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10

Let \begin{align} I(m) &= \int_0^\infty \frac{dx}{(1+x^m)(1+x^2)}\tag{x = t^{-1}}\\ &= \int_0^\infty \frac{t^{-2}\,dt}{(1+t^{-m})(1+t^{-2})}\\ &= \int_0^\infty \frac{dt}{(1+t^{-m})(1+t^2)}\\ &= I(-m). \end{align} But \begin{align} I(m) + I(-m) &= \int_0^\infty \left(\frac{1}{1+x^m} + \frac{1}{1+x^{-m}}\right)\frac{dx}{1+x^2}\\ ... 5 By the inequality of Power Means, it is sufficient to prove this for p = \frac13. Also WLOG we can assume a > b > 0 \implies x = \frac{a}b > 1. So the inequality we are left to show is, for x > 1:\frac{x-1}{\log x} < \left(\frac{x^{1/3}+1}2 \right)^3$$Simplifying using x = t^3, this is equivalent to showing for t > 1 ... 4 Under the assumption that \epsilon is constant across a, then no additional assumptions are needed. Suppose that x_a is the fixed point of f_a, and choose e>0. Then there is a \delta such that for all b within \delta of b, you have e\epsilon\geq \vert f_b(x_a)-f_a(x_a)\vert=\vert f_b(x_a)-x_a\vert. Now, repeatedly apply f_b to ... 4 If it were the exact same argument, you'd get the volume again. :) Seriously, though: "Each segment is approximately a cylinder" is the critical statement: not only is it approximately a cylinder, but the difference in volume between a true cylinder of that size and the approximating one goes down linearly as \delta gets smaller. Yeah, your calculus book ... 4 The roots of the characteristic equation are not real. They are purely imaginary since the characteristic equation is$$ \lambda^2+9=0, $$and hence the general solution of y''+9y=0 is$$ y=c_1\cos 3x+c_2 \sin 3x. $$Incorporating the initial conditions we obtain that the solution of the IVP is$$ y(x)=\sin 3x. $$3 Yes. Since K is compact, K\times K is also compact. Since g is continuous, g is uniformly continuous. Let x_0\in K. Given \epsilon>0 there is a \delta>0 such that$$|x-x_0|<\delta\implies |g(x_0,y)-g(x,y)|<\epsilon\quad\forall y\in K.$$Then$$ g(x_0,y)\le g(x,y)+\epsilon\le f(x)+\epsilon\quad\forall y\in K. $$Taking the supreme ... 3 Question 1$$ \lim_{h\rightarrow0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2} \\ $$when h = 0 is substituted numerator and denominator reduce to 0. So, applying L'Hopital's rule (differentiate wrt h)$$ \lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h} \\ \lim_{h\rightarrow0}\frac{f'(a+h)-f'(a)-f'(a-h)+f'(a)}{2h} \\ ...

3

If you want to write this as a series, write this as a series, that is, as $\sum\limits_na_nz^n$ where the sequence $(a_n)$ is independent of $z$. Here $a_n=1$ when $n$ is in $K=\{k!\,;\,k\geqslant0\}$ and $a_n=0$ otherwise, hence $|a_n|^{1/n}$ is always $0$ or $1$ and is $1$ infinitely often (since $K$ is infinite). In particular, ...

