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The first paragraph is not meant to show $\int_\sigma f \,dz=0$; its purpose is to show that there exists an analytic function $F$ in $\Delta$ such that $F'=f$. Once this is obtained, the statement about $\int_\gamma f\,dz$ follows from the fundamental theorem of calculus for line integrals: the integral of $f$ is equal to the difference of the values of $F$ ...

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Because the function $z\mapsto\dfrac{1}{z-a}$ is not analytic at $z=a$. In fact, if $a$ is in the exterior of the disk then Cauchy's theorem is applicable and the integral is $0$. It is only if $a$ is in the interior of the disk, i.e. the curve over which one integrates winds around $a$ one in a counterclockwise direction, that the integral is $2\pi i$

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Being $a$ near of far from the origin is irrerevant. The important fact is $\gamma$ enclose $a$? (the singular point)

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Yes, it is true. Define a (complex-valued) measure on $\mathbb{R}$ by $$\mu(B) := \sum_{k \in \mathbb{Z}} c_k \delta_k(B), \qquad B \in \mathcal{B}(\mathbb{R}). \tag{1}$$ Since $\sum_{k \in \mathbb{Z}} |c_k|<\infty$, $\mu$ is a finite measure on $\mathbb{R}$. From $(1)$ we see that the formula $$\int f(x)\left( \sum_{k \in \mathbb{Z}} c_k \delta_k(dx) ... 1 I guess this might be a suitable example. We know that FT of an everlasting sinusoid of frequency f is two impulses of strength (\pi) at frequencies \pm f. Now in the case of Linear Frequency Modulation, the frequency of a carrier is linearly modulated. Suppose the modulation is up-chirp, ie, the frequency is linearly increasing, say, from f_1 to ... 0 I would do it as follows; say for \overline D_\nu\mu. First, observe that for each fixed r>0, the functions x\mapsto \mu(B_r(x)) and x\mapsto \nu(B_r(x)) are Borel. There is a reference for this (not completely trivial) fact in the link you give. It follows that the set \Omega=\{ x\in\mathbb R^N;\; \nu(B_r(x))>0\;\hbox{for all}\;r>0\} is ... 0 Given an open set V\subset Y, denote \widetilde V =\{x: f(x)\subset V\}. The first definition says: if \bar x\in \widetilde V, then \bar x is an interior point of \widetilde V. The second definition says: if \bar x\in \widetilde V and x_n\to x, then x_n\in \widetilde V for all large n. Now all multi-valuedness disappears from ... 0 Given  \varepsilon> 0  is  k \in \ N  such that \varepsilon >1/k consider a 'homogeneous' partition of  [0,1] \times [0,1] , i.e., all sub- rectangles of partition has area 1/k^2, we know that  m_B = 0 and  m_B = 1 if B intection the line  y = x  is not empty set, so  S (f, P)-s (f, P) = \sum_ {B \ in P} (m_B-m_B) vol (B) = \sum_ ... 1 Let y=x^2-x. Then$$\lim_{x\to 0}\frac{x(x^2-x)}{x^2}=\lim_{x\to 0}\frac{x^3-x^2}{x^2}=0-1=-1$$which is not equal to the limit along the axes, thus the limit does not exist. 3 You don't really need to do any partial fraction decomposition. The integrand \frac{1}{z^2+1} has two poles at \pm i, both of them are inside the contour |z| = 2. Since the integrand is analytic on the rest of complex plane, one can evaluate it by deforming the contour to a large circle of radius R. We have following estimate:$$\left|\oint_{|z| = ...

