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0

Assume that $J \colon H \to \mathbb{R}$ is G-differentiable. By definition, given $u \in H$, there is a linear functional $J'(u) \in H^*$, the G-derivative at $u$. By the Riesz representation theorem, this element is isometrically identified to a vector $\nabla J(u) \in H$, called the gradient of $J$ at $u$. Now, consider thae map $$\phi \colon \tau ... 0 For a bounded domain \Omega with \,\partial\Omega\in C^1, it is obviously true if \,u\in C(\overline{\Omega}) and x_0\, is a removable singular point. Of course, it cannot be true whenever a singularity x_0\, be non-removable, e.g., take \,\Omega=\{x\in\mathbb{R}^2\,\colon\,\,|x|<1\}\, with \,x_0=0, and consider an example of \,u(x)= x_1 ... 0 The theorem you cite is correct for when the exponent of the variable is just n. Here obviously we have 2n+1. 1 This power series contains only terms of odd powers, and hence its derivative contains terms of even powers only, and since the power series starts from the first power of x, then its derivative start with the zero power of x. 1 I do not think that \mathscr D(\Omega) is sequential. On the other hand this is probably not used: By definitionn of the locally convex inductive limit topology of \mathscr D(\Omega)= \lim X_n (where X_n are the Frechet spaces of smooth functions with support in K_n for a compact exhaustion) a linear map with values in any locally convex space is ... 0 if n<m then the rank cannot be m (linear algebra), if n\geqslant m then your local representation$$ (x_1,\dots,x_n) \mapsto (x_1,\dots,x_m) $$proves the assertion 1 Consider the following statement, which does not involve any PDE: this is multivariable calculus only. If f:\mathbb R^n\to \mathbb R is C^1 smooth and f(0,0,0)=0, then for every R>0 there exists C>0 such that |f(z)|\le C|z| whenever |z|\le R. Proof: by the mean-value inequality we have |f(z)-f(0)|\le |z|\sup_{|\xi|<R} ... 0 Take all the continuous functions that satisfy f(x +2\pi) = f(x) for all real x. Make a distance on these by$$ d(f,g) = \sqrt{ \frac{1}{\pi} \int_0^{2 \pi} (f(x) - g(x))^2 dx } $$With this distance function, we can make an infinite set of functions that are distance 1 apart with$$ f_n(x) = \frac{1}{\sqrt 2} \sin (nx). $$The word "point" ... 0 Here the proof should be similar to how you show \mathbb{R} is complete. Your proof got stuck because you did not use the condition$$ |x|^{2}=\sum^{\infty}_{i=1}\langle x_{i},x_{i}\rangle<\infty $$1 This space can be defined using any set actually. It is generated by the metric called the discrete metric. Define d: M \times M \rightarrow M by$$ d(x, y) = 0 \iff x = y \;\;\text{and }d(x, y) = 1 \iff x \neq yThis is what is called the discrete metric and is defined for any set M. You can show for yourself that d actually constitutes a ... 2 Physical interpretation can be very helpful, but ultimately we need E to be such that we can infer something about dE/dt from the PDE infer what we want about the solution from E The consideration usually begins with 1. Note that u_tu_{tt} = \frac{1}{2}(u_t)^2. So, if we have a hyperbolic PDE u_{tt}=F(x,u,Du,D^2u), it is reasonable to ... 1 We know \mathbb{Q} is a countable set, so let q_1,q_2,q_3,... be a sequence that enumerates all elements of \mathbb{Q}. We know for all n \in \mathbb{N} |f_n(q_1)| \leq F(q_1) So bolzano-weierstrass theorem says there is a convergent sub-sequence, say f_{k1_n}(q1) with limit Q_{1} In the same way, for all n \in \mathbb{N} |f_{k_n}(q_2)| ... 0 Hint: The uniform convergence of a sequence of continuous functions converges to a continuous function. 0 dx and dy aren't real numbers; they are things called differential forms. Thus, you can't use the real number multiplication operation to multiply them. However, dx \, dy is a thing, and it is not terribly unreasonable to define "the multiplication of dx and dy to be dx \, dy. The trick is that you have to make the inferences in the opposite ... 2 The only continuous field isomorphisms \mathbb C\to\mathbb C are the identity and complex conjugation. Using the Axiom of Choice one can prove that additionally there exists infinitely many "wild" everywhere-discontinuous automorphisms of \mathbb C. Each of them acts either like the identity or like conjugation on all points with rational coordinates, ... 4 |e^z| = |e^{x+iy}| = |e^xe^{iy}| = e^x|e^{iy}| = e^x. 1 \begin{align*} e^z &= e^x (\cos y + i \sin y) \\ \left| e^z \right| &= \sqrt{e^{2x} \cos^2 y + e^{2x}\sin^2 y} \\ &= e^x \sqrt{\cos^2 y + \sin^2 y} \\ &= e^x. \end{align*} Also, you should be careful--\arg e^z is only equal to y when y \in (-\pi, \pi] (or whatever branch you're using for the \arg function.) 1 Try \tilde{w}_\alpha = t |z_\alpha|, \tilde{z}_\alpha = z_{\alpha} where t = \sum_{\alpha} |w_\alpha|/\sum_{\alpha} |z_\alpha|. 2 Yes. {}{}{}{}{}{}{}{}{}{}{}{} 4 As others have said dx\, dy does not represent a product of differentials. But it represents a product of measures. We have the "natural" Lebesgue measure \lambda on the x-axis, and integration with respect to this measure is signalled by writing {\rm d}x as right parenthesis of the integral. Similarly we have the "natural" Lebesgue measure ... 1 As @Claude has stated for the simpler cases here is it for 4 elements in a and in b. The correct sum is the sum over all elements of the ("outer"(?)) product C of the two vectors A^T \cdot B=C= \small \begin{array} {r|rrrr} & b_0 & b_1 & b_2 & b_3 \\ \hline a_0 & a_0b_0 & a_0b_1 & a_0b_2 & a_0b_3 \\ a_1 & ...

