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Hint: First solve the question for functions defined on a closed interval, rather than open. Then think of an open interval as the (infinite) union of closed intervals.

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Hint: Consider $f(n)=n^2$ and $g(n)=n$.

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$|b_n-a|\leq |b_n-a_n|+|a_n-a|=|a_{n+1}-a_n|+|a_n-a|\to 0+0$ because $\{a_n\}$ is Cauchy as Mhenni told. With this way you don't mess up with epsilon definition and you can solve many problems like that. Just use the fact that a Cauchy sequence is convergent in complete metric spaces.

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That looks like the right answer here.

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Since $\{a_n\}\rightarrow \alpha$ we know that for all $\epsilon>0$ there exists $N_1\in \mathbb{N}$ such that $|a_n-\alpha|<\epsilon$ for all $n\geq N_1$. Let $N=N_1-1$. Then $\forall n\ge N$, $|b_n-\alpha|=|a_{n+1}-\alpha|<\epsilon$.

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Suppose that $a_n \rightarrow a$ as $n \to \infty$ we have that, Let $\epsilon >0, \exists N_1 \in \mathbb{N}$ such that $\forall n \geq N_1, |a_n -a|<\epsilon/2$. Also $\exists N_2 \in \mathbb{N}$ , such that for all $n,m \geq N_2$ $|a_n-a_m|=|a_n-a+a-a_m| \leq |a_n-a|+|a_m-a|<\epsilon/4+\epsilon/4<\epsilon/2$ So,for $n \geq \max \{N_1,N_2 \} ... 0 Just notice that since$\left\{a_n\right\} $converges then it is a Cauchy sequence. 0 The Algorithm: Input:$n=147,a=4258$Set$x_1=1$Set$x_2=0$Set$y_1=0$Set$y_2=1$Set$r_1=n$Set$r_2=a$Repeat until$r_2=0$: Set$r_3=r_1\bmod{r_2}$Set$q_3=\lfloor\frac{r_1}{r_2}\rfloor$Set$x_3=x_1-{q_3}\cdot{x_2}$Set$y_3=y_1-{q_3}\cdot{y_2}$Set$x_1=x_2$Set$x_2=x_3$Set$y_1=y_2$Set$y_2=y_3$Set$r_1=r_2$Set$r_2=r_3$If$y_1>0$... 1 Actually each remainder in the euclidean algorithm satisfies Bézout's identity. Let's start with$r_0=4258, r_1=147$. If$r_{i+1} $is the remainder at the$i$-th step (dividing$r_{i-1}$by$r_i$), write$r_i=u_i \cdot 4258+v_i\cdot147$. Let$q_ibe the corresponding quotient. The algorithm translates into the relations: $$u_{i+1}=u_{i-1}-q_iu_i,\qquad ... 1 In your back substitution you're collapsing some things too much. In the second back-sub line you should have 1 as a combination of 5 and 142. That is$$1=5-2\cdot\left(142-28(5)\right)=5\cdot(57)-142\cdot(2)$$. In your next line you should sub in 5=147-142 and get 1 as a combination of 147 and 142. And so on. 1 Consider the function:$$ f(x)=\sum_{k=1}^n x^k=\frac{1-x^{n+1}}{1-x} $$Then:$$f'(x)=\sum_{k=1}^n kx^{k-1}=\frac{1}{x}\sum_{k=1}^n kx^{k}=\frac{{x}^{n}\,\left( n\,x-n-1\right) +1}{{x}^{2}-2\,x+1} $$So:$$\sum_{k=1}^n kx^{k}=\frac{{x}^{n+1}\,\left( n\,x-n-1\right) +x}{{x}^{2}-2\,x+1} $$(you could obtain something equivalent to the above just by asking ... 0 There's no need for induction. Note that when you start increasing any x_i, the left side increases slower (or equally) because the product of the other factors besides (x_i+1) is less or equal to 1. So without loss of generality, all the variables are 0 and the inequality holds. 2 Why not directly instead of cases?