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0

What's going wrong is that you are using the triangle equality too much. This is an alternating series which does not converge absolutely so if you lose the signs of course it will not converge. What I suggest trying is to group the terms in the partial sums "by two" so that all the terms are positive (or, all the terms are negative) and then the series ...


1


0

Hint (my idea): Let $m<n$ and define $$ f(x) = \sum_{i=m+1}^{n}\frac{x^i}{i} $$ then $|f_n-f_m|=|f(-1)|$. Note that $$ \frac{\rm d}{{\rm d}x}f(x) = \sum_{i=m}^{n-1}{x^i} =x^m\frac{x^{n-m}-1}{x-1} $$ Now try to integrate $f$ or an approximation of it then bound $f(-1)$.


0

The argument is symmetric in the $a_k$ series and the $a_k'$ series. The point is that both of these series are series of positive elements, and either can be seen as the other with their elements rearranged. This does not say anything about conditionally convergent series, since a series of positive numbers can never be conditionally convergent (as ...


0

Theorem: Let $a_i\ge0$ for any $i\in {\mathbb N}$ and $s_n = \sum_{i=1}^na_i$ for any $n\in {\mathbb N}$. If $s_i$s are convergent to $s\in{\mathbb R}$ then for any permutation $\pi :{\mathbb N}\rightarrow {\mathbb N}$, the series related to $a_{\pi(i)}$ is also convergent to $s$. Proof: Let $t_m = \sum_{i=1}^ma_{\pi(i)}$, $p_m=\max\{ ...


4

Hint: changing variables in the integral $\int_{0}^{x}e^{k}f(x-k)\,\mathrm{d}k$ by $w=x-k$, $dw=-dk$, makes it easier to differentiate. The substitution gives us: $$\int_{0}^{x}e^{k}f(x-k)\,\mathrm{d}k=\int_{0}^{x}e^{x-w}f(w)\,\mathrm{d}w.$$ Differentiating with respect to $x$ on both sides, and applying the general Leibniz rule to the RHS, we ahave: ...


0

Change $x-k$ to $t$ in your integral. Your equation is now $$f(x)=1+x+x^2+\exp(x)\int_0^ {x}\exp(-t)f(t)dt$$ Put now $F(x)=\int_0^x \exp(-t)f(t)dt$. You have $F(0)=0$, and your equation is now $$F^{\prime}(x)=(1+x+x^2)\exp(-x)+F(x)$$ And last hint: Note that $(F^{\prime}-F)\exp(-x)=(F(x)\exp(-x))^{\prime}$


1

Some context: a density means a nonnegative integrable function on $[0,1]$ with integral equal to $1$. Total boundedness is understood in the $L^1$ norm. Let $\mathcal F_C$ be the set of $C$-Lipschitz densities. I'll prove that it is totally bounded in the uniform norm $\sup|f|$, which will imply total boundedness in the $L^1$ norm (since the $L^1$ norm ...


2

Define $$ F(x)=\int_a^xf(t)\,\mathrm{d}t\tag{1} $$ Since $f\in L^1([a,b])$, $F$ is continuous. By definition, $F(a)=0$. Furthermore, by hypothesis, $\displaystyle F(b)=\int_a^bf(t)\,\mathrm{d}t=0$. For any polynomial, $p$, let $\displaystyle P(x)=\int_a^xp(t)\,\mathrm{d}t$, which is also a polynomial. Then integration by parts yields $$ \begin{align} ...


1

First, observe that $f(x) = \frac{x}{1+x}$ is an increasing function. Thus, if $a \leq b$, then $f(a) \leq f(b) \leq f(b) + f(c)$. Similarly if $a \leq c$. Now suppose $b \leq a$ and $c \leq a$. Then $$ \frac{a}{a+1} \leq \frac{b}{a+1} + \frac{c}{a+1} \leq \frac{b}{b+1} + \frac{c}{c+1} $$ As an aside, to prove my observation, let $P$ be the statement ...


