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# The complete weight enumerator of a subclass of optimal three-weight cyclic codes

A class of optimal three-weight cyclic codes of dimension 3 over any finite field was presented by Vega [Finite Fields Appl., 42 (2016) 23-38]. Shortly thereafter, Heng and Yue [IEEE Trans. Inf. Theory, 62(8) (2016) 4501-4513] generalized this result by presenting several classes of cyclic codes with either optimal three weights or a few weights. On the other hand, a class of optimal five-weight cyclic codes of dimension 4 over a prime field was recently presented by Li, et al. [Adv. Math. Commun., 13(1) (2019) 137-156]. One of the purposes of this work is to present a more general description for these optimal five-weight cyclic codes, which gives place to an enlarged class of optimal five-weight cyclic codes of dimension 4 over any finite field. As an application of this enlarged class, we present the complete weight enumerator of a subclass of the optimal three-weight cyclic codes over any finite field that were studied by Vega [Finite Fields Appl., 42 (2016) 23-38]. In addition, we study the dual codes in this enlarged class of optimal five-weight cyclic codes, and show that they are cyclic codes of length q^2-1, dimension q^2-5, and minimum Hamming distance 4. In fact, through several examples, we see that those parameters are the best known parameters for linear codes.

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07/09/2021

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## I Introduction

The problem of obtaining the weight distribution of a given code is important because it plays a significant role in determining the capabilities of error detection and correction of such a code. For cyclic codes this problem is even more important because this kind of codes possess a rich algebraic structure (they are ideals in the principal ideal ring , where is the length of the cyclic codes). In addition, it is known that cyclic codes with few weights have a great practical importance in cryptography and coding theory since they are useful in the design of secret sharing schemes and association schemes (see [1, 4]). Not long ago, a characterization of a class of optimal three-weight cyclic codes of dimension 3, over any finite field , was presented in [20], and shortly thereafter, several classes of cyclic codes with either optimal three weights or a few weights were given in [9], showing that one of these classes can be constructed as a generalization of the sufficient numerical conditions of the characterization given in [20]. On the other hand, a class of optimal five-weight -ary cyclic codes of dimension was recently presented in [14].

In this paper we take the direct sum (as vector spaces) of a one-weight cyclic code of dimension

, and two different one-weight cyclic codes of dimension in order to construct an enlarged class of optimal five-weight cyclic codes of length and dimension , over any finite field , with , that generalizes the class of optimal five-weight -ary cyclic codes presented in [14]. The codes in this class are optimal in the sense that their lengths reach the Griesmer lower bound for linear codes. In fact, we explicitly determine the weight distribution for the cyclic codes in this class. As an application of this enlarged class of optimal five-weight cyclic codes, we present the complete weight enumerator of a subclass of the optimal three-weight cyclic codes that were studied in [20]. The complete weight enumerator of a code enumerates the codewords by the number of symbols of each kind contained in each codeword. In fact, if , the complete weight enumerator contains much more information than the ordinary weight enumerator. For this reason, the determination of the complete weight enumerators of cyclic codes or linear codes over finite fields has received a great deal of attention in recent years (see for example [2], [5], [12], [13], [22], [24], [25], [26], and [27]). In this respect, it should be pointed out that, for a prime field , several classes of three-weight linear codes and their complete weight enumerators were recently presented in [11], [23], [25], [26], and [27]. In that context, we want to emphasize that the three-weight codes that we are going to present here are not only linear, but also cyclic, optimal and defined over any finite field .

In addition, we study the dual codes in our enlarged class of optimal five-weight cyclic codes, and show that, except for the binary case, they are cyclic codes of length , dimension , and minimum Hamming distance . In fact, through several examples, we see that those parameters are the best known parameters for linear codes.

This work is organized as follows: In Section II we fix some notations and recall some definitions, along with some known results to be used in subsequent sections. Section III is devoted to presenting some preliminary and general results. Particularly, we study a kind of exponential sums that will be important in order to determine the weights, and their corresponding frequencies, for the class of cyclic codes that we are interested in. This kind of exponential sums are then used in Section IV to present an enlarged class of optimal five-weight cyclic codes of dimension over any finite field, showing at the same time that, except for the binary case, the dual codes in this enlarged class have minimum Hamming distance . Examples of optimal five-weight cyclic codes belonging to this enlarged class, along with their corresponding dual codes, are presented at the end of this section. In Section V we use our enlarged class of optimal five-weight cyclic codes over any finite field in order to obtain the complete weight enumerator of a subclass of the optimal three-weight cyclic codes that were studied in [20]. Finally, Section VI is devoted to conclusions.

## Ii Preliminaries

In this section, we recall some definitions and notations along with some known results to be used in subsequent sections.

### Ii-a Basic definitions and notation

First of all we set for this section and for the rest of this work, the following:

Notation. Let be a prime, a power of , and the finite field of order . By using we will denote a fixed primitive element of . We are going to fix , and consequently note that is a fixed primitive element of . For any integer , the polynomial will denote the minimal polynomial of (see for example [17, p. 99]). For integers , such that if , will denote the cyclic code of length over , whose parity check polynomial is . In addition, for any integer , we will denote by “Tr”, the absolute trace mapping from to the prime field , and by “” the trace mapping from to .

### Ii-B Gauss Sums

The canonical additive character of is defined as follows:

 χ(x):=e2π√−1Tr(x)/p, for all x∈IFq.

On the other hand, if , then any multiplicative character of is defined by

 ψj(αl):=e2π√−1jl/(q−1), for j,l=0,1,⋯,q−2.

Commonly is referenced as the trivial multiplicative character. If

is odd, an important multiplicative character is the so-called

quadratic character which is denoted by and defined by: if is the square of an element of and otherwise. For the canonical additive character , and for any multiplicative character , of , the following two properties will be useful for us

 ∑x∈IF∗qχ(ax) = ⎧⎪⎨⎪⎩q−1if a=0,−1otherwise., and ∑x∈IF∗qψ(x) = 0. (1)

Now, for any multiplicative character of and for the canonical additive character of , the Gaussian sum is defined by

 G(ψ,χ):=∑c∈IF∗qψ(c)χ(c).

Two useful properties of are (see for example [15, Theorems 5.12 and 5.15]):

 G(η,¯χ)=η(−1)G(η,χ) and G(η,χ)2=η(−1)q. (2)

Another important property of Gaussian sums is the so-called expansion of the restriction of to in terms of the multiplicative characters of , with Gaussian sums as Fourier coefficients (see for example [15, p. 195]):

 χ(c)=1(q−1)∑ψψ(c)G(¯ψ,χ), for c∈IF∗q.

### Ii-C Griesmer lower bound

When constructing a code, from an economical point of view, it is desirable to obtain an code over whose length is minimal for given values of , and . A lower bound for the length in terms of these values is as follows. Let be the minimum length for which an linear code, over , exists. If the values of , and are given, then a well-known lower bound (see [7] and [19]) for is:

###### Theorem 1

(Griesmer lower bound) With the previous notation,

 nq(k,d)≥k−1∑i=0⌈dqi⌉,

where denotes the smallest integer greater than or equal to .

As a consequence of the previous theorem we have:

###### Lemma 1

Suppose that is a linear code over . Then is an optimal linear code in the sense that its length reaches the lower bound in the previous theorem.

