# Slides — Discrete Exterior Calculus

When it comes to computation, everything we’ve learned about differential forms and exterior calculus boils down to building just a few very simple matrices. These slides contain a bunch of examples that should help with both the written and coding part of your current assignment.

## 5 thoughts on “Slides — Discrete Exterior Calculus”

1. plav says:

I am trying to understand the integration of a 1-Form over an Edge (slide 14).

$\alpha(T) = \frac{(4xy + x^2)}{L}$
$p(s) = p_0 + \frac{S}{L}(p_1 – p_0) = (-1, 2) + \frac{S}{L}((3, 1)-(-1, 2)) = (-1, 2) + \frac{S}{L}(4,-1) = (\frac{4S}{L}-1, -\frac{S}{L}+2)$
$\alpha(T)_P(S) = 4(\frac{4S}{L} – 1)(-\frac{S}{L}+2)-(\frac{4S}{L}-1)^2$

When I try to simplify $\alpha(T)_P(S)$ I do not get to $\frac{7}{17L}(4s-L)$. Is my computation wrong or the initial formula itself wrong?

1. Josh says:

In your calculation, $-(\frac{4s}{L}-1)^2$ should be $+(\frac{4s}{L}-1)^2$. Also you seem to have dropped a factor of $1/L$.

1. plav says:

$\alpha(T) = \frac{(4xy + x^2)}{L}$
\\
$p(s) = p_0 + \frac{S}{L}(p_1 – p_0) = (-1, 2) + \frac{S}{L}((3, 1)-(-1, 2)) = (-1, 2) + \frac{S}{L}(4,-1) = (\frac{4S}{L}-1, -\frac{S}{L}+2)$
\\
$\alpha(T)_P(S) = 4\frac{(\frac{4S}{L} – 1)(-\frac{S}{L}+2)+(\frac{4S}{L}-1)^2}{L}$

Yes… it works. Thanks.

1. plav says:

1) The integration of a k-form with a k-simplex, result in values associated with the vertices, edges and faces of our mesh. Using the Stokes Theorem, we have a very simple formula to compute these values by getting the difference between the bound values. However, in the slide 20 if discrete exterior calculus, the values do not match this logic.

1. Josh says:

The 0-form, 1-form, and 2-form on slide 20 are just examples; they don’t correspond to taking the differential of the earlier ones.