = e^(ln(y) + P)
k^a = Py. I'll simplify that to y = C*k^a, where C = 1/P.
so f(k) = C*k^a.
Substitute that into s * f(k) = (e+n+x)k and solve for k to get k(n).
s*C*k^a = (e+n+x)k
k^(a-1) = (e+n+x)/sC
Take the log of both sides:
ln(k^(a-1)) = ln((e+n+x)/sC))
ln(k) * (a-1) = ln((e+n+x)/sC)
ln(k) = ln((e+n+x)/sC)/(a-1)
Raise the whole thing to the power of e (I’m not going to write that out!)
k(n) = [(e+n+x)/sC]^(1/(a-1))
Okay, now substitute that back in for k:
s * f(k) = (e+n+x)* [(e+n+x)/sC]^(1/(a-1))
so y =[ (e+n+x)* [(e+n+x)/sC]^(1/(a-1))]/s, or:
y = [(e+n+x)/s]^(a/(a-1))*Q , where Q is so I can get rid of that messy (1/C)^(1/(a-1)).
Okay! Now take the derivative.
dy/dn = (a/(a-1))*Q*[(e+n+x)/s]^(1/(a-1))/s This is just a combination of the chain rule and power rule. It looks less messy if you write it out normally.
If you divide dy/dn by y, much of the stuff cancels. Multiply by n, and you’ll be left with a*n/[(a-1)*(e+n+x)].
Um, was that what you were supposed to get?
Posted by Jhai (Member # 5633) on :
Thanks a ton Shigosei & HollowEarth! It is indeed a f prime, HollowEarth. And Shigosei, I'm not sure what the correct answer would be, but the one you came up with (I followed all the math very well!) seems plausible.
Basically (if you're interested), the equations represent a (very) simplified model of macro economic growth. y is total output per capita, k is capital per capita, s is the economy's saving rate, e is the depreciation rate of capital, n is population growth, and x is technological growth. a is the share of capital (how much capital is responsible for output, compared with labor). What I needed to calculate was the percentage change in output given a marginal percentage change of population growth (n).
Using your math, Shigosei, the answer for dy/ds*(s/y) comes out even nicer [a/(1-a)], and seems correct.
Posted by Phanto (Member # 5897) on :