Linear recurrence sequence, closed form.

**mrszone** · February 20th 2009, 04:41 AM

Hi I need help with the following –

Consider the linear recurrence sequence:

x(subscript)1 = 5, x(subscript)n+1=1.2x(subscript)n-14 (n=1,2,3,…)

Find a closed form for the sequence.

Do I need to use the geometric progression to work out the closed form?

Thanks.

**Danny** · February 20th 2009, 04:50 AM

Originally Posted by mrszone

Hi I need help with the following –

Do I need to use the geometric progression to work out the closed form?

Thanks.

The 14 messed that up. However, if we let

$x_n = y_n + k$ then $y_{n+1} + k = 1.2(y_n + k )-14$ and if we choose $k = 70$ then we have $y_{n+1} = 1.2 y_n$ which is now geometric.

**ADARSH** · February 20th 2009, 06:48 AM

The general term of the sequence is given by
$ T_n= 5(1.2)^{n-1} - \sum _{r=0} ^{n-2} {(1.2 )^r\times 14} $ ........for n>=2

$=5(1.2)^{n-1} - \frac{14((1.2)^{n-2} -1)}{0.2}$

**Danny** · February 20th 2009, 08:57 AM

Originally Posted by ADARSH

The general term of the sequence is given by
$ T_n= 5(1.2)^{n-1} - \sum _{r=0} ^{n-2} {(1.2 )^r\times 14} $ ........for n>=2

$T_n=5(1.2)^{n-1} - \frac{14((1.2)^{n-2} -1)}{0.2}$

I think you meant
$T_n=5(1.2)^{n-1} - \frac{14((1.2)^{n-1} -1)}{0.2}$

since what you have now $T_1 \ne 5$

However, I think it's easier to write the answer as

$T_n=-65 (1.2)^{n-1} +70$ .

**ADARSH** · February 20th 2009, 09:03 AM

Originally Posted by danny arrigo

I think you meant
$T_n=5(1.2)^{n-1} - \frac{14((1.2)^{n-1} -1)}{0.2}$

since what you have now $T_1 \ne 5$

However, I think it's easier to write the answer as

$T_n=-65 (1.2)^{n-1} +70$ .

Yeah it can further be simplified
but the second step was

$T_n=5(1.2)^{n-1} - \frac{14((1.2)^{n-1} -1)}{0.2}$
(as you said)
It was a typo

, Sorry for that

**Soroban** · February 20th 2009, 09:18 AM

Hello, mrszone!

Find the closed form: . $x_1 = 5,\;x_{n+1} \:=\:1.2x_n - 14$

It is not a geometric series . . . . $x_1=5,\;x_2 = \text{-}8,\;x_3 =\text{-}23.6 \;\hdots$

. . $\begin{array}{cccccc}\text{We have:} & x_{n+1} &=& 1.2x_n - 14 & {\color{blue}[1]}\\ \text{and:} & x_{n+2} &=& 1.2x_{n+1} - 14 & {\color{blue}[2]} \end{array}$

Subtract [2] - [1]: . $x_{n+2} - x_{n-1} \;=\;1.2x_{n+1} - 1.2x_n \quad\Rightarrow\quad x_{n+2} - 2.2x_{n+1} + 1.2x_n \:=\:0$

Suppose the function is exponential: . $x_n \:=\:Y^n$

. . The equation becomes: . $Y^{n+2} - 2.2Y^{n+1} + 1.2Y^n \:=\:0$

. . which factors: . $Y^n\left(Y^2 - 2.2Y + 1.2\right) \:=\:0 \quad\Rightarrow\quad Y^n(Y - 1)(Y - 1.2) \;=\;0$

. . and has nonzero roots: . $Y \:=\:1,\:1.2$

$\text{The function, }f(n)\text{, is either: }\:1^n\,\text{ or }\,1.2^n$

Form a linear combination of the two functions: . $f(n) \;=\;A + B(1.2)^n$

From the first two values of the sequence:

. . $\begin{array}{cccccc}f(1)=5\!: & A + 1.2B &=& 5 & {\color{blue}[3]} \\ f(2)=\text{-}8\!: & A + 1.44B &=&\text{-}8 & {\color{blue}[4]} \end{array}$

Subtract [4] - [3]: . $0.24B \:=\:\text{-}13 \quad \Rightarrow\quad\boxed{ B \:=\:-\frac{325}{6}}$

Substitute into [3]: . $A + (1.2)\left(-\tfrac{325}{6}\right) \:=\: 5 \quad\Rightarrow\quad\boxed{ A \:=\:70}$

Therefore: . $f(n) \;=\;70 -\frac{325}{6}(1.2)^n$

Edit: a silly typo! . . . fixed it.
.

**ADARSH** · February 21st 2009, 07:20 AM

Originally Posted by Soroban

Hello, mrszone!

It is not a geometric series . . . . $x_1=5,\;x_2 = \text{-}8,\;x_3 =\text{-}23.6 \;\hdots$

. . $\begin{array}{cccccc}\text{We have:} & x_{n+1} &=& 1.2x_n - 14 & {\color{blue}[1]}\\$ $ 

Subtract [2] - [1]: . $x_{n+2} - x_{n-1} \;=\;1.2x_{n+1} - 1.2x_n \quad\Rightarrow\quad x_{n+2} - 2.2x_{n+1} + 1.2x_n \:=\:0$

Suppose the function is exponential: . $x_n \:=\:Y^n$

. . The equation becomes: . $Y^{n+2} - 2.2Y^{n+1} + 1.2Y^n \:=\:0$

. . which factors: . $Y^n\left(Y^2 - 2.2Y + 1.2\right) \:=\:0 \quad\Rightarrow\quad Y^n(Y - 1)(Y - 1.2) \;=\;0$

. . and has nonzero roots: . $Y \:=\:1,\:1.2$

$\text{The function, }f(n)\text{, is either: }\:1^n\,\text{ or }\,1.2^n$

Form a linear combination of the two functions: . $f(n) \;=\;A + B(1.2)^n$

From the first two values of the sequence:

. . $\begin{array}{cccccc}f(1)=5\!: & A + 1.2B &=& 5 & {\color{blue}[3]} \\$ $ 

Subtract [4] - [3]: . $0.24B \:=\:\text{-}13 \quad \Rightarrow\quad\boxed{ B \:=\:-\frac{325}{6}}$

Substitute into [3]: . $A + (1.2)\left(-\tfrac{325}{6}\right) \:=\: 5 \quad\Rightarrow\quad\boxed{ A \:=\:70}$

Therefore: . $f(n) \;=\;70 -\frac{325}{6}(1.2)^m$

Just a (very) small thing
m = n

$f(n) \;=\;70 -\frac{325}{6}(1.2)^{n}$

And I think we got it same

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