7

Hint : Show inductively that $$f'(x)=\left[\cos(x)\right]\times\left[\cos(\sin(x))\right]\times\left[\cos(\sin(\sin(x))\right]\times\left[...\right]\times\left[\cos(\sin(\sin(\sin(...(x)))))\right]$$


4

If all you know about $f$ is that it's continuous, then $f$ may well fail to be constant. Take, for instance$$\begin{array}{rccc}f\colon&\Bbb Q&\longrightarrow&\Bbb Z\\&x&\mapsto&\begin{cases}0&\text{ if }x<\sqrt2\\1&\text{ otherwise.}\end{cases}\end{array}$$So, your proof cannot work, since it does not use uniform ...


4

Just to flesh out @TheSilverDoe's answer: Denoting functional composition by $\circ$, we want to differentiate $\sin^{\circ n}x$. We prove by induction$$\frac{d}{dx}\sin^{\circ n}x=\prod_{j=0}^{n-1}\cos(\sin^{\circ j}x).$$For $n=0$, this is the trivial $\frac{d}{dx}x=1$, the RHS being an empty product. If the case $n=k$ works, by the chain rule$$\begin{align}...


4

(Edit: Added a plot of $a_n$ and its asymptotics, corrected numerics. Also, a short answer: There's not really a formula, but people can work out its growth.) Denote $a_n = $ number of ways to write $n$ as a sum of distinct squares. Here we demonstrate, though not quite a formula (as you might expect), how we can compute $a_n$ by using the generating ...


4

Notice that $$\ln(x+1)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots +(-1)^{n+1}\frac{x^n}{n}=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}=-\sum_{n=1}^{\infty}\frac{(-x)^n}{n}$$ We may let $x=-1/5$, which gives $$\ln(-\frac{1}{5}+1)=-\sum_{n=1}^{\infty}\frac{(\frac{1}{5})^n}{n}=-\sum_{n=1}^{\infty}\frac{1}{n5^n}$$ which implies $$\sum_{n=1}^{\infty}\frac{1}{n5^n}=-\ln(\...


4

You are on the right track. As regards the remaining integral, note that $$\int \frac{x^2}{2\sqrt{1-x^2}}\,dx=\frac{1}{2}\int \frac{1-(1-x^2)}{\sqrt{1-x^2}}\,dx =\frac{1}{2}\int \frac{1}{\sqrt{1-x^2}}\,dx-\frac{1}{2}\int \sqrt{1-x^2}\,dx.$$ The first integral is easy, whereas for the second one you may use the substitution $x=\sin(t)$. Can you take it from ...


4

Yes, if you mean $0.\overline{9}$ by $0.999\ldots$ then it is true, since $$\lim_{x\to \infty} 1^x = 1$$ and, as you said, $0.\overline{9}$ , is just another way to write 1. If you just mean a lot of nine's, then it is not true, and the limit would go to $0$


3

Write $$f_s(x) = \frac{x^ks}{(s^2+x^2)^{(k+3)/2}}$$ Notice that any such function vanishes at infinity. Let $A$ be the subalgebra of $\textbf{C}_0((0, \infty), \mathbb{R})$ generated by the family $f_s$. All these functions are positive on $(0, \infty)$ so the family doesn't vanish at any point, and it clearly separates points. Thus by the locally compact ...


3

As other did show, it is almost always easier to work around $0$. So $$\lim_{x \to \frac 12}\frac{\tan(\pi x)}{2x - 1}+\frac{2}{\pi(2x - 1)^2}=\lim_{y \to 0}\frac{1-\pi y \cot (\pi y)}{2 \pi y^2}=\pi\lim_{z \to 0}\frac{1-z \cot (z)}{2 z^2}$$ Now, using Taylor expansions $$\frac{1-z \cot (z)}{2 z^2}=\frac {1-z\left(\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}+O\...


2

The first can be evaluated using L'Hopital's rule to get $lim_{x\to 1/2} \frac{\pi sec^2(\pi x)}{2}=\infty$, while the second term $\to \infty$ as is.