By means of a direct application of the Griesmer lower bound, we have

 = [(q−1)q−2]+[q−1]+1+1=q2−1.

### Ii-D The weight enumerator and the complete weight enumerator of a code

We recall that the weight enumerator of a code of length over a finite field is defined as the polynomial , where denote the number of codewords with Hamming weight in the code . The sequence is called the weight distribution of the code. An -weight code is a code such that the cardinality of the set of nonzero weights is . That is, .

In a similar way let be a code of length over . Denote the elements of by , in some fixed order. For each codeword in let be the monomial in the variables given by

 Z(c):=zw00zw11⋯zwq−1q−1,

where the power () is the number of components () of that are equal to . Denote by the set of all integer vectors such that and . Then the complete weight enumerator of (see for example [16] and [17, p. 141]) is the polynomial

 CWEC:=∑c∈CZ(c)=∑→t∈V(n,q)A→tZ→t, (3)

where and

 A→t:=♯{c∈C|Z(c)=Z→t}.

The sequence is called the complete weight distribution of . Obviously it coincides with the weight distribution if and contains much more information if . In fact, the complete weight enumerator has a wide range of applications in many research fields as the information it contains is of vital use in practical applications. For example, as pointed out in [3] the complete weight enumerator of Reed-Solomon codes could be helpful in soft decision decoding. As other example, the complete weight enumerator is useful in the computation of the Walsh transform of monomial functions over finite fields [8].

## Iii A class of exponential sums

It is well known that the weight distribution of some cyclic codes can be obtained by means of the evaluation of some exponential sums. The following two lemmas goes along these lines.

###### Lemma 2

Let and be respectively the canonical additive characters of and . For any integers , and , and for all and , consider the sums

 S(e1,e2,e3)(a,b,c):=∑x∈IF∗q2χ(ax(q+1)e1+bx(q+1)e2)χ′(cxe3).

If and , then

 S(e1,e2,e3)(a,b,c)=−∑z∈IF∗q∑x∈IF∗qχ(z+axe1+bxe2+z−1cq+1xe3).

Recalling that we have

 S(e1,e2,e3)(a,b,c)=q−2∑i=0χ(aδie1+bδie2)∑w∈γi⟨γq−1⟩χ′(cwe3)

and, since , we have

 ∑w∈γi⟨γq−1⟩χ′(cwe3)=∑w∈γie3⟨γq−1⟩χ′(cw)=1q−1∑w∈IF∗q2χ′(cγie3wq−1).

Let and be respectively the multiplicative character groups of and . Now, if N is the norm mapping from to and is the subgroup of order of , then note that (that is, is nothing but the “lift” of to ). Therefore, owing to [15, Theorem 5.30, p. 217], we have

 ∑w∈IF∗q2χ′(cγie3wq−1) = ∑ψ∈ˆIFqGIFq2(¯ψ∘N,χ′)ψ(N(cγie3)) = −∑ψ∈ˆIFqGIFq(¯ψ,χ)2ψ(N(cγie3)),

where the last equality arises due to the Davenport-Hasse Theorem (see [15, Theorem 5.14, p. 197]). In consequence, since and , we have

 S(e1,e2,e3)(a,b,c)=−1q−1∑x∈IF∗qχ(axe1+bxe2)∑ψ∈ˆIFqGIFq(¯ψ,χ)2ψ(cq+1xe3). (4)

On the other hand, by using the Fourier expansion of the restriction of to in terms of the multiplicative characters of , we have that for all :

 χ(z−1cq+1xe3)=1q−1∑ψ∈ˆIFqGIFq(¯ψ,χ)¯ψ(z)ψ(cq+1xe3),

and by multiplying both sides of the preceding equation by and by summing we obtain

 ∑z∈IF∗qχ(z+z−1cq+1xe3)=1q−1∑ψ∈ˆIFqGIFq(¯ψ,χ)2ψ(cq+1xe3).

Finally, by substituting the previous equation in (4) we obtain the desired result.

###### Lemma 3

With the same notation, consider the character sum of the form:

 T(e1,e2,e3)(a,b,c):=∑y∈IF∗qS(e1,e2,e3)(ya,yb,yc).

If , , and , then

Without loss of generality we assume that . Thus, from the previous lemma, and since , we have

 T(e1,e2,e3)(a,b,c)=−∑x∈IF∗q∑z∈IF∗q∑y∈IF∗qχ(z+(a+bxe2−e1)xe1y+z−1cq+1xe1+e2y2). (5)

Now, suppose that is even. Then by [15, Theorem 5.34, p. 218] we know that, for all ,

 (6)

Note that iff , with . But, since , the last equality holds iff , with . Therefore

 T(e1,e2,e3)(a,b,c) = 1−q−∑x∈IF∗q∑z∈IF∗q∑y∈IFqχ(z+(a+bxe2−e1)xe1y+z−1cq+1xe1+e2y2), = 1−q−q∑x∈IF∗q∖{ab}χ(cq+1x(a+bx)2), = 1−q−q(∑x∈IFq∖{ab}χ(cq+1x(a+bx)2)−¯δ0(a)), = 1−q+¯δ0(a)q−q∑x∈IFq∖{ab}χ(cq+1x(a+bx)2),

where if and otherwise. By making the variable substitution , with (that is , with ), we get

 T(e1,e2,e3)(a,b,c) = 1−q+¯δ0(a)q−q∑w∈IF∗qχ(wcq+1(1+awb)), = 1−q+¯δ0(a)q−q(∑w∈IFqχ(cq+1bw+cq+1abw2)−1), = 1+¯δ0(a)q−q∑w∈IFqχ(cq+1bw+cq+1abw2).

Thus, owing to (6), the case even follows from the fact that

 ∑w∈IFqχ(cq+1bw+cq+1abw2)={qif a=cq+1b,0otherwise.

Now suppose that is odd. Then by [15, Theorem 5.33, p. 218] we know that, for all with ,

 ∑y∈IF∗qχ(ρ0+ρ1y+ρ2y2)=χ(ρ0−ρ21(4ρ2)−1)η(ρ2)GIFq(η,χ)−χ(ρ0), (7)

where is the quadratic character of . Therefore, from (5), we have that is equal to

 −∑x∈IF∗q∑z∈IF∗qχ(z−(a+bxe2−e1)2x2e1(4z−1cq+1xe1+e2)−1)η(z−1cq+1xe1+e2)GIFq(η,χ)−χ(z), = 1−q−GIFq(η,χ)∑x∈IF∗q∑z∈IF∗qχ(z−(a+bxe2−e1)2(4z−1cq+1xe2−e1)−1)η(z−1cq+1xe1+e2), = 1−q−GIFq(η,χ)∑x∈IF∗q∑z∈IF∗qχ(z−(a+bxe2−e1)2(4z−1cq+1xe2−e1)−1)η(z−1cq+1xe2−e1),

where the last equality arises because . But , thus after applying the variable substitution (that is ), we have that

 T(e1,e2,e3)(a,b,c) = 1−q−GIFq(η,χ)∑w∈IF∗qη(w)∑z∈IF∗qχ(z−14w(a+bwzcq+1)2), = 1−q−GIFq(η,χ)∑w∈IF∗qη(w)∑z∈IF∗qχ(−a24w+(1−ab2cq+1)z−b2wz24c2(q+1)).

