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Consider $$f(x)=\frac{1}{\log(x+3)}$$

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The answer is that: If the imposed boundary condition leads to a well-posed weak formulation (solution exists and is unique), then no matter what this boundary condition is, we can not get a weak solution $u\in H^2_0$ in general. We simply can not find such non-trivial function satisfying: \left\{ \begin{aligned} -\Delta u &= f \quad \text{ in } U, \\ ... 1 Just note that \left(1+\frac{\sqrt2}{\sqrt {n-1}}\right)^n\ge1+n\frac{\sqrt 2}{\sqrt n} +\frac{n(n-1)}2\frac 2{n-1}>n for n>1, so1<n^{\frac1n}<1+\frac{\sqrt2}{\sqrt{n-1}}\to 1$$6 You cannot swap the n around, that is, inside and outside the limit. The step$$\left (\lim_{n \to \infty} n^{\frac{1}{n}} \right)^n=\lim_{n \to \infty} \left ( \left (n^{\frac{1}{n}} \right)^n \right )$$is wrong. The n outside the LHS is fixed, while the n inside, in the RHS, varies like the other n. The result would be correct, if you said, say, ... 2 Why would \lim a_n = L  imply \lim a_n^n = \lim L^n? 0 I guess that by "true" constraints you are meaning "active" or "blocking" constraints, that is constraints such that f(x^*,y^i)=0. Because, by convexity, if a constraint does not block -- i.e., f(x^*,y^i)<0 -- then removing it from the set of constraints does not change the optimum. To count the blocking constraints is quite easy a posteriori, that ... 0 I created a function based on root. May f_{money} be a function that depicts amount of money an agent posses, so that f_{money}(u) \in \mathbb{N}, u \in \mathbb{A} where \mathbb{A} is the amount of all agents. Furthermore money in the system m = \sum_{i=0}^{n-1}f_{money}(u_i) = 2500 Number of agents n = 5 mean (arithmetic mean) z = ... 1 By definition of m you have an<m. Note that -k<m-1 because -k<an<m and m-1<m<k. Assume an<m-1, then m is not the minimal number of the set \{j:-k<j\leq k,\; an<j\}. Contradiction, so m-1\leq an. 0 S_k := \{j\in Z:−k<j≤k \text{ and }an<j\} is finite clearly and it is non-empty because an \lt k. Then you can find the smallest integer m \in S_k such that an \lt m. Once you have this m, you want to show that an < m < bn. To do this, he says m = (m-1)+1 \le an + 1 because by definition m-1 \le an. And an + 1 \lt an + (bn ... 1 The set is finite because there are at most 2k elements in it. It is non-empty because by assumption k is in it. By the least upper bound property for \mathbb Z, since this set is bounded below, it admits a minimum. The minimum m is in that set, but by minimality, m-1 is not in that set, because otherwise m-1 would be an element of the set ... 0 This is a definition. See bottom of pg. 100 of Spivak's book. If \omega is a k-form on [0,1]^k, there is a unique f such that \omega = f dx^1 \wedge \cdots \wedge dx^k. Then define$$ \int_{[0,1]^k} \omega := \int_{[0,1]^k} f $$or$$ \int_{[0,1]^k} f dx^1 \wedge \cdots \wedge dx^k= \int_{[0,1]^k} f(x^1,\cdots,x^k)dx^1\cdots dx^k $$0 Let B = \{x\in [a,b]: |f(x)|=0\}. By continuity, this set is closed. Since f(0) = 0, the set is non-empty. We show that it is also open. Let x\in B, i.e. f(x) =0. By continuity, there exists 0<r<(2A)^{-1}, such that |f(y)|<\frac 12 for all y\in B_r(x). Fix such a y\in B_r(x). We want to show that f(y) =0. By the mean value ... 1 The differential equation f'(x) = A(x) f(x)^\beta can be solved explicitly, by separation of variables. The calculation shows that f(a)=0 is necessary, or we could write down nowhere zero solutions, by solving the equation. This bears further analysis, because superficially the problem looks like a Lipschitz condition with |f(x) - f(y)| \leq ... 1 A definition of relative homology is given for instance on page 115 of Hatcher's book on algebraic topology, which is freely available on his website. The q-th critical group seems to be defined in your text itself as H_q(\Phi^c\cap U,\Phi^c\cap U\backslash\{u\}:\mathbb{F}). 1 The Euler-Maclaurin Sum Formula gives the asymptotic approximation:$$ \begin{align} \sum_{n=1}^k\log(n)^m &\sim k\left(\log(k)^m-m\log(k)^{m-1}+m(m-1)\log(k)^{m-2}-\dots+(-1)^mm!\right)\\ &+\frac12\log(k)^m+C+\frac{m}{12k}\log(k)^{m-1}+O\left(\frac{\log(k)^{m-1}}{k^3}\right) \end{align} $$The constant C depends on m and needs to be determined ... 