# Tag Info

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Do not treat $\infty$ as a real number. If you do treat $\infty$ as a real number, you may run into some problems. For example, $\infty+2=\infty$. We treat infinity as an IDEA, not an actual number. It is so far beyond our reach that we cannot perform operations on it and manipulate it. The limit: $$\lim\limits_{x\to \infty} \dfrac{3x-1}{2x-3}$$ is not ...

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This is a general fact for any ultrametric space $(X,d)$: this is a metric space where the triangle inequality is strengthened to: for all $x,y,z \in X$ we have $d(x,z) \le \max(d(x,y), d(y,z))$. In that case, any two $r$-balls either are disjoint, or are equal. For suppose that $z \in B_r(x) \cap B_r(y)$. So $d(x,z) < r$ and $d(y,z) < r$. It follows ...

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Inverse function theorem requires continuous differentiability, which you don't assume.

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Take $\mathbb R$ to be the domain of $x$ and $y$ and let $$f(x,y) = \frac{y^2}{1+x^2},$$ $g(y) = y$, and $c = 1$.

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$f-f\chi_E = f(1-\chi_E)$, which is $0\cdot \infty = 0$ when $f(x) = f\chi_E(x)= \infty$

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Let's fix this, then. Since $1-\chi_E = \chi_{E^c}$, we can read $f-f\chi_E$ as $f\chi_{E^c}$. So, $f=f\chi_E+f\chi_{E^c}$ which implies $$\int f=\int f\chi_E+ \int f\chi_{E^c} = \lim \int f_n\chi_E + 0 = \lim \int f_n$$

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It would not. Let's say $f$ is a function such that $\partial_x f(t,0)$ is not integrable. I define $$\tilde f(t,x) = \begin{cases} f(2t,x),\quad & 0\le t\le 1/2 \\ -f(2t-1,x),\quad &1/2\le t\le 1 \end{cases}$$ Although $\partial_x \tilde f(t,0)$ is still not integrable, the function $F$ is identically zero.

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Yes. Or in more detail, $\ldots 000n$ is (shorthand for) an infinite string of digits which notates a decimal. And that decimal does represent a natural number. Note that $n$ and $\ldots 000n$ are two notations for the same decimal. A decimal has digits in every place, and our convention is that if we only write a string of finitely many digits, then all ...

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To answer the other question: Cantor diagonalization cannot be used in this case: if you try using it for "most" choices, the digit you'll have will be $0$. So infinitely many times, you'll need to pick a digit $a \neq 0$, but this would create an expression with infinitely many non-zero digits, which is not a number.

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It's a little odd to write a number that way, but it's clear. However, every natural number written that way has a "leftward tail" of zeros, so that the diagonalization argument won't actually work: the new number you create won't be a natural.

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$0's$ on the left don't affect the value of a number, so that $0....0n=n$

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For the other way: Hint: If $\sum\frac{x_n}{1+x_n}$ converges, then in particular $x_n$ must converge to $0$, and in particular there exists a constant $C\geq0$ such that $x_n\leq C$. Then $$\sum\frac{x_n}{1+x_n}\geq\sum\frac{x_n}{1+C}.$$

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As $$0 \le \frac{x_n}{1+x_n}\le x_n,\\ \sum {x_n}<\infty\Rightarrow \sum \frac{x_n}{1+x_n}<\infty$$ Now suppose that $\sum \frac{x_n}{1+x_n}<\infty$. $\frac{x_n}{1+x_n}\to 0$, and so $x_n\to 0$, so there is a $N$ such as $n>N\Rightarrow x_n<\frac 12$ and then$$\frac{x_n}{1+x_n}>\frac 23 x_n$$so $$\sum ... 2 \begin{eqnarray} a^n &=& \exp(n\log(a)), \\ &=& \exp(\log(a) + \ldots + \log(a)), \quad \mbox{n times}\\ &=& \exp(\log(a))\exp(\log(a))\ldots \exp(\log(a)) \\ &=& a a \ldots a. \end{eqnarray} 1 Let A be a dense G_\delta. Then A =\bigcap\limits_{i=1}^\infty O_i where each O_i is open. So, A^C=\bigcup\limits_{i=1}^\infty O_i^C . Each O_i^C is closed. Show, using the denseness of A, that each O_i is nowhere dense. Once you do this, it follows by definition that A^C is of first category. It then follows that A must be ... 0 Can someone find the flaw in this argument? Suppose it were; then for every \epsilon, there exists a \delta for which$$|x - y| < \delta \implies |f(x) - f(y)| < \epsilon$$However, consider \epsilon = 1; if such \delta existed and y = x + \frac{\delta}{2}, we would find that$$|x^\alpha\sin(x^\beta) - \left(x + ...

