<p>I believe there is a quicker way here. By the inverse function theorem, we have</p>

<p>$$y = f(k) \implies k = f^{-1}(y) = f^{-1}[f(k)] \implies \frac {\partial f^{-1}(y)}{\partial y} = \frac {1}{f'(k)}$$</p>

<p>Keeping all these relations in mind, we also have</p>

<p>$$\dot y = f'(k)\cdot \big[sf(k) - (n+g+\delta)\cdot f^{-1}(y)\big]$$</p>

<p>$$\implies  \frac{\partial \dot y}{\partial y} \bigg|_{y=y^*}  = \frac {\partial f'(k^*)}{\partial y}\cdot \big[sf(k^*) - (n+g+\delta)\cdot k^*\big]\\ + f'(k^*)\cdot \left[s - (n+g+\delta)\cdot \frac {1}{f'(k^*)} \right]$$</p>

<p>$$\implies \frac{\partial \dot y}{\partial y} \bigg|_{y=y^*} = 0 + sf'(k^*) - (n+g+\delta)$$</p>

<p>etc</p>

CC BY-SA 3.0
2016-01-23T04:25:47.813Z
Wow absolutely!  Reading my answer again is so messy .. Thanks @Alecos_Papadopoulos !
Sunhwa
CC BY-SA 3.0
2016-01-23T04:17:06.227Z
2016-01-23T04:17:06.227Z
Alecos Papadopoulos
<p>Okay so, in the book they looked at the first order taylor series of $\dot{k}$, so as @denesp said, we need the time here too!  So rather here we should look at the taylor series around $\dot{y}$</p>

<p>$$ \displaystyle \dot{y}(y) \simeq \left[ \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|_{y=y^*} \right] (y(k) - y(k^*)) $$</p>

<p>$$ \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|_{y=y^*} = \left( \frac{\partial \dot{y}(y)}{\partial k(t)} \bigg|_{y=y^*} \right) \left( \frac{\partial k(t)}{\partial y(k)} \bigg|_{k=k^*} \right) $$</p>

<p>First we look at:</p>

<p>$$ y=f(k) $$</p>

<p>$$\Rightarrow \dot{y} = \frac{d}{dt} f(k) = \frac{d f}{d k} \frac{dk}{dt} = f'(k) \dot{k} $$</p>

<p>We know the key equation of the Solow model is:</p>

<p>$$ \dot{k}(t) = s f(k(t)) - (n+g+\delta)k(t) $$</p>

<p>$$ \Rightarrow \dot{y} = f'(k) \left[ s f(k(t)) - (n+g+\delta)k(t) \right] $$</p>

<p>We take the derivative of this with respect to capital:</p>

<p>$$ \frac{\partial \dot{y}}{\partial k} = f''(k)\left[ s f(k(t)) - (n+g+\delta)k(t) \right] + f'(k) \left[ s f'(k) - (n+g+\delta) \right]$$</p>

<p>The value $k^*$ is the <em>golden-rule</em> level of the capital stock so:</p>

<p>$$ s f(k^*) = (n+g+\delta)k^* $$</p>

<p>And hence </p>

<p>$$ \left( \frac{\partial \dot{y}}{\partial k} \bigg|_{y=y^*} \right) = f''(k^*)*(0) + f'(k^*)\left[ s f'(k^*) - (n+g+\delta) \right] $$</p>

<p>Next we use the hint (not much of a hint, and rather misleading to begin with in my opinion)</p>

<p>$$ \left( \frac{\partial k(y)}{\partial y(t)} \bigg|_{k=k^*} \right) = \left( \frac{\partial y(k)}{\partial k(t)} \bigg|_{y=y^*} \right)^{-1} = f'(k^*)^{-1} = g'(y^*) $$</p>

<p>Plugging both of these in to our first order partial derivative:</p>

<p>$$ \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|_{y=y^*} = \left( f'(k^*)\left[ s f'(k^*) - (n+g+\delta) \right] \right) * \left( f'(k^*)^{-1} \right) $$</p>

<p>$$ = s f'(k^*) - (n+g+\delta) $$</p>

<p>since around the balanced growth path $s = (n+g+\delta)k^*/f(k^*)$ and putting</p>

<p>$$ - \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|_{y=y^*} = \lambda $$</p>

