# Thread: General Forms of Recursive Series

1. ## General Forms of Recursive Series

Hi,

I have a set of expressions which are related. The relationship is apparent in the multipliers in each line.

$\displaystyle 3rd: {\it t_1}+{\it t_3}$
$\displaystyle 4th: {\it t_1}+{\it t_2}+{\it t_3}+{\it t_4}$
$\displaystyle 5th: {\it t_1}+2\,{\it t_2}+2\,{\it t_3}+2\,{\it t_4}+{\it t_5}$
$\displaystyle 6th: {\it t_1}+3\,{\it t_2}+4\,{\it t_3}+4\,{\it t_4}+3\,{\it t_5}+{\it t_6}$
$\displaystyle 7th: {\it t_1}+4\,{\it t_2}+7\,{\it t_3}+8\,{\it t_4}+7\,{\it t_5}+4\,{\it t_6}+{\it t_7}$
$\displaystyle 8th: {\it t_1}+5\,{\it t_2}+11\,{\it t_3}+15\,{\it t_4}+15\,{\it t_5}+11\,{\it t_6}+5\,{\it t_7}+{\it t_8}$
$\displaystyle 9th: {\it t_1}+6\,{\it t_2}+16\,{\it t_3}+26\,{\it t_4}+30\,{\it t_5}+26\,{\it t_6}+16\,{\it t_7}+6\,{\it t_8}+{\it t_9}$
$\displaystyle 10th: {\it t_1}+7\,{\it t_2}+22\,{\it t_3}+42\,{\it t_4}+56\,{\it t_5}+56\,{\it t_6}+42\,{\it t_7}+22\,{\it t_8}+7\,{\it t_9}+{\it t_{10}}$

eg. t1 is always multiplied by 1

the multiplication of t2 in each case increments by 1. [0 1 2 3 4 5 6 ....]

the multiplication of t3 in each case increments using the previous increment intervals [0 1 2 3 4 5 6 ....] to give [1 1 2 4 7 11 ...].

the multiplication of t4 in each case increments using the previous increment intervals [1 1 2 4 7 11 ...] to give [0 1 2 4 8 15 26 ...]

My problem is this...

I would like to give a general solution for the Nth order case using some sort of series notation but although there is a pattern clear in the multipliers I am unsure how to go about this.

Does anyone have any experience with this type of problem or how to go about formulating a general solution.

Thanks
Alex

2. Hello, asouthern!

I have a set of expressions which are related.
The relationship is apparent in the multipliers in each line.

. . $\displaystyle \begin{array}{ccccc}\text{3rd:} &\qquad\qquad\qquad\qquad\qquad t_1+ t_3 \\ \text{4th:}& \qquad\qquad\qquad\qquad\qquad t_1 + t_2 + t_3 + t_4 \\ \end{array}$
. . $\displaystyle \begin{array}{cc} \text{5th:} & t_1 + 2t_2 + 2t_3 + 2t_4 + t_5 \\ \text{6th:} & t_1 + 3t_2 + 4t_3 + 4t_4 + 3t_5 + t_6 \\ \text{7th:} & t_1 + 4t_2 + 7t_3 + 8t_4 + 7t_5 + 4t_6 + t_7\\ \text{8th:} & t_1 + 5t_2 + 11t_3 + 15t_4 + 15t_5 + 11t_6 + 5t_7 + t_8 \\ \text{9th:} & t_1 + 6t_2 + 16t_3 + 26t_4 + 30t_5 + 26t_6 + 16t_7 + 6t_8 + t_9 \\ \text{10th:} &t_1+7t_2+22t_3+42t_4+56t_5+56t_6+42t_7+22t_8+7t_9 +t_{10} \end{array}$
I see an even stronger pattern, but . . .

You have a variation of Pascal's Triangle . . .

. . $\displaystyle \begin{array}{cccccccccccc} 3 &&1\;\;0\;\;1\\ 4 && 1\;\;1\;\;1\;\;1\\ 5 && 1\;\;2\;\;2\;\;2\;\;1\\ 6 && 1\;\;3\;\;4\;\;4\;\;3\;\;1\\ 7 && 1\;\;4\;\;7\;\;8\;\;7\;\;4\;\;1\\ 8 && 1\;5\;11\;15\;15\;11\;5\;1\\ 9 && 1\;\;6\;16\;26\;30\;26\;16\;6\;1 \\ 10 && 1\;7\;22\;42\;56\;56\;42\;22\;7\;1\end{array}$

Each number is the sum of the two numbers directly above it.

