<p>You cannot completely ignore the RHS. Starting with
<span class="math-container">$$\frac{U'(c_{t+1})}{U'(c_{t})}=RHS,$$</span>
replace <span class="math-container">$t+1$</span> by <span class="math-container">$t+\Delta t$</span> to get
<span class="math-container">$$\frac{U'(c(t+\Delta t))}{U'(c(t))}=RHS_{\Delta t},$$</span>
where <span class="math-container">$RHS_{\Delta t}$</span> is the modified version of <span class="math-container">$RHS$</span> which contains terms depending on <span class="math-container">$\Delta t$</span>, e.g. the modified discount factor. Expanding around <span class="math-container">$c(t)$</span> (and neglecting higher order terms for brevity) then gives you
<span class="math-container">$$\frac{U''(c(t))}{U'(c(t))}\dot{c}(t)=\frac{RHS_{\Delta t}-1}{\Delta t},$$</span>
and letting <span class="math-container">$\Delta t\rightarrow 0$</span>, everything should fall into place.</p>

CC BY-SA 4.0
2021-04-27T13:19:23.210Z
What do you mean by "expanding around c(t)" in this context. I usually hear that term in reference to taylor expansions.
EconJohn
CC BY-SA 4.0
2021-04-28T11:27:01.780Z
that's exactly what he/she means: write `U'` in terms of `U''` around  `dt`
PatrickT
CC BY-SA 4.0
2021-04-29T07:26:23.077Z
@EconJohn, yes, it's a Taylor expansion. The equation suppresses higher order terms, which I clarified in the term added in brackets now.
VARulle
CC BY-SA 4.0
2021-04-27T08:53:54.223Z
2021-04-29T07:24:27.823Z
2021-04-29T07:24:27.823Z
VARulle
<p>One way to see (intuitively) the connection between the left hand sides is to write the discrete case as:
<span class="math-container">$$
\frac{u'(c(t + \tau))}{u'(c(t))},
$$</span>
for <span class="math-container">$\tau = 1$</span>. Now if we generalise this to a setting where <span class="math-container">$\tau$</span> is now a variable in <span class="math-container">$\mathbb{R}$</span>, this becomes a function of <span class="math-container">$\tau$</span>. Taking the derivative with respect to <span class="math-container">$\tau$</span> and evaluating at <span class="math-container">$\tau = 0$</span>, gives:
<span class="math-container">$$
\frac{u''(c(t))}{u'(c(t))} \dot c(t).
$$</span>
which is the left hand side for the continuous case. As such the left hand side for the continuous case can be seen as the &quot;derivative&quot; of the discrete setting (when allowing the time interval to go to zero).</p>

CC BY-SA 4.0
2021-04-27T08:50:22.557Z
2021-04-27T08:50:22.557Z
tdm
<p>I have seen the euler equation in discrete time for the baseline neoclassical growth model written as:
<span class="math-container">$$\frac{U'(c_{t+1})}{U'(c_{t})}=\frac{1}{\beta(1+r)}$$</span></p>
<p>however I have also seen the euler equation for the continous time equivalent written as:
<span class="math-container">$$\frac{U''(c(t))}{U'(c(t))}\dot{c}(t)=r-f'(k)$$</span></p>
<p>Ignoring the right hand sides of these equations (which vary based on model set up) and exclusively looking at the left hand side I see two very different formulations of the same equation.</p>
<p>My question is: &quot;How does the Euler equation in Continous time represent the same thing conceptually when compared to its discrete time counter part?&quot;</p>

<macroeconomics><mathematical-economics><optimal-control>
Euler equation in Continuous time VS Discrete time
EconJohn
# Euler equation in Continuous time VS Discrete time

(Asked by: username_0 on 2021-04-27)

I have seen the euler equation in discrete time for the baseline neoclassical growth model written as:
$$\frac{U'(c\_{t+1})}{U'(c\_{t})}=\frac{1}{\beta(1+r)}$$

however I have also seen the euler equation for the continous time equivalent written as:
$$\frac{U''(c(t))}{U'(c(t))}\dot{c}(t)=r-f'(k)$$

Ignoring the right hand sides of these equations (which vary based on model set up) and exclusively looking at the left hand side I see two very different formulations of the same equation.

My question is: "How does the Euler equation in Continous time represent the same thing conceptually when compared to its discrete time counter part?"

**Tags:** macroeconomics, mathematical-economics, optimal-control


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## 2 answers

### Answer by username_3 on 2021-04-27. Score: 7 | Accepted answer

One way to see (intuitively) the connection between the left hand sides is to write the discrete case as:
$$
\frac{u'(c(t + \tau))}{u'(c(t))},
$$
for $\tau = 1$. Now if we generalise this to a setting where $\tau$ is now a variable in $\mathbb{R}$, this becomes a function of $\tau$. Taking the derivative with respect to $\tau$ and evaluating at $\tau = 0$, gives:
$$
\frac{u''(c(t))}{u'(c(t))} \dot c(t).
$$
which is the left hand side for the continuous case. As such the left hand side for the continuous case can be seen as the "derivative" of the discrete setting (when allowing the time interval to go to zero).


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### Answer by username_1 on 2021-04-27. Score: 6

You cannot completely ignore the RHS. Starting with
$$\frac{U'(c\_{t+1})}{U'(c\_{t})}=RHS,$$
replace $t+1$ by $t+\Delta t$ to get
$$\frac{U'(c(t+\Delta t))}{U'(c(t))}=RHS\_{\Delta t},$$
where $RHS\_{\Delta t}$ is the modified version of $RHS$ which contains terms depending on $\Delta t$, e.g. the modified discount factor. Expanding around $c(t)$ (and neglecting higher order terms for brevity) then gives you
$$\frac{U''(c(t))}{U'(c(t))}\dot{c}(t)=\frac{RHS\_{\Delta t}-1}{\Delta t},$$
and letting $\Delta t\rightarrow 0$, everything should fall into place.

> **Comments:**
> **username_0**: What do you mean by "expanding around c(t)" in this context. I usually hear that term in reference to taylor expansions.  
> **username_2**: that's exactly what he/she means: write `U'` in terms of `U''` around  `dt`  
> **username_1**: @username_0, yes, it's a Taylor expansion. The equation suppresses higher order terms, which I clarified in the term added in brackets now.  

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