### 数学代写|数论代写Number theory代考|MATH4304

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## 数学代写|数论代写Number theory代考|How many semitones should there be in an octave

In musical theory, the interval of an octave contains twelve semitones. Musically inclined mathematicians (or mathematically talented musicians) may have wondered if there is anything special about the number twelve. Could one work with a musical system of, say, eleven, thirteen or forty-one semitones to the octave? In this section we shall use continued fractions to see that there are very good reasons for having twelve notes in an octave. For readers who may be unfamiliar with basic musical terminology, a brief summary is given in appendix 4 at the end of this chapter.

There are coherent acoustical reasons for asserting that a combination of two musical notes at different pitches will be pleasing to the ear if the ratio of their frequencies is a “simple” rational number. The simplest ratios are $\frac{2}{1}$ and $\frac{3}{2}$; in musical terminology these correspond to the intervals of the octave and the (perfect) fifth respectively. Suppose that we take a fixed note as the basis of a tonal system, and build upon this foundation two sequences of intervals, one consisting of fifths and the other of octaves. In order to obtain a coherent system of finitely many notes rather than an infinite mess, we require these two sequences to meet again at some point. Suppose, then, that $p$ perfect fifths exactly equal $q$ octaves; in terms of frequencies, we have
$$\left(\frac{3}{2}\right)^{p}=2^{q}$$
Unfortunately, as is easily proved, this equation has no solutions in integers except for $p=q=0$, which is musically trivial. So we shall once again employ continued fractions to find the best possible approximate solutions. Rewriting the above equation to find the desired (but unachievable) value of $p / q$, and then computing its continued fraction, we obtain
$$\frac{p}{q} \approx \frac{\log 2}{\log \frac{3}{2}}=1+\frac{1}{1+} \frac{1}{2+} \frac{1}{2+} \frac{1}{3+} \frac{1}{1+} \frac{1}{5+} \frac{1}{2+} \frac{1}{23+} \frac{1}{2+} \ldots$$
whose first few convergents are
$$\frac{1}{1}, \frac{2}{1}, \frac{5}{3}, \frac{12}{7}, \frac{41}{24}, \frac{53}{31}$$

## 数学代写|数论代写Number theory代考|A “COMPUTATIONAL” TEST FOR RATIONALITY

Continued fractions can sometimes be used to give us an idea (though not necessarily a proof) that a certain number, presented as an infinite decimal, may be rational. For example, in connection with Apéry’s irrationality proof for $\zeta(3)$ it was conjectured, see $[66]$, that $\zeta(4)$ can be written as a sum
$$\zeta(4)=c \sum_{n=1}^{\infty} \frac{1}{n^{4}\left(\begin{array}{c} 2 n \ n \end{array}\right)}$$
with $c \in \mathbb{Q}$. Since it is known that $\zeta(4)=\pi^{4} / 90$, the claim is, in effect, that
$$c=\pi^{4} / 90 \sum_{n=1}^{\infty} \frac{1}{n^{4}\left(\begin{array}{c} 2 n \ n \end{array}\right)}$$
is rational. Evaluating $c$ to 10 significant figures (which can be done by taking the first 10 terms of the sum) and then calculating its continued fraction gives
$$c \approx 2.117647059=2+\frac{1}{8+} \frac{1}{2+} \frac{1}{19607842+} \ldots .$$
Now the partial quotients in the continued fraction of a “sensible” real number generally consist of fairly small integers. In fact it can be shown (see, for example, Khinchin [35], section 16) that for a “randomly chosen” real number, a proportion about
$$\log _{2}\left(\frac{a+1}{a} / \frac{a+2}{a+1}\right)$$
of the partial quotients should be equal to a given positive integer $a$; doing the appropriate calculations, we find that about $42 \%$ of the partial quotients should be 1 , about $17 \%$ should be 2 , and so on. One suspects, then, that the partial quotient 19607842 is due to numerical inaccuracy, and that the continued fraction should have terminated at the previous partial quotient. Therefore, it seems reasonable to believe that
$$c=2+\frac{1}{8+} \frac{1}{2}=\frac{36}{17} .$$
Obviously this does not constitute a rigorous proof, but in fact, it is possible to prove that $c$ has the value $\frac{36}{17}$, as conjectured.

## 数学代写|数论代写Number theory代考|FURTHER APPROXIMATION PROPERTIES OF CONVERGENTS

We know that for every irrational $\alpha$ the inequality
$$\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}$$
has infinitely many solutions, and that for certain $\alpha$ the right-hand side can be decreased by substituting for $q^{2}$ a higher power $q^{s}$. Indeed, if $\alpha$ is a Liouville number then we can choose arbitrarily large $s$ and still find infinitely many solutions; on the other hand, we know from Liouville’s Theorem, page 49 , that if $\alpha$ is a quadratic irrational then the exponent 2 cannot be increased at all.
Thus, if we want a result which is true for all $\alpha$, we cannot decrease the right-hand side of the above inequality by increasing $s$. Perhaps, however, we could replace the 1 in the numerator by something smaller. Indeed, we can, for the comment following Theorem $4.13$ shows that 1 can be replaced by $\frac{1}{2}$. Can we do even better than this?

Recall that convergents give “the best” approximations to a real number $\alpha$, and that the approximations are especially good when the next partial quotient is large. Consider what happens if the “next partial quotient” of $\alpha$ is never large. An extreme example of such a number is
$$\alpha=1+\frac{1}{1+} \frac{1}{1+} \frac{1}{1+\ldots}=\frac{1+\sqrt{5}}{2}$$
In this case it is plain that every complete quotient $\alpha_{k}$ is equal to $\alpha$. Moreover, if we calculate the first few convergents to $\alpha$ it is very easy to conjecture, and equally easy to prove by induction, that $q_{k}=p_{k-1}$ for $k \geq-1$. (It is also easy to show, though unimportant at present, that the numerators $p_{k}$ and denominators $q_{k}$ are just the Fibonacci numbers.) Therefore, from equation $(4.5)$, we have
$$\left|\alpha-\frac{p_{k}}{q_{k}}\right|=\frac{1}{\left(\alpha q_{k}+q_{k-1}\right) q_{k}}=\frac{1}{\left(\alpha+q_{k-1} / q_{k}\right) q_{k}^{2}} \approx \frac{1}{\left(\alpha+\alpha^{-1}\right) q_{k}^{2}}=\frac{1}{\sqrt{5} q_{k}^{2}}$$
Since we do not expect that any real irrational number will have worse rational approximations than $\alpha$, the following result is plausible.

## 数学代写|数论代写Number theory代考|How many semitones should there be in an octave

(32)p=2q

pq≈日志⁡2日志⁡32=1+11+12+12+13+11+15+12+123+12+…

11,21,53,127,4124,5331

## 数学代写|数论代写Number theory代考|A “COMPUTATIONAL” TEST FOR RATIONALITY

G(4)=C∑n=1∞1n4(2n n)

C=圆周率4/90∑n=1∞1n4(2n n)

C≈2.117647059=2+18+12+119607842+….

C=2+18+12=3617.

## 数学代写|数论代写Number theory代考|FURTHER APPROXIMATION PROPERTIES OF CONVERGENTS

|一个−pq|<1q2

|一个−pķqķ|=1(一个qķ+qķ−1)qķ=1(一个+qķ−1/qķ)qķ2≈1(一个+一个−1)qķ2=15qķ2

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