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When I look at the Taylor series for $e^x$ and the volume formula for oriented simplexes, it makes $e^x$ look like it is, at least almost, the sum of simplexes volumes from $n$ to $\infty$. Does anyone know of a stronger relationship beyond, "they sort of look similar"?

Here are some links:
Volume formula
http://en.wikipedia.org/wiki/Simplex#Geometric_properties

Taylor Series
http://en.wikipedia.org/wiki/E_%28mathematical_constant%29#Complex_numbers

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  • $\begingroup$ Do you mean $\sum_{n=0}^\infty \text{volume of unit-}n\text{-simplex}$? $\endgroup$
    – kennytm
    Commented Jul 23, 2010 at 17:58
  • $\begingroup$ The function e^x is the solution of functional equation exp(x+y)=exp(x)exp(y) s.t. exp'(0)=1. I wonder, if one can see that the generating function for simplex volumes satisfies this equation... $\endgroup$
    – Grigory M
    Commented Jul 23, 2010 at 18:00
  • $\begingroup$ @Kenny I messed up the question, I meant e^x, is that what is confusing you? $\endgroup$ Commented Jul 23, 2010 at 18:15
  • $\begingroup$ @Jon: No, I was asking the definition of "sum of simplexes volumes from n to infinity". $\endgroup$
    – kennytm
    Commented Jul 23, 2010 at 19:09
  • $\begingroup$ Oh yes. But not neccarily unity. Depends on what x is. Like e^1.5i could be thought of as adding and subtracting oriented volumes that are not unity ... I think $\endgroup$ Commented Jul 23, 2010 at 20:32

1 Answer 1

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The answer is, it's just a fact “cone over a simplex is a simplex” rewritten in terms of the generating function:

observe that because n-simplex is a cone over (n-1)-simplex $\frac{\partial}{\partial x}vol(\text{n-simplex w. edge x}) = vol(\text{(n-1)-simplex w. edge x})$; in other words $e(x):=\sum_n vol\text{(n-simplex w. edge x)}$ satisfies an equvation $e'(x)=e(x)$. So $e(x)=Ce^x$ -- and C=1 because e(0)=1.

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  • $\begingroup$ I think understand the basic idea. The relationship between the border of the simplex and its volume, is such it can phrased in a way that satisfy's the same functional equation that equation that e satisfies, mainly that it's own derivative? Is that close? $\endgroup$ Commented Jul 26, 2010 at 18:26
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    $\begingroup$ @Jonathan Yes, something like this (I'd say "n-dimensional simplex is constructed from (n-1)-dimensional in such way that..."). In combinatorics such things happen quite often: you write down a generating function for something and then observe that it satisfies some simple differential equation (coming from reccurence relation on that something); and when you're solving differential equation you often encounter something like e^x (because it satisfies f'=f, indeed). $\endgroup$
    – Grigory M
    Commented Jul 27, 2010 at 5:52

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