New answers tagged

0

If the step distance $k$ is a rational multiple of $π$, the stepper will (eventually) reach the same points again, and in fact will do so with period $lcm(a,b)$ for $k=a×π/b$ as a necessary outcome of the way steps are made. Any other step distance may eventually come arbitrarily close to any previously visited point, but will never actually reach any of ...


0

This answer complements the solution presented by Ron Gordon. Note that we have $$\begin{align} I&=\int_0^{2\pi}\int_0^\pi \sin(\theta)e^{\sin(\theta)(\cos(\phi)-\sin(\phi))}\,d\theta\,d\phi \tag 1\\\\ &=\int_0^{2\pi}\int_0^\pi \sin(\theta)e^{\sqrt{2}\sin(\theta)\cos(\phi)}\,d\theta\,d\phi \tag 2\\\\ &=\oint_{|r|=1}e^{\sqrt{2}x}\,dS \tag 3\\\\ ...


0

$$\frac1{1+z^2}=\frac1{2i}\left(\frac1{i-z}+\frac1{i+z}\right)=\frac1{2i}\left(-\frac1{1-i+(z-1)}+\frac1{1+i+(z-1)}\right)=$$ $$=\frac1{2i}\left(-\frac1{1-i}\frac1{1+\frac{z-1}{1-i}}+\frac1{1+i}\frac1{1+\frac{z-1}{1+i}}\right)=$$ $$=-\frac1{2+2i}\sum_{n=0}^\infty(-1)^n\frac{(z-1)^n}{(1-i)^n}-\frac1{2-2i}\sum_{n=0}^\infty(-1)^n\frac{(z-1)^n}{(1+i)^n}=$$ ...


0

Hint: if $f$ takes on an interior maximum in some point, then in that point ($\nabla f=0$) and the second derivative is $\le 0$. Assuming you know a little bit of linear algebra, this has an implication for the trace of the second derivative, which is what you have on the left hand side of your PDE.


0

I've just managed to prove it thanks to what André Nicolas said: $$\begin{align}a\le b &\implies 0\le b-a\\ &\implies 0\le (\sqrt{b}+\sqrt{a})(\sqrt{b}-\sqrt{a}) \\ &\implies 0\le \sqrt{b}-\sqrt{a}\quad\cdots\quad (1)\\ &\implies \sqrt{a}\le \sqrt{b}. \end{align}$$ $(1)$ must be true because $(\sqrt{b}+\sqrt{a})$ is already positive or zero. ...


0

Supposing that not all the $c_k$ are $0$, let $n_0=\min{\{n\geq 0, c_n\neq 0\}}$ Then $f(x)=\sum_{k=n_0}^\infty c_kx^k=x^{n_0}(c_{n_0}+\sum_{k=n_0+1}^\infty c_kx^{k-n_0})$ The power series $g(x)=\sum_{k=n_0+1}^\infty c_kx^{k-n_0}$ has the same radius as $f$ and $g(0)=0$. $g$ being continuous at $x=0$, there is a neighborhood $V$ of $0$ such that ...


0

Hint: $f$ is continuous for $|x|<R$ - conclude that $c_0=f(0)=0$. It is also differentiable there, and $f'(x)=\sum_{n=0}^\infty (n+1)c_{n+1}x^n$. - Use Rolle to find a sequence of zeroes of $f'$ converging to $0$.


0

Lemma. for $f(x)=x^k$. if $k\geq 1$ then for $a,b>0$,then $a^k+b^k\leq (a+b)^k$. if $0\leq k\leq 1$, then $a^k+b^k\geq (a+b)^k$. proof.when $k\geq 1$, then we have $(\frac{a}{a+b})^k\leq\frac{a}{a+b}$ and $(\frac{b}{a+b})^k\leq\frac{b}{a+b}$ for $\frac{a}{a+b}\leq 1$, $\frac{b}{a+b}\leq 1. $ Hence:$(\frac{a}{a+b})^k+(\frac{b}{a+b})^k\leq ...


0

Since the dual of $(L^\infty,\sigma(L^\infty,L^1))$ is $L^1$ (more precisely, every element of the dual is of the form $f\mapsto \int fg$ for some $g\in L^1$) the Hahn-Banach theorem tells you what to show: every $g\in L^1$ with $\int fg =0$ for all $f\in C_c^\infty$ is $0$. This can be done by approximating (w.r.t the $L^1$-norm) the sign $|g|/g$ by smooth ...


0

Roots of your polynomial are: $$x_1 = \frac{1-\sqrt{1-4C\varepsilon}}{2}, x_2 = \frac{1+\sqrt{1-4C\varepsilon}}{2}.$$ If $\varepsilon > \frac{1}{4C}$, you have two complex conjugate roots and $x^2 - x + C\epsilon > 0 ~\forall x$. But in this case $\varepsilon$ is not sufficiently small. If $\varepsilon \leq \frac{1}{4C}$, you have two real roots ...


