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Lemma: The algebra of operators on $\ell^2$ generated by $\{I, L, R\}$ contains all operators whose matrix has finitely many nonzero entries. Proof: Denoting the canonical basis of $\ell^2$ by $\{e_n\}_{n=0}^\infty$, notice that $$p:= I-RL$$ is the projection onto the space spanned by $e_0$. Moreover $R^mpL^n$ is the operator which sends all basis ...

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Just note that for any fixed $\epsilon \in(0,1)$ then $0\leqslant f_n(x)\leqslant \frac1{(1+x^2)\arctan (\epsilon ^2) }$ for all $x\in \mathbb{R}\setminus (-\epsilon ,\epsilon )$ and all $n\in \mathbb{N}$ so the dominated convergence theorem show us that $$\lim_{n\to\infty}\int_{\mathbb{R}\setminus (-\epsilon ,\epsilon )}f_n\mathop{}\!d \lambda =0$$ Now ...

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We know $\cap_{n=1}^\infty I_n \neq \emptyset$. Suppose there are two points, $x_1$ and $x_2$ in this set. Clearly, $a_n\leq x_1 \leq b_n$ and $a_n\leq x_2 \leq b_n$ for all $n$. Hence, $|x_1-x_2|\leq b_n-a_n$ for all $n$. If $x_1 \neq x_2$, then $|x_1-x_2|>0$. Where's the contradiction? Or you can just prove that $x_1=x_2$ directly using squeeze theorem.

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$I_{n+1}\subset I_n$ means $[a_{n+1},b_{n+1}]\subset[a_n,b_n]$ and thus $a_{n+1}\geqslant a_n$ and $b_{n+1}\leqslant b_n$ for all $n\geqslant 1$. $(a_n)_{n\geqslant 1}$ is an increasing sequence, $(b_n)_{n\geqslant 1}$ is a decreasing sequence, thus $(b_n-a_n)_{n\geqslant 1}$ is a decreasing sequence, its limit is $\inf_{n\geqslant 1}(b_n-a_n)=0$. Thus the ...

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Using change of bases $\ln n=\log_2 n / \log_2 e$ so $$2^{\ln n} =(2^{\log_2 n}) ^{1/\log_2(e)}=n^{1/\log_2 e} =n^{\log_e 2}=n^{\ln 2}$$

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$\ln n = k$ were $e^k = n$. And $\ln 2 = m$ where $e^m = 2$. As a definition there fore $2 = e^{\ln 2}$ and $n = e^{\ln n}$. And for any $W$ we will always have $e^{\ln W} = W$. SO if we replace $2$ with $e^{\ln 2}$ we can see $\frac 1{2^{\ln n}} = \frac 1{(e^{\ln 2})^{\ln n} } =\frac 1{e^{\ln 2\times \ln n}}$. But if we replace $n$ with $e^{\ln n}$ we ...

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$$2^{\ln n}=n^{\ln 2}\Leftrightarrow \ln(2^{\ln n})=\ln (n^{\ln 2})\Leftrightarrow (\ln n)(\ln 2)=(\ln 2)(\ln n)$$

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We have that by $A^B=e^{B \ln A}$ $$2^{\ln n}=e^{\ln n\cdot \ln 2}=\left(e^{\ln n}\right)^{\ln 2}=n^{\ln 2}$$

