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

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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

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

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

$2^{p-1}=p^{2}u+1|{\log }_2$

$p-1=?$

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

$e^{(p-1)\log 2}-1-u{p}^2=0.$

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

$\left(\sum _{i=0}^{\mathrm{\infty }}\frac{1}{i!}{x}^i\right)-1-u-\frac{2u}{\log 2}x-\frac{u}{(\log 2{)}^2}{x}^2=0$

or

$-u+(1-\frac{2u}{\log 2})x+(\frac{1}{2}-\frac{u}{(\log 2{)}^2})x^2+\sum _{i=3}^{\mathrm{\infty }}\frac{1}{i!}{x}^i=0$

Truncate to finite degree and solve numerically by a standard root-finding algorithm. For very small values of u the approximate equation is

$-u+(1-\frac{2u}{\log 2})x=0$

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).