Reccurence relation

**sammy28** · October 27th 2009, 02:34 AM

hi all,

Find the first five terms of these sequence:

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

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

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

The only way this seems to make sense if

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

thank sammy

**proscientia** · October 27th 2009, 02:48 AM

Try making $U_{n+1}$ the subject of the formula $U_n=\frac5{U_{n+1}}+1.$

**sammy28** · October 27th 2009, 03:10 AM

thanks for that, but thats what the second part was saying apart from spelling it out.

$5U_{2}=5 + U_{2}$

**Soroban** · October 27th 2009, 06:30 AM

Hello, Sammy!

Find the first five terms of this sequence:

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

What a stupid way to give a recurrence!

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

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

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

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

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

The first five terms are:

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

**sammy28** · October 27th 2009, 06:51 AM

thats great soroban

. i think this is the model answer they must have been looking for.

**HallsofIvy** · October 28th 2009, 08:06 AM

Originally Posted by Soroban

Hello, Sammy!

What a stupid way to give a recurrence!

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

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

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

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

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

The first five terms are:

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

But you can't use that formula to find $U_2$ from $U_1$ because we don't know if it applies! We are told that $U_n= \frac{5}{U_n}+ 1$ for $n\ge 2$ . 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!

**sammy28** · October 28th 2009, 02:48 PM

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

the answer given is

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

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

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

**CaptainBlack** · November 1st 2009, 01:43 AM

Originally Posted by sammy28

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

the answer given is

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

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

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

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

CB

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