<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