0

One intuition (perhaps idiosyncratic) I have about integration comes from motion. Suppose we know how fast someone is going over a range of times; in that case we can work out how far they went. Intuitively, you could say that you take a lot of estimates of velocity $times$ distance at different times and add them up. But mathematically what you're doing is ...


3

It is not bounded. Take for example $n=2$, $a_1=a_2=1$ and $b_1=1,b_2=1/2$. Then we can make the denominator arbitrarily closed to zero by taking $x_2=1$, and saying $x_1$ getting close to $-1/2$. Then, the numerator won't be zero, but the denominator will converge to zero making the function diverge to infinity.


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 ...


0

To prove boundedness, you need to show that no matter what your parameters ($\{a_i\}$,$\{b_i\}$) are, there exist some $f_{min}, f_{max}$ s.t. $f_{min} \leq f(x) \leq f_{max}, \forall x \in S$. In other words, you need to find a formula or set of instructions for picking $f_{min}, f_{max}$, given an arbitrary set of ($\{a_i\}$,$\{b_i\}$).


2

By hypothesis is easy to conclude that $\int pf \ dx = 0$ for all polynomials $p$ (*). Consider $\varphi $ a smooth function with compact support. By the Weierstrass approximation theorem exists a sequence $p_n$ of polynomial that converges uniformly to $\varphi$. Then given $\epsilon >0$ exists $n_o \in N$ where $|p_n (x) - \varphi (x)| < \epsilon$ ...


1

No, that need not be the case. Consider $$\Omega = \left\{ x\in \mathbb{R}^N : x_1 > 0, \sum_{i=2}^N x_i^2 < 1 \right\}.$$ Then you have $$\frac{m(\Omega\cap B(0,r))}{m(B(0,r))} \sim \frac{\tau_{N-1}r}{\tau_Nr^N} \xrightarrow{r\to\infty} 0,\tag{1}$$ and $$\mathcal{H}^{N-1}\left(\Omega \cap \partial B(0,r)\right) \to \tau_{N-1} > 0,\tag{2}$$ ...


5

Fix $\xi \in \mathbb{R}$. Recall that $$e^{\imath \, x \xi} = \sum_{n=0}^{\infty} \frac{(\imath \, x \xi)^n}{n!}.$$ As $$\sum_{n=0}^k \frac{(\imath x \xi)^n}{n!} f(x) \stackrel{k \to \infty}{\to} e^{\imath \,x \xi} f(x), \qquad x \in [a,b]$$ and $$\left|\sum_{n=0}^k \frac{(\imath x \xi)^n}{n!} f(x) \right| \leq |f(x)| e^{|\xi| \max\{|a|,|b|\}} \in ...


4

There is always an intermediate point $y = \frac{x+b}{2}$. Choose $\delta := y - x = \frac{b-x}{2}$. Update: In fact we can exploit a whole range of points between $x$ and $b$, all less than $b$: $$ y(t) = x (1-t) + b t \in [x, b) $$ for $t \in [0, 1)$. So $$ \delta(t) := y(t) - x = (b - x) t $$ will ensure $$ b - \delta(t) = b - (b-x) t = b (1-t) + x ...


3

You started your proof by saying "$g$ is obviously continuous", but this is not true in general. However, it does follow if you assume that $f$ does not attain the value $M$. So if you begin your proof by saying: Assume $f$ does not attain the value $M$. Let $g(x) = \frac{1}{M-f(x)}$. Now $g$ is continuous... You now have a valid argument. Since ...


2

Hint I correct only the first question and the other are not very different. First we should determinate the point-wise limit. It's simple to see that the sequence $(f_n)$ is point-wise convergent to the zero function $f$. Now to we should compute $$\lim_{n\to\infty}\sup_{x\in[0,1]}|f_n(x)-f(x)|$$ so notice that $$f'_n(x)=\frac{1}{(nx+1)^2}>0$$ so the ...


1

Your closed box could certainly contain the support of such an $f$, but it cannot be equal to the support. Be careful that a closed box is not compact in the topology you are looking at. Here is a "hint": if you denote by $A$ the set of all $x$ with the above property, then the sets $\{ x\}\times \mathbb R$ are open (in the topology you are looking at) and ...


