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

  • Thread starter Pirang
  • Start date
  • Tags
In summary, you have been able to show that the continued fraction sequence is increasing and converging towards a number. However, for k=0 the sequence is undefined.
  • #1
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.


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:


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

There i wrote Tn = Tn+1

and inserting this into


you get



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

x^2 - x - 1 = 0

so the exact value is:





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.
Last edited by a moderator:
Physics news on Phys.org
  • #2
You haven't finished the first problem!

You've shown that if the limit exists, then it is either equal to [itex](1 + \sqrt{5}) / 2[/itex] or [itex](1 - \sqrt{5})/2[/itex]. 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?
  • #3
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?
  • #4
Have you actually looked (numerically) at the terms in your sequence? How would you describe them?
  • #5
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 [itex](1 - \sqrt{5}) / 2[/itex] solution to the problem and leaves [itex](1 + \sqrt{5}) / 2[/itex].

  • #6
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!
  • #7
it looks to me as if the sequence is decreasing and moving towards the asymptote which i guess is [itex](1 + \sqrt{5}) / 2[/itex]
  • #8
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.
  • #9
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?
  • #10
k=-1 has a well defined limit, while k=0 does not (it oscillates), although both will probably be considered invalid.
  • #11
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:

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 [itex]-((-1 + \sqrt{5}) / 2)^2[/itex].

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.
  • #12
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...

  • #13
Ah, that's the problem. I usually see the the continued fraction defined by this sequence:

k + 1/k
k + 1/(k + 1/k)
k + 1/(k + 1/(k + 1/k)
  • #14
You could reach a compromise of

In this way, the values for k=-1 become:
[tex]-1,0,-2,[\tex]und[tex],-1.5,\to-1,-1.\overbar{6},\to-2,-1.6,\to-1.5,-1.625,\to-1.\overbar{6},...[/tex], tending towards [tex]-\frac{1+\sqrt5}{2}[/tex]
And for k=0,
Last edited:
  • #15
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
  • #16
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.
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.

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

1. What is a continued fraction?

A continued fraction is a mathematical expression that represents a real number as a sequence of nested fractions. It is written in the form of [a0; a1, a2, ..., an], where a0 is the whole number part and a1, a2, ..., an are the coefficients of the partial fractions.

2. What is the purpose of solving continued fraction problems?

The main purpose of solving continued fraction problems is to accurately approximate irrational numbers. Continued fractions allow for a more precise representation of irrational numbers compared to other numerical systems, making them useful in various mathematical and scientific applications.

3. How do you convert a decimal number into a continued fraction?

To convert a decimal number into a continued fraction, follow these steps:

  • Separate the whole number part from the decimal part.
  • Write the whole number as the first term in the continued fraction.
  • Take the reciprocal of the decimal part and write it as the next term in the continued fraction.
  • Repeat the process with the new decimal part until you reach a point of termination, where the decimal part becomes zero or starts repeating.

4. How do you solve a continued fraction problem?

To solve a continued fraction problem, you can either use the Euclidean algorithm or the continued fraction algorithm. The Euclidean algorithm involves finding the greatest common divisor of the numerators and denominators of the fractions, while the continued fraction algorithm involves manipulating the partial fractions to find a pattern and solve for the original number.

5. What are some applications of continued fractions in science?

Continued fractions have various applications in science, including:

  • Approximating irrational numbers, such as pi and square roots, for more accurate calculations in engineering and physics.
  • Generating optimal rational approximations for real numbers, which is useful in data compression and signal processing.
  • Representing complex numbers in a more concise and efficient manner for use in quantum mechanics and other areas of physics.
  • Studying the properties of quadratic irrational numbers in number theory and algebra.

Similar threads

  • Calculus and Beyond Homework Help
  • Introductory Physics Homework Help
  • Calculus and Beyond Homework Help
  • Calculus and Beyond Homework Help
  • Sticky
  • Topology and Analysis
  • Calculus and Beyond Homework Help
  • Calculus and Beyond Homework Help
  • Calculus and Beyond Homework Help