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Using your definition of boundary, "I am defining the boundary of the set as vectors in $R^d$ s.t: for all positive epsilon, the open ball around a point contained within the boundary contains points from $A$ and $A^c$", we will show that it contains its limit points. If $p$ is a limit point of the boundary then for any $\epsilon\gt 0$ the open ball of ...

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As mentioned "closed" is not the opposite of "open". Take for instance $[a,b) \subset \mathbb R$. Is it closed? Hint: In order to show that $\partial A$ is closed notice that the complement of $\partial A$ is $(\mathrm {int} A) \cup (\mathrm {int} \,M - A)$. Since we may write $M$ (space in question) as the disjoint union $$M = (\mathrm {int} A) \cup ... 1 here are some ideas i have. first we establish that$$\frac{dx}{dt} = -x + k + t^3 \tag 1$$has the solution x_k(t) =k+ ce^{-t}+ x_p(t) where x_p(t) is a particular cubic polynomial solution of (1). the solution exists for all t. now, argue that the solution x to$$\frac{dx}{dt}=-x+ \sin x+t^3 \tag 2$$lies between the solutions x_{-1}(t) and ... 0 Case I: v^2(1+\alpha)^2 - 4\alpha \geq 0, thus Z_1 is a real root. Its clear that 0 \leq Z_1, so you only need to show: Z_1 \leq 1 \iff v(1+\alpha) + \sqrt{v^2(1+\alpha)^2 - 4\alpha} \leq 2\iff \sqrt{v^2(1+\alpha)^2-4\alpha} \leq 2 - v(1+\alpha)\iff v^2(1+\alpha)^2 - 4\alpha \leq 4- 4v(1+\alpha) + v^2(1+\alpha)^2 \iff 4v(1+\alpha) \leq ... 1 Your function is a) not a function and b) is not continuous. f(0,- 1/2) = {0, -1/2} and f(0, 1/2) = {1/2, 0} have two values so it is not a function. It can be made a function by setting: f(x,y)=\begin{cases}\left(0,\frac{1}{2}+y\right) & \text{ if }(x,y)\in[-1,1]\times\left[-1,-\frac{1}{2}\right)\\(0,y) & \text{ if }(x,y)\in ... 2 According to your definition, f(0,-1/2) is being mapped to two distinct points : (0,0) and (0,-1/2). So it's not quite a function. 2 You have a function f(t) then you have to impose and study$$f'(t) > 0$$namely$$-te^{-t} + e^{-t} > 0$$id est$$e^{-t}(1 - t) > 0$$Considering that e^{-t} > 0 for every t real, then (1-t) > 0 if t < 1 so this means the function is increasing in the range$$t\in[-\infty,\ 1]$$2 On (0,1) the derivative is positive, hence the function is strictly increasing on that interval. 1 First of all: you can take c<0, although of course you can also take c>0. What you ask simply means that in a sink of a linear equation x'=Ax everything goes exponentially fast to the origin (it could go slower and still be a sink). In order to show this, it is sufficient to observe that each entry of e^{At} is bounded by p(t)e^{-dt} for ... 2 You can find a function u\in C^\infty\left(\Bbb R^N\right) and a subsequence u_{n_k} such that u_{n_k} converges uniformly to u on compact subsets of \Bbb R^N and all the derivatives D^\alpha u_{n_k} converge to D^\alpha u uniformly on compact subsets. Let E_n the closed ball of radius n. Since \{u_k\}_{k\in\Bbb N} is uniformly ... 1 Derivative at 0:$$ \lim_{h\to0}\frac{f(h)-f(0)}{h}= \lim_{h\to0}(1+2ih\mathrm{Im}(h))=1 $$So, yes, the function is differentiable at 0. It is differentiable nowhere else, because$$ f(z+h)=z+h+2i(z+h)^2\mathrm{Im}(z+h) $$and so$$ f(z+h)-f(z)= h+2i(2zh+h^2)\mathrm{Im}(z)+2i(z+h)^2\mathrm{Im}(h) $$Therefore$$ \frac{f(z+h)-f(z)}{h}= ...

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Look at a precise statement of the chain rule, such as Baby Rudin 5.5. There we read (with variable names changed) "Suppose $g$ is continuous on $[a,b]$, $g'(x)$ exists at some point $x\in [a,b]$, $h$ is defined on an interval $I$ which contains the range of $g$, and $h$ is differentiable at the point $g(x)$. If $f(t)=h(g(t))$ $(a\le t\le b)$ then $f$ is ...