2

It seems that the limit exists at $x\to\frac{1}{2}$. Lets' take $x=\frac{1}{2} -\epsilon$; we will set $\epsilon\to0$. Let's consider $f(x)=\frac{\tan(\pi x)}{2x - 1}+\frac{2}{\pi(2x - 1)^2}$ $$f(\frac{1}{2}-\epsilon)=\frac{\sin(\frac{\pi}{2}-\pi{\epsilon})}{\cos(\frac{\pi}{2}-\pi\epsilon)(-2\epsilon)}+\frac{2}{\pi(2\epsilon)^2}=$$$$\frac{\cos\pi\epsilon}{\...


1

With the help of the Laplace transform $$ Y(s) = \frac{\int_{-1}^0 e^{-s t} y(t) \, dt+e^s y(0)}{e^s s-1} $$ now considering $y(t)=0,\ \ -1\le t\le 0$ and $y(0)=1$ $$ Y(s) = \frac{1}{s-e^{-s}} $$ now taking the denominator $d(s) = s-e^{-s}$ and making $s=x + iy$ we have $$ d(s) =x - e^{-x} \cos (y)+i \left(e^{-x} \sin (y)+y\right) $$ so the denominator has ...


1

Less formal, more intuitive: If $f$ is uniformly continuous, then there is a $\delta>0$ such that $f$ varies by less than $1$ on all intervals of length $\delta$. But since it's a function to $\mathbb Z$, it then has to be constant on all those intervals. And since $\mathbb Q$ can be covered by overlapping intervals of that length, $f$ has to be constant ...


1

All the ideas are there but you've made a small logical mistake, in that the quantity $n$ has not been specified. After setting $\epsilon_0 > 1/2$, next you say: Let $m > 0$. Now you have to find a value of $x_0 > m$ such that $|f(x_0)| > 1/2$. To find that value of $x_0$ here's what you do: Choose an integer $n$ such that $2n\pi+\pi/6 > m$....


1

You can use substitution $a=x+y$, $b=x-y$. Then we can reverse it by $x=\frac{a+b}{2}$, $y=\frac{a-b}{2}$, so this is a bijection. Putting this into our inequality, we get $f(a)>f(b)f(-b)$ for any $a,\ b \in \mathbb{R}$. $b=0$ gives $f(a)>f(0)^2 \geqslant 0$, so $f$ attains only positive values. Moreover, if $f(b) \geqslant 1$ for any $b$, we will have ...


1

We are picking $k$ between the $y$-values $f(x_1)$ and $f(x_2)$. That is not the same as picking an arbitrary $x$ between $x_1$ and $x_2$ and setting $k = f(x)$. Not every $x\in [x_1,x_2]$ necessarily corresponds to a value of $f(x)$ such that $f(x) \in [f(x_1),f(x_2)]$. See the image below: the "lowest point" is not a valid choice of $k$, so ...


1

Problems in Mathematical Analysis I: Real Numbers, Sequences and Series Problems in Real Analysis: Advanced Calculus on the Real Axis by Teodora-Liliana Radulescu, Titu Andreescu, and Vicentiu Radulescu Knopp's Theory and Applications of Infinite Series Infinite Series by Earl David Rainville Bonar and Khoury's Real Infinite Series An Introduction to the ...


1

You can follow the below references: A Problem Book in Real Analysis by Asuman G. Aksoy & Mohamed A. Khamsi Advanced Calculus by Robert Wrede, Murray Spiegel Methods of Solving Sequence and Series Problems by Ellina Grigorieva Sequences and Series by Edward A. Azof All the above mentioned books contains various problems with solutions (not all). I ...


1

The sequence ${\rho_{n}}$ is bounded below by 0 and above by 1. Also $\rho_{n}$ is monotone decreasing. therefore $\rho_{n}$ converges to some limit $\rho$ as a consequence of the monotone convergence theorem for sequences. You aren't asked to find $\rho$ so you are done. So you also want to show $\rho$ is not zero. As you have done take the natural log: $ln(...


