# Tag Info

3

Hint: Use strengthenings of Bertrand's Postulate, in particular Nagura's result that for $n\ge 25$ there is always a prime between $n$ and $n\left(1+\frac{1}{5}\right)$. We can use this to show that unless $k$ is very small, there are always at least $4$ distinct primes $p$ that satisfy $\lfloor \log p \rfloor=k$.

3

First I'll consider your (bolded) question about taking only discs with centers in $S^1$. Indeed we can do this - consider your open cover $\{ U_i\}$. Each $x\in S^1$ is contained in some $U_x\in\{ U_i\}$, and $U_x$ is open, so it contains some open ball in $\mathbb{R}^2$ centered at $x$, say $B_x$. Indeed, it should be clear that the collection of open ...

2

You can do this by induction over the size of $X$ Or rather prove something stronger: If $f:Y\rightarrow Z$ is a function between sets of equal size, then $f$ is injective if and only if $f$ is surjective Base: $|Y|=|Z|= 1$ (or possibly $0$ but that is completely trivial), then every function is both surjective and injective. Induction assumption: If ...

2

Step 1. Applying the mean value theorem to the function $x \mapsto x - \arctan x$, for some $\xi \in [0, A_{k-1}]$ we have $$A_{k-1} - A_k = (A_{k-1} - \arctan A_{k-1}) = A_{k-1} \cdot \frac{\xi^2}{1 + \xi^2} \leq \frac{A_{k-1}^3}{1 + A_{k-1}^2}.$$ This shows that $$A_{k-1} - A_k \leq A_{k-1}^3 \qquad \text{and} \qquad \frac{A_{k-1}}{1+A_{k-1}^2} \leq ... 2 Your proof is perfectly correct, and there is nothing else to justify. A more rigorous way would be Let \varepsilon>0. Since \lim_{x\to\infty }f(x)=L,$$\exists M>0: x>M\implies |f(x)-L|<\varepsilon.$$Let \delta=\frac{1}{M}. Then, if y=\frac{1}{x},$$0<y<\delta\implies x>M\implies |f(x)-L|<\varepsilon\implies ...

2

We need some proposition depending on $n$ that we can prove by induction. In this case, it is $$P(n)\ \colon\ \sum_{i=1}^n i \geq \frac{n^2}{2}.$$ First check the base case: $$P(1)\ \colon\ 1\geq \frac{1}{2}.$$ This is true, so we have proved the base case $P(1)$. Next suppose that $P(n)$ is true, for some $n\geq 1$. We wish to prove that then $P(n+1)$ is ...

2

The sequence $1,6,20,48,90,132,132$ can be generated as $\displaystyle {5+i \choose i-1}\dfrac{8-i}{7}$ with $i$ running from $1$ to $7$, though it would be slightly more conventional to write $\displaystyle {n+k \choose k}\dfrac{n-k+1}{n+1}$ with $n=6$ and $k$ running from $0$ to $6$. The right hand part of the first diagram is known as Catalan's ...

2

show $f(x)=6x+4$ is a bijective function

2

I quote from Wikipedia: A tautology in first-order logic is a sentence that can be obtained by taking a tautology of propositional logic and uniformly replacing each propositional variable by a first-order formula (one formula per propositional variable). In propositional logic a tautology is a formula which evaluates to be true for every possible ...

2

About de Polignac's formula: You found yourself how many factors 5 the number n! has: The number of factors 5 is n/5 + n/25 + n/125... If you take other prime numbers, then you get very similar results: The number of factors 2 is n/2 + n/4 + n/8 ..., then number of factors 103 is n/103 + n/103^2 + n/103^3 ... and so on. $s_p(n)$ is the formula you found, ...