<p>$$ \lambda = (n+g+\delta) - \frac{(n+g+\delta)(k^*)*f'(k^*)}{f(k^*)} $$</p>

<p>With $\alpha_k = \frac{k*f'(k)}{f(k)} $ being the elasticity of output with respect to capital we obtain:</p>

<p>$$ \lambda = (n+g+\delta)(1-\alpha_k(k^*)) $$</p>

<p>So $y$ converges to its balanced growth-path value rate $\lambda$ the same as $k$ converges to $k^*$</p>

CC BY-SA 3.0
2016-01-22T17:05:52.130Z
2016-01-22T17:05:52.130Z
Sunhwa
<blockquote>
  <p>Find how quickly $y$ converges to $y^*$ in the vicinity of the balanced growth path
  $$ y=Y/AL=f(k); \; y^*=f(k^*) $$
  They suggest the hint, write $k=g(y)$, where $g(\bullet)=f^{-1}(\bullet)$.)</p>
  
  <p>$Y$ is product output; $AL$ is effective labour; $K$ is capital; lowercase letters are divided by $AL$ (e.g. $y=Y/AL$); $k^*$ is the balanced growth path of capital.</p>
</blockquote>

<p>This is problem $1.11$ in Advanced Macro by Romer.  I have tried looking at <a href="https://en.wikipedia.org/wiki/Lagrange_inversion_theorem" rel="nofollow">Lagrange Inversion Theorem</a>, and applying the above get to the first order Taylor series centered around $k^*$</p>

<p>$$ \displaystyle g(y) = k^* + \lim_{k \to k^*} \left( \frac{y-f(k^*)}{1!} \right) \frac{d^0}{dw^0} \left( \frac{k-k^*}{f(k)-f(k^*)} \right) $$</p>

<p>but since $y=f(k)$ and the zeroth derivative is the function itself, we get the following cancelation:</p>

<p>$$ = k^* + \lim_{k \to k^*} \frac{f(k)-f(k^*)}{f(k)-f(k^*)}(k-k^*) $$</p>

<p>$$ = k^* + \lim_{k \to k^*}(k - k^*) = k^* $$</p>

<p>Which doesn't tell us anything ... Do we need to look at the second order derivative?</p>

CC BY-SA 3.0
2016-01-22T10:33:55.503Z
Time as a variable does not appear in any of your equations. Perhaps you neglected to type in relevant information?
Giskard
CC BY-SA 3.0
2016-01-22T15:41:39.680Z
Ahh, okay thank you, I think I have it now .. should I post the answer?
Sunhwa
CC BY-SA 3.0
2016-01-22T16:07:58.643Z
If you have the time, sure!
Giskard
<macroeconomics>
Speed of convergence of output on the balanced growth path
Sunhwa
# Speed of convergence of output on the balanced growth path

(Asked by: username_0 on 2016-01-22)

> Find how quickly $y$ converges to $y^\*$ in the vicinity of the balanced growth path
> $$ y=Y/AL=f(k); \; y^\*=f(k^\*) $$
> They suggest the hint, write $k=g(y)$, where $g(\bullet)=f^{-1}(\bullet)$.)
>
> $Y$ is product output; $AL$ is effective labour; $K$ is capital; lowercase letters are divided by $AL$ (e.g. $y=Y/AL$); $k^\*$ is the balanced growth path of capital.

This is problem $1.11$ in Advanced Macro by Romer. I have tried looking at [Lagrange Inversion Theorem](https://en.wikipedia.org/wiki/Lagrange_inversion_theorem), and applying the above get to the first order Taylor series centered around $k^\*$

$$ \displaystyle g(y) = k^\* + \lim\_{k \to k^\*} \left( \frac{y-f(k^\*)}{1!} \right) \frac{d^0}{dw^0} \left( \frac{k-k^\*}{f(k)-f(k^\*)} \right) $$

but since $y=f(k)$ and the zeroth derivative is the function itself, we get the following cancelation:

$$ = k^\* + \lim\_{k \to k^\*} \frac{f(k)-f(k^\*)}{f(k)-f(k^\*)}(k-k^\*) $$

$$ = k^\* + \lim\_{k \to k^\*}(k - k^\*) = k^\* $$

Which doesn't tell us anything ... Do we need to look at the second order derivative?

**Tags:** macroeconomics
> **Comments:**
> **username_1**: Time as a variable does not appear in any of your equations. Perhaps you neglected to type in relevant information?  
> **username_0**: Ahh, okay thank you, I think I have it now .. should I post the answer?  
> **username_1**: If you have the time, sure!  

