# Tag Info

4

Hint: $$n!>\left(\frac{n}{2}\right )^n$$ for large enough $n$.

4

Pick $K$ as large as you want, then $n!\ge K^{n-K}$ (for $n\ge K$), so $$(n!)^{1/n}\ge K^{(n-K)/n}=K^{1-K/n}\to K\text{ as }n\to\infty.$$

4

Since $f$ is integrable, the function $F(x)=\int_0^xf(x)\,dx$ is continuous. From this, the linearity of the integral, and the fact that $F(b)=6$, it follows that there is a $t_1$ with $b>t_1>a$ such that $\int_{t_1}^b f(x)\,dx>2$. Now let $G(x)=\int_{t_1}^x f(x)\,dx$. $G$ is continuous. Also, we have $G(t_1)=0$ and $G(b)>2$. Now, you can ...

4

You're right that such a sequence cannot be convergent. It basically says "There is some $r$ such, for any $\varepsilon>0$ that the sequence $r_n$ is only in $(r-\varepsilon,r+\varepsilon)$ finitely many times". Now, what this is doing is clear if you start choosing $\varepsilon$ to be really big. Like, if we chose $\varepsilon=r$, then $r_n$ could only ...

3

Consider mechanical interpretation. $f'(t)\leq K$ means that instantaneous velocity at some interval $I$ can reach maximum K vaue. $\dfrac{f(b)-f(a)}{b-a}\leq K$ means that then average velocity at any subinterval of $I$ can't be bigger than K.

2

Note that $x(n+1-x)\ge (1)(n)$ for $1\le x\le n$. It follows that $$(n!)^2=(1)(n)(2)(n-1)(3)(n-2)\cdots (n)(1)\ge n^n$$ and therefore $(n!)^{2/n}\ge n$.

2

If you know that $C_0^\infty((0,1))$ is dense in $L^2([0,1])$, a natural approach is to do what you did: take a sequence $\phi_n$ in $C_0^\infty((0,1))$ converging to $f$ in $L^2$ and obtain, $$\int_0^1 |f|^2 = \int_0^1 (f-\phi_n)\overline{f} \le \|f-\phi_n\|_{L^2}\|f\|_{L^2}\to 0$$ hence $\int_0^1 |f|^2=0$. Without the density result, I would use a ...

2

Thomas Bayes showed that $${n \choose k}\int_0^1 x^k (1-x)^{n-k}\mathrm dx = \frac{1}{n+1}$$ by pure thought, without using calculus (for all integers $k,n$ with $0 \leq k \leq n$). His argument, known as the Bayes' billiards argument, uses two equivalent probabilistic stories about picking random points on a number line from $0$ to $1$. That's just one ...

2

Here is an example. The question is how to show that $$\binom{n}{k}^{-1}=(n+1)\int_0^1 x^k (1-x)^{n-k} \, dx.$$ To make this self-contained, I'll paste this answer below: Let's do it somewhat like the way the Rev. Thomas Bayes did it in the 18th century (but I'll phrase it in modern probabilistic terminology). Suppose $n+1$ independent random variables ...

2

There is an $M>0$ with $$|f''(x)|\leq M\qquad\forall\>x\in{\mathbb R}\ .\tag{1}$$ Claim: $\quad f'^2(x)\leq 2M \>f(x)\quad\forall x\in{\mathbb R}$. Proof. $\$ When $f(x)=0$ at some point $x\in{\mathbb R}$ then $f'(x)=0$ as well, since otherwise $f$ would assume negative values in some points. Therefore it is enough to show that $f(0)=:y_0>0$ ...

2

These two cases give a contradiction: $\forall t_1,t_2\in(a,b)$: $$F(t_1,t_2)=\int_{t_1}^{t_2}f(x)dx<2$$ so by continuity we get $$6=\lim_{(t_1,t_2)\to(a,b)}F(t_1,t_2)\le2$$ $\forall t_1,t_2\in(a,b)$: $$F(t_1,t_2)=\int_{t_1}^{t_2}f(x)dx>2$$ so we get $$0=\lim_{(t_1,t_2)\to(a,a)}F(t_1,t_2)\ge2$$ so there's $t'_1,t''_1,t'_2,t''_2$ such that ...

2

Your reindexed summation is correct. Since the sums (outer and inner) for the second series start at $n = 2$ resp. $k = 2$ while in the first they start at $0$, combining the two power series is not totally direct. One way is to split the sums, treat $n = 0$ and $n = 1$ separately, and in the remaining series, for the inner sum $k = 0$ and $k = 1$. A more ...

2

As in differential geometry, we are defining the "distance" between two points to be the inf of the "lengths" of all paths joining the points. (For example, with the usual notion of lengths of paths in $\Bbb R^2-\{0\}$, what is the distance from $(-1,0)$ to $(1,0)$? Note that you cannot use the usual line segment joining them.)

2

Since: $$\left|\sum_{k=1}^{n} a_k\right|\leq \sqrt{n}\sqrt{\sum_{k=1}^{n} a_k^2}\leq n^{\frac{3}{4}}\left(\sum_{k=1}^{n}a_k^4\right)^{\frac{1}{4}}$$ by applying the Cauchy-Schwarz' inequality twice (or the Holder's inequality once), we have: $$\frac{1}{n}\left|\sum_{k=1}^{n}a_k\right|\leq\left(\frac{1}{n}\sum_{k=1}^{n}a_k^4\right)^{\frac{1}{4}},$$ so, if ...

