# Tag Info

6

Base Case: For $n = 1$, we have: $$1 \times 5 = 5 = \frac{1}{6}(1)(1 + 1)(2(1) + 13)$$ which works. Inductive Hypothesis: Assume that the claim holds for $n' = n - 1$, where $n \geq 2$. It remains to show that the claim holds for $n' = n$. Indeed, observe that: \begin{align*} &1 \times 5 +\cdots + n(n + 4) & \\ &= [1 \times 5 + \cdots + (n - ...

4

In the simple case of $A>0$, it suffices to just look at the taylor series definition: $$1+A\leq 1+A+A^2/2!+A^3/3!+\cdots=e^A$$

4

I'll leave the details to you and go straight to the main integral. Let $$I_n = \int \frac{du}{(1+u^2)^n}.\tag{1}$$ We clearly have $I_1=\arctan u$ and: $$\begin{eqnarray*} I_{n+1} = \int \frac{(1+u^2)-u^2}{(1+u^2)^{n+1}}\,du &=& I_n -\frac{1}{2}\int ... 3 Let's do the job as suggested by @Dr. MV. Integrate by parts with u=x^{m-1} and dv=x\left(x^2-1\right)^{5}dx. This means du=(m-1)x^{m-2} and v={\left(x^2-1\right)^6\over 12}. And so$$\begin{align}I_m=&=\left[uv\right]_0^1-\int_0^1vdu\\ &=-{m-1\over 12}\int_0^1\left(x^2-1\right)^6x^{m-2}dx\\ ...

3

Since $(e^x)' = e^x$ for all $x$, and $e^x > 1$ for $x > 0$, we have, for $x > 0$, $\begin{array}\\ e^x-1 &=\int_0^x e^t dt\\ &> \int_0^x 1 dt \qquad\text{since } e^x > 1 \text{ for } x > 0\\ &=x\\ \end{array}$

2

Consider a sequence of natural numbers $n - 1$, $n$, and $n + 1$. Since the question wants you to add them up, do so. Check if this gives you a multiple of 3. The smart choice of $n - 1$, $n$, $n + 1$ is "easier" than the choice of $n$, $n + 1$, $n + 2$ in many cases, and is a useful trick to know

2

Easier: let $(a,b)$ be an arbitary open interval, and $x \in A$. Then both $(a,x)$ and $(x,b)$ are non-empty open intervals, so they both contain a rational point, say $p$ in the former, $q$ in the latter. Then $x \in (p,q) \subset (a,b)$ as required.

2

Given the prime $p\neq 3$, the integer number you are looking for is $n(p) = 3^2p^2$. It has $3^2 = 9$ factors ($1$, $3$, $p$, $3p$, $3^2p$, $3p^2$, $3^2$, $p^2$, $3^2p^2$) and $$\mbox{Factof}(n(p)) = \frac{3^2p^2}{3^2} = p^2$$ If $p=3$ then the number $n(p)=2^2 3^3 = 108$. In this case $$\mbox{Factof}(108) = \frac{2^23^3}{2^2 3} = 3^2$$

2

'A prime no. always has two factors.' So the product of two primes $p$ and $q$ will have $4$ factors: $1$, $p$, $q$, $pq$ Now if the primes are distinct then the product $pq$ is not a multiple of $4$ since $2$ is the only even prime (and hence the other prime will be odd). So, $\text{Factof}(pq)$ is not an integer. For (b) part, Number of factors of ...

2

If $\mathcal{B}'$ is a basis, then in particular every element of $\mathcal{B}$ is a union of elements of $\mathcal{B}'$. But a singleton cannot be a union of proper subsets, so $\mathcal{B} \subset \mathcal{B}'$ and $\mathcal{B}'$ has at least $n$ elements. As an alternative proof, we could observe that the number of possible unions that we can for from a ...

2

For $n\in\mathbb{N}$ and $x\gt-1$, we have $$(1+x)^n\ge1+nx$$ which is easily proved by induction on $n$: \begin{align} (1+x)^n\ge1+nx\implies(1+x)^{n+1}&=(1+x)(1+x)^n\\ &\ge(1+x)(1+nx)\\ &=1+(n+1)x+nx^2\\ &\ge1+(n+1)x \end{align} It follows that, if $A\gt0$, then $$\left(1+{A\over n}\right)^n\ge1+n\left(A\over n\right)=1+A$$ and ...

2

Is this are you looking for? (using Einstein convention) $$\left[T^a,T^b\right]^i_j=(T^aT^b )_{ij}-(T^bT^a )_{ij}$$ $$=(T^a)^i_p(T^b)^p_j-(T^b)^i_q(T^a)^q_j=\epsilon_{aip}\epsilon_{bpj}-\epsilon_{biq}\epsilon_{aqj}.$$ And then whatever you need to do you probably will need to use the following identity: ...

2

If $x$ is rational, then $f(x) >0$ and hence there are points $y$ arbitrarily close such that $f(y) = 0$. Suppose $x$ is irrational and let $\epsilon>0$. Note that $I=\{ n | {1 \over n} \ge \epsilon \}$ is finite. Choose $\delta>0$ such that $B(x,\delta) \cap \{ r_n \}_{n \in I}$ is empty. Then $f(y) < \epsilon$ for all $y \in B(x,\delta)$.

