# Tag Info

0

The first integral is equal to, in the limit as $R \to \infty$, $$e^{i \pi} \int_{\infty}^0 \frac{dx}{e^{i \pi/2} \sqrt{x} (1+x^2)} + \int_0^{\infty} \frac{dx}{\sqrt{x} (1+x^2)} = (1-i) \int_0^{\infty} \frac{dx}{\sqrt{x} (1+x^2)}$$ which is equal to $i 2 \pi$ times the residue at the pole $z=e^{i \pi/2}$, so that $$\sqrt{2} e^{-i \pi/4} \int_0^{\infty} \... 0 Take the branch cut of \sqrt{x} to be the negative real axis. We can deform the contour from (0,\infty) to (-\infty,0) on either side of the branch cut, at the cost of adding in a residue. Notice that the integrand has opposite values on either side of the branch cut. If we deform (0,\infty) to (-\infty,0) requiring that arg(z)=\pi, we have$$...

2

The last one is a well known inequalitie, i think it's the triangle inequalitie

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I would propose the following to avoid the multiple valued problem in this case: substitute $$x=u^2\implies dx=2u\,du\implies\;\text{we get the integral}\;\;$$ $$\int_0^\infty\frac{2u\;du}{u(1+u^4)}=2\int_0^\infty\frac{du}{1+u^4}=\frac\pi{\sqrt2}$$ which is your first result (This is more or less well known result, which can also be obtained by "usual", ...

0

Using the dual basis, the matrix representation of $T^+$ is given by the transpose, $T^t$.

1

Put f equal to zero everywhere but on the curve $y=x^2$, where it is 1. Try to fill in the details.

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Note that the dot product commutes, so $$f(x)g'(x)^T= \langle f(x), g'(x)\rangle = \langle g'(x), f(x)\rangle =f(x)^Tg'(x)$$

1

Let $a \in \mathbb R^2$ arbitrary and $r > 0$. Let $x \in B_1(a,r)$, then we have $\Vert x - a \Vert_1 < r$. Set $r_2 := \frac{r - \Vert x - a \Vert_1}{2}$ and let $y \in B_1(x, r_2)$. Then we have $$\Vert y - a \Vert_1 \leq \Vert y - x \Vert_1 + \Vert x - a \Vert_1 \leq \frac r 2 + \frac{\Vert x - a \Vert_1}{2} < r.$$ So we showed that $B_1(x, r_2)... 1$\dfrac{1}{n!} \geq \dfrac{1}{n^n}$and this proves the second series diverges by comparison to the harmonic series. The first one is already asked here, and I just quoted its proof:$\sqrt[n]{n} - 1 \geq c\dfrac{\ln n}{n}, c > 0$and this means it diverges by comparison test as well since the right series diverges by again comparing it to the harmonic ... 1 No, it need not be constant. Consider the Banach space$\mathbb C^2$with$\|(z_1, z_2)\| = \max(|z_1|, |z_2|)$and$f: \mathbb C \to \mathbb C^2$defined by$f(z) = (z, 1)$. Then$\|f(z)\| = 1$for$|z| \le 1$. The open mapping theorem certainly does not hold if the codomain has dimension$> 1$: no open subset of such a space can be homeomorphic to a ... 1 Using Induction on$n$,$\sum_{i=1}^{n+1}x_i$.$\sum_{i=1}^{n+1}1/x_i$$\geq n^2+x_{n+1}$$(\frac{1}{x_1}+$$\frac{1}{x_2}+....+$$\frac{1}{x_n}$)$+$$\frac{1}{x_{n+1}}$$(x_1+x_2+......+x_n)=n^2+1+(\frac{x_{n+1}}{x_1}+\frac{x_1}{x_{n+1}})+....+ (\frac{x_{n+1}}{x_n}+\frac{x_n}{x_{n+1}})$$\geq n^2+1+2n=(n+1)^2 Note: The minimum value of each term of type (\... 0 Note first that we have 0\leq 1-\exp(-u)\leq u and (u+1)\exp(-u)-1\leq 0 for u\geq 0. Some hints: a) Write for t>0 G(t)=\frac{g(0)-g(t)}{t}=\int_{\mathbb{R}}h(x,t)dx. Show that$$|h(x,t)|\leq x^4\sin(\frac{1}{1+x²})f(x)\leq x^2f(x)$$b) Suppose that g has a derivative at 0. If v>0, show that \displaystyle \frac{1-\exp(-vt)}{t} is ... 1 Expand it out as \sum_{i,j} x_i/x_j, and use t + 1/t \ge 2 for t > 0 1 Assume known that (x+y)^2\ge4xy,\forall x,y\in\Bbb R. Then suppose this inequality holds for n. Now we have$$\begin{align}(\sum\limits_{i=1}^{n+1}x_i)\cdot(\sum\limits_{i=1}^{n+1}\frac{1}{x_i})&\ge(\sum\limits_{i=1}^{n})\cdot(\sum\limits_{i=1}^{n})+(\sum\limits_{i=1}^{n}x_i)\frac{1}{x_{n+1}}+x_{n+1}(\sum\limits_{i=1}^{n}\frac{1}{x_i})+1\\ &\ge n^...

