Approximaths

Some observations on numbers

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The aim of this section is to get us used to using geometric series or continued fractions to represent numbers. We will first have a look at some geometric series:

\displaystyle 2 = 1 + 1
\displaystyle = 1 + \frac{1}{2} + \frac{1}{2}
\displaystyle = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4}
\displaystyle = \cdots
\displaystyle = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \cdots
\displaystyle = \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n

More generally, we can write a geometric series as:

\displaystyle S(x) = x^0 + x^1 + x^2 + x^3 + \cdots
\displaystyle = 1 + x + x^2 + x^3 + \cdots
\displaystyle = 1 + x (1 + x + x^2 + \cdots)
\displaystyle = 1 + x S(x)

Therefore:

\displaystyle S(x) = \frac{1}{1-x}

We have to be careful using this formula, since the radius of convergence of S(x) is (-1, 1) and \frac{1}{1-x} is defined on \mathbb{R} \setminus \{1\}. We can consider \frac{1}{1-x} as some “representation” of S(x) on (-1, 1).

This formula allows us, for example, to find that the argument of the geometric series of 3 is \frac{2}{3} (\frac{2}{3} \in (-1, 1) ensures convergence of the series):

\displaystyle 3 = \frac{1}{1-x} \implies x = \frac{2}{3}

Now we can write 3 as:

\displaystyle 3 = \sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n
\displaystyle = 1 + \frac{2}{3} + \frac{4}{9} + \frac{8}{27} + \frac{16}{81} + \frac{32}{243} + \cdots

Instead of writing numbers as a geometric series, we could also decide to write a number as a continued fraction with the form:

\displaystyle a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{a_3 + \frac{1}{a_4 + \cdots}}}}

To calculate the coefficients a_0, a_1, a_2, \ldots, we can use the continued fraction algorithm, which iteratively takes the integer part of the number and inverts its fractional part. In the case of the number \sqrt{2} \approx 1.41421356, we proceed as follows:

\displaystyle \sqrt{2} \approx 1.41421356, \quad a_0 = \lfloor \sqrt{2} \rfloor = 1
\displaystyle \sqrt{2} - 1 \approx 0.41421, \quad \frac{1}{\sqrt{2} - 1} \approx 2.41421, \quad a_1 = \lfloor 2.41421 \rfloor = 2
\displaystyle \frac{1}{\sqrt{2} - 1} - 2 = \sqrt{2} - 1 \approx 0.41421, \quad \frac{1}{\sqrt{2} - 1} \approx 2.41421, \quad a_2 = \lfloor 2.41421 \rfloor = 2
\displaystyle \frac{1}{\sqrt{2} - 1} - 2 = \sqrt{2} - 1 \approx 0.41421, \quad \frac{1}{\sqrt{2} - 1} \approx 2.41421, \quad a_3 = \lfloor 2.41421 \rfloor = 2
\displaystyle \cdots

The process repeats, leading to the coefficients a_0 = 1, a_1 = 2, a_2 = 2, a_3 = 2, \ldots. Thus, we can write \sqrt{2} as a continued fraction:

\displaystyle \sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \cdots}}}

The convergents of this continued fraction are:

\displaystyle c_0 = 1
\displaystyle c_1 = 1 + \frac{1}{2} = \frac{3}{2} = 1.5
\displaystyle c_2 = 1 + \frac{1}{2 + \frac{1}{2}} = 1 + \frac{2}{5} = \frac{7}{5} = 1.4
\displaystyle c_3 = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2}}} = 1 + \frac{5}{12} = \frac{17}{12} \approx 1.4167
\displaystyle \cdots

These convergents approach \sqrt{2} \approx 1.41421356. Note that an alternative method, such as associating the partial sums of a geometric series to a continued fraction, often leads to non-standard coefficients (e.g., negative or non-integer values like -\frac{3}{2} for 2 or -\frac{4}{3} for \frac{3}{2}), which may cause convergence issues. The continued fraction algorithm ensures integer coefficients and reliable convergence.

To find a continued fraction for \pi, we can use its decimal expansion or a series approximation. For example, we can use the following series:

\displaystyle \pi = \sum_{n=0}^{\infty} \frac{2^{n+1} n!^2}{(2n+1)!} \approx 3.1415926535\ldots

Then we proceed as follows:

  1. Let a_0 be the largest integer that does not exceed \pi, namely 3:
    \displaystyle a_0 = 3
  2. Compute:
    \displaystyle \pi \approx 3.141592
    \pi - 3 \approx 0.141592, \quad \frac{1}{\pi - 3} \approx 7.062513, \quad a_1 = \lfloor 7.062513 \rfloor = 7
  3. Continue iteratively:
    \displaystyle \frac{1}{\pi - 3} - 7 \approx 0.062513, \quad \frac{1}{0.062513} \approx 15.99659, \quad a_2 = \lfloor 15.99659 \rfloor = 15
    \displaystyle \frac{1}{0.06251} - 15 \approx 0.99659, \quad \frac{1}{0.996594} \approx 1.003417, \quad a_3 = \lfloor 1.003417\rfloor = 1
    \displaystyle \cdots

Using this technique, the continued fraction for \pi is:

\displaystyle \pi = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1 + \cdots}}}

The number \pi itself does not appear as a coefficient in this continued fraction. We could also represent \pi (or any real number) using a generalized continued fraction:

\displaystyle a_0 + \frac{b_1}{a_1 + \frac{b_2}{a_2 + \frac{b_3}{a_3 + \cdots}}}

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