The continued fraction expansion of an irrational number r...

The continued fraction expansion of an irrational number $r$ is the sequence $a_0, a_1, a_2, \ldots, a_i \in$ $\mathbf{N}$, such that (in the limit) $$ r=a_0+\frac{1}{a_1+\frac{1}{a_2+\ldots}} $$ The $a_i$ can be found as follows. Let $r_0=r$. The first term, $a_0$, is equal to the integer part of $r_0$ : $$ a_0=\left\lfloor r_0\right\rfloor \text {. } $$ The fractional part of $r_0$ is $r_0-a_0$. Its reciprocal, $r_1=1 /\left(r_0-a_0\right)$, is therefore $$ r_1=a_1+\frac{1}{a_2+\frac{1}{a_3+\ldots}} . $$ Thus, we get $$ a_1=\left\lfloor r_1\right\rfloor $$ and so on. 1. Write a function continuedFraction $\left[r_{-}, n_{-}\right.$Integer $]$that computes the first $n$ elements of the continued fraction expansion of $r$ as the list $\left\{a_0, a_1, \ldots, a_{n-1}\right\}$. You can use a loop or program recursively. 2. Write a function continuedValue[1_List] that returns the number belonging to the initial segment $l=\left\{a_0, a_1, \ldots, a_{n-1}\right\}$ of a continued fraction. 3. Write a simple definition for continuedError $[r, n]$, which finds the absolute error of the continued fraction approximation of length $n$. This error is the absolute value of the difference of $r$ and the approximation with $n$ terms. Here are the first 10 elements of $\pi$ 's continued fraction expansion. $$ \begin{aligned} & \text { In }[1]:=\text { continuedFraction }[\mathrm{Pi}, 10] \\ & \text { Out }[1]=\{3,7,15,1,292,1,1,1,2,1\} \end{aligned} $$ The numerical value of the previous continued fraction is this rational number. The numerical value of the previous continued fraction is this rational number. $$ \begin{aligned} & \text { In }[2]:=\text { continuedValue[\%] } \\ & \text { Out }[2]=\frac{1146408}{364913} \end{aligned} $$ Here is the error of the 10-term approxima- $$ \begin{aligned} & \operatorname{In}[3]:=N[\text { continuedBrror }[\mathrm{Pi}, 10]] \\ & \text { Out }[3]=1.6107110^{-12} \end{aligned} $$ Note that coth 1 has an interesting expansion. $$ \begin{aligned} & \text { In }[4]:=\text { continuedPraction }[\operatorname{Coth}[1], 12] \\ & \text { Out }[4]=\{1,3,5,7,9,11,13,15,17,19,21,23\} \end{aligned} $$

The continued fraction expansion of an irrational number $r$ is the sequence $a_0, a_1, a_2, \ldots, a_i \in$ $\mathbf{N}$, such that (in the limit)
$$
r=a_0+\frac{1}{a_1+\frac{1}{a_2+\ldots}}
$$
The $a_i$ can be found as follows. Let $r_0=r$. The first term, $a_0$, is equal to the integer part of $r_0$ :
$$
a_0=\left\lfloor r_0\right\rfloor \text {. }
$$
The fractional part of $r_0$ is $r_0-a_0$. Its reciprocal, $r_1=1 /\left(r_0-a_0\right)$, is therefore
$$
r_1=a_1+\frac{1}{a_2+\frac{1}{a_3+\ldots}} .
$$
Thus, we get
$$
a_1=\left\lfloor r_1\right\rfloor
$$
and so on.
1. Write a function continuedFraction $\left[r_{-}, n_{-}\right.$Integer $]$that computes the first $n$ elements of the continued fraction expansion of $r$ as the list $\left\{a_0, a_1, \ldots, a_{n-1}\right\}$. You can use a loop or program recursively.
2. Write a function continuedValue[1_List] that returns the number belonging to the initial segment $l=\left\{a_0, a_1, \ldots, a_{n-1}\right\}$ of a continued fraction.
3. Write a simple definition for continuedError $[r, n]$, which finds the absolute error of the continued fraction approximation of length $n$. This error is the absolute value of the difference of $r$ and the approximation with $n$ terms.