1

Rewrite the integral as $$\int_1^x dt \, \exp{\left [t \ln{\left ( 1+\frac1{t} \right )}\right ]} $$ Because $t \gt 1$ we can Taylor expand the log term to get an approximate expression for the integral: $$\int_1^x dt \, e^{\displaystyle 1-\frac1{2 t} + \frac1{3 t^2}-\cdots} = e \int_1^x dt \, e^{\displaystyle -\frac1{2 t} + \frac1{3 t^2}-\cdots} $$ ...


4

I will evaluate this integral indirectly by recognizing it for what it is: an integral over the surface of a unit sphere. First of all, for simplicity sake, I am going to replace your notation so it looks like we are integrating over solid angle: $$\int_0^{2 \pi} d\phi \, \int_0^{\pi} d\theta \, \sin{\theta} \, e^{\sin{\theta} (\cos{\phi}-\sin{\phi})} $$ ...


1

Let $\lambda$ denote the measure on $\mathbb R$ which is being used (I assume it is the Lebesgue measure, but this argument works for whatever measure you wish to use). We see \begin{align*} \lambda(f^{-1}((a,\infty))) = \lambda(\{x \in \mathbb R : f(x) > a \}) &= \int_{\{x \in \mathbb R : f(x) > a \}} d\lambda(x) \\ &\le \int_{\{x \in \mathbb ...


2

If $f$ is continuous on a compact set $K$, then there is a constant $M$ such that $|f(x)|\leq M$ for all $x\in K$. Therefore $$ \int_K|f(x)|\;dx\leq Mm(K)<\infty $$ so $f$ is integrable on $K$.


1

In the same spirit as other answers, when you have an expression such that $$S=\sum_{n=1}^\infty P(n)x^n$$ where $P(n)$ is a polynomial in $n$, first expand $P(n)$ and it will become something like $$P(n)=\sum_{i=0}^m a_i n^i$$ Now rewrite the successive powers of $n^i$ such as $$n^2=n(n-1)+n$$ $$n^3=n(n-1)(n-2)+3n(n-1)+n$$ ...


3

For completeness sake, the following are useful. For $|k|<1$ we have: $\sum\limits_{n=0}^\infty k^n = \frac{1}{1-k}$ Deriving each side with respect to $k$, we get: $\sum\limits_{n=0}^\infty nk^{n-1} = \frac{1}{(1-k)^2}$ Multiplying by $k$ yields $\sum\limits_{n=0}^\infty nk^n = \frac{k}{(1-k)^2}$ Deriving the expression by $k$ yields ...


0

prove by induction that $$\sum_{i=1}^n (2i-1)^2(\frac{1}{2})^n=2^{-n} \left(-4 n^2-12 n+17\ 2^n-17\right)$$


2

The easiest way is to use the Weirestrass M-test. The $n$-th term is uniformly bounded by $e^{-n}$, which is summable. Thus, the sum converges uniformly. Each term of the sum is continuous, so you have a uniform limit of continuous functions, and therefore the limit function is continuous.


1

In order to correctly construct confidence interval, first of all, you need a "pivot" T such that T is a function of all data $X_1,X_2,...,X_n$ T is a function of $\mu$ The distribution of T is know, and it is not a function of $\mu$ For example, $T=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ is a pivot, and $T$~$N(0,1)$ regardless of the value of $\mu$ (if the ...


0

The alternative series test that you no doubt learned in Calculus is a good tool for dealing with this problem because it gives you error estimates. Recall that if $a_n \ge 0$ satisfies $$ a_0 \ge a_1 \ge a_2 \ge a_3 \ge a_4 \ge \cdots ,\;\;\; a_n \rightarrow 0, $$ Then $\sum_{n=0}^{\infty}(-1)^{n}a_n$ converges, and the maximum error in the value ...


0

Hint: $\int_a^b f - \int_a^{b_n} f = \int_{b_n}^b f .$


1

In vol. 1, Chap.II, Sec.3.3 of the book Generalized functions by Gelfand-Shilov you will find the formula $$ F\bigl[\Vert x\Vert^\lambda\bigr](\omega)=2^{\lambda+d}\pi^{\frac{n}{2}}\frac{\Gamma\left(\frac{\lambda+d}{2}\right)}{\Gamma\left(-\frac{\lambda}{2}\right)} \Vert\omega\Vert^{-\lambda-d}, $$ or all $\lambda\neq -d,-d-2,\dotsc $, where, for ...


