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Math Help - Continued fraction

  1. #1
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    Continued fraction

    I am asked the following question..

    9. Evaluate the continued fraction x = 1 + 1/ [ 2 + 1/ [ 3 + 1/ [ 2 + 1/ [ 3 + 1/ ... ] ] ] ]
    Tabulate the first 6 convergents, and find the exact value of x.
    I am not wondering the answer, but just what is being asked here.

    What does "tabulate the first 6 convergents" mean exactly?

    Does this mean just continue the pattern up to 6 terms?
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  2. #2
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    Re: Continued fraction

    Hello, ehpoc!

    x \;=\;1 + \dfrac{1}{2 + \dfrac{1}{3 + \dfrac{1}{2 + \dfrac{1}{3 + \hdots }}}}}}

    I am not wondering the answer, but just what is being asked here.

    What does "tabulate the first 6 convergents" mean exactly?

    Does this mean just continue the pattern up to 6 terms? . Yes!

    Crank out the first six "terms".

    x_1 \;=\;1

    x_2 \;=\;1 + \frac{1}{2} \;=\;\frac{3}{2}

    x_3 \;=\;1 + \frac{1}{2 + \frac{1}{3}} \;=\;\frac{10}{7}

    x_4 \;=\;1 + \frac{1}{2 + \frac{1}{3 + \frac{1}{2}}} \;=\;\frac{23}{16}

    x_5 \;=\;1 + \frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3}}}} \;=\;\frac{79}{55}

    x_6 \;=\;1 + \frac{1}{2+\frac{1}{3 +\frac{1}{2 + \frac{1}{3 + \frac{1}{2}}}}} \;=\;\frac{181}{126}

    But check my work . . . please!

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  3. #3
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    Re: Continued fraction

    Is there any trick to solving a really messy looking fraction like x6?
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  4. #4
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    Re: Continued fraction

    Yes. It is called "using a computer". I cannot for the life of me see why someone would want to do continued fractions by hand.
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  5. #5
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    Re: Continued fraction

    I cannot for the life of me see why someone would want to do continued fractions by hand.
    What does "tabulate the first 6 convergents" mean exactly?
    Well sorroban suggested that this means calculate the first 6 convergents.

    He said..

    But check my work . . . please!
    So I assume he calculated it by hand.

    EDIT:

    I found the formula for taking the previous convergent to make solving the next convergent easier

    Term A is a rational number pn/qn

    Pn = AnP(n-1)+P(n-2)
    Qn = AnQ(n-1)+Q(n-2)
    Last edited by ehpoc; December 3rd 2011 at 03:08 PM.
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  6. #6
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    Re: Continued fraction

    Quote Originally Posted by ehpoc View Post
    I am asked the following question..



    I am not wondering the answer, but just what is being asked here.

    What does "tabulate the first 6 convergents" mean exactly?

    Does this mean just continue the pattern up to 6 terms?
    It may be useful to observe that if You want to obtain 'directly' the limit of the 'continous fraction'...

    x=1 + \frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3+...}}}  } (1)

    ... You can write...

    x=1 + \frac{1}{2 + \frac{1}{3 + x-1}} (2)

    ... and then solve (2) for x...

    Kind regards

    \chi \sigma
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  7. #7
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    Re: Continued fraction

    I am working on some more continued fractions.

    I am doing this one.

    x = 2 + 1/[2 + 1/[2 + 1/[2 + ...]]]

    I would be so so SO happy if someone could confirm if my solutions are correct.

    To tabulate the first 6 convergents I got

    x1=2
    x2=5/2
    x3=12/5
    x4=29/2
    x5=70/29
    x6=169/70

    To solve for the value to which x converges I did...

    x=2+(1/(2+1/x))

    x=2+(x/(2x+1))

    x=(5x+2)/(2x+1)

    2x^2+x=5x+2

    x^2-2x-1=0

    x=(2√8)/2

    x=√8

    Like I said I would be VERY VERY appreciative if someone could confirm if that solution is correct or not. Thanks.
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  8. #8
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    Re: Continued fraction

    Quote Originally Posted by ehpoc View Post
    I am asked the following question..



    I am not wondering the answer, but just what is being asked here.

    What does "tabulate the first 6 convergents" mean exactly?

    Does this mean just continue the pattern up to 6 terms?
    The first 6 convergents give you an idea of what x is.

    To find x

    x-1=k=\frac{1}{2+\frac{1}{3+k}}}

    \frac{1}{k}=2+\frac{1}{3+k}

    \frac{1-2k}{k}=\frac{1}{3+k}

    (3+k)(1-2k)=k\Rightarrow\ 3-6k+k-2k^2=k\Rightarrow\ 2k^2+6k-3=0

    For k>0, solve for k using the quadratic formula and deduce the value of x from x = 1+k.
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  9. #9
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    Re: Continued fraction

    The first 6 convergents give you an idea of what x is.