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You might consider looking at "Analysis in Euclidean Space" by Kenneth Hoffman. The book concentrates attention on $\mathbb{R}^n$, and has some good exercises. Generalizations refer to normed linear spaces, generally $\mathbb{R}^n$ with some norm, or $\mathbb{C}$, the complex numbers. This is great preparation if you eventually decide to get into functional ...

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I suggest two real classics. The first one is Introduction to Calculus and Analysis Volume I and II from Richard Courant. The first volume is based on $\mathbb{R}$, the second treats Analysis on $\mathbb{R^d}$. Metric spaces are not considered. If you are looking for lots of good examples from Analysis without (explicitely) using the concept of metric ...

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It does not follow. For example $$u_n=2^n+(-2)^n$$ has characteristic equation with roots $2$ and $-2$; however $u_n=0$ whenever $n$ is odd, so $u_n$ does not tend to $\infty$.

2

$P(z)= a^z = E(A(z))$, where $E(z)=\exp(z)$ and $A(z)=z \log a$. Both $E$ and $A$ are continuous functions and so is $P = E\circ A$, being a composition of continuous functions. This argument assumes you have proved that $\exp$ is continuous, which should come from its power series definition. The function $A$ is simply a scaling.

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please do not up- or downvote this answer. It's just impossible to do this by pure commenting. I copy pasted your calculus and added some remarks. $V_{w(x)_n} =\pi\displaystyle\int_{\alpha}^{\beta} (w(x)_n)^2dx = \pi\displaystyle\int_{\alpha}^{\beta} (\displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) )^2dx =$ $\pi\displaystyle\int_{\alpha}^{\beta} ... 0 I agree with you for (a), (c), (d), (g), (h) For (b) I would say "all functions" because$\delta$is allowed to be negative. (e) is the empty set because you can take$x_1=x_2$. For (i), I would say "nondecreasing and uniformly continuous". If$f$satisfies (i), then it is nondecreasing because$x_1-x_2\leq 0\implies f(x_1)-f(x_2)<\varepsilon$for any ... 1 A classical method, going back to Newton himself, is to expand the integrand into a power series and integrate term by term. If the resulting series converges, it converges to the value of the integral (this is easy to show if the interval of integration is within the open interval of convergence; in your case, since$1$is right at the edge of convergence, ... 1 If you know the Ruffini identity, which is connected to the Horner scheme, then you know that there exists a polynomial$q(a,x)$such that $$p(x)=p(a)+(x-a)q(a,x)$$ Now you need only to argue that$q(a,x)$is bounded if the distance from$x$to$a$is bounded to get the estimate $$|p(x)-p(a)|\le M(r)\,|x-a| \quad\text{for}\quad |x-a|<r.