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I have proved it as follows. Taking the expression for $f'(x)$, multiplying through by $(1+x^{2} \sin^{4} \frac{1}{x})$, we need to prove the following inequality, call it (a): $$\left(1+x^{2} \sin^{4} \frac{1}{x}\right) \arctan(x \sin^{2} \frac{1}{x}) + x\sin^{2}\frac{1}{x} > 2\sin\frac{1}{x} \cos\frac{1}{x} \qquad \mbox{(a)}$$ Observe that since ...

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Since $$\left\|\frac{f(x+hy)-f(x)}{h}\right\|\ge c\|y\|,$$ for $y\ne0$ you get $$\|d_xfy\|=\left\|\lim_{h\to0}\frac{f(x+hy)-f(x)}{h}\right\|\ge c\|y\|\ne0.$$ So $d_xf$ is invertible.

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In my view it is not really the definition of the radius of convergence. Rather the definition of the radius of convergence of $$\sum_{n = 0}^{\infty} a_n(z - c)^n$$ is $R=\sup\{|z_0 - c | \colon \sum_{n = 0}^{\infty} a_n(z_0 - c)^n \text{ converges}\}$. With this definition, it is clear that this is the largest disk where the convergence could ...

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Sorry to take so long. I've been busy. On the interval in question, $$\lfloor x \rfloor = \begin{cases}1, & 1 \le x < 2\\2, & x = 2\end{cases}$$ Given $n$, the intervals are $\left[1 + \frac {k-1}n,1 + \frac{k}n\right], k = 1, \ldots, n$. So the sums are $$L_n = \sum_{k=1}^{n}\left\lfloor 1 + \frac {k-1}n \right\rfloor\frac1n = ... 1 Getting a finite open cover K consisting of metric balls seems to be a red herring. Your assumptions imply there is an open U containing K over which f is injective. Let \epsilon=d(K,\mathbb R ^n \setminus U). All you have to show is \epsilon>0, because clearly \{x\in \mathbb R^n : d(x,K) < \epsilon \}\subseteq U. Suppose ... 1 This is good. Alternatively, for a direct proof, note A is non-empty and bounded below. Every lower bound c of A must satisfy c \notin \mathbb{R_+} since for all \epsilon \in \mathbb{R_+} there is some n such that \epsilon > \frac{1}{n} and \frac{1}{n} \in A. Since \max \mathbb{R} \setminus \mathbb{R_+} = 0, and 0 is a lower bound of ... 0 I'd like to point out that$$\frac b2<b\leq\text{inf}(A).$$This being a contradiction implies that b>\inf(A). 1 Your answer is fine. Just take some care with the assumption that "since b \in A, b can not be a lower bound of A". For example: A = [0, 1]. The infimum of this set is also 0 and 0 \in A. 1 Yes, but you have to verify this. Suppose x=\sum f(v)v\in\operatorname{span}\beta, where v\in\beta and the sum is finite. Show that \Vert Tx\Vert^2=\Vert x\Vert^2= \sum|f(v)|^2 (use orthonormality of \beta). Thus, if Tx=0, we must have x=0, which means that T is injective. 2 Yes. This is even true for a real analytic section of a real analytic bundle over a real analytic manifold. To prove this, suppose f=g on a set U with nonempty interior (which we will henceforth also call U); if U is not everything, then there is a point x \in \partial U. Work instead with f-g; we wish to show it vanishes in a neighborhood of ... 3 Actually, it depends on what you call a continuous curve. If you mean an object defined in a parametric way t \mapsto (x(t),y(t)) where x and y are two continuous functions on an interval I then x+y is also a continuous function on I. But by the intermediate value theorem, if you require x+y to take values in \mathbf{Q} then it forces this ... 0 If you know a bit about connected components: The set P=\left\{(x,y):x+y\in\mathbb{Q}\right\}, which is the union of all lines y=-x+q for q rational, has precisely these lines as connected components, so these are in fact all "curves" which satisfy the equation x+y\in\mathbb{Q}. 