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0

I don't know if this is quite what you're looking for, but I know I have used this one is class before and if'd you'd like to look I've also used it in a question that i posted on here. Hope it helps! $$\sum_{k=1}^nku^k={u\over (1-u)^2}\bigg[nu^{n+1}-(n+1)u^n+1\bigg]\forall u\ge 1$$ P.S.- I know it's not for $k=0$ to $n$, but i thought it might help still. ...


5

Note that for $|x| < 1$, we have $$\sum_{n=0}^\infty nx^n = x \sum_{n=0}^\infty nx^{n-1} = x \sum_{n=0}^\infty \frac{d\ }{dx} x^n = x \frac{d\ }{dx} \sum_{n=0}^\infty x^n = x \frac{d\ }{dx} \frac{1}{1-x} = \frac{x}{(1-x)^2}$$ You can now write down an expression for your sum.


2

Differentiate the first expression with respect to $x$, then multiply by $x$, then add missing initial term(s)?


1

$f$ is continuous is for every open set $V \in F$, $f^{-1}(V)$ is open in $E$. Equivalently, $f$ is continuous if for every $V \in \theta$, $f^{-1}(V)$ is in $\tau$. Note that $$f^{-1}(V) = \{x \in E \mid f(x) \in V\}.$$ So, look at all $V \in \theta$. Is $f^{-1}(\varnothing)$ open in $\tau$? Well, $f^{-1}(\varnothing) = \varnothing$, and $\varnothing \in ...


0

$$I=\int_0^1\frac{\log(u)-\log(1-u)}{u}\,dx = \int^1_0\frac{\log(u)}{u}\,du+\zeta(2)$$


0

Consider $$\lVert T(x-x_0)\rVert=\lVert T(x)-T(x_0)\rVert$$ So if $\lVert x-x_0\rVert<\delta$ we have that $\lVert T(x)-T(x_0)\rVert=\lVert T(x-x_0)\rVert<\epsilon$ by continuity at $0$.


0

Suppose $f$ is continuous at some $x_0\in X$ and let $x_n\to x^*$, for some $x^*\in X$. Then the sequence $x_n-x^*+x_0\to x_0$ and thus $\|f(x_n)-f(x^*)\|=\|f(x_n-x^*+x_0)-f(x_0)\|\to 0$. Hence $f$ is continuous at $x^*$ as well.


1

The easier way is to note that $x_n(t)$ is bounded between $-1$ and $1$. Therefore $(x_n)$ is bounded in the sup norm. If $x_n$ converges uniformly to some function $f$ then it is easy to see that $f$ must be zero, by taking the limit pointwise. Thus it suffices to look at the maximum of $|x_n|$ on $[0,1]$. You can find that the maximum is taken in ...


0

We also have $x_n(t) = t^{2n} - t^{3n} = t^{2n}(1 - t^n), \tag{1}$ whence, for all $t \in [0, 1]$, $\vert x_n(t) \vert = \vert t^{2n} \vert \vert 1 - t^n \vert; \tag{2}$ taking suprema over $[0, 1]$ yields $\Vert x_n(t) \Vert_\infty \le \Vert t^{2n} \Vert_\infty \Vert 1 - t^n \Vert_\infty, \tag{3}$ and since $\Vert t^{2n} \Vert_\infty = 1 \tag{4}$ ...


4

We have $$||x_n||_\infty\le \sup_{t\in[0,1]}|t^{2n}|+\sup_{t\in[0,1]}|t^{3n}|=2$$ hence $(x_n(t))$ is bounded on $[0,1]$.


1

I’m going to assume that by stable sequence you mean a sequence that is eventually constant: there is an $m\in\Bbb N$ such that $x_n=x_m$ for all $n\ge m$. I expect that you can prove that such a sequence converges in any space. Now suppose that $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $E$ that converges to $x$. HINT: Let $A=\{x_n:x_n\ne x\}$. ...


1

About compact: take the open cover (and prove it is such) $$\left\{\;\left(\frac1n\;,\;\;\sqrt2-\frac1n\right)\cap\Bbb Q\;\right\}_{n\in\Bbb N\setminus\{1\}}$$ Resuming (see the comments): it is closed, bounded and not compact.


0

It is not closed (in $\Bbb R$) because of what you said. It is not compact because not closed in this space (and compactness is not related to the ambient subspace). It is obviously bounded. But what you wrote on compactness refers not to compactness, but connectedness. edit: let us prove it is closed in $\Bbb Q$: consider a sequence of rational ...