$$ 0 $$\frac \beta{k^\alpha} - \frac \beta{(k+1)^\alpha} \sim \frac {\alpha\beta}{k^{\alpha + 1}} = \frac 1{\sqrt{k}}$$if$\alpha = -1/2$and$\beta = -2$now as$\sum \frac 1{\sqrt{k}} = \infty$: $$\sum_{k=1}^N \frac 1{\sqrt{k}}\sim \sum_{k=1}^N \frac \beta{k^\alpha} - \frac \beta{(k+1)^\alpha} \sim -\frac \beta{N^\alpha} = 2\sqrt{N}$$ 1 OK, check (as you mentioned, by induction, or whatever method you prefer) that for all$k$,$a_n^k\to a^k$. For this, use that the limit of a product of the product of the limits. Now, if$a_n^k\to a^k$, then also$ca_n^k\to ca^k$for any constant$c$. Finally, note that any polynomial is a sum of terms of that form, and use that the limit of a sum is the ... 1 Clearly, (as you know continuity), the polynomial$P$is a continuous function mapping from$\mathbb{R}$to$\mathbb{R}$. Sequential continuity coincides with continuity in this case, then$a_n\to a$implies$P(a_n)\to P(a)$. See sequential continuity and continuity 1 Given$\varepsilon$, choose$n_0$such that for all$n\ge n_0$we have$|a_n-a|<\min\bigg(1 \frac{\varepsilon}{|a|^2+|a|(1+|a|)+(1+|a|)^2}\bigg)$so$|a_n| \le1+|a|\begin{align}|a_n^3-a^3|=|a-a_n||a^2+aa_n+a_n^2|\le |a_n-a|(|a|^2+|a||a_n|+|a_n|^2)\\ \le|a_n-a|\big[|a|^2+|a|(1+|a|)+(1+|a|)^2\big]\\<\varepsilon \end{align} 3 By looking at upper and lower Riemann sums on this decreasing function, you can bracket it by $$\int_1^{n} x^{-\frac 12}\ dx \lt \sum_{k=1}^nk^{-\frac 12} \lt \int_0^{n-1} x^{-\frac 12}\ dx\\2\sqrt n-2\lt \sum_{k=1}^nk^{-\frac 12} \lt2\sqrt {n-1}$$ 0 Since\alpha_n \to \alpha$there is an$n_0$such that$|\alpha_n - \alpha| < \sqrt[3]{\epsilon}$for every$n \geq n_0$. Now raise both sides to the$3$rd power, which is allowed as the function$x \mapsto x^3$is strictly increasing. 1 I can't post comments, but i guess you got a typo there. If not, this is obviously wrong. Consider any ball in$\Bbb R^2$with the standard scalar product, and think about it geometrically. (Please feel free to delete the answer again after an update from the OP) 1 This is a point of definitions and naming conventions. My preferred definition is that when $$S_n = \sum_{i=0}^n a_i$$ we say that the sequence$S_n$is the series of term$a_n$. Both$S_n$and$a_n$are sequences. While $$S = \lim_{n\to \infty} S_n = \sum_{i=0}^\infty a_i$$ is the sum of the series of term$a_n$.$S$is a number (or$\pm \infty$). 3 No, unless you impose some extra constraints on$f$. For example, let$a=0$and let$f$be the function: $$f(x)=\left\{\begin{array}{ll} x-1 & \textrm{if }x\in \{1,2,3,\ldots\} \\ x & \textrm{otherwise} \end{array}\right.$$ Then$f$is actually a bijection (one-to-one and onto) from$\mathbb{R}\setminus\{0\}$to$\mathbb{R}$. 1 Not at all... the range of a one-to-one function like that can be any more than countable set. For example it could be:$[0,1]$,$\mathbb R$, a cantor set... 1 As others have pointed out, you should prove this by using the triangle inequality. I also think you should try to understand the problem intuitively so I drew a picture:$|a-b|$represents the distance between the points$a$and$b$on the number line. If$c$is between$a$and$b$, or is equal to either$a$or$b$, then the distance from$a$to$c$, which ... 1 You could use the triangle inequality and the fact that you can wright$|a-b|=|(a-c)+(c-b)|$1 Yes: If$r\in\Bbb Q$, then$\forall n\in\Bbb N$:${rn\over n+\sqrt2}\in{\Bbb Q}^c$and $$\lim_{n\to\infty}{rn\over n+\sqrt2}=r.$$ 0 For any rational number$x=\frac{p}{q}$with$\gcd(p,q)=1$, just consider: $$x_n = \frac{p}{q}\cdot\frac{n}{\sqrt{n^2+1}}.