0

You seem to be looking for domains $\Omega$ for which $W^{1,p}_0(\Omega) = W^{1,p}(\Omega)$. This will be the case if $\partial \Omega$ has $p$-capacity equal to $0$. Look e.g. in Measure Theory and Fine Properties of Functions by Evans and Gariepy, or Function Spaces and Potential Theory by Adams and Hedberg.


-1

In Brezis, Functional analysis, Sobolev spaces and partial differential equations on p. 277 you can find the following: Assume that $\Omega$ is of class $C^1$, and let $u\in W^{1,p}(\Omega)$ with $1\leq p <\infty$. Then there exists a sequence $(u_n)_{n\in \mathbb{N}} \subseteq C_c^{\infty}(\mathbb{R})$ such that $u_n |_{\Omega} \rightarrow u$. There is ...


0

The improper integral is convergent since we have for all $x >0$ $$\left|\frac{\sin x}{x}e^{-tx}\right| \leqslant e^{-tx},$$ and for $t > 0$ $$\int_0^\infty e^{-tx} \, dx = \frac{1}{t} < \infty.$$ To prove differentiability with respect to $t$, it is sufficient to show uniform convergence of $$\int_0^\infty \frac{\partial}{\partial ...


1

For the first part, since $g$ is of compact support, it is bounded. Hence $\|g\|_{\infty} < \infty$, so $$\int |f_ng| = \int |f_n| |g| \le \int |f_n| \|g\|_{\infty} = \|g\|_{\infty} \int |f_n| \to 0$$ For the second, note that: $\overline{C^{0}_{cpt}} = L^1$ (wrt $\|\cdot \|_1$), so for $g \in L^1$ and $\epsilon > 0$, there exists a continuous ...


1

Essentially the same example as given by copper.hat, but perhaps written in a simpler way. Let $X=\mathbb{N}$ and let $d$ be the discrete metric given by $d(m,n)=1$ if $m\neq n$ and $d(m,n)=0$ if $m=n$. Every subset is closed and bounded by $1$. But only finite subsets admit an $1/4$-net.


0

I'll give a few hints. You have that $S = \{ (x,y,z) \in \Bbb R^3 \mid F(x,y,z) = 6 \}$, where $F\colon \Bbb R^2 \times \Bbb R_{\geq 0} \to \Bbb R$ is given by $F(x,y,z) = x+g(x)+y+g(y)+z+g(z)$. We have that $F(1,1,1) = 6$. Now, compute $D_z F(1,1,1)$. If this is not zero, then the Implicit Function Theorem applies, giving us $z$ as a function of $x$ and ...


2

Take $l_\infty$ and $S = \{e_k\}_{k \in \mathbb{N}}$. The $S$ is closed, bounded, but not totally bounded since $\|e_i - e_j\| = 1$ for all $i \neq j$ (hence there can be no finite $\epsilon$-net for $\epsilon <1$).


0

You would also need that there is an isometric isomorphism from $X$ to $Y$ that fixes $M$ (equivalently, $d_X,d_Y$ agree on $M$). Otherwise, the set intersection of $X$ and $Y$ is not very meaningful, since it's changed by a simple relabelling of elements. For instance, for $X=\mathbb R$, you could replace each negative number $x$ with $(0,x)$, and apply ...


1

Perhaps the simplest counterexample is the discrete metric [ $d(p,q)=1$ when $p\ne q$ ] on any infinite set $X.$ Then every subset of $X$ is bounded and open and closed, but only the finite subsets of $X$ are compact.


2

If $(X,d)$ denotes a metric space then we can construct a metric $d'$ on $X$ by stating: $$d'(x,y)=\min(d(x,y),1)$$ This function can be shown to be a metric and induces the same topology as the original $d$. However every subset of $X$ is bounded with respect to the constructed metric.


1

It is not true in general, the closed unit ball of an infinite dimensional Banach space is not compact


0

Take $m=2$ and two points in $\mathbb{R}^2$. Their convex hull is open line segment: $\left\{\lambda x + (1- \lambda)y \hspace{2 pt} | \hspace{2 pt} 0<\lambda<1 , x,y \in \mathbb{R}^2 \right\}$. But this set is not open w.r.t. $\mathbb{R}^2$.


2

The definition you give is correct, but rather confusingly written. What it means is the set of values $\sum_{i=1}^n \lambda_i x_i$ where $\lambda_1,\ldots,\lambda_n$ are positive real numbers with sum $1$ and $x_1,\ldots,x_n$ are taken among $M$. Now in fact we can replace "positive" by "nonnegative" in this definition, because we can always remove those ...


