$lim_{n rightarrow infty} frac{(-1)^{n}sqrt{n}sin(n^{n})}{n+1}$. Am I correct?

I have to find this limit:

begin{align} lim_{n rightarrow infty}
frac{(-1)^{n}sqrt{n}sin(n^{n})}{n+1} end{align}

My attempt:

Since we know that $-1leq sin (x) leq 1$ for all $x in mathbb{R}$ we have:

begin{align}
lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdot(-1)}{n+1}underbrace{leq}_{text{Is this fine?}}&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{(1)^{n}cdotsqrt{n}cdot(1)}{n+1}\ \ lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{n}}{n+1}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{sqrt{n}}{n+1}
end{align}

My doubt in the inequeality is because of the $(-1)$ terms. If everything is correct, then we have

begin{align}
lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{frac{1}{n}}}{1+frac{1}{n}}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{sqrt{frac{1}{n}}}{1+frac{1}{n}}\ \ Rightarrow 0 leq &lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq 0 \ \ therefore& lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1}=0
end{align}

Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!

Problem proving $lim_{n rightarrow infty} int_S(f(x))^{frac{1}{n}}=mu(S)$

Statement:
Given a measurable set $S subseteq mathbb{R}$ and a measurable function $f:S longrightarrow mathbb{R}$ with $f(x)gt1$ for all $x in S$. Show that

$$lim_{n rightarrow infty} int_S(f(x))^{frac{1}{n}}=mu(S)$$

I have searched that there are some similar questions here and here, but they do not start from the same hyphothesis.The sequence $f_n = (f(x))^{frac{1}{n}}$ is decreasing and I have tried to solve it by dominated/monotone convergence, but i believe that there is something missing about the integrability of $f$. I do not know how to continue.

nt.number theory – Generalization of $lim_{n rightarrow infty} prod_{i=1}^{n}frac{2n-1}{2n}$ for a character $chi:mathbb{Z}/s mathbb{Z} rightarrow mathbb{C}^*$

Playing with some infinite products I came up with this problem, that I’m not able to figure it out by myself. Moreover in the internet it doesn’t seem to appear anywhere.
Maybe it is just an easy consequence of properties of characters that I’m not aware of, anyway thank you in advance for any help/answers/suggestions.

All of us know (it is fairly easy to see) that $$lim_{n rightarrow infty} frac{1 cdot 3cdot dots cdot (2n-1)}{2 cdot 4 cdot dots cdot (2n)}=0$$

Now this fact could be reformulated in this fashion: let $$chi:mathbb{Z}/2 mathbb{Z} rightarrow mathbb{C}^*$$
the only non trivial character of $mathbb{Z}/2 mathbb{Z}$, then the expression $$lim_{n rightarrow infty}(prod_{n in mathbb{N}}n^{chi(n : text{mod} 2mathbb{Z})})^{-1}=0$$
More generally, we can perform this construction for every $s in mathbb{N}$.

Indeed all we have to do is to consider a non trivial character $$chi:mathbb{Z}/s mathbb{Z} rightarrow mathbb{C}^*$$
and consider the limit
$$lim_{n rightarrow infty}(prod_{n in mathbb{N}}n^{chi(n : text{mod} smathbb{Z})})^{-1}$$

Now it is true that:

  1. The value of the limit is finite for every $s$ and every non trivial character $chi$?
  2. If so the value of the limit depends only on $s$ or also on $chi$?
  3. The limit is always a real number? (Possibly $0$?)

linear algebra – Is the transformation $T:V$($mathbb{R}$) $rightarrow$ $mathbb{S}$ ONTO?

I was given the map $T:V$($mathbb{R}$) $rightarrow$ $mathbb{S}$ which is the map from the vector space of real-valued functions defined on $mathbb{R}$ to the space of signals defined by sending each function to the signal obtained by evaluating at integers. (ie. $T(f)$ = {${u_k}$} = $f(k)$).

I’ve been asked to determine if this map is onto or not; I understand that a transformation is onto if there is at least one x in $V$($mathbb{R}$) for every b in $mathbb{S}$ such that $T$(x) = b. I’m a little stuck on what a vector in $mathbb{S}$ would be though; would it be a series of values for a function evaluated by only integers?

