I am given the sequence 1, 7, 21, 43, 73, ... and can derive without any assumption about the nature of the sequence that the ultimate difference is 8, I'm interested in understanding what this difference allows you to deduce about the original polynomial which generates this sequence; and whether it is possible to find this polynomial?

wasangagac4

wasangagac4

Answered question

2022-10-26

What does the common difference of a sequence describe
I am given the sequence 1, 7, 21, 43, 73, ... and can derive without any assumption about the nature of the sequence that the ultimate difference is 8, I'm interested in understanding what this difference allows you to deduce about the original polynomial which generates this sequence; and whether it is possible to find this polynomial?
I can see that 1, 7, 21, 43, 73, ... has a difference of 6, 14, 22, 30, ... which is generated by the arithmetic sequence 6 + 8(n-1), but trying to rationalize how the difference of a sequence can then be non-constant (1, 7, 21, 43, 73) it doesn't really make sense.
Can anyone shed some light on how to go about solving this intuitively?

Answer & Explanation

amilazamiyn

amilazamiyn

Beginner2022-10-27Added 14 answers

Step 1
You have to go through 2 levels of differences to get to the ultimate difference, so your polynomial will be of degree 2. (If you know some calculus, this corresponds to taking 2 derivatives to reach a constant function.)
Because the ultimate difference is 8 and it appears at degree 2, the first coefficient is 8 2 ! = 4 (the 2! is again a factor that appears if you differentiate; if you don't know any calculus, I'm not sure if there's an intuitive explanation for it, but you can derive it by a method similar to Axion004's answer), so the first term of the polynomial is 4 x 2 .
The first term of the polynomial will get in the way of determining the other terms intuitively, so at this point it's easiest to just subtract it out. Subtracting 4 x 2 from each term (and assuming the terms are indexed starting at x = 0) gives
1,3,5,7,9,…
Step 2
Now we go through the same process again. We know the degree will now go down one level to 1, and we can find that the new ultimate difference is 2, so the next coefficient is 2 1 ! = 2, and so far our polynomial is 4 x 2 + 2 x. Subtracting the 2x from our last sequence gives
1,1,1,1,1,…
so we have reached the constant sequence. Our generating polynomial is 4 x 2 + 2 x + 1.
faois3nh

faois3nh

Beginner2022-10-28Added 4 answers

Step 1
The difference sequence is 6,14,22,30, etc., for which, as you noted, the nth term is 6 + 8 ( n 1 ).
When you evaluate the sum for the mth term of the sequence, 1 + n = 1 m 1 [ 6 + 8 ( n 1 ) ] ,
the answer is 1 + 8 n = 1 m 1 n n = 1 m 1 2 = 1 + 8 ( m 1 ) m 2 2 ( m 1 ) = 4 m 2 6 m + 3..
Step 2
Just as when you integrate a k-th degree polynomial, the answer is a polynomial of degree k + 1, when you repeatedly sum a kth degree (in this case, linear) polynomial in n from 1 to m 1, the answer is a k + 1 t h degree (in this case, quadratic) polynomial in m.

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