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I'll begin by saying that when you want to prove an identity like $L=R$ you cannot multiply both sides by any value, since then it wouldn't be an identity. Anyway, Lets say that $L$ is LHS, and $R$ is RHS. Let $t=\tan(A)$. We can then say that $\frac{1}{t}=\cot(A)$. Therefore, RHS becomes $$R = \frac{t^3 + \frac{1-t}{\frac{1}{t}}}{\frac{1-\frac{1}{t}}{t} ... 4 note that i\sin iz=\frac{e^{-z}-e^z}{2}=-\sinh z so \sin iz=i \sinh z hence the equation is \sin z=\sin iz or 2\sin ((z-iz)/2) \cos ((z+iz)/2)=0 so either z−iz=2k\pi or z+iz=π+2k\pi and simplifying the cases reduces to z=(1+i)k \pi or z=(1-i)(\pi/2 +k\pi) 0$$e^{iz}-e^{-iz}=-(e^z-e^{-z})\implies e^{(i-1)z}=1\implies(i-1)z=2m\pi i$$1$$t^4\left(\dfrac{1 -t^{-1}}{t} +t^{-3}\right)=t^3 + \dfrac{1 - t}{t^{-1}}$$is$$t^3-t^2+t=t^3+t-t^2.0 Let t=tan(A). Multiply both sides by rhs denominator and let cot(A)=1/t. Then lhs =t^4(1-1/t)(1/t)+t=t^3-t^2+t, while rhs =t^3+(1-t)t=t^3+t-t^2. Therefore equality. 1 1. \ \cot^3A = \dfrac{1}{\tan^3A}\\ 2. \ \dfrac{1-\cot A}{\tan A } = \dfrac1{\tan A}-\dfrac{1}{\tan^2A}\\ 3. \ \dfrac{1-\tan A}{\cot A} = \tan A - \tan^2A So, the RHS becomes, \begin{align}\dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac1{\tan A}-\dfrac{1}{\tan^2A} +\dfrac{1}{\tan^3A}} &= \dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac{1}{\tan^3A}\left(\... 0 The conclusion is wrong. For example, let a=-1,f(x)=0 and then the equation becomes y'-y=0. $$This one has solution y=ce^x which is unbounded in the whole real line for c\neq0. 2 In your proof, \delta_1 depends on y. So, it may happen that \delta_1 shrinks to 0, as y\to y_0. Consider the function f\colon \mathbb{R}^2\to \mathbb{R}, f(x,y)=\frac{x^2y^2}{x^2y^2+(x-y)^2} for x,y\neq 0 and h(0,0):=0. Let (x_0,y_0):=(0,0). Now, it is easy to see, that (in your notation) l_{x_0}=l_{y_0}=l_{x_0 y_0}=0, but f(\delta,\... 0 By the mean value theorem for Integrals, for each n there is c_n \in [a_n, b_n] such that$$\frac{1}{b_n-a_n} \int_{a_n}^{b_n} f(x) \, dx=f(c_n).$$From a_n \to c and b_n \to c we get c_n \to c. Since f is continuous, f(c_n) \to f(c). 1 Hint :$$\frac{1}{b_n-a_n} \int_{a_n}^{b_n} f(x) \, dx = \frac{F(b_n)-F(a_n)}{b_n-a_n}$$where F is a primitive of f over \mathbb{R}. 0 In your post and the answer provided we are making two assumptions : f is integrable and \mu(A)<\infty This post provides a counterexample to your claim : https://mathoverflow.net/questions/149936/are-measurable-functions-almost-surely-constant-on-atoms So let us consider only integrable functions. Then only the case \mu(A)=\infty remains. WLOG ... 0 If the metrics are induced by some norms, I think I have found a sufficient condition (although I cannot say that this condition is reasonably weak). Proposition. Suppose \|\cdot\|_1 and \|\cdot\|_2 are two norms on a linear space X. If$$\left\{ \|\mathbf x\|_1: \mathbf x\in X,\ \|\mathbf x\|_2=1 \right\}is unbounded above, the open unit ball under ... 1 There are multiple applications of the concept of a Fréchet derivative. Think of it as an enhanced tool to find extremums (i.e. maximizers/minimizers) just like in ordinary calculus, but here, the Fréchet derivative allows us to solve optimization problems where our sets are infinite dimensional vector spaces (i.e. L^p spaces). A very famous example is ... 0 Assume that f is entire and that for each z \in \mathbb{C} we have |f(z)| \leq 1+ |z|^{1-\alpha} for an \alpha \in (0,1). Fix z_0 \in \mathbb{C}. We will show that f'(z_0) = 0. Let C_R = \{ z \in \mathbb{C} | |z-z_0| = R \}. By Cauchy's formula we have \begin{align*}|f'(z_0)| &= \frac{1}{2\pi} \bigg| \int_{C_R} \frac{f(z)}{(z-z_0)^2} \, dx \... 1 I may be missing some part of your question... But strange that you have never seen usage of Fréchet derivative. For example, imagine that V is a Hilbert space equipped with the inner product \langle \cdot, \cdot \rangle. What is the derivative off: x \mapsto \sqrt{\langle x, x \rangle}?$$What if V = L^2([0,1], \mathbb R) equipped with following ... 2 Suppose that there is t_0>0 such that$$ \lim_{t\to t_0^-} x(t)=\infty. $$Namely,$$ x(t) \text{ is not left continuous at $t=t_0$.} \tag{*} $$Letting t\to t_0^- in$$ \frac{1}{2}y^2(t) + \frac{1}{x^4(t)} = C$$gives$$ y^2(t_0)=2C. $$So x'(t_0)=y(t_0) is finite and hence x(t) is continuous at t=t_0, which is against (*). 0 Let y_0 = 1 and y_n = \prod\limits_{k=1}^n x_n for n > 0, we have x_n = \frac{y_n}{y_{n-1}} for all n \ge 0. In terms of y_n, the recurrence relation become$$x_{n+1} = 1 + \frac{1}{x_n} \iff \frac{y_{n+1}}{y_n} = 1 + \frac{y_{n-1}}{y_n}\iff y_{n+1} = y_n + y_{n-1}$$This is the recurrence relation for the Fibonacci numbers. Since y_1 = y_0 = ... 1 Let's try to apply the Fixed Point Theorem. Consider I=[\frac{3}{2},2] and f(x)=1+\frac{1}{x} defined in I. f is continuous in I f(I)\subset I once f'<0 f is Lipschitz continuous because as f is a function of class C^1 you have that |f'(x)|=\frac{1}{x^2}\leq \frac{1}{(\frac{3}{2})^2}=(\frac{2}{3})^2<1. Then for any x,y \in I \... 0 Hint: if it does converge, the number L that it converges to must satisfy$$ L = 1 + \frac{1}{L}. 1 Let f(n) = 1 + 1/n So the sequence is just x_{n+1} = f(x_n) Clearly our sequence is always positive, so f is continuous. So the limit of our sequence must be a fixed point of f. The only fixed points of f are \frac{1 \pm \sqrt{5}}{2}, and since our sequence is always positive, we can discount the negative fixed point as a potential limit. So the only ... 1 (X^2-1)^n is a polynomial of degreee 2n, with first coefficient 1. Thus, its 2n derivative is a constant polynomial, with coefficient (2n)(2n-1)\cdots 2\cdot 1 = (2n)!. It is because when you differentiate X^{2n} you get 2nX^{2n-1},then 2n(2n-1)X^{2n-2}, etc. 0 Hint : (x^2-1)^n is a polynomial of degree 2n whose dominant coefficient is x^{2n}. 1 Answer : \sqrt{n^3} (\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3}) =\sqrt{n^3}\frac{(2\sqrt{n^2 +3n+2}-2\sqrt{n^2 +3n})}{\sqrt{n+1}+\sqrt{n+2}+\sqrt{n}+\sqrt{n+3}}=\frac{4 \sqrt{n^3} }{(\sqrt{n^2 +3n+2}+\sqrt{n^2 +3n})(\sqrt{n+1}+\sqrt{n+2}+\sqrt{n}+\sqrt{n+3})}=\frac{4 \sqrt{n^3} }{n \sqrt{n + 3} + \sqrt{n^3 + 4 n^2 + 5 n + 2} +\sqrt{n^3 + 5 n^2 + 8 n + 4} + \... 1 `You have to use Taylor's expansion at order 2: \begin{align} &\phantom{=}\sqrt{n^3}\bigl(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3}\bigr)\\&=n^2\Bigl(\sqrt{1+\frac1n}+\sqrt{1+\frac2n}-1-\sqrt{1+\frac3 n}\Bigr) \\ &=n^2\biggl(1+\frac1{2n}-\frac1{8n^2}+1+\frac1n-\frac1{2n^2}-1-1-\frac 3{2n}+\frac9{8n^2}+o\Bigl(\frac1{n^2}\Bigr)\biggr)\\ &=n^2\... 2 It's a perfectly nice counterexample. The statement that d_1 is not finer than d_2 boils down to an existence statement, as you rightly noted, and that is what you proved. What you did is a perfectly standard method for proving an existence statement, namely: Write down an example and prove that it satisfies the required properties. It's true, not all ... 1 Another way to multiply by conjugates:\begin{align} \sqrt{n+1}+\sqrt{n+2}-\sqrt n-\sqrt{n+3}&={(n+1+2\sqrt{(n+1)(n+2)}+n+2)-(n+2\sqrt{n(n+3)}+n+3)\over\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3}}\\ &={2(\sqrt{n^2+3n+2}-\sqrt{n^2+3n})\over\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3}}\\ &={4\over(\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3})(\sqrt{n^2+3n+2}...