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Using AM-GM $$\sum_{k=1}^n x_k\ge n\sqrt[n]{x_1\cdot...\cdot x_n}$$ and $$\sum_{k=1}^n \dfrac{1}{x_k} \geq n\sqrt[n]{\frac1{x_1}\cdot...\cdot \frac1{x_n}}$$ Then $$\displaystyle \sum_{k=1}^n x_k\cdot \displaystyle \sum_{k=1}^n \dfrac{1}{x_k} \geq n^2$$

1

Let $A = \displaystyle \sum_{i=1}^n x_i, B = \displaystyle \sum_{i=1}^n \dfrac{1}{x_i}\implies P_{n+1} = \left(A+x_{n+1}\right)\left(B+\dfrac{1}{x_{n+1}}\right)= P_n+\dfrac{A}{x_{n+1}}+Bx_{n+1}+1\geq n^2+2\sqrt{AB}+1\geq n^2+2n+1 = (n+1)^2$.

0

The integrand is nonnegative, so by Tonelli's theorem the value of the double integral is insensitive to the order (i.e. switching the order of integration is justified). Just consider the region of integration as a two-dimensional set: the set of points $(\tau, \eta)$ that comprises the domain of integration for $\int_0^\xi \int_\eta^\xi d\tau d\eta$ is $$... 1 Yes this is true. In fact we can even say more; this is actually an if and only if statement (i.e. any metrizable topological space is first countable). This follows because the topology induced by the metric is translation invariant, hence if a countable neighborhood basis exists at 0, then one must exist at every point of the space. See "Köthe, G.-- ... 0 You over complicate things. Simply take the left hand limit and the right hamd limit seperately. You will find the left hand limit to be -1 and the right hand limit to be 0. Therefore, there is jump discontinuity at x=0. 2$$x^{ log_{ 2 }x }+\frac { 16 }{ x^{ log_{ 2 }x } } =17\\ { x }^{ 2\log _{ 2 }{ x } }-17x^{ log_{ 2 }x }+16=0\\ \left( x^{ log_{ 2 }x }-16 \right) \left( x^{ log_{ 2 }x }-1 \right) =0\\ x^{ log_{ 2 }x }=16\Rightarrow \log _{ 2 }{ x^{ log_{ 2 }x } } =\log _{ 2 }{ 16 } \Rightarrow { \left( \log _{ 2 }{ x } \right) }^{ 2 }=4\Rightarrow \log _{ 2 }{ x } =\pm ...

2

let $$y=x^{log_2x}$$ your equation becomes, $$y+\frac{16}y=17$$ solve it, you get two solutions: 1 and 16. Now it becomes less horrible, $$x^{log_2x}=1 ~or~ 16$$ This leads to solution $x =1$, $2^2$ and $2^{-2}$

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Well ordering principle is a direct consequence of the Principle of Mathematical Induction and here we need the second version of Principle of Mathematical Induction stated below: Principle of Mathematical Induction Second Version: If $A$ is a subset of $\mathbb{N}$ such that $1 \in A$ If $1, 2, \ldots, n \in A$ then $(n + 1) \in A$ then $A = \mathbb{N}$...

1

You cannot ignore the exponent or the multiplication. In this case, set temporarily $t=\log_3x$ so the equation becomes $$t^2-3t+2=0$$ which has roots $t=1$ and $t=2$. Thus you get the two equations $$\log_3x=1$$ and $$\log_3x=2$$ Can you finish them? Your method would be sound, too, provided you did the decomposition right.