    To find x









    For k>0, solve for k using the quadratic formula and deduce the value of x from x = 1+k.
    Thanks. But that was my old question. Could you check my work in question 7. I am really anxious to see if I solved this one right.
    Last edited by ehpoc; December 17th 2011 at 05:48 PM.
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  10. #10
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    Re: Continued fraction

    Quote Originally Posted by ehpoc View Post
    I am working on some more continued fractions.

    I am doing this one.

    x = 2 + 1/[2 + 1/[2 + 1/[2 + ...]]]

    I would be so so SO happy if someone could confirm if my solutions are correct.

    To tabulate the first 6 convergents I got

    x1=2
    x2=5/2
    x3=12/5
    x4=29/2
    x5=70/29
    x6=169/70

    To solve for the value to which x converges I did...

    x=2+(1/(2+1/x))

    x=2+(x/(2x+1))

    x=(5x+2)/(2x+1)

    2x^2+x=5x+2

    x^2-2x-1=0

    x=(2√8)/2

    x=√8

    Like I said I would be VERY VERY appreciative if someone could confirm if that solution is correct or not. Thanks.
    You have typos on x_4

    and in solving your quadratic.

    x_1=2

    x_2=2+\frac{1}{2}=\frac{5}{2}

    x_3=2+\frac{2}{5}=\frac{12}{5}

    x_4=2+\frac{5}{12}=\frac{29}{12}

    x_5=2+\frac{12}{29}=\frac{70}{29}

    x_6=2+\frac{29}{70}=\frac{169}{70}

    A quicker way to solve for x (as it is simpler than your first example) is

    x=2+\frac{1}{x}=\frac{2x+1}{x}

    x^2-2x-1=0

    x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{4+4}}{2}

    and as x is positive

    x=\frac{2+\sqrt{8}}{2}=1+\frac{\sqrt{8}}{\sqrt{4}}  =1+\sqrt{2}
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  11. #11
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    Re: Continued fraction

    see I think my biggest problem is recognizing exactly what I need to have as the equation that x equals.

    in my example i had

    x=2+(1/(2+1/x))

    and you had

    x=2+(1/x)

    So I guess first before I worry about solving the equation I must first be able to recognize it properly.

    I have a list of practice examples. Can someone confirm if I am recognizing the equation that x equals properly when finding what x converges to?



    1. x = 1 + 1/[1 + 1/[1 + 1/[1 + ...]]]
    x=1+1/x?

    2. x = 2 + 1/[2 + 1/[2 + 1/[2 + ...]]]
    x=2+1/x?

    3. x = 1 + 1/[2 + 1/[3 + 1/[2 + 1/[3 + ...]]]]
    this one confuses me because a 1 shows up in the first term and never again.

    Is it...

    x=1+1/[2+1/[3+1/x]]?

    or

    x=2+1/[3+1/x]?

    4. x = 1 + 1/[2 + 1/[4 + 1/[1 + 1/[3 + 1/[1 + 1/[3 + ...]]]]]]
    This one confuses me so much!!

    is it?

    x=2 + 1/[4 + 1/[1 + 1/[3 + 1/[1 + 1/[3 + 1/x]]]]?

    Any help would be greatly appreciated as I need to nail this before I can move on to solving the fractions right?
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  12. #12
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    Re: Continued fraction

    Could you write out number 4 above again please?

    x=1+\frac{1}{\left(1+\frac{1}{1+......} \right)}

    Notice the expression in parentheses is also x, so this is

    x=1+\frac{1}{x}


    x=2+\frac{1}{\left(2+\frac{1}{2+....}\right)}

    x=2+\frac{1}{x}


    x=1+\left(\frac{1}{2+\frac{1}{3+\left(\frac{1}{2+ \frac{1}{3+....}\right)}}}\right)

    The terms in brackets are recurring, so start by subtracting 1 from both sides.

    x=1+k

    k=\frac{1}{2+\frac{1}{3+k}}

    Solve for k, then for x.
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  13. #13
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    Re: Continued fraction

    thanks SO much for the clarification!!!!


    PS. WHy do you want me to write out 4 again? Nothing will change LOL.
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  14. #14
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    Re: Continued fraction

    Ok, it's just a bit deeper then, though you have written it twice and they differ.

    If it's

    x=1+\frac{1}{2+\frac{1}{4+\frac{1}{\left(1+\frac{1  }{3+ \frac{1}{\left(1+\frac{1}{3+.....}\right)}}\right)  }}}

    then you get

    \frac{1}{x-1}=2+\frac{1}{4+\frac{1}{k}}

    so use that to write the relationship between x and k,
    and use the "nesting" to write

    k=1+\frac{1}{3+\frac{1}{k}}

    to solve for k
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  15. #15
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    Re: Continued fraction

    I am a little confused. Do I simplify for x first equation. Then simplify for k, then sub k into the other equation?
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