Wikipedia: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

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

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

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

$p-1=\ ?$

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).