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## Hot answers tagged real-analysis

6

Hint 1: What can you say about $f(y)$ if $y$ is in the image of $f$? Hint 2: The intermediate value theorem will be helpful. Note that $999$ is in the image of $f$.

4

Let $x\in [0,1]$, $\epsilon >0$. Then there is $N \in \mathbb N$ large so that $$\frac 1N <2\epsilon.$$ Now we know that there is $m \in \mathbb Z$ so that $0< \{ m\alpha\} <\frac 1N$. Let $K \in \mathbb N$ so that $K\{m\alpha\} <1$ and $(K+1)\{m\alpha\} \ge 1$. Then the set $$C = \{ \{m\alpha\}, \{2m\alpha\}, \{3m \alpha\}, \cdots, \{ K ... 3 You meant O((x-x_0)^2). If f is differentiable and g=f' then f and g need not satisfy f(x)=f(x_0)+g(x_0)(x-x_0)+O((x-x_0)^2) (that will hold under stronger conditions on f.) But if f:I\to\Bbb R then yes, f is differentiable if and only if there exists a function g with$$f(x)=f(x_0)+g(x_0)(x-x_0)+o(x-x_0)\quad(x\to x_0)$$(where the ... 3 Since the other answers already address your idea, I will try to focus on solving the problem. We will abuse of connectedness. Define the equivalence relation: x \sim y \text{ if } y \text{ and } x \text{ can be connected by a broken line}. This is obviously an equivalence relation. Now, since the set is open, every equivalence class is easily seen to ... 3 Since [[1/x]]=[1/x]. Consider two sequnces x_n=\frac{1}{2n} and x'_n=\frac{1}{2n+1} and you get two different limits. 2 Your idea is good but needs a bit of modifying. First of all, if you take open ball and take intersection of the arc with the balls boundary for the center of the next ball, who can guarantee that the radii won't converge to 0 too fast so you will never be able to reach \gamma(b)? But, for any x\in[a,b] you can choose an open ball B_x = ... 2 First of all a comment: I think what you mean by a "broken line" is a continuous, piecewise straight path. However, to define a straight line, you need to be in Euclidean space (or at least a Riemannian/Finsler manifold). So let's assume (X,d) is \mathbb{R}^n with the Euclidean metric. Your idea looks fine, but you have to be careful with the statement ... 2 Your first argument actually showed already that f(m) =m for all m. Thus there are no other choices for f, f has to be the identity function. But concerning your argument, let me just point out that You are using the fact that if a sequence \{a_n\} converges to b, then any subsequence a_{n_k} converges to b. In your calculation you fixed ... 2 You cannot use the mean value theorem for vector valued functions. But you can argue as follows: Let v = F (b)-F (a) . Then consider the function$$ f (x) := \langle F (x), v \rangle $$for which you can apply the mean value theorem. For the resulting bound, apply the Cauchy-Schwarz inequality. Note though that the above assumes that you use the ... 2 So your question is satisfactorily answered if you are given initial conditions for Newton's method which find each root. Here I will assume that two roots exist, i.e. p^2-4q>0. Given an initial condition not exactly on the vertex, the Newton iteration will stay on that side of the vertex, because of the fact that there is only one turning point. Also, ... 2 Hint: Using t=e^{-x}\ge 0, we see that this series is a usual power series in t where all the common techniques apply. 2 The coefficients of the series for a rational function obey a linear recursion equation that is defined by the coefficients of the denominator. Thus any coefficient series that has zero gaps of increasing length can not belong to a rational function. 2 In your answer, you must show for any \epsilon > 0, you can find an n such that 1-\epsilon < 1-\dfrac{1}{n}, and this means n > \dfrac{1}{\epsilon}. Thus choose n > \dfrac{1}{\epsilon} is sufficient. 2 Hint Use Intermediate Value Theorem between \frac{1}{999} and 999. 500 belongs in this interval. 2 The comments lead to a proof that f(m)=m for all m: By induction on k you find from the given functional equation that f(m^k)=f(m)^k for all m\in\Bbb N, k\in\Bbb N. For m>1, the sequence m, m^2,m^3,\ldots is a subsequence of 1,2,3,\ldots, hence$$\lim_{k\to\infty}\frac{\log f(m^k)}{\log (m^k)}=\lim_{n\to\infty}\frac{\log f(n)}{\log ...

