Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

May 5

How to prove this?

Given a natural number $n$, we determine $f(n)$ by the following algorithm:

1. Set $s=0.$
2. For $i=1,...,n,$

   if $s\geq i,$ set $s=s-i,$

   otherwise, set $s=s+i.$

3. The result $f(n)=s.$

In words, we subtract when we can, but we are not allowed to go negative, so if we can't subtract we add. For example, to find $f(12)$ we compute:

$0+1=1$; $1+2=3$; $3-3=0$; $0+4=4$; $4+5=9$; $9-6=3$; $3+7=10$; $10-8=2$; $2+9=11$; $11-10=1$; $1+11=12$; $12-12=0.$

The outcome is that $f(12)=0.$ So $12$ is a zero of function $f.$ I am interested in the sequence of zeros. The first few are $0,3,12,39,120,363,1092,3279,9840,$ in general the numbers of the form ${\tfrac {3}{2}}(3^{k}-1).$ While it is not very hard to prove this in a somewhat laborious way, I suspect there is some elegant proof, but it escapes me.  --Lambiam 19:23, 5 May 2023 (UTC)

I'd guess this is what you did but just to check. The sequence of adds and subtracts after each zero is add followed by pairs of add and subtract till the next zero. The values of n for which f is zero are 3, 3+3^2, 3+3^2+3^3 etc.The two adds after each of these are ${\tfrac {3}{2}}(3^{k}-1)+1$ and ${\tfrac {3}{2}}(3^{k}-1)+1+{\tfrac {3}{2}}(3^{k}-1)+2=3^{k+1}$, and this last number is the number of steps from the zero till the next zero. The values then continue in pairs going down and up by one till zero is reached. And the laborious part is just showing this is true. NadVolum (talk) 20:24, 5 May 2023 (UTC)

It might help to generalize a bit. Start with any given k and assume g(k)=0, and g(i+1)=g(i)-(i+1) if g(i)>i, g(i)+(i+1) else. So if g(10)=0 then g(11)=11, g(12)=23, g(13)=10, and so on. It shouldn't be too hard to prove by induction that g(k+1)=k+1, g(k+3)=k, g(k+5)=k-1, ... g(k+2s+1)=k+1-s, and g(k+2)=2k+3, g(k+4)=2k+4, g(k+2+2s)=2k+3+s for s≤k+1. I'm pretty sure you'd have to do this by induction and it would be a bit messy, but the pattern is fairly simple so the proof should be straightforward. The upshot would be that if g(k)=0 then the next 0 of g is 3k+3. Starting with k=0 this produces 0, 3, 12, 39, ... . But if you start with 10 you get 10, 33, 69, ... . In any case, at this point it amounts to turning the recursion z(i+1)=3*z(i)+3 into an explicit formula. More generally, if z(0)=0, z(i+1)=a*z(i)+b, then the sequence of z's is 0, b, b(1+a), b(1+a+a²), ... , and in general the kth term is b(a^k-1)/(a-1). So I don't know about elegant, perhaps generalizing makes it more messy in the long run. If you start with different initial values for g, say g(k)=n, then the behavior of g is more complex; it takes a while to get into the range [0, 2k] and then it starts following the predictable up-down pattern. A complete description including the "transient" would be rather complex. --RDBury (talk) 22:29, 5 May 2023 (UTC)

Is the laborious part getting to $f'(k+1)=3(f'(k)+1)$ or the proof by induction? fiveby(zero) 23:33, 5 May 2023 (UTC)

iff s = 0 you have to add twice, if you've added twice you have to subtract, if you just subtracted you have to add, if you add then subtract you reduce s by one, so ++-+-+-+-...until you get back to 0

if f(i) = 0 then f(i+1) = i+1 and f(i+1 + 2(i+1)) = 0

the induction part is ${\tfrac {3}{2}}(3^{0}-1)=0$ and ${\tfrac {3}{2}}(3^{k+1}-1)=3({\tfrac {3}{2}}(3^{k}-1)+1)$

fiveby(zero) 02:26, 6 May 2023 (UTC)

The proof appears to proceed in two phases, each of which requires induction. Define the sequence $s$ by $s_{0}=0$ and $s_{z}=s\odot z$ for $z\geq 1,$ where $s\odot z=s-z$ if $s\geq z,$ and $s\odot z=s+z$ otherwise. In the first phase we prove a formula for $s_{z{+}d},$ where $1\leq d\leq 2z{+}3,$ under the assumption that $s_{z}=0.$ This involves case distinctions based on the parity of $d.$ We then observe that $s_{z{+}d}>0$ if $1\leq d<2z{+}3,$ whereas $s_{z{+}d}=0$ if $d=2z{+}3.$ So if $s_{z}=0,$ the next occurrence of a $0$ is $s_{3z{+}3}=0.$ If you write this all out neatly, it seems rather a lot for such a simple thing. But the proof is not convincing unless you also verify the low-level stuff. The second phase, proving the formula ${\tfrac {3}{2}}(3^{k}{-}1),$ goes smoothly.  --Lambiam 19:32, 6 May 2023 (UTC)

May 7

Interpretation of p-values from the Kruskal-Wallis test with Dunn's multiple comparisons test applied post-hoc

See the discussion here about an issue in a recent paper about Wikipedia in "Big Data & Society". Thoughts welcome, in particular about what might be a proper way to form such conclusions from this data. Regards, HaeB (talk) 20:57, 7 May 2023 (UTC)