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This is not yet a full answer, but too much for a comment-entry. let $a=c e^c$ and $b = d e^d$ (so $\small c=\operatorname{LambertW}(a)$) then the two expressions go like this $$ce ^ {c de ^{dce ^{c de ^d}}}$$ and $$de ^ {dc e ^{cde ^{dc e ^c}}}$$ so the intermediate expressions are all of the form $\small cde^x$ and only the ...

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For any point $a\neq 1$, choose $\varepsilon=|a-1|/2$. This ought to work. The reason you can't do it at $1$ is that this function is actually continuous at $x=1$. This idea also works for the case $f(x)=0$ for irrationals because then $\varepsilon=|a-0|/2$ works, as you pointed out.

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Let $U$ be an open set contained in $A$; if $U\ne\emptyset$, then $E\setminus U$ is finite. However, from $U\subseteq A$ it follows $E\setminus U\supseteq E\setminus A$. Hence…

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Both of your sequences have convergent subsequences because they are infinite and take values in a compact set $[-1,1]$ (This is a theorem about compactness in ${\mathbb R}^d$). I believe that in fact both of your sequences are dense in that range. The first one certainly is, and the second is harder to see but I believe it still is dense. If you like, I can ...

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Given $f(z)$ holomorphic in some $\Omega$ (feel free to take any condition you want on $\Omega$), what is the relationship between the root of $f(z)$ and the roots of the truncated expansion of $f(z)$? Let $$f(z) = \sum_{k=0}^{\infty} a_k z^k$$ and define $$p_n(z) = \sum_{k=0}^{n} a_k z^k.$$ If $f$ has a zero $z_0$ of multiplicity $m$ then ...

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When you have an equality of sets you almost always have to prove the two inclusions $\subseteq$ and $\supseteq$. We have $[\frac{1}{n},n]\subseteq (0,+\infty)$ for all $n$ which implies $\bigcup [\frac{1}{n},n]\subseteq (0,+\infty)$. Now for the other inclusion take $x\in (0,+\infty)$. If $x\geq 1$ there is a $n$ such that $n>x$ and $x\in ... 1 What want to show two things: (1) If$x$is a number in the interval$(0,\infty)$then it is in the set$\cup_{n=1}^\infty [1/n,n]$. (2) If$x \in \cup_{n=1}^\infty [1/n,n]$then$x \in (0,\infty)$. The second one is easier. Take$x \in \cup_{n=1}^\infty [1/n,n]$. Then in particular there is some$N$such that$x \in [1/N, N]$because this is what set ... 2 I think there's just a typo in the solution; the final inequality should be$>$, not$<$. From the solution you gave, we have: $$a_k - \frac{a_k}{1+a_k} < \frac{a_k}2$$ Now move the$\frac{a_k}{1+a_k}$term to the right-hand side and the$a_k$to the left to get: $$\frac{a_k}2 < \frac{a_k}{1+a_k}$$ Taking partial sums, the left side is ... 0 Hint: Condition on whether$x \in (0, \infty)$is less than or equal to or greater than$1$, and then use the Archimedes principle. 3 The solution is flawed. The conclusion in the final sentence does not follow from the preceding line (an upper bound won't do; we need a lower bound). But if we're assuming$0<a_k<1$, then$\dfrac{a_k}{1+a_k} > \dfrac{a_k}2$, and so we have bounded our series below by a divergent series. The solution presented heads off in a far more complicated ... 1 If$ v_k=\dfrac{a_k}{1 +a_k} $then$a_k=\dfrac {v_k }{1-v_k } $Then if$v_k $converge to zero ,$ a_k $converge also to zero. 0$f_n(x)=n$on$[-1, 1]$it is uniformly continuous since it has bounded derivation but not uniformly convergence (for given$\epsilon$let$x=(2\epsilon)^{\frac{1}{n}}). $Note that: theorem: uniform convergence of a sequence of functions imply uniform continuity. 0 Hint:$\sum_{n=1}^{\infty}\frac{1}{\epsilon} = \frac{1}{\epsilon}\sum_{n=1}^{\infty}1 = \frac{1}{\epsilon}*{\infty}$Well, as$\epsilon$is always positive$\frac{1}{\epsilon}$is always positive so$\frac{1}{\epsilon}*{\infty}$is always$\infty$. So ... 1 Notice that: $$\sum_{n=1}^{\infty}\frac{1}{\epsilon} = \lim_{N \to +\infty} \frac{N}{\epsilon} = +\infty.$$ At the end: $$\lim_{\epsilon \to 0}\sum_{n=1}^{\infty}\frac{1}{\epsilon} = +\infty.$$ 0 We say that a function$f$is uniformly continuous on a metric space$X$if for all$\epsilon>0$there exists$\delta>0$such that for all$x,y \in X$,$d(f(x),f(y))<\epsilon$whenever$d(x,y)<\delta$. 3 The roots go off to infinity. One can show the roots of$\displaystyle f(z) = \sum_{k=0}^{N-1} \frac{z^k}{k!}$tend to a curve$|z e^{1-z}| = 1$if you rescale by a factor of$N$. Please see MathOverflow: Roots of truncations of$e^x - 1$Math.SE: Approximating roots of the truncated Taylor series of$\exp$by values of the Lambert W function arXiv: ... 1 Using Abel's summation we have $$\sum_{n=2}^{N}\log\left(\log\left(n\right)\right)=\sum_{n=2}^{N}1\cdot\log\left(\log\left(n\right)\right)=\left(N-1\right)\log\left(\log\left(N\right)\right)-\int_{2}^{N}\frac{\left\lfloor t\right\rfloor -1}{t\log\left(t\right)}dt$$ where$\left\lfloor t\right\rfloor $is the floor function and using$\left\lfloor ...