0

To show $\| \cdot \|$ is a norm on $\mathscr{C}\big( [0,2] ; \mathbb{R} \big)$ : Clearly, for any $f\in \mathscr{C}\big( [0,2] ; \mathbb{R} \big)$, $\| f \|$ is nonnegative. Since for $x\in[0,2]$, $x(2-x)\vert f(x) \vert \geq 0$, then $\| f\| = 0$ implies $$0 = \int_0^2 x(2-x) \big\vert f(x) \big\vert\, dx \Longrightarrow f \equiv 0$$ that is, ...

2

Just try with two terms. So, you have $$(\sum_{i=0}^1 a_i)(\sum_{i=0}^1 b_i)=a_0 b_0+a_1 b_0+a_0 b_1+a_1 b_1$$ $$\sum_{i=0}^1 \sum_{j=0}^1 a_jb_{i-j}=a_0 b_0+a_1 b_0+a_0 b_1$$

4

The OP listed non-standard analysis as one of the tags. It is therefore not inappropriate to point out that, just as one can think of the derivative as a true ratio $\frac{dy}{dx}$ modulo an infinitesimal error (eliminated by applying the shadow), so also one can think of a single-variable integral as an infinite sum of infinitesimal terms of type $dx$ ...

3

This is certainly not true: $$\left(\sum_{i=0}^n a_i\right)\left(\sum_{i=0}^nb_i\right) = \sum_{i=0}^n \sum_{j=0}^i a_jb_{i-j}$$ So it's not clear why you'd expect the limit as $n\to\infty$ is the same. If definitely requires a much more careful argument. The actual equality is: $$\left(\sum_{i=0}^n a_i\right)\left(\sum_{i=0}^nb_i\right) = \sum_{i=0}^n ... 0 Relevant facts: For infinitely many n, we have$$ \varphi(n) < \frac{n}{e^\gamma \log \log n} \tag{1} $$See here. But the constant e^\gamma doesn't really matter. In fact, the simpler condition given in spin's answer suffices; i.e. that for any 0 < c < 1, there are infinitely many n such that$$ \varphi(n) < cn \tag{2} $$Claim: The ... 0 Here is a more precise bound in case you find it useful, even though the question is already answered. It was shown by Nicolas, that for infinitely many positive integers n we have \displaystyle \frac{n}{\varphi(n)}>e^{\gamma}\log \log n where \gamma is the Euler constant. You can find details in J.-L. Nicolas. Petites valeurs de la fonction ... 2 Let 0 < c < 1. Then there are infinitely many n such that \varphi(n) < cn, and thus the line y = c(x-1) + 1 (x \geq 1) is included in the convex hull. The convex hull is hence the triangle-shaped area determined by the lines y = x, (x \geq 1) and y = 1. EDIT: Proof for the claim above. If P_k = p_1 p_2 \ldots p_k is the product ... 0 For simplicity we can take the case of R^2. We want to go from (a,b) to (c,d) parallel to coordinate axes, but not get out of the given set. We move from (a,b) to (a,c) and test if the path (a,b)-(a,c) is in the set. If yes, we than move from (a,c) to (b,c) and test if (a,c)-(b-c) is in the set. If yes, we are done. But if (a,b)-(a,c) is not in the set, ... 0 First note that for any cube C=[-r,r]^n\subseteq\mathbb R^n any point c=(c_0, c_1,\ldots c_{n-1})\in C is polygonally connected to the center of C along the axes.