Wikipedia:Reference desk/Archives/Mathematics/2008 October 24

= 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

Help with problem
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 3w is positive, so you'd be adding positive numbers and getting zero, which is impossible. However, supposing u = 3x, 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
 * $$ u = \frac{-1}{3}\text{ or }u = -9. $$
 * $$ u = \frac{-1}{3}\text{ or }u = -9. $$

So we need As I said, if x is a real number, then 3x is positive, so let us seek solutions that are non-real complex numbers. Now
 * $$ 3^x = u = -9\text{ or }3^x = u = \frac{-1}{3}. $$
 * $$ 3^x = u = -9\text{ or }3^x = u = \frac{-1}{3}. $$
 * $$ 3^x = e^{x\ln 3}.\, $$
 * $$ 3^x = e^{x\ln 3}.\, $$

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 &minus; 28x + 9 = 0, since that would work out much more neatly, with real numbers):
 * $$ 3^x = e^{(u + iv)\ln 3} = e^{u\ln3}(\cos(v\ln3) + i\sin(v\ln3)
 * $$ 3^x = e^{(u + iv)\ln 3} = e^{u\ln3}(\cos(v\ln3) + i\sin(v\ln3)

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

So you'll need 3u = 9, cos = &minus;1, and sin = 0. That gives us u = 2, and v ln 3 = &pi; ± 2&pi;n., 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)