<p>The discrete time law of motion is given by
<span class="math-container">$k(t+1) = (1-\delta)k(t) + s f(k(t))$</span>
This can be rewritten as:
<span class="math-container">$$
(k(t+1)-k(t)) = s f(k(t)) - \delta k(t).
$$</span>
Now, take a Taylor expansion of order 1 of <span class="math-container">$k(t+1)$</span> around <span class="math-container">$t$</span>.
<span class="math-container">$$
k(t+1) \approx k(t) + \frac{\partial k}{\partial t}(t),\\
\to k(t+1) - k(t) \approx \frac{\partial k}{\partial t}(t).
$$</span>
This gives:
<span class="math-container">$$
\frac{\partial k}{\partial t}(t) = sf(k(t)) - \delta k(t)
$$</span></p>
<p>So we are basically exchanging the discrete difference <span class="math-container">$\Delta k(t) = k(t+1) - k(t)$</span> by its continuous approximation <span class="math-container">$\dfrac{\partial k}{\partial t}(t)$</span>.</p>

CC BY-SA 4.0
2023-09-15T06:40:10.250Z
I wonder why at the beginning (()) is steeper than (), but at some point, it starts flatter than ()?
Huy Lê Thanh
CC BY-SA 4.0
2023-09-15T05:10:16.267Z
2023-09-15T05:10:16.267Z
tdm
<p>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-δ)k(t) + sf(k(t)) and in continuously time function use: dy/dt = s*f(k) - δk. Are they different? (i think they are not)</p>

<macroeconomics><microeconomics><mathematical-economics><economic-growth><solow>
Solow Model in disrectly and continuously
Huy Lê Thanh
# Solow Model in disrectly and continuously

(Asked by: username_0 on 2023-09-14)

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-δ)k(t) + sf(k(t)) and in continuously time function use: dy/dt = s\*f(k) - δk. Are they different? (i think they are not)

**Tags:** macroeconomics, microeconomics, mathematical-economics, economic-growth, solow

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## 1 answers

### Answer by username_1 on 2023-09-15. Score: 3

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 $\dfrac{\partial k}{\partial t}(t)$.

> **Comments:**
> **username_0**: I wonder why at the beginning (()) is steeper than (), but at some point, it starts flatter than ()?  

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