Wikipedia: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 $z\in \mathbb {R}$ such that $|z|<1$, and let $z_{n}=z^{n}$ for each $n\in \mathbb {N}$. Prove that $z_{n}\rightarrow 0$. 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 $u_{n}\rightarrow 0$.

To complete the above proof, you need to establish that $z_{n}\rightarrow 0$, and that if $|r|\geq 1$, 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)

(Click the "show" button at the right to see a proof relevant to the above post or the "hide" button to hide it.)

Result

Let $z\in \mathbb {R}$ such that $|z|<1$, and let $z_{n}=z^{n}$ for each $n\in \mathbb {N}$. Then $z_{n}\rightarrow 0$.

Solution

We must show that for all $\epsilon >0$, there exists $N\in \mathbb {R}$, such that for all $n>N$ ($n\in \mathbb {N}$), $|z_{n}|<\epsilon$ (we can assume without loss of generality that $z\geq \epsilon$). If $\epsilon >0$, let $N=\log _{|z|}(\epsilon )$. Since the function $f(x)=|z|^{x}$ is decreases for $|z|<1$ and $x\geq 1$, we have:

$n>N$

$n>\log _{|z|}(\epsilon )$

$|z|^{n}<\epsilon$ (we assumed without loss of generality that $z\geq \epsilon$)

$|z^{n}|<\epsilon$

$|z_{n}|<\epsilon$

Since the last statement above is true for all $n>N$ and $\epsilon >0$ was arbitrary, the result follows.

PST 05:36, 9 March 2010 (UTC)

Collapsed off-topic discussion

Wiktionary does not recognize the word furthur. Did you mean further? -- Meni Rosenfeld (talk) 09:08, 9 March 2010 (UTC)

Of course that is what he meant. Don't be a jerk. StatisticsMan (talk) 18:12, 9 March 2010 (UTC)

Please assume good faith. I was seriously considering the possibility (however remote) that in fact this is an alternative spelling which has not made it to Wiktionary, and did not want to come out a fool if this turned out to be the case.

I should also add that this is the third time I have noticed PST making this mistake, so this being a simple typographical error was out of the question. Either this was, indeed, a correct spelling, or PST mistakenly thought that it was correct. In the latter case I thought he would appreciate noting this error. -- Meni Rosenfeld (talk) 19:38, 9 March 2010 (UTC)

"Your English inflexion is almost perfect, Oberleutnant PST " ;-) --pma 08:48, 10 March 2010 (UTC)

This is no laughing matter. I find it repugnant that StatisticsMan would insult me like this and not bother to follow up on the discussion. -- Meni Rosenfeld (talk) 20:49, 11 March 2010 (UTC)

My excuses, Meni. I still don't know what "jerk" means nor if it has offensive connotations (I am possibly the only one here who would need explanations about English); in any case I'm quite sure SM didn't meant to insult you. As to my comment, it was not addressed to anybody in particular; the episode just reminded me of some old good Bogart style movie. --pma 21:28, 11 March 2010 (UTC)

That is no problem - I just didn't want the lighthearted reference to overshadow how seriously I take the matter.

See also the third definition of jerk. Presumably, StatisticsMan was under the impression that my post was sarcastic and intended to ridicule PST. Usually the word isn't terrible, but in the current context StatisticsMan's comment was incredibly offensive. -- Meni Rosenfeld (talk) 07:38, 12 March 2010 (UTC)

I do not believe that Meni intended to ridicule my use of the English language with his comment, but this conclusion is certainly influenced by the fact that I have interacted with Meni at the reference desk quite often recently. Had I been reading an identical discussion between two people whom I did not know, I might have seen the comment as an insult. I think that this scenario is relevant to the present context.

StatisticsMan has not commented at the reference desk frequently in the recent past (he only contributed to the reference desk five times last month - [1]); of course, that does not necessarily give any indication as to how often he visits the reference desk. Nonetheless, in StatisticsMan's case, I believe that he does not visit the reference desk very frequently either. Therefore, StatisticsMan's failure to follow up on the discussion may simply be because he has not visited the reference desk since posting his most recent comment, in which case I do not think that StatisticsMan can be considered repugnant in the context subsequent to his most recent response. Furthermore, knowing StatisticsMan, I do not think that he would be under the impression that he is "too good" to follow up on this sort of discussion.

That said, I shall emphasize once more that I do not believe Meni intended to ridicule me with his comment; personally, it seems unlikely (if not impossible) that he would behave in that manner, having had some experience interacting with him. StatisticsMan, on the other hand, may not have significantly interacted with Meni in the past, and as I mentioned earlier, if the name "Meni" had not been attached to Meni's comment, I might have interpreted it as an insult (but as I shall clarify once more, I definitely do not think that Meni is a jerk, in that it is very unlikely that his comment is an intended insult).

With regards to the usage of "jerk", I think that sometimes it is considered appropriate for someone to deliver an insult to a "definite troll" (for instance, if a troll came here and vandalized the reference desk, and demanded a response to his behavior, it would be considered OK (by some) to dismissively insult him (I certainly would not object to this use of an insult if the nature of the vandalism were extreme, but that is a different matter in a different context)). Perhaps StatisticsMan decided that Meni's comment was "in the manner of a jerk" and dismissed it with what he considered appropriate. As I said, whether this was appropriate or not is "relative", but I think that anyone who knows Meni Rosenfeld well would see the comment as an incorrect portrayal of his character.

Regarding my English, sometimes I have a habit of making minor misspellings when I type; had I written that comment by hand, I do not think that I would have made that error. Perhaps the first "u" in "further" influences my brain to believe that the "u" repeats, in much the same way as in other words having a "repeating vowel", and perhaps this influence is more significant while I type. I don't really know. But thanks to Meni for noting that; I might have noticed it in the past when editing my typed text, but I never really knew that I was repeating the mistake (it would be nice if Wikipedia had a "spell-check" like Microsoft Word; perhaps I should paste my text into Microsoft Word and run "spell-check", and then paste the corrected text into Wikipedia ;)). PST 09:41, 12 March 2010 (UTC)

My post was intended to be as diametrically opposite to "insulting" or "jerky" as possible, in the same way that "you are wrong" is inferior to "I think you may be wrong". The former is arrogant and rude and assumes the speaker is infallible while the target is error-prone (unless, of course, the speaker is an authority on the subject).

Prior to StatisticsMan's comment, it never crossed my mind that my post could have been interpreted as an insult. Furthermore, if I had been asked "do you think your post could be interpreted as an insult?" my answer would have been a definite "no". I would not be able to conceive of anything more polite than presenting the evidence I have that a mistake has taken place, offering a correction and acknowledging that I might be wrong myself.

Now that I know it can be interpreted negatively, I have certainly noted to myself how to avoid such problems in the future (primarily "don't be so polite nobody would believe you are that polite and assume you are rude instead"). I will also be happy to hear any additional insight on what I should have said instead in this particular situation.

I believe that StatisticsMan, even without deep familiarity with me, had enough information that with a little WP:AGF, he should not have concluded that I am a jerk.

"[maybe] because he has not visited the reference desk since posting his most recent comment" - this is factually false, as he has posted here since my reply to him. Obviously he has no obligation to visit here at any frequency, but I, for one, when I refer to someone as a jerk (not very often), I sure don't wait >2.5 days to check back if he has something to say for himself. Or, if I anticipate a period of unavailability, I don't enter heated debates to begin with.

What browser do you use? Firefox has a spellchecker. -- Meni Rosenfeld (talk) 11:31, 12 March 2010 (UTC)

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)