1 For each x,y\in\mathbb R define f_x and f^y on \mathbb [0,1] by y\mapsto f(x,y) and x\mapsto f(x,y), respectively. Since \mathbb Q has zero Lebesgue measure, f_x=0 a.e. for all x so$$0=\int_0^1 f_x(y)\ \mathsf dy=\int_0^1\int_0^1f(x,y)\ \mathsf dy\ \mathsf dx.$$However, f is not Riemann integrable on [0,1]^2. Let (x,y)\in[0,1]^2 ... 1 Use a "diagonal": Given n, divide the interval [-n,n] in 2n^2 intervals of length 1/n^2, namely I^n_i=[n+(i-1)/n^2,n+i/n^2], 1\leq i\leq 2n^2, and define "rectangular functions" f^n_i=n1_{I^n_i}. Then \int f^n_i=1/n. (1_A denotes the characteristic function of A). The sequence ... 1 If you have such a sequence (f_n)_{n\geq0} on [0,1], then g_n(x)=1/(\lfloor x\rfloor^2)f_n(\{x\}) works on \mathbb{R}. 0 For each x,y\in\mathbb R define f_x and f^y on \mathbb [0,1] by y\mapsto f(x,y) and x\mapsto f(x,y), respectively. Since \mathbb Q has zero Lebesgue measure, f_x=0 a.e. for all x so$$0=\int_0^1 f_x(y)\ \mathsf dy=\int_0^1\int_0^1f(x,y)\ \mathsf dy\ \mathsf dx.$$A similar argument may be used to show that the other iterated integral is ... 0 Simple, use derivative, not approximations. You get: (x\arctan(x\sin^2(\frac{1}{x})))'=\arctan(x\sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(-2x\sin\frac{1}{x}\cos\frac{1}{x}(-\frac{1}{x^2})) so all are positive for x \geq 1. 3 Check for m = 0 and n = 0. Then (n \neq 0 \wedge m \neq 0) divide numerator and denominator by mn \neq 0.$$ A = \left\lbrace \frac{1}{4 \frac{m}{n} + \frac{n}{m}} \left| m \in \mathbb{Z}\smallsetminus \lbrace 0 \rbrace \wedge n \in \mathbb{N}\smallsetminus \lbrace 0 \rbrace\right.\right\rbrace \cup \lbrace 0 \rbrace $$Now let q = \frac{m}{n}. q ... 1 You made a mistake.$$|f|=\sqrt{(\langle f|f\rangle)}.$$Angle depends on the definition of inner/dot product. You have to define the inner product first (i.e. including fixing your a and b, different a and b will give you different angle) before computing its angle.It has nothing to do with intersection point. 0 The domain of f is broken into two parts: the triangle A above the diagonal in [0,1]^2 and the triangle B below the diagonal. The parts of f are continuous (being a composition of continuous functions) on the interiors of A and B, respectively. What you have to check is that f is still continuous on the boundaries \partial A and \partial ... 0 I think you are confused with the definition of cover and compactness. Let's deal only with subsets of \mathbb{R} to make things clearer: Definition: A cover of a subset A of \mathbb{R} is a collection \mathcal{C} of subsets of \mathbb{R} such that A\subseteq \bigcup\left\{B:B\in\mathcal{C}\right\}, that is, the union of all elements of ... -1 It's also worth noting that this representative of is a strong bias towards analysis. An algebraist generally cares minimally about \mathbb{R}, generally field extensions of \mathbb{Q} and \mathbb{F}_p are far more interesting. The badly named so-called Real Numbers are very much unimportant in Combinatorics and Number Theory as well, but ... 1 First (\epsilon,1)_{\epsilon>0} is an open cover of (0,1) because: for any \epsilon>0, (\epsilon,1) is open (0,1)\subset \bigcup_{\epsilon>0}(\epsilon,1) (in fact it is equal) Now assume that this open cover has a finite subcover. This means that there exists \epsilon_1,\dots,\epsilon_n such that (0,1)\subset ... 3 You write some \epsilon as close as possible to zero But there is no such \epsilon! For any positive \epsilon, there is a strictly smaller positive \epsilon - for instance, \epsilon/2. And the interval ({\epsilon/2}, 1) will contain some points in (0, 1) that (\epsilon, 1) does not (for instance, \epsilon itself). While no specific ... 1 Denote f_k(x)=f(kx) then we have$$\lim_{k\to\infty}f_k(x)=r\;\text{ a.e.