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## New answers tagged real-analysis

0

First, we prove the fact $T(t)=T\left(\frac{t}{n+1}\right)^{n+1}$ by induction. First, we show the property hold for the case $n=1$, i.e. $T(t)=T\left(\frac{t}{2}\right)^{2}$. The right hand side is \begin{equation*} T\left(\frac{t}{2}\right)T\left(\frac{t}{2}\right)=T(t) \end{equation*} equals the left hand side. Now assume it is true for $n=k$, where ...

1

we have $$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+2+h}-\sqrt{x+2}}{h}=\frac{x+2+h-x-2}{h(\sqrt{x+2+h}+\sqrt{x+2})}$$ and then you can compute the Limit for $h$ tends to zero.

4

You don't need to start from the definition of a limit. Just use the definition of the derivative of a function $$f'(x) := \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ and the rules for evaluating limits. To see how this is done explicitly scroll over the region below.

0

Another solution: Suppose $\mathcal{P} = x_{0}, x_{1}, \ldots, x_{N}$. We're going to show that $U( \operatorname{id}, \mathcal{P}) - U(f, \mathcal{P}) < \epsilon$ for every $\epsilon > 0$, and so $U(f, \mathcal{P}) = U(\operatorname{id}, \mathcal{P}) \geq \frac{1}{2}$, where $\operatorname{id}$ is the identity map on $[0, 1]$. Since $f \leq ... 0 It does not matter if it is decreasing. Given the set$A=\{1/2,...,1/n\}$, define your partition$P=\{x_0,x_1...,x_{2n+2}\}$such that for every$k\in\{1,...,n\}$, $$\frac{1}{k}-\frac{\varepsilon}{2n}\in P$$ and $$\frac{1}{k}+\frac{\varepsilon}{2n}\in P,$$ where$0<\varepsilon < 1$. Of course,$0\in P$and$1\in P$. Giving$x_i$a proper order, we have ... 0 Let$\mathcal P_0$be a partition of$[0,1]$. Refine this partition if necessary to a partition$\mathcal P$containing$n+1$points so that$x_{i}-x_{i-1}=1/n.\ $Note that$x_0=0$,$x_{n}=1$and in general$x_i=i/n$. We have$U(\mathcal P_0)\geq U(\mathcal P)=\frac{1}{n}\sum_{i=1}^{n}f(x_i^{*})$. Now let$\epsilon >0$and choose, using density of the ... 1 Another way: Let$P = \{x_0,x_1,\ldots,x_n\}$be a partition of$[0,1]$, and let$M_i$denote the supremum of$f$on$[x_{i-1},x_i]$. There must exist a rational number$q$in the interval$[(x_{i-1} + x_i)/2,x_i]$, and so it follows that$M_i \geq (x_{i - 1} + x_{i})/2$. This gives: $$U(f,P) = \sum_{i = 1}^{n}M_i(x_i - x_{i - 1}) \geq \sum_{i = ... 1 Given any small positive real \varepsilon > 0, for any partition P = \{x_0, x_1, \ldots, x_n\} of [0, 1] such that 0 = x_0 < x_1 < \cdots < x_n = 1, since \mathbb{Q} is dense in \mathbb{R}, for each i, there exists q_i \in \mathbb{Q} such that x_i - \varepsilon < q_i < x_i, it then follows that$$M_i = \sup_{I_i} f(x) - ... 2 If you're thinking Riemannly: What if your$k^\text{th}$partition is$\left\{\frac{j}{k^k} \mid 0 \leq j < k^k \right\}$? If you're thinking Lebesguely: What do you think of$[0,1] = \{0\} \cup \bigcup_{i \in \Bbb{N}} \left( (\frac{1}{i+1},\frac{1}{i}) \cup \{\frac{1}{i}\}\right)$? 0 Your proof looks good to me. An outline for an alternative proof: define$g\ :\ \Bbb R \times \Bbb R^k \to \Bbb R^n\ :\ (t, x) \mapsto f_t(x)$.$g$is then continuous.$T := [-1,1]\times S^{k-1}$is compact. Since for all$t, f_t$is injective and$f_t(0) = 0$, we know that$0 \notin f_t(S^{k-1})$, so$0 \notin g(T)$. Since$g(T)$is compact, it is closed, ... 1 No, it's not even closed. Using the interval$[-1,1]$instead of$[0,1]$, one can see that the sequence $$f_n(x) = |x|^{2+1/n}\operatorname{sign} x$$ converges in$C^1$norm to$f(x)=|x|^2\operatorname{sign} x$which is not in$C^2$. Indeed,$f_n'(x) = (2+1/n)|x|^{1+1/n}$converges uniformly to$f'(x) = 2|x|$. The second derivatives ... 1 To add to @Joey Zou's answer, the hint from the book is If$h>0$, Taylor's theorem shows that $$f'(x)=\frac1{2h}(f(x+2h)-f(x)]-hf''(\xi)$$ for some$\xi\in(x,x+2h)$. Hence $$|f'(x)|\leqslant hM_2 + \frac{M_0}h.$$ From there it follows that $$h^2M_2 - h|f'(x)| + M_0 \geqslant 0,$$ and since this is a quadratic in$h$which is strictly nonnegative, ... 