0

No. Let the metric space be $\mathbb{N}$ with this metric: $$d(m,n)=\begin{cases}\displaystyle\frac1{m\wedge n}=\frac1m\vee\frac1n&m\ne n,\\0&m=n\end{cases}$$ (where $\wedge$ stands for minimum and $\vee$ for maximum). Now let $\xi\colon\mathbb{N}\to\mathbb{N}$ be given by $\xi(n)=n+1$.


0

Let's define $E_n$ as follows: $$ E_n = \begin{cases} \{(x,y): n+1\leq d(x,y)<n+2\}, & n\geq0 \\ \{(x,y): \frac{1}{|n|+1}\leq d(x,y)<\frac1{|n|}\}, & n\leq-1 \end{cases} $$ then if $(x,y)\in E_n$ then $(\xi(x),\xi(y))\in E_{n-1}$ and in particular $\lim_n d(\xi^n(x),\xi^n(y)) = 0$ for all $x,y\in X$. From that, I guess you can proceed ...


0

Hint: Let $\xi ^n = \xi^{n-1}\circ\xi\ $ where $\xi^1 = \xi$. Let $x_0 = x\in X$, and $x_n = \xi(x_{n-1})$, i.e. $x_n = \xi^n(x)$. What can you say about $d(x_n, x_{n-1})$ as $n$ aproaches infinity? Note that since $X$ is a complete metric space, then $\tilde{x}:=\left(\displaystyle\lim_{n\to\infty} x_n \right)\in X$. Bonus: Now you can also prove that ...


3

Consider $$ \int_1^2 \frac{\log (x-1)}{x(x-1)}dx = \int_0^1 \frac{\log u}{(1+u)u}du \sim \int_0^1 \frac{\log u}{u}du, $$ which is divergent, since $$\frac{d}{du} \frac12 \log^2 u = \frac{\log u}{u}.$$ Hence the whole integral is divergent. By the way, the integral "at infinity" converges.


0

Hints: Let $t=\log(x-1)$. Then $x=1+e^t$


1

Quadratic spline is a piecewise continuous curve where each segment is a quadratic polynomial. Quadratic interpolation means given a set of data points find a quadratic spline that goes thru all the points.


-1

Let $D=\|f-g\|_\lambda$; then $|f(t)-g(t)|\le e^{\lambda t}D$, hence $$ \rho e^{-\lambda x} \int_0^x |f(t)-g(t)|\;dt \le \rho D e^{-\lambda x} \int_0^x e^{\lambda t} \;dt =\frac{\rho D}{\lambda}e^{-\lambda x}(e^{\lambda x}-1)<\frac{\rho D}{\lambda} $$ so you just need $\rho/\lambda<1$.


2

A linear operator is injective iff it's kernel is zero. The direction you want follows from linearity: If $T(x)=T(y)$ then $T(x-y)=0$ by linearity and so $x-y=0$.


0

Every continuous increasing function on the real line is quasiconvex, but not every such function is convex, so there are tons of counterexamples, e.g., $f(x) = x^3$.


0

Here is an example showing that the assumption that $H$ is Hausdorff is needed. Let $H$ be the set of natural numbers $1,2,3,\ldots$ in the cofinite topology (the topology in which a subset is closed iff it is finite or the whole space). And consider $f(n)=n+1$. Here are some facts: a) The space is $T_1$ (in fact, the minimal $T_1$-space on this set) ...


2

Begin by defining $$S(1;9) := \sum_{i=2}^9 \frac{1}{i}$$ Let $a = S(1;9)$. Then $a < 2$. Let $S(10;99)$ be the sum over the fractions of all numbers between $10$ and $99$ that don't contain a $1$ in their depiction. Prove that $S(10;99) < \frac{9}{10}a$. Similarly, define $S(100;999)$ to be the sum over the fractions of all numbers between $100$ ...


0

Under the $L^p$ metric, the completion of the space of continuous functions from $\mathbb{R}^n$ to $\mathbb{C}$ is $L^p(\mathbb{R}^n)$ for any $p$ such that $1 \le p < \infty$.


0

The above comments take two approaches, either use that (by definition) a bounded set is contained in a ball, or alternatively, that each coordinate is bounded. Assuming that a set $X$ is bounded (by definition) if there is $M>0$ such that $||x||\le M$ for each $x\in M$ (where $x=\{x_1,...,x_k\}$ and $||x||=\sqrt{x_1^2+...+x_k^2}$ ). Then $X$ is ...


