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0

Here's another point of view. For each $n\in\mathbb N$ choose an integer $k(n)\in\{0,1,\ldots,n\}$ such that the following two properties are satisfied: $$\lim_{n\to\infty}\frac{k(n)}{n}=1\qquad\text{and}\qquad\lim_{n\to\infty}\left(\frac{k(n)}{n}\right)^n=0.\tag{$\ast$}$$ Some possible choices are: $$k(n)=\lfloor n-\log{n}\rfloor\qquad\text{or}\qquad ...


2

Write $$ f_{-1}(x)=0;\qquad f_n(x) = x^{f_{n-1}(x)} $$ so that $f_0(x)=1$ and the higher ones are the same as before. Then $$ \frac{d}{dx} f_n(x) = \sum_{k=1}^n x^{\left(-1+\sum_{j=n-1-k}^{n-1}f_j(x)\right)}\log^{k-1}(x) $$


0

By translating we may assume $x=0$ for convenience, and then by rescaling we have $$\frac{1}{c_nr^n}\lambda_n(B_r(0)\setminus B_r(y)) = \frac{1}{c_n} \lambda_n(B_1(0)\setminus B_1(y/r)).$$ Now observe $y/r\to 0$ as $r\to\infty$ and say "dominated convergence theorem". The dominated convergence theorem is overkill. To be more elementary, $B_1(0)\cap ...


3

My approach would be to try to find derivatives inductively. First we have $f_1'(x)=1$. Then differentiating $f_2(x)=x^{f_1(x)}$ we have $$ \log(f_2(x))=f_1(x)\log x\implies\frac{f_2'(x)}{f_2(x)}=f_1'(x)\log x +\frac{f_1(x)}{x}=1+\log x $$ so that $f_2'(x)=f_2(x)(1+\log x)=x^x(1+\log x)$ and more generally $$ f_n'(x)=f_n(x)\cdot\left(f_{n-1}'(x)\log ...


0

Let us analyse the upper sum: $U_{f,P} = \sum_{i=1}^n(x_i - x_{i-1})M_i$ where $M_i = \sup_{x\in[x_{i-1}, x_i]} f(x)$. So, we have to look for a $\sup$ for any interval $[x_{i-1}, x_i]$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, any interval contains rational numbers, and hence, $\sup f(x)$ is equal to $x_i$ (even if $x_i$ is irrational, there is a ...


0

Any interval in any subdivision of $[a,b]$ must contain a rational and irrational point, since the intervals are not singletons nor empty. This is because the irrationals and rationals are both dense in $\mathbb{R}$. We then have that the supremum of $f$ on any non-empty, non-singleton closed interval $[x,y]$ is $\sup[x,y] = y$ and the infimum of $f$ on any ...


0

Let $v(x) = x_1^2 + \cdots +x_n^2$. Then $v \ge 0$ and $\Delta v = 2n$. If $u \ge 0$ and $\Delta u = -1$ then $u + v/2n \ge 0$ and $\Delta(u + v/2n) = 0$. A nonnegative harmonic function is constant, so that $u + v/2n = C$ for some constant $C$. This leads rather quickly to a contradiction since $v(x) \to \infty$ as $|x| \to \infty$.


2

Hint $$\sup_{x\in[0,a]}\left|\frac{\cos(nx)(nx)^{1/2}}{(n+1)^2}\right|\le\sqrt a\frac{1}{n^{3/2}}$$ and the series $\sum \frac{1}{n^{3/2}}$ is convergent.


2

I don't know where you're getting the $(2k+1)$-form. If you wedge $\omega$ with itself $k$ times, you get — as you yourself observed — a $2k$-form. Let's try an example: With $n=2$ and $\omega = dx_1\wedge dx_2+dx_3\wedge dx_4$, we'll have $\omega\wedge\omega = 2 dx_1\wedge dx_2\wedge dx_3\wedge dx_4$. Since $2$-forms commute with one another, you should be ...


0

A piecewise constant function lies below $f$ if and only if it is nonpositive everywhere. So the supremum of the lower sums of $f$ is zero. Among piecewise constant functions lying above $f$, for any $\delta>0$ we have a function which is $1$ on $[-\delta,\delta]$ and zero elsewhere. The integral of this function is $2 \delta$, which can be made ...