}$$so by the dominated convergence theorem we have$$\lim_{k\to\infty}\int_0^a f_k(x)dx=\int_0^a rdx=ra$$0 If not, then you immediately have a disconnection of X as$$f^{-1}((-\infty,y))\cup f^{-1}((y,\infty))$$(can you see why these two sets are disjoint, nonempty, open, and exhaust X?). 0 (1). The set of congruences of squares, modulo 10, is A=\{0,1,4,5,6,9\} And the set of congrences, modulo 10, of \{2 x: x\in A\} is B=\{0,2,8\}. And A\cap B=\{0\}. So, modulo 10, we have$$(2). \quad n^2-2 m^2\equiv 0\iff n^2\equiv 2 m^2\equiv 0\iff ((n\equiv 0 \land (m\equiv 0\lor m\equiv 5)).$$(3). Observe that n\equiv 0\pmod {10}\iff ... 2 The continuous image of a connected space is connected, and the only connected spaces in \;\Bbb R\; are intervals, so since \;f(X)\subset\Bbb R\; is connected it is an interval. Try now to take it from here. 0 the signs depend on the solutions of k^x=x^k, which cannot be solved by elementary means. Rewrite$$x=\left(\sqrt[k]k\right)^x=e^{x\ln(k)/k}$$and set u=-x\ln(k)/k. Then$$e^{-u}=-\frac k{\ln(k)}u,\\ ue^u=-\frac{\ln(k)}k,\\ u=W\left(-\frac{\ln(k)} k\right),\\ x=-\frac k{\ln(k)}W\left(-\frac{\ln(k)} k\right).$$where W denotes the Lambert function. ... 0 By definition the product measure is the outer measure defined using covers by countable unions of "rectangles". If$$\Delta\subset\bigcup_jA_j\times B_j$$then it's not hard to show that$$\sum\mu(A_j)c(B_j)=\infty;$$hence (\mu\times c)(\Delta)=\infty. 0 You are dealing with$$\frac{\exp((\ln n)^2)}{n!}$$when n is large. (If I'm not misinterpreting it, you mean (\ln n)^{\ln n} rather than \ln(n^{\ln n}).) For an elementary approach, two facts, which every sensible calculus textbook should include, might be helpful here: 1). For any a>0 (however terribly small), and any \epsilon>0, there ... 1 It's easy to see that \ln n\lt n/2 and n!\gt(n/2)^{n/2} for n\gt1. Thus$${(\ln n)^{\ln n}\over n!}\lt{(n/2)^{\ln n}\over(n/2)^{n/2}}={1\over(n/2)^{(n/2)-\ln n}}$$and the latter tends to 0 for any number of reasons. 1 For 0\le m\le n, let S_m:=\sum_{k=1}^m a_k-\sum_{k=m+1}^na_k. If S_m=0 for some m, we are done. Hence assume S_m\ne 0 for all m. As S_n=-S_0, there are m such that S_m is positive and others such that S_m is negative. Hence there exists m>0 such that S_{m-1} and S_{m} have opposite sign. But S_{m}-S_{m-1}=2a_m (or equivalently ... 0 It is easy to show that \ln (n!) > \frac{1}{3} n \ln n.^{(\dagger)} Using this, rewrite$$ 0 < \frac{(\ln n)^{\ln n}}{n!} < \frac{(\ln n)^{\ln n}}{e^{ \frac{1}{3} n \ln n}} = e^{(\ln n)\ln \ln n - \frac{1}{3} n \ln n} = e^{(\ln n)( \ln \ln n - \frac{1}{3} n )} \xrightarrow[n\to \infty]{} 0 $$where the limit follows from observing that the ... 1$$a_{e^n}=\frac{n^n}{\Gamma(1+e^n)}=\frac{n}{e^n}\frac{n}{e^n-1}...\frac{n}{e^n-n+1}\frac{1}{\Gamma(1+e^n-n)}\le\left(\frac{n}{e^n-n+1}\right)^n\frac{1}{\Gamma(1+e^n-n)}$$Each factor converges to 0. 0 Let$$a_n=\frac{\left(\ln n\right)^{\ln n}}{n}$$Then$$\ln a_n=\ln\frac{\left(\ln n\right)^{\ln n}}{n}=\ln\left(\left(\ln n\right)^{\ln n}\right)-\ln n=\ln n\cdot\ln\ln n-\ln n=\ln n\cdot\left(\ln\ln n-1\right)$$It is easy to see that \ln a_n\to\infty, so also a_n\to\infty EDIT: I just noticed that$$a_n=\frac{\left(\ln n\right)^{\ln n}}{n!}$$I ... 1 Yes. Def'n : A linear order <_S on a set S is order-dense iff \forall x,y\in S\;(x<_S y\implies  \exists z\;(x<_Sz<_Sy)). Theorem.(Cantor). If S is countably infinite and <_S is an order-dense linear order on S with no end-points (no <_S max or min) then there is an order-isomorphism from S to Q . So let S be the ... 7 If f_1 and f_2 differ in values just at let us say in the middle of the interval, thus at a point, this would not change the integral value, it would vanish, but f_1 and f_2 would be considered different. 