Saturday, February 9, 2013

a_(n+1) = S * a_n + T

Hi everyone!


First a correction. In my Numerical CAS section, posted December 2012, I typed an extra quotation mark in the POLYBINE program for the Casio Prizm - that has been corrected. Much thanks to Ryan Maziarz who pointed that out to me.

Here is a link to the corrected post:
http://edspi31415.blogspot.com/2012/12/numeric-cas-part-2-binomial-expansion.html?m=1


Today we are finding the general formula for this recursion formula:

a_(n+1) = S * a_n + T

with the initial condition a_0 = A.

I could try using a characteristic polynomial, but with some observation we may be able to detected a pattern which leads us to a general formula.

Observe that:

a_0 = A

a_1 = S * a_0 + T
a_1 = S * A + T

a_2 = S * a_1 + T
a_2 = S * (S * A + T) + T
a_2 = S^2 * A + S * T + T
a_2 = S^2 * A + T * (S + 1)

a_3 = S * a_2 + T
a_3 = S * (S^2 * A + T * (S + 1)) + T
a_3 = S * A^3 + T * (S^2 + S + 1)

a_4 = S * a_3 + T
a_4 = S^4 * A + T * (S^3 + S^2 + S + 1)

From the pattern, observe that:

a_n = S^n * A + T * Σ(S^k, k=0, n-1)

With Σ(t^k, k=0, n-1) = (t^n - 1)/(t - 1) :

a_n = S^n * A + T * (S^n - S)/(S - 1)


An Example:

S = 3, T = 1, A = -2

Using the HP 39gii to generate the sequence (the 39gii uses a_1 as its initial term):


The general formula is:

a_n = (3^n) * (-2) + 1 * (3^n - 1)/(3 - 1)
a_n = (-2) * 3^n + (3^n - 1)/2
with a_0 = -2. (The 39gii has u_1 = a_0)

Example:

a_2 = u_3 = (3^2) * (-2) + (9 - 1)/2 = -14


That is all for now, see you next time!

Eddie

This blog is property of Edward Shore. 2013

5 blog entries away from the 200th blog entry.

HHC 2025 Videos

  HHC 2025 Videos The talks from the HHC 2025 conference in Orlando, Florida are starting to be up on hpcalc’s YouTube page within th...