After the application of (7) again, we get that is equal to

 1−q−GIFq(η,χ)∑w∈IF∗qη(w)[ χ(−a24w−(1−ab2cq+1)2(−b2wc2(q+1))−1)× η(−b2w4c2(q+1))GIFq(η,χ)−χ(−a24w)].

But, by considering (2), . On the other hand, note that (see (II-B)) and if , then . Thus

 −GIFq(η,χ)∑w∈IF∗qη(w)(−χ(−a24w)) = ¯δ0(a)GIFq(η,χ)∑w∈IF∗qη(a24w)χ(−a24w), = ¯δ0(a)GIFq(η,χ)GIFq(η,¯χ)=¯δ0(a)q.

Therefore

 T(e1,e2,e3)(a,b,c)=1−q+¯δ0(a)q−q∑w∈IF∗qχ(−a24w+(1−ab2cq+1)2(c2(q+1)b2w)).

Applying the variable substitution (that is ), we have

 T(e1,e2,e3)(a,b,c) = 1−q+¯δ0(a)q−q∑v∈IF∗qχ(−a2bv4cq+1+(1−ab2cq+1)2(vcq+1b)), = 1−q+¯δ0(a)q−q∑v∈IF∗qχ((cq+1b−a)v).

Finally, the result follows from the fact that (see (II-B))

 ∑v∈IF∗qχ((cq+1b−a)v)={[]clq−1if a=cq+1b,−1otherwise.
###### Corollary 1

Assume the same notation as in the previous lemma. If , and , then the character sum takes six different values according to the following seven cases:

 T(e1,e2,e3)(a,b,c)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(q−1)(q2−1)if c=a=b=0,\rm Case 1,1−q2if c=0 and a or b is zero but not both,\rm Case% 2,q+1if c=0 and a,b∈IF∗q,\rm Case 3,1−qif c≠0 and a=b=0,\rm Case 4,1if c≠0 and a or b is zero but not both,\rm Case % 5,−q2+q+1if c≠0 and a,b∈IF∗q and a=cq+1b,\rm Case 6,q+1if c≠0 and a,b∈IF∗q and a≠cq+1b,\rm Case 7.

Clearly .

Case 2: Without loss of generality we assume that and . Then

 T(e1,e2,e3)(0,b,0)=∑x∈IF∗q2∑y∈IF∗qχ(bx(q+1)e2y)=−(q2−1).

Case 3:

 T(e1,e2,e3)(a,b,0) = ∑x∈IF∗q2∑y∈IF∗qχ(ax(q+1)e1y+bx(q+1)e2y), = (q+1)∑x∈IF∗q∑y∈IF∗qχ(axe1y+bxe2y).

Applying the variable substitution (that is ), we have

 T(e1,e2,e3)(a,b,0)=(q+1)∑v∈IF∗qχ(av)∑x∈IF∗qχ(bvxe2−e1).

But , then

 T(e1,e2,e3)(a,b,0)=(q+1)∑v∈IF∗qχ(av)∑x∈IF∗qχ(bvx)=(q+1).

Case 4:

 T(e1,e2,e3)(0,0,c) = ∑x∈IF∗q2∑y∈IF∗qχ(TrIFq2


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#### annabeljxde

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##### Re: 3U Maths Question Thread
« Reply #4155 on: July 27, 2019, 12:01:47 am »
0
Hey!

Need a bit of help with this question. The solutions went straight from (1-sin2θ) to the solution for tan2θ, which has left me confused. Do you use the condition 0 < θ < π/4 anywhere in the working out?

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#### DrDusk

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##### Re: 3U Maths Question Thread
« Reply #4156 on: July 27, 2019, 01:28:58 am »
+2
Hey!

Need a bit of help with this question. The solutions went straight from (1-sin2θ) to the solution for tan2θ, which has left me confused. Do you use the condition 0 < θ < π/4 anywhere in the working out?

Start off with
$LHS = \sqrt{\frac{1 - sin2\theta}{1 + sin2\theta}}$
$= \sqrt{\frac{sin^2\theta + cos^2\theta - 2sin\theta cos\theta}{sin^2\theta + cos^2\theta + 2sin\theta cos\theta}}$
$\text{Using}\hspace{2mm}(sin\theta + cos\theta)^2 = sin^2\theta + cos^2\theta + 2sin\theta cos\theta$
$LHS = \frac{\pm(sin\theta - cos\theta)}{\pm(sin\theta + cos\theta)}$
$\text{Multiplying top and bottom by}\hspace{2mm}\frac{1}{cos\theta}\hspace{2mm}$$\text{and obseving that taking the negative one for the top one will give the desired answer we get}$
$LHS = \frac{-(tan\theta - 1)}{tan\theta + 1} = \frac{1 - tan\theta}{1 + tan\theta}$
$= RHS$
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#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4157 on: July 27, 2019, 08:55:00 am »
+3
$LHS = \frac{\pm(sin\theta - cos\theta)}{\pm(sin\theta + cos\theta)}$
$\text{Multiplying top and bottom by}\hspace{2mm}\frac{1}{cos\theta}\hspace{2mm}$$\text{and obseving that taking the negative one for the top one will give the desired answer we get}$
Note: Not really the recommended justification here. Firstly (and somewhat less importantly) note that saying $\sqrt{a^2} = \pm a$ is technically incorrect, because that means you're claiming that $\sqrt{a^2}$ could potentially equal two different values. This would be problematic when saying something like $\sqrt{5^2} = \pm 5$. In general, the safest bet is to always say that $\sqrt{a^2} = |a|$ when doing these problems correctly.

But it's less of an issue here because you attempted to justify your way around it.

The justification behind the "$\pm$'s cancelling out into a negatives" isn't sufficient. The subtlety is that you haven't convinced me between which "$\sqrt{\frac{a^2}{b^2}}=\frac{\sqrt{a^2}}{\sqrt{b^2}}=\frac{\pm a}{\pm b}$" or "$\sqrt{\frac{a^2}{b^2}}=\frac{\sqrt{a^2}}{\sqrt{b^2}}=\frac{\pm a}{\mp b}$" is the favourable one. Saying that "observing that taking the negative one for the top one will give the desired answer" is not convincing because it looks like you're assuming what you're trying to prove,

A better justification here would be that $\sqrt{\frac{(\sin\theta-\cos\theta)^2}{(\sin\theta+\cos\theta)^2}} = \left| \frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta} \right| = \left| \frac{\tan\theta-1}{\tan\theta+1} \right| = -\frac{\tan\theta-1}{\tan\theta+1}$, justifying the last equality here by the fact that $\boxed{0\leq\theta\leq\frac\pi4}$.

