# Tag Info

34

$\LaTeX$ is not only free like in beer, but free like in speech. There are enormous advantages to using open source standards. Right now you like Mathematica and are happy with its software. Suppose that five years from now the company decides to put out a new version and break backwards-compatibility. Or the company goes out of business. And suddenly ...

31

I use both Mathematica and $\LaTeX$ extensively. While it is possible to make nice looking documents in each, they really have two different aims. Mathematica is mainly about mathematical computing and interactive content. $\LaTeX$ is about typesetting and publication-ready articles with fine language-based (as opposed to WYSIWYG) control of layout. As ...

10

OP asks: is there something significant I can do in LATEX/TEX that I can't do in Mathematica? LaTeX far better at handling print page layout … for example, 2 column layout, or page breaks, footnotes etc some journals/conferences require submissions in LaTeX format I am sure there are other benefits! — but much steeper learning curve … though I believe ...

9

Since $n$ is even, we can write $n = 2k$ for some integer $k$. Hint: $$n(n^2 + 20) = 2k((2k)^2 + 20)= 2k(4k^2 + 20) = 8k(k^2 + 5)$$ Hence, $8$ is a factor. Note further that one of $k$ or $k^2 + 5$ must be even, and hence divisible by $2$. Why? So now we know that $8\cdot 2 = 16$ is a factor. All that remains to be shown is that $3$ is also a factor.

8

Hint: Consider the function $g(x)=f(x)-f(-x)$. Since $g(0)=0$ and $\lim\limits_{x\to\infty}g(x)=0$, $g$ is either identically $0$ or it has a local maximum or minimum at $x_0\in(0,\infty)$. In the same way it was shown for Rolle's Theorem, we have that $g'(x_0)=0$. Thus, $$f'(x_0)+f'(-x_0)=0$$ If $f'$ is not identically $0$ and $x_0\ne0$, apply the ...

8

From a group of $n$ people, we want to select $r$ people to go on a trip, including the trip leader. We count the number of ways of doing the choosing in two different ways. (i) We can choose the leader in $n$ ways. For each of these ways, we can choose the rest of the people in $\binom{n-1}{r-1}$ ways, for a total of $n\binom{n-1}{r-1}$. (ii) We can ...

7

You might be interested in the essay Halmos, How to write mathematics and the lecture Serre, How to write mathematics badly. Both suggest to avoid starting a sentence with a symbol. In my personal experience from reading mathematical texts and having my own texts corrected by professors, they all agree with this convention. It's just not pleasing to read a ...

6

By definition functions have domains, proving injectivity without considering the domain is something devoided of sense, the domain of a function is part of its essence. In practice there can be times in which whatever the domain is, the proofs will look the same (if the functions are in fact injective), but they are necessarily different because the ...

6

Following the hint given by the OP, it suffices to show that $$\sum_{k=0}^n {n \choose k}^{\!2}=\sum_{k=0}^n {n \choose k}{n \choose {n-k}} = {2n \choose n}.$$ The last equality is a consequence of the following more general identity (known as Vandermonde's identity) $$\sum_{j=0}^k \binom{n-m}{k-j}\binom{m}{j}=\binom{n}{k},$$ where $n\ge m,k\ge 0$, which ...

5

The answer of Ishfaag is right, but I want to review some points about this problem. There are a lot of attempts for generalizing the concept of abelian groups and one of them is $n$-abelian groups. A group $G$ is said to be $n$-abelian if $(xy)^n = x^ny^n$ for all $x,y \in G$. It is also easy to see that a group $G$ is $n$-abelian if and only if it is ...

4

Basically, Mathematica is designed for calculations, and can do some typesetting too. $\LaTeX$ is designed for typesetting, and can do some calculations too. So the main strength of $\LaTeX$ here is that if and when you want to add some other common document feature to your math papers, like cross-referencing, bibliography, numbered lists, etc., you'll ...

4

6 divides $n(n+1)(n+2)$ if both 2 and 3 divide it. One of every three consecutive numbers is divisible by 3, and one of every two consecutive numbers is divisible by 2. Alternatively, by induction: it works for $n=1$. Assume the case of $n-1$, we have $$6\mid (n-1)(n)(n+1).$$ Then $$(n)(n+1)(n+2)=(n)(n+1)(n-1+3)=(n-1)(n)(n+1)+3(n)(n+1).$$ 6 divides the ...

4

My two-pennyworth ... Starting sentences with lower case letters from the Roman alphabet "looks wrong" in general. We don't expect sentences to start that way, and this habit of the well-trained eye (if we can call it that!) naturally carries over to some extent -- doesn't it? -- to the case where the lower case letters are italic letters used as ...

4

Think of it this way. On the right side, you're choosing $n$ objects from $2n$ objects. On the left side, it's equal to $\sum\binom{n}{k}\binom{n}{n-k}$. So, divide the $2n$ objects into 2 groups, both of $n$ size. Then, the total number of way of choosing $n$ objects is partitioning over how many elements you choose from one group, and the remaining ...

4

Generally, derangements come in pairs: the inverse of a derangement is a derangement. The exceptions are the fixed-point-free involutions, i.e., the permutations in which every cycle is a $2$-cycle. Thus the parity of the number of derangements is the same as the parity of the number of fixed-point-free involutions. If $n$ is odd, there are no ...

4

For me, the beauty of this statement is not in its proof, its the whole path it takes for you to get there. First learning what elementary functions such as $\sin$, $\cos$ and $\exp$ are, learning Taylor's series, expanding the functions to the complex plane, then learning of the equation $$e^{a+bi} = e^a (\cos b + i\sin b).$$ You feel really cool about ...

