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7

Hint. Consider the set $S=\{k+2\pi u | (k,u) \in \mathbb{Z}^2\}$. S is an additive subgroup of the reals. And you probably know that the additive subgroups of the reals are either closed (and have a least positive element) or dense. $S$ cannot be closed as this would imply that $\pi$ would be rational. So $S$ is dense. As $\sin$ is continuous, $\sin(S)$ is ...


5

So we know that: $$ I_{2n}=\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta = \frac{\pi}{2}\cdot\frac{(2n-1)!!}{(2n)!!}=\frac{\pi}{2}\cdot\frac{1}{4^n}\binom{2n}{n}.\tag{1}$$ If, in the same way, we prove: $$ I_{2n+1}=\int_{0}^{\pi/2}\cos^{2n+1}(\theta)\,d\theta = \frac{(2n)!!}{(2n+1)!!}\tag{2} $$ then we have: $$\lim_{n\to +\infty} n\,I_{2n}\, I_{2n+1} = ...


5

Since $f:\mathbb{N} \rightarrow \mathbb{N}$ is injective $$ \prod_{k=1}^{n} f(k)\geqslant n!$$ And $$ \sum_{n=1}^{\infty} \left(\frac{1}{n+1!} \prod_{k=1}^{n} f(k)\right)\geqslant \sum_{n=1}^{\infty}\dfrac1{n+1}\to\infty $$


5

We may prove for first that: $$ a_{n+1}^2 = a_{n}^2 + a_{n}\leq \left(a_n+\frac{1}{2}\right)^2 \tag{1}$$ from which it follows that $\{a_n\}_{n\geq 1}$ is increasing and $a_n\leq \frac{n+1}{2}$. $(1)$ also gives: $$ a_{n+1}-a_n = \frac{a_n}{a_n+a_{n+1}}=\frac{1}{2}\left(1-\frac{a_{n+1}-a_{n}}{a_n+a_{n+1}}\right)\tag{2} $$ but since ...


5

If $z(t)=x_1 (t) + \Bbb i x_2 (t)$ then $- z^2 = (x_2 ^2 - x_1 ^2) - 2 x_1 x_2 \Bbb i$. Therefore, your equation can be rewritten as $\dot z = - z ^2$ or, equivalently, $\frac {\dot z} {z^2} = -1$. Integrate this with respect to $t$, getting $\frac 1 z = t + C$, or $z(t) = \frac 1 {t+C}$, with $C=a + b \Bbb i$ an integration constant. To get back to $x_1$ ...


5

Observe how the right hand side looks like the real and the imaginary term of a squared complex number. In fact, if you write $$z^2=x_1^2-x_2^2+i (2x_1x_2)$$ you can see, that if you multiply the second equation by $i$ and add it to the first one, you will get $$\dot{z}=-z^2$$ which is trivially solvable. This is a very handy trick that can be used in ...


4

If $z>y>0$, then $$\lim \int_y^z x^a dx = \lim \frac {z^{a+1} - y^{a+1}}{a+1} \\ = \lim \left(\frac {\exp ((a+1)\log z) - \exp 0}{a+1} - \frac{\exp((a+1)\log y) - \exp 0}{a+1} \right) = \log z - \log y$$ by definition of what $\exp'(0)=1$ means. So $\lim \int_y^z x^a dx = \int_y^z x^{-1} dx$ and so the function $a \mapsto \int_y^z x^a dx$ is ...


4


4

$$a_n = \sum_{k=1}^{n}\frac{1}{n+k} = H_{2n}-H_n $$ gives an increasing sequence since: $$ a_{n+1}-a_n = \sum_{m=n+2}^{2n+2}\frac{1}{m}-\sum_{m=n+1}^{2n}\frac{1}{m} = \frac{1}{2n+1}-\frac{1}{2n+2}=\frac{1}{(2n+1)(2n+2)}\geq 0$$ and by summing both sides of the previous line for $n=0,1,2,\ldots$ we get: $$ \lim_{n\to +\infty} a_n = \sum_{n\geq ...


