continued fractions
Created | Updated Feb 19, 2003
A continued fraction, rather than an ordinary fraction, differs from the latter in that you can have a series of fractions, all under each other. This is hard to illustrate and it's going to be no easier here. Here's an example of a continued fraction: 1+1/(2+1/3)
This continued fraction is called 'simple' as all the divisors are equal to 1. You can have the divisors being something different all the way, but here we will only consider simple continued fractions.
The above example, by the way, is equal to 10/7. A shorthand to represent simple continued fraction is to say [a0; a1, a2, ... , an], and a0 is the first number before a reciprocal is taken: which is the integral value of the continued fraction. So the above example is [1;2,3]. To get you started we'll use the full notation in the below examples.
If you have access to a math formatting system, you will see what I am trying to show you much clearer.
Here's another example of a continued fraction. I'll show you how you can find the value of this one too.
Let's say
k=2+1/(3+1/(5+1/8))
Like our order of operations tell us, we do the brackets first. Doing the most innermost one, 5+1/8=41/8
So now
k=2+1/(3+1/(41/8))
Simplifying
k=2+1/(3+(8/41))
Again, working out the brackets, 3+8/41 is 131/41
So now
k=2+1/(131/41)
k=2+41/131=303/131
That seemed like a lot of work! But continued fractions aren't just little fun things to use to change numbers into other forms. Here's some other interesting examples.
You're all familiar with such numbers as pi, and e, and the square root of 2. Let's look ath the square root of two, first.
We're going to extract the continued fraction representation of a number from it's decimal expansion. You can do this for any number.
We know sqrt(2) is approximately 1.4142135...
First we need the integer at the front, obviously this is 1
Subtract that and we're left with:
.4142135...
Find the reciprocal of this and we have
2.414213...
Take the integer off the front, 2, and repeat the process, subtract:
.4142135...
find the reciprocal
2.414213...
take the integer off the front, 2...
So, if we continue like this, we go on forever: the continued fraction for the square root of 2 is 1+1/(2+1/(2+1/(2+1/(2......)). The fact is, that every quadratic irrational (that is every square root in the form (a+sqrt(b))/c)has a continued fraction that has a repeating block!
As an exercise, try doing the same with e, pi, and the irrational (1+sqrt(5))/2, or the golden ratio (for mor einformation on the golden ratio see http://www.bbc.co.uk/dna/h2g2/alabaster/A643844). See what you find!
Things get meaty when you look at things called convergents. These are successive continued fractions which allow us to approximate any irrational number by a series of rational ones. Let's look at our sqrt(2) example. Our convergents will be (using the shorter notation) [1], [1;2], [1;2,2], [1;2,2,2..] and so on. If you work these out, and obtain their decimal representation, you'll find that they are successively closer to 1.4142135... . Where this becomes very useful is where the pattern of convergents can be expressed in algebraic terms: what does this mean? A convergent will be in the form a0/b0, since it's a continued fraction which we've simplified down like above. Now, we'll call the next convergent a1/b1, and so on. Now, if the series a_k and b_k are series we can describe by a function, then we can describe how a_k/b_k behaves as k gets large: lim a_k/b_k=R, and R becomes the number that the continued fraction represents.
This continued fraction is called 'simple' as all the divisors are equal to 1. You can have the divisors being something different all the way, but here we will only consider simple continued fractions.
The above example, by the way, is equal to 10/7. A shorthand to represent simple continued fraction is to say [a0; a1, a2, ... , an], and a0 is the first number before a reciprocal is taken: which is the integral value of the continued fraction. So the above example is [1;2,3]. To get you started we'll use the full notation in the below examples.
If you have access to a math formatting system, you will see what I am trying to show you much clearer.
Here's another example of a continued fraction. I'll show you how you can find the value of this one too.
Let's say
k=2+1/(3+1/(5+1/8))
Like our order of operations tell us, we do the brackets first. Doing the most innermost one, 5+1/8=41/8
So now
k=2+1/(3+1/(41/8))
Simplifying
k=2+1/(3+(8/41))
Again, working out the brackets, 3+8/41 is 131/41
So now
k=2+1/(131/41)
k=2+41/131=303/131
That seemed like a lot of work! But continued fractions aren't just little fun things to use to change numbers into other forms. Here's some other interesting examples.
You're all familiar with such numbers as pi, and e, and the square root of 2. Let's look ath the square root of two, first.
We're going to extract the continued fraction representation of a number from it's decimal expansion. You can do this for any number.
We know sqrt(2) is approximately 1.4142135...
First we need the integer at the front, obviously this is 1
Subtract that and we're left with:
.4142135...
Find the reciprocal of this and we have
2.414213...
Take the integer off the front, 2, and repeat the process, subtract:
.4142135...
find the reciprocal
2.414213...
take the integer off the front, 2...
So, if we continue like this, we go on forever: the continued fraction for the square root of 2 is 1+1/(2+1/(2+1/(2+1/(2......)). The fact is, that every quadratic irrational (that is every square root in the form (a+sqrt(b))/c)has a continued fraction that has a repeating block!
As an exercise, try doing the same with e, pi, and the irrational (1+sqrt(5))/2, or the golden ratio (for mor einformation on the golden ratio see http://www.bbc.co.uk/dna/h2g2/alabaster/A643844). See what you find!
Things get meaty when you look at things called convergents. These are successive continued fractions which allow us to approximate any irrational number by a series of rational ones. Let's look at our sqrt(2) example. Our convergents will be (using the shorter notation) [1], [1;2], [1;2,2], [1;2,2,2..] and so on. If you work these out, and obtain their decimal representation, you'll find that they are successively closer to 1.4142135... . Where this becomes very useful is where the pattern of convergents can be expressed in algebraic terms: what does this mean? A convergent will be in the form a0/b0, since it's a continued fraction which we've simplified down like above. Now, we'll call the next convergent a1/b1, and so on. Now, if the series a_k and b_k are series we can describe by a function, then we can describe how a_k/b_k behaves as k gets large: lim a_k/b_k=R, and R becomes the number that the continued fraction represents.
