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0

Try Peter D Lax’s Multivariable calculus. I took a sophomore level multivariable calculus courses at an American university under a European professor and he used this book. This was the hardest math class I ever took as this book introduces multivariable calculus using rigorous proofs and introducing techniques for analysis at the same time. The class was ...


0

Put more formally, prove that, for $L, K, n\ge 1 \in \mathbb{Z}$, $$ \tag{1}\label{proposition} (L < K) \wedge (L < m \le K) \wedge (K/n \text{ is upper bound}) \wedge (L/n \text{ is no upper bound}) \Rightarrow (\exists m) (m/n \text{ is upper bound}) \wedge ((m-1)/n \text{ is no upper bound}) $$ Proof is by induction over $K - L$. Base case: $K-L ...


0

Any continuous function $f$ on $\mathbb R^2$ that satisfies $\lim_{|z|\to \infty}f(z)=0$ is uniformly continuous on $\mathbb R^2.$


0

For real numbers the equation $x^n = b$ will, if $b\ne 0; b\ne 1$ 1) If $b >0$ and $n$ is even have exactly two solutions $x = c$ for some $c > 0$ and $-c$. 1b) If $b > 0$ and $n$ is odd then there will have exactly one solution $x = c$ for som $c > 0$. 2) If $b < 0$ and $n$ is odd then there will be exactly one solution $x = c$ for some $c ...


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Just wanted to cover up the $\implies$ case, where the other answers doesn't addressed, and the OP just explained vaguely "differentiating and using the chain rule gives that the required derivatives of $f$ vanish". As stated in a similar question, it is not just that simple. Let's show that if $f \circ \psi^{-1}\colon \widetilde{\mathbb{R}}\to \mathbb{R}$ ...


2

Name $X=(x,y)$. Then $$f(x,y)=f(X)= \frac{1}{\Vert X\Vert^2 +1}.$$ Where $\Vert \cdot \Vert$ is the Euclidean norm. We have $$\begin{aligned}\vert f(X_1)-f(X_2) \vert &= \left\vert \Vert X_1\Vert - \Vert X_2 \Vert\right\vert \left\vert \frac{\Vert X_1\Vert + \Vert X_2\Vert}{(\Vert X_1\Vert +1)(\Vert X_2\Vert +1)}\right\vert\\ &\le 2\left\vert \...


0

Note that $\displaystyle \|\nabla f(x,y)\| = 2\sqrt{\frac{x^2+y^2}{(x^2+y^2+1)^4}}$ and the function $z\mapsto \frac{z}{(z+1)^4}$ is bounded for non-negative $z$, so $\nabla f$ is bounded and $f$ is Lipschitz, hence uniformly continuous.


2

For all $x\in\mathbb{R}$ we can write $$x=\lfloor x\rfloor+\{x\}$$ where $\{x\}$ denotes the fractional part of $x$. Thus we have that $$f(x)=\lfloor x\rfloor+\{x\}^2$$ I will now prove that for any $y\in\mathbb{R}$ there exists some $x\in\mathbb{R}$ such that $y=f(x)$. Note that this is equivalent to $$y=\lfloor y\rfloor+\{y\}=\lfloor x\rfloor+\{x\}^2=f(x)$$...


1

$$\frac1h\int_a^b\left(f(x+h+t)-f(x+t)\right) dt\stackrel{u:=x+t}=\frac1h\int_{x+a}^{x+b}\left(f(u+h)-f(u)\right)\,du$$ and since $\;f\;$ is continuous it has a primitive function, say $\;F\;$ , so $$\frac1h\int_{x+a}^{x+b}\left(f(u+h)-f(u)\right)\,du=\frac1h\left(F(x+b+h)-F(x+a+h)-F(x+b)+ F(x+a)\right)=$$ $$=\frac{F(x+b+h)-F(x+b)}h-\frac{F(x+a+h)-F(x+a)}...


2

Hint: use a substitution, and the fundamental theorem of calculus.