We know that over this domain, $0 \leq \tan\theta \leq 1$. Hence on one hand, it is clear that $\tan\theta+1 > 0$. On the other hand, rearranging $\tan\theta \leq 1$ gives $\tan\theta - 1\leq 0$. Therefore $\frac{\tan\theta-1}{\tan\theta+1} \leq0$ (non-positive number divided by positive number), so $\left|\frac{\tan\theta-1}{\tan\theta+1}\right| = -\frac{\tan\theta-1}{\tan\theta+1}$. This now relies on the actual definition of the absolute value
$|x| = \begin{cases}x&x\geq 0\\ -x&x<0\end{cases}$
instead of intuition that cannot be justified.
« Last Edit: July 27, 2019, 06:08:57 pm by RuiAce »

#### mani.s_

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##### Re: 3U Maths Question Thread
« Reply #4158 on: July 27, 2019, 07:08:34 pm »
0
Hello, I just wanted some help on my assignment. We have to make a graph of desmos using parametric equations: x = cos 2t and y = sin t, (0 ≤ t ≤ 2pi).
We then have to choose 6 points on the graph including critical points. Now, this is the part which I don't understand. They ask us for a reason for each coordinate and how we obtained that coordinate from the parametric equations. I don't understand what that means, like what are we supposed to do???

Thanks

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4159 on: July 27, 2019, 07:56:51 pm »
+2
Hello, I just wanted some help on my assignment. We have to make a graph of desmos using parametric equations: x = cos 2t and y = sin t, (0 ≤ t ≤ 2pi).
We then have to choose 6 points on the graph including critical points. Now, this is the part which I don't understand. They ask us for a reason for each coordinate and how we obtained that coordinate from the parametric equations. I don't understand what that means, like what are we supposed to do???

Thanks
Without risking giving away a lot, this is my interpretation. (However note that usually it's worth double checking with your teachers if you don't get the wording of an assignment.)

Since this is a parametric plot, every point on the plot corresponds to at least one value for $t$. For example, the point represented by $(\cos (2\times 0), \sin 0)$ corresponds to at least the value $t=0$. It may be possible that it corresponds to another value of $t$, but that can be something for you to investigate.

So my guess would be that relating certain values of $t$ to the points on the curve may be something important here.

Judging by your assignment's context, however, $t$ may as well be a specific parameter that represents some sense of "time". So potentially you could talk about how at a certain time $t$, we are at this point. (Note that since the question also asks you to examine critical points, it is likely that they also want to see some, let's say, specific values of $t$ in your analysis.)

#### iktimal

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##### Re: 3U Maths Question Thread
« Reply #4160 on: July 27, 2019, 11:31:18 pm »
0
Hey,
Can I please have some help for this binomial question?

#### blyatman

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##### Re: 3U Maths Question Thread
« Reply #4161 on: July 28, 2019, 08:21:23 pm »
+1
Hey,
Can I please have some help for this binomial question?
Since nobody else has posted a solution, I'll post mine.

Out of curiosity, is this a HSC question? Because I've never seen a binomial question involving square roots. It can be done, just that I've never seen it asked in the HSC.

First, multiply the top and bottom of the LHS by $1+x$, so that
$\sqrt{\frac{1+x}{1-x}}=\frac{1+x}{\sqrt{1-x^2}}$
We then use the following binomial expansion:
$
(1+x)^n = 1 + nx + O(x^2)
$

where $O(x^2)$ means terms of order $x^2$ and higher. The assumption of $|x|\ll1$ then implies that:
$(1+x)^n\approx1+nx$
Hence,
$(1-x^2)^{-1/2}\approx1+\frac{1}{2}x^2$
The LHS is now
$\sqrt{\frac{1+x}{1-x}}=\frac{1+x}{\sqrt{1-x^2}}\approx(1+x)\left(1+\frac{1}{2}x^2\right)$
which, when expanded, gives
$\sqrt{\frac{1+x}{1-x}}\approx1+x+\frac{1}{2}x^2$
which is the required second-order approximation.
« Last Edit: July 29, 2019, 01:04:42 pm by blyatman »
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#### mani.s_

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##### Re: 3U Maths Question Thread
« Reply #4162 on: July 28, 2019, 08:26:22 pm »
0
I just wanted to double-check a solution. I have to convert the following parametric equation to a cartesian equation: x = cos 2t y = sin t, 0<_ t <_ 2pi.

x = cos (2t)
cos (2t) = x
arccos (x) = 2t
t = arccos (x)/2

so the cartesian equation would be:
y = sin ( arccos (x) ) / 2

but apparently its:
y = plus-minus sine( arccos (x) ) / 2

Why is it plus-minus???

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4163 on: July 28, 2019, 08:30:30 pm »
+2
Is this a HSC question? Because I've never seen a binomial question involving square roots. It can be done, just that I've never seen it asked in the HSC.
I consulted with the poster later on and we found that the source of the question was well beyond the HSC. The question can be buzz-killed using their method, which is actually the generalised binomial theorem. (The generalised binomial theorem can be used, and then the usual method of equating coefficients is sufficient.)

They were just trying to look for hard binomial theorem questions for practice, and unintentionally stumbled somewhere too far. We do not need to worry about it.

#### blyatman

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##### Re: 3U Maths Question Thread
« Reply #4164 on: July 28, 2019, 08:42:59 pm »
0
I consulted with the poster later on and we found that the source of the question was well beyond the HSC. The question can be buzz-killed using their method, which is actually the generalised binomial theorem. (The generalised binomial theorem can be used, and then the usual method of equating coefficients is sufficient.)

They were just trying to look for hard binomial theorem questions for practice, and unintentionally stumbled somewhere too far. We do not need to worry about it.

Yeh I thought it didn't seem like HSC material. As you said, I used the generalised binomial theorem in the above proof in the expansion of (1+x)^n, which is more or less the Taylor series expansion.

An interesting problem nonetheless.
« Last Edit: July 28, 2019, 08:48:55 pm by blyatman »
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Research background: General Relativity (Gravitational Astrophysics Research Group, Sydney Institute for Astronomy, USYD)
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M.Phil. (Aerospace Engineering)
B.Eng. (Aerospace Engineering, Honours Class I)

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4165 on: July 28, 2019, 08:59:18 pm »
0
I just wanted to double-check a solution. I have to convert the following parametric equation to a cartesian equation: x = cos 2t y = sin t, 0<_ t <_ 2pi.

x = cos (2t)
cos (2t) = x
arccos (x) = 2t
t = arccos (x)/2

so the cartesian equation would be:
y = sin ( arccos (x) ) / 2

but apparently its:
y = plus-minus sine( arccos (x) ) / 2

Why is it plus-minus???

Without the plus-minus, sin(arccos(x))/2 is only positive. The range of arccos x is 0<=y<=pi, and hence the range of sin(arccos(x))/2 is 0<=y<=1/2. The plus-minus effectively remedies this situation by 'making' the range -1/2<=y<=1/2, since for values of t between +-2pi, sine and cosine can be both positive and negative, effectively mapping out the cartesian equation across the four quadrants instead of the two the inverse trig functions are usually limited to.

Hope this helps
Failing everything, but I'm still Flareon up.

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#### mani.s_

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##### Re: 3U Maths Question Thread
« Reply #4166 on: July 28, 2019, 09:01:37 pm »
0
Without the plus-minus, sin(arccos(x))/2 is only positive. The range of arccos x is 0<=y<=pi, and hence the range of sin(arccos(x))/2 is 0<=y<=1/2. The plus-minus effectively remedies this situation by 'making' the range -1/2<=y<=1/2, since for values of t between +-2pi, sine and cosine can be both positive and negative, effectively mapping out the cartesian equation across the four quadrants instead of the two the inverse trig functions are usually limited to.

