# Tag Info

9

To put that into a more reasonable (though less prone to enthusiastic upvotes) form: We have $x_1=e^{1/e}$ and $x_{n+1}=e^{x_n/e}$, and we're looking for $\lim_{n\to\infty}n\,(e-x_n)$. Letting $y_n=1-x_n/e$, some elementary algebra gives $$y_{n+1}=1-e^{-y_n}.$$ The precise value of $y_1$ is not so important, as long as it is positive. Then, $y_n$ is monotone ...

8

Presume such a function $f$ existed. Then the function $g(x)=x^2f(x)$ would be another continuous function on $[0,1]$ such that $\int_0^1 x^n g(x)dx=0$ for all $n=0,1,2\ldots$. Now, that would imply that $\int_0^1p(x)g(x)dx=0$ for every real polynomial $p$, and so, due to density of all polynomials, you would have $g(x)=0$. Thus, $f(x)=\frac{g(x)}{x^2}=0$ ...

6

The only difficulty here is justifying switching the order of integration and the summation in $$\int_0^1 \frac{dx}{1+x^3} = \int_0^1 \sum_{n=0}^\infty (-x^3)^n\, dx = \sum_{n=0}^\infty \int_0^1 (-x^3)^n\, dx = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1}$$ We don't have uniform convergence of the series on the interval $[0,1]$ since the series diverges at $x=1$. ...

5

For a really simple proof, @saulspatz gave you a nice little observation. We have $$j^2 + 1 > j^2 - 1$$ and so $$\frac{j^2 + 1}{j+1} > \frac{j^2-1}{j+1} = \frac{(j+1)(j-1)}{(j+1)} = j-1$$ so $$\frac{j^2+1}{j+1} > j-1$$ since ${\log}$ is increasing, we have $$\log\left(\frac{j^2+1}{j+1}\right) \geq \log\left(j-1\right)$$ can you finish it ...

5

\begin{align} \det(AB-I_3) &= (-1)^3\det(I_3 - AB)\\ &= - \det(I_2 - BA)\\ &= - (-1)^2 \det(BA-I_2) \end{align} The second equality holds due to Sylvester determinant identity. In general, if $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{n \times m}$, then we have \det(AB-I_m) = (-1)^{n+m} \det(BA-I_n)= (-1)^{n+m \pmod{2}} \det(BA-I_n)... 5 First approach. By the beta integral \begin{align*} \frac{{\Gamma (x)}}{{\Gamma \left( {x + \tfrac{1}{2}} \right)}} & = \frac{1}{{\Gamma \left( {\frac{1}{2}} \right)}}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}\int_0^1 {\frac{{t^{x - 1} }}{{\sqrt {1 - t} }}dt} \\ & = \frac{1}{{\... 4 Hints: (1) The function is continuous on any compact interval [-a,a]. (2) The derivative is bounded on semi-infinite intervals (-\infty,-a] and [a,\infty) where a > 0. 4 Hint: Rewrite the DE as\frac{dx}{dy}-\frac{x}{y}=y$$which is easy by integrating factor method 4 We have$$ \zeta ( - 1,n + 1) = - \frac{1}{2}n^2 - \frac{1}{2}n - \frac{1}{{12}} and \begin{align*} \zeta ^{(1,0)} ( - 1,n + 1) = \zeta ^{(1,0)} ( - 1,n) & + n\log n = - \frac{1}{4}n^2 + \left( {\frac{1}{2}n^2 + \frac{1}{2}n + \frac{1}{{12}}} \right)\log n + \frac{1}{{12}} \\ & - \int_0^{ + \infty } {\left( {\frac{1}{{e^t - 1}} - \frac{1}{t} ... 4 Following up on a line of reasoning suggested by MikeG and Danny Pak-Keung Chan: Let \mu be the law of the X_n (which is also the law of X). Fix \epsilon, \eta > 0. By Lusin's theorem applied to the measure \mu, there is a compact set K \subset \mathbb{R} with \mu(K) > 1-\eta on which which f|_K is continuous. It is thus even ... 4 You can use the fact that a^m-1=(a-1)(a^{m-1}+a^{m-2}+\cdots+1), with a=\sqrt[m]{P(x)+1}. You will get that\frac{\sqrt[m]{P(x)+1}-1}x=\frac{P(x)}{x\left(\sqrt[m]{P(x)+1}^{m-1}+\sqrt[m]{P(x)+1}^{m-2}+\cdots+1\right)}.Can you take it from here? 4 The magic words are: Stirling's formula. (check out the wikipedia article on the gamma function). 4 Let's go back to basics, where M = [0,1]. I'm sure you can construct a non-zero function f: M \to \Bbb R such that \int_M f(x) \ \mathrm dx = 0. 3 As mentioned in the last comment, you need to find an r such that B(x,r)\subset [0,3), whenever x\in [0,3). Hint: Instead of x, you can write 0\le x=3-\epsilon<3, where 0<\epsilon\le 3. Namely, you can express every element in [0,3) as 3-\epsilon, for some \epsilon\in(0,3]. 3 It comes from computing the sum of a geometric series:\begin{align}1+\frac1{n+1}+\frac1{(n+1)^2}+\cdots&=\frac1{1-1/(n+1)}\\&=\frac{n+1}n.\end{align}So,\frac1{(n+1)!}\left(1+\frac1{n+1}+\frac1{(n+1)^2}+\cdots\right)=\frac{n+1}{(n+1)!n}=\frac1{n!n}.$$3 Simply False. Take a_n=(-1)^n and b_n=\frac{1}{n}. Then a_nb_n \to 0 !! 3 Take a_n=n and b_n=\frac{1}{n}. The product converges 3 Using the density of irrationals and rationals and the monotonicity of x \mapsto \cos^2x on [0,\pi/2], we have for any subinterval [x_{j},x_{j+1}] of a partition P,$$\sup_{x \in [x_{j},x_{j+1}]} f(x) = \cos ^2 x_{j}, \quad \inf_{x \in [x_{j},x_{j+1}]} f(x) = 0$$Immediately we see that the lower Darboux sum is L(P,f) =0 and the upper Darboux sum ... 3 Your approach is correct. The obvious choice of \mathcal{C} would of course be \mathcal{M}. Hint: \mathcal{M} must contain all unions of the form \bigcup_{n=1}^\infty A_n, where A_1,A_2,\dots \in \mathcal{F}. Is \{\bigcup_{n=1}^\infty A_n \: | \: A_1,A_2,\dots \in \mathcal{F}\} a \sigma-algebra? 3 There are a few problems with your approach: when you write that you have to show that$$\overline{S_\sigma}<\varepsilon\ \forall\varepsilon>0,\tag1$$you don't tell was which partition \sigma is, but then you acto as if it was a concrete partition. For instance, how do you know that \frac\varepsilon2 belongs to the partition. Besides, what you need ... 3 It should read: \left|\dfrac{x-a}{(1+x)(1+a)}\right| < |x-a| < \epsilon if you let \delta = \epsilon > 0. 3 When you solve the recurrence a_n=a_{n-1}+a_{n-2}, you find that the solutions all have the form$$a_n=A\varphi^n+B\hat\varphi^n\,,$$where \varphi=\frac12(1+\sqrt5) and \hat\varphi=\frac12(1-\sqrt5); the values of A and B are determined by the initial values a_0 and a_1. And \varphi\hat\varphi=-1, so,$$\begin{align*} \frac{a_{n+1}}{a_n}&...

