What is the generalized formula for the exact value of a continued fraction?

[Pirang]

Hello, I was not sure if this belonged in the precalculus or calculus section so i hope no one minds that i posted it here.

This is a problem that has completely lost me.


http://img76.imageshack.us/img76/9191/continuousfractionrg3.gif

T1 = 1 + 1

T1 = 1+ (1/1+1)

T3 = 1+ (1/(1+1/1+1)

and so on

I have been able to draw a generalized formula for Tn+1 in terms of Tn and that is:

[PLAIN]http://img76.imageshack.us/img76/3978/continuousfraction2fg8.gif

i realized that as n increases to infinity, (Tn+1) - Tn converges towards 0

There i wrote Tn = Tn+1

and inserting this into

[PLAIN]http://img76.imageshack.us/img76/3978/continuousfraction2fg8.gif

you get

http://img76.imageshack.us/img76/1003/continuousfraction3nu9.gif

Therefore

An exact value for the continued fraction will be considering Tn = x :

x^2 - x - 1 = 0

so the exact value is:

http://img76.imageshack.us/img76/3245/continuousfraction4tw6.gif

Now,

Considering

http://img161.imageshack.us/img161/691/continuousfraction5az5.gif

For any value of k, I am supposed to determine a generalized statement for the exact value of any such continued fraction. For which values of k does this hold true and how do i know?

Can anyone please help me with this, i got so far and do not want to give up. Any tips? anything.

 

[Hurkyl]

You haven't finished the first problem!

You've shown that if the limit exists, then it is either equal to (1 + \sqrt{5}) / 2 or (1 - \sqrt{5})/2. You haven't shown that the limit does, in fact, exist... nor have you figured out which of those values it equals.


As for the second problem... have you tried doing exactly the same thing that you did for the first problem?

 

[Pirang]

NO! don't tell me i did not finish the first problem... haha. So how can i show that the limit exists or which of the values it equals?

 

[Hurkyl]

Have you actually looked (numerically) at the terms in your sequence? How would you describe them?

 

[Pirang]

well the sequence for the first 10 terms goes like this

1/2, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/35, 89/55, 144/89

from what i can tell these number are converging towards a number and since the numerator is always greater than the denominator, the number they are converging towards must be > 1 .

So that eliminates the (1 - \sqrt{5}) / 2 solution to the problem and leaves (1 + \sqrt{5}) / 2.

Correct?

 

[Hurkyl]

Importantly, it's a increasing sequence, right? And increasing, bounded sequences converge!

You have a guess as to what the upper bound is -- so all you have to do is prove it starts below that upper bound, and then each time you iterate, the value increases, but stays below that upper bound!

 

[Pirang]

it looks to me as if the sequence is decreasing and moving towards the asymptote which i guess is (1 + \sqrt{5}) / 2

 

[Hurkyl]

Er, wait. Silly mistake. It's an alternating sequence! I knew that increasing sounded wrong.

So you just need to show the magnitude of the differences is a decreasing sequence.

In fact, that's the essential property of a continued fraction: the terms in any such sequence (called convergents) form an alternating sequence whose differences are decreasing. Wikipedia says a lot about them.

 

[Pirang]

I did some calculations and found that the only numbers for k that make the fraction invalid are 0 and -1.

What do you say?

 

[lotrgreengrapes7926]

k=-1 has a well defined limit, while k=0 does not (it oscillates), although both will probably be considered invalid.

 

[Hurkyl]

Okay, so you got that it works for positive integers k, which agrees with the usual theory, so that's good!


The sequences is, indeed, not well defined for k=0. The first few convergents would be:
0/1
1/0
0/1
...

so every other term is undefined.


For k = -1, they are:
-1 / 1
0 / -1
-1 / 2
1 / -3
-2 / 5
3 / -8
...
which appear to be converging to -((-1 + \sqrt{5}) / 2)^2.


I'm pretty sure the theory in the Wikipedia link I gave will actually work for any domain (not just the integers). Well, I broke down and grinded it out! Let r and s be the two roots to:

x^2 - kx - 1

then you can rigorously prove that the n-th convergent is

(r^(n+1) - s^(n+1)) / (r^n - s^n)

If |r| = |s| (i.e. when k = 0), the denominator is zero, which gives you problems. Otherwise, assume |r| > |s|, (and so |r| > 1) so the n-th convergent is

(r^(2n+1) - (-1)^n s) / (r^(2n) - (-1)^n)

which converges to r.

 

[Pirang]

I don't quite think i understand this.

As i see it,

for k = -1

1st term = -1 + 1 = 0
2nd term = -1+1/-1+1 = undefined
3rd term = undefined
4th term undefined

This happens since the bottom part is always -1 + 1 = 0 and you can't divide by 0.
What you guys are saying is considering what I posted right? not any other continued fraction?

What is this convergents and things you guys are talking about? Sorry, but I don't understand the theorems. This is the first time i see a continued fraction hehe. Maybe you could explain it a bit more? I need to understand this stuff for Monday...

Thanks

 

[Hurkyl]

Ah, that's the problem. I usually see the the continued fraction defined by this sequence:

k
k + 1/k
k + 1/(k + 1/k)
k + 1/(k + 1/(k + 1/k)
...

 

[lotrgreengrapes7926]

You could reach a compromise of
k
k+1
k+1/k
k+1/(k+1)
k+1/(k+1/k)
k+1/(k+1/(k+1))
...

In this way, the values for k=-1 become:
-1,0,-2,[\tex]und,-1.5,\to-1,-1.\overbar{6},\to-2,-1.6,\to-1.5,-1.625,\to-1.\overbar{6},..., tending towards -\frac{1+\sqrt5}{2}<br /> And for k=0,<br /> 0,1,[\tex]und,1,\to0,1,[\tex]und,1,\to0,1,[\tex]und,1,\to0,1,[\tex]und,1,\to0,...

 

[20010958]

so what is the general formula?? i need help quick!

and i don't understand how you guys went from x^2 - x - 1=0 to (root5 +1)/2

help please

 

[Thor1123]

The general formula is (K+sqrt((K^2)+4))/2 for k>0 and (k-sqrt((K^2)+4))/2 for k<0. Let x be the continued fraction.
x-k=1/(K+1(...)
1/(x-k)=k+(1/(K+...)=x
1=x^2 - xk
0=x^2 - xk - 1
Using the quadratic formula, x= (k+sqrt((k^2)+4))/2
If k<0, x<0 so you subtract the sqrt instead of adding.