1

$$y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$$ So $$y^{'}(x)=\sum_{n=0}^{\infty}na_{n}x^{n-1}$$ import these power series to main differential equation $y^{'}(x)=1+xy(x)$to receive to $$\sum_{n=0}^{\infty}na_{n}x^{n-1}=1+\sum_{n=0}^{\infty}a_{n}x^{n+1}$$ by simplifying this equation: $$a_{1}=1 , a_{3}=\frac{1}{3}, a_{5}=\frac{1}{15}, \cdot\cdot\cdot,a_{2n-1}=...


1

If the conditions of the Picard–Lindelöf theorem are satisfied then the function sequence $(y_n)$ defined iteratively by $$ y_0(x) = 0 \, , \\ y_{n+1}(x) = Ty(x) = x + \int_0^x t y_n(t) \, dt $$ converge to a solution of the initial value problem. This is called Picard-iteration. The first iterates are $$ \begin{align} y_0(x) &= 0 \\ y_1(x) &= x + ...


0

Yes. Let $g:[-1,1]\to\mathbb{R}$ be a given smooth function. Let $g_n:[-1,1]\to\mathbb{R}$ be a sequence of polynomial that converges uniformly to $g$ and let $h:[-1,1]\to\mathbb{R}$ be a bounded nowhere differentiable function. Then the sequence $$v_n (t)=g_n (t) +n^{-1} h(t) $$ is a sequence of nowhere differentiable functions that converges to $g$ ...


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Hint: Remember that $p = \sup B$ if and only if for every $ε > 0$ there is an $x ∈ B$ with $x > p − ε$, and $x ≤ p$ for every $x ∈ B$. And $p = \inf B$ if and only if for every $ε > 0$ there is an $x ∈ B$ with $x < p + ε$, and $x ≥ p$ for every $x ∈ B$. It's clear that $$\forall x \in B,\, b^2 \ge x$$ But you have to show that $$ \forall \...


1

For a coherent answer, you really have to go via complex numbers. The answer is that $a^c = \exp(c \log(a))$, where $\log(a)$ is a number $b$ such that $\exp(b) = a$. The problem is that this is multi-valued, because $\exp(2\pi i)=1$. Now $\exp(x+iy) = \exp(x) (\cos(y) + i \sin(y))$ (where $x$ and $y$ are real) is real if and only if $\sin(y)=0$, i.e. $y = ...


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If you want to refer to Wikipedia, you have to identify the proper variables. Here, you have $z(y)=f(y)$ and $y(x,s)=x+s$. Hence $\frac{dz}{dy}= \frac{df}{dy}$ and $\frac{\partial y}{\partial s}= 1$. Which leads to $$\frac{\partial f(x+s)}{\partial s}= \frac{df}{dy}(x+s)= f^\prime(x+s).$$


2

You should write, since $f$ is a function of a single variable $$\frac{\partial f(x+s)}{\partial s}=\frac{\mathrm df}{\mathrm d x}(x+s)\cdot\frac{\partial (x+s)}{\partial s}=\frac{\mathrm df}{\mathrm d x}(x+s).$$ Note the two positions of $(x+s)$: on the left side, $f(x+s)$ implicitly defines a function of two variables $g(x,s)$, whereas in the first factor ...


0

Of the ODE is locally Lipschitz with a Lipschitz constant $L$, then the Grönwall lemma tells us that for two solutions (and in a time interval that has them in the set where $L$ is valid) $$ |y_1(t)-y_2(t)|\le e^{L|t-s|}|y_1(s)-y_2(s)|. $$ The usual conclusion is that the solution depends continuously on the initial point. And also that two solutions that ...