3

Hint use stolz $$I=\lim_{n\to\infty}\dfrac{a_{n}}{n}=\lim_{n\to\infty}(a_{n+1}-a_{n})$$ and Note $$a_{n}=S_{n}-S_{n-1}=a^2_{n+1}-a^2_{n}$$ so $$a_{n+1}=\sqrt{a^2_{n}+a_{n}}$$ it is not easy to prove $a_{n}\to+\infty,n\to\infty$ so $$I=\lim_{n\to\infty}(\sqrt{a^2_{n}+a_{n}}-a_{n})=\lim_{x\to+\infty}(\sqrt{x^2+x}-x)=\dfrac{1}{2}$$


2

The proof you have given is acceptable within the context you're working in assuming you are allowed to use the special case of the intermediate value theorem about sign changes implying the existence of a zero (as your proof shows, this special case is logically equivalent to the seemingly more general version). If you have not been given the "special case" ...


2

This is the rather well known theorem about the derivative of an inverse function. Let us put $\;g:=f^{-1}\;$ , for simplicity, and observe that both $$\begin{cases}\;y\to b\implies& \;\;(y=)\;\color{red}{f(x)\to f(a)}\;(=b)\\{}\\f(x)\to f(a)\implies&\;\; (g(f(x))=)\;\color{red}{x\to a}\;(=g(f(a)))\end{cases} \;\;\;\text{(why?)}$$ , so: ...


2

Hint: Use the chain rule $$(f^{-1} \circ f )(x) = x \implies (f^{-1} \circ f ) ' (x) = 1 \implies (f^{-1})'(f(x)) f'(x) = 1$$ Apply it at $a$. Conversely, if $f'(a) \neq 0$ then any sequence of points $y_n = f(x_n) \in Y - \{b\}$ with $\lim y_n = b$ the continuity of $g$ at $b$ yields $\lim x_n = a$ then $$\begin{align}g'(b) = \lim \frac{g(y_n) - ...


2

We can look at this from another angle. What we need to show is that $$\int_a^b (f(x)-g(x)) dx = 0.$$ The function $h(x) = f(x)-g(x)$ is zero except at the one point that I will relabel as $p$. Now suppose that $\delta >0$ and small enough so that $(p-\delta, p+\delta) \subset (a,b)$. Then $$\int_a^b (f(x) - g(x)) dx = \int_{p-\delta}^{p+\delta} (f(x) - ...


2

$x\mapsto \frac{x^2}{9}$ is a bijection of $\Bbb{R}^+$. So if you show that a statement holds for every $\epsilon$ "of the form" $\frac{\epsilon'^2}{9}$, you show that it holds for ALL $\epsilon>0$.


2

If I understood you well, your $\ell^0$ is the space of complex sequences that are eventually vanishing. If that is right, consider the sequence $(c_n) \in (\ell^0)^{\mathbb N}$ defined by $c_n(k) = \frac{1}{k}$ for $1 \le k \le n$ and $c_n(k) = 0$ for $k > n$. $(c_n)$ is a Cauchy sequence as for $n < m$ $$\Vert c_n - c_m \Vert^ 2=\sum_{k=n+1}^m ...


2

I'll answer your second question first. You have $f:K\times K\to\mathbb R$. The space $K\times K$ is the product of two compact sets, and is therefore compact. This is not hard to prove, and is a very simple result of the Tychonoff Theorem. Now, in order to apply the extreme value theorem, we have to show that $f$ is continuous. To do this, pick ...


2

Let $x = (\xi_n)$ and $y = (\omega_n)$. We note that $$ p(x + y) = \lim_{n \to \infty} \sup_{m \geq n} (\xi_m + \omega_m) \leq \lim_{n \to \infty} \left(\sup_{m \geq n}\xi_m + \sup_{m \geq n}\omega_m\right) = p(x) + p(y) $$ To further expand on the inequality: we note that $A \subset B \implies \sup A \leq \sup B$. This allows us to state that $$ ...


2

We can use compactness of the product $K\times K$ without mentioning it. There is a sequence $(x_m,y_m)\in K\times K$ such that $$ \lim_{m\to\infty}|x_m-y_m|=\sup_{x,y\in K}|x-y| $$ by the properties of the supremum. Since $K$ is compact, the sequence $(x_m)$ has a convergent subsequence, say $(x_{m_h})$. Similarly, the sequence $(y_{m_h})$ has a convergent ...


