You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
linear functions: (x,y) = (1, 8), (2, 13), (3, 19), (4, 26),
- Thread starter darlene
- Start date
Hello, Darlene!
It is usually very difficult to find the generating function "by inspection".
\(\displaystyle \text{We want a quadratic function: }\:f(n) \;=\;an^2 + bn + c\)
\(\displaystyle \text{Use the first three terms of the sequence:}\)
. . \(\displaystyle \begin{array}{ccccccc} f(1) = 8\!: & a + b + c &=& 8 & [1] \\ f(2) = 13\!: &4a + 2b + c &=& 13 & [2] \\ f(3) = 19\!: & 9a + 3b + c &=& 19 & [3] \end{array}\)
. . \(\displaystyle \begin{array}{cccccc}\text{Subtract [2] - [1]:} & 3a + b &=& 5 & [4] \\ \text{Subtract [3] - [2]:} & 5a + b &=& 6 & [5] \end{array}\)
. . \(\displaystyle \text{Subtract [5] - [4]: }\;2a \:=\:1 \quad\Rightarrow\quad \boxed{a \:=\:\tfrac{1}{2}}\)
. . \(\displaystyle \text{Substitute into [4]: }\;3\left(\tfrac{1}{2}\right) + b \:=\:5 \quad\Rightarrow\quad\boxed{ b \:=\:\tfrac{7}{2}}\)
. . \(\displaystyle \text{Substitute into [1]: }\;\tfrac{1}{2} + \tfrac{7}{2} + c \:=\:9 \quad\Rightarrow\quad\boxed{ c \:=\:4}\)
\(\displaystyle \text{Therefore: }\;f(n) \;=\;\tfrac{1}{2}n^2 +\tfrac{7}{2}n + 4 \quad\Rightarrow\quad \boxed{f(n)\;=\;\tfrac{1}{2}(n^2 + 7n + 8)}\)
It is usually very difficult to find the generating function "by inspection".
\(\displaystyle \begin{array}{|c|c|}\hline \text{Input} & \text{Output} \\ \hline 1 & 8 \\ 2 &13 \\ 3 & 19 \\ 4 & 26 \\ 5 & 34 \end{array}\)
What is the rule (function)?
I notice that the change in numbers is increasing by 1.
So I am thinking that x should have a square in the solution somewhere. . . . . Right!
\(\displaystyle \text{We want a quadratic function: }\:f(n) \;=\;an^2 + bn + c\)
\(\displaystyle \text{Use the first three terms of the sequence:}\)
. . \(\displaystyle \begin{array}{ccccccc} f(1) = 8\!: & a + b + c &=& 8 & [1] \\ f(2) = 13\!: &4a + 2b + c &=& 13 & [2] \\ f(3) = 19\!: & 9a + 3b + c &=& 19 & [3] \end{array}\)
. . \(\displaystyle \begin{array}{cccccc}\text{Subtract [2] - [1]:} & 3a + b &=& 5 & [4] \\ \text{Subtract [3] - [2]:} & 5a + b &=& 6 & [5] \end{array}\)
. . \(\displaystyle \text{Subtract [5] - [4]: }\;2a \:=\:1 \quad\Rightarrow\quad \boxed{a \:=\:\tfrac{1}{2}}\)
. . \(\displaystyle \text{Substitute into [4]: }\;3\left(\tfrac{1}{2}\right) + b \:=\:5 \quad\Rightarrow\quad\boxed{ b \:=\:\tfrac{7}{2}}\)
. . \(\displaystyle \text{Substitute into [1]: }\;\tfrac{1}{2} + \tfrac{7}{2} + c \:=\:9 \quad\Rightarrow\quad\boxed{ c \:=\:4}\)
\(\displaystyle \text{Therefore: }\;f(n) \;=\;\tfrac{1}{2}n^2 +\tfrac{7}{2}n + 4 \quad\Rightarrow\quad \boxed{f(n)\;=\;\tfrac{1}{2}(n^2 + 7n + 8)}\)
Methinks you're complicating it, Soroban.
Using leftmost number of each line before 0:
8 + 5(n - 1) + 1(n - 1)(n-2) / 2
= 8 + 5n - 5 + (n^2 - 3n + 2) / 2
= (6 + 10n + n^2 - 3n + 2) / 2
= (n^2 + 7n + 8) / 2
General case; keeping the typing short (L? = 1st number in that line):
L1 + L2(n-1)/1 + L3(n-1)(n-2)/(1*2) + L4(n-1)(n-2)(n-3) / (1*2*3) + ......
Code:
8 13 19 26 34 ....
5 6 7 8 ....
1 1 1 ....
0 0 ...8 + 5(n - 1) + 1(n - 1)(n-2) / 2
= 8 + 5n - 5 + (n^2 - 3n + 2) / 2
= (6 + 10n + n^2 - 3n + 2) / 2
= (n^2 + 7n + 8) / 2
General case; keeping the typing short (L? = 1st number in that line):
L1 + L2(n-1)/1 + L3(n-1)(n-2)/(1*2) + L4(n-1)(n-2)(n-3) / (1*2*3) + ......