1

Since $\| \alpha v \|$ = $\vert \alpha \vert \cdot \| v \|$ (it's the property from definition of norm or length) then $\| \hat{e}_j(P) \|= \left \| \frac{\frac{\partial \vec{r}}{\partial q_i}}{ \left \| \frac{\partial \vec{r}}{\partial q_i} \right \|} \right \| = \frac{\left \| \frac{\partial \vec{r}}{\partial q_i} \right \|}{\left \| \frac{\partial ... 1 Use $$\frac{a-b}{ab}=\frac1b-\frac1a$$ to obtain a telescoping sum. 1 go through the definition of outer measure. that'll help you out. it's infimum over the covers that covers$I$. now as that is infimum hence there'll be a cover such that$\sum_{n=1}^{\infty}l(I_n) \leqslant m^*(I)+\frac{\epsilon}{2}$1 Consider$g(x) = e^{-x}f(x)$so that$g'(x) = e^{-x}\{f'(x) - f(x)\} > 0$for all$x$. And clearly$g(0) = f(0) = 0$. Since$g$is strictly increasing it follows that$g(x) > g(0) = 0$for all$x > 0$. Thus$f(x) = e^{x}g(x) > 0$for all$x > 0$. 1 This holds in an arbitrary hausdorff space$X$: If$X$is compact, every infinite subset has a limit point. Suppose$X$is not finite. Then there exists an infinite subset$A$. By what is stated above, you can choose a limit point$x$of$A$. Take it away from$A$if it is in$A$, and denote the new set by$\tilde{A}$. It is still infinite. By assumption, ... 1 You correctly shows that$1$is an upper bound for$S$. To complete the proof that$1$is the least upper bound for$S$, you need to show tha tthere is no upper bound strictly smaller than$S$. You can do so by contradiction: Assume$a$is an upper bond for$S$and$a<1$. Recall that we want to exhibit and element$1-\frac1n$of$S$that is$>a$. ... 1 I think that it would be good example. Moreover you must prove it strictly. $$\sum \limits_{k=1}^{\infty}{x^{2^k}}\notin \mathbb{Q}[x]$$ 1 By mean value theorem, there is a$|c_x-c|<|x-c|$such that $$\frac{f(x)-f(c)}{x-c}=f'(c_x),$$ and thus $$f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c}f'(c_x)= L.$$ 1 What you want to prove is that$f'(c)=L$. But $$f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}$$ and, since$f$is continuous at$c$(because it's differentiable at$c$), we can use l'Hôpital's theorem: $$f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c}f'(x)$$ provided the last limit exists. 1$f$is the sum of three functions. You might be able to show that each of them is convex, in which case the sum is also convex. Notice that the first function$(a,b,c)\mapsto {1\over a+b}$is the composition of the linear function$(a,b,c)\mapsto a+b$with the convex funtion$0<t\mapsto 1/t$. 1 Let$\epsilon > 0$, be a positive number To find a$\delta$such that$|x-3|<\delta$implies$|f(x)-f(3)|<\epsilon$. Case-I if$x\in \mathbb Q$Then$|f(x)-f(3)|=|x^2-9|=|(x-3)(x+3)|$Therefore we choose our$\delta$to be$<1$, then$|x-3|<1$implies$-1<x-3<1$, adding 6 on both sides we get$ x+3<7$. ... 1 For your first problem, (the case where$x \in \mathbb Q$) take$x= \frac 3 7$and calculate$a_1, a_2, a_3, \cdots$. You will see a pattern. For example$\sin \left( \frac{21 \pi}{7} \right) = 0$. Try to generalize for any rational$x$. On the other hand, when$x \in \mathbb Q^c$, observe that$y:= \pi x$is also irrational, and now you will have to show ... 1 Abel's Theorem states that if$\sum_n^\infty a_n$is a convergent series of constants, then the power series$\sum_n^\infty a_nx^n$converges uniformly for$0\le x\le 1$. In addition, we have that if a sequence of functions$f_n(x)$converges uniformly to$f(x)$on$x\in[a,b]$, then $$\lim_{n\to \infty}\int_a^b f_n(x)\,dx=\int_a^b f(x)\,dx$$ ... 1 Answering$1$and$2$simultaneously: It is equivalent due to the fact that$\overline{\mathbb{R}}$is first countable: that is, for every point$x$there exists a countable local basis. You can find more info on definitions here. Note that this is valid for an arbitrary metric space. However, the definition you state is not equivalent in general ... 1 While Aloizio Macedo correctly states that these two definitions are equivalent because of the first countability of the reals, here is a short proof: ($\Rightarrow$) Assume that$f$is continuous at$x\in T$with respect to the$\varepsilon-\delta$convention. Let$\{x_n\}$be a sequence in$T\setminus\{x\}$converging to$x$. Then fix$\varepsilon>0\$. ...

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