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$f(x)=\log\log x$ is a concave function on $[2,+\infty)$, since its derivative is $f'(x)=\frac{1}{x\log x}$ and its second derivative is $f''(x)=-\frac{1}{x^2}\left(\frac{1}{\log x}+\frac{1}{\log^2 x}\right)$. The Hermite-Hadamard inequality hence gives: $$\sum_{n=2}^{N}\log\log n \leq \int_{3/2}^{N+1/2}\log\log x\,dx = \left.\left(x\log\log ... 2 Using geometric and arithmetic mean and Stirlings formula you get$$ \left(\prod_{k=2}^n\ln(k)\right)^{1/(n-1)}\le\frac{\sum_{k=2}^n\ln(k)}{n-1} =\frac{\ln(n!)}{n-1}\le\frac{\ln(\sqrt{2\pi(n+1)})+n·(\ln(n)-1)}{n-1} $$5 Looks like it's impossible for two distinct numbers to be "power equivalent". Really, a_{n+1}=a^{b_n} and b_{n+1}=b^{a_n}. Now, if we use the limit definition and choose n large enough, a_{n+1} is close to b_{n+1}, which means for a_n that it is close to {\log a\over\log b}b_n, but at the same time it is close to b_n, which is impossible ... 3 let f=y'+\alpha y hence y is a solution of the ODE y'+\alpha y=f. By resulving this equation we abtain y(t)=\lambda e^{-\alpha t}+[\int_0^tf(s)e^{\alpha s}ds]e^{-\alpha t}. The first terme is Ok. for the second fixe \epsilon>0 and A\geq 0 s.t |f(s)|\leq \epsilon /2 for all s\geq A. And "cut your integral " ... 1 The other direction: Suppose X is compact, and f \in C(X). As \mathrm{supp}\, f is closed in X and X is compact, \mathrm{supp}\, f is compact. Therefore f \in C_c(X). 2 From the first condition, f(1)=1 and thus$$ 1 = f(1) = f((e^{2\pi i/n})^n)=f(e^{2\pi i/n})^n , $$but this contradicts the second requirement. 4 Related approach: If n(n+1) is a square, then 4(n^2+n)=4n^2+4n is a square. But so is 4n^2+4n+1=(2n+1)^2. The difference of two square is 1 when... 3 (n+1)^2 > n(n+1) > n^2 taking square roots n+1 > \sqrt{n(n+1)} > n, and there is no integer between n and n+1. 13 Despite being about n\in\Bbb{N}, induction doesn't work too great here. Instead, how about trying to identify the nearest integers... in particular, you should be able to show$$n<\sqrt{n(n+1)}<n+1$$for n\in\Bbb{N} 1 The space A=[0,\infty) with the norm d(x,y)=|\sqrt{x}-\sqrt{y}| is complete. Here's a more elementary proof. Let (a_n) be a Cauchy sequence in (A,d), and \varepsilon a positive real number. Then, there is a positive integer N_\varepsilon such that$$ d(a_n,a_m)\le \varepsilon \quad \forall m,n\ge N_\varepsilon. $$For \varepsilon=1, we get$$ ...