3

This is an inhomogeneous Euler equation. At first we need to carry out the transformation $$y(x)=z(\log x),$$ which turns the equation into an equation of constant coefficients. In particular, $$y'(x)=z'(\log x)\frac{1}{x}, \quad y''(x)=z''(\log x)\frac{1}{x^2}-z'(\log x)\frac{1}{x^2}$$ and hence $$xy'=z', \quad x^2y''=z''-z'.$$ So $$... 3 In a double integral, you are actually integrating a differential two-form:$$\int_R \mathrm{f}(x,y) \ \mathrm{d}x \wedge \mathrm{d}y$$Here, \mathrm{d}x and \mathrm{d}y are the basis differential one-forms and \mathrm{d}x \wedge \mathrm{d}y is their exterior product. 2 In plain english, this means that \alpha is the limes superior of the sequence a_n exactly if for every \epsilon > 0 infinitely many of the a_n are larger than \alpha - \epsilon. but only finitely many of the a_n are larger than \alpha + \epsilon In other words, for every non-empty interval (\alpha-\epsilon,\alpha+\epsilon) around ... 2 If f is differentiable, we can write f(x+{1 \over n}) = f(x) + {1 \over n} f'(x) + r({1 \over n}), where \lim_n {r({1 \over n}) \over {1 \over n}} = 0. In particular, for any \epsilon>0 we can find some N such that if n \ge N then -\epsilon < {r({1 \over n}) \over {1 \over n}} < \epsilon, and so  {1 \over n} (f'(x) - \epsilon) < ... 2 Let us assume that \lvert f(z)\rvert\le 1, for all \lvert z\rvert=1, and show that in such case f has to be equal to z^n. Cauchy Integral formula implies that$$ n!=f^{(n)}(0)=\frac{n!}{2\pi i}\int_{|z|=1}\frac{f(z)\,dz}{z^{n+1}}=\frac{n!}{2\pi }\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt, $$and thus,$$ 1=\frac{1}{2\pi }\int_0^{2\pi} ...

2

I would do it using approximations. First suppose that $g \in C_c(\mathbb R^n)$. Construct a sequence of measures $\mu_k$ that are absolutely continuous w.r.t. Lebesgue measure (or alternatively are finite linear combinations of dirac delta functions) such that $\mu_k$ converges weakly to $\mu$. Then for each $x \in \mathbb R^n$, we have $\mu_k*g(x) \to ... 2 Consider the function$e:\mathbb{R}\cup\{0\} \to\mathbb{R}$;$e(s)=\dfrac{x^s-y^s}{s}$, when$s\neq 0$, and$e(s)=\ln(x)-\ln(y)$, when$s=0$(where,$x,y>0$). Note that$e$is continuous on$\mathbb{R}\cup\{0\}$. Further,$e(s)=\int_y^x v^{s-1}\,dv$, Since, for arbitrary$a,b,s,t\in \mathbb{R}$,$a^2e(s)+2abe(\frac{s+t}{2})+b^2e(t)=\int_y^x ...

2

Here is my proof. I feel like I made a huge leap at the end. I was not sure how to embed my LaTex code, it would not work. So I took screenshots. The last two lines, I have a gut feeling that I am missing a key step that links the two.

2

We can view $\bar{x}_a$ as a minimizer of the continuous function $x\mapsto \|f_a(x)-f_a(f_a(x))\|$. If $f$ is jointly continuous as a function of $\mathbb{R}^n\times\mathbb{R}$, then the argmin correspondence that maps $a$ to the set of minimizers of this functions is upper hemi-continuous by Berge's maximum theorem (one has to show that locally all ...

2

You can use the contraction mapping estimates directly. You have the estimate $\|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} \|f_a(x_0) - x_0\|$, so we can see that if we let $B = \sup_{a \in B(\hat{a},1)} \|f_a(x_0) - x_0\|$, then $\|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} B$ for all $a \in B(\hat{a},1)$. So, ...

2

Chain Rule implies that the composition of $C^1$ functions is $C^1$. Note this applies to multivariable functions too. Consider $M(u,v)=uv$, the multiplication function, and $S(u,v)=u-v$, the subtraction function. These are $C^1$ smooth (they are polynomials after all). So, the composition of $x\mapsto (f(x),g(x))$ with $(u,v)\mapsto uv$ is also $C^1$; and ...

