New answers tagged

0

Notice simply that $$ \left\|e^{At}x_0 + \int\limits_0^t e^{A(t-s)}b(s)ds\right\|\le \|e^{At}\|\left(\|x_0\| + \int\limits_0^\infty \|e^{-As}\|\cdot\|b(s)\|ds\right). $$ Since $\|e^{At}\|$ tends to zero exponentially when $t\to\infty$, the expression inside parenthesis is bounded (because $b$ is bounded) and so the right-hand side tends to zero exponentially ...


1

The idea: if $$f(x)\approx f(\hat x) + f'(\hat x)(x-\hat x)\qquad\text{and}\qquad g(x)\approx g(\hat x) + g'(\hat x)(x-\hat x)$$ then $$f(x)g(x)\approx f(\hat x)g(\hat x) + g(\hat x)f'(\hat x)(x-\hat x) + f(\hat x)g'(\hat x)(x-\hat x) + f'(\hat x)(x-\hat x)g'(\hat x)(x-\hat x) $$ with the last term much smaller than the other terms.


0

Looking at each $x_i$ individually, we know that the derivative of $x\mapsto |x_i-x|$ is $x \mapsto \left\{ \begin{array}{ll} -1 & (x< x_i) \\ 1 & (x> x_i) \\ \end{array} \right.$ Therefore, we know that the solution isn't at $]-\infty,x_i[$ or at $]x_i,+\infty[$, as $|x_i-x|$ will grow indefinitely in those regions. Let's ...


0

Your weak formulation is $\int\limits_{\Omega}{\nabla u\cdot\nabla vdx}+\int\limits_{\Omega}{Vuvdx}=\int\limits_{\Omega}{fvdx},\,\forall v\in H_0^1(\Omega)$. So if you denote $$a(u,v)=\int\limits_{\Omega}{\nabla u(x)\cdot\nabla v(x)dx}+\int\limits_{\Omega}{V(x)u(x)v(x)dx}$$ you have $$a(u,u)\ge C\|u\|_{H_0^1(\Omega)}+\int\limits_{\Omega}{Vu^2dx}\ge ...


1

Squeeze theorem (like the OP wrote) we know that that $lim_{y\to 0} {siny\over y} = 1$ Using the epsilon delta definition (am going to use it in for $lim_{y\to 0} {siny\over y} = l, l \in \mathbb R$) gives the following: Given a $\epsilon$ > 0 $\exists \delta$ such that $|{siny\over y} - l| < \epsilon$ for y $\in$ $[y_0-\delta, y_0+\delta], l \in ...


1

Let $u = x \ln x$. We have $$\lim_{x \to 0^+} u = 0,$$ and $$\lim_{u \to 0}\frac{\sin u}{u} = 1.$$ You can prove both limits with L'hôpital's rule. For the first one, apply the rule to $\ln x/(1/x)$. Therefore $$\lim_{x \to 0^+}\frac{\sin(x \ln x)}{x \ln x} = 1.$$


0

Expanding on André Nicolas's comment, $\begin{array}\\ (x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2 &=((x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2) \dfrac{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{(x_nn^4 + y_nn^2)- x_nn^4}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{ y_nn^2}{(x_nn^4 + y_nn^2)^{1/2} + ...


0

For 2.: Around each point $t\in[0,1]$ such that $f(t)\in V:=\mathbb{R}^2\setminus(F_1\cup F_2)$, you can take a maximal open interval $(c_1,c_2)$ such that $f((c_1,_2))\subseteq V$ (since $V$ is open). By maximality, we have that $f(c_1),f(c_2)\in F_1\cup F_2$. Now take $a$ to be the $\sup$ of all the endpoints $c$ such that $f(c)\in F_1$. This must be less ...