0 Let \{a_n\} a sequence. For each n define$$s_n=\sup\left\{a_n,a_{n+1},\ldots,\right\}$$so$$\limsup a_n = \inf\left\{s_n:n\in \Bbb N\right\}=\lim_{n\to\infty} s_n\tag{1}because the sequence \{s_n\} is decreasing. There are two cases: There exist an n such that s_n=\infty. This means that the set \left\{a_n,a_{n+1},\ldots,\right\} is ... 1 First note that f_- = 0, as if f_-(x^*) > 0, then for each x \ge x^*, we have f(x) \le f(x^*) < 0, which implies that f isn't integrable. Now choose x^* such that for x \ge x^* we have f(x) \le 1. Then \begin{align*} \int_1^\infty f(x)^p\, dx &= \int_1^{x^*} f(x)^p\, dx + \int_{x^*}^\infty f(x)^p\, dx\\ &\le ... -1 If your function is monotone and decreasing on [1,\infty[ then your function must be positive (if it's negative at any moment, it will not be integrable). So I don't know (yet) the way to proceed, but I'm not sure you're on the right direction. I hope this could help. 1 Zero is an eigenvalue. Let x(t)=1 to see this. Whether or not zero is a natural number depends on your definition of \mathbb N which is not standardized. Either convention can be used. 0 NO. Let x=t+d, y=t-d, then we shall consider f(t+d)+f(t-d)-2f(t)=O(d^2). Note that\lim_{d\to 0}\frac{f(t+d)+f(t-d)-2f(t)}{d^2}=f''(t),$$so basically we need f'' is bounded in (0,1), by calculation you will find this is not the case. 2 One way to think about problems of such type is in terms of linear interpolation. Define your function f(n)=a_n>0 let f to be linear between [n,n+1]. Then, f\in L_1 is equivalent to \sum_{n=1}a_n<\infty whereas f'\in L_1 is equivalent to \sum_{n=1}|a_{n+1}-a_n|<\infty. Now it is clear that the standard example of series mentioned by ... 3 Here's a construction of a counterexample. The idea is based on a simple construction for series that is given on p. 350 of the centenary edition of G. H. Hardy's A Course of Pure Mathematics. The idea in Hardy's example is to construct a sequence \{a_n\} of positive numbers such that \sum a_n < \infty but na_n \not\to 0 as n\to\infty; to do this ... 0 Hint: \displaystyle\sin(x)\approx x-\frac{x^3}{6} so$$f(x)\approx x^2\left(\frac{1}{x^2}-\frac{1}{6x^6}\right)=1-\frac{1}{6x^4}$$(this is assuming x is meant to be large) 0 I'm not exactly sure what the actual question is, but I'll try to give an answer nevertheless. First, a matter of notation. Usually, an integral of f over some set M (no matter in how many dimensions) is written simply as \int_M f\,d\mu, where \mu indicates the measure that is to be used. For pure riemann integrals, just writing \int_M f suffices, ... 0 To estimate xf(x) we consider$$\int^{a}_{0} f(x)dx=xf(x)|^{a}_{0}-\int^{a}_{0}xf'(x)$$We know the first term is finite since f\in L^{1}, so if \lim_{a\rightarrow \infty}af(a)=0 we must require \int^{\infty}_{0}xf'(x) be finite as well. To be more precise we need$$\int^{\infty}_{0}f(x)dx=-\int^{\infty}_{0}xf'(x)$$because the middle term goes to ... 1 why convergence in probability doesn't imply convergence almost surely? For the definitions of them are really the same thing. No, they are not. For a counterexample, assume (X_n)_{n\geqslant1} are independent Bernoulli with P[X_n=1]=1/n and P[X_n=0]=1-1/n. Then X_n\to0 in probability (why?) while the event [X_n\to0] has probability zero ... 0 You have a collection (T_i)_{i\in I} of bounded operators, and (e_j)_j a collection of directions. Given a direction, the "orbit" \{T_ie_j,i\in I\} is bounded. This means that there is a bounded set S_j such that applications of the bounded operators to e_j cannot make you leave this set. If the space is "without holes", this was because the ... 1 The strategy of the proof is to convert the statement about smooth maps between manifolds into a statement about linear maps between vector spaces. At this early point in the book, the authors have not yet presented a way to define d(\phi^{-1}), because \phi^{-1}: V \rightarrow U has a domain V that is an open subset of a manifold, whereas they have ... 1 The idea is that as long as you consider a sufficiently large compact set you can apply Weierstrass Theorem. the fact that the function goes to 0 with |x| \to \infty implies that for each \varepsilon > 0 there exists M > 0 such that |f(x)| < \varepsilon for all x such that |x| > M. If you now take, for example, \varepsilon = ... 2 HINT: One direction is very easy. For the other, assume that the sets E[f\le c] and E[f\ge c] for c\in\Bbb R are closed. Then the sets \{x\in F:f(x)>c\} and \{x\in F:f(x)<c\} are relatively open in the subspace topology on F for each c\in\Bbb R. This implies that if a,b\in\Bbb R with a<b, then$$\{x\in F:a<f(x)<b\}=\{x\in ...

