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(Note: This is a what's-in-the-literature question, not a what's-mathematically-true question, but I believe both are considered valid kinds of MathOverflow question.)

I saw this amusing derivation on the blackboard at MSRI a few months ago (I'm paraphrasing and reformatting slightly, though my attempts at formatting may not work as intended):

"Problem: Solve $x = ax + b$ for $x$.
Solution: $$x = a(ax+b) + b = a^2 x + ab + b = a(a(ax+b)+b)+b= a^3 x + a^2 b + ab + b = \cdots$$ (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.

It would be embarrassing to publish this and then receive letters saying "This argument appears almost word-for-word in Littlewood's Miscellany" (or something like that).

James Propp
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    This seems to be inspired by the standard proof of Picard-Lindelöf (notice the step which can be phrased as an appeal to the Banach fixed point theorem). – Qiaochu Yuan Jul 17 '12 at 05:26
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    This reminds me the classical fact that if the spectral radius of $A$ is less than $1$, then $I_n-A$ is non-singular and equals the sum of the geometrical series. – Denis Serre Jul 17 '12 at 05:33
  • Sorry, the sum in "i" in the formula should be up to "n" not to \inf , is not it ? – Alexander Chervov Jul 17 '12 at 06:06
  • Why not to solve quadratic equation in this difficult way ?:) Actually any equation... – Alexander Chervov Jul 17 '12 at 07:35
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    A silly and obvious observation is that it is of course a cute reversal of the usual hand-waving way to sum the geometric series: if $S=\sum_{k\ge0}ba^k$, we instantly see that $S=b+aS$, so $S=\frac{b}{1-a}$. – Vladimir Dotsenko Jul 17 '12 at 08:47
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    Indeed I do not see anything unusual in the computation of the question. For some numerical values, this is also the numeric solution (e.g. having to evaluate $x=8/93$, in elementary school they learn, or used to learn, to write $(100-7)x=8$ or $x=0.08+0.07x=0.08+ 0.07(0.08+0.07x) $ and get $x\sim $0,086. Also, it is also how they learn to switch from fractions to decimal expansions, and conversely, in the representations of rational numbers. – Pietro Majer Jul 17 '12 at 11:45
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    What Qiaochu says is the key point: the map $f: f(x) = ax +b$ is a contraction mapping on $\mathbb{R}$ as long as $|a| < 1.$ Hence the sequence $(f^{n}(x))$ is Cauchy and converges to the unique fixed point of $f$ for every real number $x.$ The fact that $f$ happens to be semilinear is a secondary issue. – Geoff Robinson Jul 17 '12 at 12:05
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    Maybe I'm being dumb, but I can't understand this sudden interest for $2=1+1/2+1/4+1/8+\dots$, at MO! Why not $(a+b)(a-b)=a^2-b^2$ then? – Pietro Majer Jul 17 '12 at 21:59

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Actually this calculation has a formal sense in every ring, by working in the ring of formal power series in $a$ (here $1-a$ is invertible with inverse $\sum_{i \geq 0} a^i$). There are many "pseudo-analytic" proofs in ring theory (one was discussed here). I've made this CW because I cannot answer the question whether this has appeared in the literature, but I am pretty sure that it has.

Community
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Martin Brandenburg
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