2

Since $\cos(1/\|x\|)$ approaches all values of $[-1,1]$ when $x\to0$, it is necessary and sufficient for differentiability at $0$ that $(\sin\|x\|)^p/\|x\|\to0$ when $x\to0$. But you know that $$\frac{(\sin\|x\|)^p}{\|x\|^p}\to1$$ when $x\to0$. Since $$ \frac{(\sin\|x\|)^p}{\|x\|}=\frac{(\sin\|x\|)^p}{\|x\|^p}\|x\|^{p-1}, $$ you find readily that it ...


0

It is not possible to have all the minimums from the second part. As you asked I am just giving hint. To have all the minimums from the second part the following inequalities hold $$ t-\mid x-y\mid\leq \mid x-y\mid\\ t-\mid x-z\mid \leq \mid x-z\mid \\ t-\mid z-y\mid\leq \mid z-y\mid $$ Now these three imply all differences and $x,y,z\in[0,t)$ would give ...


1

Note that if you replace $x^2$ by $x'$ and $y^3$ by $y'$, then the expression reduces to $\frac{x'y'}{x'^2 + y'^2}$. If you now take $x' = y'$, then you get $1/2$. This shows that the limit does in fact not exist, as you found correctly that some approaches lead to $0$. If you do not like the replacement: when you approach via $(t^3, t^2)$ you get ...


2

There is no limit. Along the semicubical parabola $x=t^3$, $y=t^2$ the limit is equal to 1/2. S.G.


0

Hint: Let $\mathcal P _{\mathbb Q}$ be the set of polynomials with rational coefficients. Let $f \in C^1.$ Then as you probably know, we can approximate $f'$ in $C^0$ by polynomials in $\mathcal P _{\mathbb Q}.$ We should then be able to approximate $f$ in $C^1$ polynomials of the form $$r+ \int_0^x p(t)\,dt,\,\,r\in \mathbb Q, \, p \in \mathcal P _{\mathbb ...


0

For ones who read German, I strongly recommend Harro Heuser's 'Lehrbuch der Analysis Teil I'. There is also 'Teil II'. I tried couple of other German text books, but gave up continuing due to many errors or lack of completeness, etc. Then a person recommended me this book. This book is self-contained and proofs are quite error-free as well as well-written ...


1

Starting like A. Jiménez, $$D_uf(0,0) = \frac{u^2_1u_2}{u_1^2+u_2^2}.$$ But can be easily proved that if $f$ is differentiable in $(0,0)$: $$D_uf(0,0) = Df(0,0)u,$$ and this is impossible because in this case $D_uf(0,0)$ is a nonlinear funcion of $u$.


1

If it were Frechet, then there would be a (bounded) linear operator $A$ so that $$ \|v\|=A(v)+o(\|v\|). $$ Now insert $-v$ and compare.


5

To check whether a function $f: \Bbb{R^2}\to \Bbb{R}$ is differentiable at $(0,0)$, we need to find a linear map $Df(x_0)$, called differential. Now, let's take $u=(u_1,u_2)\in \Bbb{R^2-\{(0,0)\}}$. The directional derivatives along $u$ are: $$D_uf(0,0)=\lim_{t\to 0}\frac{f(tu_1,tu_2)-f(0,0)}{t}=\lim_{t\to 0}\frac{f(tu_1,tu_2)}{t}=\lim_{t\to ...


1

Write functions in $C^1$ as $$ f(x)=f(0)+\int_0^xf'(t)dt $$ with $f'\in C^0$ to find that this provides a parametrization of $C^1$ by $\Bbb R×C^0$. Show that this is an homeomorphism. Only then can you use the separability of $C^0$.