2

In situations like this, it is more convenient to write the coefficients of a power series as $\dfrac{c_n}{n!}$. Then $$f(x) = \sum_{n = 0}^{\infty} \frac{a_n}{n!}x^n\quad\text{and}\quad g(x) = \sum_{n = 0}^{\infty} \frac{b_n}{n!}x^n$$ yields $$f(x)g(x) = \sum_{m = 0}^{\infty} \biggl(\sum_{k = 0}^m \frac{a_k}{k!}\cdot \frac{b_{m-k}}{(m-k)!}\biggr)x^m = ... 1 You add a zero every time that you multiply by 10. Since the only prime factors of 10 are 2 and 5, then clearly the trailing number of zeros in a number is the minimum of the two exponents in the prime factorization of that number. To relate this to the formula you found, note that when computing a factorial, you will add a zero to the end every ... 1 To tell whether the formula is true in every interpretation, the first step is to think through what each side of the formula says about an interpretation. The left side$$ (\forall x)[P(x) \land Q(x)] $$says that P and Q hold of every object x in the interpretation. The right side$$ (\forall x)[P(x)] \land (\forall x)[Q(x)] $$says that P holds ... 1 Since this field K is contained in \mathbb{R}, it does not contain one root, \beta of the polynomial x^7 - 5. There is a homomorphism from K to \mathbb{C} which sends \alpha to \beta and fixes \sqrt{5}. This homomorphism does not map K to itself, so K is not a normal extension of \mathbb{Q}. Hence it is not a splitting field. 1 What you wrote in the second line is incorrect. To show that n(n+1) is even for all nonnegative integers n by mathematical induction, you want to show that following: Step 1. Show that for n=0, n(n+1) is even; Step 2. Assuming that for n=k, n(n+1) is even, show that n(n+1) is even for n=k+1. [Added:] In Step 2, what you really need to ... 1 No, there is (provably) no universal method to determine whether a formula in first-order logic is logically valid. We say that first-order logic is undecidable. If the formula is logically valid, it will have a formal proof (in each of the various proof systems for first-order logic). On the other hand, if it is not logically valid, then there will be a ... 1 No, the mean value theorem is not necessary to prove these. Neither of the following arguments uses the mean value theorem per se. The first relies on the concavity of x^t for 0<t<1, and the second uses that for positive x,h and t>1, (x+h)^t > x^t + tx^{t-1}h. Both these facts can be demonstrated without using the MVT explicitly. ... 1 Take the 1-to-1 function f of the open interval ]0,1[ in \mathbb R defined by f(x)= representation of x in the numerical system of base 2 This is enough. 1 by taylor's theorem with lagrange reminder gives you$$f(a+h) = f(a) + f'(a) h + h^n f^{(n)}(a+\theta h), \space 0 < \theta < 1. use the fact that $f^{(n)}(a)\neq 0$ and continuity of $f^{(n)}$ to conclude that $f^{(n)}(a+\theta h) \neq 0.$ now for $n$ odd $h^nf^{(n)}(a + \theta h)$ changes sign as $h$ goes from negative to positive implying that ...

1

For the first one, we have $F_0=0$ and $F_1 = 1$ with $F_n=F_{n-1}+F_{n-2}$ Keep in mind that in arithmetic we have that $E+E=E$, $E+O=O$ and $O+O=E$ with $O$ being some odd number and $E$ being an even number. If there are finitely many even numbers that means that at some $M$ we have that for all $n>M$ that $F_n$ is odd, but then we have that ...

1

Imagine connecting your "inlets" together to the left of your city and the "outlets" at the bottom, like this (shown for 4 of each instead of 7 of each): * | +-->: | ::. +-->:::: | :::::. <- dotted area corresponds to your drawing +-->::::::: | ::::::::. +-->:::::::::: v v v v | | | | +--+--+--+--* Then each route ...

1

Suppose that $f(a)<f(b)$ we show that $f$ is strictly increasing. Let $x\in [a,b]$ firstly, we show that $f(x)>f(a)$ if $f(x)\leq f(a)$, since $f$ is continuous, the image of $[x,b]$ by $f$ is an interval, since it contains $f(x)\leq f(a)$ and $f(b)>f(a)$ it contains every element $u$ such that $f(x)\leq u\leq f(b)$ in particular, there exists an ...

1

Posting this separately as an answer to the following side question, which is not directly related to the main one covered in my previous answer. How did the mathematicians found the values of these sine expressions $(sin \frac{\pi}{96})$? (Please don't tell me that series expansions were used because they give approximate results and that too is not ...

1

This is your exam review, so you should really solve it yourself. Doing so will help you remember what's going on in the exam. I will give you some hints on how to go about proving this: Prove if $det(A)$ = $0$, then columns of $A$ are linearly dependent. (By going through the definition of $det(A)$) Based on contrapositive version of 1, if columns of $A$ ...

1

You need to prove $P \iff Q \iff R$ . This mean $P\iff Q$ and also $Q \iff R$, thus this mean $P \iff R$ and $Q \iff R$ etc... One way to prove this type of statement is proving $P \implies Q$, then $Q \implies R$, then $R \implies P$.

1

Yes, exactly. You have three equivalent statements. Finally, this gives a proof that $HK$ is a subgroup if and only if $HK=KH$. This is a standard exercise in group theory - see question $7$ in this homework here.

1

You should not write for $i= 1,2, \dots$ since you do not know that $\mathcal{I}$ is countable (or even consists of integers - perhaps $\mathcal{I}=\mathbb{Q}$ or $\mathcal{I}=\mathbb{R}^2$ or some set that doesn't even contain numbers). In the reverse direction, you jump from $y \notin A_i$ for all $i \in \mathcal{I}$ to $y \in (\bigcup_{i \in ... 1 Yes it should be fine if you use the previous proof/construction as a "subroutine". For the second question, let's use${rev}$rather than$r$as a superscript to denote the reverse operation on strings and languages. Note that the definition of$L_2$amounts to$ \langle M \rangle\in L_2 \iff L(M)^{rev}\subseteq L(M). \$ But in fact equality holds, ...

Only top voted, non community-wiki answers of a minimum length are eligible