0

Fix $1> \epsilon > 0$, and consider the step function $$ \varphi(x) = \begin{cases} 1 &: -\epsilon \leq x \leq \epsilon \\ 0 &: \text{ otherwise} \end{cases} $$ Then $$ \int_{-1}^1 \varphi(x)dx = 2\epsilon $$ Hence $$ \int_{-1}^{1^{\ast}} f(x)dx = 0 $$ Now for any step function $\psi \leq f$, then choose a partition $-1 < x_1 < x_2 < ...


2

Another method : Thanks to some known properties of digamma function, especially the asymptotic expansion, the limit for $n$ tending to infinity is obtained $=\frac{\ln(2)}{3}+\frac{\pi\sqrt 3}{9}$ Of course, this method is arduous. but it is a nice exercise for the lovers of special functions.


5

Your integral is correct, since the terms of your sum can be rewritten as $$\frac{n^2}{n^3+i^3} = \frac{1}{n} \cdot \frac{1}{1+(i/n)^3}$$, which means that your sum is a Riemann sum for the corresponding integral. If you are having a hard time getting started on evaluating the integral, you can factor $x^3+1 = (x+1)(x^2-x+1)$, and then focus on rewriting ...


1

For a less pathological example than the Cantor set, consider $$\{0\}\cup\bigcup_{n=1}^\infty\left\{\frac1n\right\}$$


2

(Some of this has already been pointed out by David G.Stork) I think a problem with how abstract algebra is taught many times is we as students lack a lot of motivation or background as to where groups come from. So we have very little idea of what the people who developed and axiomatized the field had in their mind at the the time. This can cause us to ...


-5

As the course name says, it is very abstract. So there is not a lot of visualization, but more combinatorials skills; that is, we have to play with the concepts and do some examples to convince us. However, the mathematical analysis is a concept that requires more visualization. We can refer to the graph (eg $f (x) = e ^ x$) when we want to compute ...


9

There is a large body of psychological evidence that mathematical skill falls into two general categories: formal and abstract visual and geometric Most mathematicians excel in one of these domains (e.g., Ramanujan, formal; Desargues, geometry), but a few in both (Cox). To speak from my own experience: I am certainly a visual/geometer and can ...


1

Start with $\ln (1+x) =\int_1^{1+x} \frac{dt}{t} =\int_0^{x} \frac{dt}{1+t} $ for $x \ge 0$. The finite series expansion for $\frac1{1+t}$ is, for any $n$, $\frac1{1+t} =\sum_{k=0}^{n-1} (-1)^k t^k + \frac{(-1)^n t^n}{1+t} $. Substituting this, $\begin{array}\\ \ln (1+x) &=\int_1^{1+x} \frac{dt}{t}\\ &=\int_0^{x} \frac{dt}{1+t}\\ &=\int_0^{x} ...


6

You could convert yours into a direct proof easily. As you have, let $$f(x) = \ln(1+x) - x + \dfrac{x^2}2 - \dfrac{x^3}3$$ We then have $$f'(x) = \dfrac1{1+x} - 1 + x - x^2 = \dfrac{-x^3}{1+x} < 0 \text{ since }x \in (0,\infty)$$ Hence, $f(x)$ is a decreasing function, which means $$f(x) < f(0) \implies \ln(1+x) - x + \dfrac{x^2}2 - \dfrac{x^3}3 < 0 ...


0

Since it is an approximation of the function around zero you use Maclaurin series.


0

A little different (again no L'Hopital): Set $g(x) = [\ln (1+x)]/x = 1-x/2+x^2/3 - \cdots .$ Our expression can be written $$-\frac{e^{g(x)}-e^{g(0)}}{x-0} \to -(e^g)'(0) = -e^{g(0)}g'(0)= -e\cdot (-1/2) = e/2.$$


3

Without using l'Hospital (I tend to avoid it as much as possible, as it always looked like a heavy hammer to me): $$ (1+x)^{1/x} = e^{\frac{1}{x}\ln(1+x)} = e^{\frac{1}{x}(x-\frac{x^2}{2}+o(x^2))} = e^{1-\frac{x}{2}+o(x))} = e(1+\frac{x}{2}+o(x)) = e-\frac{xe}{2}+o(x) $$ using the Taylor expansions of $\ln(1+x)$ and $e^x$ when $x\to0$. Plugging it back, the ...