In row 3, we have: $\displaystyle [1\:0\:1]$ which represents the polynomial: $\displaystyle x^2 + 1$

Multiply by $\displaystyle (x+1)\!:\;\;(x+1)(x^2+1) \:=\:x^3 + x^2 + x + 1$
. . and its coefficients are in row 4.

Multiply by $\displaystyle (x+1)\!:\;\;(x+1)(x^3+x^2+x+1) \:=\:x^4 + 2x^3 + 2x^2 + 2x + 1$
. . Its coefficients are in row 5.

Multiply by $\displaystyle (x+1)\!:\;\;(x+1)(x^4 + 2x^3 + 2x^2 + 2x + 1) \:=\:x^5 + 3x^4 + 4x^3 + 4x^2 + 3x + 1$
. . Ites coefficients are in row 6.

So I see the powerful pattern embedded in the triangle,
. . but so far, I haven't been able to express this in a closed form.

3. Originally Posted by asouthern
Hi,
I have a set of expressions which are related. The relationship is apparent in the multipliers in each line.

$\displaystyle 3rd: {\it t_1}+{\it t_3}$
$\displaystyle 4th: {\it t_1}+{\it t_2}+{\it t_3}+{\it t_4}$
$\displaystyle 5th: {\it t_1}+2\,{\it t_2}+2\,{\it t_3}+2\,{\it t_4}+{\it t_5}$
$\displaystyle 6th: {\it t_1}+3\,{\it t_2}+4\,{\it t_3}+4\,{\it t_4}+3\,{\it t_5}+{\it t_6}$
$\displaystyle 7th: {\it t_1}+4\,{\it t_2}+7\,{\it t_3}+8\,{\it t_4}+7\,{\it t_5}+4\,{\it t_6}+{\it t_7}$
$\displaystyle 8th: {\it t_1}+5\,{\it t_2}+11\,{\it t_3}+15\,{\it t_4}+15\,{\it t_5}+11\,{\it t_6}+5\,{\it t_7}+{\it t_8}$
$\displaystyle 9th: {\it t_1}+6\,{\it t_2}+16\,{\it t_3}+26\,{\it t_4}+30\,{\it t_5}+26\,{\it t_6}+16\,{\it t_7}+6\,{\it t_8}+{\it t_9}$
$\displaystyle 10th: {\it t_1}+7\,{\it t_2}+22\,{\it t_3}+42\,{\it t_4}+56\,{\it t_5}+56\,{\it t_6}+42\,{\it t_7}+22\,{\it t_8}+7\,{\it t_9}+{\it t_{10}}$

...

Alex
From where are you getting the $\displaystyle t_i$ values?
Could you list ALL of the t values at the start?
for example:

$\displaystyle t_0 = -1$
$\displaystyle t_1 = 1$
$\displaystyle t_2 = 3$
$\displaystyle t_3 = 5$
$\displaystyle t_4 = 7$
$\displaystyle t_5 = 9$

which would give:
$\displaystyle S_0 = ?$
$\displaystyle S_1 = ?$
$\displaystyle S_2 = ?$
$\displaystyle S_3 = t_1 + t_3$ = 6
$\displaystyle S_4 = t_1 + t_2 + t_3 +t_4$ = 16
$\displaystyle S_5 = t_1 + 2 t_2 +3 t_3 + 2 t_4 + t_5$ = 45

Soroban shows a modified Pascal Triangle for the coefficients,
and I'm sure he can modify the binomial expansion to give coefficients for any n.

Do you have method that generates the t values or are they already permanently defined?

Is it possible that the $\displaystyle t$ values and the coefficients are related?

4. ## The relation between the t terms

Hi,

Thanks for your replies, I agree the pascal triangles do seem the strongest relationship. As for the definition of the t values. Well,

$\displaystyle t_i$ is shorthand for $\displaystyle r\cos{\theta_i}$

r is constant, the angle changes with every order N, so that,

$\displaystyle \theta_i = \frac{\pi\cdot(1+2(i-1))}{2\cdot{N}}$

i = 1,2...,N

I didn't include the terms for when N =1 and N =2 as they are too simple.

Thanks
Alex

5. Also,

In my mind the general representation could be,

$\displaystyle =\sum_{i = 1}^N(a_i\cdot{t_i})$

e.g.
for N = 4,

$\displaystyle a_i =[1,1,1,1]$

e.g.
for N = 5,

$\displaystyle a_i =[1,2,2,2,1]$

So some way to form the series $\displaystyle a_i$ for any order N, would be sufficient.

Any advice on this would be greatly appreciated.

Thanks Again
Alex