Wikipedia:Reference desk/Archives/Mathematics/2008 October 24
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October 24
editMatroid morphism
editHow 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
editI 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 as 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 (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)
- 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)
Mdob, you seem to think 33 = 28. That is incorrect. Michael Hardy (talk) 02:05, 25 October 2008 (UTC)
- Uh? , 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:
and this is a quadratic equation. Plugging the three coefficients into the quadratic formula and simplifying, we get
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
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):
So you'll need 3u = 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)