2

Notice that $\mathbb{sin}A \mathbb{cos}B = \frac{1}{2}(\mathbb{sin}(A+B)+\mathbb{sin}(A-B))$. Also notice that $ax^2+bx=(\sqrt{a}x+\frac{b}{2\sqrt{a}})^2-\frac{b^2}{4a}$ and $ax^2-bx=(\sqrt{a}x-\frac{b}{2\sqrt{a}})^2-\frac{b^2}{4a}$. You will get two Fresnel integrals in which you will have to make the changes of variables $\sqrt{a}x+\frac{b}{2\sqrt{a}}=y$ ...


4

Compute them together: Add $i$ times the first integral and the second, to obtain $$\int_0^\infty e^{iax^2}\cos (2bx)\,dx.$$ The integrand is even, so that is $$\frac{1}{2}\int_{-\infty}^\infty e^{iax^2}\cos (2bx)\,dx.$$ $\sin$ is odd, so the integral of $e^{iax^2}\sin (2bx)$ vanishes, hence we get $$\frac{1}{2} \int_{-\infty}^\infty ...


1

A very important property of a transformation is bijectivity, i.e. if we consider a map $T: U \to V$, $U \subseteq \mathbb{R}^d$, $V \subseteq \mathbb{R}^n$, then for any $v \in V$ there should exist exactly one $u \in U$ such that $T(u)=v$. For the spherical coordinates we see that for $$V := \mathbb{R}^3 \qquad \quad U_1 := [0,\infty) \times [0,\pi] ...


0

In your first step, it should be: $m=y_1 -y_2$ $$\int_0^1\int_0^1g(y_1-y_2)\Bbb{1}_{\{y_1>y_2\}}dy_1dy_2 = \int_0^1\int_{y_2}^{1}g(y_1-y_2)dy_1dy_2=\int_0^1\int_0^{1-y_2}g(m)dm dy_2=\int_0^1\int_0^1 g(m)I_{(0<m<1-y_2)}dmdy_2=\int_0^1\int_0^1 g(m)I_{(0<m<1-y_2)}dy_2 dm \quad=\int_0^1\int_0^1 g(m)I_{(0<y_2<1-m)}dy_2 dm =\int_0^1 ...


5

I agree with J. Loreaux's commrent above (“Your argument doesn't really work”), and I'd go farther: to me, your argument makes no sense whatever, for several reasons: You say “If the every compact set on a metric space is not bounded, then …”. This is at least confusingly stated. You want to prove that every compact set is bounded. You seem to be trying ...


0

Your argument is essentially correct but pay attention to the essentially. I would phrase it in words like so: If not every compact set on a metric space is bounded then there exists at least one compact set on a metric space that is not bounded. Since it's unbounded we can construct an infinite open covering which doesn't have a finite subcover thus ...


0

Yes, in the given situation, one could just use $r = \operatorname{dist}(A,B)$, and then have $$N_r(A) = \bigcup_{x\in A} B_r(x) = \{ y\in X : \operatorname{dist}(y,A) < r\}$$ as a convex open neighbourhood of $A$ disjoint from $B$. But of course, to use that result, $\operatorname{dist}(A,B) > 0$ must have been proved. In the setting of a normed ...


1

This is another path. Theorem. Let $x$ be a real number such that $-1<x<1$ and set $$ Z(x):=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\frac{(i-1)! (j-1)!}{(i+j)!} x^{i+j}. $$ Then $$ Z(x)= 2 \:\mathrm{Li_2} (x) - \mathrm{Li_2} \left( x(2-x) \right) \quad \quad (1) $$ where $\mathrm{Li_2}$ is the dilogarithm function such that ...


6

Your argument is correct, if you note that the left hand side of the equality is a continuous function of $f\in L^1$. Then the denseness perpetuates the equality from the simple functions to all of $L^1$. But there is an easier argument, change the order of integration: $$\begin{align} \int_\mathbb{R} \left(\int_x^{x+h} \lvert f(t)\rvert\,dt\right)\,dx ...