Here’s how I tried to prove that $T$ is onto: If I let $f(x) = sin(x pi)$, then $T(f) = f(k) = 0$ for all integers $k$. Since integers are in $mathbb{R}$, will always be an x in $V$($mathbb{R}$)for every b in $mathbb{S}$. Is this correct?

How could I prove that this map is (or isn’t) onto? Any help is GREATLY appreaciated!

real analysis – $frac{1}{t}e^{F(t)} to infty Rightarrow t F'(t) to infty$?

Let $F:(0,infty)to(0,infty)$ be a non-decrasing continuously differentiable function satisfying $F(t)toinfty$ as $ttoinfty$. Condition the following statements:

  1. $t F'(t) to infty$, as $ttoinfty$.
  2. $frac{F(t)}{log t} to infty$, as $ttoinfty$.
  3. For each $epsilon>0$, $frac{1}{t}e^{epsilon F(t)} to infty$, as $ttoinfty$.
  4. For each $epsilon>0$, $F'(t)e^{epsilon F(t)} to infty$, as $ttoinfty$.

I can derive that
1 $Rightarrow$ 2 $Rightarrow$ 3 and 1 $Rightarrow$ 4 $Rightarrow$ 3, as follows.

1 $Rightarrow$ 2: Straightforward application of L’Hôpital’s rule.

2 $Rightarrow$ 3: $frac{1}{t}e^{epsilon F(t)} = exp( epsilon F(t) – log t ) = expleft( left(epsilon frac{F(t)}{log t} – 1 right) log t right)$.

1 $Rightarrow$ 4: Using 1 and 3, $F'(t)e^{epsilon F(t)} = t F'(t) cdot frac{1}{t}e^{epsilon F(t)}$.

4 $Rightarrow$ 3: Straightforward application of L’Hôpital’s rule.


So here are my questions:

  • Are the above derivations correct?
  • Can we prove that 3 $Rightarrow$ 1? If so, then we get two closed loops 1 $Rightarrow$ 2 $Rightarrow$ 3 $Rightarrow$ 1 and 1 $Rightarrow$ 4 $Rightarrow$ 3 $Rightarrow$ 1, so that all the four statements are equivalent. If not, any counterexamples?

TIA…

linear algebra – Given $f:mathbb{R}^m supset U rightarrow mathbb{R}^n$ differential. Relation of rank of $f’$ and the property of image of $f$?

Let $f:mathbb{R}^m supset U rightarrow mathbb{R}^n$ a differential map, $U$ open, and $a in U$. Are the followings true:

  1. If $m > n$ and $f(U)$ is open set in $mathbb{R}^n$ then $text{rank}f’_a = n$?

  2. If $m < n$ and $f(U)$ is open set in $mathbb{R}^m$ then $text{rank}f’_a = m$?

I don’t how to get information of Jacobian of $f$ to know about its rank. I try proof by contradiction as follows:

In 1’s problem, If $text{rank}f’_a = k < n$ then $f'(a):mathbb{R}^mrightarrow mathbb{R}^k$. We have, when $hrightarrow 0$, $mathbb{R}^n ni f(a + h) – f(a)rightarrow f’_a(h)in mathbb{R}^k$. But, I don’t see any reason to forbid $f(a + h) – f(a)in mathbb{R}^n$ close to a point $mathbb{R}^k$. So may be the conclusions are wrong.

If they are wrong, could you give an example?

$X_n xrightarrow{p} X Rightarrow existshspace{0.2em} X_{n_j} xrightarrow{a.s.}X$

i need a check about this demonstration

suppose that $X_n xrightarrow{p} X$ and take some decreasing ${epsilon_m}_{m in mathbb{N}}$ who satisfy $epsilon_mrightarrow0$ and define:

  • $A_{j,epsilon}= { omega in Omega : vert X_j(omega)-X(omega)vert <epsilon}$
  • $S_{n,epsilon}= bigcap_{j geq n} A_{j,epsilon}$
  • $S_{epsilon}=bigcup_{n in mathbb{N}}S_{n,epsilon}=liminf A_{j,epsilon}$
  • $S:= bigcap_{m in mathbb{N}}S_{epsilon_m}$

is easy to demonstrate that $S ={ omega in Omega: lim_{jto infty}X_j(omega) = X(omega)}  $

now fixing $m$ i can build an increasing $n : mathbb{N} to mathbb{N}$ who satisfy $$mathbb{P}(vert X_{n_j}-Xvert>epsilon_m)leq2^{-j}, forall jin mathbb{N}$$