1

You are right. The problem statement is wrong. For example, $$f(x)=\begin{cases}(x+1)^2&x\in\Bbb Q\\0&0>x\notin \Bbb Q\\\frac1{x^2}&0<x\notin\Bbb Q\end{cases}$$ is differentiable at $x_0=-1$, but $g(x)=x^3f(x^2)$ is nowhere differentiable (in fact, nowhere continuous). The problem statement should ask for differentiability at $\sqrt{x_0}$ (...


1

Alternate approach: Worth looking at the concrete equation $x^2 - 2 = 0$ we are dealing with. We know that there are methods to solve this equation. Among them, there is the Newton approximation (going on the tangent), and the secant method. In this case, since we want values less than the solution, we use the secant method. So consider the points on the ...


1

Your answers for 1), 3) and 4) are corect but you should also know why 2) is false. You need a counter-example to show that 2) is false. Take two disjoint sets, say $A=(0,1)$ and $B=(1,2)$. Let $A_n=A$ for $n$ even and $A_n=B$ for $n$ odd. Take $x \in A$. Then you can check that $g(x)=1$ and $h(x)=0$. But there is no $m$ such that $x \in A_n$ for all $n \...


1

What year was your edition of Zorich published? A copy of the Russian edition from 2012 is here. It has the expansion you write about on the bottom of page number 77 and the calculations there match your correction.


1

Just for your curiosity since you already received good answers. We could even compute the partial sums for $x<1$ $$S_p=\sum_{n=1}^{p} \frac{x^n}{n}=-\log(1-x)- \sum_{k=1}^\infty \frac{x^{p+k}}{p+k}$$ Using, as in your case, $x=\frac 15$ $$S_p=\log \left(\frac{5}{4}\right)-\sum_{k=1}^\infty \frac{1}{(p+k)\,5^{p+k}}$$ and the summation converges very ...


1

Everything's fine except #2. Just because it's not in the set doesn't mean it isn't an upper bound; indeed, if $s > k$ for every $k \in K$ then it's still an upper bound. Also, it may be that $s$ is between components of $K$, or it could be less than every element of $K$ . . . being in the exterior of $K$ doesn't give you any info about if it's an upper/...


1

Since the question is tagged complex-analysis, here is such an argument: If $a=0$ the statement is trivial, so assume that $a\ne0$. Now $f(z)+af(z)^2=z$ is equivalent to $$\left(f(z)+\frac{1}{2a}\right)^2=\frac{z}{a}+\frac{1}{(2a)^2}$$ Let $h(z)=\frac{z}{a}+\frac{1}{(2a)^2}$. Since $h(0)\ne0$ there is some neighborhood of $0$ in which $h$ has a square root $...


1

If $x=y$, any truth about $x$ is a truth about $y$—both $x$ and $y$ represent the same object. Sometimes the notation we use can distract us from this fact. So as Kinheadpump has already noted, since $$ \lim_{x \to \infty}1^x = 1 $$ it follows that $$ \lim_{x \to \infty}(0.\overline{9})^x=1 \, . $$ Indeed, this is actually the same statement repeated twice. ...


1

Whenever you have $a=1$, since $1^x:=e^{x\ln 1}=e^0=1$, you have $a^x=1^x$ (ultimately by the axiom of equality) and thus $$ \lim_{x\to\infty}a^x=\lim_{x\to\infty}1^x=\lim_{x\to\infty}1=1 $$ Depending on how you define the set of real numbers, there are several equivalent definitions to the string $0.99999\ldots$; it is equal to $1$ by any one of the ...


1

You can simply take $f(x)=\sin\left(\frac1x\right)$. It is not uniformly continuous because, for any $\delta>0$, there are $x,y\in(0,1]$ such that $|x-y|<\delta$ and that $\bigl|f(x)-f(y)\bigr|=2$.


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