5

I'll assume $S$ not empty. Since the function $f\colon S\to\mathbb{R}$, $f(p)=d(p_0,p)$ is continuous, when $S$ is compact its image is compact, hence closed and bounded; therefore the image of $f$ contains its minimum. If $S$ is only assumed to be closed and bounded, but not compact, the statement is not generally true. Consider $X=\{0\}\cup (1,2]$, with ...


0

So your Professor's hint is pretty useful! Do you know what does Stokes' theorem says? If $M$ is a $k+l+1$ manifold then the theorem applies to exact $k+l+1$-forms, such as $d(\omega \wedge \eta)$. That's the left-hand side of the equation your professor gave you. If you substitute the right-hand side of the equation your professor gave you into Stokes' ...


0

For $$c^2\gt 4d_j$$ $$\sqrt{c^2-4d_j}=c\sqrt{1-\frac{4d_j}{c^2}}\approx c\left(1-\frac{2d_j}{c^2}\right)=c-\frac{2d_j}c,$$ with a relatively good approximation. Summing, $$\sum_{j=1}^N\left(c\pm\left(c-\frac{2d_j}c\right)\right)=2c.$$ With $M$ positive signs, $$Mc-\frac{d}c=c,\\c=\pm\sqrt{-\frac{d}{M-1}},$$ where $d=\sum_{j=1}^N\pm d_j$. This might ...


1

$n=3$ is rather easy: $w_i$ satisfies a quadratic polynomial. For $n=4$, each $w_i$ satisfies a rather nasty polynomial of degree $8$ (but involving only even powers). Thus there is a solution in terms of radicals, but it won't be pleasant. For $n=5$, it seems each $w_i$ satisfies a polynomial of degree $22$. A solution in radicals is not to be expected. ...


1

A few more digits $$ 1.6798002778544903357 $$ (where the last digit shown has rounded up). The Inverse Symbolic Calculator doesn't recognize it, so you're probably out of luck.


3

The function $x\mapsto d(p_0,x)$ is continuous on $X$, hence attains its minimum on $S$ since $S$ is compact.


3

Hint. When $x$ is positive and $\frac{x}{1+x}$ is sufficiently small, we have $0 < x < 2\frac{x}{1+x} $.


2

Suppose that such series converges. Note then that since $a_n+1> 1$, we must have that $a_n\to 0$. But then the quotient of $a_n$ with $a_n/(1+a_n)$ tends to $1$.


1

Your bound is wrong, but here is an other approach similar : $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\dots>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\dots=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\dots$$ More formaly if ...


0

Make groups . ie $1+(\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})....$ so sum of first parantheses is $1$ second is $1/2$ then third is $7/12$ now you ll see that $7/12>1/2$ so sum of such grouping will atleast be greater than $1/2$ so we can create as much large numbers as we want using these halves . Thus the series diverges.


1

Fix $x \in [a,b]$. If $x+h \in [a,b]$, and $h \neq 0$, then $$\frac{1}{h}(\Phi(x+h) - \Phi(x)) = \int_c^d \frac{f(x+h,y)-f(x,y)}{h}dy.$$ For each $c \leqq y\leqq d$, we can apply the mean value theorem to the map $t \mapsto f(t,y)$ to obtain $0<\theta_y<1$ such that $$f(x+h,y) - f(x,y) = h\frac{\partial f }{\partial x}(x+\theta_yh,y).$$ Let $\epsilon ...


1

Yes. A corollary to the Rellich Kondrachov theorem says that for all $p\ge 1$, we have $W^{1,p}$ compactly contained in $L^p$. Evans PDE book has a nice treatment of this. EDIT: Actually, if I recall correctly, for $\Omega \subseteq \mathbb R^n$, to prove that $W^{1,p}(\Omega)$ is compactly contained in $L^p(\Omega)$, we only need the Rellich Kondrachov ...


0

For the first part just integrate with respect to $x_1$ and $x_2$, you can find $\displaystyle f=\arctan(\frac{-x_1}{x_2})$ satisfies the condition. For the second part let $v=(F_1,F_2)$, a vector function. Suppose such an $f$ exist, $v$ is the gradient of some scalar so the integrate around unit circle(closed curve in $\mathbb{R}-(0,0)$) is zero. ...


0

One may consider $$ \sum_{n=1}^\infty e^{-(1-|x|)n^2-n} $$ If $x \in [-1,1]$ then $$\left| e^{-(1-|x|)n^2-n}\right|\leq e^{-n}$$ and the initial series converges by comparison. If $x \notin [-1,1]$ then, as $n \to \infty$, using $-(1-|x|)n^2-n=n((|x|-1)n-1)\to +\infty$, one gets $$ \lim_{n \to +\infty}e^{-(1-|x|)n^2-n}=+\infty \neq0$$ and the ...



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