4

$$\lim_{x\to0}\frac{e-(1+x)^{\frac{1}{x}}}{x}\stackrel{l'H}=\lim_{x\to 0}\left(\frac1{x^2}\log(1+x)-\frac1{x(1+x)}\right)(1+x)^{1/x}=$$ $$=\lim_{x\to 0}\color{red}{(1+x)^{1/x}}\frac{(1+x)\log(1+x)-x}{x^2(1+x)}$$ Now, on the problematic factor above: $$\lim_{x\to ...


0

I think I've fully understood this question. Yes, Przemysław Scherwentke's answer here is correct. However, [$[a,b] - \mathop {\cup}^{n}_{j = 1} I_j$] is more complicated than the general title coz $I_j$ and [a,b] both are intervals here(distinct by open and closed) that means they all have interiors that are not empty. Since {$j$} is finite, that means if ...


1

For 1 see the asymptotic $f_1(x) \sim \ln(x^2) = 2\ln |x|$ and note that $1+x^2 \ge 1$ so $f_1(x) \ge 0$ and opens up like $\ln$ to both sides (resembles a smooth bucket) Similarly, the shape of $f_2$ is like the shape of $\frac1{1+x^2}$, somewhat like a bell with its only maximum at $0$ with value $4$. For 3 just try some algebraic manipulations: ...


1

Let $a = (a_n)_{n\in \Bbb N}$ and $b = (b_n)_{n\in \Bbb N}$ be in $\ell^p$. If $\|a + b\|_p = 0$, the inequality is clear, so suppose $\|a + b\|_p > 0$. Since $$|a_n + b_n|^p = |a_n + b_n| |a_n + b_n|^{p-1} \le (|a_n| + |b_n|)|a_n + b_n|^{p-1}$$ for all $n\in \Bbb N$, then $$\|a + b\|_p^p = \sum_{n = 1}^\infty |a_n + b_n|^p \le \sum_{n = 1}^\infty ...


0

How about this: let $\mu$ be the counting measure on $\mathbb N$. Then $$\int_{\mathbb N} |a| \, d\mu = \sum_{n=0}^\infty |a_n|$$ for any sequence $a = (a_n)_{n \in \mathbb N}$. If $a,b \in L^p(\mathbb N,\mu)$, the Minkowski inequality gives you $$ \|a+b\|_{L^p(\mathbb N,\mu)} \le \|a\|_{L^p(\mathbb N,\mu)} + \|b\|_{L^p(\mathbb N,\mu)}$$ which says that $$ ...


0

The definition for the limit of the function says: If a limit $$\lim_{x \rightarrow a}f(x)=L$$ exists, then given any real number $\epsilon>0$, there exists another real number $\delta>0$ such that if $0< |x-a| <\delta $, then $|f(x)-L|<\epsilon$. Now let us consider the first case, that is: $$\lim_{x \rightarrow a}f(x)=L ...


2

Let $f^{(1)}(x)=\cos x$ and $f^{(n+1)}(x)=\cos\left(f^{(n)}(x)\right)$. For any $x\in I=[0,\pi/2]$ we have: $$ \lim_{n\to +\infty} f^{(n)}(x)=\xi \tag{1}$$ where $\xi$ is the only root of $x-\cos x$ over $I$, i.e. $\xi=0.739\ldots$. Quite trivially $\xi$ is also a root of $x-f^{(n)}(x)$ over $I$, and it is the only root, since the existence of another ...


1

In order to apply Banach, you need a $q<1$ such that $|f(x)-f(y)|\le q|x-y|$. For $\cos$ on $(0,\pi)$ we do not have such $q$ (we cannot pick $q$ smaller than $\lim_{x\to\frac\pi2}\cos'(x)=1$). Also, $(0,\frac\pi2)$ is not complete. However, any $x$ we are looking for is certainly $\in[0,1]$ and here we can let $q=\sin 1<1$ for $\cos$ itself and this ...


