Wikipedia:Reference desk/Archives/Mathematics/2012 July 5

= July 5 =

Cartesian product of an EVEN number of nonorientable manifolds
Is the product orientable?--Richard Peterson76.218.104.120 (talk) 05:36, 5 July 2012 (UTC)

NO, AxB is orientable iff both are.--刻意(Kèyì) 07:16, 5 July 2012 (UTC)
 * Thanks.76.218.104.120 (talk) 05:13, 7 July 2012 (UTC)

Solving an equation for a variable
I am a bit lost. I want to solve the equation $$\frac{2^{p-1}-1}{p}=pu$$ for p? I guess what I have to do is

$$\frac{2^{p-1}-1}{p}=pu \qquad \Big| \cdot p$$

$$2^{p-1}-1=p^2\cdot u \qquad \Big| +1$$

$$2^{p-1}=p^2 u + 1 \qquad \Big| \log_{2}$$

$$p-1=\ ?$$

How do I continue, ie how do I apply the binary logarithm to the right-hand side of the equation? -- Toshio Yamaguchi (tlk−ctb) 09:41, 5 July 2012 (UTC)

I am not even sure, whether I am on the right track. What I want to do is expressing p as a function of u, so that I have something like $$p=\ ?$$ with only u on the right-hand side. -- Toshio Yamaguchi (tlk−ctb) 10:17, 5 July 2012 (UTC)


 * You seem to be searching for Wieferich primes. There are only two known primes p with this property, and there is no known formula for generating other values for p. Gandalf61 (talk) 14:14, 5 July 2012 (UTC)


 * I think that might be too sophisticated an answer. The basic answer is that the equation cannot be solved in closed form -- there is no simple algebraic expression for p as a function of u. Looie496 (talk) 16:36, 5 July 2012 (UTC)


 * Yepp, Gandalf is right, I am in fact looking at this equation due to my interest in Wieferich primes. -- Toshio Yamaguchi (tlk−ctb) 09:39, 6 July 2012 (UTC)


 * The Lambert W function is often useful for expressing the solution to equations involving both an exponential and a polynomial. Not this equation though. -- Meni Rosenfeld (talk) 18:54, 5 July 2012 (UTC)

The equation
 * $$2^{p-1}=p^2 u + 1$$

is written
 * $$e^{(p-1)\log 2}-1-up^2=0.$$

Substitute
 * $$x=(p-1)\log 2$$
 * $$p=1+\frac x{\log 2}$$

get the equation
 * $$e^x-1-u-\frac {2u}{\log 2}x-\frac u{(\log 2)^2}x^2=0$$

Expand the exponential function as a power series
 * $$\left(\sum_{i=0}^\infty \frac{1}{i!}x^i\right)-1-u-\frac {2u}{\log 2}x-\frac u{(\log 2)^2}x^2=0$$

or
 * $$-u+\left(1-\frac {2u}{\log 2}\right)x+\left(\frac{1}{2}-\frac u{(\log 2)^2}\right)x^2+\sum_{i=3}^\infty \frac{1}{i!}x^i=0$$

Truncate to finite degree and solve numerically by a standard root-finding algorithm. For very small values of u the approximate equation is
 * $$-u+\left(1-\frac {2u}{\log 2}\right)x=0$$

having the solution
 * $$x=\frac {u\log 2}{\log 2-2u}$$

such that
 * $$p=1+\frac {u}{\log 2-2u}$$

Bo Jacoby (talk) 08:56, 6 July 2012 (UTC).