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11

What seems incorrect is because the language seems informal. Formally, one would write $$ x_0=5, \quad x_n = \sqrt{x_{n-1}} $$ to describe the iterations of the square roots. One computes easily that $$ x_n = x_0^{1/{2^n}} $$ and since $x_0^{1/n} \to 1$ for all $x_0 > 0$, so does $x_n$. (You can prove this by showing that the logarithm of the sequence ...


10

Assume the series $\sum_{k=0}^\infty a_kz^k$ converges for some $z\in{\mathbb C}$. Then it converges absolutely for any $z'$ that is nearer to the origin than $z$; see the proof below. From this fact it follows by mere logic, that when the series diverges at some $z\in{\mathbb C}$ it cannot converge at any point $z''$ farther away from the origin. Proof. ...


10

Expanding on comments above to give an intuitive idea of this result: write $f(x) = \frac{1}{1+x}$ and set $n = 4$. A rough diagram: Then, for example, the area of the first (left most) rectangle is $$f(1+1/4) \cdot \Delta x = \frac{1}{1+1/4} \cdot \frac{1}{4}$$ The area of all four rectangles is $$\sum_{k=1}^4 f(1 + k/4) \Delta x = \frac{1}{n} ...


9

The function $f: x \mapsto (x+\frac2x)^2$ is convex on $(0,\infty)$, so by Jensen's inequality we have $$ \frac{\left(a+\frac2a\right)^2+\left(b+\frac2b\right)^2}{2} = \frac{f(a)+f(b)}{2} \geq f\left( \frac{a+b}{2} \right) = f\left(\frac12\right) = \left( \frac92 \right)^2 = \frac{81}{4}. $$


5

The Smith-Volterra-Cantor set fits the bill. It is a closed subset of $[0, 1]$ (so it compact), with empty interior and measure $\frac{1}{2}$. More generally, you can modify the construction of the Cantor set to obtain so-called 'fat Cantor sets' which have the desired properties.


4

1) Your function is clearly differentiable at $t$ if $t\ne 0$, since cosine and $t^2$ are differentiable. To show differentiability at $t=0$ use the limit-of-the-difference-quotient definition of the derivative, and use the fact that $|\gamma(t)|\le t^2$. This will quickly show that the derivative at $t=0$ is defined and is zero. 2) Consider that ...


4

We solve the "dual" problem of finding the smallest square containing the semicircle $H:=\{(x,y)|x^2+y^2\leq 1, \ x\geq0\}$. Drawing supporting lines to $H$ having slopes $\phi$ and $\phi+{\pi\over2}$ with respect to the horizontal we see that the smallest square containing $H$ and having sides with these slopes has side length ...


4

As $ T $ is bounded and self-adjoint with norm $ 1 $, its spectrum $ \sigma(T) $ is a compact subset of $ [-1,1] $, and its spectral radius $ r(T) $ equals $ 1 $. Hence, either $ -1 \in \sigma(T) $ or $ 1 \in \sigma(T) $. If $ -1 \in \sigma(T) $, then $ 0 \in \sigma(I + T) $, and so $ I + T $ is not invertible. If $ 1 \in \sigma(T) $, then $ 0 \in \sigma(I ...


4

I guess you mean this: $$\lim_{n\to\infty}5^{1/2^n}=1$$


3

Heuristically, $\sin x \approx. x$ for small $x$. Thus, $$\int_{1/(n+1)}^{1/n} \frac{\sin x }{x^3}dx\approx. \int_{1/(n+1)}^{1/n} \left(\frac{1}{x^2} \right)dx\ =1$$ We can make this argument rigorous by writing $$\frac{\sin x}{x^3}=x^{-2}+\sum_{k=1}^{\infty} \frac{x^{2k-2}}{(2k+1)!}$$ whereupon $$\int_{1/(n+1)}^{1/n} \frac{\sin x ...


3

Yes. Try $f(x)=\sqrt{x}\sin(1/x)$. This curve has unbounded variation, so in particular it is not rectifiable.


3

Maybe you want to prove: $$ \sum_{k=\color{red}{1}}^{n}\frac{1}{1+\frac{k}{n}}\leq n\log 2\tag{1}$$ that follows from: $$ \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}\leq\int_{0}^{1}\frac{dx}{1+x}=\log 2 \tag{2}$$ since $f(x)=\frac{1}{1+x}$ is a Riemann-integrable, decreasing function over $[0,1]$. If you do not like the integral approach, you may also ...


2

You have $\{v_1,v_2,v_3,v_4\}$. An affine space containing those points is $v_1+\text{ span }\{v_2-v_1,v_3-v_1,v_4-v_1\}$ which is $v_1$ plus an ordinary subspace. Find a basis for that subspace.