1

If $(a_n)\in\ell_1$, then it is bounded. There is a constant $M>0$ such that $|a_n|\le M$ for all $n$. Then for any $p>1$ we have $$ \sum|a_n|^p\le M^{p-1}\sum|a_n|<\infty, $$ proving that $\ell^1\subset\ell^p$.


2

By using polar coordinates: $$ \iint_{x^2+y^2\leq R^2}\frac{\sin(x^2+y^2)}{(x^2+y^2)^{\alpha}}=2\pi\int_{0}^{R}\frac{\sin(\rho^2)}{\rho^{2\alpha-1}}\,d\rho = \pi\int_{0}^{R^2}\frac{\sin(z)}{z^\alpha}\,dz$$ and by bounding $|\sin z\,|$ with $\min(1,|z|)$ we have that $\frac{\sin z}{z^{\alpha}}\in L^1(\mathbb{R}^+)$, provided that $\alpha\in(1,2)$.


1

The statement $x_i=x_m({a_i \over a_m})^{1\over q-1}$ holds for all $i$ including $i=m$, so: $$ A=\sum_{i=1}^{n}a_ix_m({a_i\over a_m})^{1\over q-1} ={x_m\over a_m^{1\over q-1}} \sum_{i=1}^na_ia_i^{1\over q-1} $$ Now add the exponents on $a_i$: $1 + {1\over q-1}={q\over q-1}$.


2

As user40276 has mentioned, using Taylor’s Theorem, we can write $$ \forall \mathbf{x} \in \mathbb{R}^{n}: \quad f(\mathbf{x}) = f(\mathbf{a}) + \sum_{i = 1}^{n} \frac{\partial f}{\partial x_{i}} \Bigg|_{\mathbf{x} = \mathbf{a}} \cdot (x_{i} - a_{i}) + R(\mathbf{x}), $$ where $ R: ...


1

I finally got to a conclusion and I decided to share it. First I'll rewrite the corollary in a precise form. For this, in what follows I'll consider the continuous maps $\pi:\mathbb R^m\longrightarrow \mathbb R^n$ and $\jmath:\mathbb R^n\longrightarrow \mathbb R^m$ given by $$\pi(x_1, \ldots, x_m):=(x_1, \ldots, x_n)\quad \textrm{and}\quad \jmath(x_1, ...


0

Note that $$\left|a-a_n\right|<\varepsilon$$ Implies: $$\left|a-a_n\right| \leq \varepsilon$$ On the other hand: $$\left|a-a_n\right| \leq \varepsilon$$ Implies: $$\left|a-a_n\right| < 2\varepsilon$$


0

Paraphrasing https://proofwiki.org/wiki/Lagrange_Polynomial_Approximation Let $p$ be the interpolating polynomial for $f$ at $x_0<x_1<…<x_n$ of degree $n$. Let $R(x)=f(x)-p(x)$ be the error. Define $$ g(t)=(x-x_0)…(x-x_n)·R(t)-(t-x_0)…(t-x_n)·R(x) $$ Then $g$ has at least $n+2$ roots defining $n+1$ bounded intervals, $g'$ has at least $n+1$ roots, ...


0

It seems there is no such formula. Vaguely speaking even for simple cases this problem is NP-hard. For details see this paper


1

$\boldsymbol{\langle x,y\rangle=0\implies\|x+\alpha y\|^2\ge\|x\|^2}$ $$ \begin{align} \|x+\alpha y\|^2 &=\|x\|^2+2\mathrm{Re}\left(\langle x,\alpha y\rangle\right)+|\alpha|^2\|y\|^2\\ &=\|x\|^2+2\mathrm{Re}\left(\overline{\alpha}\langle x,y\rangle\right)+|\alpha|^2\|y\|^2\\ &=\|x\|^2+|\alpha|^2\|y\|^2\\ &\ge\|x\|^2 \end{align} $$ ...


