<img src="./images/cubical_lattice.png" align="center" style="width: 300px";/>
If you put a positive test charge at the center of this cube of charges, could it be in stable equilibrium?
1. Yes
2. No
3. ???
Note:
* CORRECT ANSWER: B
### Method of Relaxation
<img src="./images/convergence_relax.png" align="center" style="width: 700px";/>
Consider a function $f(x)$ that is both continuous and continuously differentiable over some domain. Given a step size of $a$, which could be an approximate derivative of this function somewhere in that domain? $df/dx \approx$
1. $f(x_i+a) - f(x_i)$
2. $f(x_i) - f(x_i-a)$
3. $\frac{f(x_i+a) - f(x_i)}{a}$
4. $\frac{f(x_i) - f(x_i-a)}{a}$
5. More than one of these
Note:
* Correct Answer: E (C and D)
If we choose to use:
$$\dfrac{df}{dx} \approx \dfrac{f(x_i+a) - f(x_i)}{a}$$
Where are we computing the approximate derivative?
1. $a$
2. $x_i$
3. $x_i + a$
4. Somewhere else
Note:
* Correct Answer: D (it's halfway between)
Taking the second derivative of $f(x)$ discretely is as simple as applying the discrete definition of the derivative,
$$f''(x_i) \approx \dfrac{f'(x_i + a/2) - f'(x_i - a/2)}{a}$$
Derive the second derivative in terms of $f$.
To investigate the convergence, we must compare the estimate of $V$ before and after each calculation. For our 1D relaxation code, $V$ will be a 1D array. For the kth estimate, we can compare $V_k$ against its previous value by simply taking the difference.
Store this in a variable called ``err``. What is the type for ``err``?
1. A single number
2. A 1D array
3. A 2D array
4. ???
Note:
* Coreect Answer: B