4

should be $$\left( \log _{ 3 }{ x } -1 \right) \left( \log _{ 3 }{ x } -2 \right) =0$$ $$\log _{ 3 }{ x } =1\Rightarrow \quad x=3\\ \log _{ 3 }{ x } =2\Rightarrow x=9$$

1

Okay, so to find extremum within the interior, you can set the gradient equal to $0$: $$\nabla f(x,y)=\left(3x^2+y, 3y^2+x\right)=0$$ As you noted, the two solutions to this equation are $(0,0)$ and $\left(-\frac{1}{3},-\frac{1}{3}\right)$. Now, for the extremum on the boundary, we can split the boundary up into four different sections. Right Edge: On ...

5

Your point about successive extensions of a basic number system is very well taken. We use the successive extensions $$\mathbb{N}\hookrightarrow\mathbb{Z}\hookrightarrow\mathbb{Q}\hookrightarrow\mathbb{R}$$ to enable easier solution of problems in algebra and geometry. The Greeks had to do everything in terms of proportions referring to natural numbers ...

0

Hints Notice the general identity $$\psi \nabla ^2 \phi=\nabla \cdot (\psi \nabla \phi)-\nabla \psi \cdot \nabla \phi \tag{1}$$ Integrate over the domain Use the divergence theorem Put $\psi=\phi$

2

Ideally you need to find points where these suprema are attained. This is possible for $f_2$ as $\exists x \in C[a,b]$ such that $\|x\| = 1$ $$x(a) = \text{sgn}(\alpha) \text{ and } x(b) = \text{sgn}(\beta)$$ where $\text{sgn}(z) = z/|z|$ (This follows from Urysohn's lemma, if you like, although one can just draw the graph of such a function on $[a,b]$). ...

1

A multilinear k-form on an $F$-vector space $V$ is just a function from $V^k$ to $F$ which is linear in each component. These can be symmetric, alternating, or neither. Here, a bilinear form and a two-form are the same thing. In $\Bbb(R)^n$, a bilinear form can be represented by a nxn matrix, based on how it treats the basis elements. A differential k-form ...

0

I came up with proof now: Let A be a non-empty subset of the set of natural numbers N, we let as assume that A does not contain a least element. Set B=N-A, meaning B is a subset of N. Let predicate P(n) be defined as follow: P(n): for all elements n of B, n in B implies that n+1 is in B. Let T be a truth set of P(n): Basis Step: P(1) is true as 1 in N ...

