solving linear equations made difficult - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T08:52:38Z http://mathoverflow.net/feeds/question/102423 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102423/solving-linear-equations-made-difficult solving linear equations made difficult James Propp 2012-07-17T04:13:26Z 2012-07-17T12:24:07Z <p>(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.)</p> <p>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):</p> <p>"<em>Problem</em>: Solve $x = ax + b$ for $x$.<br> <em>Solution</em>: $$x = a(ax+b) + b = a^2 x + ab + b = a(a(ax+b)+b)+b= a^3 x + a^2 b + ab + b = \cdots$$ (assuming $|a| &lt; 1$) $$= \lim_{n \rightarrow \infty} a^n x + b \sum_{i=0}^{\infty} a^i = 0 + b/(1-a).$$ This also holds by analytic continuation for all $a \neq 1$."</p> <p>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.</p> <p>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.</p> <p>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).</p> http://mathoverflow.net/questions/102423/solving-linear-equations-made-difficult/102441#102441 Answer by Martin Brandenburg for solving linear equations made difficult Martin Brandenburg 2012-07-17T12:24:07Z 2012-07-17T12:24:07Z <p>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 <a href="http://mathoverflow.net/questions/31595/how-would-you-solve-this-tantalizing-halmos-problem" rel="nofollow">here</a>). 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.</p>