# Tag Info

## Hot answers tagged analysis

10

For the limit $\lim_{n\to+\infty}\sin^2\sqrt{n^2+n}$, since $\{e^{in}\}_{n\in\mathbb{N}}$ is dense in the unit circle (it is a consequence of the irrationality of $\pi$ and the Dirichlet box principle), the same holds for the sequence $\{e^{i(n+1/2)}\}_{n\in\mathbb{N}}$ (a translation of the unit circle preserves density). By taking imaginary parts (the ...

9

The case $n=2$ is simple. If $z_i = 0$ for any $i$ then we can prove it by induction. So we suppose that $z_i \neq 0$ for each $i$. We choose $R > 0$ such that $|Rz_i| \leqslant 1$ for each $i$. Then $\displaystyle{ \sum_{m=0}^{\infty} (xz_i)^m}$ converges. We have \displaystyle{ \prod_{i=1}^n \sum_{m=0}^{\infty} (xz_i)^m = \prod_{i=1}^n ... 9 The idea in the following is to simplify using logarithmic identities, and then to get rid of the nasty integration term using the symmetry of the limits. \begin{align} I&=\int_{0}^{\Large\frac\pi2}\ln \left ( \frac{(1+\sin x)^{1+\cos x}}{1+\cos x} \right )\,dx \\&=\int_{0}^{\Large\frac\pi2}[(1+\cos x)\ln(1+\sin x)-\ln({1+\cos x})]\,dx ... 8 It's a good start. But there are three problems. There's an assumption that integrals and areas are the same. It's true that the integral was defined to try to generalize the standard area computations, but are you certain that it succeeds? If not, then proving things about areas doesn't prove things about integrals (which involve lower and upper sums, ... 8 Use polar coordinates, We know thatr^2 = x^2 + y^2$$So our double integral becomes$$\int_{0}^{2\pi} \int_0^{1}r^2\cdot rdrd\theta$$Now solve. EDIT I see that your computation is correct, I am simply offering another alternative and more easier way to solve this double integral. 7 Aside from showing that two limits differ as you do for y=x and y=x^2-x, you can also do this in one fell swoop: Let y=x^3-x, then you get$$\lim_{(x,y)\to (0,0)}{x^2\over x^3}$$which clearly diverges to infinity. 6 We can show the result along the lines you sketched. First, we note that it suffices to consider monotonically increasing f, for the transformation g \mapsto 1-g is an isometry that preserves polynomials. Now we need to approximate the homeomorphism f by continuously differentiable homeomorphisms. For that, extend f to \mathbb{R} by setting ... 6 If you use polar coordinates you will get the following$$ \int_{0}^{2\pi} \int_{0}^{1} r^2 r dr \ d\theta $$5 Domains are open, so not compact. If you throw in the boundary to make it compact, it will no longer be a manifold (though it will be a manifold with boundary, since \Gamma is smooth). 5 Expanding the fraction \frac{x}{1-x} as the series \sum\limits_{k\geqslant1}x^k for each x=\frac1{10^n} yields$$\sum_{n\geqslant1}\frac1{10^n-1}=\sum_{n\geqslant1}\frac{\frac1{10^n}}{1-\frac1{10^n}}=\sum_{n,k\geqslant1}\frac1{10^{kn}}.$$In the RHS, each \frac1{10^i} appears as many times as one can decompose the integer i\geqslant1 as i=kn, ... 4 I don't have enough reputation for a comment, so I'll write this in answer: Consider Does the everywhere differentiablity of f imply it is absolutely continuous on a compact interval? I think that answers your question. 4 Notice that$$(\sin(\pi(\sqrt{n^{2}+n}-n)))^{2}=((-1)^{n}\sin(\pi\sqrt{n^{2}+n}))^{2}=(\sin(\pi\sqrt{n^{2}+n}))^{2}$$and that$$\sqrt{n^{2}+n}-n=\frac{n}{\sqrt{n^{2}+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\to\frac{1}{2}$$as n\to\infty. So ... 4 Apply \exp to both, express trig functions in terms of exponentials, and take the difference. I get$$ \left( \dfrac12\,{{\rm e}^{C_{{3}}+iC_{{4}}}}-4\,{{\rm e}^{C_{{2}}}}C_{{1} } \right) {r}^{2}+{\frac {\dfrac12\,{{\rm e}^{C_{{3}}-iC_{{4}}}}-{{\rm e}^{ C_{{2}}}}}{{r}^{2}}} $$For this to be 0 for all r, you need both coefficients to be 0. This ... 3 Answer assuming looking for "at least one closed interval": Define the sequence a_{n}=\frac{1}{3^{n}}. Then s_{k}=\sum\limits_{i=k}^{\infty}a_{i}=\frac{3}{2\times 3^{k}}. Hence \lim\limits_{k\rightarrow\infty}\frac{s_{k}}{(\frac{1}{2^{k+1}})}=\lim\limits_{k\rightarrow\infty}3(\frac{2}{k})^{k}=0. Define a collection of function ... 3 Assuming X,Y are CW-complexes, this is just the quotient of \pi_1(X \times Y,p \times q) modulo the normal closure of the image of \pi_1(X,p) * \pi_1(Y,q) \hookrightarrow \pi_1(X \times Y,p \times q). You can prove this by applying Van Kampen's theorem, because the Smash product of (X,p) and (Y,q) is homotopy equivalent to the space obtained from ... 3 There are different ways to define integrals named after different people. What you teacher described is an informal explanation of the Riemann integral. You can see rigorous construction under the link, but it amounts to subdividing the interval of integration into subintervals of smaller and smaller lengths, and replacing the area under the graph with the ... 