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Calculating the residues, $i$ and $-i$ which are inside the contour. Both are simple poles... This should give that $\Sigma Res = 1/(2i) + 1/(-2i) = 0$. We verify this now. Let $z = 2e^{it}$ and $dz = i z dt$, which gives $1/i\cdot \int_{[0,2\pi]} 1/(2z)\cdot \frac{i}{z+i} + \frac{-i}{z-i} = 1/2\cdot \int_{[0,2\pi]} \frac{1}{z^2+iz} - \frac{1}{z^2-iz} = i/2 ... 3 Let's try that decomposition again. In particular, we can find$A,B$such that $$\frac{1}{z^2+1} = \frac A{z-i} + \frac B{z+i} \implies\\ 1 = A(z+i) + B(z-i)$$ There are different ways to solve form this point. I like to solve by plugging in the roots of the denominator for$z. $$1 = A(-i + i) + B(-i -i) \implies 1 = -2Bi\\ 1 = A(i + i) + B(i -i) ... 0 \lim a_n = 0 and a_n decreasing \Longrightarrow b_n=\frac{1}{a_n} increasing and unbounded. For the Cesàro-Stolz Theorem, if \lim_{n\to\infty}\dfrac{(n+1)-n}{\frac{1}{a_{n+1}}-\frac{1}{a_{n}}}=L (*), so \lim \dfrac{n}{\frac{1}{a_{n}}}=L. So, we have to proof (*). Notice that \lim_{n\to\infty}\dfrac{(n+1)-n}{\frac{1}{a_{n+1}}-\frac{1}{a_{n}}}=\lim ... 0 You put "if" in the wrong place: There exists an n ∈ ℕ such that if 1≤x≤1+1/n then 1≤x≤2 should be If there exists an n ∈ ℕ such that 1≤x≤1+1/n then 1≤x≤2. 1 I think it is that straightforward. The reverse implication is trivially true. If there exists an n \in \mathbb{N} such that 1 \leq x \leq 1+1/n, then x must necessarily lie in [1,2], since \sup_{n \in \mathbb{N}}{(1+1/n)}=2. Note that for any n \neq 1, x must still lie in [1,2] even though x\neq 2. 5 Because \gamma(t)=e^{it},\;t\in[0,2\pi]. Hence$$ \frac{1}{2\pi i}\int_0^{2\pi}\frac{\gamma'(t)}{\gamma(t)}\,dt =\frac{1}{2\pi i}\int_0^{2\pi}\frac{ie^{it}}{e^{it}}\,dt=1$$0 I suspect you need f''(x_*) \neq 0. If f(x) = x^3, then { f(x) \over f'(x) } = {x \over 3}, and the update becomes x_{n+1} = x_n - 2{x_n \over 3} = { 1 \over 3} x_n, which is not quadratic. 0 We know that$$\liminf \frac{a_{n+1}}{a_n}\le \liminf (a_n)^{1/n}\le \limsup(a_n)^{1/n} \le \limsup \frac{a_{n+1}}{a_n}$$if (a_n) is a bounded sequence of positive real numbers. Take a_n = 1/n and we have \lim n^{1/n}=1 0 Be carefully with negations! We know that$$ \lim_{x \to a} f(x) = l \Leftrightarrow \forall \epsilon > 0 \,:\, \exists \delta \,:\, \forall x \,:\, |x-a| < \delta \Rightarrow |f(x) - l| < \epsilon \text{.} $$Let's negate that now, using that \lnot \exists x\, \phi = \forall x\, \lnot \phi and that \lnot \forall x\, \theta = \exists x \, \lnot ... 1 Up to a rotation of \mathbb R ^{2n} you ask for which n the space \mathbb R^{2n} - \mathbb R^{n} is path connected, where \mathbb R^{n} sits in the first n coordinates. But contracting the first n coordinates is a homotopy equivalence ending in \mathbb R^{n} - 0. 0 The negation reads as follows: There exists \varepsilon_0>0 such that, for every \delta>0, there are points x in the domain with the property that 0<|x-x_0|<\delta and yet |f(x)-l| \geq \varepsilon_0. Fix \varepsilon_0 and choose \delta=\frac{1}{n} with n \in \mathbb{N}. This yields a sequence \{x_n\}_n as you need. 0 Nitpicks: The negation must be cleaned up in the following manner. "There exists \epsilon \gt 0 such that for any \delta \gt 0 there is a real number x such that 0\lt |x - a| \lt \delta but |f(x) - l| \ge \epsilon ". And when you figure this out you should be able to define a sequence which tends to a and fulfills your condition. For an ... 0 Related with the comment of Andrew D., you can be interested in read about the Runge phenomenon, in the section of constrained minimization. Maybe what you want is to limit the (absolute value of the) derivatives to ve constrained in certain interval. 1 One measure of "spikiness" (or maybe "wobblyness" would be more fitting) is the total variation. The formal definition is that$$ TV_a^b(f) = \sup_{a = x_1<x_2<\ldots x_n=b} \sum_{k=1}^n \left|f(x_{n+1}) - f(x_n)\right| \text{,} $$meaning that you look at arbitrary fine partitions [a,b] = [x_0,x_1] \cup [x_1,x_2] \cup \ldots \cup [x_{n-1},x_n] and ... 1 Your example is wrong. You can't extend \dfrac1x to a continuous function on \mathbb R. (and e.g. -\dfrac1x would be a counterexample on \mathbb R\setminus\{0\}) The idea of a proof is this: Since f(x) is continuous, then f(x)-x is continuous too. If f(x)-x\ne0\Longleftrightarrow f(x)\ne x everywhere, then f(x) is entirely below or above the ... 