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Math Help - Help with Calculating Convergents of a continued fraction (Semi-Urgent)

  1. #1
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    Help with Calculating Convergents of a continued fraction (Semi-Urgent)

    Hi, Say i have the continued fraction Sqrt(23) = [2,1,3,1,8] (1,3,1,8 reoccuring)

    How do i calculate it's convergents? I can do it for a normal continued fraction but it appears to be done differently..

    The solution has a table which reads as follows

    4 4 1
    1 5 1
    3 10 4
    1 24 5

    Column 1 is obviously the continued fraction.. where do the convergents come from? It isn't at all obvious from looking at it to me...

    Thank you
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  2. #2
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    Hello, Bruce!

    Say i have the continued fraction: . \sqrt{23}\;=\;[4,\overline{1,3,1,8}]

    How do i calculate it's convergents?

    The solution has a table which reads as follows

    \begin{array}{ccc}4 & 4 & 1 \\1 & 5 & 1 \\3 & 19 & 4 \\1&  24 & 5 \end{array}

    Column 1 is obviously the continued fraction.
    Where do the convergents come from?
    I'll try to explain the routine that I was taught . . .


    Place the numbers in a row of a horizontal chart.

    \begin{array}{ccccccccccccc}n &|& 1 &|& 2 &|& 3 &|& 4 &|& 5 &| & \hdots \\ \hline a_n &|&  4 & | & 1 &|& 3 &|& 1 &|& 8 &|& \cdots \\ \hline p_n & | &  & | &  &|&   &|&     &|&   &|& \cdots \\ \hline q_n & | &  & | &  &|& &|&  &|& &|& \cdots \\\end{array}


    p_1 = a_1,\;q_1 = 1

    p_2 = a_2p_1 + 1,\;q_2 = a_2

    The table looks like this:

    \begin{array}{ccccccccccccc}n &|& 1 &|& 2 &|& 3 &|& 4 &|& 5 &| & \hdots \\ \hline a_n &|&  4 & | & 1 &|& 3 &|& 1 &|& 8 &|& \cdots \\ \hline p_n & | & 4 & | & 5 &|&   &|&     &|&   &|& \cdots \\ \hline q_n & | & 1 & | & 1 &|& &|&  &|& &|& \cdots \end{array}



    The rows can be completed with this strange move . . .

    \begin{array}{ccccccccccccc} a_n &|&  4 &  & 1 & & 3 &\quad & 1 &\quad & 8 \\ & & & & &\swarrow \\ p_n & |& 4 & \leftarrow  & 5 & & {\color{blue}19}   \end{array}

    This is from: . 3 \times 5 + 4 \:=\:19


    The next p comes from:

    \begin{array}{ccccccccccccc} a_n &|&  4 &  & 1 & & 3 &\quad & 1 &\quad & 8 \\ & & & & & & & \swarrow \\ p_n & |& 4 &  & 5 & \leftarrow & 19 & & {\color{blue}24}    \end{array}

    From: . 1 \times 19 + 5 \:=\:24



    The q rows are filled with an identical move . . .

    \begin{array}{ccccccccccccc}a_n &|&  4 &  & 1 & & 3 &\quad & 1 &\quad & 8 \\ & & & & &\swarrow \\ q_n & |& 1 & \leftarrow  & 1 & & {\color{blue}4} \end{array}

    From: . 3 \times 1 + 1 \:=\:4


    Then:

    \begin{array}{ccccccccccccc}a_n &|&  4 &  & 1 & & 3 &\quad & 1 &\quad & 8 \\ & & & & & & & \swarrow \\ q_n & |& 1 &  & 1 & \leftarrow & 4 & & {\color{blue}5} \end{array}

    From: . 1 \times 4 + 1 \:=\:5



    And the table can be completed:

    \begin{array}{ccccccccccccc} n &|& 1 &|& 2 &|& 3 &|& 4 &|& 5 &| & \hdots \\ \hline a_n &|&  4 & | & 1 &|& 3 &|& 1 &|& 8 &|& \cdots \\ \hline<br />
p_n & | & 4 & | & 5 &|&19 &|& 24 &|& 211 &|& \cdots \\ \hline  q_n & | & 1 & | & 1 &|& 4 &|& 5 &|& 44 &|& \cdots \end{array}

    The convergents are of the form: \frac{p_n}{q_n}


    The convergents are:

    . . \begin{array}{ccc}\frac{4}{1} & = & 4.000000000 \\ \\ \frac{5}{1} & = & 5.000000000 \\ \\ \frac{19}{4} & = & 4.750000000 \\ \\ \frac{24}{5} & = & 4.800000000 \end{array}

    . . \begin{array}{ccc} \frac{211}{44} & = & 4.795454545 \\ \\ \frac{235}{49} & = & 4.795918367 \\ \\ \frac{916}{191} & = & 4.795811518 \\ \\ \vdots & & \vdots\end{array}


    Note that the convergents are alternately above and below
    . . the exact value of: . \sqrt{23}\;=\;4.795831523...

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