Hello everyone, I am trying to obtain the requested solution as shown in the image (last equation of the image attached), however, after I calculate the law of motion, I only obtain only part of the equation that I am requested to show (i.e I only get n+d/sz part of the equation). What am I missing to do in this case?


2 Answers 2


Part 2 of the solution

I post a step by step solution, slow passages are the best way to avoid mistakes (I hope).

In the previuos answer we calculated the intensive production function:

$$f(k)=z[\alpha k^\rho+ (1-\alpha)]^{1/\rho}\;\;\;\; (1)$$

The fundamental equation, set equal to $0$ in order to find the steady state $k$, is:

$$ \Delta k= sf(k) -(n+d)=0 \;\;\;\; \;\;(2)$$

Substituting $(1)$ into $(2)$ we have:

$$sz [\alpha k^\rho+ (1-\alpha)]^{1/\rho} - (n+d) k=0 \Leftrightarrow$$

$$sz [\alpha k^\rho+ (1-\alpha)]^{1/\rho} = (n+d) k \Leftrightarrow$$(raising to $\rho$-th power) $$s^\rho z^\rho [\alpha k^\rho+ (1-\alpha)] = (n+d)^\rho k^\rho \Leftrightarrow$$ $$s^\rho z^\rho \alpha k^\rho+ (1-\alpha)s^\rho z^\rho = (n+d)^\rho k^\rho \Leftrightarrow$$ $$k^\rho (s ^\rho z^\rho \alpha -(n+d)^\rho)= -(1-\alpha) s ^\rho z^\rho \Leftrightarrow $$ $$k^\rho= \frac {-(1-\alpha) s^\rho z^\rho}{s^\rho z^\rho \alpha -(n+d)^\rho} \Leftrightarrow$$ $$k= \left( \frac {-(1-\alpha) s^\rho z^\rho}{s^\rho z^\rho \alpha -(n+d)^\rho}\right)^{1/\rho}=$$ $$= (1-\alpha)^{1/\rho} \left( \frac {- s^\rho z^\rho}{s^\rho z^\rho \alpha -(n+d)^\rho}\right)^{1/\rho}=$$

$$=\left(\frac{1}{1-\alpha}\right) ^{-1/\rho} \left(\frac{s^\rho z^\rho \alpha- (n+d)^\rho}{- s^\rho z^\rho}\right)^{-1/\rho}=$$

$$=\left(\frac{1}{1-\alpha}\right) ^{-1/\rho} \left( \left( \frac {n+d} {sz}\right)^\rho - \alpha \right)^{-1/\rho}$$


  • 1
    $\begingroup$ Thanks for such a great and well explained answer. The more I do these exercises, the more I get better with my first time immersion into algebra. $\endgroup$ Commented Dec 6, 2022 at 16:52
  • $\begingroup$ It is just question of practice. Thanks to you! $\endgroup$ Commented Dec 6, 2022 at 17:01

Part 1 of the solution

The fundamental equation of Solow model is (neglecting the $t$ subscripts):

$$ \Delta k= sf(k) -(n+d),$$

where $k= K/N$, $\Delta k= k_{t+1}-k_t$ and $f(k)$ is the intensive production function. And this equation must be set equal to $0$ in order to find the steady state value of $k$.

But the first step is to calculate the intensive production function $f(k)$.

As in your result you didn't obtain the part $(\frac {1} {1-\alpha}) ^{-1/\rho}$, I suppose that you lost somewhere in your calculations $(1-\alpha)$. As $(1-\alpha)$ is in the production function, I suppose that your mistake could be in the calculation of the intensive production function.

Let's calculate the intensive production function from the production function in your question, which is (again neglecting the $t$ subscripts):

$$Y=z[\alpha K^\rho + (1-\alpha)N^\rho) ^{1/\rho}.$$

Dividing both sides by $N$ we have:

$$\frac{Y}{N}= f(k)= z[\alpha \frac {K^\rho}{N^\rho}+(1-\alpha)\frac {N^\rho}{N^\rho} ]^{-1/\rho}=$$ $$\;\;\;=z[\alpha k^\rho+ (1-\alpha)]^{1/\rho}.$$

Using this last equation, making the calculations and rearranging, I found the result in your question.

You can try, verifying if you manage to obtain it.$^1$

$^1$ I didn’t post all the passages because they are enough long, but enough trivial algebraic passages, and I don’t know if it is worthwhile to post them, if you reach the result yourself. But I can post them subsequently, if you prefer.

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    $\begingroup$ Thanks for the great answer, your time is much appreciated. I am working on it, however, I started by plugging f(k) into the sf(k)-(n+d) equation (i.e change in capital), and then setting it equal to 0. I am note sure whether this step is incorrect, or is it my algebra that is failing, but I obtain a result that is almost similar, but not similar to the one requested in the setup. For example, my results in regards of the alpha, as well as the (1-alpha) are different than on the proof. $\endgroup$ Commented Dec 4, 2022 at 16:44
  • $\begingroup$ It's easy to make mistakes in this kind of calculations. I have posted in a separate answer a step by step solution. $\endgroup$ Commented Dec 5, 2022 at 12:49

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