1 This is the definition of the derivatve. Let f(x)=-e^{-x}. Then$$ \lim_{x\rightarrow 0}\frac{-e^{-1}+1}{x}=\lim_{x \rightarrow 0} \frac{f(0+x)-f(0)}{(0+x)-0}=f'(0)= e^{-x}|_{x=0}=1. $$2 This one is perhaps easier than the arbitrary function f(x) in the example in the comments. First note that:$$ \begin{aligned} &\int_0^1 \int_0^1 \ldots \int_0^1 \sin \bigg(\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 \ldots \,dx_n\\&=\Im{\left[\int_0^1 \int_0^1 \ldots \int_0^1 \text{exp} \bigg(i\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 ... 1 Sincee^y=\sum_{n=0}^\infty{\frac{y^n}{n!}}$, simplify$(1-e^{-x})/x$to see that the limit is 1. 0 The angle$\theta$that corresponds to$x$is $$\theta=-2\tan^{-1}\left(\frac{1-x}{1+x}\right)^2,$$ so $$\frac{1-x}{1+x}=\sqrt{-\tan (\theta/2)}.$$ Solving for$x$we get $$x=f^{-1}(\theta)=\frac{1-\sqrt{-\tan (\theta/2)}}{1+\sqrt{-\tan (\theta/2)}}, -\pi<\theta<0.$$ 1 Yes it means exactly that. Note that the graph of$f$on$[a,b]$is a closed bounded set, because of the continuity. So is the graph of the zero function$g=0$on$[a,b]$. But if$2$compacts have their distance$0$, they necessarily intersect. (this happens if there's no such$\varepsilon>0$) 1 Yes, it means that either there is an$x \in \left[a,b\right]$such that$f\left(x\right)=0$or you have the other condition. Hint: use that you have both maximum and minimum for a continuous function on a closed interval. Then you have two cases: either maximum and minimum have the same sign (and then you will find your$\varepsilon$), or they have ... 1 For the first entry, by the quotient rule:$(\frac{u}{\sqrt{u^2+v^2+1/4}})'=\frac{(\sqrt{u^2+v^2+1/4})u'-u(\sqrt{u^2+v^2+1/4})'}{u^2+v^2+1/4}$Evaluating this at zero we get:$\frac{(1/2)u'(0)}{(1/4)}=2u'(0).$Now you can do the other entry. 3 Consider the series $$f(x) = \sum_{n=0}^{\infty} (-1)^n x^{4n} = \dfrac1{1+x^4}$$ then $$\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{4n+1}}{4n+1} = \int_0^x f(t)dt = \int_0^x \dfrac{dt}{1+t^4} = \int_0^x \dfrac{dt}{(t^2+1)^2-(\sqrt2 t)^2}$$ and some brute force partial fraction to your rescue. Proceed along similar lines for the second problem. 6 Well, to attack the first sum, if we call it$f(x), then $$f'(x) = \sum_{n=0}^{\infty} (-1)^n x^{4 n} = \frac1{1+x^4}$$ so that $$f(x) = \int_0^x \frac{dt}{1+t^4}$$ You can use partial fractions to deduce that \begin{align}f(x) &= \frac1{2 \sqrt{2}}\int_0^x dt \, \left [\frac{t+\sqrt{2}}{t^2+\sqrt{2} t+1}-\frac{t-\sqrt{2}}{t^2-\sqrt{2} t+1} ... 0 You can look up a result of Isserlis, which is proved by him in "On a formula for the product-moment coefficient of any order of a normal frequency distribution in any number of variables". Biometrika 12: 134–139. Essentially, if X_1,\ldots,X_n follow a zero mean multivariate gaussian distribution, E[X_1\ldots X_n] = \sum_{\text{Partitions of ... 1 Yes your proof looks rigorous and correct. You can also use the fact that a function with finitely many discontinuities in an interval satisfies that the integral of the function is equal to the sum of integrals of the function over the subintervals in-between the discontinuities, which is basically what you just proved. In your case you would get \int_0^2 ... 3 This is a bit of a trick, but:a_kb_k = \tfrac12\big(a_k^2 + b_k^2 - (a_k-b_k)^2\big) \ge -\tfrac12(a_k-b_k)^2 \to 0$$(Maybe it's worth comparing to the proof that multiplication is well-defined on reals — I suspect it must use some trick like this.) 0 Yes, there's such a function. Let c\neq 0. Define \overline{\gamma}(x)=\gamma(f(x/c)) where f(x)=\varphi(\mu(\varphi^{-1}(x))). It works because cf(x) \leq f(x) then \gamma(cf(x)) \leq \gamma(f(x)) which is in fact,$$\gamma(c\varphi(\mu(\varphi^{-1}(x))))\leq \gamma(\varphi(\mu(\varphi^{-1}(x)))=\overline{\gamma}(cx)$$and after changing the ... 1 To formalize your idea for the first part, note that if we define$$F(x) = \int_0^x f(t) dt$$then F is continuously differentiable (since f = F' is continuous). Now f is the sum of continuously differentiable functions. To find solutions, differentiate, and notice that$$f' = 5 + 2f$$which is a first-order ODE that can be solved with any number ... 2 HINT: If a_k\sim b_k then without loss of generality you can assume that a_k=b_k. Therefore, a_kb_k is now (a_k)^2. Use the fact that x^2\geq 0 for rational numbers (or prove that fact first), and conclude that a limit of squares cannot be a negative number. 2 For the computation in part (4), just treat h(t) as some black boxed function. So, using the chain rule and product rule, since \gamma(t) \not= a for any t,$$\frac{d}{dt} e^{-h(t)}(\gamma(t) - a) = -e^{-h(t)}h'(t)(\gamma(t) - a) + e^{-h(t)}(\gamma'(t)) = e^{-h(t)}\left(\gamma'(t) - \frac{\gamma'(t)}{\gamma(t) - a}(\gamma(t) - a)\right) = 0.\$ To ...

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