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0

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.


0

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


0

$|b_n-a|\leq |b_n-a_n|+|a_n-a|=|a_{n+1}-a_n|+|a_n-a|\to 0+0$ 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.


0

That looks like the right answer here.


2

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 $\forall n\ge N$, $|b_n-\alpha|=|a_{n+1}-\alpha|<\epsilon$.


0

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 $q_i$ be 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)


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 ...


1

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 $f(z)−f(z_0)−w$ ...


1

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1

The proof works with any denominator because $L\gt 0$ yields $\frac{L}{n}\gt 0$ Chosing $2$ is just picking one that works


0

Not really an answer, but I can't comment. You won't be able to prove that $F$ is Lipschitz continuous. Consider the Cantor function to see an example of a CDF which is not Lipschitz continuous. http://en.wikipedia.org/wiki/Cantor_function#Properties


0

$f$ is uniformly continuous on $\mathbb{R}\iff\forall\epsilon>0,\exists\delta>0$ s.t. $\forall x,y\in\mathbb{R},(|x-y|<\delta\implies|f(x)-f(y)|<\epsilon)$ If not, then: $\qquad$ $\exists\epsilon>0$ s.t. $\forall\delta>0,\exists x,y\in\mathbb{R}$ s.t. $|x-y|<\delta$ and $|f(x)-f(y)|\ge\epsilon$. $\qquad$ Substitute $\delta=1/n$ ...


1

Given a bounded monotone function $f:\mathbb{R}\to\mathbb{R}$, $\lim_{x\to-\infty} f(x)=a$ and $\lim_{x\to\infty} f(x)=b$ exists, by monotone convergence theorem. Given a continuous function $f:\mathbb{R}\to\mathbb{R}$ s.t. $\lim_{x\to-\infty} f(x)=a$ and $\lim_{x\to\infty} f(x)=b$ exists, then $f$ is uniformly continuous. Proof ...



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