-2

Hint: Things will be clearer if you make the change of variables $t=N u$.


0

The statement we are trying to prove is: $$|s|e^{xs}\leq C(e^{-\epsilon s}+e^{\epsilon s}),\qquad\mbox{for $x\in[-\frac{\epsilon}{2},\frac{\epsilon}{2}]$}$$ Note that $C$ must be independent of $x$ and $s$, but is allowed to depend on $\epsilon$. Make the following choice: $$ C=\max_{t>0}te^{-\epsilon t/2}<\infty. $$ To prove the bound, consider the ...


2

The general gaussian integration formula is $$\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty} e^{-\frac12 x^T Ax}dx_1\ldots dx_n=\frac{(2\pi)^{\frac{n}{2}}}{\sqrt{\operatorname{det}A}}.$$ In your case, $n=2$ and $A=\left(\begin{array}{cc} 1 & -1/2 \\ -1/2 & 1\end{array}\right)$.


2

Note that the Lebesgue outer measure is usually constructed as follows: Let $m : \mathcal{R}(\mathbb{R}^n) \to \overline{\mathbb{R}}$ where $\mathcal{R}(\mathbb{R}^n)$ is the set of all $n$ dimensional rectangles so that $$ [a_1, b_1] \times [a_2, b_2] \times \cdots \times [a_n, b_n] \mapsto \prod_{i=1}^n (b_i - a_i) $$ then we put the Lebesgue outer ...


11

Recall: $$x^2-xy+y^2=(x-\frac{1}{2}y)^2+\frac{3}{4}y^2$$ With that you get: $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}(x^2-xy+y^2)}dxdy=\\\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left((x-\frac{1}{2}y)^2+\frac{3}{4}y^2\right)}dxdy=\\ \int_{-\infty}^\infty\int_{-\infty}^\infty ...


3

An outer measure is a set function which is not necessarily a measure as it is monotonic and subadditive, but not necessarily additive. Outer measures are usually defined an all subsets of the underlying set. For example, the Lebesgue outer measure is defined on all subsets of the set of reals. However, as I mentioned, it is only a subadditive function of ...


0

Inspired from Nate's answer, the answer is NO in general, and here is one example. Let $X = Y = [0,1]$, and $f(x,y) = I_{\{x = y\}} I_{\{y\in M\}}$, where $M$ is some non-Lebesgue-measurable set and $I$ is the indicator. $f$ is Lebesgue measurable since $\{f=1\} \subset \{(x,y): x = y\}$ is zero set. However, $g(x) = I_M(x)$, which makes it ...


0

When we consider integrability, estimates within a constant factor are enough. Then the sum of two positive numbers is as good as their maximum, because $$\max(a,b)\le a+b\le 2\max(a,b)$$ So, replace $x^2+y^a$ by $\max(x^2,y^a)$. Then we only need to integrate $$ \iint \chi_{\{x^2<y^a\}} y^{-a} \,dx\,dy + \iint \chi_{\{x^2>y^a\}} x^{-2} \,dy\,dx $$ ...


3

Summation by parts seems to work fine. For $j=1$ we have: $$\sum_{i=1}^{+\infty}\frac{H_{i+1}}{i(i+1)}=2,$$ since $\sum_{i=1}^{N}\frac{1}{i(i+1)}=1-\frac{1}{1+N}$ and so: ...


0

No, the limit function, $F$, need not even be differentiable if the original sequence $\phi_n \rightarrow F$ pointwise. For an example on $[0,1]$, take $F = 1_{[1/2,1]}$ (indicator function) and mollify it with an approximation to the identity $\eta_n$. This yields a sequence $\phi_n = F \ast \eta_n$ such that $\phi_n \rightarrow F$. $\phi_n$ are analytic ...


1

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, ...