Hope this helps
So is my working out and answer correct???

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4167 on: July 28, 2019, 09:13:23 pm »
+3
So is my working out and answer correct???
cos (2t) = x
arccos (x) = 2t
It is incorrect (only partially correct) as this line is unjustified. You haven't done a quadrants analysis on your condition $0\leq t \leq 2\pi$.
$\text{Note that consequently }0\leq 2t \leq 4\pi.$
That is, we have solutions for $2t$ in the 1st to 8th quadrants.
$\text{First suppose }x \geq 0.\\ \text{Then we need to consider the 1st, 4th, 5th and 8th quadrants.}$
$\text{Solving }\cos 2t = x\text{ then gives}\\ \boxed{2t = \arccos x, \, 2\pi - \arccos x, \, 2\pi + \arccos x, \, 4\pi - \arccos x}.$
You can then plug each of these back into $y=\sin t$. You will find that two of the solutions give $y = -\sin \left( \frac{\arccos x}{2} \right)$.

A similar thing happens for the case $x < 0$.

#### mani.s_

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##### Re: 3U Maths Question Thread
« Reply #4168 on: July 30, 2019, 08:15:56 pm »
0
Hi, I'm having trouble converting this parametric equation: x= sin t and y= cos 2t, where 0<=t<=2pi. I'm supposed to use the double angle formula for cos 2t: cos 2t = 1 - 2sin^2 t.

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4169 on: July 30, 2019, 08:51:59 pm »
+1
Hey there!

It's a lot simpler than you think If x=sin t, and you have y=cos 2t = 1-2sin^2 t, then logically y=1-2x^2, substituting in the first thing.

Hope this helps
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

format: plain


text:
# block.contact.normal

Syntax

v := block.contact.normal(bcp)

Get the unit normal of a contact.

Returns: v - unit normal bcp - block contact pointer

Component Access

f := block.contact.normal (bcp)->x

f := block.contact.normal.x(bcp)

Get the $$x$$-component of the unit normal of a contact.

Returns: f - $$x$$-component of the unit normal vector bcp - block contact pointer

Access other vector components ($$y$$, $$z$$) by substituting the component name where $$x$$ appears above. See Member Access Operator for information about accessing members from data types using ->.

Deprecated Component Access

Component access by adding an optional integer in the function arguments (block.contact.normal(bcp,<int>)) is deprecated. It remains available but will not be supported in future versions. See Component Access with Integers in FISH on the Deprecated Commands and FISH page for details.

This is hidden. Added to include remaining vector component access functions for inline help.

f = block.contact.normal.y(bcp)
f = block.contact.normal.z(bcp)


text:
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# If f(x) = (5x+2)/(3x−5) and g(x) = x^2 – 2x – 1, then the value of g(f

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If f(x) = (5x+2)/(3x−5) and g(x) = x^2 – 2x – 1, then the value of g(f  [#permalink]

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If $$f(x) = \frac{5x+2}{3x−5}$$ and $$g(x) = x^2 – 2x – 1$$, then the value of $$g(f(f(3)))$$ is:

A. 1/3
B. 2/3
C. 1
D. 2
E. 6

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Re: If f(x) = (5x+2)/(3x−5) and g(x) = x^2 – 2x – 1, then the value of g(f  [#permalink]

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02 Apr 2020, 10:18
If $$f(x) = \frac{5x+2}{3x−5}$$ and $$g(x) = x^2 – 2x – 1$$, then the value of $$g(f(f(3)))$$ is:

A. 1/3
B. 2/3
C. 1
D. 2 --> correct: f(3)=$$\frac{5*3+2}{3*3−5}$$ = $$\frac{17}{4}$$, f(f(3))=f(17/4)=$$(5*(17/4)+2)/(3*(17/4)−5)$$ =$$\frac{85+8}{51−20}$$ =$$\frac{93}{31}$$ = 3, so g(f(f(3))) = g(f(17/4)) = g(3) = 3^2-2*3-1=9-6-1=2
E. 6
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Re: If f(x) = (5x+2)/(3x−5) and g(x) = x^2 – 2x – 1, then the value of g(f  [#permalink]

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02 Apr 2020, 10:22
f(x)=5x+2/3x−5 and g(x)=x2–2x–1
f(3)=(5*3+2)/(3*3−5) = 17/4
now, f(17/4) =(5*17/4+2)/(3*17/4−5) = 73/31 = 3
therefore, g(f(f(3))) = g(3) =3^2–2*3–1 = 2

Re: If f(x) = (5x+2)/(3x−5) and g(x) = x^2 – 2x – 1, then the value of g(f   [#permalink] 02 Apr 2020, 10:22


text:
# Low Dimensional Topology

## July 28, 2015

### Tangle diagram crossings and quantum entanglement

Filed under: Uncategorized — dmoskovich @ 1:55 pm

In low dimensional topology we speak of tangles, while quantum physics speaks of entanglement. Similar words, but is there a deeper connection? Kauffman conjectured that the answer is yes (and I think he’s right, although maybe for other reasons). Glancing through arXiv this morning, I came across the following recent preprint:

Alagic, G., Jarret M., and Jordan S.P. Yang-Baxter operators need quantum entanglement to distinguish knots
arXiv:1507.05979

Their result is what it says in the title. Namely, we comb the knot into a braid, and assign R-matrices to crossings. An R-matrix underlies a linear operator $V\otimes V \rightarrow V \otimes V$. The authors prove that if the this operator maps product states to product states, then it gives rise (via a certain “taking the normalized trace of the operator the braid gives” procedure) to a trivial quantum knot invariant. Thus, entanglement is an essential part of being a nontrivial quantum invariant. Very cool!

There’s a suggestive picture in my head. Entanglement is all about nonlocality, where two non-interacting objects (an overstrand and an understrand?) cannot be described as separate systems (crossing?), but are inseparably intertwined in that they share some sort of coordination. It’s the entanglement which allows the overstrand to “communicate” to the understrand that it is there, making it possible to construct a nontrivial quantum invariant.

I suspect there’s a lot more to this story. Well done Alagic, Jarret, and Jordan!


text:
# The Pure Theory Behind Support Vector Machines

I’m extremely agitated today. I dunno why. Maybe because there was some convulsion in the peaceful tidings of the house I live in, or the fact that I’m kinda hungry at the moment. Anyways, I don’t have time for chitchat. Let’s get to the studying.

The following is taken from Foundations of Machine Learning by Rostamyar, et al.

Support Vector Machines are the most theoretically well motivated and practically most effective classification algorithms in modern machine learning.

Consider an input space $\mathcal{X}$ that is a subset of $\mathbb{R}^N$ with $N \geq 1$, and the output or target space $\mathcal{Y}=\{-1, +1\}$, and let $f : \mathcal{X} \rightarrow \mathcal{Y}$ be the target function. Given a hypothesis set $\mathcal{H}$ of functions mapping $\mathcal{X}$ to $\mathcal{Y}$, the binary classification task is formulated as follows:

The learner receives a training sample $S$ of size $m$ drawn independently and identically from $\mathcal{X}$ to some unknown distribution $\mathcal{D}$, $S = ((x_1, y_1), \ldots, (x_m, y_m)) \in (\mathcal{X}\times\mathcal{Y})^m$, with $y_i = f(x_i)$ for all $i \in [m]$. The problem consists of determining a hypothesis $h \in \mathcal{H}$, a binary classifier, with small generalization error: The probability that hypothesis set is not the target function is our error rate.