0

Clearly $A$ is bounded. Let $g : \mathbb{R}^n \to \mathbb{R}^{n+1}$ be given by $$g(x_1, \ldots, x_n) = (x_1, x_2 - x_1, x_3 - x_2, \ldots, x_n - x_{n-1}, x_n)$$ Then $g$ is continuous, so the inverse image of the closed set $$[-1,1] \times [0,2] \times [0,2] \times \cdots \times [0,2] \times [-1,1]$$ is closed. But this is just $A$. We have shown that $A$ ...


0

Let $\{a_i\}$ be a sequence in $A$ which converges in $\mathbb{R}^n$. Let $a$ be the limit of this sequence. If we can show that $a \in A$, then we have shown that $A$ is closed (in fact sequentially closed, but this is enough as we are in a metric space). Letting $a_i = (a_i^1, \dots, a_i^n)$ and $a = (a^1, \dots, a^n)$, as $a_i \to a$, we know that $a_i^j ...


0

Let $\big(x^i_1,\cdots,x^i_n\big)_{i\in\mathbb{N}}$ be a sequence of members $A$ which converges to $(l_1,\cdots,l_n)$. It's equivalent to this fact that for each $j$, we have : $x^i_j\rightarrow l_j$, as $i$ tends to $\infty$. Obviously, for each $j$ we have : $-1\le l_j\le1$ . So It's sufficient to prove $l_1\le l_2\le \cdots\le l_n$. And this is also ...


0

The set $A$ is compact in $\mathbb{R}^n$ since it is closed and bounded. It is bounded since it is contained in the ball $B[0,\sqrt{n}].$ Indeed: $$ \sum_{i=1}^n x_i^2\le \sum_{i=1}^n1=n \implies (x_1,\cdots,x_n)\in B[0,\sqrt{n}].$$ To show that it is closed we will show that its complement is open. Let $(x_1,\cdots,x_n)\in A^c.$ By definition of $A$ it is ...


0

All you have to do is show that given any $\epsilon>0$, you can find a $\delta>0$ such that for all $x,y \in (0,\infty)$ where $\left|x-y \right|<\delta$ then $\left|f(x)-f(y) \right|<\epsilon$


3

$|a_m-a_n| = \left|\displaystyle \int_{n}^m \dfrac{\cos x}{x^2}dx\right| \leq \displaystyle \int_{n}^m \dfrac{|\cos x|}{x^2}dx \leq \displaystyle \int_{n}^m \dfrac{1}{x^2}dx = \dfrac{1}{n} - \dfrac{1}{m} < \dfrac{1}{n}$. Thus $\{a_n\}$ is a Cauchy sequence, hence converges.


0

Take $$B_M = \left\{x\in[0,1] \colon |f(x)|<M\right\}.$$ If $f$ is not infinity almost everywher, then $\mu(B_M)>0$ for some $M$.


1

Let $B_n = \{x \in [0,1] \mid |f(x)| \leq n\}$. Then $[0,1] = \bigcup_n B_n$ (why?) with $B_n \subset B_{n+1}$. Now use "continuity of the measure from below" to conclude $\mu(B_n) \to \mu([0,1]) = 1$ to conclude that for every $1>\varepsilon > 0$, there is some $B_n$ with $\mu(B_n) > 1-\varepsilon > 0$. Hence, you can even come arbitraryly ...


2

One of the characteristics of an outer measure is: $$A\subset B\Rightarrow \mu^*(A)\leq\mu^*(B)$$ If $\mu^*$ coincides with $m$ on $C$ then it cannot have that property since: $$\left\{ a\right\} \subset\left\{ a,b\right\} \wedge m\left(\left\{ a\right\} \right)=2>1=m\left(\left\{ a,b\right\} \right)$$


0

First prove that $M$ is well-defined, start with showing that $c:=\inf(f([a,b]))> -\infty$. If $c=-\infty$ then we could take a sequence of $x_n\in[a,b]$ such that $f(x_n)<-n$. The $x_n$ have a convergent subsequence $x_{n_k}\to r$ (for some $r\in[a,b]$) and clearly then $f(x_{n_k})\to -\infty$, but this is a contradiction since $f$ is continuous and ...


0

Let $I\subset \mathbb R$ a connexe and compact set. Let show that $f(I)\subset \mathbb R$ is also compact. Consider $\mathcal U$ a recovery of open set of $f(I)$. We have that $$f^{-1}\left(\bigcup_{U\in\mathcal U}U\right)=\bigcup_{u\in\mathcal U}f^{-1}(U)$$ By continuity of $f$ and by the compactness of $I$, $$I\subset \bigcup_{U\in\mathcal ...