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If you have two functions $f$, $g$ that are defined and continuous on $\mathbb{R}$, then $f=g$ if the two agree a.e. with respect to Lebesgue measure. That may be how the question was meant to be interpreted, because that would be the default for a.e. on $\mathbb{R}$ if no measure is mentioned. Assuming this interpretation: If the exceptional set is $E$, ...

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Let $(r_n)_{n \in \mathbb N}$ be an enumeration of $\mathbb Q \cap [0,1]$, which is possible as $\mathbb Q$ is countable. Then $$\{(r_n-\frac{1}{2^{n+1}},r_n+\frac{1}{2^{n+1}}) \cap \mathbb Q \ : \ n \in \mathbb N\}$$ is an open cover with no finite subcover.

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This cover has no finite subcover : $$\Bbb Q \cap [0,1] = \bigcup_{n} \left( \Bbb Q \cap \left( \left[0,\ln(2)-\frac{1}{n}\right[\;\cup\; \left]\ln(2)+\frac{1}{n},1\right] \right)\right)$$ Edit : forgot to add the $\frac{1}{n}$

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In general, if a differentiable function is Lipschitz of rank $L$, then we have $\|Df(x)\| \le L$. Since any norm satisfies $| \|x\| - \|y\| | \le \|x-y\|$ (using the triangle inequality), we see that $\|A\| \le 1$. We also have that the directional derivative satisfies $df(x,h) = \lim_{t\to 0} {f(x+th)-f(x) \over t} = Df(x) h$. Letting $n(x) = \|x\|$, we ...

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Actually the problem has nothing to do with square roots except for the property of $x\mapsto\sqrt x$ to be a bijection. If $f:X\to(Y,d)$ is any injective function, where $X$ is a set and $Y$ is a space with metric $d$, and if we equip $X$ with the metric $d_f(x,x')=d(f(x),f(x'))$, then $f$ becomes an isometry $(X,d_f)\to(Y,d)$, i.e. a map which preserves ...

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Suppose $i,j,k$ is a permutation of $1,2,3$. Then we have \begin{eqnarray} \bar{z}_i (z_i+z_j+z_k) &=& |z_i|^2+ \bar{z}_i z_j+\bar{z}_i z_k= 0 \\ \overline{\bar{z}_j (z_i+z_j+z_k)} &=& |z_j|^2+ \bar{z}_i z_j+z_j \bar{z}_k= 0 \end{eqnarray} Subtracting and using the fact that $|z_i|=|z_j|$ gives $\bar{z}_i z_k = z_j \bar{z}_k$. Since ...

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Pick $x$ with $\|x\|=1$ and let $\phi(t) = \|tx\| = |t|$. Since $\phi$ is not differentiable at zero, $x \mapsto \|x\|$ cannot be either.

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From the point of view of complex addition and absolute value the complex numbers behave like vectors. If the lengths of the vectors are the same and if the resulting vector is $0$ then the vectors (the first one is originated at $0$) form a triangle. A triangle with sides of equal length cannot be anything else but an equilateral triangle.

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HINT: If $x,y\in X$, and $x\ne y$, then $\lceil d\rceil(x,y)\ge 1$, $\{x\}$ is always open (why?). Is this necessarily true in $\langle X,d\rangle$?

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Hint: Consider the interval $[0,1]$ with the usual metric. What is the topology under the ceiling metric?

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think of the points $z_1,z_2,z_3$ as points $P,Q,R$ on the circumference of a circle centered at the origin. a well-known theorem of Euclidean geometry gives: $$Q\hat O R = 2 Q\hat PR$$ from which the result follows

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Assume $X\ne \{0\}$. Otherwise it would be unnecessarily pathological. Remember that $$\lim_{h\to 0, h\ne 0} \frac{ \|x_0 + h\| - \|x_0\| - Ah }{\| h \|} = 0$$ implies $$\lim_{t\to 0, t > 0} \frac{ \|x_0 + t h_0\| - \|x_0\|}{t} = A h_0$$ for any fixed $h_0\ne 0$, by setting $h = t h_0$. Hints: "$x_0 \ne 0$": Let $h_0 \ne 0$. Show that ...

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Let $f(x)=\exp(-x^2)$. By the MVT, there is $z$ between $x$ and $y$ such that $$f(y)-f(x)=(y-x)f'(z)=(y-x)[-2z\exp(-z^2)].$$ Now using $\exp(z^2)\geq z^2+1=|z|^2+1\geq2|z|$, we have $$|2z\exp(-z^2)|=\frac{2|z|}{\exp(z^2)}\leq\frac{2|z|}{|z|^2+1}\leq 1.$$ Thus, $|f(y)-f(x)|\leq|y-x|$ and the claim follows.

