## 数学代写|数论作业代写number theory代考|PMATH641

2022年7月6日

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## 数学代写|数论作业代写number theory代考|COMPUTING THE CONTINUED FRACTION OF AN ALGEBRAIC IRRATIONAL

Let $\alpha$ be a root of a known irreducible polynomial $f$ with degree $n \geq 2$ and integral coefficients. We shall assume that $\alpha>1$, and that $f$ has no other roots $\beta>1$. In this case there is a very simple algorithm to find the zeroth partial quotient $a_{0}=\lfloor\alpha\rfloor:$ calculate $f(1), f(2), f(3), \ldots$ until a change of sign occurs; then $a_{0}$ is the last argument before the change of sign. Since $\alpha$ is irrational we have $a_{0}<\alpha1$ since $0<\alpha-a_{0}<1$; so $f_{1}$ has a real root greater than 1 . Conversely, if $\beta$ is any such root of $f_{1}$, then $a_{0}+1 / \beta$ is a root of $f$ and hence $a_{0}+1 / \beta=\alpha$. Because $f_{1}$ is a polynomial having integral coefficients and a unique real root $\alpha_{1}>1$, the procedure can be iterated to find the sequence of partial quotients of $\alpha$. Observe that the complete quotients $\alpha=\alpha_{0}, \alpha_{1}, \alpha_{2}, \ldots$ need never be calculated, so we do not have the problem of calculating decimal expansions to many places: all our calculations will be performed in terms of integer arithmetic, and so the process will be free of rounding errors.

Example. We can find the continued fraction for $\sqrt[3]{2}$ by starting with the polynomial $f(z)=f_{0}(z)=z^{3}-2$. We have
\begin{aligned} f_{0}(z)=z^{3}-2, & a_{0}=1 ; \ f_{1}(z)=-z^{3}+3 z^{2}+3 z+1, & a_{1}=3 ; \ f_{2}(z)=10 z^{3}-6 z^{2}-6 z-1, & a_{2}=1 ; \ f_{3}(z)=-3 z^{3}+12 z^{2}+24 z+10, & a_{3}=5 \ f_{4}(z)=55 z^{3}-81 z^{2}-33 z-3, & a_{4}=1 ; \ f_{5}(z)=-62 z^{3}-30 z^{2}+84 z+55, & a_{5}=1 ; \ f_{6}(z)=47 z^{3}-162 z^{2}-216 z-62, & a_{6}=4 \end{aligned}
and so
$$\sqrt[3]{2}=1+\frac{1}{3+} \frac{1}{1+} \frac{1}{5+} \frac{1}{1+} \frac{1}{1+} \frac{1}{4+\ldots}$$

