This is a question also for those with a good expertise in micro. For micro guys who wanna go streight to the question, just jump to equation $(1)$

I'm studying the Solow growth model.

Let's write the law of motion of capital in intensive form (with technology normalized to one):

$\dot{k}(t)=s f(k(t)) -(\delta + n)k(t) \equiv \phi (k(t))$

where $k(t) \equiv \frac{K(t)}{L(t)}$, $f(k(t))=\frac{F(K(t),L(t),1)}{L(t)}$,

$\delta$ denotes the depreciation rate of capital, and $n$ is the population growth rate.

Then, taking a linear approximation of $\phi (k)$ around $k^\ast$

$\dot{k}(t) \approx \ \phi(k^\ast) + \phi'(k^\ast) (k(t)-\phi(k^\ast))= \phi'(k^\ast) (k(t)-\phi(k^\ast))$

Solving this linear differential equation, we get

$k(t) = k^\ast + e^{-\lambda t} (k(0)-k^\ast)$, with $\lambda \equiv - \phi'(k^\ast) >0$

The speed of convergence is determined by $\lambda$

$\lambda \equiv - \phi'(k^\ast) = - \frac{ s f'(k^\ast)k^\ast}{k^\ast} + (\delta +n)$

Which boils down into

$\lambda= (1- \frac{f'(k^\ast)k^\ast}{f(k^\ast)})(\delta +n) >0$ $~~~~~~(1)$

The slides says:

The speed of convergence determined by (i) the degree of concavity in the production function and (ii) the effective depreciation rate.

Can you tell me why $\frac{f'(k^\ast)k^\ast}{f(k^\ast)}$ describes the degree of concavity of the production function ?



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