0 The sum$s = \sum_{2^n<a} 2^n$evaluates to$2^n$where$n$is the smallest integer such that$2^n\ge a$. Hence,$s/2\le a\le 2s$. The above observation made, the proof consists of introducing $$g=\sum_{n}2^n \chi_{\{f>2^n\}}$$ and using the inequalities $$g/2 \le f \le 2g$$ 2 Your problem is similar to Problem 5.15 from Rudin's Principles of Mathematical Analysis, where the problem is to show that$M_1^2\le 4M_0M_2$for$f:(a,\infty)\rightarrow\mathbb{R}$which are bounded and have two bounded derivatives, where$a\in\mathbb{R}$. In the problem statement, Rudin mentions that equality can be achieved, e.g. for $$f(x) = ... 0 (i) (1) follows because the sum of two norms is a norm. (2) is not a norm; \|f'\|_\infty = 0 does not imply f \equiv 0. (3) is a norm, even though it the sum of a norm and a non-norm. (4) I do not understand your question "What is |f(0)|?" Anyway, this is a norm: Show |f(0)| +\|f'\|_1 = 0 \implies f' \equiv 0 \implies f is constant; the condition ... 0 The trivial upper bound is n! \le n^n and so n! = O(n^n). Wikipedia on Stirling's approximation tells us that n! \le e\ n^{n+1/2}e^{-n} and so n! = O(n^{n+1/2}e^{-n} ). 0 Disclaimer:Not my solution. One elegant answer was posted by Robert Isreal on here on here. Apparently a holistic answer is not known. 1 By archimedian property for all -x>0\exists n_x\in\mathbb N s.t. 0<\frac1{n_x}<-x\implies x<-\frac1{n_x}<0. I hope you can get the rest of the thing. 0 It's not a bad idea to keep track of your variable dependencies, especially when first learning chain rule problems. Specifically, your statement \tan u = f is clearer when written$$\tan u(x,y) = f(x,y).$$Now you can take the derivative of both sides with respect to x.$$\sec^2 (u(x,y)) \frac{\partial u}{\partial x}(x,y) = \frac{\partial f}{\partial ... 0 Hint: The mean value theorem implies$f'(s_n) = 0 $for a sequence$s_n\to t_0.$2 Let$x$be any negative real number. Then$-x$is positive, and by the Archimedean property there is a natural number$n$such that$0<-x<n$. Then$-\frac 1n$is a negative rational number, and $$x<-\frac 1n<0$$ Thus$x$is not greater than all negative rational numbers. Since$x$are arbitrary, there is no negative real number greater that ... 1 If we use Taylor's theorem with an integral remainder we get: $$1+x+\frac{x}{2}+\ldots+\frac{x^{2n}}{(2n)!} = e^x-\frac{1}{(2n)!}\int_{0}^{x}t^{2n}e^{x-t}\,dt \tag{1}$$ hence in order to prove the non-negativity of the LHS it is enough to show that: $$\int_{0}^{x} t^{2n}e^{-t}\,dt \leq (2n)! \tag{2}$$ holds for any$x\in\mathbb{R}$.$(2)$is trivial if ... 2 If it's uniformly continuous on$[0, 1]$, then for every epsilon, there is$\delta$such that for all$x \in [0, 1]$, something holds. Can you show that that$\delta$works as a bound on all of$\mathbb{R}$, not just$[0, 1]$? 0 Let$P_n(x)$your polynomial. The Taylor-Lagrange formula gives for$x\in \mathbb{R}$that there exists$c$such that$\exp(x)=P_n(x)+\frac{x^{2n+1}}{(2n+1)!}\exp(c)$for some$c$depending on$x$. Now suppose that there exists$x\in \mathbb{R}$such that$P_n(x)=0$. Obviously, we have$x<0$. Then$\exp(x)=\frac{x^{2n+1}}{(2n+1)!}\exp(c)$, and ... 10 By induction, assume that $$f_{2n}(x)=1+x+\frac{x^2}2+\cdots\frac{x^{2n}}{(2n)!}>0.$$ Then the antiderivative $$f_{2n+1}(x)=1+x+\frac{x^2}2+\cdots\frac{x^{2n+1}}{(2n+1)!}$$ is a growing function and has a single root. Then the antiderivative $$f_{2n+2}(x)=1+x+\frac{x^2}2+\cdots\frac{x^{2n+2}}{(2n+2)!}$$ has a single minimum, which occurs at this ... 1 No. Take$f(x)=|x|$it's not differentiable at$0$but$\lim \limits_{h\to 0}\dfrac{f(h)-f(-h)}{h}$exists. 0 Let's consider two problems here. Problem (for calculus students): If$f:\mathbb R\to\mathbb{R}$is a continuous function that has a symmetric derivative at a point$x_0$, i.e. $$SD\, f(x_0) = \lim_{h\to 0} \frac{f(x_0+h)-f(x_0-h)}{h}$$ exists, does it follow that$f'(x_0)$exists? Problem (for analysis students): If$f:\mathbb R ...

