Wikipedia:Reference desk/Archives/Mathematics/2008 October 24#Matroid morphism

How do you define matroid homomorphism (matroid map, morphism of matroids)? 212.87.13.70 (talk) 10:09, 24 October 2008 (UTC)

There are two candidates: strong map relation and weak map relation. The latter is defined by:

The inverse image of any independent set is an independent set.

For strong map relation see http://www.sciencedirect.com/science/article/pii/S0012365X05004231

I need HELP: 3^(2x+1)+(28*3^(x))+9=0 —Preceding unsigned comment added by 72.205.199.12 (talk) 23:09, 24 October 2008 (UTC)

:Rewrite $3^\{2x+1\}\; +\; 28\; *\; 3^x\; +\; 9\; =\; 0$ as $3^\{2x+1\}\; +\; 3^\{3x\}\; +\; 3^x\; +\; 3^2\; =\; 0$Mdob | Talk 00:04, 25 October 2008 (UTC)

::Could you explain? —Preceding unsigned comment added by 72.205.199.12 (talk) 00:21, 25 October 2008 (UTC)

::Mdob, do you mean $3^\{2x+1\}\; +\; 3^\{3+x\}\; +\; 3^x\; +\; 3^2\; =\; 0$ (note, 3+x not 3x)? --Tango (talk) 00:29, 25 October 2008 (UTC)

:::Ooops! Yeah, I botched the equation....:P Thanks again, Tango.

:::There seems to be something odd with this equation... if we extract the logarithm in base 3 of the lhs we get a logarithm of zero in the rhs. What am I doing wrong? Mdob | Talk 01:02, 25 October 2008 (UTC)

:::: It makes no sense at all to take logarithms, since on the left side you get a logarithm of a sum. Please. Don't be that clumsy. Michael Hardy (talk) 02:31, 25 October 2008 (UTC)

:::::Doh! You are correct, of course. mea culpa. thanks. Mdob | Talk 11:34, 25 October 2008 (UTC)

Mdob, you seem to think 3³ = 28. That is incorrect. Michael Hardy (talk) 02:05, 25 October 2008 (UTC)

:Uh? $27*3^x+3^x\; =\; 28*3^x$, I've checked this with the Eigenmath CAS, so my expansion of the middle term is correct. Mdob | Talk 11:34, 25 October 2008 (UTC)

::but the rest of your solution is correct, of course. Mdob | Talk 11:34, 25 October 2008 (UTC)

There are no solutions that are real numbers, since if w is real then 3^w is positive, so you'd be adding positive numbers and getting zero, which is impossible. However, supposing u = 3^x, then we have:

:

: $$

\begin{align}

3^{2x+1} + 28 \cdot 3^x + 9 & = 0 \\

(3^x)^2\cdot 3 + 28\cdot 3^x + 9 & = 0 \\

3u^2 + 28u + 9 & = 0,

\end{align}

:

and this is a quadratic equation. Plugging the three coefficients into the quadratic formula and simplifying, we get

:

: $u\; =\; \backslash frac\{-1\}\{3\}\backslash text\{\; or\; \}u\; =\; -9.$

:

So we need

:

: $3^x\; =\; u\; =\; -9\backslash text\{\; or\; \}3^x\; =\; u\; =\; \backslash frac\{-1\}\{3\}.$

:

As I said, if x is a real number, then 3^x is positive, so let us seek solutions that are non-real complex numbers. Now

:

: $3^x\; =\; e^\{x\backslash ln\; 3\}.\backslash ,$

:

Now suppose u and v are the real and imaginary parts of x respectively. Then (OK, at this point I'm really wondering if maybe you misread 3x² − 28x + 9 = 0, since that would work out much more neatly, with real numbers):

:

: $3^x\; =\; e^\{(u\; +\; iv)\backslash ln\; 3\}\; =\; e^\{u\backslash ln3\}(\backslash cos(v\backslash ln3)\; +\; i\backslash sin(v\backslash ln3)$

= 3^u(\cos(v\ln3) + i\sin(v\ln3).\,

:

So you'll need 3^u = 9, cos = −1, and sin = 0. That gives us u = 2, and v ln 3 = π ± 2πn., etc. Michael Hardy (talk) 02:26, 25 October 2008 (UTC)

:Thanks! —Preceding unsigned comment added by 72.205.199.12 (talk) 17:45, 25 October 2008 (UTC)