0

The absolute value of the expression is $$\left |\frac{\arctan (xy)}{\sqrt {x^2+y^2}}\right| = \left |\frac{\arctan (xy)}{xy}\right|\cdot |y|\cdot \left |\frac{x}{\sqrt {x^2+y^2}}\right |\le \left |\frac{\arctan (xy)}{xy}\right|\cdot |y|\cdot 1.$$ Because $\lim_{u\to 0}(\arctan u)/u =1$ and $|y|\to 0,$ the desired limit is $0.$


2

From where you left off: Assume $$n \geq m \Rightarrow |x_n-x_m| \leq M\left(|q|^{m+1}+|q|^{m+2} + ...+ |q|^n\right)= M|q|^{m+1}\left(1+ |q|+ |q|^2+...+|q|^{n-m-1}\right)= \dfrac{M|q|^{m+1}\left(1-|q|^{n-m}\right)}{1-|q|} < \dfrac{M|q|^{m+1}}{1-|q|}$$. Thus let $\epsilon > 0$ be given, and choose N$ \in \mathbb{N}$ such that if $m > N$ we have: ...


2

Use polar coordinates and L'Hospital's rule: $$\lim\limits_{(x,y) \to (0,0)} \frac{\arctan(xy)}{\sqrt{x^2+y^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{\sqrt{(r \cos(\theta))^2+(r \sin(\theta))^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{r} = \lim\limits_{r \to 0} \frac{2r \cos(\theta) ...


3

$-\frac{x^2+y^2}{2}\leq xy\leq \frac{x^2+y^2}{2}$ and $\arctan$ is a monotonic increasing function, so $$\frac{\arctan\frac{-r^2}{2}}{r}\leq \frac{\arctan{(xy)}}{\sqrt{x^2+y^2}}\leq \frac{\arctan\frac{r^2}{2}}{r}$$ where $r=\sqrt{x^2+y^2}$. Now, using L'Hospital's rule you get that both limits in the sandwich are $0$. Or, you can use the inequality Oliver ...


7

Observe that you have $$ \left|\arctan u\right|\leq|u|, \quad |u|\leq1, $$ then, switching to polar coordinates with $r=\sqrt{x^2+y^2}$, as $r\to 0$, you get $$ \left|\frac{\arctan(xy)}{\sqrt{x^2+y^2}}\right|=\frac{\left|\arctan(r^2 \sin \theta \cos \theta)\right|}{r}\leq \frac{\left|r^2 \sin \theta \cos \theta\right|}{r}\leq r. $$ The sought limit ...


6

A set $U\subset \mathbb R$ is open if and only if for every $x\in U$, there exists some $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon)$ is a subset of $U$. For $U=\mathbb Z$, this is clearly not the case: Take $x=0$ Take any $\epsilon > 0$. Then, $\min\{x+\frac\epsilon2, x+\frac12\}$ is an element of $(x-\epsilon, x+\epsilon)$, but it is not an ...


2

$\mathbb{Z}$ is not open in $\mathbb{R}$. One way to see this is that given any $n\in \mathbb{Z}$ we have for every $\epsilon>0$ that $(n-\epsilon, n+\epsilon)$ is not contained in $\mathbb{Z}$.


1

We have that If a sequence of functions $(f_n)_{n\geq 1}$ converges pointwise to a function $f$ and uniformly to a function $g$, then $f=g$. and If a sequence of continuous functions $(f_n)_{n\geq 1}$ converges uniformly to a function $f$, then $f$ is continuous. Prove convergence pointwise to the function $f\colon[-1,1]\mapsto ...


1

The two are unrelated mathematical structures. However, every topological space can be given a particular useful $\sigma$-algebra, called the Borel $\sigma$-algebra. This is the smallest $\sigma$-algebra containing all the open sets of the space. One useful property is that continuous real-valued functions are measurable with respect to this ...


1

Notice that the series is defined as $$\sum_{n=0}^{\infty}{a_n} = \lim_{n \to \infty} \sum_{k=0}^{n}{a_k}.$$ Your example does not fit into the definition, because you would have $$a_k = \binom{n}{k}x^k$$ and your $a_k$ would depend on $n$, but $n$ is different for different partial sums. So this is not description of infinite series and you cannot use ...


