1
$\begingroup$

what is the differences between solow model in disrectly time and continuously time? Why in disrectly time function use the equation: k(t+1)=(1-ฮด)k(t) + sf(k(t)) and in continuously time function use: dy/dt = s*f(k) - ฮดk. Are they different? (i think they are not)

$\endgroup$

1 Answer 1

3
$\begingroup$

The discrete time law of motion is given by $k(t+1) = (1-\delta)k(t) + s f(k(t))$ This can be rewritten as: $$ (k(t+1)-k(t)) = s f(k(t)) - \delta k(t). $$ Now, take a Taylor expansion of order 1 of $k(t+1)$ around $t$. $$ k(t+1) \approx k(t) + \frac{\partial k}{\partial t}(t),\\ \to k(t+1) - k(t) \approx \frac{\partial k}{\partial t}(t). $$ This gives: $$ \frac{\partial k}{\partial t}(t) = sf(k(t)) - \delta k(t) $$

So we are basically exchanging the discrete difference $\Delta k(t) = k(t+1) - k(t)$ by its continuous approximation $\dfrac{\partial k}{\partial t}(t)$.

$\endgroup$
1
  • $\begingroup$ I wonder why at the beginning ๐‘ ๐‘“(๐‘˜(๐‘ก)) is steeper than ๐›ฟ๐‘˜(๐‘ก), but at some point, it starts flatter than ๐›ฟ๐‘˜(๐‘ก)? $\endgroup$ Sep 15, 2023 at 6:40

Your Answer

By clicking โ€œPost Your Answerโ€, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.