$$\prod_{k=1}^n(1+x_k)=\prod_{k=1}^{n-1}(1+x_k)\cdot(1+x_n)\stackrel{\text{ind. hyp.}\,+\,(1+x_n)\ge0}\ge(1+x_1+\ldots+x_{n-1})(1+x_n)==1+\sum_{k=1}^n x_k+\sum_{k=1}^{n-1}\overbrace{x_kx_n}^{\ge 0}\ge1+\sum_{k=1}^n x_k$$and we're done 4 We know that$$(1+x_1)\cdots(1+x_n) \geq 1+x_1+\ldots +x_n$$If we now multiply the inequality above by (1+x_{n+1}) \geq 0 we obtain$$(1+x_1)\cdots(1+x_n)(1+x_{n+1}) \geq (1+x_1+\ldots +x_n)(1+x_{n+1})\\= (1+x_1+\ldots +x_n) + x_{n+1} + x_{n+1}(x_1+\ldots +x_n) \geq 1 + x_1 + \ldots + x_{n+1}$$where the last inequality follows from the fact that x_i ... 0 Some remarks concerning your proof: You write: Since x \in X is an adherent point, then there exists a sequence (a_n) such that a_n \in X and converges to x. Note that, by definition, we have to consider any two sequences (a_n)_n, (b_n)_n in X which are equivalent. This means in particular that (a_n)_n need not be convergent and it ... 1 You found |b_n-\beta|<\beta-B. Writing this in expanded form gives -\beta+B<b_n-\beta<\beta-B. We add \beta to both sides giving B<b_n. 1 Remember \lvert b_n-\beta\rvert=\max\{b_n-\beta,\beta-b_n\} so the inequality \lvert b_n-\beta\rvert <\beta-B implies \beta-b_n<\beta-B, which implies b_n>B, so that B wouldn't be an upper bound for b_n. 1 Let A be an element of F\setminus F_n (i.e., A has at least n+1 elements and may even be infinite). Pick n+1 distinct points a_0,\ldots, a_{n}\in A and let r=\min_{0\le i<j\le n}p(a_i,a_j)>0. Now let B be any element of F_n. Then for any b\in B there exits at most one i with p(a_i,b)<\frac r2, hence by pigeon-hole there ... 0 To prove that F_n is closed. Take an element K not in F_n, then we can take n+1 points x_1,x_2,...,x_{n+1} in K. Let 2r be the minimum distance between any two of these n+1 points. Then$$B(K,r)$$is disjoint from F_n. The reason is that any element J of B(K,r) ought to have a point in the n+1 balls (in X) with centers ... 2 Since a_n and b_n are equivalent sequences, we can say that for every rational 1 \geq \epsilon >0, there exists N \geq 0 such that |a_i-b_i| < \epsilon for all i \geq N. ...Then b_n for n \geq N will be bounded by M' := M+1. So b_n is bounded by \max \{|b_1|,\dots,|b_{N-1}|, M+1\} 0 Clearly we must have x_3<4, x_2<16, and x_1<256, so the largest possible value of x_1 is 255. Since x_3>1, we know that x_3\ge 2 and hence that x_2\ge 4. This implies that x_1\ge 16. Thus, the range is from 16 through 255, the difference being 239. 1 We can work backwards more easily than we can work forwards. Firstly, what does a number x need to satisfy to have \lfloor x \rfloor = 1? Easy. We need:$$1\leq x < 2.$$Well, suppose that \sqrt{y}=x or, equivalently, y=x^2. Well, obviously we just square the above equation (as all of its terms are positive):$$1\leq x^2 < 2^2.$$Suppose ... 0 Since g is continuous on [0,1], it is bounded by some number K. Given \varepsilon > 0, let N be a positive integer greater than or equal to \frac{K}{\varepsilon}. Let n \ge N and x\in [0,1]. By the mean value theorem, |\sin x| \le x, hence |\sin^n x| \le x^n. Therefore$$|f_n(x)| \le \frac{|g(x)|x^n}{1 + nx} \le K\frac{x^n}{1 + nx}.$$... 