Testwiki:Reference desk/Archives/Mathematics/2010 March 9

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March 9

Homework question

For what values of x is the sequence Un = $(x-2{)}^n$ convergent to 0?

The answer is obviously 1 < x < 3 but I have no idea how to prove that. Could someone lead me in the right direction? --124.171.116.21 (talk) 04:38, 9 March 2010 (UTC)

Let $z\in ℝ$ such that $|z|<1$, and let $z_n=z^n$ for each $n\in \mathbb{N}$. Prove that $z_n\to 0$. Once this has been established, write $z=x-2$, and note that $|x-2|<1$ if and only if $|z|<1$, and therefore, if $1<x<3$, $|z|<1$, $u_n=z^n$, and thus $u_n\to 0$.

To complete the above proof, you need to establish that $z_n\to 0$, and that if $|r|\ge 1$, and we let $r_n=r^n$ for each $n\in \mathbb{N}$, the sequence $\{r_n{\}}_{n\in \mathbb{N}}$ does not converge to zero. I have hidden the proof of the former below; if you feel that no furthur attempts at the former will be at least slightly productive, you may see the proof. However, even if you do see the proof of the former, I have not included a proof of the latter (which essentially follows from the line of thinking required to prove the former). If you feel that you cannot solve the latter, even after seeing the hidden proof of the former below, post here again, and I (or another volunteer), will give you additional hints. Hope this helps (and the proof of the former is hidden below). PST 05:36, 9 March 2010 (UTC)

PST 05:36, 9 March 2010 (UTC) Template:Hidden Alternatively, if you are OK with 1/n → 0 and with the sandwich theorem, you may prove that if 0 ≤ a < 1 then 0 ≤ aⁿ ≤ C/n with C := a/(1-a). Start from the Bernoulli inequality with x := (1-a)/a .--pma 09:38, 9 March 2010 (UTC)

Divisibility In Digit-by-Digit Inter-Base Conversion

I've been trying to get my head around the meaning of some empirical results that don't seem to have an obvious explanation. Up to some very large number at least, there is no number that generates only numbers not divisible by any of {2, 3, 5} when one takes its representations in bases 2 through 5 and translates these into greater bases up to 6 (Ten numbers output for each input). Can anyone prove the impossibility of finding such a number, explain reasonably precisely why the first such number would be so inordinately large that it may not be computable, or demonstrate an example which does give ten numbers relatively prime to 30?Julzes (talk) 08:40, 9 March 2010 (UTC)

What's wrong with the number 1? Dmcq (talk) 09:45, 9 March 2010 (UTC)

20821 - outputs are 5321791, 285282577, 6348063151, 80542674037, 267781, 1973131, 10134727, 94531, 328147, 58633; outputs modulo 30 are 1, 7, 1, 7, 1, 1, 7, 1, 7, 13. Gandalf61 (talk) 11:51, 9 March 2010 (UTC)

Ok, that means my most recent program was in error. Anyway, can anyone find a number that generates 10 primes or suggest to me a stronger sieve than simply checking numbers congruent to 1 modulo 60? I know why the first such number is likely to be huge. In fact, just checking numbers like mentioned has another program that hopefully is not also defective up to around 190 billion without finding such a number.Julzes (talk) 16:36, 9 March 2010 (UTC)

Well, if anyone comes up with anything profound on this, please let me know.

Off-topic slightly, but here is a brief description of the granddaddy of coincidences that I discovered since the last time I posted anything of the nature here: List the primes that translate as primes twice in going from base 2 to base 10 and also translate twice as primes in going from base 3 to base 10. The 4th of these is the first to also generate a prime once in going from base 4, the 44th is the first to do it twice, both begin with the digits 234 in base 10, and the tenth on the list begins with 365. Furthermore, the 4th on the list is more special in that it translates twice as primes from base 5, does not translate once as a prime again until base 20 (allowing 'digits' greater than 9, and the tautology at base ten isn't counted) at which base it translates 5 times as a prime, and then it also does it 4 times at base 22 and twice at base 25 (not at bases 21, 23, or 24) to boot. Don't ask me how I found it.Julzes (talk) 03:42, 10 March 2010 (UTC)