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I am not sure, because of your notation, which is exactly the function that you mean, but as a rule of thumb: functions that depend on $\overline{z}=Re(z)-Im(z)$ are not derivable. As your function depends in a single place of on $Im(z)=\frac{z-\overline{z}}{2}$, your function have a dependence on $\overline{z}$ and therefore it is not derivable.

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The functions $f_n(x)$ defined to be 1/n for -1/n< x or x> 1/n, $f_n(x)=-x$$for -1/n< x< 0, f_n(x)= x for 0< x< 1/n converge to f(x)= 0 as n goes to infinity but not uniformly. 3 It depends on what you already "know". First, there's another condition you probably meant to add in. You probably want to assume that a_m > 0. Otherwise, you can have things like (-2)^{1/2}=\sqrt{2}i appear. If you assume that a_m>0 for all m, then a_m\to a>0 follows as well. If you know continuity of the exponential and natural log ... 5 Yes.$$ \left|\sqrt {a^2+x^2} - \sqrt {a^2+y^2}\right| =\frac{\lvert x^2-y^2\rvert}{\left|\sqrt {a^2+x^2} + \sqrt {a^2+y^2}\right|} = |\sqrt {x^2} - \sqrt {y^2}|\cdot \frac{|\sqrt {x^2} + \sqrt {y^2}|}{\left|\sqrt {a^2+x^2} + \sqrt {a^2+y^2}\right|} \leq |\sqrt {x^2} - \sqrt {y^2}| $$2 This is true. You can see this by assuming x>y without losing generality and then differentiating $$f(a)=\sqrt{a^2+x^2}-\sqrt{a^2+y^2}$$ with respect to a. Derivative is negative hence f is decreasing function of a and is maximized at 0. 0 Make y=(x-a)(b-x), then your function trasforms to exp(-\frac{1}{y}) and it's infinitely differentiable except points where y = 0, e.g. x = a or x=b. But a and b aren't included in interval (a,b) so your function is infinitely differentiable. 1 Take the log. \log(\frac{e^{\sqrt{k}}}{k^c}) =\sqrt{k}-c\log(k) \to \infty since \log(k) = O(k^{\epsilon}) for any \epsilon > 0. \log(\frac{e^{\sqrt{\log(k)}}}{k^c}) =\sqrt{\log(k)}-c\log(k) \to -\infty since \sqrt{x}-x \to -\infty for any x \to \infty . 0 See this answer for a proof that for f\in \mathscr R[a,b] and given any \epsilon>0, one can find g\in\mathscr C[a,b] such that$$\int |f-g|<\epsilon. $$Now note that$$|\int f^2|\le|\int (f^2-fg)|\le \int |f^2-fg|\le \sup |f|\int |f-g|\le \sup |f|\epsilon. $$By our arbitrary choice of \epsilon it follows that$$\int f^2=0. $$Take it from ... 1 Let \lambda \perp \rho, and let E \subset \mathbb{R} be such that \lambda(E)=0 and \rho(E^c)=0. Writing \lambda(E)=\int_E f \mathop{d\mu} and \rho(E^c)=\int_{E^c} g \mathop{d\mu}, what does this imply about the value of f on E (\mu-almost everywhere)? and of g on E^c (\mu-almost everywhere)? Can you show the converse? Let \lambda ... 1 Hint: Show that if f is not 0 at a continuity point, then f(x) > 0 or f(x) < 0 on some interval [c_1,c_2]. Now choose$$h(x) = \begin{cases} (x-c_1)(c_2-x), \,\, &x \in [c_1,c_2],\\ 0, \,\, &\mbox{otherwise}\end{cases}$$2 Separation of variables works on regions that are rectangular in some particular coordinate system. The underlying operator must be separable in that coordinate system. For the Laplacian, that generally means an orthogonal coordinate system such as spherical coordinates, cylindrical coordinates, elliptic coordinates, etc.. A rectangular region in spherical ... 