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Math Help - Reccurence relation

  1. #1
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    Reccurence relation

    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
    Last edited by sammy28; October 27th 2009 at 03:13 AM. Reason: typo
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  2. #2
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    Try making U_{n+1} the subject of the formula U_n=\frac5{U_{n+1}}+1.
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  3. #3
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    thanks for that, but thats what the second part was saying apart from spelling it out.

    5U_{2}=5 + U_{2}
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  4. #4
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    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}

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  5. #5
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    thats great soroban . i think this is the model answer they must have been looking for.
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  6. #6
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    Quote Originally Posted by Soroban View Post
    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!
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  7. #7
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    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}
    Last edited by sammy28; October 28th 2009 at 03:02 PM. Reason: added reasoning
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  8. #8
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    Quote Originally Posted by sammy28 View Post
    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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