2

If you have: $$ |x| = \frac{\pi}{2}-\frac{4}{\pi}\sum_{n\geq 0}\frac{\cos((2n+1)x)}{(2n+1)^2}\tag{1}$$ by setting $x=0$ it follows that: $$ \sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}.\tag{2}$$


2

$$\cos\dfrac x2=\pm\dfrac1{\sqrt2}\iff\cos x=2\cos^2\dfrac x2-1=0$$ $x=(2m+1)\dfrac\pi2$ where $m$ is any integer


2

Your solution of $x/2 = \pm \pi/4 + 2 \pi k$ is equivalent only to $cos(x/2) = 1/\sqrt2$. For $cos(x/2) = -1/\sqrt2$, $x/2 = \pm 3 \pi /4 + 2 \pi k$. Combining them, we get $x/2 = \pm \pi /4 + \pi k$ ,or $x/2 = \pi /4 + \pi k/2$ ,or $x = \pi /2 + \pi k$ PS: Try imagining/drawing angles in Quadrants of Cartesian coordinates if you are confused in any ...


2

It depends on how the Fourier series and the power series are related. When you ave a real-analytic $2\pi$-periodic function $f:\>{\Bbb R}\to{\Bbb C}$ then it has a well convergent Fourier series $$f(t)=\sum_{k=-\infty}^\infty c_k\>e^{ikt}\quad(t\in{\Bbb R})\ ;\tag{1}$$ and on the other hand it can be developed into a Taylor series near $0$, say, of ...


2

Let $x=\sqrt{\sqrt{...\sqrt{5}}}$. $$x^2 = \sqrt{\sqrt{...\sqrt{5}}} = x$$ $$x = 0 \, or \, 1 $$ Verify:$$\sqrt{5}=2.24$$ $$\sqrt{\sqrt{5}}=1.50$$ $$...$$ From the pattern we can see $x$ tends to $1$, so $x = 0$ is rejected.


2

$ f(x) = \frac {x^2-y^2}{x^2+y^2} $ is not continuous at $(0,0) $ , Because if we approach through real axis $ f(x,0)=1 $ and through imaginary axis $f(0,y)=-1$


2

This is because of Abel's lemma, which is the basis for analytic functions: Abel's lemma. – If $r_0$ is positive real number such that the sequence $(\lvert a_n\rvert r_0^n)$ is bounded from above, then the series $\sum a_nz^n$ is absolutely convergent for $\lvert z\rvert<r_0$. Hence if $R=\sup\bigl\{\lvert z\rvert\:;\: \sum ...


2

If $(a_n)_{n\in\Bbb N}$ is a sequence of complex numbers such that the series $\sum_{i=0}^na_n$ is convergent in the usual sense then certainly the sequence $(a_n)_{n\in\Bbb N}$ is bounded (it must even converge to$~0$); it follows that for any $z\in\Bbb C$ with $|z|<1$ the series $\sum_{i=0}^na_nz^n$ is absolutely convergent, and therefore convergent. ...


1

For power series, the area of convergence really is the interior of a circle, perhaps with some points on the boundary. Here is a quote from Rudin's Principles of Mathematical Analysis, second edition (also called Baby Rudin), page 60, referring to power series of the complex number $z$. More specifically, with every power series there is associated a ...


1

I think that in this case, an example is worth more than a detailed answer. Consider the power series $$ \sum_{i=1}^\infty\frac{(-1)^{i+1}}{i2^i}x^i $$ (this is $\log(1+x/2)$ for $|x|<2$). Now, apply the ratio test (but keep the $x$) to get $$ ...


1

Yes, you are right. More formally, using your estimates, you can write: $$\left|\frac{5y - 5x}{(2x - 1)(2y - 1)}\right|\leq \left|5y - 5x\right| = 5|y - x|\leq 5\delta.$$ So, take $$\varepsilon = 5\delta\Rightarrow \delta = \frac{\varepsilon}{5}.$$ Since this estimate doesn't depend on the choice of $x$ and $y$, your continuity is uniform.


1

$$n=\sqrt{...\sqrt{\sqrt{\sqrt{5}}}}$$ Since there's an infinite cascade of roots, we can add 1 (or as many as we want): $$n=\sqrt{...\sqrt{\sqrt{\sqrt{\sqrt{5}}}}}$$ Replacing by the $n$ from the first equation: $$n=\sqrt{n}$$ or $$n^2=n$$ Since square roots of a number greater than 1 never can become less than 1, we can discard the solution $n=0$, ...


1

$\sqrt{...\sqrt{\sqrt{\sqrt{5}}}}=\lim x_n$, where $x_n=5^{(1/2^n)} (n\in\mathbb{N})$


1

This might seem redundant but also seems correct to me. If we take the sequence $(S_n)$ to be the constant sequence of $\textbf{Identity operators}$ then this sequence converges strongly to the Identity operator. Since $(T_n)$ converges to $T$ weakly and in this case the sequence $T_nS_n$ is nothing but the sequence $(T_n)$, which clearly fails to converge ...


1

HINT: Maybe using this simple inequality and then squeeze it? $$x-\frac{x^3}{6}\le \sin(x) \le x \ , x\ge0$$ The limit is $1$.


1

Let's generalize. Why? Because this kind of limit doesn't really depend on power series or anything fancy. Suppose we have a continuous $f$ on $(0,1)$ with $\lim_{x\to 0^+}f(x) = 1.$ Then $$\int_{1/(n+1)}^{1/n}\frac{f(x)}{x^2}\,dx \to 1$$ as $n \to \infty.$ (In our problem we have $f(x) = (\sin x)/x.$) Proof: Let $\epsilon>0.$ Choose $\delta > 0$ ...


1

Hint: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$



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