0

I am proving this result for positive rationals; proof for negative rationals will be similar. For convenience, I am using $(a)$ to denote the fractional part of $a$. Let $\frac{p}{q}$ be a rational. Let $r=\frac{p}{q}$. Now we write the function as $$f(r)=\sum_{n=1}^{\infty}\frac{(nr)}{n^2}=\sum_{n:\ q\mid n}\frac{(nr)}{n^2}+\sum_{n:\ q\nmid ...


0

(I don't think you have to use Weierstrass M-test, since one cannot conclude that a series is not uniformly convergent from that test) Hint: Calculate the series, both for $x\neq 0$ and $x=0$. Do you have any knowledge of what kind of function a series of continuous functions that converge uniformly must converge to? Update If you have a series of ...


0

Yet another formulation for topological spaces: If $f:X \to Y$ continuous and $f(x_\iota)$ is a net in $f(X)$, then $x_\iota$ has a converging subnet, say $x_\tau \to x$. Then $f(x_\tau) \to f(x)$, hence $f(x_\iota)$ has a converging subnet, so $f(X)$ is compact.


0

I was recently in your situation and encountered this book http://www.amazon.com/Calculus-Algebra-Differential-Unified-Approach/dp/0971576653/ref=sr_1_1?ie=UTF8&qid=1455361217&sr=8-1&keywords=vector+calculus+linear+algebra+and+differential+forms+a+unified+approach It is written at a beginner level and is very easy to read for some with your ...


0

Starting from $e^x\ge 1+x$ and thus also $e^{-x}\ge 1-x$ one gets $$ -\ln(1-x)\ge x \ge \ln(1+x) $$ Now set $x=\frac1n$ to get $$ \ln(n)-\ln(n-1)\ge\frac1n\ge\ln(n+1)-\ln(n) $$ The left and right terms are telescoping in sums, so $$ \ln(n)-\ln(m)\ge\sum_{k=m+1}^n\frac1k\ge\ln(n+1)-\ln(m+1) $$ Since $\ln(n)\to\infty$, this shows that the sequence of partial ...


0

For your first question, $C_c(X)$ is a subspace of $L^1(\mu)$, and $\phi$ is dominated by the $L^1(\mu)$ norm on $C_c(X)$. By the Hahn-Banach theorems, $\phi$ extends to some continuous linear functional $\Lambda$ on $L^1(\mu)$. For any $f\in L^1(\mu)$, we can find a sequence $\{f_n\}$ in $C_c(X)$ such that $\|f_n-f\|_1 \rightarrow 0$ as $n\rightarrow ...


1

However, since there exists an element $cx \in \{xn\}$, and $cx>c$, it is a contradiction. This is false, and I cannot figure out how you contrived this claim. $\def\nn{\mathbb{N}}$ Instead, since $c = \sup(\{x_n:n\in\nn\}) \ge x > 1$, let $y \in \{x_n:n\in\nn\}$ such that $y>\frac{c}{x}$ since $\frac{c}{x} < c$. Then $xy > c$, which is ...


2

Note first that $x>0$. Moreover $p\neq 0$ because this is only possible when $x=1$, which would not comply with the 3 roots assumption. Next, we have $$ p = \frac{|\log (x)|}{x} $$ When $0<x<1$, the right hand side derivative w.r.t $x$ is $$ \frac{\log (x)-1}{x^2} $$ Now, $\log(x)$ is negative, thus this function is increasing to the left of ...


0

According to the hint in "Problems in Mathematical Analysis", Biler and Witkowski, problem 4.112. Define $g_n$ as the function which is $0$ on the intervals $(-\infty, 0]$ and $[2/n,\infty)$, $1$ at $1/n$, and linear otherwise. Then $g_n$ converges pointwise to $0$ but is not uniform in any interval containing $0$. Now define $$f_n(x) = ...


1

It helps to prove that a function is continuous in that point iff the oscilation is zero. Also you can aproach as much as you want to the discontinuity, and measure the skip.