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It's obvious that there exists $p(x)>1$ such that $$\frac1{(1+|x|)^{p(x)}}=\frac12\frac1{1+|x|}\quad\quad(x\ne0).$$


0

Writing $$\frac{1}{a^4\left(\left(\frac{x}{a}\right)^4+1\right)}$$ now substitute $$t=\frac{x}{a}$$ so $$dt=\frac{1}{a}dx$$ and factorize $$t^4+1=t^4+2t^2+1-2t^2=(t^2+1)^2-2t^2=(t^2+1-\sqrt{2}t)(t^2+1+\sqrt{2}t)$$


0

One such condition would be: there is a countable set $C$ such that$$(\forall x\in\mathbb R\setminus C):f(x)\geqslant g(x).$$


1

$A - B$ is defined as $\{c| c+B \subset A\}$ this is equivalent to: $$A-B=(A^c+(-B))^c$$ (I define $(-B)$ as $\{-b|b \in B\}$) The equivalence holds since $$(A-B)^c=\{c| c+B\not\subset A\}=\{c|\exists b \in N \text{ with } c+b \in A^c\}= \{A^c-b| b\in B\}=\{A^c+b| b\in (-B)\}=A^c +(-B)$$ Since $B$ and $A^c$ are closed it follows that $A^c +(-B)$ is closed....


0

Note that as $h \to 0$, $$\frac{f(x+h)+f(x-h)-2f(x)}{h^2} = \frac{g(x)h+a(x,h)+g(x)(-h)+a(x,-h)}{h^2} \to 0,$$ that is, $f''(x) = \lim \frac{f(x+h)+f(x-h)-2f(x)}{h^2} = 0$ for all $x$. This already completes the proof.


0

As zwh as noted in his comment, there are some hidden assumptions in your question. Precisely, you define (Definition 2 above) a closed set as a set which contains all its accumulation points, while in your (completely correct) proof of the sufficient part you use the fact that if $S$ is open then $S^{c}$, which is the (or better one of the) "classical ...


1

By the triangle inequality, $$d(y_{m_k},z_{m_k}) \leq d(y_{m_k},y) + d(y,z) +d(z,z_{m_k})\tag{1}$$ and also $$d(y,z) \leq d(y, y_{m_k}) + d(y_{m_k},z_{m_k}) +d(z_{m_k},z)\tag{2}$$ Combining $(1)$ and $(2)$ gives $$|d(y,z)-d(y_{m_k},z_{m_k})|\leq d(y, y_{m_k}) +d(z_{m_k},z)$$ Now, the right-hand side converges to zero, by assumptions.


1

It is $$\Gamma \left(\frac{6}{5}\right)$$ so the integral does converge.


4

Compare the integrand with a simpler function: for all $x\geq 1$, we have $e^{-x^5}\leq e^{-x}$, and $$\int_1^{+\infty}e^{-x}dx=\frac 1 e <+\infty$$ so the integral is convergent.


4

Since $\int_0^1e^{-x^5}$ converges, you can compare with ${1\over x^5}$ on $[1,+\infty)$


0

If I differentiate both sides of the equation I get $$\frac{d^2y}{{dx}^{2}}= \frac{dy}{dx}(1-y) -\frac{dy}{dx}y$$ $$\frac{d^2y}{{dx}^{2}}= y(1-y)(1-y) -y(1-y)y$$ So the second derivative must also be $0$ at $y=1$ You can continue this process to get that the $n$th derivative is $0$ at $y=1$ and if you assume that the solution is analytic, this implies that ...


0

Reaching $1$ and staying constant, or touching $1$ at any point at all cannot happen unless it is $1$ throughout, this ODE is locally lipschitz continuous so that a "solutions can't cross" lemma applies: formally if $x:I \rightarrow \mathbb{R}$ and $y: J \rightarrow \mathbb{R}$ are two solutions that agree at some point in $I \cap J$ then they agree on all ...


0

Your argument is correct. I only have a minor suggestion (which is perhaps a bit nitpicking). You do not explicitly define what is means that a map with range $A \subset \mathbb R^n$ smooth if $A$ is not known to be open. The solution is of course to regard $f$ as a map from $U$ to $\mathbb R^n$. Thus your statement could be reformulated as follows: Let $f : ...


1

In such proofs, you can always use $\frac{1}{n}$ instead of $\varepsilon$, because of Archimedean property. If some property is satisfied by $\frac{1}{n}\forall n\in \mathbb N$ then choosing an $\varepsilon>0$, we can choose $\frac{1}{n}<\varepsilon$. Then, since the property is satisfied for all $\frac{1}{n}$ where $n\in\mathbb N$, it is satisfied by $...