1 Observe that$$ f'(z)=zf(z)\,\,\Rightarrow\,\,\mathrm{e}^{-z^2/2}\big(f'(z)-zf(z)\big)=0 \,\,\Rightarrow\,\,\big(\mathrm{e}^{-z^2/2}f(z)\big)'=0, $$and thus \mathrm{e}^{-z^2/2}f(z) is constant. In particular$$ \mathrm{e}^{-z^2/2}f(z)=\mathrm{e}^{-0^2/2}f(0)=1, $$and thus$$ f(z)=\mathrm{e}^{z^2/2}. $$2$$\sum_{n=1}^\infty\frac{(-1)^n}{n\sqrt{2}}$$this may be the series that you are looking for it has all irratiaonal 1 A subset in \;\Bbb R^n\; is closed iff it contains all its limit points, so suppose \;\{x_n\}\subset K:=\{x\in\Bbb R^n\;;\;f(x)=b\}\; converges:$$\lim_{n\to\infty}x_n=x_0\implies b=\lim_{n\to\infty}b=\lim_{n\to\infty}f(x_n)\stackrel{\text{cont.}}=f(x_0)\implies x_0\in K$$5 You should probably know that for all b \in \mathbb{R}^n, \{b\} is closed. Then we can write$$\{x \in \mathbb{R}^n : f(x) =b\} = f^{-1}(\{b\}).$$Since f is continuous, the conclusion follows. (The inverse image of a closed subset by a continuons fonction is closed) 0 Can you solve this ODE?: y'=xy, y(0)=1\,. 2 I want to expand upon Brevan Ellefsen's comment and my comments a little. I assume that dots in OP's case mean x derivative, since y(x) depends only on x. Then we can easily get rid of \alpha and \beta.$$ \frac{d^2 y}{dx^2}+\alpha \left(\frac{d y}{dx} \right)^2+\beta y=0 $$Replacing:$$ x \rightarrow \frac{x}{\sqrt{ \beta }}~~~~~y \rightarrow ... 1 The second curve is$y=1/x$and is not$sin(1/x)$. Make sure you type the parenthesis correctly. 0 OK, here is my idea, just a set up as you requested: Multiply by$-i$to get$z^2+(-2-2i)z+2-2i(\sqrt{3}-1)=0$Keeping the$z$terms on the left, we have$z^2+(-2-2i)z=-2+2i(\sqrt{3}-1)$Completing the square:$(z-1-i)^2=-2+2i(\sqrt{3}-1)+(-1-i)^2$When you work out left hand side and setting$z-1-i=t$we get the equation$t^2=-2+2i\sqrt{3}$Can you figure ... 5 One need not to know the whole dual space of$C[0,1]$to see the lack of reflexivity. Simply note that$C[0,1]^*$, opposed to$C[0,1]$, is non-separable because for each$t\in [0,1]$the map $$\langle f, \delta_t\rangle = f(t)\quad (f\in C[0,1])$$ is a norm-one linear functional on$C[0,1]$and$\|\delta_t - \delta_s\| =2$for distinct$s,t$. To see this, ... 1 Let$y=(y_1,...,y_n).$$\text { We have }\quad |y|^2-\|y\|^2=[\;\sum_{i=1}^n|y_i|\;]^2-\sum_{i=1}^n y_i^2=\sum_{1\leq i<j\leq n}2|y_i|.|y_j|\geq 0.$$ $$\text { We have }\quad 0\leq \sum_{1\leq i<j\leq n}(y_i-y_j)^2=(n-1)\sum_{i-1}^ny_i^2-2\sum_{1\leq i<j\leq n}y_iy_j=$$ $$=n\|y\|^2-\sum_{i=1}^ny_i^2-2\sum_{1\leq i<j\leq ... 2 Hint: for any nonnegative number a and b you have$$ (a+b)^2=a^2+b^2+2ab\ge a^2+b^2 $$and$$ 2(a^2+b^2)-(a+b)^2=a^2+b^2-2ab=(a-b)^2\ge0. $$Hence,$$ a+b\ge\sqrt{a^2+b^2}\quad\text{and}\quad \sqrt2\sqrt{a^2+b^2}\ge a+b. $$0 Generalized intervals at least partially answer your question. Basically, the idea is to allow intervals [a,b] to be a pair of numbers, without requiring a < b, and then to define adequate operations on them. See for example the webpage http://msse.gatech.edu/research/interval/WhyGeneralizedInterval.html and the references therein. 2 Problem Let f:(a,b)\to\mathbb R be a continuous function such that$$\limsup\limits_{n\to \infty}\frac{f(x_n)-f(x_0)}{|x_n-x_0|}\leq 0$$for every x_0\in (a,b) and every sequence x_n converging to x_0 such that x_n\not= x_0 for all n. Does this imply that f is a constant on (a,b)? Note the absolute values in the denominator. By ... 3 For x\gt0, repeatedly integrating from 0 to x gives$$ \cos(x)\le1\implies\sin(x)\le x\implies1-\cos(x)\le\frac{x^2}2\implies x-\sin(x)\le\frac{x^3}6 $$Noting that both sides are odd, we get$$ \left|x-\sin(x)\right|\le\frac{\left|x^3\right|}6 $$Since \frac{n}{n^2+k^2}\le\frac1n,$$ \begin{align} \sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\!\right) ... 