Franz Scheerer (Olbers)

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Hello, Franz Scheerer (Olbers), and welcome to Wikipedia! Thank you for your contributions, especially what you did for Magnetic field. I hope you like the place and decide to stay. Here are a few links to pages you might find helpful:

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Please remember to sign your messages on talk pages by typing four tildes (~~~~); this will automatically insert your username and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or ask your question on this page and then place {{help me}} before the question. Again, welcome! RockMagnetist (talk) 17:31, 29 October 2013 (UTC)

October 2013

Welcome to Wikipedia. We welcome and appreciate your contributions, including your edits to Magnetic field, but we cannot accept original research. Original research also encompasses combining published sources in a way to imply something that none of them explicitly say. Please be prepared to cite a reliable source for all of your contributions. Thank you. RockMagnetist (talk) 17:32, 29 October 2013 (UTC)

Collatz conjecture

Your edits here, so far, might possibly be suitable for Wikibooks or Wikiversity; they are not suitable here, because there is no source outside of your edits for the material. The (3/2)^k analysis, as I said on the article talk page, might be suitable here if a reputable mathematician stated it. Unfortunately, some of your posts and websites clearly indicate that mathematician is not anyone named "Franz Scheerer". — Arthur Rubin (talk) 14:41, 8 January 2014 (UTC)

Good morning Arthur Rubin, I tried something new. If one start the some number s the numbers n alway can be written as

$n=a\cdot s+b$

with a,b fractional numbers. If known which operations (3n+1)/2 or n/2 is performed the new a,b can be calculated as follows.

$a'=(3/2)a\,or\,(1/2)a$

Finally a gets the value

$a_{final}=(3^{c}/2^{l})$

and as well b

$b'=(3/2)b+1/2\,\,or\,(1/2)b$

Now we can ask wether a cycle occurs

$a_{final}\,s+b_{final}=s$

or

$s={\frac {b_{final}}{a_{final}-1}}$

Using $a_{final}=(3^{c}/2^{l})$ we obtain

$s={\frac {2^{l}\cdot b_{final}}{2^{l}-3^{c}}}$

If the first operation is (3n+1)/2 and the other (3n+1)/2 operations follow directly, we can derive

$b_{final}={\frac {3^{c}-2^{c}}{2^{l}}}$

and final get

$s={\frac {3^{c}-2^{c}}{2^{l}-3^{c}}}$

For c=1, l=2 we get s=1.

If the (3n+1)/2 don't follow directly on each other or we don't start with the minimum still

$s>={\frac {3^{c}-2^{c}}{2^{l}-3^{c}}}={\frac {1-2^{c}/3^{c}}{2^{l}/3^{c}-1}}={\frac {1-(2/3)^{c}}{2^{l-c}(2/3)^{c}-1}}$.

A minimum is, starting with (n/2) divisions

$s<=2^{l-c}{\frac {3^{c}-2^{c}}{2^{l}-3^{c}}}=2^{l-c}{\frac {1-2^{c}/3^{c}}{2^{l}/3^{c}-1}}=2^{l-c}{\frac {1-(2/3)^{c}}{2^{l-c}(2/3)^{c}-1}}$.

Franz Scheerer (Olbers) (talk) 21:40, 22 January 2014 (UTC)