3

By the Mean Value Theorem, for a $x\in(c-\epsilon,c)\cup(c,c+\epsilon)$, there is $t_x\in(c-\epsilon,c)\cup(c,c+\epsilon)$ such that $$\frac{f(x)-f(c)}{x-c}=f'(t_x).$$ Can you take it from here?


5

Assume WLOG $f$ is non-decreasing. First of all, Claim As an $\mathbb R\to\mathbb R$ function, $f$ cannot be continuous everywhere. proof. Suppose $f$ is continuous on $\mathbb R$. Then we find a sequence $\{a_n\}$ in $\mathbb Q$ such that $a_n\nearrow\sqrt{2}$. By the surjectiveness (when considered as a $\mathbb R\to\mathbb Q$ function) we can find ...


0

Enough to show $\forall r,f(r)=f(0)+rg(0)$. First, let me simply the question a bit. If there exist $f,g$ satisfying the condition and $f$ is nonaffine with $f(0)=a$ and $f(1)=a+b$, we may consider $f^*(x)=\frac{f(x)-a}{b},g^*(x)=\frac{g(x)}{b}$, where $f^*$ is not affine and $(f^*,g^*)$ also satisfies the condition. Furthermore, if there exist such $f,g$...


1

Corrections for Answer 1: ... That is the ball $B_\epsilon(x)$ contains infinitely many points of $(x_n)$ (for $\forall n > N)$), that is $(x_n)_{n = N}^{\infty} \subset B_\epsilon(x)$. So $B_\epsilon (x) \cap A \neq \emptyset$. Your stated conclusion does not immediately follow because your sequence elements $(x_n)_{n = 1}^{\infty}$ are from $A'$ and ...


0

For (2) you could say: Since $x_n \to x$ theres a $k$ such that $|x_k-x|=\delta<\epsilon/2$. If $\delta=0$ we have $x_k=x$ and therefore x is a limit point. Lets consider the case with $\delta>0$. Since $x_k$ is a limit point theres a sequence in A converging to $x_k$. Let this sequence be $ (m_n)_{n\in \mathbb{N}}$. Then theres a $N$ such that $n\...


2

Mathworld's claim is false. This issue is actually discussed by Kouba (1995), who gives this counterexample: Define $f:\mathbb R \rightarrow \mathbb R$ by: $$f\left(x\right)=\begin{cases} x^{3}+x^{4}\sin\frac{1}{x} & \text{ for }x\neq0,\\ 0 & \text{ for }x=0. \end{cases}$$ Kouba shows that $f$ is continuous at the stationary point $0$. Also, $f'(...


2

We use the max norm $$ || T(y) - T(z) ||_{\infty} = \max_{x \in [-1, 1]} | T(y)(x) - T(z)(x) | $$ For any $x \in [-1, 1]$, we write $$ | T(y)(x) - T(z)(x) | = \left| \int_{0}^{x} t( y(t) - z(t) )dt \right| $$ and $$ | T(y)(x) - T(z)(x) | \leq \left| \int_{0}^{x} t dt \right| ||y-z ||_{\infty} \leq \frac{1}{2} ||y-z ||_{\infty} $$


1

There exists $r<1$ such that $|z| \leq r$ for all $z \in K$. Hence $|\frac {z^{2}} {n^{2}-z^{2}}| \leq \frac 1 {n^{2}-r^{2}} \leq \frac 2 {n^{2}}$ for $n >\sqrt 2 r$. [First few terms do not have an effect on unform convergence].


4

Okay, a little late, but I just figured how to go about it myself so I'll post an answer for completeness: $|\frac{x^2 - \pi^2}{\sqrt{x^2 + 5} + \sqrt{\pi^2 + 5}}| <|\frac{x^2 - \pi^2}{x + \pi}| < |x - \pi| < \delta$ and choosing $\delta = \epsilon$ completes the proof!


