Eikonal Blog

2010.07.13

Diferentiating primes

Filed under: mathematics — Tags: , — sandokan65 @ 15:50

Source:

  • How to Differentiate a Number; by Victor Ufnarovski and Bo Ahlander; Journal of Integer Sequences, Vol 6 (2003), Article 0334
  • T24661

Definition: On the set of integer numbers {\Bbb N}_0=\{0,1,2,\cdots\} introduce following unary operation:

  • 0' = 0,
  • 1' = 0,
  • p' = 1, \ \ \ (\forall p \in {\Bbb P}),
  • (ab)' = a'b+a b' \ \ \ \ (\forall a, b \in {\Bbb N}) \ \ (\hbox{Leibnitz rule}).

Examples:
\begin{tabular}{ | c | c c c c c c c c c c c c c c c c c c |} \hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\ \hline  n' & 0 & 1 & 1 & 4 & 1 & 5 & 1 & 12 & 6 & 7 & 1 & 16 & 1 & 9 & 8 & 32 & 1 & 81 \\ \hline n'' & 0 & 0 & 0 & 4 & 0 & 1 & 0 & 16 & 5 & 1 & 0 & 32 & 0 & 6 & 12 & 80 & 0 & 10 \\ \hline n''' & 0 & 0 & 0 & 4 & 0 & 0 & 0 & 32 & 1 & 0 & 0 & 80 & 0 & 5 & 16 & 176 & 0 & 7 \\  \hline n'''' & 0 & 0 & 0 & 4 & 0 & 0 & 0 & 80 & 0 & 0 & 0 & 176 & 0 & 0 & 32 & 368 & 0 & 0 \\ \hline \end{tabular}

Note:

  • (a+b)' \ne a' + b',
  • (a b)'' \ne a''b+2a'b'+a b''.

Examples:

  • (2k)'  = k + 2 k' \ge 2k',
  • (3k)' = k+3k' \ge 3k',
  • (4k)' = 4k + 4k' \ge 4k',
  • (5k)' = k + 5k' \le (2k)' + (3k)'.

Note: The only pairs (a,b) s/t a\le b \le 100 with gcd(a,b)=1 and (a+b)'=a'+b' are:
\{(1,2), (4,35), (4, 91), (8, 85), (11,11), (18, 67), (26,29), (27, 55), (35, 55), (36, 81), (38, 47), (38, 83), (50, 79), (62, 83), (95,99)\}.

Theorem: n'\ge n \Rightarrow (kn)' \ge kn, \ \ \ \ (\forall k > 1).

Theorem: If n = \prod_{i=1}^k {p_i}^{n_i} then n' = n \sum_{i=1}^{k} \frac{n_i}{p_i}.

Theorem: For n = m p^p, (p\in {\Bbb P}, m> 1, \in {\Bbb N})

  • n' = p^p (m+m'),
  • \lim_{k\rightarrow \infty} n^{(k)} = \infty.

Extensions:

One can extend the differentiation operation to wider sets of numbers. For example, by setting $(-1)’=0$ one gets extension to the whole set of integer numbers {\Bbb Z}:

  • (-1)' = 0, ,
  • (-p)' = -1, \ \ \ (\forall p \in {\Bbb P}),
  • (-x)' = - x', \ \ \ (\forall x \in {\Bbb Z}).

One can also infer that:

  • i' = 0,
  • \omega' = 0, \ \ \ (\forall \omega^m = 1, \ \ m\in {\Bbb N}),
  • \omega' = 0, \ \ \ (\forall \omega^\alpha = 1, \ \ \alpha\in {\Bbb C}\ \{0\}).
  • \left(\sqrt{p}\right)' = \frac{\sqrt{p}}{2p},
  • \left(\frac1{p}\right)' = -\frac{1}{p^2},
  • \left(p^\alpha\right)' = -\alpha p^{\alpha-1},
  • \ln(p)' = \frac1{p},
  • \left(\frac{n}{m}\right)' = \frac{n'm- n m'}{m^2}.

Definition:

  • \pi(n) :\equiv \ (\# \ of \ p \le n),
  • \pi_1(n) :\equiv \pi(n) - \pi(n-1).

Lemma:

  • \pi_1(n) = 1 if n=p,
  • \pi_1(n) = 0 if n\ne p.

Lemma:

  • \pi_1(n) = 1 \Leftrightarrow  n'=1,
  • \pi_1(n) = 0 \Leftrightarrow n'\ne 1.

1 Comment »

  1. okay thankyou for the list
    hope i can use this

    Like

    Comment by retnolaras — 2010.07.14 @ 23:42


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