$$(0,0,0,\ldots,0)\to(c_0,0,0, \ldots, 0)\to(c_0,c_1,0,\ldots0)\to\ldots\to(c_0,c_1,\ldots c_{n-1})$$Let G be any nonempty open connected set in \mathbb R^n and let a\in G. Now set ... 1$$ f(a+h)-f(a)=f'(a)h+\frac{1}{2}f''(a)h^2+o(h^2) $$and$$ f'(a+h\theta) = f'(a)+f''(a)h\theta + o(h\theta). $$In the last equation I just used the definition of f''(a). Hence$$ f'(a)h+\frac{1}{2}f''(a)h^2 + o(h^2)=f'(a)h+f''(a)\theta h^2 + o(h^2\theta). $$Dividing by h^2 we deduce$$ \frac{1}{2}f''(a) = f''(a) \theta +o(1). $$If f''(a) \neq 0, ... 0 Multiply$$ \cos a+\cos 2a+\cos 3a+...+\cos(Na) $$with 2\sin b to obtain$$ \sum_{k=1}^N 2\sin b\cos ka=\sum_{k=1}^N (\sin(ka+b)-\sin(ka-b)) $$Using either b=a or better b=\frac a2, you get a cancellation of inner terms, in other words, a telescoping sum. To the original question, how to make the denominator real,$$ \frac{e^{ia}-1}{e^{ib}-1} ...

2

$$1+S=\sum_{k=0}^M\cos\frac{\pi k}{M+1}={\rm Re}\sum_{k=0}^Me^\frac{\pi k}{M+1}={\rm Re}\frac{e^{i\pi}-1}{e^{i\pi/(M+1)}-1}=0.$$ Thus, $S=-1.$

3

We should have $f(0)=0$ hence $f(-z)=-f(z)$ for any $z\in\mathbb C$. In particular, $f(-1)=f(i)^2=-f(1)=-1$ (because $f(1)^2=f(1)$ and $f(1)\neq 0$). We thus have $f(i)=i$ or $f(i)=-i$. In the first case, we get the identity map, and in the second one the complex conjugation.

0

1) Steps are correct. 2) Hint: The values are roots of unity, and satisfy the equation $x^{M+1} -1 = 0$. By Vieta's, the sum of all the roots will be 0 (for $M \geq 1$). There are $M+1$ roots of unity, but we only have $M$ values, so which root is missing? Hence, what is the sum of our $M$ values?

1

The earlier accepted answer here (by Priyatham) had a subtle flaw (namely repeated use of L'Hospital's Rule without checking conditions of its applicability) but it has been fixed now after my comments. I am sure this question is solved elsewhere on MSE, but I am not able to find link. Here is the right approach to this question. 1) We will use ...

0

The fact that $\| \pi(x) \| \le \|x\|$ for all $x \in X$ tells you that $\|\pi\| \le 1$ as $\|\pi\| := \sup\{\|\pi(x)\| : \|x\| \le 1 \}$. If you pick a point $z \in M$ then $\|\pi(z)\| = \|z\|$ and so $\|\pi\| = 1$.

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You have $\pi^2 = \pi$, so $||\pi|| = ||\pi^2|| \le ||\pi||^2$, implying that $||\pi|| \ge 1$. Your statement implies that $||\pi|| \le 1$.

1

Any number not in your set $S$ is a real number $r$ that you're looking for. Can you establish that there are infinitely many real numbers not in $S$, once you know that $S$ is countable? In fact, every number in your set $S$ is algebraic, and the algebraic numbers are a countable subset of $\mathbb{R}$. Any transcendental number would work as $r$, so if ...

