[math-fun] solving linear equations made difficult
I saw this amusing derivation on the blackboard at MSRI a couple of months ago (I'm paraphrasing and reformatting slightly): "Problem: Solve x = ax + b for x. Solution: x = a(ax+b) + b = a^2 x + ab + b = a(a(ax+b)+b)+b= a^3 x + a^2 b + ab + b ... = (assuming |a| < 1) lim_{n \rightarrow \infty} a^n x + b sum_{i=0}^{\infty} a^i = 0 + b/(1-a). This also holds by analytic continuation for all a neq 1." Has anyone seen this before? I took a photograph of the blackboard, and I am inclined to submit it to Mathematics Magazine, but first I want to know the provenance. Curt McMullen was in residence at MSRI at the time, and he seemed a likely culprit, but when I pointed it out to him he seemed amused, and he denied authorship, so I don't have any suspects at present. Jim Propp
="James Propp" <jamespropp@gmail.com> I saw this amusing derivation on the blackboard at MSRI [...]
"Problem: Solve x = ax + b for x. Solution: x = a(ax+b) + b = a^2 x + ab + b = a(a(ax+b)+b)+b= a^3 x + a^2 b + ab + b ... = (assuming |a| < 1) lim_{n \rightarrow \infty} a^n x + b sum_{i=0}^{\infty} a^i = 0 + b/(1-a). This also holds by analytic continuation for all a neq 1."
Has anyone seen this before? [...]
Heh, haven't seen it, but that IS amusing. However--just for fun doubtless taking this far more seriously than intended--I really like how it casts the problem as finding a fixpoint rather than just expression shuffling, and how it leverages infinite symbolic objects. I'd even advocate dropping that "(assuming |a| < 1)" and blithely proceed symbolically (trusting in the spirit of Euler) and only lastly go back and ponder what happens when we bind the resulting abstract forms in different concrete domains. For example it shows, as recently discussed here, why binary ...111 = -1 (heh, justifying two's complement arithmetic by analytic continuation?!) The pattern also generalizes to, say, evaluating continued fractions, etc.
For example it shows, as recently discussed here, why binary ...111 = -1 (heh, justifying two's complement arithmetic by analytic continuation?!)
That's also how negative integers are represented as 2-adics. Indeed, one can view the [field?] of 2-adics as a generalisation of two's complement capable of representing rationals (and other things, as well). For example, we can divide ...111 by 111 (decimal 7) to obtain the 2-adic expansion for minus one-seventh: ...001001001001001 and obtain one-seventh by taking the two's complement: ...110110110110111. Sincerely, Adam P. Goucher
participants (3)
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Adam P. Goucher -
James Propp -
Marc LeBrun