# Reccurence relation

• Oct 27th 2009, 02:34 AM
sammy28
Reccurence relation
hi all,

Find the first five terms of these sequence:

$\displaystyle U_{1} = 5, U_{n} = \frac{5}{U_{n+1}} + 1$ for $\displaystyle n \geq 2$

how does this work given that the first term is 5 and for n to be greater or equal to 2 then

$\displaystyle U_{2} = \frac{5}{U_{3}} + 1$

The only way this seems to make sense if

$\displaystyle 5=\frac{5}{U_{2}} + 1$ but then wouldnt this be out of range?

thank sammy
• Oct 27th 2009, 02:48 AM
proscientia
Try making $\displaystyle U_{n+1}$ the subject of the formula $\displaystyle U_n=\frac5{U_{n+1}}+1.$
• Oct 27th 2009, 03:10 AM
sammy28
thanks for that, but thats what the second part was saying apart from spelling it out.

$\displaystyle 5U_{2}=5 + U_{2}$
• Oct 27th 2009, 06:30 AM
Soroban
Hello, Sammy!

Quote:

Find the first five terms of this sequence:

$\displaystyle U_{1} = 5,\;\;U_{n} \:=\: \frac{5}{U_{n+1}} + 1\;\text{ for }n \geq 2$

What a stupid way to give a recurrence!

We have: .$\displaystyle U_n \;=\;\frac{5}{U_{n+1}} + 1$

Multiply by $\displaystyle U_{n+1}\!:\quad U_n\!\cdot\!U_{n+1} \;=\;5 + U_{n+1}$

Rearrange: .$\displaystyle U_n\!\cdot\!U_{n+1} - U_{n+1} \;=\;5$

Factor: .$\displaystyle U_{n+1}\left(U_n - 1\right) \;=\;5$

Hence: .$\displaystyle U_{n+1} \;=\;\frac{5}{U_n-1}$

The first five terms are:

. . $\displaystyle \begin{array}{ccccc} U_1 &=& 5 &=& 5\\ \\[-3mm] U_2 &=& \dfrac{5}{5-1} &=& \dfrac{5}{4} \\ \\[-3mm] U_3 &=& \dfrac{5}{\frac{5}{4}-1} &=& 20 \\ \\[-3mm] U_4 &=& \dfrac{5}{20-1} &=& \dfrac{5}{19} \\ \\[-3mm] U_5 &=& \dfrac{5}{\frac{5}{19} - 1} &=& \text{-}\dfrac{95}{14} \end{array}$

• Oct 27th 2009, 06:51 AM
sammy28
thats great soroban (Clapping). i think this is the model answer they must have been looking for.
• Oct 28th 2009, 08:06 AM
HallsofIvy
Quote:

Originally Posted by Soroban
Hello, Sammy!

What a stupid way to give a recurrence!

We have: .$\displaystyle U_n \;=\;\frac{5}{U_{n+1}} + 1$

Multiply by $\displaystyle U_{n+1}\!:\quad U_n\!\cdot\!U_{n+1} \;=\;5 + U_{n+1}$

Rearrange: .$\displaystyle U_n\!\cdot\!U_{n+1} - U_{n+1} \;=\;5$

Factor: .$\displaystyle U_{n+1}\left(U_n - 1\right) \;=\;5$

Hence: .$\displaystyle U_{n+1} \;=\;\frac{5}{U_n-1}$

The first five terms are:

. . $\displaystyle \begin{array}{ccccc} U_1 &=& 5 &=& 5\\ \\[-3mm] U_2 &=& \dfrac{5}{5-1} &=& \dfrac{5}{4} \\ \\[-3mm] U_3 &=& \dfrac{5}{\frac{5}{4}-1} &=& 20 \\ \\[-3mm] U_4 &=& \dfrac{5}{20-1} &=& \dfrac{5}{19} \\ \\[-3mm] U_5 &=& \dfrac{5}{\frac{5}{19} - 1} &=& \text{-}\dfrac{95}{14} \end{array}$U[sub]2[/sup]

But you can't use that formula to find $\displaystyle U_2$ from $\displaystyle U_1$ because we don't know if it applies! We are told that $\displaystyle U_n= \frac{5}{U_n}+ 1$ for $\displaystyle n\ge 2$. Assuming that U_2= \frac{5}{U_1- 1}[/tex] assumes that the formula applies to n= 1 which we are NOT told!

I thought that was the whole point of sammy28's original post!
• Oct 28th 2009, 02:48 PM
sammy28
hi HallsofIvy, yes that was my original question. i should have checked the answer section sorry.

the answer given is

$\displaystyle 5,2,\frac{7}{2},\frac{17}{7},\frac{52}{17}$

its a bit past my bedtime at the minute so ill look into it tomorrow

do you think its a typo in the question? perhaps it should be

$\displaystyle U_{n-1}$ rather than $\displaystyle U_{n+1}$
• Nov 1st 2009, 01:43 AM
CaptainBlack
Quote:

Originally Posted by sammy28
hi HallsofIvy, yes that was my original question. i should have checked the answer section sorry.

the answer given is

$\displaystyle 5,2,\frac{7}{2},\frac{17}{7},\frac{52}{17}$

its a bit past my bedtime at the minute so ill look into it tomorrow

do you think its a typo in the question? perhaps it should be

$\displaystyle U_{n-1}$ rather than $\displaystyle U_{n+1}$

It did seem likely that there was a typo right from the word go, the quoted solution confirms it.

CB