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April 23rd, 2016, 06:42 AM  #1 
Newbie Joined: Apr 2016 From: Hayes Posts: 1 Thanks: 0  Sequence  Term to term help required
What are the missing numbers in the following, where X is a missing number: 2, X, 17, 23, 52, X, What is the pattern please?
Last edited by skipjack; April 30th, 2016 at 11:44 AM. 
April 23rd, 2016, 06:46 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,331 Thanks: 2457 Math Focus: Mainly analysis and algebra 
It could be literally anything.

April 23rd, 2016, 09:53 AM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,245 Thanks: 887  Quote:
Now, a sequence is simply a list of number there does not have to be any pattern (When I was a freshman, my Calculus professor gave, as an example of a sequence, "32, 30, 28, 24, 22" and asked "what is the next number". Everyone answered, of course, "20" but the correct answer was "30". Why? Those were the exit numbers on the limited access highway on which he drove to work and at "22" he turned off onto another highway! What you are really being asked here is "what is the simplest arithmetic pattern that includes those numbers?" The simplest would be an "arithmetic" or "geometric" sequence where we just add or multiply by the same number to go from one term to the next. But those obviously don't work. To go from 17 to 23 we add 6 but adding 6 to 23 gives 29, not 52. Similarly to go from 17 to 23 we could multiply by 23/17 but 23/17 times 23 is not 52. Here is one plausible answer using "Lagrange's divided difference" method. Given any n points, there exist a unique n1 polynomial giving those values. "Lagrange's divided difference" method is a way of finding that polynomial. If we think of these as numbers in a sequence $\displaystyle a_0= 2$, $\displaystyle a_1= X$, $\displaystyle a_2= 17$, $\displaystyle a_3= 23$, $\displaystyle a_4= 52$, and $\displaystyle a_5= X$ then the successive differences are $\displaystyle a_1 a_0= X 2$, $\displaystyle a_2 a_1= 17 X$, $\displaystyle a_3 a_2= 6$, [math]a_4 a_2= 29[math], and $\displaystyle a_5 a_4= X 52$. The "second differences" are $\displaystyle (17 X) (X 2)= 19 2X$, $\displaystyle 6 (17 X)= X 11$, $\displaystyle 29 6= 23$, and $\displaystyle X 52 29= X 81$. The "third differences" are $\displaystyle (X 11) (19 2X)= 3X 30$, $\displaystyle 23 (X 11)= 34 X$, and $\displaystyle X 81 23= X 104$. The "fourth differences" are $\displaystyle 34 X (3X 30)= 64 4X$ and $\displaystyle X 104 (34 X)= 2X 168$. We could continue to the "fifth difference" 2X 169 (64 4X)= 6X 233 to find the fourth degree polynomial (depending on X) that will give those values for any X but we can get the simplest, third rather than fourth degree, polynomial by choosing X so that 2X 168= 0. That is, X= 168/2= 84. If we do that then the top row in all those differences are 2, X 2= 82, 19 2X= 149, 3x 30= 222, and 64 4X= 272. Then the fourth degree polynomial is given by 2+ 82n (149/2)n(n1)+ (222/6)n(n1)(n 2) (272/24)n(n1)(n2)(n3).  
April 30th, 2016, 10:12 AM  #4 
Banned Camp Joined: Apr 2016 From: Australia Posts: 244 Thanks: 29 Math Focus: horses,cash me outside how bow dah, trash doves 
interpolation polynomial (LaGrange polynomial also called I think) for the sequence mapped to indexed data point pairs, {[0,2],[3,X],[4,17],[5,23],[6,52],[7,X]} will not be able to compute as it is because there is a multiple root, ie, the same value X occurs more than once. But since you have more than two data points, so you could generate the maximum number subsets of the maximum size( ie 5 pairs ) s.t X only appears exactly once, eg {[0,2],[3,X],[4,17],[5,23],[6,52]},{[0,2],[4,17],[5,23],[6,52],[7,X]} then I guess take the average of the two polynomials that fit these two subsets and it will be as close as youll ever get knowing what you know, could be represented in a number of different ways but power series from polynomial interpolation would be me if I already knew there is meant to be an algebraic relation / solution Last edited by Adam Ledger; April 30th, 2016 at 10:19 AM. 

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