2 In fact,F$is uniformly continuous in$\mathbb R$. To see this, let$\varepsilon>0$and set$g_n(x)=\min\{\lvert\, f(x)\rvert,n\}$. By virtue of the Lebesgue Dominated Convergence Theorem, we have that$\|g_n-\lvert\, f\rvert\|_{L^1}\to 0$. Fix$N>0$, such that$\|g_N-\lvert\, f\rvert\|_{L^1}<\varepsilon/2$. Then $$\int_{\mathbb R}g_N\,dx\le ... 1 As a consequence of the fact that \,\,\int_0^1 f(x)\,dx=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\big(\frac{k}{n}\big), we obtain$$ \sum_{k=1}^n\frac{n}{n^2+k^2}=\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^2}\to\int_0^1\frac{dx}{1+x^2}=\tan^{-1}1=\frac{\pi}{4} $$Next, Taylor expansion of \sin x, provides that 0\le x-\sin x\le ... 9 We can use Riemann sums to evaluate the limit. We have$$\sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\right)=\sum_{k=1}^n\,n\sin\left(\frac{1/n}{1+(k/n)^2}\right)\frac1n\to \int_0^1\frac{1}{1+x^2}\,dx=\pi/4$$since we have$$\left(\frac{1}{1+(k/n)^2}\right)-\frac{1}{6n^2}\left(\frac{1}{1+(k/n)^2}\right)^3\le n\sin\left(\frac{1/n}{1+(k/n)^2}\right)\le ... 0 The best book (in my opinion) to learn properly the$\delta-\varepsilon$definitions of continuity (and convergence of sequences etc) is Michael Spivak, Calculus. 1 Use Euler-Maclaurin to approximate the sum as $$\int_0^n \sin\left(\frac{n}{n^2+x^2}\right)dx\ .$$ Then change variables$x=ny$and use$\sin(\alpha/n)\sim\alpha/n$for$n\to\infty$, to obtain the result$\int_0^1 dy\frac{1}{1+y^2}=\pi/4$. 1 The second point is (one of) the definition(s) of the exponential function. The last one is obtained from the second by taking$z=-1$. 1 No it is not. The dual of$C[0,1]$is the space$\mathfrak{M}([0,1])$of complex (or signed) regular Borel measures on$[0,1]$. The dual of$\mathfrak{M}([0,1])$contains the set$\mathcal L^\infty([0,1])$of bounded Borel functions on$[0,1]$. For example,$\chi_{\{0\}}$, the function which vanishes everywhere, except at$x=0$, where it is equal to$1$, ... 2 Your functional assigns to every test function the value$0$. Since $$\mathrm{e}^{-x}x^2\varphi(x)=0,$$ if$x$is sufficiently large. 0 Noise is usually modeled as white noise, the sum of all audio frequencies. The example from wiki page modeled it as a sum of two frequencies. That will never fit. Of course fitting random data takes time! 1 Let us consider the family of operators $$T_\ell ((x_n)_n) = x_\ell \cdot e_1,$$ where$e_1 = (1, 0,0,\dots) \in \ell^2$is the first basis vector. This family satisfies the first property, since for$x = (x_n)_n \in \ell^2$with$\| x\| = 1$, we have $$\sum_{\ell=1}^\infty \|T_\ell x\|^2 = \sum_\ell |x_\ell|^2 = \|x\|^2 \leq 1.$$ But it is not too ... 2 Note that this is a differential equation. The equivalent Picard integral equation is $$f(x)=f(0)+\int_0^x g(f(t))\,dt$$ From here it is trivial to observe that if$f$is$C^n$or better, then the composition$g\circ f$is also at least$C^n$and thus the anti-derivative$C^{n+1}$. Which gives that$f$is also$C^{n+1}$and so on. 4 Suppose$f \in C^k$, then$f^{(k+1)} = f'^{(k)} = (g \circ f)^{(k)}$. Since$g \in C^\infty$and$f \in C^k$, we have that$g \circ f \in C^k$, and thus that$f^{(k+1)} = (g \circ f)^{(k)} \in C^0$, so it follows that$f \in C^{k+1}$. QED Or even more succinctly as John Ma pointed out below: Suppose$f \in C^k$, then since$g \in C^\infty$, we have that ... 4 Notation: For any function$h$let$h^{(0)}=h$, and let$h^{(n)}$be the$n$th derivative of$h$for$n\geq 1.$Suppose for some$n\geq 0,$there is a polynomial$ P_n$in$(2 n+2)$variables such that $$f^{(n+1)}=P_n(g^{(0)}(f),...,g^{(n)}(f),\;f^{(0))},...,f^{(n)}).$$ Then every term occurring in the polynomial on the RHS is differentiable (because$g\$ is ...

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