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By the Bolzano-Weierstrass theorem, any bounded sequence has a convergent subsequence. So, you are correct. As @lulu pointed out, you can get a subsequence that converges to any decimal. Can you enumerate them all? No. Here's why: For any convergent subsequence, you can mix in values from any other sequence, provided that you are only mixing in a finite ...

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You're missing the fact that disjoint implies that you cannot have more than one non-zero term in that sum. Your assumption on uncountability is not actually quite equivalent, and when we reprhase it more carefully, the ostensible problem disappears, you have $$\mu\left(\coprod_{i=1}^\infty A_i\right) = \begin{cases} 0 & \forall i \;|A_i|\le \aleph_0 ... 2 So your question is satisfactorily answered if you are given initial conditions for Newton's method which find each root. Here I will assume that two roots exist, i.e. p^2-4q>0. Given an initial condition not exactly on the vertex, the Newton iteration will stay on that side of the vertex, because of the fact that there is only one turning point. Also, ... 0 Either use the quadratic formula (since your equation is quadratic) or let f(x)=x^2+px+q=0, then$$f^{\prime}(x)=2x+p$$and by Newton-Rapson method$$x_{n+1}=x_n-\cfrac{f(x_n)}{f^{\prime}(x_n)}$$Starting with x_0=1 this will be$$x_{n+1}=x_0-\cfrac{x^2+px+q}{2x+p}=x_{n+1}=1-\cfrac{1+p+q}{2+p}Normally, when we want to chose a starting value we seek ... 2 Hint: |z-1| \le |z-|z|| + ||z|-1|. Now the points z,|z| lie on the circle centered at 0 of radius |z|. So the straight line distance between these points is no more than the arc length between them on this circle. 1 We can split the sum \begin{align} \frac{1}{n}\sum_{k = 1}^n n^{\frac{1}{k}} &= 1 + \frac{1}{n}\sum_{2 \leqslant k \leqslant n^{1/3}} n^{\frac{1}{k}} + \frac{1}{n}\sum_{n^{1/3} < k \leqslant n} n^{\frac{1}{k}}\\ &\leqslant 1 + \frac{1}{n}\cdot \sqrt{n}\cdot n^{1/3} + \frac{n - \lfloor n^{1/3}\rfloor}{n}\cdot n^{1/(n^{1/3})}\\ &\leqslant 1 + ... 0 First, assume that f is an indicator function, say f=\mathbf 1_A where A is a measurable. You can find Borel sets B,C such that B\subseteq A\subseteq C and C\setminus B has measure 0; then take g:= \mathbf 1_B and h:=\mathbf 1_C. Next, assume that f is a simple function, say f=\sum_{i=1}^N \alpha_i\mathbf 1_{A_i}, where the sets A_i ... 2 Not only on an intuitive level: Define the map \Phi\colon L^2(\Omega)\otimes H\longrightarrow L^2(\Omega;H),\,f\otimes \xi\mapsto f(\cdot)\xi. This is an isometric isomorphism: \Phi is isometric:\langle f(\cdot)\xi,g(\cdot)\eta\rangle_{L^2}=\int_\Omega f(\omega)g(\omega)\langle \xi,\eta\rangle_H\,dP=\langle \xi,\eta\rangle_H\int_\Omega ...

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is this what you mean? If not please correct me. Please, anyone... Since $f$ is uniform continuous, then given any $\epsilon^*>0$, there exists a $\delta^*>0$ such that for any $y$, $x\in D$, we have $$|y-x|<\delta^*\,\,\Longrightarrow\,\,|f(y)-f(x)|<\epsilon^*$$. Now take $\epsilon^*=\frac{1}{2}\epsilon$. Let $\delta^*=\frac{\delta}{2}$ Then ...

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My first thought did not use the identity in the question, but used identity $(7)$ proven in this answer, $$\sum_{k\in\mathbb{Z}}\frac1{z+k}=\pi\cot(\pi z)\tag{1}$$ so that  \begin{align} \sum_{k\in\mathbb{Z}}\frac1{1+6k} &=\frac16\sum_{k\in\mathbb{Z}}\frac1{\frac16+k}\\ &=\frac16\pi\cot\left(\frac\pi6\right)\\[6pt] &=\frac\pi6\sqrt3\\[6pt] ...

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