4

The simplest way of deriving/proving this is through Euler's formula: Take $e^{i(\theta+\alpha)}$ which, through basic exponent rules we know is equal to $e^{i\theta +i\alpha} = e^{i\theta}\cdot e^{i\alpha}\ .$ Now expanding this (using Euler's formula) we get: \begin{align} e^{i\theta}\cdot e^{i\alpha} &= ... 4 Your line is, after multiplying by 6 and moving things to the left, 3x+2y-6=0, whose distance to the origin is |-6|/\sqrt{3^2+2^2}=6/\sqrt{13}\approx 1.6641. See this page for the distance from point to line formula, with several proofs of it. Added: to proceed via the contrapositive, assume in fact that x^2+y^2 \le 1. Then we have |x|\le 1 and ... 3 Hint: Try the other way: x/2+y/3=1, so, \frac32x+y=3, yielding y=3-\frac32 x. Then compute x^2+y^2. Or, even better: try geometrically: the set of points (x,y) on the plane that satisfy x/2+y/3=1 is a line. This line contains (2,0) and (0,3). Draw it and draw also the disk x^2+y^2<1. 3 \angle ABE = 60\,^{\circ} + 90\,^{\circ} = 150 \,^{\circ}. Since, ABE is isosceles, \angle AEB = \frac{180\,^{\circ} -150\,^{\circ}}{2} = 15 \,^{\circ}. So, \angle AED = 90\,^{\circ}- 15\,^{\circ} = 75\,^{\circ}. Since, AED is isosceles, \angle EAD = 180\,^{\circ} -(75\,^{\circ} + 75\,^{\circ}) = 30\,^{\circ}. So, circumradius of ADE:R = ...

3

Hint: You can use one round of recursion to define add: add(x,0) = x add(x, Sy) = S(add(x,y)) Then you can use another round of recursion to define mult: mult(x,0) = 0 mult(x,Sy) = add(x, mult(x,y)) To show that composition alone does not suffice, you'll want to first show that any function defined from the basic elements in an expression of length k ...

3

Is it normal that I have the hardest time when I'm trying to prove statements that are blatantly obvious on a visual and/or intuitive level? Yes, this is quite common. The Jordan curve theorem is a classic example of a geometrically obvious theorem that is true, but quite hard to prove. The idea that there do not exist space-filling curves is a ...

3

Well, if you buy or have access to (through previous learning) $e^{i\theta} = \cos \theta + i \sin \theta$, then setting $\theta = a - b$ one has $e^{i(a - b)} = \cos(a - b) + i\sin(a - b). \tag{1}$ But also, $e^{i(a - b)} = e^{ia}e^{-ib} = (\cos a + i \sin a)(\cos b - i\sin b). \tag{2}$ The imaginary part of the product on the right of (20) is $\sin a ... 3 For me it is the proof with differential equation: Consider a unit circle on$\mathbb{C}$. It can be viewed as a function$f(t)$of some$t$. The tangent to a circle is orthogonal to the radius. Orthogonality on$\mathbb{C}$means multiplication by$i$, so we can write$f'(x)=if(x)$. Solving it gives$\dfrac{f'}f=i$, integrating gives$\ln(f)=ix+c$and ... 3 It is possible to find the parity from the formula $$\sum_{i=0}^n (-1)^i \frac{n!}{i!}=\sum_{i=0}^n (-1)^i n(n-1)\cdots(i+1).$$ Working modulo$2$: If$n$is even, then there is one non-zero contribution to the sum is when$i=n$, so we have an odd number. If$n$is odd, then there are two non-zero contributions to the sum, when$i=n$and when$i=n-1$, so ... 3 Make a list of (supposedly) all the primes, then write the product $$Q = p_1 p_2 p_3 \cdots p_k,$$ and write $$N = 12 Q^2 + 1.$$ You need to know how to prove this much: if we have a prime$q \equiv 5 \pmod 6,$and $$3 u^2 + v^2 \equiv 0 \pmod q,$$ then both $$u,v \equiv 0 \pmod q.$$ 3 The question is equivalent to proving that, for any integer$m$,$6$divides$m(m^2+5)$, because for$n=2m$the expression is$8m(m^2+5)$. Divisibility of$m(m^2+5)$by$2$is obvious, because$m^2\equiv m\pmod{2}$, so $$m(m^2+5)\equiv m(m+1)\equiv m^2+m\equiv 2m\equiv 0\pmod{2}.$$ Divisibility of$m(m^2+5)$by$3$follows similarly, because$m^3\equiv ...

3

A function $f:\ A\to B$ is tantamount to a subset $G_f\subset A\times B$ having certain properties. In most cases $A$ and $B$ are clearly specified in advance, and one can then start right away to investigate whether $f$ is injective or not. Very often $A$ and/or $B$ have to be surmised from the context. While the exact envisaged range $B$ (e.g., ${\mathbb ... 2 Hint: It suffices to show: $$\left(zw\right)^{\frac{n}{2}} = 1.$$ If that is the case, then (using the fact that conjugate of unity root is its inverse): $$\overline{\left(z+w\right)^{\frac{n}{2}}} = \left(z+w\right)^{\frac{n}{2}}.$$ Edit: We have:$\alpha \in \mathbb{R}$iff$\overline{\alpha} = \alpha$. Also, for a root of unity$u$we have: ... 2 Suppose for contradiction that$ f(c) \neq 0 $. Then by continuity, there exists$ \delta > 0 $such that$ f(x) \geq \frac{f(c)}{2} > 0 $when$ x \in [c-\frac{\delta}{2}, c+ \frac{\delta}{2} ] $. The integral$ \int_a^b f(x) dx > \int_{c-\delta/2}^{c+\delta/2} f(x) dx > \frac{f(c)}{2} \delta > 0 \$.

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