2

If $(x,y) \ne 0$ then $\|y\|\ne 0$, and the following is an orthogonal decomposition: $$ x = \left[x-\frac{(x,y)}{(y,y)}y\right]+\frac{(x,y)}{(y,y)}y. $$ Hence, $$ \|x\|^{2} = \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2}+\frac{|(x,y)|^{2}}{\|y\|^{2}} > \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2}. $$


1

Your intuition is correct: insert a $T=T_n f(x;a)$ in your fraction as you say. You'll get $\lim \limits_{x \to a} \space \frac {f(x) - T(x)} {(x-a)^n} + \lim \limits_{x \to a} \space \frac {T(x) - P(x)} {(x-a)^n}$. The first limit is $0$ by the definition of the Taylor polynomial. This means that the second limit must be $0$ too. Let $T-P = a_m (x-a)^m + ...


0

For the converse, if $\langle x,y \rangle \ne 0$ then in line 2 of your argument you can choose $\alpha \in \mathbb{R}$ to make the r.h.s. smaller than $||x||^2$, since for any constants $a \ne 0,b$ (in this case $a=2\langle x,y\rangle$ and $b=||y||$) you can choose $\alpha$ such that $a\alpha + b\alpha^2 < 0$. (the parabola $y=ax+bx^2$ intersects but ...


0

Assumig \begin{align*} \Vert x \Vert^2 + 2 \Re \bar{\alpha} \langle x, y \rangle + \vert \alpha \vert^2 \ \Vert y \Vert^2 \ge \Vert x \Vert^2, \end{align*} we get \begin{align*} 2 \Re \bar{\alpha} \langle x, y \rangle + \vert \alpha \vert^2 \ \Vert y \Vert^2 \ge 0 \end{align*} and \begin{align} \Re \bar{\alpha} \langle x, y \rangle \ge -\frac{1}{2} \vert ...


0

Follow the hint given in the question: use the argument in Sec. 5.11, use the Banach-Steinhaus Theorem in exactly the same way with the change of the linear functionals to ${\Lambda _n}f = \frac{1}{{{\lambda _n}}}{s_n}(f;0)$. The proof is almost word for word there. Again the hint suggest a better estimate for the 1-norm of the Dirichlet kernel Dn. In the ...


2

If there is no restrictions on the metric $e$ then the claim can be false, even when $Y=\{y_0\}$ is a one-point set. For instance, let $X$ be the real line, $e((x’,y_0),(x’’,y_0))=|x’-x’’|$ and $d_X(x’,x’’)=\operatorname{arctan}|x’-x’’|$ for each points $x$ and $x’$ of the space $X$. For each $n$ put $x_n=n$. Then a sequence $\{x_n\}$ is $d_X$-Cauchy, but a ...


3

Let $v = u^4$ and rewrite the PDE in terms of $v$ to get: $$ v^{-3/4} v_t = \Delta v. $$ Let $\;v = A(t) B(x,y,z)\;$ and separate variables to get: $$ A' = 4 A^{7/4} c \quad \mbox{and} \quad B^{3/4} \Delta B = c \quad \mbox{for a constant } c . $$ The DE in $A$ has the solution $\;A(t)^{3/4} = -\frac{4/3}{4 c t + k}\;$ for a constant $k$. So when $t$ goes to ...


0

since $$\dfrac{a^{i/n}}{n+1}<\dfrac{a^{i/n}}{n+1/i}<\dfrac{a^{i/n}}{n}$$ so $$\sum_{i=1}^{n}\dfrac{a^{i/n}}{n+1}<\sum_{i=1}^{n}\dfrac{a^{i/n}}{n+1/i}<\sum_{i=1}^{n}\dfrac{a^{i/n}}{n}$$ and Note ...


0

If the convergence were uniform, we would have $f_n(x)<1/3$ on $[0,1)$ for all sufficiently large $n.$ But note $f_n(1-1/n) = (1-1/n)^n \to 1/e > 1/3,$ contradiction


1

If $x\geq5$, then $e^{-x}\leq\displaystyle\frac{1}{x^3}$. Since for large enough $n$, we have $y_\min\geq5$, then: $$0<\sum_{i=1}^n e^{-y_i}\leq \sum_{i=1}^n \frac{1}{y_i^3}\leq \sum_{i=1}^n\frac{1}{y_\min y_i^2}\leq \frac{1}{y_\min}$$ Since $\displaystyle\frac{1}{y_\min}\rightarrow 0$ as $n\rightarrow \infty$, then so does $\sum_{i=1}^n e^{-y_i}$



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