$R_{\mathcal{D}} = \underset{x\sim\mathcal{D}}{\mathbb{P}} [h(x) \neq f(x)].$

Different hypothesis sets $\mathcal{H}$ can be selected for this task. Hypothesis sets with smaller complexity provide better learning guarantees, everything else being equal. A natural hypothesis set with relatively small complexity is that of a linear classifier, or hyperplanes, which can be defined as follows:

$\mathcal{H}= \{x \rightarrow sign(w.x+b) : w \in \mathbb{R}^N, b \in r\}$

The learning problem is then referred to as a linear classification problem. The general equation of a hyperplane in $\mathbb{R}^N$is $w.x+b=0$ where $w\in\mathbb{R}^N$ is a non-zero vector normal to the hyperplane $b\in\mathbb{R}$ a scalar. A hypothesisol. of the form $x\rightarrow sign(w.x+b)$ thus labels positively all points falling on one side of the hyperplane $w.x+b=0$ and negatively all others.

From now until we say so, we’ll assume that the training sample $S$ can be linearly separated, that is, we assume the existence of a hyperplane that perfectly separates the training samples into two populations of positively and negatively labeled points, as illustrated by the left panel of figure below. This is equivalent to the existence of $(\boldsymbol{w}, b) \in (\mathbb{R}^N – \boldsymbol{\{0\}}) \times \mathbb{R}$such that:

$\forall i \in [m], \quad y_i(\boldsymbol{w}.x_i + b) \geq 0$

But, as you can see above, there are then infinitely many such separating hyperplane. Which hyperplane should a learning algorithm select? The definition of SVM solution is based on the notion of geometric margin.

Let’s define what we just came up with: The geometric margin $\rho_h(x)$ pf a linear classifier $h:\rightarrow \boldsymbol{w.x} + b$ at a point $x$ is its Euclidean distance to the hyperplane $\boldsymbol{w.x}+b=0$:

$\rho_h(x) = \frac{w.x+b}{||w||_2}$

The geometric margin of $\rho_h$ of a linear classifier h for a sample $S = (x_1, …, x_m)$ is the minimum geometric margin over the points in the sample, $\rho_h = min_{i\in[m]} \rho_h(x_i)$, that is the distance of hyperplane defining h to the closest sample points.

So what is the solution? It is that, the separating hyperplane with the maximum geometric margin is thus known as maximum-margin hyperplane. The right panel of the figure above illustrates the maximum-margin hyperplane returned by SVM algorithm is the separable case. We will present later in this chapter a theory that provides a strong justification for the solution. We can observe already, however, that the SVM solution can also be viewed as the safest choice in the following sense: a test point is classified correctly by separating hyperplanes with geometric margin $\rho$ even when it falls within a distance $\rho$ of the training samples sharing the same label: for the SVM solution, $\rho$ is the maximum geometric margin and thus the safest value.

We now derive the equations nd optimization problem that define the SVM solution. By definition of the geometric margin, the maximum margin of $\rho$ of a separating hyperplane is given by:

$\rho = \underset{w,b : y_i(w.x_i+b) \geq 0}{max}\underset{i\in[m]}{min}\frac{|w.x_i=b}{||w||} = \underset{w,b}{max}{min}\frac{y_(w.x_i+b)}{||w||}$

The second quality follows from the fact that, since sample is linearly separable, for the maximizing pair $(w, b), y_i(w.x_i+b)$ must be non-negative for al $i\in[m]$. Now, observe that the last expression is invariant to multiplication of $(w, b)$ by a positive scalar. Thus, we can restrict ourselves to pairs $(\boldsymbol{w},b)$ scaled such that $min_{i\in[m]}(\boldsymbol{w}.x_i+b) = 1$:

$\rho = \underset{min_{i\in[m]}y_i(w.x_i+b)=1}{max}\frac{1}{||w||} = \underset{\forall i \in[m],y_i(w.x_i+b) \geq }{max}\frac{1}{||w||}$

Figure below illustrates the solution $(w, b)$ of the maximization we just formalized. In addition to the maximum-margin hyperplane, it also shows the marginal hyperplanes, which are the hyperplanes parallel to the separating hyperplane and passing through the closest points on the negative or positive sides.

Since maximizing $1/||w||$ is equivalent to minimizing $\frac{1}{2}||w||^2$, in view of the equation above, the pair $(\boldsymbol{w}, b)$ returned by the SVM in the separable case is the solution of the following convex optimization problem:

$\underset{w, b}{min}\frac{1}{2}||w||^2$$\text{subject to}: y_i(\boldsymbol{w}.x_i+b) \geq 1, \forall i \in[m]$

Since the objective function is quadratic and the constraints are affine (meaning they are greater or equal to) the optimization problem above is in fact a specific instance of quadratic programming (QP), a family of problems extensively studied in optimization. A variety of commercial and open-source solvers are available for solving convex QP problems. Additionally, motivated by the empirical success of SVMs along with its rich theoretical underpinnings, specialized methods have been developed to more efficiently solve this particular convex QP problem, notably the block coordinate descent algorithms with blocks of just two coordinates.

So what are support vectors? See the formula above, we note that constraints tare affine and thus qualified. The objective function as well as the affine constrains are convex and differentiable.

We introduce Lagrange variables $\alpha_i \geq 0, i\in[m]$, associated to the m constrains and denoted by $\boldsymbol{\alpha}$ the vector $(\alpha_1, \ldots, \alpha_m)^T$. The Lagrangian can then be defiend for all $\boldsymbol{w}\in\mathbb{R}^N,b\in\mathbb{R}$, and $\boldsymbol{\alpha}\in\mathbb{R}_+^m$ by:

$\mathcal{L}(\boldsymbol{w},b,\boldsymbol{\alpha} = \frac{1}{2}||w||^2 – \sum_{i = 1}^{m}\alpha_i[y_i(w.x_i+b) -1]$

Support vectors fully define the maximum-margin hyperplane or SVM solution, which justifies the name of the algorithm. By definition, vectors not lying on the marginal hyperplanes do not affect the definiton of these hyperplanse – in their absence, the solution the solution to the SVM problem is unique, the support vectors are not. In dimensiosn $N, N+1$ points are sufficient to define a hyperplane. Thus when more then $N+1$ points lie on them marginal hyperplane, different choices are possible for $N+1$ support vectors.

