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Continuos fractions

  1. Nov 27, 2005 #1
    Hi there all smart people!
    I'm doing some work on continued fractions of this type:
    [​IMG]
    I'w worked out an formula for the exact value of tn and I'm now looking for limitations for that formula...
    K≠-1 is one limitation since it will give dev. by 0.
    My question now is:
    Is k=0 a possible value, it gives the same value for all tn, i.e. 1. Though, it does not generate the same pattern as other values of k.
    Is a value betwen o and -1 a possible value for k? These does not give the same pattern as other values.
    Note that I get the right answer for the value of tn using the formual, my question is simply if they are a part of the continuos fraction even if they don't follow the same pattern on the graph?


    Please answer asap!
    //Martin
     
  2. jcsd
  3. Nov 27, 2005 #2

    shmoe

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    Is [tex]t_n[/tex] supposed to be the nth convergent? How are you defining the nth convergent here anyway? The usual way would make this undefined for k=0, the sequence of convergents usually looks like:

    [tex]k[/tex]

    [tex]k+\frac{1}{k}[/tex]

    [tex]k+\frac{1}{k+\frac{1}{k}}[/tex]

    etc.

    So what is your [tex]t_n[/tex]?
     
  4. Nov 27, 2005 #3
    tn is the nth value of the continued fraction.

    tn+1 is defined as:
    tn+1=k+(1/tn)


    The problem is that I don't know if k=0 gives an continuous fraction since it only gives one value for tn, independent of the n value.
     
  5. Nov 27, 2005 #4

    shmoe

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    So [tex]t_0=k[/tex]? This is the usual way, but then with k=0 you have [tex]t_0=0[/tex], and [tex]t_1=k+\frac{1}{t_0}=0+\frac{1}{0}[/tex], which is undefined.
     
  6. Nov 27, 2005 #5
    Our continuous fraction starts at t1...

    So that t1=k+1
    so if k=0, the first value would be 1
    but the thing is that all the others would also equal 1

    There is only one undefined value for k, which is -1, since it gives dev by 0.
     
  7. Nov 27, 2005 #6

    shmoe

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    Alright, that's why I was asking what [tex]t_n[/tex] was. It looked like you had a different definition for the nth convergent and I just wanted to be sure.

    That shouldn't be a problem. A continued fraction is said to be convergent if the sequence of nth convergents is convergent. The constant sequence is convergent, so no problem.
     
  8. Nov 27, 2005 #7
    Ok, thank you really much!! :)

    Finally I'm done with my 16 pages and 2728 word long work about this...now I only need someone who can proofread it.
     
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