Continuous Fractions: Solving for k Limitations - Martin

[Calavera]

Hi there all smart people!
I'm doing some work on continued fractions of this type:
http://viitanen.se/cf.gif
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 between 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

 

[shmoe]

Is $$t_n$$ 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:

$$k$$

$$k+\frac{1}{k}$$

$$k+\frac{1}{k+\frac{1}{k}}$$

etc.

So what is your $$t_n$$?

 

[Calavera]

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.

 

[shmoe]

So $$t_0=k$$? This is the usual way, but then with k=0 you have $$t_0=0$$, and $$t_1=k+\frac{1}{t_0}=0+\frac{1}{0}$$, which is undefined.

 

[Calavera]

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.

 

[shmoe]

Our continuous fraction starts at t1...

So that t1=k+1

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

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

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.

 

[Calavera]

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.