1

Here is another approach. Since $x >1$, we have $x=1+t$ for some $t>0$. Then $x^n = (1+t)^n = \sum_k \binom{n}{k} t^k \ge 1 + n t$. Since $1+nt \to \infty$, we have the desired result.


5

It's the closure of $C^\infty_c(\Omega)$ with respect to $ ||.||_{1,2}$ and is thought of as the space of Sobolev functions with zero boundary value. As closure of a linear subspace in a Hilbert space it is a Hilbert space, too. See, e.g., this Wikipedia entry


0

YES. An infinite sum of irrational numbers can be rational. PROOF: Let the set A be all the positive irrational numbers and the set B be the negative irrational numbers. Take each positive irrational number and add it to the matching negative irrational number to get 0. The sum of all these 0 numbers is 0 which is a rational number.


2

They are equivalent. Suppose your first statement holds, then $\forall y_1,y_2 \in S$, $$ d(y_1,y_2) \leq d(x,y_1)+d(x,y_2) < 2r. $$ Conversely, suppose your second statement holds, then $\forall x \in S$, $S\subset B_x(2r)$. Your second definition is more commonly seen in literature. There is, however, a different definition of boundedness in general ...


6

We use the definition of Cauchy sequence to show the sequence $(a_n)$ is not Cauchy. Let $\epsilon=1/2$. We will show that there does not exist an $m$ such that for any $n\gt m$, we have $|a_n-a_m|\lt \epsilon$. For let $m$ be given, and let $2^k$ be the smallest power of $2$ that is $\gt m$. Let $n=2^{k+1}-1$. Then $$a_n-a_m\ge ...


0

Hint: Showing it doesn't converge (specifically that it goes to infinity) would help.


2

The problem is, the proof is proceeding by contradiction. That is why it says "Then ${m_1+1\over N}>b$" -- because it's assumed there's no number of the form $m/N$ between $a$ and $b$. It then proves that this leads to a contradiction, just as you proved that in fact $(m_1+1)/N$ in your situation was actually $<b$ and not $>b$ as assumed. The ...


1

I am assuming you want $V$ to actually be the image of $U$. In this case, there is no such map satisfying your second condition. If $m = n$, this follows from invariance of domain, since the image of $U$ will necessarily be open. If $m < n$, there is no continuous injective map from $\mathbb{R}^n$ to $\mathbb{R}^m$ (let alone to a closed subset). You ...


0

If possible , let, f is not uniformly continuous.Then there exists a Cauchy sequence {$x_n$} in $\mathbb R$ such that {$f(x_n)$} is not a Cauchy sequence. (If not, i.e for every Cauchy sequence {$x_n$}, {$f(x_n)$} is a cauchy sequence, then f is uniformly continuous, which contradicts our hypothesis) So there exists $\epsilon > 0$ corresponding to a ...


1

If you want to analyze differentiability away from $(0,0)$, you need not worry about the $f(0,0)$ part, just about the formula. In general, writing $x=(x_1,x_2)$ (I'm using notation different from your $(x_1,y_1)$ to make indexing easier, so I can compute all the partial derivatives at once), we have that: $$\frac{\partial}{\partial x_i}(\|x\|) = ...


1

You shouldn't start with what you're trying to prove. Note that the result you're trying to prove is intuitively obvious if you divide both sides by $(z-y)(z-x)$ : The average value over $[y,z]$ is larger than the average value over $[x,z]$ because the function is bigger over $[x,y]$ than $[y,z]$. Basically this is like saying the average of 2, 4, and 5 is ...


1

Here is why there always exists a sequence in $K$ converging to $\sup K$, provided its existence (which is guaranteed by the fact that bounded sets always have a least upper bound in $\Bbb R$). Let $A\subseteq \Bbb R$ and let $s=\sup A<+\infty$. If there were a $n>0$ such that $\left(s-\frac1n,+\infty\right)\cap A=\emptyset$, then $s-\frac1{2n}$ ...



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