New answers tagged

2

The monotone convergence theorem says if $0 \leq c_n \leq d_n$, and $\sum d_n$ converges, then so does $\sum c_n$. Assume $\sum a_n$ converges absolutely. This means that $\sum\limits_n |a_n|$ converges. To show that, say, $\sum\limits a_{2n+1}$ converges, define $b_n$ by $$b_n = \begin{cases} a_n & \textrm{ if $n$ is odd} \\ 0 & \textrm{ if $n$ ...


1

Yes. Each sum is an increasing function, bounded above by the overall sum, so it converges to its supremum. The series may not be monotone, but the sum is.


4

First of all, if you do direct substitution, then yes, $\sec 0 = 1,$ but you still have to consider the arc sine function. And $\sin^{-1} (1) = \frac\pi2.$ So the result of direct substition by $x=0$ is $\sin^{-1} (\sec x) = \frac\pi2.$ But the direct substitution method agrees with the limit only if both of the following conditions are met: The limit ...


2

Direct substitution does not always work. It only works when the given function is continuous about $x_0$ (the number x approaches, in this case 0). Remember the idea of limits is that when x gets "close" to $0, f(x)$ ought to get "close" to some number. In the case of this problem, $\sin^{-1}(x)$ is only defined on the domain $[-1, 1]$, but the range of $...


3

Once again, it doesn't matter that you arrive at $\epsilon$ precisely! If for $|x-a|<\delta$ you can show that $|x^3-a^3|<K\epsilon$ with $K$ a constant (i.e. independent of $x$) then it's a win. Cutting epsilons is just a matter of aesthetic and IMHO just confuse the beginner. It's crap to let people think it is mandatory to have to go for the ...


1

Yes, try with cases : if $\{x\}=\{x'\}$ or $\{x\}=\{x',y'\}$ and when $\{x,y\}=\{x'\}$ or is $\{x,y\}=\{x',y'\}$.


2

I don't have the book, so I don't know how Tao defines these sums, but I suppose it's something like the least upper bound of the set of all finite sums. Then, as you say, the theorem is intuitively clear, but that's not a proof. When you allow negative summands, you quickly get non-intuitive results. For example, a conditionally convergent sequence can ...


2

Let $A_k=[a_k-2^{-k-1},a_k+2^{-k-1}]$ for all $k\in \mathbb N$, then $\mu(A_k)=\frac1{2^k}$, and thus $\sum\mu(A_k)<\infty$. By Broel-Cantelli lemma, $\mu(\limsup A_k)=0$, which means $$\mu(\{x\in \mathbb R: x \text{ belongs to infinitely many }A_k\})=0.$$ So for almost every $x\in\mathbb R$, the sum $$\sum_{k=1}^{\infty}2^k{\chi_{[a_k-2^{-k-1},a_k+2^{-...


0

Since $x$ is Lipschitz, it is absolutely continuous, and $x(t) = x(0) + \int_0^t \dot{x}(s)\; ds$. Noting that $\dot{x}(s) = (1 - 2 x(s)) I_E(s)$ for almost every $s$ (where $I_E$ is the indicator function of $E$), we have $$ x(t) = x(0) + \int_0^t (1 - 2 x(s)) I_E(s)\; ds $$ EDITED: Let $$m(t) = \int_0^t I_E(s)\; ds$$ i.e. the Lebesgue measure of $[0,t] \...


0

Fot this kind of problems, I think that defining first the implicit function makes life easier. For your case, consider $$F(x,y)=y-\cos(xy)=0$$ and compute the partial derivatives $$\frac{\partial F(x,y)}{\partial x}=y \sin(xy)\qquad \qquad \qquad \frac{\partial F(x,y)}{\partial y}=1+x \sin(y)$$ Now, using the implicit function theorem $$\frac{dy}{dx}=-\...


0

Hint: the induction step is like adding $1$ in radix $q$ arithmetic. The analogy is clearer with $2$ digits $r_i$. By induction $\,n\!-\!1\, =\, m\, q^2 + r_1 q + r_0,\ $ all $\,r_i < q$ when $\,r_0 < q\!-\!1\,$ then adding $\,1\,$ yields $\,r_0 < q\,$ so we're done. else $\ \ \ r_0 = q\!-\!1\,$ then adding $\,1\,$ we get a carry into $\,r_1,\,$ ...


