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  • Now by using the above relations repeatedly, we find .../math> for <math>n\ge 13</math>. For smaller values, it is easy to use the relations to compute that the terms are powers of <math>2</math>, although we note th
    9 KB (1,491 words) - 00:23, 26 December 2022
  • ...>th crawl (as it will be forced to move somewhere else.). Thus, we get the recurrence relation <cmath>a_n=3^{n-1}-a_{n-1}.</cmath> ...ity that the bug is on some other vertex. We have the following recurrence relations. <cmath>A_n = \frac{1}{3}O_{n-1}</cmath> <cmath>O_n = A_{n-1} + \frac{2}{3}
    19 KB (3,128 words) - 20:38, 23 July 2024
  • ...maller cases, which is something that we don't like -- we want to keep our relations as simple as possible. Therefore, seeing that we can't work forwards anymor We thus have established a recurrence relation -- since the first house either gets mail or it doesn't, and canno
    15 KB (2,590 words) - 20:17, 29 November 2024
  • ...ations]] (ordinary or partial), [[difference equations]], and [[recurrence relations]]. Almost every technical field requires the consideration of dynamical sys
    789 bytes (107 words) - 20:52, 18 October 2017
  • recurrence relations:
    11 KB (1,889 words) - 12:45, 4 July 2013
  • For the sake of convenience in determining recurrence relations, we define another type of board with two <math>1\times n</math> boards whe We can determine reccurence relations for <math>s_n</math>, <math>t_n</math>, and <math>u_n</math> in terms of ea
    14 KB (2,076 words) - 19:29, 10 July 2023
  • ...ve for <math>p</math>, <math>q</math>, and <math>r</math> by iterating the recurrence to obtain <math>x_1=180^\circ-2x_0</math>, <math>x_2=4x_0-180^\circ</math>, ...ath> assuming our inductive hypothesis holds for <math>n</math>. Using the recurrence relation, we have <cmath>\begin{align*}
    5 KB (933 words) - 23:22, 3 November 2024