## 数学代写|数论作业代写number theory代考|THE CONTINUED FRACTION OF e

We shall determine the continued fractions for a class of numbers related to the exponential constant $e$. To do so, we first consider the functions defined by the infinite series
$$f(c ; z)=\sum_{k=0}^{\infty} \frac{1}{c(c+1) \cdots(c+k-1)} \frac{z^{k}}{k !}$$
here the parameter $c$ is any real number except $0,-1,-2, \ldots$, and it is easy to show that the series converges for all $z$. To simplify the notation we write $c^{(k)}=c(c+1) \cdots(c+k-1)$, with the understanding that $c^{(0)}=1$; thus
$$f(c ; z)=\sum_{k=0}^{\infty} \frac{1}{c^{(k)}} \frac{z^{k}}{k !} .$$
The expression $c^{(k)}$ is referred to as ” $c$ rising factorial $k$ “. It satisfies the two important recurrences
$$c^{(k+1)}=c^{(k)}(c+k)=c(c+1)^{(k)},$$
both of which are instances of the more general relation $c^{(k+m)}=c^{(k)}(c+k)^{(m)}$.
Lemma 4.18. Let $c$ be a positive real number, $z$ a non-zero real number and $k$ a non-negative integer; then
$$\frac{c}{z} \frac{f\left(c ; z^{2}\right)}{f\left(c+1 ; z^{2}\right)}=\left[\frac{c}{z}, \frac{c+1}{z}, \ldots, \frac{c+k-1}{z}, \frac{c+k}{z} \frac{f\left(c+k ; z^{2}\right)}{f\left(c+k+1 ; z^{2}\right)}\right]$$
Proof. First, observe that under the stated conditions $f\left(c+k+1 ; z^{2}\right)$ is given by a series of positive terms, so it does not vanish and the last term in the continued fraction makes sense. From (4.14) the rising factorials satisfy
$$\frac{1}{c^{(k)}}=\frac{c+k}{c^{(k+1)}}=\frac{1}{(c+1)^{(k)}}+\frac{k}{c^{(k+1)}},$$
and hence
$$f(c ; z)=\sum_{k=0}^{\infty} \frac{1}{(c+1)^{(k)}} \frac{z^{k}}{k !}+\sum_{k=0}^{\infty} \frac{k}{c^{(k+1)}} \frac{z^{k}}{k !} .$$
The first series on the right-hand side is evidently $f(c+1 ; z)$. The second may be written
$$\sum_{k=1}^{\infty} \frac{1}{c^{(k+1)}} \frac{z^{k}}{(k-1) !}=\sum_{k=0}^{\infty} \frac{1}{c^{(k+2)}} \frac{z^{k+1}}{k !}=\frac{z}{c(c+1)} \sum_{k=0}^{\infty} \frac{1}{(c+2)^{(k)}} \frac{z^{k}}{k !},$$
and we have the second-order recurrence
$$f(c ; z)=f(c+1 ; z)+\frac{z}{c(c+1)} f(c+2 ; z) .$$

## 数学代写|数论作业代写number theory代考|COMPUTING THE CONTINUED FRACTION OF AN ALGEBRAIC IRRATIONAL

$$f_{0}(z)=z^{3}-2, a_{0}=1 ; f_{1}(z)=-z^{3}+3 z^{2}+3 z+1, \quad a_{1}=3 ; f_{2}(z)=10 z^{3}$$

$$\sqrt[3]{2}=1+\frac{1}{3+} \frac{1}{1+} \frac{1}{5+} \frac{1}{1+} \frac{1}{1+} \frac{1}{4+\ldots}$$

## 数学代写|数论作业代写number theory代考|THE CONTINUED FRACTION OF e

$$f(c ; z)=\sum_{k=0}^{\infty} \frac{1}{c(c+1) \cdots(c+k-1)} \frac{z^{k}}{k !}$$

$$f(c ; z)=\sum_{k=0}^{\infty} \frac{1}{c^{(k)}} \frac{z^{k}}{k !}$$

$$c^{(k+1)}=c^{(k)}(c+k)=c(c+1)^{(k)}$$

$$\frac{c}{z} \frac{f\left(c ; z^{2}\right)}{f\left(c+1 ; z^{2}\right)}=\left[\frac{c}{z}, \frac{c+1}{z}, \ldots, \frac{c+k-1}{z}, \frac{c+k}{z} \frac{f\left(c+k ; z^{2}\right)}{f\left(c+k+1 ; z^{2}\right)}\right]$$

$$\frac{1}{c^{(k)}}=\frac{c+k}{c^{(k+1)}}=\frac{1}{(c+1)^{(k)}}+\frac{k}{c^{(k+1)}}$$

$$f(c ; z)=\sum_{k=0}^{\infty} \frac{1}{(c+1)^{(k)}} \frac{z^{k}}{k !}+\sum_{k=0}^{\infty} \frac{k}{c^{(k+1)}} \frac{z^{k}}{k !}$$

$$\sum_{k=1}^{\infty} \frac{1}{c^{(k+1)}} \frac{z^{k}}{(k-1) !}=\sum_{k=0}^{\infty} \frac{1}{c^{(k+2)}} \frac{z^{k+1}}{k !}=\frac{z}{c(c+1)} \sum_{k=0}^{\infty} \frac{1}{(c+2)^{(k)}} \frac{z^{k}}{k !},$$

$$f(c ; z)=f(c+1 ; z)+\frac{z}{c(c+1)} f(c+2 ; z) .$$

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