Continued Fractions?

[fbellman]

Hi there! I need some help with continued fractions..

So I was given the following information:

and t sub1= (1+1)
t sub2= 1+ (1/(1+1))
t sub3= 1+ (1/(1+(1/(1+1))))
...

And here are my objectives:
1) Determine a generalized formula for t sub(n+1) in terms of t sub(n).
So I don't know if this is what they're talking about, but I figured out this: t sub1= 2/1. so t sub2 will equal 2+1 divided by two (aka 3/2). And then t sub3 will equal 3+2 divided by 3 (5/3). So it's basically the sum of the two numbers divided by the numerator. Is this helpful to me at all, other than in calculating the decimal equivalents?


2) Compute the decimal equivalents of the first ten terms [I've done that]. Enter the terms into a data table and plot the relation between n and t sub(n). What do you notice? What does this suggest about the value of t sub(n) - t sub(n+1) as n gets very large?

I notice that the equivalents get closer and closer together.. that's all I've got.

3. What problems arise when you try to determine the 200th term?

4. Use the results of step 1 and 2 to establish an exact value for the continued fraction. Doesn't it go on for infinity? How can there be an exact value???

Thanks a lot for any help!!!

 

[stapel]

Here's a trick:


. . . . .\(\displaystyle x\.=\,1\,+\,\frac{1}{1\,+\,\frac{1}{1\,+\,\frac{1}{...}}}\)


Note that:


. . . . .\(\displaystyle x\,=\,1\,+\,\frac{1}{x}\)


Then:

. . . . .x² = x + 1

. . . . .x² - x - 1 = 0

Solve using the Quadratic Formula.

Eliz.

 

[galactus]

Wow, this is fun. Something a little different.

This is one of the simplest and slowest to converge continued fractions.

As you can see, here are the partial quotients:

1, 1+1=2, \(\displaystyle 1+\frac{1}{1+1}=\frac{3}{2}\), \(\displaystyle 1+\frac{1}{1+\frac

{1}{1+1}}=\frac{5}{3},...............\), Without listing more, I can

tell the next few will be 8/5, 13/8, 21/13,................

The numerators and denominators of these partial quotients are the

Fibonacci numbers. See?. Do you know the Fibonacci numbers?.

1,1,2,3,5,8,13,21,34,55,...................

See the ratio?. Do you know what the ratio of the Fibonacci numbers

converges to?. That is, the ratio \(\displaystyle r_{n}=\frac{f_{n+1}}{f_{n}}\)

\(\displaystyle L=\lim_{n\to\infty}\frac{f_{n+1}}{f_{n}}\)

Knowing the limit exists, we can find it.

Start with \(\displaystyle f_{n+2}=f_{n+1}+f_{n}\)

Divide through by \(\displaystyle f_{n+1}\) and get the relation:

\(\displaystyle r_{n+1}=1+\frac{1}{r_{n}}\)

Let n go to infinity:

\(\displaystyle L=1+\frac{1}{L}\)

So we have a quadratic to solve:

\(\displaystyle L^{2}-L-1=0\)

Find the roots and thus your convergence.

 

[fbellman]

Ah! Thank you both very much, it makes a lot more sense now (and seems a lot easier)!

Thanks again!