Wikipedia:Reference desk/Archives/Mathematics/2012 July 5#Solving an equation for a variable

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)

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

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

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

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

$p-1=\backslash \; ?$

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 $p=\backslash \; ?$ 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

:$e^\{(p-1)\backslash log\; 2\}-1-up^2=0.$

Substitute

:$x=(p-1)\backslash log\; 2$

:$p=1+\backslash frac\; x\{\backslash log\; 2\}$

get the equation

:$e^x-1-u-\backslash frac\; \{2u\}\{\backslash log\; 2\}x-\backslash frac\; u\{(\backslash log\; 2)^2\}x^2=0$

Expand the exponential function as a power series

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

or

:$-u+\backslash left(1-\backslash frac\; \{2u\}\{\backslash log\; 2\}\backslash right)x+\backslash left(\backslash frac\{1\}\{2\}-\backslash frac\; u\{(\backslash log\; 2)^2\}\backslash right)x^2+\backslash sum\_\{i=3\}^\backslash infty\; \backslash frac\{1\}\{i!\}x^i=0$

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

:$-u+\backslash left(1-\backslash frac\; \{2u\}\{\backslash log\; 2\}\backslash right)x=0$

having the solution

:$x=\backslash frac\; \{u\backslash log\; 2\}\{\backslash log\; 2-2u\}$

such that

:$p=1+\backslash frac\; \{u\}\{\backslash log\; 2-2u\}$

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