1

Note that the base case is trivial, that is, for $n=1$ and any positive integer $q$ we have that $n=0.q+1$ if $q>1$ ($n=1.q+0$ otherwise). Assume that the claim holds for every natural number $n\leq N$. (To be more explicit, it means that given a natural number $n\le N$ and a positive integer $q$ we have $n=mq+r$ such that $0 \le r < q$.) We need ...


1

That is not a valid argument because the series $\sum_{n=0}^\infty M\lvert R\rvert$ actually diverges (unless $M=0$), and therefore you did not prove that it follows from the Weierstrass $M$-test that your series converges.


1

We need only assume that $f$ is once differentiable and the existence of higher derivatives follows because with $x=\frac12$, $$f'(y)=f(y+\tfrac12)-f(y-\tfrac12) $$ and the right hand side is differentiable so that by induction $$f^{(n+1)}(x)=f^{(n)}(x+\tfrac12)- f^{(n)}(x-\tfrac12).$$ By differentiating the functional equation with respect to $x$, we ...


0

the answer may be easily obtained using the general relation: $$ \frac{d}{dx} = \frac{\partial}{\partial x} + \frac{dy}{dx}\frac{\partial}{\partial y} $$ since it is easy to feel a bit confused by the apparent abstraction of this formula, you may check by working from "first principles". then the calculation is straightforward using infinitesimals of the ...


1

You should get that $$\frac{dy}{dx}=-\sin (xy)\left(y+x\frac{dy}{dx}\right),$$ or $$\frac{dy}{dx}=-\frac{y\sin xy}{1+x\sin xy}.$$ You must've made a mistake in your implicit differentiation. You might need to think a bit more carefully about how to properly implicitly differentiate $\cos xy$.


0

Since this equation can be put in the form $y'+a(x)y=b(x)$, you can solve by using the integrating factor $e^{\int a(x)\operatorname dx}$.


1

I made a spreadsheet taking $X$ to be $1$ second. The dog only makes three turns before the boys meet, so it runs $3$ seconds less than $1$ hour and covers $10\cdot \frac {3597}{3600} \approx 9.99$ km. Shortening the turn to $\frac 12$ second just made one more turn, so the dog covers just about $10$ km for any reasonable turn around time.


0

The equation $$y'+\left(-1+\frac{a-1}{x}\right)y+b x -c=0$$ is linear. You have $$y= y_h + y_p$$ where $y_h$ is the solution to the homogeneous equation, $$y'+\left(-1+\frac{a-1}{x}\right)y=0$$ and $:y_p $. is a particular solution to the in-homogeneous equation. Both solutions are found easily because for the first one the equation is separable and for ...


0

According to Maple, the solutions can be expressed with the help of Whittaker functions:


1

It's much easier to try to prove that the collection of cylinder sets is closed under finite intersections. Since you already know that it is closed under complements, this will be sufficient to see that it is an algebra. If $$A = \{x \in X: (f_1(x), \dots, f_n(x)) \in A_0\} \qquad B = \{x \in X: (g_1(x), \dots, g_m(x)) \in B_0\}$$ then we have that $$A \...


1

$x_0 \in \partial B$ so $|x_0| =R$. Hence $R^{2}-|x|^{2}$ can be made as small as wish by making $|x-x_0|$ small.


2

$|\Delta f(x)|\leq \sum_{i=1}^3 |\partial^2_{x_i} f(x)|$ by the triangle inequality and clearly, $\sum_{i=1}^3 |\partial^2_{x_i} f(x)|\leq \sum_{i,j=1}^3 |\partial_{x_i}\partial_{x_j} f(x)|,$ since the latter sum simply has some new, added terms. Now, apply linearity of integrals.


2

In set-theoretic foundations, the statement is false. The problem is the distinction between logical functions and set-theoretic functions. If $f$ is a (unary) logical function (i.e. a function symbol in our theory), then $f(x)$ is always defined.1 To this end, your question wouldn't make sense. However, it is clear that you intend a set-theoretic function (...


4

$V$ is already closed in $L^2(\mathbb{T}^n)$, since it is the orthogonal complement of the function $1$. Now $H^1$ is a strictly stronger topology than $L^2$, so $V$ is also closed in $H^1$, and thus not dense.


4

Note that this result cannot be true because, as noted in the comments, convergence in $H^1(\mathbb{T}^N)$ implies convergence in $L^1(\mathbb{T}^N)$ and hence if $u_n \in V$ and $u_n \to u$ in $H^1(\mathbb{T}^N)$ then $$0 = \int u_n dx \to \int u dx$$ and so $u \in V$ also. This shows that $V$ is a closed subset of $H^1(\mathbb{T}^N)$ which is not equal to ...


