# Thread: Trouble with exponents and differentiation

1. ## Trouble with exponents and differentiation

I think I have an error in my calculation, because I should get $\displaystyle s= \alpha$ at the end but I do not. Can someone please check to see where I've gone wrong?

Suppose you have: $\displaystyle c=(1-s) \left[ \frac{s}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}$

Rewrite as

$\displaystyle c=(1-s)s^\frac{\alpha}{1-\alpha} \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}$

Expand the brackets:

$\displaystyle c=(s^\frac{\alpha}{1-\alpha}-s^\frac{1}{1-\alpha}) \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}$

To find golden rule consumption, take the partial:

$\displaystyle \frac{\partial c}{\partial s}=\left( \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha}-\frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha} \right) \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}$

Set that beastie equal to zero so we can find the local max:

$\displaystyle \frac{\partial c}{\partial s}=\left( \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha}-\frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha} \right) \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha} = 0$

Divide through by $\displaystyle \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}$ to get

$\displaystyle \left( \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha}-\frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha} \right)= 0$

$\displaystyle \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha} = \frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha}$

Multiply both sides by $\displaystyle 1-\alpha$

$\displaystyle \alpha s^\frac{2 \alpha -1}{1-\alpha} = s^\frac{\alpha}{1-\alpha}$

Raise both sides to the power of $\displaystyle 1-\alpha$

$\displaystyle \alpha s^{2 \alpha -1} = s^{\alpha}$

Multiply both sides by $\displaystyle s^{-\alpha}$

$\displaystyle \alpha s^{\alpha -1} = 1$

$\displaystyle s^{\alpha -1} = \frac{1}{\alpha }$

And finish by raising both sides by $\displaystyle \frac{1}{\alpha -1}$

$\displaystyle s = \left( \frac{1}{\alpha } \right)^\frac{1}{\alpha -1}$

We can rewrite as

$\displaystyle s = \alpha^\frac{1}{1- \alpha}$

But that hardly makes it look closer to $\displaystyle s=\alpha$, which I think is the right answer. What have I done wrong?

Originally Posted by garymarkhov
Raise both sides to the power of $\displaystyle 1-\alpha$

$\displaystyle \alpha s^{2 \alpha -1} = s^{\alpha}$
When you raise a function ab^c to the power of d, you should get:
$\displaystyle (\alpha b^c)^d = \alpha ^d*b^{cd}$
but you have:
$\displaystyle (\alpha b^c)^d = \alpha *b^{cd}$
Note that the $\displaystyle \alpha$ term is missing an exponent.

3. Dear Lord,

Thank you for Mr. ebaines. Amen.

Love,

Gary