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## 2 answers

### Answer by username_2 on 2016-01-23. Score: 1 | Accepted answer

I believe there is a quicker way here. By the inverse function theorem, we have

$$y = f(k) \implies k = f^{-1}(y) = f^{-1}[f(k)] \implies \frac {\partial f^{-1}(y)}{\partial y} = \frac {1}{f'(k)}$$

Keeping all these relations in mind, we also have

$$\dot y = f'(k)\cdot \big[sf(k) - (n+g+\delta)\cdot f^{-1}(y)\big]$$

$$\implies \frac{\partial \dot y}{\partial y} \bigg|\_{y=y^\*} = \frac {\partial f'(k^\*)}{\partial y}\cdot \big[sf(k^\*) - (n+g+\delta)\cdot k^\*\big]\\ + f'(k^\*)\cdot \left[s - (n+g+\delta)\cdot \frac {1}{f'(k^\*)} \right]$$

$$\implies \frac{\partial \dot y}{\partial y} \bigg|\_{y=y^\*} = 0 + sf'(k^\*) - (n+g+\delta)$$

etc

> **Comments:**
> **username_0**: Wow absolutely!  Reading my answer again is so messy .. Thanks @Alecos_Papadopoulos !  

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### Answer by username_0 on 2016-01-22. Score: 3

Okay so, in the book they looked at the first order taylor series of $\dot{k}$, so as @denesp said, we need the time here too! So rather here we should look at the taylor series around $\dot{y}$

$$ \displaystyle \dot{y}(y) \simeq \left[ \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|\_{y=y^\*} \right] (y(k) - y(k^\*)) $$

$$ \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|\_{y=y^\*} = \left( \frac{\partial \dot{y}(y)}{\partial k(t)} \bigg|\_{y=y^\*} \right) \left( \frac{\partial k(t)}{\partial y(k)} \bigg|\_{k=k^\*} \right) $$

First we look at:

$$ y=f(k) $$

$$\Rightarrow \dot{y} = \frac{d}{dt} f(k) = \frac{d f}{d k} \frac{dk}{dt} = f'(k) \dot{k} $$

We know the key equation of the Solow model is:

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

$$ \Rightarrow \dot{y} = f'(k) \left[ s f(k(t)) - (n+g+\delta)k(t) \right] $$

We take the derivative of this with respect to capital:

$$ \frac{\partial \dot{y}}{\partial k} = f''(k)\left[ s f(k(t)) - (n+g+\delta)k(t) \right] + f'(k) \left[ s f'(k) - (n+g+\delta) \right]$$

The value $k^\*$ is the *golden-rule* level of the capital stock so:

$$ s f(k^\*) = (n+g+\delta)k^\* $$

And hence

$$ \left( \frac{\partial \dot{y}}{\partial k} \bigg|\_{y=y^\*} \right) = f''(k^\*)\*(0) + f'(k^\*)\left[ s f'(k^\*) - (n+g+\delta) \right] $$

Next we use the hint (not much of a hint, and rather misleading to begin with in my opinion)

$$ \left( \frac{\partial k(y)}{\partial y(t)} \bigg|\_{k=k^\*} \right) = \left( \frac{\partial y(k)}{\partial k(t)} \bigg|\_{y=y^\*} \right)^{-1} = f'(k^\*)^{-1} = g'(y^\*) $$

Plugging both of these in to our first order partial derivative:

$$ \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|\_{y=y^\*} = \left( f'(k^\*)\left[ s f'(k^\*) - (n+g+\delta) \right] \right) \* \left( f'(k^\*)^{-1} \right) $$

$$ = s f'(k^\*) - (n+g+\delta) $$

since around the balanced growth path $s = (n+g+\delta)k^\*/f(k^\*)$ and putting

$$ - \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|\_{y=y^\*} = \lambda $$

$$ \lambda = (n+g+\delta) - \frac{(n+g+\delta)(k^\*)\*f'(k^\*)}{f(k^\*)} $$

With $\alpha\_k = \frac{k\*f'(k)}{f(k)} $ being the elasticity of output with respect to capital we obtain:

$$ \lambda = (n+g+\delta)(1-\alpha\_k(k^\*)) $$

So $y$ converges to its balanced growth-path value rate $\lambda$ the same as $k$ converges to $k^\*$


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