0

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1

You have to be careful with $\tan \, x$ because it is undefined at $\pm \frac {\pi} 2$ and $\pm \frac {3\pi} 2$. Consider the interval $(\frac {\pi} 2, \frac {3\pi} 2)$. Show that in this interval $\tan \, x -x$ takes all values from $-\infty$ to $\infty$. [$\lim_{x \to \pi /2+} \tan \, x -x=-\infty $ and $\lim_{x \to 3\pi /2-} \tan \, x -x=\infty $]. Hence ...


2

Let $R >0.$ $b(R)=R(1-2M/R)^{-1/2}.$ Binomial expansion for rational exponent $\alpha$: $(1+x)^{\alpha}=$ $ 1+\alpha x+(\alpha (\alpha-1)/2!)x^2+ O(x^3)$ converges for $|x|<1.$ Hence for large $R$: $b(R)=R(1+M/R +O((M/R)^2))$ https://brilliant.org/wiki/fractional-binomial-theorem/


0

$$\frac{\sqrt{\frac{r^3}{r - 2M}}}{r+M}=\sqrt{\frac{r^3}{(r-2M)(r+M)^2}}=$$ $$\sqrt{\frac{r^3}{r^3\left(1-\frac{2M}r\right)\left(1+\frac Mr\right)^2}}=\sqrt{\frac{1}{\left(1-\frac{2M}r\right)\left(1+\frac Mr\right)^2}}\to 1$$ as $r\to\infty$.


4

This is just a Taylor expansion $$b(r) = \sqrt{\frac{r^3}{r - 2M}}=r \sqrt{\frac{r}{r - 2M}}$$ Use long division or Taylor series $$\frac{r}{r - 2M}=1+\frac{2 M}{r}+\frac{4 M^2}{r^2}+O\left(\frac{1}{r^3}\right)$$ $$\sqrt{\frac{r}{r - 2M}}=1+\frac{M}{r}+\frac{3 M^2}{2 r^2}+O\left(\frac{1}{r^3}\right)$$ $$b(r)=r+M+\frac{3 M^2}{2 r}+O\left(\frac{1}{r^2}\right)=...


1

I am posting this to get more information about the exact meaning of both $dq$ as well as the product in $\dfrac1q?dq$. There is some ambiguity about the nature of this integral. Your sources should make clear what $dq$ really means here, and that should resolve the ambiguity. I give somewhat plausible interpretations, but cannot be certain :-( If this ...


4

The answer (under the conventional metrics, etc.) is yes! Any sequence $\Bbb Z^+\to Y$ is continuous on its entire domain, or at every $a\in\Bbb Z^+.$ However, we'd typically say "at every $n\in\Bbb Z^+,$" for a few reasons. In this particular case, one will often denote the sequence of points $a_n$ by $a,$ rather than by $(a_n),$ so saying that such a ...


0

for what it's worth, I finally put together a pretty thorough reworking of Rudin's proof. I'd be happy to receive feedback, but mostly I just thought I'd share in case anyone cares. Here's a link to the PDF, because I don't know how to make my LaTeX macros compatible with this website (sorry).


3

From the inequality we know that $f$ is invertible and that $f^{-1}$ is Lipschitz continuous. Take any closed set $A\subset K$, then it is compact because of compactness of $K$, and $f^{-1}(A)$ is then also compact, hence closed, so that $f$ is continuous. Hence $f$ and $f^{-1}$ are continuous, and so $f$ is a homeomorphism between $K$ and some subset of $K$....


5

When you solve the Schrodinger equation, you solve for the possible energies of the system. These are the eigenvalues of the Hamiltonian operator. Just like a single eigenvalue can correspond to multiple eigenvectors, a single energy can correspond to multiple eigenfunctions of the Hamiltonian. This is called degeneracy. An eigenvalue (energy) is called ...


3

$f(x)=e^{-\frac 1 {1-x}}$ for $x<1$ and $0$ for $x \geq 1$ defines a smooth function which is positive on $(-\infty,1)$ and $0$ outside it. So $f(\|x\|^{2})$ is a smooth function on $\mathbb R^{2}$ which is positive for $\|x\|<1$ and $0$ elsewhere. For any other disk in $\mathbb R^{2}$ use an appropriate affine transformation.


3

Define$$\eta(z)=\begin{cases}e^{\frac1{\lvert z\rvert^2-1}}&\text{ if }\lvert z\rvert<1\\0&\text{ otherwise.}\end{cases}$$


0

One can just use the map $(x,y) \mapsto (x,y,1)$ and then normalize the resulting 3d element. This should do the trick.