0

This is indeed true. One way that immediately comes to mind is to use Schwarz Lemma. But first we need to ensure that $f(0) = 0$. Here we can use an automorphism of the unit disk such that if $f(0) = a$, then both $\phi_a \circ f (0) = \tilde{\phi}_a \circ f(0) = 0 $ where $$ \phi_a (z) = \frac{a-z}{1-\bar{a}z} ; \qquad \tilde{\phi}_a(z) = ...


1

I can't leave comments (not enough reputation), so I'm writing here. If I understand you correctly (which I'm not sure I am), I believe your solution in incorrect. For example, for positive integers it seems that $f(2n-1)=f(2n)=n$ (e.g. $f(3)=f(4)=2$), so it can't be true that $f(x)$ is asymptotically $x\ln2$. I think that a useful observation is that ...


1

Suppose that $$(m,n)\simeq(p,q)\quad\hbox{and}\quad (p,q)\simeq(r,s)\ .$$ According to your definition this means that $$m+q=n+p\quad\hbox{and}\quad p+s=q+r\ .$$ Adding these equations, $$m+q+p+s=n+p+q+r\ ;$$ cancelling $q$ and $p$ gives $$m+s=n+r\ .$$ By definition this means $$(m,n)\simeq(r,s)\ .$$


1

Let $X=\{0,1\}$ and $\mu$ be the counting measure. Then $L^2(X,\mu) = \{ f:X \to \mathbb{R} \}$ (that is, essentially $\mathbb{R}^2$). If we let $e_k = 1_{\{k\}}$, then $e_0,e_1$ forms a basis for $L^2(X,\mu)$, in particular, if $f \in L^2(X,\mu)$, we can write $f = f(0) e_0 + f(1) e_1$. Define $T:L^2(X,\mu) \to L^2(X,\mu)$ by $T e_k = e_0+e_1$, and so ...


0

I am not sure but let $y_n=\tan[x_n-n\pi]$. Then we have $$\tan x_n - n\pi = x_n - n\pi.$$ Thus, using $\tan(x_n)=\tan(x_n-n\pi)$, $$y_n - n\pi = \arctan(y_n).$$ Now use $\arctan(y)+\arctan(1/y)=\pi/2$ to obtain $$y_n+\arctan(1/y_n)=n\pi + \pi/2.$$ Now, because $y_n \rightarrow \infty$, you can use a Taylor expansion of $\arctan$ around 0. This should give ...


2

@Vladmir asks if $T$ is bounded or closed, since this does not hold in general for linear operators. If the domain of your operator $T$ is all of $H$ and it is closed, then it is also bounded. For bounded linear operators, it is certainly true. That is because these operators are continuous and $$Tx = T\left(\lim_{N\to\infty} \sum_{n=0}^N \langle x, e_n ...


1

We want to show that for every $\epsilon\gt 0$ there is a $\delta\gt 0$ such that if $0\lt |x-c|\lt \delta$ then $|f(x)-f(c)|\lt \epsilon$. Let $\epsilon\gt 0$ be given. We produce a suitable $\delta$. We have $$|f(x)-f(c)|=|x^2-c^2|=|x+c||x-c|.\tag{1}$$ We want to show that the expression on the right side of (1) can be made "small" if we take $|x-c|$ ...


1

To use the product rule in part (c), it suffices to show that $$ F(t)=\int_0^te^{-s}u(s)ds $$ is a differentiable function of $t$, since we know that $e^t$ is differentiable. Depending on what you're "allowed" to use, this is simply a part of the fundamental theorem of calculus. If you cannot invoke FTOC, simply consider the difference quotient: $$ ...


1

Assuming that $u(x)$ is at least $C^2(\mathbb{R})$ and that your domain is $\mathbb{R}$ (connected) you have: $$u'(x)=1-u(x)$$ So $u'(x)=0 \iff u(x)=1$. Also $u'(0)=1$ so we just have to prove that $u'(x) \geq 0 \quad \forall x$ But $u'(x)$ is continuous so $u'(x) \geq 0 \quad \forall x$ (it can't go to a negative value because it should pass by zero and ...



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