Here are the first 10 elements of $\pi$ 's continued fraction expansion.
$$
\begin{aligned}
& \text { In }[1]:=\text { continuedFraction }[\mathrm{Pi}, 10] \\
& \text { Out }[1]=\{3,7,15,1,292,1,1,1,2,1\}
\end{aligned}
$$
The numerical value of the previous continued fraction is this rational number.
The numerical value of the previous continued fraction is this rational number.
$$
\begin{aligned}
& \text { In }[2]:=\text { continuedValue[\%] } \\
& \text { Out }[2]=\frac{1146408}{364913}
\end{aligned}
$$
Here is the error of the 10-term approxima-
$$
\begin{aligned}
& \operatorname{In}[3]:=N[\text { continuedBrror }[\mathrm{Pi}, 10]] \\
& \text { Out }[3]=1.6107110^{-12}
\end{aligned}
$$

Note that coth 1 has an interesting expansion.
$$
\begin{aligned}
& \text { In }[4]:=\text { continuedPraction }[\operatorname{Coth}[1], 12] \\
& \text { Out }[4]=\{1,3,5,7,9,11,13,15,17,19,21,23\}
\end{aligned}
$$

Key Concepts

Continued Fraction

A continued fraction expresses a real number as the sum of its integer part and the reciprocal of another number, which in turn is written in a similar way. This representation is infinite for irrational numbers and provides insights into the number's properties and structure.

Continued Fraction Expansion Algorithm

The expansion process involves taking the floor of the number to extract its integer part, removing this part to obtain the fractional remainder, and then taking the reciprocal of the remainder. This procedure is repeated iteratively to generate the sequence of coefficients that define the continued fraction.

Convergents

Convergents are the finite approximations obtained by truncating an infinite continued fraction at some term. Each convergent is a rational number that approximates the original real number, and these approximations generally converge rapidly to the true value as more terms are included.

Error Estimation in Rational Approximations

The approximation error in continued fractions is measured by the absolute difference between the original real number and its convergent. This error decreases as the number of terms in the continued fraction increases, reflecting the improved accuracy of the rational approximation.

Transcript

00:01 We're given a sequence of rational numbers, and in part a, we were asked to verify some facts about this sequence.

00:15 So the sequence is described as follows.

00:18 1 over 1, 3 over 2, 7 over 5, 17 over 12, and so on, with the general term a over b, and the next term being a plus 2b over a plus b.

00:36 And we're told that the numerator and denominator also form a sequence so that xn and yn are the sequences such that the nth term in this sequence, rn, is equal to xn over yn.

01:23 And in part a, we're asked to verify that x1 squared minus 2y1 squared equals negative 1.

01:38 Well, we have, from our sequence, this is the same as 1 over 1 squared minus 2 times, or sorry, this is 1 squared minus 2 times 1 squared, which is 1 minus 2, which is negative 1, as we wanted to show.

02:00 And x2 squared minus 2 y 2 squared is positive 1.

02:06 So x2 squared minus 2 y2 squared.

02:17 Well, we have the x2 is going to be 3.

02:25 So 3 squared minus 2 times, and y2 is 2.

02:31 So this is 2 squared, which is 9 minus 8, or a positive 1, which is what we wanted to show.

02:41 And we're asked to show that if a squared minus 2b squared equals, negative 1 or positive 1, then a plus 2b squared minus 2, a plus b squared equals positive 1 or negative 1.

02:55 So we're going to suppose that a squared minus 2b squared equals first minus 1 and then it follows that a plus 2b squared minus 2, a plus b squared, is equal to.

03:34 This is going to be a squared plus 4ab plus 4b squared minus 2a squared minus 4ab minus 2b squared minus 2b squared and this reduces to a squared minus 2a squared is negative a squared the 4abes cancel out and then 4b squared minus 2b squared is plus 2b squared this is the same as the opposite of a squared minus 2b squared, which we have from our assumption is negative 1, so that we get that this is equal to positive 1, as we wanted to show.

04:29 And likewise, if we were to suppose that a squared minus 2b squared was equal to positive 1, then by the same process, we would have that a plus 2b squared minus 2, a plus b squared, would be equal to the opposite of positive one, which is negative one, which is also what we wanted to show.

05:05 And in part b, we're told the fractions, rn approach a limit as n increases, and we're asked to find that limit.

05:42 So a hint suggests that we find a way to write rn in terms of yn.

05:54 So, recall from the previous problem, we have that from part a, since a plus 2b squared minus 2, a plus or minus 1, when a squared minus 2b squared was plus or minus 1, we have that by induction, we have that the numerator xn squared minus 2 times the denominator, yn squared, is always is going to be equal to plus or minus 1...

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