<p>The steady-state value $k^*$ must be a fixed point :</p>

<p>$(1+g)(1+n)k^* = s(k^*)^{\alpha} +(1 - \delta) k^*$</p>

<p>Taking the difference between this equation and the dynamic one :</p>

<p>$(1+g)(1+n)(k_{t+1} - k^*) = s((k_{t+1})^{\alpha} - (k^*)^{\alpha}) +(1 - \delta) (k_t - k^*)$</p>

<p>Now if you denote by $d_t = k_t - k^*$ the distance to steady-state, this gives :</p>

<p>$(1+g)(1+n)d_{t+1} = s((k_{t+1})^{\alpha} - (k^*)^{\alpha}) +(1 - \delta) d_t$</p>

<p>But :</p>

<p>$(k_{t+1})^{\alpha} = (k^* + d_{t+1})^\alpha = (k^*)^\alpha + \alpha d_{t+1} (k^*)^{\alpha - 1} + o (d_{t+1})$</p>

<p>(Here was the Taylor development.) So that finally, this equality holds approximately around steady state ($d_t$ small):</p>

<p>$(1+g)(1+n)d_{t+1} = \alpha s (k^*)^{\alpha - 1} d_{t+1} +(1 - \delta) d_t$</p>

<p>From the steady state condition you can also compute (assuming that $ng &lt;&lt; n+g$):</p>

<p>$(k^*)^{\alpha -1} = \frac{n+g+\delta}{s}$</p>

CC BY-SA 3.0
2017-02-25T17:03:27.480Z
2017-02-25T17:03:27.480Z
user11629
<p>In my Advanced Macro script, the professor says take TSA1 of the following equation:</p>

<p>$(1+g)(1+n)k_{t+1} = sk^\alpha_{t+1} + (1-\delta)k_t$</p>

<p>where $g$ is technological progress, $n$ population growth, $k$ capital per effective labour (AL). </p>

<p>In the class we work with discrete time. The results don't seem to make sense to me - could anyone guide me how to take the TSA in this particular case?</p>

<p>The $k_{t+1}$ means subscript and thus capital per AL at time $t+1$.</p>

<p>EDIT: TSA around the steady state.</p>

<p>Many thanks!</p>

<macroeconomics><steady-state><solow>
Taylor Series Approximation around steady state in Solow
Karel
# Taylor Series Approximation around steady state in Solow

(Asked by: username_0 on 2017-02-25)

In my Advanced Macro script, the professor says take TSA1 of the following equation:

$(1+g)(1+n)k\_{t+1} = sk^\alpha\_{t+1} + (1-\delta)k\_t$

where $g$ is technological progress, $n$ population growth, $k$ capital per effective labour (AL).

In the class we work with discrete time. The results don't seem to make sense to me - could anyone guide me how to take the TSA in this particular case?

The $k\_{t+1}$ means subscript and thus capital per AL at time $t+1$.

EDIT: TSA around the steady state.

Many thanks!

**Tags:** macroeconomics, steady-state, solow

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## 1 answers

### Answer by username_1 on 2017-02-25. Score: 2

The steady-state value $k^\*$ must be a fixed point :

$(1+g)(1+n)k^\* = s(k^\*)^{\alpha} +(1 - \delta) k^\*$

Taking the difference between this equation and the dynamic one :

$(1+g)(1+n)(k\_{t+1} - k^\*) = s((k\_{t+1})^{\alpha} - (k^\*)^{\alpha}) +(1 - \delta) (k\_t - k^\*)$

Now if you denote by $d\_t = k\_t - k^\*$ the distance to steady-state, this gives :

$(1+g)(1+n)d\_{t+1} = s((k\_{t+1})^{\alpha} - (k^\*)^{\alpha}) +(1 - \delta) d\_t$

But :

$(k\_{t+1})^{\alpha} = (k^\* + d\_{t+1})^\alpha = (k^\*)^\alpha + \alpha d\_{t+1} (k^\*)^{\alpha - 1} + o (d\_{t+1})$

(Here was the Taylor development.) So that finally, this equality holds approximately around steady state ($d\_t$ small):

$(1+g)(1+n)d\_{t+1} = \alpha s (k^\*)^{\alpha - 1} d\_{t+1} +(1 - \delta) d\_t$

From the steady state condition you can also compute (assuming that $ng << n+g$):

$(k^\*)^{\alpha -1} = \frac{n+g+\delta}{s}$


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