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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 ...

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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 $f \circ \psi^{-1}\colon \widetilde{\mathbb{R}}\to \mathbb{R}$ ...

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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 writex=\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)}...

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

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$$y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$$ 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 + ...

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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 $$\forall \... 1 For a coherent answer, you really have to go via complex numbers. The answer is that a^c = \exp(c \log(a)), where \log(a) is a number b such that \exp(b) = a. The problem is that this is multi-valued, because \exp(2\pi i)=1. Now \exp(x+iy) = \exp(x) (\cos(y) + i \sin(y)) (where x and y are real) is real if and only if \sin(y)=0, i.e. y = ... 0 If you want to refer to Wikipedia, you have to identify the proper variables. Here, you have z(y)=f(y) and y(x,s)=x+s. Hence \frac{dz}{dy}= \frac{df}{dy} and \frac{\partial y}{\partial s}= 1. Which leads to$$\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)|. $$The usual conclusion is that the solution depends continuously on the initial point. And also that two solutions that ... 0 It's obvious that there exists p(x)>1 such that$$\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$$So the second derivative must also be 0 at y=1 You can continue this process to get that the nth derivative is 0 at y=1 and if you assume that the solution is analytic, this implies that ... 0 Reaching 1 and staying constant, or touching 1 at any point at all cannot happen unless it is 1 throughout, this ODE is locally lipschitz continuous so that a "solutions can't cross" lemma applies: formally if x:I \rightarrow \mathbb{R} and y: J \rightarrow \mathbb{R} are two solutions that agree at some point in I \cap J then they agree on all ... 0 Your argument is correct. I only have a minor suggestion (which is perhaps a bit nitpicking). You do not explicitly define what is means that a map with range A \subset \mathbb R^n smooth if A is not known to be open. The solution is of course to regard f as a map from U to \mathbb R^n. Thus your statement could be reformulated as follows: Let f : ... 1 In such proofs, you can always use \frac{1}{n} instead of \varepsilon, because of Archimedean property. If some property is satisfied by \frac{1}{n}\forall n\in \mathbb N then choosing an \varepsilon>0, we can choose \frac{1}{n}<\varepsilon. Then, since the property is satisfied for all \frac{1}{n} where n\in\mathbb N, it is satisfied by ... 3 By the Mean Value Theorem, for a x\in(c-\epsilon,c)\cup(c,c+\epsilon), there is t_x\in(c-\epsilon,c)\cup(c,c+\epsilon) such that$$\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.)

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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 ...

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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. ...

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$x_{n+2} = \sqrt{x_{n+1} x_n}$. 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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