1

I think the point of the question is that $f(x)$ may not be continuous at $a$ or $b$. So the best answer is $(A)$. The reason is that the existence of these limits \eqalign{ & f(a + ) = \mathop {\lim }\limits_{x \to a + } f(x) \cr & f(b + ) = \mathop {\lim }\limits_{x \to b + } f(x) \cr} does not guaarantee that $f(x)$ is defined at $a$ ...

1

A Baire category argument would be recommended here, rather than a compactness argument like the nested interval attack. For each integer $m=1,2,3, ...$ define the set $$E_m = \bigcap_{n=1}^\infty \{x \in [a,b]: |f_n(x)| \leq m\}.$$ Since each $f_n$ is continuous, each set $E_m$ is closed. Also, since each sequence $\{f_n(t)\}$ is bounded for a fixed $t\in ... 0 I would use equivalents, via an asymptotic development:$\sin\dfrac1{n^\alpha}=\dfrac1{n^\alpha}+o\Bigl(\dfrac1{n^{2\alpha}}\Bigr)$,$\dfrac1{n^\alpha+1} =\dfrac1{n^\alpha}\biggl(1-\dfrac1{n^\alpha}+\dfrac1{n^{2\alpha}}+o\Bigl(\dfrac1{n^{2\alpha}}\Bigr)\biggr)$. Hence $$\sin\dfrac1{n^\alpha}-\dfrac1{n^\alpha+1} ... 0 Heine-Borel and Bolzano-Weierstrass theorems are equivalent in the sense that their proofs can be derived from each other. In fact, there are other axioms and results such as completeness axiom, the nested interval property, the Dedekind cut axiom of continuity and Cauchy’s general principle of convergence which are equivalent to these theorems, stated as: ... 1 You can indeed do Taylor expansions for \frac{1}{n^\alpha}, as long as \frac{1}{n^\alpha} \xrightarrow[n\to\infty]{} 0. (That is, for any \alpha > 0. Note that the case \alpha \leq 0 is straightforward). Doing so, you obtain:$$ \sin \frac{1}{n^\alpha} = \frac{1}{n^\alpha} + o\left(\frac{1}{n^{2\alpha}}\right) $$and$$ \frac{1}{n^\alpha+1} = ... 1 Hint. Using, as$x$tends to$0$, the expansions $$\sin x = x+o(x^2)$$ $$\frac{1}{1+x}=1-x+O(x^2)$$ you get $$n^2 \Big(\sin(\frac{1}{n^{\alpha}})-\frac{1}{n^{\alpha}+1}\Big)=n^2 \Big(\frac{1}{n^{\alpha}}-\frac{1}{n^{\alpha}}(1-\frac{1}{n^{\alpha}})+O(\frac{1}{n^{2\alpha}})\Big)=O(\frac{1}{n^{2\alpha-2}})$$ Then you are led to consider $$\sum ... 1 Hint: Write \displaystyle \sin(x)-\frac{x}{1+x}=x^2+O(x^3). 5 This can be transformed to$$\sum_{n=1}^{\infty} \frac{2}{(2n-1)2^{2n-1}}$$Let$$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)}$$Then, we have f'(x)=\sum_{n=1}^{\infty} x^{2n-2} = \frac{1}{1-x^2}. Therefore, we have$$f(x)=\int \frac{1}{1-x^2} = \frac{1}{2} \ln \frac{x+1}{1-x}+C$$It is clear that C=0. Now plugging x=\frac{1}{2} in this ... 2 There is a connection but to make it clear you have to be more precise about the boundaries of the inner integral on the right hand side of your second equation. The primitive function F of f is only determined up to a constant. Assuming that it is chosen so that F(a)=0, then we have$$\int_a^bFg'=\int_{x=a}^b\int_{t=a}^xf(t)dt\ g'(x)dx$$The ... 2 As Daniel fischer noted this argument only works for the special case of non negative terms b_i . Hint: if the set \{n: b_n \geq 1\} is infinite then you should know what to do. However if that is finite,then try to make use of the following fact: for every n such that b_n<1 we have$$\frac {b_n}{1+b_n} > \frac {b_n}{2} $$0 I have an answer when the limit of b_{n} does exist and b_{n}>0. We know that \sum_\limits{n\in\mathbb{N}}{b_n} is divergent so If \lim_\limits{n\to \infty}b_{n}\neq 0 so \lim_\limits{n\to \infty}\dfrac{b_n}{1+b_n}\neq 0. And if \lim_\limits{n\to \infty}b_{n}= 0 so we apply the limit test for series \lim_\limits{n\to ... 