Why is $\frac{\pi^2}{12}=ln(2)$ not true ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T11:29:18Z http://mathoverflow.net/feeds/question/27592 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27592/why-is-frac-pi212ln2-not-true Why is $\frac{\pi^2}{12}=ln(2)$ not true ? Max Muller 2010-06-09T15:50:09Z 2010-07-08T12:44:59Z <p>This question may sound ridiculous at first sight, but let me please show you all how I arrived at the afore mentioned 'identity'.</p> <p>Let us begin with (one of the many) equalities established by Euler:</p> <p>$$\displaystyle f(x) = \frac{sin(x)}{x} = \prod_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\pi^2}\Big)$$</p> <p>as $(a^2-b^2)=(a+b)(a-b)$, we can also write: (EDIT: We can not write this...)</p> <p>$$\displaystyle f(x) = \prod_{n=1}^{\infty} \Big(1+\frac{x}{n\pi}\Big) \cdot \prod_{n=1}^{\infty} \Big(1-\frac{x}{n\pi}\Big)$$ </p> <p>We now we arrange the terms with $(n = 1 \land n=-2)$, $(n = -1 \land n=2$), $( n=3 \land -4)$ , $(n=-3 \land n=4)$ , ..., $(n = 2n \land n=-2n-1)$ and $(n=-2n \land n=2n+1)$ together . After doing so, we multiply the terms accordingly to the arrangement. If we write out the products, we get:</p> <p>$$f(x)=\big((1-x/2\pi + x/\pi -x^2/2\pi^2)(1+x/2\pi-x/\pi - x^2/2\pi^2)\big)...$$ $$...\big((1-\frac{x}{(2n)\pi} + \frac{x}{(2n-1)\pi} -\frac{x^2}{(2n(n-1))^2\pi^2})(1+\frac{x}{2n\pi} -\frac{x}{(2n-1)\pi} -\frac{x^2}{(2n(2n-1))^2\pi^2)})\big)$$</p> <p>Now we equate the $x^2$-term of this infinite product, using <a href="http://en.wikipedia.org/wiki/Newton%27s_identities" rel="nofollow"> Newton's identities</a> (notice that the'$x$'-terms are eliminated) to the $x^2$-term of the Taylor-expansion series of $\frac{sin(x)}{x}$ . So,</p> <p>$$-\frac{2}{\pi^2}\Big(\frac{1}{1\cdot2} + \frac{1}{3\cdot4} + \frac{1}{5\cdot6} + ... + \frac{1}{2n(2n-1)}\Big) = -\frac{1}{6}$$</p> <p>Multiplying both sides by $-\pi^2$ and dividing by 2 yields</p> <p>$$\sum_{n=1}^{\infty} \frac{1}{2n(2n-1)} = \pi^2/12$$</p> <p>That (infinite) sum 'also' equates $ln(2)$, however (According to the last section of <a href="http://www.stat.purdue.edu/~dasgupta/publications/tr02-03.pdf" rel="nofollow"> this </a> paper).</p> <p>So we find $$\frac{\pi^2}{12} = ln(2)$$ .</p> <p>Of course we all know that this is not true (you can verify it by checking the first couple of digits). I'd like to know how much of this method, which I used to arive at this absurd conclusion, is true, where it goes wrong and how it can be improved to make it work in this and perhaps other cases (series). </p> <p>Thanks in advance, </p> <p>Max Muller</p> <p>(note I: 'ln' means 'natural logarithm) (note II: with 'to make it work' means: 'to find the exact value of)</p> http://mathoverflow.net/questions/27592/why-is-frac-pi212ln2-not-true/27596#27596 Answer by Helge for Why is $\frac{\pi^2}{12}=ln(2)$ not true ? Helge 2010-06-09T15:58:52Z 2010-06-09T15:58:52Z <p>You cannot split (1-(x/n)^2) into (1 -x/n) (1 + x/n), since the products no longer converge.</p> http://mathoverflow.net/questions/27592/why-is-frac-pi212ln2-not-true/27679#27679 Answer by Franz Lemmermeyer for Why is $\frac{\pi^2}{12}=ln(2)$ not true ? Franz Lemmermeyer 2010-06-10T11:33:30Z 2010-06-10T11:33:30Z <p>Eisenstein defined elliptic functions by working with conditionally convergent series. In particular he studied how a series changes when you rearrange the terms in a specific way. You can find a lot about his work in this direction in Weil's beautiful book <a href="http://books.google.de/books?id=voR95sDdb_MC&amp;printsec=frontcover&amp;dq=weil+eisenstein&amp;source=bl&amp;ots=nhaL-B34rr&amp;sig=K4P27Xpb3gOFMEZiWlMblemC9JY&amp;hl=de&amp;ei=kswQTPG5OJOoOLzG0M0H&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=5&amp;ved=0CDMQ6AEwBA#v=onepage&amp;q&amp;f=false" rel="nofollow">Elliptic Functions according to Eisenstein and Kronecker</a>. An analogous question would be what happens to your product formula when you use a different way of pairing positive and negative indices. I do not know whether this has been studied before . . . A look into Weil's book will convince you (if you didn't know that already) that some functions are most interesting at those places where convergence fails.</p> http://mathoverflow.net/questions/27592/why-is-frac-pi212ln2-not-true/27688#27688 Answer by Gerald Edgar for Why is $\frac{\pi^2}{12}=ln(2)$ not true ? Gerald Edgar 2010-06-10T12:59:30Z 2010-06-10T12:59:30Z <p>It is a common trick, found in many elementary calculus texts: take a conditionally convergent series, and rearrange it to have any sum you like.</p>