2

You know the denominator is always greater than or equal to $\sqrt{\pi^2+5}$, so observe $\big|\frac{x^2-\pi^2}{\sqrt{x^2+5}+\sqrt{\pi^2+5}}\big|\le\big|\frac{x^2-\pi^2}{\sqrt{\pi^2+5^2}}\big|$. Then choose $\delta>0$ such that $|x^2-\pi^2|<\epsilon\sqrt{\pi^2+5^2}$.


0

Here is as elementary a proof as I can come up with that the limit exists and the limit is $\sqrt[3]{x_1^2 x_0} $. $x_{n+2} = \sqrt{x_{n+1} x_n} $, so, taking logs, $\begin{array}\\ \log x_{n+2} &= \log\sqrt{x_{n+1} x_n}\\ &= \frac12 \log(x_{n+1} x_n)\\ &= \frac12 (\log x_{n+1} +\log x_n)\\ &= \frac12 \log x_{n+1} +\frac12 \log x_n\\ \end{...


3

The existence of such a sequence is impossible. In fact, if $\{x_n\}$ is a sequence of real numbers diverging to $\infty$, then $x_n$ has a subsequence which, for almost every $r$, is uniformly distributed mod $r$. To see this, pass to a subsequence $\{x_n'\}$ of $\{x_n\}$ to get $|x_n'-x_m'|>1$ for all $n\neq m$ and apply the following Theorem and ...


2

The inequality arises due to a much simpler reason: $$ |x_{in} - x^\ast_{i}|^2 = \color{red}{(x_{in} - x^\ast_{i})^2} \leq \color{blue}{(x_{1n} - x^\ast_{1})^2} + \cdots + \color{red}{(x_{in} - x^\ast_{i})^2} + \cdots + \color{blue}{(x_{mn} - x^\ast_{m})^2} $$ The LHS is less than the RHS simply because the RHS includes the LHS $\color{red}{\text{red }(x_{in}...


1

Regarding the inequality $$|x_{in} - x_i^*| \le \sqrt{(x_{1n} - x_1^*)^2 + (x_{mn} - x^*_m)^2},$$ try squaring both sides. (No triangle inequality needed.)


1

By "invertible" it specifically means there is a smooth inverse. ("A $C^\infty$ map $f$... is locally invertible... if $f$ has a $C^\infty$ inverse...") The cubing map $x\mapsto x^3$ from $\mathbb{R}^1$ to itself is topologically invertible with zero Jacobian at $x=0$, but the inverse is not smooth, specifically it is not differentiable at 0. If there is ...


1

There is a disagreement between introductory calculus and real analysis. The Calculus definition is: "If $a$ lies in some open interval within the domain of $f(x)$, we say that $\lim_{x\to a} f(x)=L$ provided that $f(x)$ gets close to $L$ as $x$ gets close to $a$". Note that it is phrased in a way for a "first year" student to be able to understand it. ...


0

$x_{n+2} = \sqrt{x_{n+1} x_n} $. Suppose $x_n =x_0^{a(n)}x_1^{b(n)} $ with $a(0) = 1, b(0) = 0, a(1) = 0, b(1) = 1 $. Then $x_0^{a(n+2)}x_1^{b(n+2)} =\sqrt{x_0^{a(n+1)}x_1^{b(n+1)}x_0^{a(n)}x_1^{b(n)}} =x_0^{(a(n+1)+a(n))/2}x_1^{(b(n+1)+b(n))/2} $ so that $a(n+2) =(a(n+1)+a(n))/2, b(n+2) =(b(n+1)+b(n))/2 $. Both $a(n)$ and $b(n)$ are of the form $ru^n+sv^...


1

Let $M$ be in $\mathcal{M}$ and let $S$ be any $\sigma$-algebra containing $\mathcal{A}$. If $M$ is countable it's a countable union of singletons, all of which are in $S$ and so $M \in S$. Otherwise it's the complement of a countable set (which is in $S$ as saw) and so also in $S$ (as $S$ is closed under complements). This show the left to right inclusion. ...


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