# Tag Info

3

If you are to derive this directly from the definition $\cos(x) \equiv \sum_{n=0}^\infty\frac{x^{2n}(-1)^{n}}{(2n)!}$ and $\sin(x) \equiv \sum_{n=0}^\infty\frac{x^{2n+1}(-1)^{n}}{(2n+1)!}$ then we can do this without having to perform the product by first using term-by-term differentiation to get $$\cos'(x) = -\sin(x),~~~~~\sin'(x) = \cos(x)$$ If we now ...

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

Let's get rid of $\sqrt{2}$ by rewriting the inequality as $$f(x)=2x-\sqrt{2(x^2+1)}-\ln x\ge0$$ We have $\lim_{x\to0}f(x)=\infty$. For computing the limit at $\infty$, we do the substitution $t=1/x$, so the limit becomes $$\lim_{t\to0^+}\frac{2}{t}-\frac{\sqrt{2(t^2+1)}}{t}+\ln t= \lim_{t\to0^+}\frac{2-\sqrt{2(t^2+1)}+t\ln t}{t}=\infty$$ Thus we know ...

2

It's clear. My suggestion is to also write them mathematically, as this will be more concise. In the following, $|A|$ denotes the size of the set $A$. $$f(a,x,p) = \left|\{(k,k+2)\in\mathbb{Z}^2\ |\ k\geq x,\ k+2\leq 30a+x,\ \gcd(k^2+2k,p\#) = 1\}\right|,$$ and $$g(a,b,x,p) = \min_{0\leq j<b,\ j\in\mathbb{N}} f(a,x+30j,p).$$

2

Let $f(x)=\cos ^2 x+ \sin ^2x$ , then $f'(x)= -2 \sin x \cos x+2 \sin x\cos x=0$. As $f(0)=1, f(x)=1$

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 $$f(x)g(x) = \sum_{m = 0}^{\infty} \biggl(\sum_{k = 0}^m \frac{a_k}{k!}\cdot \frac{b_{m-k}}{(m-k)!}\biggr)x^m = ... 2 well, you can start with a more simpler solution:$$ \sin(x) = 1/2i\ *\ (e^{ix} - e^{-ix}) \\ \cos(x) = 1/2\ *\ (e^{ix} + e^{-ix}) \\ \sin^2(x) = -1/2\ *\ (e^{2ix} + e^{-2ix} -2)\\ \cos^2(x) = 1/2\ *\ (e^{2ix} + e^{-2ix} +2)\\ \sin^2(x) + \cos^2(x) = -1/2\ *\ (e^{2ix} + e^{-2ix} -2) + 1/2\ *\ (e^{2ix} + e^{-2ix} +2) = 1 $$2 Your sequence is$$a_n=\frac{n(-1)^n-2^{-n}}{n},\qquad n\in\mathbb{N}$$Or, rewritten to$$a_n=(-1)^n-\frac{1}{n2^n}$$The first term (-1)^n alternates between 1 and -1, and notice that \frac{1}{n2^n} is always positive, and never greater than one. So it is true that for all n\in\mathbb{N}, -2\leq a_n< 1, i.e. (a_n) is bounded. 2 I quote from Wikipedia: A tautology in first-order logic is a sentence that can be obtained by taking a tautology of propositional logic and uniformly replacing each propositional variable by a first-order formula (one formula per propositional variable). In propositional logic a tautology is a formula which evaluates to be true for every possible ... 2 About de Polignac's formula: You found yourself how many factors 5 the number n! has: The number of factors 5 is n/5 + n/25 + n/125... If you take other prime numbers, then you get very similar results: The number of factors 2 is n/2 + n/4 + n/8 ..., then number of factors 103 is n/103 + n/103^2 + n/103^3 ... and so on. s_p(n) is the formula you found, ... 1 You add a zero every time that you multiply by 10. Since the only prime factors of 10 are 2 and 5, then clearly the trailing number of zeros in a number is the minimum of the two exponents in the prime factorization of that number. To relate this to the formula you found, note that when computing a factorial, you will add a zero to the end every ... 1 To tell whether the formula is true in every interpretation, the first step is to think through what each side of the formula says about an interpretation. The left side$$ (\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)] $$says that P holds ... 1 Hint: Try using the triangle inequality to break up the fraction into two terms, and showing each term is bounded. Answer: 1 Since this field K is contained in \mathbb{R}, it does not contain one root, \beta of the polynomial x^7 - 5. There is a homomorphism from K to \mathbb{C} which sends \alpha to \beta and fixes \sqrt{5}. This homomorphism does not map K to itself, so K is not a normal extension of \mathbb{Q}. Hence it is not a splitting field. 1 No, there is (provably) no universal method to determine whether a formula in first-order logic is logically valid. We say that first-order logic is undecidable. If the formula is logically valid, it will have a formal proof (in each of the various proof systems for first-order logic). On the other hand, if it is not logically valid, then there will be a ... 1 What you wrote in the second line is incorrect. To show that n(n+1) is even for all nonnegative integers n by mathematical induction, you want to show that following: Step 1. Show that for n=0, n(n+1) is even; Step 2. Assuming that for n=k, n(n+1) is even, show that n(n+1) is even for n=k+1. [Added:] In Step 2, what you really need to ... 1 No, the mean value theorem is not necessary to prove these. Neither of the following arguments uses the mean value theorem per se. The first relies on the concavity of x^t for 0<t<1, and the second uses that for positive x,h and t>1, (x+h)^t > x^t + tx^{t-1}h. Both these facts can be demonstrated without using the MVT explicitly. ... 1 For the first one, we have F_0=0 and F_1 = 1 with F_n=F_{n-1}+F_{n-2} Keep in mind that in arithmetic we have that E+E=E, E+O=O and O+O=E with O being some odd number and E being an even number. If there are finitely many even numbers that means that at some M we have that for all n>M that F_n is odd, but then we have that ... 1 by taylor's theorem with lagrange reminder gives you$$f(a+h) = f(a) + f'(a) h + h^n f^{(n)}(a+\theta h), \space 0 < \theta < 1.$$use the fact that f^{(n)}(a)\neq 0 and continuity of f^{(n)} to conclude that f^{(n)}(a+\theta h) \neq 0. now for n odd h^nf^{(n)}(a + \theta h) changes sign as h goes from negative to positive implying that ... 1 Let f(x)=x\sqrt{2}-\sqrt{x^{2}+1} -\dfrac{\sqrt{2}}{2}\ln{(x)}. Then,$$f'(x)=\sqrt{2}-\frac{x}{\sqrt{x^2+1}}-\frac{1}{x\sqrt{2}}=\frac{2x\sqrt{x^2+1}-x^2\sqrt{2}-\sqrt{x^2+1}}{x\sqrt{2(x^2+1)}}.$$For x>0, the denominator is positive, so we just need to check the sign of the numerator: ... 1$$f(x):=x\sqrt{2}-\sqrt{x^2+1}-\frac{\sqrt{2}}{2}\ln x$$Which is continuously differentiable on (0,\infty)$$f'(x)=-\frac{x}{\sqrt{x^2+1}}-\frac{1}{\sqrt{2} x}+\sqrt{2}=\frac{-2 x^2+2 \sqrt{2} \sqrt{x^2+1} x-\sqrt{2} \sqrt{x^2+1}}{2 x \sqrt{x^2+1}}$$We want to solve$$ -2 x^2+2 \sqrt{2} \sqrt{x^2+1} x-\sqrt{2} \sqrt{x^2+1}=0 ...

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

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 ...

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