3

$\Rightarrow)$ Suppose $A'$ is dense in $X$. Fix $x\in X$ and $\varepsilon>0$. By assumption, there is some $b\in A'$ such that $0<d(x,b)<\frac{\varepsilon}{2}$. Since $b\in A'$, there is some $a\in A$ with $0<d(a,b)<\frac{\varepsilon}{2}$. Note $d(x,a)<\varepsilon$ by the triangle inequality. This shows that $A$ is dense. The other ...

3

$$\int_{-\infty}^{\infty} |\frac {\sin (nx) \sin x}{ x^{2}}|dx$$ $$\geq \int_0^{1} |\frac {\sin (nx) \sin x}{ x^{2}}|dx$$ $$\geq \frac 2 {\pi} \int_0^{1} |\frac {\sin (nx) }{ x}|dx$$ since $\sin x\geq \frac 2 {\pi} x$ for $0 <x<1$. Now put $y=nx$ and use the fact that $\int_0^{\infty} |\frac {\sin y} y| dy =\infty$.

3

Let us try to understand the computer-based answer of Oliver Oloa. Rational decomposition. Is known the rational decomposition of the trigonometric functions in the forms of $$\tan z = \sum\limits_{n=0}^\infty\dfrac{8z}{\pi^2\left(2n+1\right)^2 - 4z^2},\tag{R1}$$ $$\sec z = \sum\limits_{n=0}^\infty (-1)^n\dfrac{4\pi(2n+1)}{\pi^2\left(2n+1\right)^2 - 4z^2},\... 3 There is such a notion, which is used often in multivariable calculus before one introduces the Lebesgue integral. The motivation is the same - we want the volume under the graph of f: \mathbb R^n \longrightarrow \mathbb R. The approach of summing the area of rectangles with Riemann sums generalizes to summing the volume of higher dimensional rectangles. ... 3 This is about feedback on your proof. Your goal is to find a partition \sigma such that \overline{S}_{\sigma} (f) <\underline{S} _{\sigma} (f) +\epsilon  but instead you are trying to use that f is integrable on [\epsilon/2,1] and prove that f is integrable on [0,\epsilon/2]. This is not what you want. You want a partition \sigma which ... 3 You mean x_n = \sqrt[4]{n}, then on [m, 2m], m > 1, f(x) = \sqrt[4]{x} is differentiable thus using MVT yields: \left|x_m - x_{2m}\right| = \left|\sqrt[4]{m} - \sqrt[4]{2m}\right| \ge |2m - m|\cdot \dfrac{1}{4\sqrt{m}} = \dfrac{\sqrt{m}}{4} > \dfrac{1}{4} = \epsilon. Thus it’s not Cauchy ( sequence ). 3 Since proving that$$\lim_{n\to\infty} (\sin(n\alpha))^{1/n} = 1$$may be difficult, and for some \alpha it's just not true, I'd first use the comparison test. We have$$ 0 \le \left|\frac{\sin(n\alpha)}{(\ln (10))^n}\right| \le \frac{1}{(\ln (10))^n} $$Since$$ \sum_{n=1}^\infty \frac{1}{(\ln (10))^n}  is convergent (by the root test), then it means ...

3

We can do a proof by “ contradiction “. Assume $s_n + t_n \to L < \infty$, then $s_n + t_n$ is bounded, say by $M$. Thus $|s_n| = |s_n + t_n - t_n| \le |s_n + t_n| + |t_n| \le M + T$. So $s_n$ is bounded, but it is unbounded since $s_n \to \infty$. Therefore $s_n + t_n \to \infty$.

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