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Solving $ \cos (\cos (\cos (\cos(x))))=\sin (\sin (\sin (\sin (x)))) $

Let $f(x)=\cos(\cos(\cos(\cos x)))-\sin(\sin(\sin(\sin x)))$ For $x\in[\pi,2\pi], f(x)>0$ As cos term positive and sin term negative ! Also $f(x)=f(\pi-x)\implies \text{it is sufficient to solve it ...
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Union of intervals of the form $[i/n,j/n)$

Note that the union only makes sense for $i\neq j$. Thus you want to determine $$\bigcup_{i,j\in \{0,...,n-1\}}\left[\frac{i}{n},\frac{j}{n}\right).$$ You can for example decide to first think about a ...
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2 votes

Finding the value of $\lim_{a\to \infty}\int_0^1 a^x x^a \,dx$

There has been a bit of activity on this question recently, so I considered an approach significantly different from my previous answer; different enough to warrant another answer rather than amending ...
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1 vote
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Solve integral resembling a gaussian

In general, an integral of this form can be re-written using algebraic manipulations by "adding zero" in the following way: $$ \int xe^{-(cx-a)^2/2\sigma^2}\,dx = \int(x-a/c)e^{-(x-a/c)^2/2(\...
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find $\int_0^\infty \dfrac{|\cos (\pi x)|}{4x^2 - 1}$

I was writing my answer and the other answers came up. So I just solved for the undefinite case. Well, we are trying to solve: $$\mathcal{I}\left(x\right):=\int\frac{\left|\cos\left(\pi x\right)\...
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find $\int_0^\infty \dfrac{|\cos (\pi x)|}{4x^2 - 1}$

$$I=\int_0^{\frac{1}2} \frac{\cos(\pi x)}{4x^2-1}dx-\int_{\frac{1}2}^{\frac{3}2} \frac{\cos(\pi x)}{4x^2-1}dx+\int_{\frac{3}2}^{\frac{5}2} \frac{\cos(\pi x)}{4x^2-1}dx-\int_{\frac{5}2}^{\frac{7}2} \...
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  • 5,261
1 vote

find $\int_0^\infty \dfrac{|\cos (\pi x)|}{4x^2 - 1}$

For starters, Fourier series give $$ \left|\cos(\pi x)\right| = \frac{2}{\pi}+\frac{4}{\pi}\sum_{n\geq 1}\frac{(-1)^{n+1}}{(2n-1)(2n+1)}\cos(2\pi n x)=1+\frac{8}{\pi}\sum_{n\geq 1}\frac{(-1)^n}{(2n-1)(...
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1 vote

find $\int_0^\infty \dfrac{|\cos (\pi x)|}{4x^2 - 1}$

\begin{align} \int_0^\infty \dfrac{|\cos (\pi x)|}{4x^2 - 1}\,dx &= \frac12\int_0^\infty |\cos (\pi x)|\bigg(\frac{1}{2x-1} -\overset{x\to -x}{\frac1{2x+1}}\bigg)dx\\ &= \frac12\int_{-\infty}^\...
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1 vote

Absolutely convergent double series which can not be arranged in a convergent series

Yes, it will converge. Indeed, we have $\sum |a_n| < \infty$ and absolute convergence. Let $I(n)$ be the largest $i$ that appears as an index of $b$ in $a_1, a_2, \ldots, a_n$, and similarly $J(n)$ ...
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1 vote
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continuity of incremental ratio (different quotient)

Consider $A := \{(x,y) \in (a,b)^2 : x < y\}$. Then the function $$g: A \to \mathbb{R}, \quad g(x,y) = \frac{f(x)-f(y)}{x-y}$$ is continuous. As $A$ is connected, $g(A)$ is necessarily an interval. ...
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Let $a_n=1+\frac{1}{2n-1}$, $b_n=\tan^{-1}n,n\in\mathbb N$. Then which of the following is/ are true?

