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ForumsDiscussion Forum → Quick Math help (logs, derivatives) ?
Quick Math help (logs, derivatives) ?
2010-06-02, 8:44 AM #1
Hey all, just working on some econ stuff before my final and am running into a sticking point with this equation (that apparently wolfram can't solve?) First, I'll give the original equation, then the equation with logs (which apparently is the first step). The answer to the question I KNOW for 100% fact is a/(1-a) (it is the elasticity of steady state output (y) with respect to savings rate (s). For the record, x and n are depreciation/growth rates of capital and population, respectively)

y = (s/(x+n))^(a/(1-a))

ln(y)= (a/(1-a))ln(s) - (a/(1-a))ln(x+n)

For the record if it helps: formula elasticity is for example, using X and Y, (X/Y)*(dY/dX), so if that helps you work with the first equation feel free.

Now from here, apparently just taking the derivative with respect to ln(s) should reveal the answer a/(1-a) but I'm just not getting it. Plus I'm plugging it into Wolfram to try and figure it out and it's not coming up with (a/(1-a)). I tried the original equation as well just with respect to s and it just doesn't compute. I also tried with the log equation multiplying by e everywhere, but I also do not end up getting a/(1-a).

Any helps or tips are appreciated
"His Will Was Set, And Only Death Would Break It"

"None knows what the new day shall bring him"
2010-06-02, 9:01 AM #2
I'm not quite understanding what the answer you're attempting to get to is.

Do you have to use ln, because if I remember correctly (a/(1-a)) could technically equal Log[s/[x+n]] (y). :P

[Dear god I hate maths displayed as text.]
nope.
2010-06-02, 9:04 AM #3
a/(1-a) apparently is the elasticity of output per worker with respect to changes in the savings rate

And no, we do not HAVE to use ln, but it seems odd that it was the first step recommended by my prof to figure it out.
"His Will Was Set, And Only Death Would Break It"

"None knows what the new day shall bring him"
2010-06-02, 10:05 AM #4
Oh, so you think you're too cool for IRC but not for math? I see how it is.
2010-06-02, 11:00 AM #5
These math threads always make me feel like a dullard.
"Honey, you got real ugly."
2010-06-02, 1:52 PM #6
Figured it out! Got some help on a math forum in the right direction. I was doing a simple algebraic error messing me up. Just some careful derivatives doing the (S/Y)(dY/dX) method and I got the answer desired.

Also, someone recommended me the way of just kind of replacing it all generally with y=Ax^n, (subbing the x for (s/x+n)), doing a simple derivative, and some exponent manipulation to find the answer as well, for those who care :)
Attachment: 23965/Screen shot 2010-06-02 at 9.57.42 PM.png (16,927 bytes)
"His Will Was Set, And Only Death Would Break It"

"None knows what the new day shall bring him"

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