2

No, in general you cannot expect this. Simply consider $(X,\|\cdot\|)=(\mathbb{R},|\cdot|)$, and set $x=0$. Then the inequality reads $$|y|^3 \leq \frac{1}{2} |y|.$$ Obviously, this is in general not true for $y>1/\sqrt{2}$. Yes. The equality \begin{align*} \|x\|^2 x - \|y\|^2 y &= (x-y)\|x\|^2 + y (\|x\|^2-\|y\|^2) \\ &= (x-y)\|x\|^2 + y ... 2 I think you misunderstand the basic concepts of the subject. The expression \nabla \chi_{E_c} is not "a function v:\partial E_c\to\mathbb R^2", but a distribution. If your sublevel set E_c happens to be a Caccioppoli set (which is not guaranteed; the sublevel set of a Lipschitz function can be any closed set) then \nabla \chi_{E_c} is a ... 2 You can do it geometrically. Take a point in your set, and suppose it is in the first quadrant. Drop lines parallel to the axes to the line x+y=1, and mark half the distance to the intersection. Then the circle with that radius is entirely contained in the square, because the diagonal is strictly longer that the radius. 2 There are several ways to show it: Let the function f: \mathbb R^2\to\mathbb R, with f(x_1,x_2)=\lvert x_1\rvert+\lvert x_2\rvert. $$Clearly f is continuous and as (-\infty,1) is open in \mathbb R, so is its inverse image:$$ f^{-1}(-\infty,1)=\big\{(x_1,x_2)\in\mathbb R^2: \lvert x_1\rvert+\lvert x_2\rvert<1 \big\}. $$2 The \rho_{\alpha,\beta} are norms on \mathcal{S}(\mathbb{R}^n). That they are seminorms is straightforward to verify, and \rho_{\alpha,\beta}(f) = 0 only for f \equiv 0 follows since x^\beta is only zero on a nowhere-dense subset (the union of finitely many coordinate hyperplanes \{x_i = 0\}), so \rho_{\alpha,\beta}(f) = 0 \Rightarrow ... 2 This is a question of topology, in which these are taken to be axioms defining open and closed sets. To see why they are reasonable, take a few examples: A_n = (-\frac{1}{n},\frac{1}{n}) is an open set for every n. What happens if we take an infinite intersection?$$\bigcap_{n=1}^\infty A_n = \{0\}$$. If we allow infinite intersections of open sets to ... 1 If v\ne 0, then$$ \frac{\partial_t v}{v\lvert v\rvert^p}=-it^q $$or$$ \frac{1}{p}\partial_t\left(|v|^{-p}\right)=i\frac{1}{q+1}\partial_t\big(t^{q+1}\big) $$or$$ \frac{1}{p}|v|^{-p}=i\frac{1}{q+1}t^{q+1}+c, $$for some c constant. Note that we have used that (|x|^k)'=k|x|^{k-2}x. 1 Let a_n = f(\tfrac{1}{a^n}). Then your equation tells you a_{n+1} = \frac{a_n}{b}. Because a_0 = f(1) you get a_n = \tfrac{f(1)}{b^n}. Since b>1 you get a_n \overset{n → ∞}{\longrightarrow} 0. Okay, one has to show that the limit exists, so here is a full argument in the same spirit: Let x ∈ [0..1]. Arguing just as above, you also get ... 1$$ f(a+h)-f(a)=f'(a)h+\frac{1}{2}f''(a)h^2+o(h^2) $$and$$ f'(a+h\theta) = f'(a)+f''(a)h\theta + o(h\theta). $$In the last equation I just used the definition of f''(a). Hence$$ f'(a)h+\frac{1}{2}f''(a)h^2 + o(h^2)=f'(a)h+f''(a)\theta h^2 + o(h^2\theta). $$Dividing by h^2 we deduce$$ \frac{1}{2}f''(a) = f''(a) \theta +o(1). $$If f''(a) \neq 0, ... 1 If \liminf \frac{a_{k+1}}{a_k} = 0, the first inequality is clear. Thus let us suppose it is strictly positive. Let 0 < c < \liminf \frac{a_{k+1}}{a_k}. Then there is a K_c such that for all k \geqslant K_c we have \frac{a_{k+1}}{a_k} > c, and hence$$a_{K_c+n} = a_{K_c}\prod_{j=0}^{n-1} \frac{a_{K_c+j+1}}{a_{K_c+j}} > a_{K_c}\cdot ...

1

These inequalities are addressed in $\S$ 11.5.3 of these notes. (More precisely the middle inequality is clear, and the two outer inequalities are very similar, so the notes carefully prove one of them.) As Daniel Fischer has pointed out, the implication you ask about at the end of your question is false in general. What is true is \$\lim_{n \rightarrow ...

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