# The Unapologetic Mathematician

## Another way to use the FToC

Let’s look back at that diagram that encapsulated the FToC. I want to point out something that’s going to seem really silly at first. Bear with me, though, because it turns out to be a lot deeper than it appears.

We used this diagram before to connect antidifferentiation to integration. We noted that we can easily find the boundary of an interval, and if we’re lucky we can find an antiderivative of the function to be integrated. Then we can move from the right side of the diagram to the left.

Today, I want to go the other way. Let’s say we’ve got a function $F$ and a collection $S$ of signed points. The signed sum of values of $F$ on the points in $S$ lives on the left side of the diagram. We want to move this over to the right.

We can differentiate $F$ to move it to the right, but now the difficult bit comes along the bottom. We need to find a collection of intervals whose boundary is $S$, just like before we needed to find a new function whose derivative was our integrand. As an example, if $S=\{-1^+,0^-,3^-,8^+\}$, we might choose the intervals $\left[0,-1\right]$ and $\left[3,8\right]$. Or we could choose $\left[0,8\right]$ and $\left[3,-1\right]$. Then we can integrate the derivative of $F$ over our collection of intervals, and get the same answer as if we’d added up the (signed) values of $F$ over the points of $S$.

Let’s look at this “antiboundary” process a little more closely. We can use the sign convention for intervals to write the latter intervals in our example as $\{\left[0,8\right],\left[-1,3\right]^-\}$. And then we can split up intervals to write $\{\left[0,3\right],\left[3,8\right],\left[-1,0\right]^-,\left[0,3\right]^-\}$. But when we integrate, the two traversals of $\left[0,3\right]$ in opposite directions will cancel each other out, leaving $\{\left[0,-1\right],\left[3,8\right]\}$. And so it doesn’t matter which “antiboundary” we choose for $S$, since the integrals will be the same anyway. A similar analysis shows that any choice of intervals is just as good as any other, no matter what $S$ is.

The question left hanging, though, is which collections of points arise as boundaries? It’s not too hard to see that any boundary has as many positively-signed points as negatively-signed ones. It turns out that this is also sufficient. Just take a positive and negative pair and use the interval between them. It doesn’t matter which pair you start with, because any collection of intervals with the same boundary is just as good as any other. As long as there are as many positive points as negative points, we can keep going, and we won’t have any points left over at the end.

Really, the dual of this question was there before: which functions show up as derivatives? What we proved was that functions which only have a finite number of discontinuities (none of which are asymptotes) are all in the image of the differentiation operator. So most of the functions we care about can be moved from right to left, and we didn’t really think much about that step. But now there’s a clear obstruction to finding an “antiboundary” for a collection of signed points: the sum of the signs. If this isn’t zero, we can’t move from left to right across the diagram.

In the end, though, this obstruction doesn’t really affect much, because moving from a finite sum to an integral is rather silly. Still, it’s worth noting that there’s a certain duality here in our diagram. Differentiation of functions and boundaries of intervals are somehow related more deeply than they might first appear to be.

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February 21, 2008 - Posted by | Analysis, Calculus

## 2 Comments »

1. Now I understand why Green’s theorem generalises the fundamental theorem of analysis/calculus. Thanks.

Comment by Tommi | May 21, 2009 | Reply

2. No problem. I’ll be getting back to that subject once I return to multivariable calculus.

Comment by John Armstrong | May 21, 2009 | Reply