now because $sum_{j=1}^{infty}mathbb{P}(vert X_{n_j}-Xvert>epsilon_m)leqsum_{j=1}^{infty}2^{-j}<infty$, by borel-cantelli lemma

$$mathbb{P}(limsup A_{j,epsilon_m})=0 Rightarrow mathbb{P}(S_{epsilon_m})=1$$
so this demonstate that $mathbb{P}(S_{epsilon_m})=1$ do not depend by $m $

now $mathbb{P}(bigcap_{m=1}^{infty}S_{epsilon_m})=lim_{mtoinfty}mathbb{P}(S_{epsilon_m})=1$, this is because $$epsilon < epsilon’ Rightarrow S_{epsilon} subset S_{epsilon’} Rightarrow S_{epsilon_{m+1}}subset S_{epsilon_{m}}, forall m in mathbb{N}$$

i think that all i used for demonstrate this theorem is legal

isometry – Isometric isomorphism from $mathbb{C} rightarrow mathbb{R}^2$.

$mathbb{C} cong mathbb{R} times mathbb{R}$ are identical from set theoretical viewpoint, even as vector spaces or groups. The difference would then be the one considering it as a field, as $mathbb{R} times mathbb{R}$ is certainly not a field.
Both underly the same metric, as by the complex modulus we get for two complex numbers $z_1 = x_1 + iy_1$ and $z_2 = x_2 + i y_2$ the metric:
begin{equation}
|z_1-z_2| = sqrt{(x_1-x_2)^2 – (y_1-y_2)^2}.
end{equation}

Clearly this is an Euclidean metric. Thus the complex plane is a $2$-dimensional Euclidean space. I wonder if one can construct an isometric isomorphism from $mathbb{C} rightarrow mathbb{R}^2$ and how this isomorphism looks like?

real analysis – Questions about a proof of $(forall a)[a in mathbb{F} rightarrow -(-a) = a]$

Proposition:

$(forall a)(a in mathbb{F} rightarrow -(-a) = a)$

proof.

Suppose that $a in mathbb{F}$. By the additive inverse property, $-a in mathbb{F}$ and $-(-a) in mathbb{F}$ is the additive inverse of $-a$; i.e. $$-(-a) + (-a) = e$$ Since $-a$ is the additive inverse of $a$, $$(-a) + a = a + (-a) = e$$ which also justifies that $a$ is the additive inverse of $-a$. From the uniqueness of additive inverses, we conclude that $-(-a) = a$.

Are we allowed to state that $-(-a) + (-a) = e$ implicitly using commutativity of the reals? I ask that because I thought that we were supposed to state that $-(-a) + (-a) = (-a) + (-(-a)) = e$ by definition. Also, is the statement “also justifies that $a$ is the additive inverse of $-a$ ” simply saying that $(-a) + a = a + (-a) = e$ justifies that $-a$ is the additive inverse of $a$ as well as justifying that $-a$ is the additive inverse of $a$?

abstract algebra – Any morphism $phi:G rightarrow A$ to an abelian group $A$ factors uniquely through the projection $G rightarrow G /C$

I’m doing this exercise 7(b) in textbook Algebra by Saunders MacLane and Garrett Birkhoff. Could you please verify if my attempt is fine or contains logical mistakes?

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Let $G$ be a group and $C$ its commutator subgroup. Prove that


My attempt:

For $a,b in G$, we have $aC, bC in G/C$. It follows from $b^{-1}a^{-1}ba in C$ that $C =
(b^{-1}a^{-1}ba)C$
. Then $(aC)(bC) = (ab)C = (ab)(b^{-1}a^{-1}ba)C=(ba)C = (bC)(aC)$. Hence $G/C$ is abelian.

Next we prove that $phi(C) = {1}$. For $x = b^{-1}a^{-1}ba in C$, we have $phi(x) = phi(b^{-1}a^{-1}ba) = phi(b)^{-1} phi(a)^{-1} phi(b) phi(a)$. On the other hand, $A$ is abelian and thus $phi(a)^{-1} phi(b) = phi(b) phi(a)^{-1}$. Hence $phi(x) = 1$.

To sum up, we have $C trianglelefteq G$ and $phi:G rightarrow A$ a group morphism and $phi(C) = {1}$. Then the result follows from Theorem 26.

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