6 There's no way to do justice to "Why is mathematics about real numbers?" within the length constraints of a Math.SE post, but here are some relatively philosophical observations and opinions (meant to be a bit provocative, in the spirit of answering a soft question). First, as multiple people have commented, the real numbers are not universally regarded as ... 1 You are pretty confused. They are used as such in the definition. When you substitute x = F^{n}(y), it changes to y - F^{-n}y. Since f and g commutes, their lifts too. Only that property is being exploited here. 2 Suppose that there exist (a,b,c,d,u,v) such that$$ad-bc=u^2+v^2a+d=2u$$where a,b,c,d\ge 0 with v\not=0. Since u=(a+d)/2, we have$$ad-bc=\left(\frac{a+d}{2}\right)^2+v^2,$$Multiplying the both sides by 4 gives$$4ad-4bc=a^2+2ad+d^2+4v^2$$i.e.$$-4bc=(a-d)^2+4v^2$$The LHS is non-positive and the RHS is positive. This is a contradiction. 1 Hint: Multiply the second equation by a (or d, it doesn't matter) and substitute into the first equation. 0 What have you tried so far? And, as for being one-to-one, can you think of examples where a function maps, say, every open interval to R? what would such a function look like? is it one-one? onto? 3 Writing x_n = \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}, we see:$$x_n>x_{n-1}\iff \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}>\frac{1-p^{n}}{1-p^{n-1}}\frac{n-1}{n}\iff(1-p^{n-1})(1-p^{n+1})n^2>(1-p^n)^2(n^2-1)\iff (1-p^n)^2 > n^2[(1-p^n)^2-(1-p^{n-1})(1-p^{n+1})]=n^2[p^{n-1}+p^{n+1}-2p^n]$$... 1 Since f is C^1, existence and uniqueness of a local solution is guaranteed by the Existence and Uniqueness Theorem (a.k.a. Picard's theorem,...) What you are asked to prove is that the solution is defined on [0,\infty). For this it is enough to prove that the solution does not blow-up, that is, it is bounded on any finite interval [0,T] (with a bound ... 0 The idea behind the proof is quite simple once you get read of the useless verbosity. You start by using the fact that \{n: |u_n-t|<1\} is infinite, take n_1 in it. Then you can use the fact that \{n: |u_n-t|<\dfrac12\} is infinite, take n_2 in it strictly bigger than n_1. Proceeding as this by induction, for each k you can choose n_k ... 0 @ChristopherCarlHeckman : it is a Laplace Transform (see remark at the end) Precisely, it is the Laplace Transform (LT) of a very important function, the cardinal sine (sinc). This LT can be found in most LT tables as$$\int_0^\infty {\sin(k)\over k}\exp(−sk)\,dk=\frac\pi2-\arctan(s)$$from which it is easy to deduce by an elementary change of ... 0 In every metric space (X,d) a sequence x_n converges to x if and only if every subsequence x_{n_k} has a further subsequence converging to x (easy proof by contradiction). All you need to now is thus that convergence in measure is convergence in a metric space (e.g. d(f,g)=\int \min\lbrace 1,|f(x)-g(x)|\rbrace \, d\mu(x) is a suitable metric on ... 2 Your equation can be changed to$$5(5^{x-1}-1)=24(3^{x-1}-2^{x-1})$$At least x=0, x=1 and x=3 are solutions. If your question is a Diophantine problem: With n_1,n_2\in\mathbb{N} you have the conditions$$3^{x-1}-2^{x-1}=5n_15^{x-1}-1=24n_2$$EDIT: x\in\mathbb{R} o.k. But with the values 0,1,3 you can check ... 3 We assume x>0 and s>0. Then by differentiating the following identity with respect to s,$$ f(s)=\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk $$one may write$$ f'(s)=-\int_0^\infty \exp(−sk)\sin(kx)\,dk=-\frac{x}{x^2+ s^2} $$giving$$ f(s)=-\arctan \left( \frac{s}x\right)+C.  Observing that, as $s \to \infty$, $f(s) \to 0$, we then obtain ...

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