5

A rational number is a number of the form $\dfrac n m$ where $n$ and $m$ are integers. A question is: what does the decimal expansion of a rational number look like? Consider a concrete example: $\dfrac n m = \dfrac{55}{148}$. To find the decimal expansion, do long division: $$\begin{array}{ccccccccccccccc} & & & 0 & . & 3 & 7 &... 1 If b > 1, then b = 1+c where c > 0, so b^n =(1+c)^n \ge 1+nc \gt 1  by Bernoulli's inequality. If b^{n/m} = a, then b^n = a^m. Since b^n > 1, a^m > 1 so that a > 1. Then apply the definition b^x =\sup_{r \le x} b^r . 0 Don't need continuity but need concept of convergence. First realize that you actually don't have any frigging idea what b^x  actually means. It's a bit of a shock to me but... it's true. We have no definition that makes any sense. But that's okay. This is a perfect opportunity to define it. Now that we've read chapter 1 of Rudin we actually know ... 2 How exactly do you define b^w for irrational w? If we follow the first chapter of Rudin, then we might use the definition as presented in exercise 6 of chapter 1, which defines$$ b^w = \sup_{\substack{p \leq w, \\ p \in \mathbb{Q}}} b^p.$$Then you should show that b^w is an increasing function for rational w, and therefore by extension also for ... 1 The way to finish that argument is to notice that the function x\mapsto b^x for x>0 is an increasing function. You can prove this by elementary calculus (just compute derivative) or simply from the definition (recall that b^x:=e^{x\log b}). To finish your argument then observe that$$r<w\Rightarrow b^r<b^w.$$0 If you can prove that Df(p)=0 (which intuitively seems true to me, though I can't think of how to prove it), the result follows: Suppose the statement is false, so there exists x \in \mathbb R^n such that f^n(x) \not\to p. So there exists \epsilon > 0 such that f^n(x) is not eventually in the neighbourhood B(p,\epsilon). Then, since f^{t_n}(... 1 You appear in effect to be assuming that \succsim is the same as ordinary \ge. This need not be the case. In fact, the whole point of the argument is that the lower contour sets for any continuous preorder on X are closed, so that in this sense all continuous preorders on X behave like the familiar natural order \ge. Of course the steps of the ... 1 Since \{f_n\} converges, \{f_n\} is Cauchy. In probability theory, we have to deal with different types of convergence, and therefore - in order to avoid any confusion - it is always good to mention which kind of convergence you are talking about, e.g. "Since f_n converges in L^p(\mathbb{R}^n), \{f_n\} is an L^p-Cauchy sequence." Define a ... 0 In my opinion, you have essentially restated the axiom by making the same assumption in a more subtle fashion (assumptions about the naturals). Generally, axioms are axioms because they cannot be proven independently using the other axioms. For instance, if you are familiar with analysis, the existence of suprema in the reals is typically axiomatic, but the ... 4 As the other answers indicate, compactness and connectedness have little to do with one another. You may be thinking in terms of compact cubes, so you will have to think of more general sets. For example, [0,1]\cup[2,3] is compact but not connected, and is made up of compact intervals. 1 As a sort of repeat to what the others have said: in \mathbb{R}^n, compactness is equivalent to closed and bounded (Heine-Borel theorem). So, with that in mind, it should be easy to construct counter-examples. 3 No. For example, a set consisting of two points is not connected but is compact. 5 Finite sets are compact, and never connected unless they have one point (or none). The Cantor set is disconnected (totally disconnected even), or more simply: take two disjoint compact sets and take their union: this is still compact but always disconnected. Etc. So there is no relation. 2 Hint:Consider \{A_n^c\}_{n=1}^{\infty } is the open covering of V. 1 Hint: Consider the set \mathcal{U}:=\{U_n |\ n \in \Bbb{N}, U_n:=\Bbb{R}^d\setminus A_n\}. Each member of \mathcal{U} is open since A_n is closed for each n. Furthermore, since V\cap (\bigcap_\Bbb{N} A_n) = \emptyset, we have$$V\subset \Bbb{R}^d\setminus \bigcap_\Bbb{N} A_n= \bigcup_\Bbb{N} \Bbb{R}^d\setminus A_n = \bigcup_\Bbb{N} U_n.$$Then \... 0 Basically, what you want to show is that any decreasing sequence of nonvoid compact sets has a nonvoid intersection. Let \{F_n\} be such a sequence. Pick x_n\in F_n for all n\in\mathbb{N}. This is a sequence lying in a compact set. It has a subsequential limit x. You can now show x\in\cap_n F_n. 2 What you have done seems Ok. It is important that \alpha>0 to ensure the convergence of the integral. Concerning the closed form of the integral, you may perform a change of variable$$ u=\sqrt{x},\quad du=\frac1{2\sqrt{x}}dx, $$obtaining$$ I_{\alpha }=\int _0^{\infty }\left(\frac{e^{-\alpha x}}{\sqrt{x}}\right)\:dx=2\int _0^{\infty }e^{-\alpha u^2}\:...

3

By the intermediate value property of continuous functions, there are $0=x_0,x_1,\ldots,x_{n-1},x_n=1\in[0,1]$ such that: $$f(x_i) = \frac{i}{n}$$ for every $i\in[0,n]$, and we may assume WLOG $x_0<x_1<\ldots<x_n$. Moreover, there is some $\xi_i$ in the interval $(x_{i-1},x_i)$ such that: $$\frac{f(x_{i})-f(x_{i-1})}{x_{i}-x_{i-1}}=f'(\xi_i)$$ ...

1

As you suspect, the tool you want to use here is the Banach fixed point theorem. Consider the space $C[0, 1]$ of real valued continuous functions on $[0, 1]$ with the sup norm, which makes $C[0, 1]$ into a complete metric space with metric $d$. Define $$f\colon C[0, 1] \to C[0, 1] \text{ by } f(y(x)) = e^{x} + \frac{y(x^{2})}{2}$$ Then for any $y_{1}, y_{2} \... 6 Here is a rigorous and systematic way to analyze the existence and value of the limit. (miracle173's answer didn't prove existence but uses essentially the same method for finding the value if it exists.) Let$c_n = \frac{a_n}{b_n}$(for each$n \in \mathbb{N}$). Then$b_{n+1} c_{n+1} = b_n + 2 b_n c_n + 14$. And$b_{n+1} = 9 b_n + 2 b_n c_n + 70$. Thus$...

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