1 Let \varepsilon > 0. Since g is continuous with g(1) = 0, there exists a \delta, 0 < \delta < 1 such that for all x, 1 - \delta < x \le 1 implies |g(x)| < \varepsilon. Let N be a positive integer greater than \log(\varepsilon)/\log(1 - \delta). If n\ge N, then 1.|f_n(x)| \le (1 - \delta)^n < \varepsilon if x\in [0, ... 0 First, if g(x)=0 for all x\in [0,1], then there is nothing to prove. Now, because the function g is continuous, then it sends compact set to compact set, so g([0,1]) is compact, hence its bounded; from this (and from the fact that g(1)=0) it follows that for every x\in [0,1], lim_{n} f_n(x)=0. Hence, f_n converge pointwise to the zero ... 2 Let B=\{e_\lambda\mid\lambda\in \Lambda\}\subset\mathbb{R} be a \mathbb{Q}-basis of \mathbb{R}. It will be helpful for you to pick out a countable-dimensional \mathbb{Q}-subspace V of \mathbb{R} with basis B'=\{f_1,f_2,\ldots\} where f_i=e_\lambda for some lambda and f_i\neq f_j for all i\neq j. Can you find a surjective ... 2 If x\in\pi\mathbb{Z} we are just summing zeroes, so there is little to prove. If x\not\in\pi\mathbb{Z}, from:$$\sum_{n=1}^{N}\sin(nx)=\frac{\sin\frac{Nx}{2}\sin\frac{(N+1)x}{2}}{\sin\frac{x}{2}}\in\left[-\frac{1}{2}\tan\frac{x}{4},\frac{1}{2}\cot\frac{x}{4}\right]$$and the fact that \frac{H_n}{n} is eventually decreasing and converging to zero, we ... 2 Note that$$\frac{n}{(n+1)^{2}}-\frac{1}{(n+2)}=\frac{n(n+2)-(n+1)^2}{(n+1)^{2}(n+2)}=\frac{n(n+2)-(n+1)^2}{(n+1)^{2}(n+2)}=-\frac{1}{(n+1)^{2}(n+2)}$$Now, since \dfrac{1}{(n+1)^{2}(n+2)}<\dfrac{1}{n^3} when n\ge 1 ... 1$$\displaystyle \sum_{n=0}^{\infty}\left(\frac{n}{(n+1)^2} - \frac{1}{n+2}\right) = \sum_{n=0}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+2}\right)-\sum_{n=0}^{\infty}\frac{1}{(n+1)^2}$$Now it should be obvious. This also gives:$$\displaystyle \sum_{n=0}^{\infty}\left(\frac{n}{(n+1)^2} - \frac{1}{n+2}\right) = \color{red}{1-\frac{\pi^2}{6}}.$$0 In general: if a_n = O(b_n) (Check this is you case) so either both diverge or converge (this follows from comparison test) 5 Use this inequality$$\frac{n+1}n |a_n|\le 2|a_n|$$2 Since the series$$\sum_{n\ge0} a_n$$is divergent then the Radius R_a of the given power series is less or equal 1. Moreover since a_n\le A_n then R_a\ge R_A where R_A is the radius of convergence of$$\sum_{n\ge0}A_nx^n$$Finally we have$$\frac{A_{n}}{A_{n+1}}=1-\frac{a_{n+1}}{A_{n+1}}\xrightarrow{n\to\infty}1$$so by the ratio criteria we ... 0 Ah sorry you're right. Replace g(x) by g(x)=-x^4. I understand the test in that way, that one has to consider f+g if g is bounded from above. f cannot be bounded from above by coerciveness. Edit: I wanted to post it under your comment. 1 You have just proved limit exists for a specific convergent sequence x_n\to x_0, you have to show then the limit is independent of the convergent sequence. That is if x_n\to x_0,y_n\to x_0, then \lim f(x_n)=\lim f(y_n), which should be easy if you make use of the uniform continuity of f on X and pass to limit. 