3 You should have gotten \displaystyle\int_{0}^{1-a}[(1-a)x-x^2]\,dx = \left[\dfrac{1-a}{2}x^2-\dfrac{1}{3}x^3\right]_{0}^{1-a} = \dfrac{(1-a)^3}{6}. Then, you have \dfrac{(1-a)^3}{6} = \dfrac{9}{2} \leadsto (1-a)^3 = 27 which should be easy to solve. 3 I would say here thart the difficulty is more in the notations than in the mathematics. I present below a proof by induction on n, where the result for \lbrace z_1,z_2,\ldots,z_n \rbrace is deduced from the result for \lbrace z_1,z_2,\ldots,z_{n-1} \rbrace and the result for \lbrace z_1,z_2,\ldots,z_{n-2},z_n \rbrace. This proof, although completely ... 3 Corrected answer: Regarding f and \mu as tempered distributions, you are looking to solve \check{f} = \mu. A bounded linear functional on C_c(\mathbb R) (which is included the Schwarz class) may be represented by a bounded complex Borel measure. This is the Riesz Representation Theorem. In the present context, it suffices for there to exist a ... 3 This may or may not be useful, but Bochner's theorem says that a function f is the Fourier transform of a positive bounded measure \mu if and only if f is continuous and positive definite. (A simple way to describe a positive definite function is this: given any x_1, \dots, x_n \in \mathbb{R}, the n \times n symmetric matrix whose ij entry is ... 3 Using k=\frac{1}{2} in the second rule, we get$$\begin{align} \int\limits_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx &=2\int\limits_{-1}^{1} (2x-3) \sqrt{4-4x^2} \ dx \\&=2\int\limits_{-1}^{1} 2(2x-3) \sqrt{1-x^2} \ dx \\&=4\int\limits_{-1}^{1} (2x-3) \sqrt{1-x^2} \ dx \\&=8\int\limits_{-1}^{1}x\sqrt{1-x^2} \ dx-12\int\limits_{-1}^{1}\sqrt{1-x^2}\ ... 3 Yes, in fact they both come from two standard results in analysis: Theorem 1: A subset of\mathbb{R}^n$is compact if and only if it is closed and bounded. Theorem 2: Continuous image of a compact set is a compact set. Corollary: If$f:\mathbb{R}^n\to \mathbb{R}$is continuous and$A\subset \mathbb{R}^n$is closed and bounded. By Theorem 1,$A$is ... 2 The classical operator $$L = -\frac{d^{2}}{dx^{2}}+V,\;\;\; a \le x \le b,$$ is different if$V$is very singular. If$V \in L^{1}[a,b]$, then things are nice because there are 2 linearly-independet classical solutions of$Lf = \lambda f$for every$\lambda$. That is, such solutions are continuous on$[a,b]$, their first derivatives are ... 2 Perhaps a sledehammer: The following holds: If$\ \ \ $1)$f$is continuous on$[a,b]$,$\ \ \ $2)$f'$exists and is finite for all but a countable subset of$(a,b)$, and$\ \ \ $3)$f'$is Lebesgue integrable, then $$f(x)-f(a)=\int_a^x f'(t)\,dt$$ for all$a\le x\le b$. In particular,$f$is absolutely continuous. This is Excercise ... 2 The very anticlimactic answer here is, that the Bohr-Mollerup-Theorem only includes real-valued functions, see here for example. Otherwise, for general complex-valued functions, you also would have problems to evaluate the condition$\frac{d^2}{dx^2}\ln f(x)>0$. 2 Generally, the idea that "both$a$and$b$are small" is expressed by an inequality with the maximum function (=MAX(...) in Excel). When the quantities are measured on the same scale, you'd just take the maximum of both. When they are on different scales, you introduce some normalizing factors. So, let's say that$T$is the number such that time$<T$is ... 2 Let us start by putting things in a nicer form. $$\begin{eqnarray*}A_{n,j}&=&3(-1)^j 2^{n-j+1}\frac{(2n-2j-4)!}{(n-j)!(n-j-2)!}\binom{j+2}{2}\frac{n^{5/2}}{8^n}\\&=&\frac{6\cdot 2^{n-j}}{8^n}(-1)^j\frac{(n-j)(n-j-1)n^{5/2}}{(2n-2j)(2n-2j-1)(2n-2j-2)(2n-2j-3)}\binom{2n-2j}{n-j}\binom{j+2}{2}\end{eqnarray*}$$ but due to Stirling's formula, ... 2 Adjusting the four parameters of the function $$f(r,a,b,c):=-\left(a-b\,(x-r)^4\right)\;e^{-c\,(x-r)^2}$$ ($b\,(x-r)^2$seemed too weak) I obtained for$f(2.02,1,12,2.4)$: Still not perfect but it may provide additional inspiration! 2 If$\mathbb{D}\subset f(\mathbb{D})$, since$f$is injective, the function $$h = f^{-1}\lvert_\mathbb{D}$$ satisfies all conditions with$h(\mathbb{D}) \subset \mathbb{D}$. So the second follows from the first. 2 The theorem asserts that$A'$exists and equals$f$. The argument given appears to assume$A'$exists, rather than treating that as a part of what was to be proved. You can argue as follows. Suppose$f$is continuous. $$\frac{A(x+h)-A(x)}{h} = \frac 1 h \int_x^{x+h} f(t)\,dt \tag 1$$ Given$\varepsilon>0$, we can say that$f(t)\$ stays between ...

Only top voted, non community-wiki answers of a minimum length are eligible