2

A proof without induction: $$\sum_{k=1}^{n}k(k+4)=\sum_{k=1}^{n}k^2+4\sum_{k=1}^{n}k= \frac{n(n+1)(2n+1)}{6}+4\cdot\frac{n(n+1)}{2} =\frac{n(n+1)}{6}(2n+13)$$

1

Considering the case when $n = 6$ only shows that $2^n - 1$ is not prime for all composite $n$ (i.e. at least one of $2^n - 1$ is composite). However it does not show that $2^n - 1$ is not prime for every composite $n$ (i.e. all of $2^n - 1$ are composite), which is what is required in the contrapositive form of the question. By invoking the stated ...

1

It looks like you're managing to confuse yourself slightly by not representing the quantifiers on $A$ and $B$ explicitly. Apparently you know that it is not the case that $$\tag{1} \forall A,B : \overline A\cap \overline B \subseteq \overline{A\cap B}$$ However, you seem to be confusing that fact with $$\tag{2} \forall A,B : \neg(\overline A\cap ... 1$$1 \times 5+2\times6+3\times7 +\cdots +n(n + 4) = =\sum_{i=1}^{n}i^2+\sum_{i=1}^{n}4i=\frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}$$1 The proof is essentially OK. Minor quibbles: you don't need the \{\emptyset\}, in the definition of \mathcal{A}_\mathcal{B}, because we get it for free using \mathcal{C} = \emptyset which has as its union \emptyset as well. I suppose you already know that \mathcal{A}_\mathcal{B} is a topology, either by the definition of a base or a theorem, and ... 1 HINT: Show that the elements (\sigma^i)^k(1) are distinct for k=1,\ldots,n; it follows that the permutations (\sigma^i)^k must be distinct for k=1,\ldots,n and hence that \sigma^i is an n-cycle. To do this, suppose that (\sigma^i)^k(1)=(\sigma^i)^\ell(1) for some k,\ell such that 1\le k<\ell\le n, and let d=\ell-k. Show that ... 1 Hint on how to compute this integral: Write x (I assume x>0) as [x]+\{x\} and now$$\int_0^x=\int_0^1+\dots+\int_{[x]-1}^{[x]}+\int_{[x]}^{x}$$Observe that on each interval [k,k+1), k\in\mathbb{Z}, you have [t]^2=k^2. 1 The notation is already defined in your "theorem". What you start with would thus be unmangling the definitions used in the theorem. It says The function \ln : \mathbb R_{>0} \to \mathbb R is surjective (=onto). (note \mathbb R_{\ge 0} is not the domain of \ln since \ln 0 is undefined) Thus you need to find two definitions: \ln is ... 1 So, you need to show f(x)=\ln(x) for (x>0) is onto. That means, you need to show that for every y\in \mathbb{R}, you can find an x\in \mathbb{R}_{>0} such that f(x)=\ln x=y. This tells me x=e^{\ln x}=e^{y}. Here's how I'd write the proof. Fix y\in\mathbb{R}. Let x=e^{y}. Since e^{y}>0 for all y\in\mathbb{R}, it follows that ... 1 The first sentence will be a definition of "surjective" in this particular context. For example: We need to show that for every y \in \mathbb{R}, there is x \in \mathbb{R}^{>0} such that \ln(x) = y. 1 The (\Rightarrow:) part is almost complete, except where you are not sure. We pick x \in X and y \in Y, and we know that there is a path between them. Let the vertices in this path be x, a_1, a_2, \dotsc, a_n, y. The first vertex is in X, and the last vertex is in Y; all vertices are either in X or Y and never in both. Consequently, there ... 1 Perhaps you would like to consider the following: the inequality$$\frac{2n^2}{n^3 + 3}< \frac{2n^2}{n^3}=\frac{2}{n}. In this case, if $N$ is an integer larger than or equal to $\frac{2}{\epsilon}$, the sequence would converge to $0$.

1

Your argument is correct, but it could be presented more clearly. Suppose that $h(x)=h(y)$. Then $x^2=y^2$ and $g(x)=g(y)$, so in particular $x=y$ or $x=-y$. If $x=y$, we’re done, so suppose that $x=-y$ and $x\ne y$. Suppose without loss of generality that $x\le y$. Then $x\le 0\le y$, so $g(x)=0\le g(y)$, and we must have $g(x)=0=g(y)$ and hence ...

1

You proof is basically correct. For the sake of proving that $h$ is injective, you don't need to say anything about why $f$ and $g$ are not. I would write "let $x,y\in \mathbb{R}$ such that $h(x)=h(y)$". Writing $(x,y)\in\mathbb{R}^2$ causes confusion. if $x=-y$ then $0=g(x)=g(y)=y$ or $x=g(x)=g(y)=1$ thus if $x=0$ then $x=y=0$ if $y=0$ then ...

1

I disagree with Hagen that you have the right idea. It is a serious logical error you made concerning induction, and I encourage you to read and fully grasp http://matheducators.stackexchange.com/a/10034/1550 before attempting induction problems. After you read that then read my answer. I give you any tree $T$ and graph $G$ such that every vertex in $G$ has ...

1

You have the right idea, but you should not modify $G$, esp. you cannot expect to obtain a given degree-$e$ graph from an unspecified degree-$(e-1)$ graph. Instead try the following: Let $T=(V_T,E_T)$ be a tree with $|E_T|=e$ edges (and hence $|V_T|=e+1$ vertices) and let $G=(V_G,E_G)$ be a non-empty, simple graph be given such that all vertices of $G$ ...

1

Archaick has already given you a simple exposition of your correct approach. I would like to point out an interesting fact, which is that it is actually crucial to use the Eulerian property of $G$, which proceeds via the theorem that every finite graph with even vertex degrees has an Eulerian loop, and hence any two vertices are in some cycle. The reason is ...

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