2

Let $(x_n)$ be a sequence in some linearly ordered set $(X,<)$. Call $n \in \mathbb{N}$ a peak of the sequence if for all $m > n$ we have $x_n > x_m$, i.e. $x_n$ is larger than all subsequent terms of the sequence. Case 1: There are infinitely many peaks: suppose $n_1 < n_2 < n_3 < \ldots$ are infinitely many peaks of the sequence. Then ...

1

It's clear that $u_k\ge1$ for all $k$ so we get (I think there's a typo in the question) $$u_n=\left(\sum_{k=1}^{\color{red}{n-1}}u_k\right)^{1/2}\ge \sqrt{n-1}$$ hence we get the desired limit of $u_n$, moreover we have $$v_n(u_{n+1}+u_n)=u_n\iff \frac1{v_n}=1+\frac{u_{n+1}}{u_n}\xrightarrow{n\to\infty}2$$ so we get the limit of $v_n$ and ...

1

This will not work: Consider a sequence $p_0, p_1, …$ made by zipping $1/n$ and $1/2^n$ – still converging to zero (and only from above), but not monotonically. You can fix this by working with rings instead of balls. Take a limit point $L$ of the sequence $a_1, a_2, …$. You may assume that it has no constant subsequences. So either you will find an ...

1

Suppose wlog that the points $p_{i_0}, p_{i_1}, ...$ strictly less than $L$ are infinitely many and converges to $L$. Then you can build up this sequence. $a_0 = p_{i_0}$, $a_n \in (a_{n-1}, L)$ (of course taking $a_k$ in increasing order with respect to the indexes of $p_i$ sequence). This new sequence should work.

1

You're almost done. Note that $g'(x)$ dominates $\sqrt{x} g'(x)$ on this interval, and the integral of $g'(x)$ is less than or equal to $g(1)-g(0)$. This doesn't require absolute continuity for $g$. See Proposition 22 here.

1

I think if you consider $f(x,y,z,n)$ as a sequence $f_n(x,y,z)$ of functions, then your statement says that $f_n$ converges pointwise in $\{x\}\times Y$ and uniformly wrt $z$ to $0$. This can be expressed like so: \begin{align*} \exists x\in X:\forall y\in Y:\lim_{n\to\infty}\sup_{z\in Z}f(x,y,z,n)=0. \end{align*} Since this is true for each $y\in Y$ we have ...

1

In the beginning of the argument you have $|f(x) - f(x_0) - L(x - x_0)| \le \epsilon_1 |x - x_0|$ for small enough $\delta$. You can't later assert a value for $\epsilon_1$, and especially not a negative value. Your argument can be repaired. Let $\epsilon > 0$ be given. Choose $\delta_1$ so that $$|x - x_0| < \delta_1 \implies |f(x) - f(x_0) - L(x - ... 1 Function f:\mathbb R\rightarrow\mathbb R prescribed by: x\mapsto x if x\notin\mathbb{N} and x\mapsto x+1 otherwise. To avoid confusion let us say explicit that 0\notin \mathbb N. Note that f(0)=0 and f(1)=2 f is injective (straightforward). f is not continuous (if x_n\notin \mathbb N with x_n\rightarrow 1 then f(x_n)=x_n\rightarrow ... 1 Using CS, you can show, (\sum_{i} a_i/n)^2\le (\sum_{i}a_i^2/n) and similarly (\sum_{i}a_i^2/n)^2\le (\sum_{i}a_i^4/n). So using the given condition the result follows. 1 Your proof, at least as written, is wrong. We do not know that |x| = x. Since most of what you wrote for (A) depends on that claim, it's not useful. You seem to be using = to equate inequalities, which is incorrect: |x| \leq y = x \leq y means that all three of the statements |x| \leq y, y = x, and x \leq y are true. Try using words to ... 1 You can do that with the command \limits. Works too for \sum \prod etc. \bigcap\limits_{u\in \mu} u produces:$$\bigcap\limits_{u\in \mu} u\sum\limits_{k=1}^n k$$1 \bigcap_{u \in \mu} produces$$ \bigcap_{u \in \mu} $$\underset{a \in A}{\bigcap_{u \in \mu}} produces$$ \underset{a \in A}{\bigcap_{u \in \mu}} $$1 with y=y(x) we get e^x-e^yy'+y+xy'=0 thus we obtain$$y'=-\frac{e^x+y}{x-e^y}$$if x\neq e^y 1 It is not so messy if you cascade the problem for successive derivatives. Let$$F=e^x - e^y + xy = 0$$So, using implicit differentiation,$$F'_x=e^x+yF'_y=-e^y+x$$and then$$y'=\frac{e^x+y}{e^y-x} Now, compute $y''$ deriving the rhs and the result will be a function of $x,y,y'$; in the same way, $y'''$ will be a function of $x,y,y',y''$ Just added ...

1

The definition given is $\bar{d}(p_1,p_2) = \inf\{\bar{L}(\gamma) : \gamma(0) = p_1, \gamma(1) = p_2\}$. The condition $\gamma(0) = p_1$, $\gamma(1) = p_2$ means the curve $\gamma(t)$ starts at point $p_1$ and ends at point $p_2$. Since $\bar{L}(\gamma)$ is the length of the curve $\gamma$, $\{\bar{L}(\gamma) : \gamma(0) = p_1, \gamma(1) = p_2\}$ is the ...

1

does this imply that if $f:A→B$ is injective that any mapping $g:{\rm Im}(f)→A$ is bijective? No. It only means that $f: A \to f(A) = {\rm Im}(f)$ is bijective. So you can consider the inverse, but with its domain restricted to the image of the initial function. So you can have more than one left inverse. The big theorem is that if exists both the ...

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