<p>I think that this question would be more suitable for <a href="http://math.stackexchange.com">math.SE</a>, since it is purely mathematical and not specific to economics.</p>

<p>In your notation, the first derivative would be given by
$$
\partial_A v_i = \frac{v_{i+1} - v_i}{\Delta}.
$$
In order to compute the second derivative, we need to apply this formula with $\partial_A v_i$ itself in place of $v$. This gives
$$
\partial_{AA} v_i = \frac{ \frac{v_{i+2} - v_{i+1}}{\Delta} - \frac{v_{i+1} - v_i}{\Delta}}{\Delta}.
$$
A bit of algebra reveals that this is the same as
$$
\partial_{AA} v_i = \frac{v_{i+2} - 2v_{i+1} + v_i}{\Delta^2}.
$$
Now if your grid is fine enough, you can lower the index $i$ on the right-hand side without changing the result significantly. This gives the equation that you came across. It is generally preferred due to the "symmetry" that it displays between $i-1$ and $i+1$.</p>

<p>For more on this topic, see e.g. Wikipedia at <a href="https://en.wikipedia.org/wiki/Finite_difference" rel="nofollow">finite difference</a>.</p>

CC BY-SA 3.0
2015-03-02T00:16:13.487Z
Perfect! Short addendum: From what I know, the difference in "lowering the index" or not is whether one takes the forward or the backwards difference, where you proceeded twice with forward, and apparently they did once forward and once backwards (perhaps that's goods against rounding errors, or perhaps it won't matter at all).
FooBar
CC BY-SA 3.0
2015-03-02T00:02:38.450Z
2015-03-02T00:02:38.450Z
Tobias Fritz
<p>Let's say we want to approximate a second order differential:</p>

<p>$$ \partial_{aa} V(a)$$</p>

<p>Now, say we have a grid of values for $A$, where the step size is $\Delta$, $i$ is the index on this A grid, and we define the short hand notation</p>

<p>$$V_i \equiv V(A_i)$$</p>

<p><a href="http://www.princeton.edu/~moll/HACTproject/HACT_Numerical_Appendix.pdf" rel="nofollow">Achdou et al</a> (page 13, equation 35) then compute the approximation to the second order differential on the grid at a point $i$ as</p>

<p>$$\partial_{AA} v_i =  \frac{v_{i+1} - 2v_i + v_{i-1}}{\Delta^2}$$</p>

<p>They don't explain why, and I can't follow here. Could someone provide me with the rough ansatz for why this is the case?</p>

<mathematical-economics>
Approximation to a second order partial derivative
FooBar
# Approximation to a second order partial derivative

(Asked by: username_0 on 2015-03-01)

Let's say we want to approximate a second order differential:

$$ \partial\_{aa} V(a)$$

Now, say we have a grid of values for $A$, where the step size is $\Delta$, $i$ is the index on this A grid, and we define the short hand notation

$$V\_i \equiv V(A\_i)$$

[Achdou et al](http://www.princeton.edu/~moll/HACTproject/HACT_Numerical_Appendix.pdf) (page 13, equation 35) then compute the approximation to the second order differential on the grid at a point $i$ as

$$\partial\_{AA} v\_i = \frac{v\_{i+1} - 2v\_i + v\_{i-1}}{\Delta^2}$$

They don't explain why, and I can't follow here. Could someone provide me with the rough ansatz for why this is the case?

**Tags:** mathematical-economics


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## 1 answers

### Answer by username_1 on 2015-03-02. Score: 2 | Accepted answer

I think that this question would be more suitable for [math.SE](http://math.stackexchange.com), since it is purely mathematical and not specific to economics.

In your notation, the first derivative would be given by
$$
\partial\_A v\_i = \frac{v\_{i+1} - v\_i}{\Delta}.
$$
In order to compute the second derivative, we need to apply this formula with $\partial\_A v\_i$ itself in place of $v$. This gives
$$
\partial\_{AA} v\_i = \frac{ \frac{v\_{i+2} - v\_{i+1}}{\Delta} - \frac{v\_{i+1} - v\_i}{\Delta}}{\Delta}.
$$
A bit of algebra reveals that this is the same as
$$
\partial\_{AA} v\_i = \frac{v\_{i+2} - 2v\_{i+1} + v\_i}{\Delta^2}.
$$
Now if your grid is fine enough, you can lower the index $i$ on the right-hand side without changing the result significantly. This gives the equation that you came across. It is generally preferred due to the "symmetry" that it displays between $i-1$ and $i+1$.

For more on this topic, see e.g. Wikipedia at [finite difference](https://en.wikipedia.org/wiki/Finite_difference).

> **Comments:**
> **username_0**: Perfect! Short addendum: From what I know, the difference in "lowering the index" or not is whether one takes the forward or the backwards difference, where you proceeded twice with forward, and apparently they did once forward and once backwards (perhaps that's goods against rounding errors, or perhaps it won't matter at all).  

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