1 First of all, using 2xy \le x^2 + y^2,$$|1-xy| \le 1+|xy| \le 1+ \frac{x^2}{2} +\frac{y^2}{2} \le 1+x^2 +y^2\le (1+x^2)(1+y^2),$$so the function value is less than one. The above equality holds only when x = y=0, so the function value is strictly less then one (as x\neq y by requirement). On the other hand, let y = 0, it becomes$$\frac{1 + ... 0 Regarding the functional equation, it can be proved with the Poisson Summation Formula and the fact that the Fourier transform of a Gaussian is also a Gaussian. 1 You are right, you just did not finish the proof: $$\epsilon|w|-\epsilon|z|\leq\epsilon|z-w|= \epsilon\delta$$ Hence $$\epsilon |w| + \delta |x|\leq\epsilon|z| + \delta |x|+\epsilon\delta$$ 1 You need to use a cover of the real numbers that consists of bounded sets only, like the one suggested by Mirko. Compactness implies that a finite number of these bounded sets cover$A.$0 See this Wikipedia page on the Weierstrass function$W_{\alpha}(x)$. It is continuous everywhere, but differentiable nowhere. So let $$f(x,y)=W_{\alpha}(x)$$ Then$\frac{\partial f}{\partial x}$does not exist anywhere, and since$\frac{\partial f}{\partial y}=0$the mixed derivative$\frac{\partial^2 f}{\partial x\partial y}$exists everywhere (and is also ... 0 This is true. The problem becomes more transparent if we use the following characterization of$K$-Lipschitz functions:$f$is$K$-Lipschitz if and only if both$Kx-f(x)$and$Kx+f(x)$are nondecreasing. Thus, it suffices to prove that any branch of a countable family of nondecreasing functions$\psi_n$is also nondecreasing. Suppose that$f$is a ... 2 The function$f: R^2 \to R$where$f(x, y) = |x|+xy$is an example See that for the following function $$f_{yx} = 1$$ for every$(x, y) \ \epsilon \ R^2$But$f_x$doesn't exist at the following set of points$\{ (0, y) \ \epsilon \ R^2 \ | \ y \ \epsilon \ R \} $Thus for the set$\{ (0, y) \ \epsilon \ R^2 \ | \ y \ \epsilon \ R \} \subset R^2$, ... 0 Another useful things: Let$f$is well-defined in some neighborhood of the point$c\in \mathbb{R}$and$f''(c)$does exist. Then $$\lim_\limits{h\to 0}\dfrac{f(c+h)+f(c-h)-2f(c)}{h^{2}}=f''(c),$$ and $$\lim_\limits{h\to 0}\dfrac{f(c+2h)-2f(c+h)+f(c)}{h^{2}}=f''(c).$$ proof: \begin{eqnarray} \lim_\limits{h\to 0}\dfrac{f(c+h)+f(c-h)-2f(c)}{h^{2}} ... 4 No, the complement of an unbounded set need not be bounded. For instance, if$A = [0,\infty)$then$\mathbb R \setminus A = (-\infty,0)$is also unbounded. 1 Hint: For all we know,$f$is continuous everywhere but at$c$and if$\alpha(x)$(where$\alpha(x)$is the monotonically increasing function that we integrate$f$with respect to) is continuous at every discontinuity of$f$(for$f$with finitely many discontinuities) then$f\in\mathcal{R}(\alpha)$. To avoid this, we must pick an$\alpha$that is ... 6 Let$f$only depend on$x$, being continuous but not differentiable. Then$f_y=0$, hence$f_{xy}=0$. (Assuming that this means first differentiating with respect to$y$) 0 You're right about it being not bounded above, but you probably need to show that more rigorously. Suppose we have a supremum$s=\frac{a}{a-b}$for$b\in(0,a)$. Then let$c=\frac{a+b}{2}$. You should be able to show fairly easily$s'=\frac{a}{a-c}>s$, but$s'\in S_a$which is a contradiction. Also this idea of using$c=\frac{a+b}{2}\$ should work as ...

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