3

$\mu ^{+}(B)=\mu(B \cap P) \leq |\mu(B \cap P)| \leq m(B \cap P)$ and $\mu ^{-}(B)=\mu(B \cap N) \leq |\mu(B \cap N)| \leq m(B \cap N)$. Hence $|\mu|(B)=\mu ^{+}(B)+\mu ^{-}(B)\leq m(B \cap P)+m(B \cap N)=m(B)$.


0

The energy of the second-order wave equation $u_{tt} = u_{xx}$ is the sum of the total kinetic energy and of the total potential energy: $E = \frac12 \int_{\Bbb R} (u_t)^2 + (u_x)^2\, \text d x$. For smooth solutions, we have $$ \begin{aligned} \frac{\text d}{\text dt} E &= \int_{\Bbb R} u_t u_{tt} + u_x u_{xt}\, \text d x \\ &= \int_{\Bbb R} u_t ...


1

$$\sum_{cyc}\frac{1}{a^6-a+3}\leq1$$ it's $$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^6-a+3}\right)\geq0$$ or $$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^6-a+3}-\frac{5}{9}\ln{a}\right)\geq0.$$ Let $f(x)=\frac{1}{3}-\frac{1}{x^6-x+3}-\frac{5}{9}\ln{x}.$ Thus, $$f'(x)=\tfrac{(1-x)(5x^{11}+5x^{10}+5x^9+5x^8+5x^7-5x^6-29x^5-29x^4-29x^3-29x^2-24x-45)}{9x(x^6-x+3)^2}.$...


1

As pointed out in the comment, under your hypothesis $(f_n(x))_{n\in\mathbb{N}}$ could diverge. But if you add another condition that every point of $E$ is an accumulation point, we can say $f_n \to f $ pointwisely. Proof $\,\,$ Suppose for contradiction, then we have $\exists$ $\varepsilon_0 >0$, $\forall$ $N$, $\exists$ $n\ge N$, such that $...


1

I have no idea if your claims are true, and I am not a specialist of the two systems above. Nevertheless, one can make the following observations. For the first system ("shallow water"), the eigenvalues of the flux's Jacobian matrix are $\lambda_\pm = q \pm \sqrt{\rho + q^2}$. Therefore the system is strictly hyperbolic if $\rho + q^2 > 0$. The gradient ...


3

You can rewrite your condition as $$f(x) \leq f(x_0) - C(x - x_0)$$ This just says that the graph of $f(x)$ lies below the line of slope $-C$ containing $(x_0, f(x_0))$. For any $C$ it's easy to draw a graph of such a function that is not decreasing... just add a few "wiggles" to a decreasing function.


3

For example, with $x_0=0$, $f(x_0)=0$, and $C=1$, this merely states $f(x)\le -x$. We cannot conclude that $f$ is decreasing. Try $f(x)=-x+\cos(1000x)-1$.


0

Without loss of generality, $0\le a<b.$ By the mean value theorem, there is a $c\in (x,y)\subset [a,b]$ such that $|f(x)-f(y)|=2c|x-y|<2b|x-y|.$


2

Since $x,y\in [a,b],$ it follows that $|x|,|y|\leq \max\{|a|,|b|\}.$ Hence, $$|x|+|y|\leq 2\max\{|x|,|y|\}\leq 2\max\{|a|,|b|\}.$$


2

Since we'll be working with both real and imaginary parts, it's convenient to consider them as components of a vector: $$ V_k = [\cos(X_k), \sin(X_k)] $$ Note that these have mean $0$ and covariance matrix $$ \Sigma = \pmatrix{1/2 & 0\cr 0 & 1/2\cr}$$ By the multivariate version of the Central Limit Theorem, if $S_n = \sum_{k=1}^n V_k$, $ S_n/\sqrt{...


3

For $2k\ge 2$ $$\prod_{n=2}^\infty (1-\frac{1}{n^{2k}})=\prod_{m=1}^{2k} \prod_{n=2}^\infty (1-\frac{e^{2i \pi m/(2k)}}{n}) = \prod_{m=1}^{k}\prod_{n =-\infty, |n|\ge 2}^\infty (1-\frac{e^{2i \pi m/(2k)}}{n}) = \prod_{m=1}^{k} f(e^{2i \pi m/(2k)})$$ where $f(x) =\frac{\sin(\pi x)}{\pi x(1-x^2)} $ and in those products the order of summation is meant to be $...


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