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$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})=$$ $$=\sqrt{n^3}\left(\frac{2}{\sqrt{n+2}+\sqrt{n}}-\frac{2}{\sqrt{n+3}+\sqrt{n+1}}\right)=$$ $$=\frac{2\sqrt{n^3}\left(\sqrt{n+3}-\sqrt{n+2}+\sqrt{n+1}-\sqrt{n}\right)}{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+1}+\sqrt{n+3})}=$$ $$=\tfrac{2\sqrt{n^3}\left(\frac{1}{\sqrt{n+3}+\sqrt{n+2}}+\frac{1}{\sqrt{n+1}+\sqrt{... 0 Finite dimensional metric spaces are "topologically equivalent" (that is, they are homeomorphic to each other (that is any open ball in one is properly contained in an open ball of the other ) ) because of the Lipschitz inequality between norms that is c_1|x|_p \leq |x|_q \leq |x|_p \leq c_2 |x|_q for any two norms p \leq q, where c_1 and ... 2 Assume \{X_n\} is a family of i.i.d \mathcal N(0,1) random variables. Fix \varepsilon > 0. We want to show \limsup_n \frac{X_n}{\sqrt{2\ln(n)}} \ge 1- \varepsilon  To this end, it would be enough to show that series \sum_{n=1}^\infty \mathbb P(X_n \ge (1-\varepsilon)\sqrt{2\ln(n)})  is divergent. Note that for t > 0 we have$$\mathbb P(X_n ...

