# Continuos fractions

1. Nov 27, 2005

### Calavera

Hi there all smart people!
I'm doing some work on continued fractions of this type:

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?

//Martin

2. Nov 27, 2005

### 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 $$t_n$$?

3. Nov 27, 2005

### 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.

4. Nov 27, 2005

### 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.

5. Nov 27, 2005

### 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.

6. Nov 27, 2005

### shmoe

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.

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.

7. Nov 27, 2005

### Calavera

Ok, thank you really much!! :)