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Set $y = -x$. Thus, we have: $x^x = (-y)^{-y}$ when $y \ge 0$ Or equivalently, $$x^x = \frac{1}{(-y)^{y}}$$ Which can be simplified to: $$x^x = \frac{1}{(-1)^{y} y^y}$$ Therefore, the domain of $x^x$ consists of both reals and complex numbers depending on the value of $(-1)^y$ or to be more precise depending on the value of $y$.

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$x^x$ is well defined as a real function for $$(0,\infty) \cup \{ -\frac{m}{2n+1}| m, n \in {\mathbb N} \}$$

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For negative values of $x$, when $x$ is not an integer, you run into surly problems involving complex numbers. These entail a study of the complex log function and its branches. Clearly, $x\mapsto x^x$ makes sense for positive $x$. It also makes sense for negative integer values of $x$. It is not defined at $0$.

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Let $u: \overline {U}\to \mathbb{R}$ be a harmonic function, i.e. $$\tag{1}\Delta u=0\ in\ U$$ Mutliply $(1)$ by $h\in H_0^2(U)$ in both sides and then integrate: $$\tag{2}\int_U\Delta u\cdot h=0,\ \forall\ h\in H_0^2(U)$$ Use the generalized Green identity to conclude from $(2)$ that $$\tag{3}0=\int_U\Delta u\cdot h=-\int_U\nabla u\nabla h=\int_Uu\Delta ... 0 This is a nice application of the slogan "convex functions lie above their tangents and below their secants", or rather its mirror image "concave functions lie below their tangents and above their secants". Specifically this slogan says that for any x\in[a,b] we have \frac{f(a)(b-x) + f(b)(x-a)}{b-a}\leq f(x) \leq f\left(\frac{a+b}{2}\right) + ... 3 This result may help you: Let \mathbb{F}:(a, b)\rightarrow \mathbb{R} that is continuous on the bounded open interval (a, b) then the two limits given by F(a +) = \lim_{x\to a^{+}} F(x), F(b -) = \lim_{x\to b^{-}} F(x) exists iff F is uniformly continuous on (a, b). This result has been given in the book "The calculus integral by Brian S. ... 3 Sure, take a function which is continuous on a finite closed interval, and remove the endpoints. And there are many functions defined on the whole real line that are uniformly continuous. 1 You already know that function is concave down (f'' \leq 0), therefore area of trapezoid is less than the integral.$$\frac{(b-a)}{2}(f(a) + f(b)) \leq \int_a^b f$$Now the second part: You can write the integral as:$$\int _{ a }^{ b }{ f } =\int _{ a }^{ (a+b)/2 }{ f } +\int _{ (a+b)/2 }^{ b }{ f }$$The function is increasing (f'\ge0), therefore ... 3 Note that \frac{1}{\sin x} blows up as x approaches 0 from the right. So the limit does not exist. Even if we allow infinite limits, the limit does not exist, for our function approaches 1 as x approaches 0 from the left. Remark: If the question is changed to \lim_{x\to 0^-} (limit as x approaches 0 from the left), then the limit does ... 