2

If $f$ is differentiable, we can write $f(x+{1 \over n}) = f(x) + {1 \over n} f'(x) + r({1 \over n})$, where $\lim_n {r({1 \over n}) \over {1 \over n}} = 0$. In particular, for any $\epsilon>0$ we can find some $N$ such that if $n \ge N$ then $-\epsilon < {r({1 \over n}) \over {1 \over n}} < \epsilon$, and so ${1 \over n} (f'(x) - \epsilon) < ... 1 $$\log\left[\left(\dfrac{f\left(x+\dfrac{1}{n}\right)}{f(x)}\right)^n \right]= n\log \dfrac{f(x) + \frac 1n f'(x)+\epsilon(\frac 1n)\frac 1n}{f(x) } = n\log \left[1+ \frac 1n\dfrac{ f'(x)}{f(x)}+\epsilon\left(\frac 1n\right)\frac 1n\right]$$ with$\lim_0\epsilon = 0$. $$\sim n \frac 1n\dfrac{ f'(x)}{f(x)} = \dfrac{ f'(x)}{f(x)}$$so the limit is ... 3 Hint: take logarithms, and compare what you get to the difference-quotient definition of the derivative. 1 I think you're concluding what you want to prove without actually proving it. You might resort to epsilon-delta proofs. You will have a "left" delta and a "right" delta, so you will just need to let the "two-sided" delta be their minimum. You're right, sometimes the obvious things are surprisingly elusive as proofs go. 0 First assume that for any sequence$x_n\in \Omega\setminus \{x_0\}$with$x_n\to x_0$, we have that $$\liminf u(x_n)>0$$ hence, by the maximum principle$u(x)>0$in$\Omega\setminus \{x_0\}$. On the other hand, if for any sequence$x_n\in \Omega\setminus \{x_0\}$with$x_n\to x_0$, we have that $$\limsup u(x_n)<0$$ then, by the maximum principle ... 0$x \in S$is not the same as$\forall x \in S$.$\in$means "is an element of," so$x \in S$simply denotes the membership of$x$to$S$, while$\forall x \in S$, meaning "for all x that is an element of$S$" must modify an action like "Let$f(x)=x^2, \forall x \in S$." The two statements are not logically equivalent in many cases. For instance, you cannot ... 1 I am wondering if there is a reference or book that clearly translates all English forms of logical quantifiers to mathematical quantifiers. Any good introductory logic text should help you out here. I confess a sneaking admiration for P*t*r Sm*th's Introduction to Formal Logic which, I'm told, students find particularly helpful on such matters of ... 2 The$\rho_{\alpha,\beta}$are norms on$\mathcal{S}(\mathbb{R}^n)$. That they are seminorms is straightforward to verify, and$\rho_{\alpha,\beta}(f) = 0$only for$f \equiv 0$follows since$x^\beta$is only zero on a nowhere-dense subset (the union of finitely many coordinate hyperplanes$\{x_i = 0\}$), so$\rho_{\alpha,\beta}(f) = 0 \Rightarrow ...

0

Let $R' = 1/R$ and $α' = 1/α$ to not confuse things too much. Rephrase your question: Let $(A_n)_n$ be a real sequence and $R' = \limsup A_n$. Let $α' > R'$. Why is there $n_0 ∈ ℕ$ such that $A_n < α'$ for any $n ≥ n_0$? Then the statement becomes a bit more obvious if you define/interpret $\limsup A_n$ to be the largest limit point of a sequence ...

0

One good definition of $\limsup_{n \to \infty} a_n = \sup E$ where the set $E$ is all limits of subsequences of the sequence $(a_n)$. Now suppose $\frac{1}{R} = \limsup A_n$ and let $\alpha < R$. Then $\frac{1}{\alpha} > \frac{1}{R} = \limsup A_n$. Now $\epsilon > 0$, then I claim that there exists an $N \in \mathbb{N}$ such that $A_n < ... 3 Question 1 $$\lim_{h\rightarrow0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2} \\$$ when$h = 0$is substituted numerator and denominator reduce to$0$. So, applying L'Hopital's rule (differentiate wrt$h$)$\$ \lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h} \\ \lim_{h\rightarrow0}\frac{f'(a+h)-f'(a)-f'(a-h)+f'(a)}{2h} \\ ...

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