$$ Clearly any$x_n$belong to$\mathbb{R}\setminus\mathbb{Q}$and we have$\lim_{n\to +\infty} x_n = x$. 20 Assume your number is$\frac{p}{q}$. Then the sequence $$a_n=\frac{\pi}{n}+\frac{p}{q}$$ converges to the given number and is irrational (any irrational number in the place of$\pi$would do). 12 Yes, take a sequence consisting of your sequence of irrationals converging to$0$plus your desired rational limit. 0 As per Daniel Fischer's hint, $$0 \leq s_n \leq \sum_{k = 1}^{m_n}\left[\left(t^{(n)}_k - t^{(n)}_{k - 1}\right)\underbrace{\max_{i \in \left\{1, \dots, m_n\right\}}\left(t^{(n)}_i - t^{(n)}_{i - 1}\right)}_{=: M_n}\right] = M_n\sum_{k = 1}^{m_n}\left(t^{(n)}_k - t^{(n)}_{k - 1}\right) = M_nt$$ Since,by assumption,$\lim_{n \rightarrow \infty} M_n = 0$, ... 0 Result: Let$f$be a mapping from a metric space$\big( M , d \big)$to a topological space$X$, then$f$is continuous at$x\in M$if and only if given any sequence$\big\{ x_n\,,\, n\in\mathbb{N}^\ast \big\}$convergent to$x$,$f(x_n)\to f(x)$in$X$. Proof. Assume that$f$is continuous at$x\in M$and let$N$be a neighborhood of$f(x)$, then ... 1 Note that 1$\implies$2$\implies$3, where: For every$(u,v)$,$|\sin u-\sin v|\leqslant|u-v|$For every$x$and every$(f,g)$,$|F(f)(x)-F(g)(x)|\leqslant|f(x)-g(x)|$For every$(f,g)$,$\|F(f)-F(g)\|_\infty\leqslant\|f-g\|_\infty$1$f$need not be uniformly continuous on$(0,3)$. Take for instance$x\mapsto \frac1x$. if it is continuous on$(0,3)$,$f$is in particular continuous on$[1,2]$. So apply Heine-Cantor theorem to get that$f$is uniformly continuous on$[1,2]$, hence on$(1,2)$. 2 The function$f$is continuous on the interval$(0,3)$then it's continuous on the compact$[1,2]$so by the Heine-Cantor theorem$f$is uniformly continuous on$[1,2]$. 0$\sin$is uniformly continuous on$[0,1]$. If we have any two functions$f$and$g$(not necessarily continuous) which are appropriately "uniformly close", then$\sin(f)$and$\sin(g)$are "uniformly close" as well. And you're correct about the role of continuity being that the supremum is well-defined. Your proof seems correct to me (you should only be a ... 0 No. Take, for example the function $$f(x)=\cos(\log x )$$ for$x>0$. This function is crearly$C^\infty$and bounded in its domain. We have: $$f'(x)=-\frac1{x}\sin\left(\log x \right)$$ which vanishes at infinity. However, since$\log x\rightarrow\infty$when$x\rightarrow\infty$,$f(x)$has no limit at infinity. Just in case os you need a proof of ... 1 My recipe for creating counterexamples of this sort is to design a function that oscillates between$-1$and$1$. In this case, you want the derivative to converge to$0$. This is easy to arrange piecewise: start at$0$and have the function increase to$1$, then decrease to$-1$more gradually, then increase to$1$even more gradually, then decrease to ... 4 Counterexample:$f(x) = \sin(\ln(x))$0 Idea : Making use of successive Integration by parts ,$\int P(x)f^{(3)}(x)\,dx = P(x)f^{(2)}(x)-P^{(1)}(x)f^{(1)}(x)+P^{(2)}(x)f(x)-\int P^{(3)}(x)f(x)\,dx$Consider the two third degree monic polynomials,$P_1(x)$and$P_2(x)$. Now, we compute the difference of the definite integrals:$\int\limits_{\frac{a+b}{2}}^b ...

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