But the points in the space are not always separable. In most practical settings, the training data is not linearly separable, which implies that for any hyperplane $\boldsymbol{w.x}+b=0$, there exists $x_i \in S$ such that:

$y_i[\boldsymbol{w.x_i}+b] \ngeq 1$

Thus, the constrains imposed on the linearly separable case cannot be hold simultaneously. However, a relaxed version of these constraints can indeed hold, that is, for each $i\in[m]$, there exists $\xi_i \geq 0$ such that:

$y_i[\boldsymbol{w.x_i}+b] \ngeq 1-\xi_i$

The variables $\xi_i$ are known as slack variables and are commonly used in optimization to define relaxed versions of constraints. Here, a slack variable $\xi_i$ measures the distance by which vector $x_i$ violates the desires inequality, $y_i(\boldsymol{w.x_i} + b) \geq 1 . This figure illustrates this situation: For hyperplane$y_i(w.x_i+b) = 1 $, a vector$x_i$with$x_i > 0$can be viewed as an outlier. Each$x_i\$ must be positioned on the correct side the appropriate marginal hyperplane. Here’s the formula we use to optimize the non-separable cases:

$\underset{w, b, \xi}{min} \frac{1}{2}||w||^2 + C\sum_{i=1}^{m}\xi_i^p$$\text{subject to} \quad y_i(w.x_i+b) \geq 1-\xi_i \wedge \xi_i \geq 0, i\in[m]$

Okay! Alright! I think I understand it now. That’s enough classification for today. I’m going to study something FUN next. Altough I’m a bit drowsy… No matter! I have some energy drinks at home. Plus I have some methamphetamine which I have acquired to boost my eenrgy… Nah, kidding. I’m a cocaine man!


text:
# Does old light contain clues to its age?

Light from celestial objects is old. In the case of galaxies, it's millions of years old. It seems plausible to me that light might show signs of its age.

I was surprised that a Google search only turned up one study in this area: Measurement of the speed of light from extraterrestrial sources. It looked at the speed of light from several bright stars: Aldebaran, Capella, and Vega. The results showed that the speeds were different!

My question is, have there been other studies by physicists that looked at old light versus new? It would be so interesting to view in an interferometer light that is a million years old. I can think of many other tests, and I'm sure physicists could think of more. Why hasn't this been done or has it? Could we probably find an age marker by looking closely at the old light?

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Nov 6 '17 at 16:50
• IMO, an ideal answer to this question would be (pointing out the obvious, to which standing answers at this instant, have done justice to) + (addressing how the linked article is compatible with this). I am really looking forward to the second part. :) – 299792458 Nov 6 '17 at 17:11
• A lot of celestial light is not very old: xkcd.com/1342 – Henry Nov 7 '17 at 12:45
• Relevant, but not a duplicate: physics.stackexchange.com/q/69448/26076 – Selene Routley Nov 8 '17 at 6:14

Light does not "experience" time, the concept "age" does not apply to light in a meaningful way (with respect to human experience). [As background; recall clocks slow for objects as they near the speed of "light" reaching a theoretical 0 if full light speed were attainable.] A thought experiment clock on a photon would therefore stand still. A photon's source does have an "age" in the traditional (human experience) sense, and it is standard that we say the light is as old as it's source. That "age" does not then carry with it the traditional effect of aging.

While the light source ages in a traditional fashion and may in fact be completely burned out though we can observe it today from our distant position in space, any photon from an object no matter how old the source is in no way different than a newly created photon presuming it is the same wavelength. As I view it you could not discern the "age" of light without knowledge of its source, because light is in reality timeless.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Nov 6 '17 at 16:50

The oldest light in the universe is the Cosmic Microwave Background, which is only about 380,000 years younger than the universe. It shows signs of its age in the redshift that has occurred as the universe expanded- the CMB would have been visible when it was created, and is now in the microwave spectrum. Besides that, however, the results are consistent with it acting like normal "young" light.

No experiment can prove that light does not evolve in some way over some timescale- we can only really test specific ways we expect light might evolve, and set limits on the timescales they must occur over. So the idea that light might evolve over time in some way is not really falsifiable. If a grad student were to work on this, I'd definitely say it should be in the context of specific, falsifiable theories regarding how light could evolve over time.

As far as I know, all models that have the photon evolving at all require the photon to have a non-zero rest mass. If photons have a non-zero rest mass, they could decay into other particles over some long timescale, they could oscillate like neutrinos do to another particle, and they would slow down over time (very slightly) as they were redshifted. The Particle Data Group lists current limits on the photon mass as $m_\gamma < 10^{-18}~\rm eV$. Note that this implies that the speed of a CMB photon is within $2\cdot10^{-4}\rm\frac{meters}{universe\ age}$ of the nominal speed of light, so if it exists the slowdown would be far too small to measure directly.

• "You can't prove a negative" is simply wrong. Every statement can be written in a negative form. Is it impossible to prove anything? "You can't prove a negative" is also a negative. How did you prove it? :) See departments.bloomu.edu/philosophy/pages/content/hales/… – Eric Duminil Nov 5 '17 at 10:14
• @tfb: Please note the difference between "Here's a negative we cannot prove yet" and "You can't prove a negative". Also, I'd argue that we cannot prove anything in physics. We merely have models that are considered "good enough" for the time being. – Eric Duminil Nov 5 '17 at 18:20
• If you can't prove a negative, I'm gonna have to hand back my math degree. – Todd Wilcox Nov 5 '17 at 18:40
• @EricDuminil I was referring not to a law of logic, but the general principle in science that it is impossible to show the nonexistence of something sufficiently open ended, and the general asymmetry in difficulty between proving the existence of something that exists and proving the nonexistence of something that does not. Still, I see your point and have removed the offending phrase ;) – Chris Nov 5 '17 at 21:17
• @EricDuminil We have no evidence that light does not decay on the timescale of $t\sim 10^{100}~\rm yr$, for instance. Or $10^{10^{10^{10}}}\rm yr$. Or someone could make a theory that light oscillates to a non-SM gauge boson that is identical to the photon except it couples to the Higgs boson slightly differently. Or any number of crazy theories that always have "some place to run" whenever an experiment excludes part of their parameter space. – Chris Nov 5 '17 at 21:53

One can ask the question "how old is this piece of rock" and get good answers from studying the changes in the lattice structures with time, the dispersion of other atoms in the lattice which also depends on time, etc. We are able to do this because the basic lattices of the rock are constant in time. They might break down at a certain rate, interact with extra atoms at a certain rate etc., but the base is constant in time.

Light is made up of zillions of photons in superposition (not interaction), the photons' quantum mechanical wavefunctions adding in such a way as to produce the observed light. In contrast to atoms in a lattice, the structure is instantaneous, i.e. a slice in this classical light wave solution perpendicular to its direction of motion

will contain zillions of photons running along, keeping no history except the history of the frequency of the wave which is their energy $E=h\nu$. This can be Doppler shifted and spectral lines can give a history of the travel time and the origin. So it is possible to find a history in the wavetrains coming from afar for special situations, (like the Doppler shift) and where there are intermittent signals as also with pulsars, one might get extra information from the variations of the amplitude of the light arriving, as with neutron star mergers which give the history of the merge in electromagnetic radiation (recently also with gravitational radiation, but that is another story).

Actually this completely new research on time crystals might open a window in the future of something similar to the contaminations time induces in a normal crystal where a history can be unraveled, but I am not holding my breath. No natural time crystals have been proposed which might have sent wavetrains that could be used to extract a history from the light arriving.

Light emitted by cosmic sources usually has spectral lines of known elements, such as hydrogen. If the light source moves from us or light lost its energy en route, the spectral lines get shifted towards red end. Given the universe expands, so that all light loses its energy when travelling, we can roughly entimate what distance light has travelled by measuring the shift of the spectral lines.

