<p>Taking a Taylor expansion of <span class="math-container">$e^\hat x$</span> gives around <span class="math-container">$0$</span> gives:
<span class="math-container">$$
e^\hat x \approx e^0 + e^0 \hat x = 1 + \hat x.
$$</span>
Substitution gives:
<span class="math-container">$$
\begin{align*}
&amp;(c + \alpha g) + c \hat c_t + \alpha g \hat g_t \approx \frac{1}{\lambda}(1 - \hat \lambda_t),\\
\iff &amp; \lambda(c + \alpha g) + \lambda c \hat c_t + \lambda \alpha g \hat g_t \approx 1 - \hat \lambda_t,\\
\iff &amp; 1 + \frac{c}{c + \alpha g} \hat c_t + \frac{\alpha g}{c + \alpha g} \hat g_t \approx 1 - \hat \lambda_t,\\
\iff &amp; -\frac{c}{c + \alpha g} \hat c_t- \frac{\alpha g}{c + \alpha g}\hat g_t \approx \hat \lambda_t
\end{align*}
$$</span></p>

CC BY-SA 4.0
2021-10-16T11:20:49.623Z
Why the $ \frac{1}{\lambda(c+\alpha g)}$ term doesn't affect RHS?
guest
CC BY-SA 4.0
2021-10-18T05:26:32.977Z
$\lambda = \frac{1}{c + \alpha g} $ so $\frac{1}{\lambda(c + \alpha g)} = 1$.
tdm
CC BY-SA 4.0
2021-10-16T11:04:47.513Z
2021-10-16T11:04:47.513Z
tdm
<p>Define <span class="math-container">$\ln\frac{c_t}{c}=\hat{c_t}$</span> which means % deviation from steady state.<br />
Then <span class="math-container">$\frac{1}{c_t +\alpha g_t}=\lambda_t$</span> is equivalent as<br />
<span class="math-container">$\frac{1}{c e^{\hat{c_t}} + \alpha g e^{\hat{g_t}}}=\lambda e^{\hat{\lambda_t}}$</span> <br />
<span class="math-container">$c e^{\hat{c_t}} + \alpha g e^{\hat{g_t}}=\frac{1}{\lambda} e^{\hat{-\lambda_t}}$</span> <br />
<span class="math-container">$-\frac{c}{c +\alpha g} \hat{c_t} - \frac{\alpha g}{c+\alpha g}\hat{g_t}\approx \hat{\lambda_t}$</span><br />
How to derive the last equation with approximation?</p>

<macroeconomics>
Approximation around steady state calculation
guest
# Approximation around steady state calculation

(Asked by: username_0 on 2021-10-16)

Define $\ln\frac{c\_t}{c}=\hat{c\_t}$ which means % deviation from steady state.  
Then $\frac{1}{c\_t +\alpha g\_t}=\lambda\_t$ is equivalent as  
$\frac{1}{c e^{\hat{c\_t}} + \alpha g e^{\hat{g\_t}}}=\lambda e^{\hat{\lambda\_t}}$   
$c e^{\hat{c\_t}} + \alpha g e^{\hat{g\_t}}=\frac{1}{\lambda} e^{\hat{-\lambda\_t}}$   
$-\frac{c}{c +\alpha g} \hat{c\_t} - \frac{\alpha g}{c+\alpha g}\hat{g\_t}\approx \hat{\lambda\_t}$  
How to derive the last equation with approximation?

**Tags:** macroeconomics


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## 1 answers

### Answer by username_1 on 2021-10-16. Score: 1 | Accepted answer

Taking a Taylor expansion of $e^\hat x$ gives around $0$ gives:
$$
e^\hat x \approx e^0 + e^0 \hat x = 1 + \hat x.
$$
Substitution gives:
$$
\begin{align\*}
&(c + \alpha g) + c \hat c\_t + \alpha g \hat g\_t \approx \frac{1}{\lambda}(1 - \hat \lambda\_t),\\
\iff & \lambda(c + \alpha g) + \lambda c \hat c\_t + \lambda \alpha g \hat g\_t \approx 1 - \hat \lambda\_t,\\
\iff & 1 + \frac{c}{c + \alpha g} \hat c\_t + \frac{\alpha g}{c + \alpha g} \hat g\_t \approx 1 - \hat \lambda\_t,\\
\iff & -\frac{c}{c + \alpha g} \hat c\_t- \frac{\alpha g}{c + \alpha g}\hat g\_t \approx \hat \lambda\_t
\end{align\*}
$$

> **Comments:**
> **username_0**: Why the $ \frac{1}{\lambda(c+\alpha g)}$ term doesn't affect RHS?  
> **username_1**: $\lambda = \frac{1}{c + \alpha g} $ so $\frac{1}{\lambda(c + \alpha g)} = 1$.  

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