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diff --git a/solve_poisson.m b/solve_poisson.m
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-function u = solve_poisson(integerN, f)
-  ##
-  ## Numerically solve the poisson equation,
-  ##
-  ##   -u_xx(x) = f(x)
-  ##
-  ## in one dimension, subject to the boundary conditions,
-  ##
-  ##   u(0) = 0
-  ##   u(1) = 1
-  ##
-  ## over the interval [0,1]. It is assumed that the function `f` is at
-  ## least continuous on [0,1].
-  ##
-  ## INPUTS:
-  ##
-  ##   * ``integerN`` - An integer representing the number of subintervals
-  ##     we should use to approximate `u`.
-  ##
-  ##   * ``f`` - The function on the right hand side of the poisson
-  ##     equation.
-  ##
-  ## OUTPUTS:
-  ##
-  ##   * u - The function `u` that approximately solves -u_xx(x) = f(x).
-  ##
-
-  [xs,h] = partition(integerN, 0, 1);
-
-  ## We cannot evaluate u_xx at the endpoints because our
-  ## differentiation algorithm relies on the points directly to the left
-  ## and right of `x`.
-  differentiable_points = xs(2:end-1)
-
-  ## Prepend and append a zero row to f(xs).
-  ## Oh, and negate the whole thing (see the formula).
-  F = - cat(2, 0, f(differentiable_points), 0)
-
-  ## These are the coefficient vectors for the u(x0) and u(xn)
-  ## constraints. There should be N zeros and a single 1.
-  the_rest_zeros = zeros(1, integerN);
-  u_x0_coeffs = cat(2, 1, the_rest_zeros);
-  u_x4_coeffs = cat(2, the_rest_zeros, 1);
-
-  ## Start with the u(x0) row.
-  A = u_x0_coeffs;
-
-  for x = differentiable_points
-    ## Append each row obtained from the forward Euler method to A.
-    u_row = forward_euler(2, xs, x);
-    A = cat(1, A, u_row);
-  end
-
-  ## Finally, append the last row for x4.
-  A = cat(1, A, u_x4_coeffs);
-
-  ## Solve the system. This gives us the value of u(x) at xs.
-  U = A \ F';
-
-  plot(xs,U)
-end