Hints: 1) $\frac 1 {2n-1}-\frac 1 {2n+1}=\frac 2 {(2n-1)(2n+1)}$ By Mean Value Theorem $\tan^{-1}(n+1)-\tan^{-1} n = [(n+1)-n] (tan^{-1})' x$ for some $x$ between $n$ and $n+1$. This gives $|\tan^{-...
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  • 5,588
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Prooving that T(n)=2∗T(⌊n/2⌋+17)+n is O(nlg(n))

Concerning the recurrence $$ T(n) = 2T(⌊n/2⌋+17)+n $$ making $n = 2^m$ and assuming $m >> 1$ we have $$ T(2^m) = 2T(2^{m-1}+17)+2^m\approx 2T(2^{m-1})+2^m $$ now solving the recurrence $$ R(m) = ...
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4 votes
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If $g$ is continuous and $A$ is a closed set and $A\subset \mathbb{R}$, then $g^{-1}(A)$ is closed

The function $g$ that you have described is not defined on all of $\Bbb R$. It is, in fact, only defined on $(0, 1)$. So this "counterexample" is not a counterexample to the given claim, ...
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  • 188k
5 votes

Suppose that $g$ is an even function and let $h=f \circ g$. Is $h$ always an even function?

(A) We are given that $g(-x)=g(x)$ (Even) Let $h(x)=f(g(x))$ , then $h(-x)=f(g(-x))=f(g(x))=h(x)$ (Even) (B) Where you are going wrong: $\sqrt{x^{2}}=|x|$ where we have to take the Positive root.
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  • 750
2 votes
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Does Continuity of Composite Functions hold for both ways?

$p$ doesn't imply $q$ : $f(x) =\begin{cases} \frac{1}{x} &x\neq 0 \\0& x=0\end{cases}$ Then $f\circ f=\operatorname{id}$ $f\circ f$ continuous at $0$ but $f$ is not continuous at $0$. For ...
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  • 9,360
0 votes

Proving an expression for the limit of a certain sequence

Let's suppose that $L<0$. Then, $-L>0$ and for $\epsilon=-L>0$ we would have that there is $n_{0}\in{\mathbb{N}}$ such that for every $n\ge n_{0}$ we would have that: $a_n-L\leq |a_n-L|< -...
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Proving an expression for the limit of a certain sequence

You can treat them in two cases: $(1) L=0$, you don't need a lower bound for $a_n$ $(2) L>0$, when $n>N$ is sufficiently large, you can have a lower bound $a_n>\frac{L}2$
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  • 5,261
3 votes

Finding the value of $\lim_{a\to \infty}\int_0^1 a^x x^a \,dx$

Here is another method: Substituting $y = x^a$, or equivalent, $x = y^{1/a}$, the integral boils down to \begin{align*} I_a := \int_{0}^{1} a^x x^a \, \mathrm{d}x &= \int_{0}^{1} a^{y^{1/a} - 1} y^...
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limit $\lim_{k \to \infty} \int_{0}^{1} \frac{kx^k}{1+x} dx$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} ...
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  • 84.5k
1 vote

Rational function $F(x)^c=G(x)$

Let $F$ and $G$ be non-constant rational functions with $F(z)^c = G(z)$. Since $F$ is non-constant, it must have at least one zero or pole, and thus $a\in \mathbb C$ such that as $z \to a$, $F(z) \...
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1 vote

Finding the value of $\lim_{a\to \infty}\int_0^1 a^x x^a \,dx$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} ...
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1 vote

Sequential definition of continuity implies delta-epsilon definition

Proof by contradiction. We shall prove prove that: $$ \text{Not}\,(2) \quad\Longrightarrow\quad \text{Not}\,(1). $$ If not (2), then there exists an $\varepsilon>0$, such that, for all $\delta>0$...
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Fixed-point iteration and continuity of parameters

Suppose $X$ is metric (with metric $d_X$), and consider the following parameterized family of optimization problems: $$ \max_{x \in X} \, -d_X\big(x, f(x,a)\big) \tag{$\ast$} $$ By construction, $d_X\...
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3 votes

Show $\mathrm{span}\{1, x^2, x^4, \cdots \}$ is not dense in $C([-1,1])$

Let $\varphi$ be the linear functional defined on $C[-1,1]$ by the formula $$\varphi(f)=f(1)-f(-1)$$ Clearly this functional is bounded $(\|\varphi\|=2)$ and nonzero, as $\varphi(x)=2.$ Thus $\ker \...
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4 votes

Show $\mathrm{span}\{1, x^2, x^4, \cdots \}$ is not dense in $C([-1,1])$

All the functions in $\{1, x^2,x^4,\dots\}$ are even. Therefore so are all their linear combinations, i.e., all the functions in their span. Finally, any limit (with respect to the supremum norm, or ...
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0 votes
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Decimal Expansions - Pugh Exercise 1.18