2

Let $y=x^{2}$. Consider $f(x,x^{2})=\frac{x^{4}}{2x^{4}}=\frac{1}{2}$.So it's not continuous at $(0,0)$. (Even it does not have a limit, you can plug $y=0$)


2

I think I can confirm your suspicion that this doesn't necessarily hold if the target space is non-Hausdorff, assuming I haven't made a mistake somewhere... Let $\mathbb{R}$ be the real line in its standard topology. Let $\mathbb{R}_0$ be the real line with the topology whose non-empty open sets $U$ are precisely the standard open sets $U \subseteq ...


1

It's not possible. Think of small $h> 0.$ Then by the MVT, $$\tag 1 [f(3R/4)-f(\,(R/2)+h\,)]/(\,(R/4)-h) = f'(c_h)$$ for some $c_h \in (R/2+h,3R/4).$ But $f(3R/4)=0,$ and because $f(R/2)=1,f'(R/2) = 0,$ we have $f((R/2)+h) = 1+o(h).$ So the left side of $(1)$ equals $$-(1+o(h)\,)/(\,(R/4)-h\,).$$ Now check that the absolute value of the above is ...


1

The usual strategy is to let the open cover of the compact set $K$ be $\mathcal B=\{B(x,1):x\in K\}$. Since each point in $K$ has its own ball, $\mathcal B$ definitely covers $K$. Since $K$ is compact, there exists a finite subcover.


1

$x^{2001} + (\frac{1}{2} - x)^{2001} = 0 \implies x^{2001} + \sum_0 ^{2001} \binom{2001}{k} (\frac{1}{2})^{2001 - k} (-x)^k = 0$... What does Vieta's formula say at this point?


1

A two variable function can be thought of as a one-variable function on a product space. For example, if $X,Y,Z$ are spaces, and $f(x,y)$ is a two-variable function taking values in $Z$, for $x\in X,y\in Y$, then we think of $f$ as a function $f\colon X\times Y\to Z$. This leaves the matter of what metric/topology to put on the product $X\times Y$. We ...


1

We are given $$I_{2n}=\frac{\pi}{2}\frac{(2n-1)!!}{(2n)!!} \tag 1$$ and $$I_n\sim \sqrt{\frac{\pi}{2n}} \tag 2$$ From $(2)$ is trivial to see that $$I_{2n}\sim \frac12 \sqrt{\frac{\pi}{n}} \tag 3$$ Then, using $(1)$ and $(3)$, we find that $$\begin{align} \frac{(2n+1)!!}{\sqrt{n}(2n)!!}&=\frac{(2n+1)(2n-1)!!}{\sqrt{n}(2n)!!}\\\\ ...


1

I'll assume you want integrability over $[0, 1]$, but it's not difficult to see that it's integrable over any interval. First things first, I hope you can see that, if anything, the integral should be $1$, since its graph is $1$ almost everywhere along a length $1$ interval. It's important to be able to see this, as this will affect how we go about the ...


1

I will prove a different estimate, but it should help you get started and give you a feeling of how to use your knowledge. Notice that since $\eta\in C^1_0(B_R)$ and $u\in W^{1,2}(B_R)$, you have $\phi\in W^{1,2}_{B_R}$. Therefore $\phi$ is a valid test function. Note also that it follows from your ellipticity estimate that $a_{ij}X_iX_j\leq\Lambda|X||Y|$ ...


1

You have $$a_n=\frac1n\sum_{k=1}^n\frac1{1+\frac kn},$$which is a Riemann sum for $\int_0^1\frac1{1+t}\,dt.$ This shows that $a_n\to\log(2)$. Dunno about the monotonicity.


1

You've got the right idea. We want, for any choice of $\epsilon$, to be able to find an $M$ such that the inequality holds. The process you worked through (i.e. taking $M > \frac{1}{\epsilon^2}$) gives us a way to find/construct our $M$ in such a way that it works for any $\epsilon$ we like. The proof lies in being able to find an $M$ for any ...



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