2 That is not possible. If$$ h(s) = f(g(s)) = f(\sqrt[3]s) $$is differentiable at zero then$$ f(x) = h(x^3) if differentiable at zero by the chain rule. 2 \frac{x^2+x+1}{4} \leq \frac{3}{4} on the given interval, and \sin(nx) is bounded, so the integrand converges uniformly to zero and so the limit is zero. 1 Your polynomial is of Brieskorn-type, i.e., f = \sum_{i = 1}^n z^{a_i}_i, so by a formula due to Pham, \mu(f) = \prod_{i = 1}^n (a_i - 1). Thus, the answer to your question is 16 not 5. To compute the Milnor number using the local algebra approach, as you mention, observe \begin{align} \langle \partial_x f, \partial_y f \rangle = \langle 1, x, x^2, x^3, ... 1 The identity just follows from:\frac{t e^{t/2}}{e^t-1}+\frac{t}{e^t-1}=\frac{t}{e^{t/2}-1}=2\cdot\frac{t/2}{e^{t/2}-1}.$2 A preliminary comment: a mathematical proof is a piece of expository prose, consisting of (paragraphs of) sentences. You’ll be much clearer if you use more words and fewer symbols for the ‘connective tissue’ of your argument — things like if ... then, therefore, etc. Starting out by supposing (to get a contradiction) that there are points$x,y$, and$m$... 1 The key tool here is the mean value theorem. Let$h \in \mathbb{R}^m$such that$\|h|| = 1$and consider$\phi_h(t) = g(t)-g(0)-\left( h^T (f(t)-f(0)) \right)$. Note that$\phi_h(0) = 0$and$\phi_h$is differentiable and$\phi_h'(t) = g'(t)-h^T f'(t)$. Note that$g'(t) \ge 0$and so the mean value theorem shows that$g(t) \ge g(0)$for all$t \in [0,1]$... 1 I was actually going to ask the same question... and in particular if the result would follow as the consequence of any hard, still open conjecture. From the MO thread mentioned by lhf (not the same as the one mentioned by mixedmath) I found out that Schanuel's conjecture would imply it. On the Mathworld page for$e$there's a bit of info on numerical ... 1$\newcommand{\Reals}{\mathbf{R}}$In the usual sense of "depend",[*] no, the Gauss curvature, mean curvature, and shape operator of a (locally oriented) regular surface in$\Reals^{3}$do not depend on parametrization; that's what's meant by saying these are "geometric" data. :) Depending on your definition of the shape operator (e.g., O'Neill's: If$U$is a ... 0 Thanks to all who answered here. I thought of a solution, and whether it is correct or not, I leave it for you to judge.$F$is increasing, continuous, so consider any sequence$\{x_n\}\subset[0,1]$such that$x_n\to0$. Then, we will get a positive divergent sequence$\{L_n\}$such that$[x_n,1-x_n]\subset F([-L_n,L_n])$. So$F(L_n)\geq1-x_n$and ... 0 The def of limit "applied" to the above case is :$\lim_{x\to -\infty}f'(x) = L \ $iff$ \ \forall \epsilon > 0 \ \exists N < 0 \ (x < N \rightarrow |f'(x) - L| < \epsilon)$. Thus, assuming for contradiction that$L > 0$, we have that : for all$x < N : - \epsilon < f'(x) - L < \epsilon$, i.e. :$L - \epsilon ...

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Inside a small circle centered at $z_0$, $f(z)-f(z_0)-w$ has at least two zeros $z_1, z_2$.(Note that if $w\ne 0$, then $z_1\ne z_0, z_2\ne z_0$.) If $z_1=z_2$, then $f(z)-f(z_0)-w=(z-z_1)^2h(z)$.This means $f^\prime (z_1)=0$, which contradicts the fact that $f^\prime (z)≠0$ for all $z≠z_0$ but sufficiently close to $z_0$. Thus the roots of $f(z)−f(z_0)−w$ ...

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