0

the apparent triviality of the problem as posed suggests there may have been a copying error? the following version seems to offer a more appropriate elementary question concerning the geometry of complex numbers: $$ |z+u|=|u| \\ |z|=|u+1| $$


3

From the first equation, since absolute values are real and non-negative, you can derive that $u\in\mathbb R$ and $u\geq0$. This implies that $u=\lvert u\rvert$, hence the second equation becomes $z=u+1$, yielding $z\in\mathbb R$ and $z\geq0$. Substitute $z$ in the first equation to get $$u=(u+1)+u\text,$$ whose only solution is $u=-1$. But this contradicts ...


0

If you wrote that accurately then the numbers are real non-complex: $$u=a+bi\;,\;\;z=x+yi\implies\begin{cases}a+bi=|z|+|u|\implies a= |z|+|u|\;,\;\;b=0\\{}\\x+yi=1+|u|\implies x=1+|u|\;,\;\;y=0\end{cases}$$ and then $$a=|z|+|u|=(1+|u|)+|u|=1+2|u|=1+2|a|\implies a-2|a|=1\implies\begin{cases}a-2a=1\iff a=-1\;,\;\;\text{if}\;\;\;\;a\ge 0\\{}\\a+2a=1\iff ...


2

Let $(A_n)_n$ be as in your question. Then $$\infty = \mu (X) = \mu \left( \bigcup_n A_n \right) = \lim_n \mu (A_1 \cup \dots \cup A_n)$$ so, by the definition of limit, for all $C>0$ there exists $n$ such that $\mu (A_1 \cup \dots \cup A_n) > C$. Finally, note that $\mu (A_1 \cup \dots \cup A_n) < \infty$ since the $A_n$'s have all finite measure. ...


2

Suppose that $g$ has compact support, say $[-1,1]$. Then $$\begin{align} \phantom{-}V_{g}f(x,w)&=\int_{x-1}^{x+1}f(t)\,g(t-x)\,e^{-2\pi itw}\,dt,\\ V_{g}f(-x,w)&=\int_{-x-1}^{-x+1}f(t)\,g(t+x)\,e^{-2\pi itw}\,dt. \end{align} $$ If $|x|\ge1$, $V_{g}f(x,w)$ and $V_{g}f(-x,w)$ take into account values of $f$ on disjoint intervals, so that in general ...


3

Hint: can be the complementary (rationals of interval) countable intersection of open sets? Also: Baire Category Theorem.


1

$f_n'(x) = nx^{n-1} - 2nx^{2n-1} = 0 \iff nx^{n-1}\left(1-2x^n\right) = 0 \iff x = 0, 2^{-1/n} \to x^n = 1/2, 0 \to \text{sup}\{|f_n(x) - 0|: x \in [0,1]\} = 1/4 \neq 0, \forall n \geq 1.$. Thus the convergent is pointwise but not uniform. To show the pointwise limit of $0$, observe that $f_n(0) = f_n(1) = 0$, and for $0 < x < 1, x^n, x^{2n} \to 0 \to ...


0

This fact is essentially a very well-known and important result called Hopf's lemma: If $\Omega$ is a smooth domain, $u\in C^1(\overline{\Omega})\cap C^2(\Omega)$ is a harmonic function, and $u(x)< u(y)$ for every $x\in \Omega$ and some $y\in \partial \Omega$, then $$ \frac{\partial u}{\partial \nu} (y) >0, $$ where $\nu$ denotes the outer unit ...


0

Hint: it's a simple function, also for simple functions integrals are sums.


1

Of course false!!!!! Consider for example the function tan(x) in $(-\pi/2, \pi/2)$ which tends to $-\infty$ at $(-\pi/2)^+$ and tends to $+\infty$ at $(\pi/2)^-$. None continuous function $g$ defined on $[-\pi/2,\pi/2]$ can extend $f=\tan$. (otherwise $g(-\pi/2)=+ \infty $ for example which impossible)


2

Hint: What about $f:(0,1)\to\mathbb{R}$ defined by $f(x)=1/x$?


1

The statement is true, but I am unsure of your proof. You should use the Intermediate Value Theorem. Since, for example, $-5$ and $3$ are in the range of $f$, then there are $x,y$ such that $f(x) = -5$ and $f(y) = 3$. Assume that $x<y$. Then, by the IVT, for every number $c$ with $-5 < c < 3$, there exists some $z, x<z<y$ such that $f(z) = c$. ...


1

Yes. The coefficients of $(xI-A)^{-1}$ are polynomials on the coefficients of $xI-A$, which depend continuously on $x$. And since $z$ is not an eigenvalue, $\det(xI-A)$ is also continuous around $z$. Thus $(xI-A)^{-1}$ is such that each entry is, at $z$, a continuous function.



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