Recall the definition of $x_1$. It is the largest integer less than or equal to $10(x-N)$. So any integer larger than $x_1$ must be greater than $10(x-N)$. In particular this is true for $x_1 + 1$, i....
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Infinite countable subset such that $\int_a^b f(x) dx=0$, whenever $a,b \notin S$

By contradiction, suppose that $f\neq 0$. It exists $c \in \mathbb R$ such that $f(c) \neq 0$. Without loss of generality, we can suppose $f(c) \gt 0$. As $f$ is supposed to be continuous, we can find ...
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4 votes
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Show $\mathrm{span}\{1, x^2, x^4, \cdots \}$ is not dense in $C([-1,1])$

Hint: Remember something in the span is a finite combination of elements in the set. So, let $2n$ be the largest power of $x$ present in a generic element of the span, so a generic function looks ...
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5 votes
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Infinite countable subset such that $\int_a^b f(x) dx=0$, whenever $a,b \notin S$

Since $S$ is countable the set $\mathbb{R}\setminus S$ is dense in $\mathbb{R}$ (each nonempty open interval is uncountable, hence hits $\mathbb{R}\setminus S$). Fix any $a \notin S$. Then $F(x):= \...
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  • 3,016
0 votes
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Definition of radial projection

Yes, it is correct. Radial projection onto the unit sphere would be the map $\pi_0(y) = \frac{y}{|y|}$, while radial projection onto the sphere of radius $r$ centered at the origin would be $\pi(y) = ...
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5 votes
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$f$ is continuous, no constant and periodic $\implies$ f is bounded

You want to show that $f(\mathbb{R})$ is a bounded set. Now observe that by periodicity, there is a compact interval $[a,b]$ such that $f(\mathbb{R}) = f([a,b])$. What do we know about the image of ...
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  • 7,034
0 votes

Let $f$ be a nonconstant polynomial of degree $k$ and let $g:\mathbb R\to \mathbb R$ be a bounded continuous function.

If $k$ is even, then $\lim_{x\to\pm\infty}f(x)=+\infty$, so there exists $\lambda=\min_{\Bbb R} f$. Then it is enough to take $g\equiv\lambda-1$ to see that a) and c) fail. Then if you take $g\equiv\...
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  • 11.5k
1 vote
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Prove that $\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$

My try : Lemma : Let $x>0$ then define : $$f(x)=\ln(x)$$ Then a trivial consequence is : $$f'''(x)>0$$ Now the problem is with the constraint of the OP: $$\ln\left(\frac{{x_1}^2+{x_2}^2+\cdots+{...
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  • 3,120
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In space $C[0,1]$ find distance from point $x(t)=t^{2}+1$ to the subspace $L_{1}=\{x\in C[0,1]: x(0)=0\}$

I'am pretty sure that (a) is meant the following way: $C[0,1]$ is endowed with the maximum norm $\|\cdot\|_\infty$. $L_1$ is a subspace of $(C[0,1],\|\cdot\|_\infty)$, thus $L_1$ carries the maximum ...
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  • 3,016
1 vote
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In space $C[0,1]$ find distance from point $x(t)=t^{2}+1$ to the subspace $L_{1}=\{x\in C[0,1]: x(0)=0\}$

If it if unspecified, the usual distance between two elements $x,y \in C[0,1]$ is $$\| x-y\| = \sup_{t\in [0,1]} \vert x(t)- y(t)\vert.$$ The distance from an element $x \in C[0,1]$ to a subset $Y\...
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  • 14.7k
2 votes

Prove $|\prod_{i=1}^n a_i - a_n^n|\leq 2n\delta$ if $0 \leq a_i \leq 1$ and $|a_i - a_{i+1}| \leq \delta$

The proposition is false. Take $a_1 = \frac{1}{2},\ a_2 = \frac{1}{2},\ n=2, \delta = 0,$ so that the condition $|a_i - a_{i+1}| \leq \delta$ for all $i<2$ is satisfied. Then, $$ \Big|\prod_{i=1}^n ...
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2 votes
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Problem in proving that the set of integers $n_k:k\geq1$ is not bounded

If $n_k$ is bounded, then because $\dfrac{m_k}{n_k} \in (0, 1)$ there are only finitely many possibilities for $r_k=\dfrac {m_k}{n_k}$. That means there are only finitely many differences between the ...
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  • 17.6k
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Verification of proof of convergence of the series $\displaystyle \sum_{n \in \mathbb{N}} (-1)^n \dfrac{n!}{n^n}$

Your proof is fine. You may also notice that for any $n\geq 1$ $$ \frac{n!}{n^n} = \int_{0}^{+\infty} n z^n e^{-nz}\,dz $$ so $$ \sum_{n\geq 0}(-1)^n\frac{n!}{n^n} x^n = 1-x\int_{0}^{+\infty}\frac{z e^...
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2 votes
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Is there a standard rule for $\arctan$ substitutions?