3 This is called the Ahmed integral.$$ \int_{0}^{1} \frac{\tan^{-1}\sqrt{x^2 + 2}}{(x^2 + 1)\sqrt{x^2 + 2}} \, dx = \frac{5\pi^2}{96} . $$You can see a solution here. 4 One can define a function f:[a,b]\to\Bbb R to be regulated if there exists a sequence \{s_n\} of step functions such that s_n\to f uniformly. Recall that if s is a step function with constants c_i and intervals of partition [x_{i-1},x_i] for say i=1,\dots, r we define its integral as over [a,b] as$$\int_a^b s=\sum_{i=1}^r c_i\Delta x_i$$... 2 I think you are trying to ask for the distinction between defining the Riemann integral as a limit of supremum and infimum partitions vs just using a fixed family of partitions, in this case something akin to the dyadic intervals as Osgood does. The point is that both methods yield precisely the same result when the function f is continuous on the whole ... 0 I think so. Assume that f is nonnegative and smooth in some open set S. Now test against \Delta h where \Delta h=1 in an open set \Omega\subset S with homogeneous Dirichlet boundary conditions. Is this ok? Now, one can use the density of C_c^\infty in L^2 and some approximation argument, isn't it? 0 I think that you can't solve this equation in H^2_0. If a function u is in H^2_0 then you have u=0 and \nabla u=0 at the boundary and these are too much boundary conditions for this equation. Let me think how one can prove the nonexistence. 0 Hint: |c_n|\le |a_n| for all n, hence \limsup\sqrt[n]{|c_n|}\le \limsup\sqrt[n]{|a_n|} 0 You can note that the function 10^x is increasing less rapidly than e^{x^2} so that the limit is absolutely \infty 1 The original problem could be handled more directly, but to pick up after what you have done so far: If you have shown that \lim_{x\to\infty}\frac{\ln(f(x))}{\ln(g(x))}=\infty, then you have shown that for any M, there is an x_M such that for any x>x_M, that$$\ln(f(x))>M\cdot\ln(g(x))$$This implies$$f(x)>g(x)^M$$and that ... 1 Set k=ln6, then 10^x=e^{kx}. The limit you want is \lim\limits_{x \to \infty}e^{x^2-kx}. With this your limits can be easily calculated. 3 You can take the natural logarithm of the fraction and use your idea.$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}}=e^{ \lim_{x\to +\infty} \ln\dfrac{e^{x^2}}{10^{x}}}$$A much simpler solution is to observe that 10 <e^3 and hence$$ \dfrac{e^{x^2}}{10^{x}}> \dfrac{e^{x^2}}{e^{3x}} Added: To clarify why I can interchange the exponential and ...

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Therefore, the series $\sum\limits_{n=0}^{\infty}\dfrac{1}{1+a^{n}}x^{n}$ is convergent for $|x| < a$ if $a>1$ and for $|x|<1$ in the case $a\leqslant{1}.$ Can you check convergence for $|x|=a, \;\; a>1$ and for $|x|=1 \text{ if } a\leqslant{1}?$

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