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Is there an theorem that says that you can't compute e from pi? Maybe this is just the second term/iteration of that algorithm. I'd be interested in seeing ot

• knz42 12y ago
  e^(i * pi) + 1 = 0
• nhaehnle 12y ago
  A related mathematical concept is that of algebraic independence. Two numbers x and y are algebraically dependent (over the rationals) if there is a rational bivariate polynomial p such that p(x,y) = 0. As far as I know, pi and e are not known to be algebraically independent, though I guess most people would expect them to be. On the other hand, it is true that there must be some (actually infinitely many) infinite series in x with rational coefficients such that plugging x = pi gives e in the limit, but in itself, that's a trivial statement (just build the coefficients of the series in such a way that you force an approximation of e). It is also not a very interesting statement, because e is the limit of a series more or less by definition (what I mean here is that plugging 1 into the series that defines the exponential function). A much more interesting relationship is Euler's identity, e^(i pi) = -1. There are many more formulas in which e and pi appear together in more or less interesting ways once you start going deeper into mathematics, but they tend to be not of the form discussed in this submission (I'm partial to Stirling's approximation as the "next" step after Euler's identity).
  • tgb 12y ago
    Wow, as someone who is studying math I'm surprised to hear that that isn't known. Wikipedia even says that it's not known whether e + pi is irrational.