It's by a simple factorization (and experience). We have \begin{align} \int\frac{du}{a^2u^2+b^2}&=\int\frac{du}{b^2\left(1+\left(\frac{au}{b}\right)^2\right)} \end{align} This immediately suggests ...
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1 vote
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Contraction Mapping Principal in $C^1[a,b]$

Let $$F(f)(x) = 1 + \frac15\int_{0}^{x}\sin (tf(t))\, \mathrm dt$$ \begin{align} \left\|F(f) - F(g)\right\| &= |F(f)(0) - F(g)(0)| + \sup\limits_{t\in \left[0,\frac\pi2\right]} \left|F(f)'(t) - F(...
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  • 5,549
1 vote
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$\binom{n}{k}p^kq^{n-k}\sim\exp(-x^2_k/2)/\sqrt{2\pi npq})$, $x_k=(k-np)/\sqrt{npq}$.

Be careful $e^{ab} \neq e^ae^b$ (I think that you did the mistake in your last transition). However if you write $a_n = 1 + b_n$ when $b_n \to 0$, $$\log\phi(n,k) = \left(1+b_n\right) \left(-\frac{x_k^...
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  • 5,549
0 votes

A function with two center of symmetry - Prove that there does not exist an irrational number $q$ such that $f(ql-x)=f(ql+x)$.

Proposition $(2)$ is false. Suppose $l=1$. If either one of $\alpha,\beta$ is irrational, then the proposition is false since you can take $q=\alpha,\beta$. It is the case for $f(x)=\cos{\left(2\pi(x-\...
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  • 1,147
2 votes
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Is this proof of monotone convergence theorem circular?

The usual proof of the MCT uses the fact that the monotone limit of a sequence of nonnegative simple functions satisfies the limit-integral swaparoo property. I believe this is what is being used to ...
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Function defined through integral not continuous

Let $\varepsilon > 0$ and define $h(y)= \max\limits_{x \in (-\varepsilon, \varepsilon)} f(x,y)$ the minimal dominating function. We see that for any $y$, the maximum of $f(x,y)$ is reached for $x = ...
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  • 315
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What's the derivative of: $ \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$?

Clarification to solve the question The question posed is calcule Derivative $\sqrt{x+\sqrt{x+\sqrt{x+....}}}$ we put $y=\sqrt{x+\sqrt{x+\sqrt{x+....}}} $ Before calculating the derivative we ...
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Misunderstanding of the definition of an almost upper bound

Hint: given $z$, how can $z = x$ hold for infinitely many $x$?
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1 vote

Misunderstanding of the definition of an almost upper bound

$A$ is a set, so by definition it contains just one instance of any of its elements. There can't be "infinitely many $x \in A$" all equal to $z$. There is at most one.
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2 votes
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Let $\varphi_n, \psi$ be simple functions such that $(\varphi_n)$ is increasing and $\psi\le\lim_n\varphi_n$. Then $\int\psi\le\lim_n \int \varphi_n$

If you know the monotone convergence theorem, this is simple: By the monotone convergence theorem, we have $\lim_{n \to \infty} \int \varphi_n = \int \lim _{n \to \infty}\varphi_n$. Since $\psi \leq \...
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  • 8,158
1 vote
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Bounding super exponential functions with factorial functions

Let's take logarithms of both sides, we then want to have $p(n) \cdot \log_2 q(n)! > 2^n$. We can use very rough estimation, $k! < k^k$, to get that left side is less than $p(n) \cdot q(n)\cdot \...
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  • 8,901
6 votes
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Borel Cantelli-Type problem from Billingsley's Probability and Measure

It seems to be simpler to show the equivalence of the negation of the assertions, that is, we will prove that $$ \mathbb P\left(\limsup_{n\to\infty}A_n\right)<1\Leftrightarrow \mbox{ there exists }...
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