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Let $e_i \in \mathbb R^n$ be the vector whose $j$-th entry is a $1$ if $j=i$ and $0$ if $j \neq i$. Then: \begin{align}\|x\| &= \|x_1e_1 + \cdots + x_ne_n\| \\ &\leq \|x_1e_1\| + \cdots + \|x_ne_n\| \\ &= |x_1|\|e_1\| + \cdots + |x_n|\|e_n\| = |x_1| + \cdots + |x_n|. \end{align} Also, since $|x_i| \leq \|x\|_M$ for all $i$, we have $|x_1| + \... 5 Always? No. E.g. when you are interested in isometry. But if up to homeomorphism only then yes. Consider any metric space$(X,d)$and introduce a new metric on$X$: $$d'(x,y)=\min\big(d(x,y), 1\big)$$ First of all note that$d'(x,y)\leq 1$. Furthermore a subset is open in$(X,d)$if and only if it is open in$(X,d')$. Ergo the identity $$id:X\to X$$ $$id(x)=... 0 Let a_n=1/n when n is even, a_n=n when n is odd. This clearly doesn't converge, but 0 is the only accumulation point. 1 How about 1, 2, \dfrac13, 4, \dfrac15, 6, \dfrac17, 8, \dfrac19, ...? 0 Yes; one can just set g(t) = f(x+t). 2 I think your examples suffer because they depend on notational conventions. AFter all, why do we denote a single number by both \frac12 and \frac24? Because rationals are defined by equivalence classes, but that's hidden in our familiarity with the notation. The same goes for modular arithmetic. I suggest the following. Let's break the integers into ... 0 You can factor the z like this$$z (z^3 - (1-i)^7) = 0.$$Then the solutions are z=0, and the solution of z^3 = (1-i)^7. You can use the polar form to solve that. 3 Note (1-i)^8=(-2i)^4=(-4)^2=16. The roots of z(z^3-16)=0 are 0 and 2\sqrt[3]{2} times the three roots of unity. I'll leave the rest to you. 2 Let x be any non-zero vector and let M be the one-dimensional subspace spanned by x. Define A(cx)=c\|x\|. This defines a linear map on M with norm 1. By Hahn - Banach Theorem there exists a linear map from \mathbb R^{n} to \mathbb R which extends this map without increasing the norm. But \|A\| =1 so the extended linear map B has norm ... 2 Probably too advanced. Since you already received good answers, let me try to go beyond the limit itself. Let y=kx which makes$$x\cdot (x+y)\cdot (x+2y) \dots (x+(n-1)y)=(k x)^n\,\frac{ \Gamma \left(n+\frac{1}{k}\right)}{\Gamma \left(\frac{1}{k}\right)}$$and so$$A_n=kx \left(\frac{\Gamma \left(n+\frac{1}{k}\right)}{\Gamma \left(\frac{1}{k}\right)}\... 0$|x|_s\leq n|x|_M$is obvious, since$\sum_{i=1}^{n}|x_i|\leq n\max_{i}|x_i|$.$|x|_M\leq|x|_s$is also obvious, since$\max_{i}|x_i|\leq\sum_{i=1}^{n}|x_i|$. But I don't know how you define$|x|$. Your definition of it is quite vague. Once you are clear on this, you can prove the set of inequalities. You should update a clear definition of it, and I can ... 2 Note that $$n^{2/n} \leqslant (n^2+1)^{1/n} \leqslant (2n^2)^{1/n} = 2^{1/n} n^{2/n}$$ By the squeeze theorem it follows that$\lim_{n \to \infty} (n^2+1)^{1/n} = 1$, since$2^{1/n} \to 1$and$n^{2/n} \to 1$as$n \to \infty$. That$2^{1/n} \to 1$is well-known and easy to prove. To show that$n^{2/n}\to 1$, we have$n^{2/n} \geqslant 1$for all$n \in \...

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We know for a seuence $\{x_n\}_n^\infty$ of positive reals, if $\lim \frac{x_{n+1}}{x_n}=L \in [0,\infty]$ then $\lim {x_{n}}^{\frac1 n}=L$. So, $$\lim \frac{(n+1)^2+1}{n^2+1}=\lim \frac{(1+\frac1 n)^2+\frac 1 {n^2}}{1+\frac1 {n^2}}=1\\ \implies \lim (n^2+1)^{\frac 1 n}=1$$

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