• yeah, but how to tell if it's an ancient photon from a distant star, or a young one from a light source in the same room. you can't ask it for a birth certificate. – Jasen Nov 6 '17 at 8:33
• @Jasen it cannot be a young one from the same room because it has hydrogen spectral lines shifted. – Anixx Nov 6 '17 at 14:39
• @Jasen The key is that you're not measuring one photon, you're measuring a population of photons that you know came from a star (but may not be sure how far away the star is) – Random832 Nov 6 '17 at 16:20

In my opinion, light slowly loses a small amount of energy as it travels very long distances. This manifests itself as a red shift in light from distant galaxies. This energy is not lost, but a result of the light particle causing a push of the light source and the light destination away from each other. This could be considered an example of the electromagnetic force pushing two galaxies away from each other as light bounces back and forth between them. Although the current scientific consensus is that this red shift is caused by space itself expanding, I believe that such an expansion of space would have caused the large scale structures of the universe such as clusters of galaxies to dissipate long ago. However, light travelling between galaxies would cause a force pushing the galaxies away from each other. This would have the same effect as space expanding, although without the need for "dark energy" or a violation of the law of conservation of mass and energy. Whether you share my opinion or strictly follow the scientific community, you can get a general idea of how old light is by measuring its red shift relative to the expected frequency.

• That’s an interesting theory. Maybe you could conduct some experiments to see if nature supports your hypothesis. – Lambda Nov 12 '17 at 14:45
• At close range, when a photon bounces back and forth between two electrons, it causes the electrons to accelerate away from each other. This acceleration of the electron causes a red shift of the photon due to the difference in the relative speed of the electrons before and after the photon strikes the electron. – Russell Hankins Nov 12 '17 at 20:41


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# What is the quotient $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3})$?

I am currently doing a past paper and it asks the following:

Prove that for $I=(1+2\sqrt{3})$ we have $\mathbb Z[\sqrt{3}]/I$ a field with $11$ elements.

If I assume standard algebraic number theory then it would only be a few lines:

We know $N(I)=|\mathcal O/I|$ where $\mathcal O=\mathbb Z[\sqrt{3}]$. And the norm of a principal ideal is equal to the absolute value of the norm of its generator, i.e. $N(I)=|N(1+2\sqrt{3})|=|1-3\cdot 4|=11$. So $|\mathcal O/I|=11$ and hence $\mathcal O/I$ must be the finite field with $11$ elements.

But this paper is for a normal Commutative Algebra module and I cannot assume any of this. Is there any other way to approach this?

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Notice that: $(-1+2\sqrt{3})(1+2\sqrt{3})=11\in I$ Also $(1+2\sqrt{3})\sqrt{3}=6+\sqrt{3}\in I$. Thus $a+b\sqrt{3}+I=a+b(-6)+I=a-6b+I=[(a-6b)\mod\,11]+I$. Hence: $$\mathbb{Z}[\sqrt{3}]/I=\{a+I:0\leq a\leq 10\}$$

Now Consider $\phi:\mathbb{Z}_{11}\rightarrow \mathbb{Z}[\sqrt{3}]/I$ that sends $x$ to $x+I$. Verify that $\phi$ is a surjective group homomorphism. Thus, $|\mathbb{Z}[\sqrt{3}]/I|$ divides $11$ ......

Finally, show that $|\mathbb{Z}[\sqrt{3}]/I|\not=1$ by showing that $1\not\in I$ using a norm argument.

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Would you mind giving some details as to why $a-6b+I=(a-6b)$ mod $(11+I)$ ? –  Tom May 8 '13 at 0:14
@Thomas The notation looks weird I will edit it –  Amr May 8 '13 at 0:14
@Tthomas Is this better now ? –  Amr May 8 '13 at 0:15
@Tthomas $a-6b=11k+(a-6b)\mod\,11$. Since $11\in I$, therefore $11k\in I$. Tthus, $(a-6b)\mod\,11+I=a-6b+I$ –  Amr May 8 '13 at 0:16
Is this because 11 is the smallest positive integer in $I$? –  Tom May 8 '13 at 0:22

There is a general method for computing these quotients, which is quite straight forward when you feel comfortable with polynomials.

Since $\mathbb{Z}[\sqrt{3}] \cong \mathbb{Z}[x]/(x^2-3)$, we have $Q:=\mathbb{Z}[\sqrt{3}]/(1+2 \sqrt{3}) \cong \mathbb{Z}[x]/(x^2-3,1+2x)$. In $Q$ we have $0=(2x+1)(2x-1)=4 x^2-1=11$. Therefore

$Q \cong \mathbb{F}_{11}[x]/(x^2-3,2 \cdot 6 + 2 \cdot x) = \mathbb{F}_{11}[x]/(x^2-3,6+x)=\mathbb{F}_{11}/(5^2-3)=\mathbb{F}_{11}.$

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In questions like this it can help to know the order of the quotient ring a priori. For this, one can use the following argument:

• One has the chain of ideals $(11) = (1+2\sqrt{3})(1-2\sqrt{3}) \subset (1+2\sqrt{3}) \subset \mathbb Z[\sqrt{3}].$

• Multiplication by $(1+2\sqrt{3})$ induces an isomorphism between $Z[\sqrt{3}]/(1-2\sqrt{3})$ and $(1+ 2\sqrt{3})/(11)$.

• Galois conjugation (swapping $\sqrt{3}$ and $-\sqrt{3}$) gives an isomorphism betweem $\mathbb Z[\sqrt{3}]/(1-2\sqrt{3})$ and $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3}).$

Putting all this together, we find that the order of $\mathbb Z[\sqrt{3}]/(11)$ is equal to the square of the order of $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3}).$ Since the former is isomorphic (as an abelian group) to $\mathbb Z/(11) \times \mathbb Z/(11)$, it has order $11^2$, and so $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3})$ has order $11$.

In this particular case, we are done, since the quotient ring has order $11$ which is prime, and so is necessarily the prime field of order $11$.

In general, one can show that if $a$ and $b$ are coprime, then $\mathbb Z[\sqrt{3}]/(a + b \sqrt{3})$ is isomorphic to $\mathbb Z/(a^2 - 3b^2)$, and it is a good exercise to write down the details, using the same argument as in Amr's answer (combined with the obvious generalization of the above argument, to see that order of the quotient ring is equal to $|a^2 - 3b^2|$).

If you want to see another phrasing of essentially the same argument (in the context of the ring $\mathbb Z[i]$, but it goes exactly the same way) see this answer.

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text:
# Enzyme Kinetics - Problems

1. Oct 6, 2004

### marksju

So, we were given a set of data of (uM) and the Velocity (umol/minute) of an inhibitor and no inhibitor.

we are supposed to find the values of Vmax and Km in the absence of inhibitor and in the presence.

do I use 1/Vo = 1/Vmax + Km/Vmax - 1/?

but what is Km and the intial velocity?

I'm sure this is fundamental for many, but the help would be appreciated.

2. Oct 6, 2004

### marksju

$$\frac{1}{V{o}}=\frac{1}{V{max}}+\frac{K{m}}{V{max}}-\frac{1}{}$$

Last edited: Oct 6, 2004
3. Oct 6, 2004

### marksju

Using that equation, wouldn't we need to know what either $$K{m}$$ or $$V{max}$$ is?