# Tag Info

5

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 $y$)

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 ...

5

Consider $f(x)=|x|$. At $x=0$ the limit is just $0$, although the function isn't differentiable there. Away from $x=0$ is not hard, just write it out.

5

The answer is false. Consider $f(x)= \begin{cases} \hfill 0 \hfill & x\neq 0 \\ \hfill 1 \hfill & x=0 \\ \end{cases}$. Clearly this function is not differentiable at $x=0$ because it is not continuous at that point but for any $h\neq0$, $f(0+h)=f(0-h)=0$ so the limit does exist. The problem is basically that the limit ...

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.

3

Hint: Does this limit existing guarantee that your function is continuous at $x$?

3

Are you sure you are trying to "prove" this? An alternate approach is \begin{align*} \mu_X +\mu_Y &= E[X]+E[Y]\\ &= E[X+Y]\\ &=\iint(x+y)f_{X,Y}(x,y)dxdy\\ &=\mu_{X+Y}. \end{align*} Obviously, this is not a valid answer if they really wanted you to do all the integration. Maybe they just wanted you to see that connection.

2

Wait!!! If you have two independent random variables, $X$ and $Y$, with distribution $f_X(x)$ and $f_Y(y)$, then the joint distribution is: $$f_{X,Y}(x,y) = f_X(x)f_Y(y)$$ Then: $$\mu_{X+Y} = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}(x+y)f_{X,Y}(x,y)dxdy = \\ = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}(x+y)f_{X}(x)f_{Y}(y)dxdy = \\ = ... 2 Note that C_n is bounded, say |C_n| \le C for all n. You want to show, given \epsilon>0, that there exists N such that$$\forall n,m \ge N, \ |D_n - D_m| < \epsilon.$$So manipulate the expression. For n>m (without loss of generality)$$|D_n - D_m| = | \sum_{k=m+1}^n C_n Q^{-k} |.$$There is a little more work that I haven't included ... 2 There are only finitely many functions from J_n to itself. Therefore, whenever the length of any list of such functions exceeds the total number of such functions, then two members of that list must be equal. For example, there are 256 functions from \{1,2,3,4\} to itself, so two of the functions f, f^2, f^3,\ldots,f^{257} must be equal to each ... 2 Hint: You have that g and g \circ f are both bijective. Use the following facts about bijective functions: Every bijective function h: C \to D has an inverse h^{-1}: D \to C so that h \circ h^{-1} = id_D and h^{-1} \circ h = id_C. If you compose a bijective function with an injective function (in either order), you get an injective function. If ... 2 Here is a direct approach to finding all selfadjoint extensions of \Delta (sorry if it doesn't answer your questions). Let V=\mathcal S(\mathbb R^+) be the space of the smooth functions f on [0,\infty) that converge to 0 as x\to\infty, together with all its derivatives, faster than any power. For f,g\in V we have ... 2 The deficiency indices are the same because of how \Delta commutes with complex conjugation and its domain is closed under complex conjugate. g \in \mathcal{N}(A^{\star}-iI) iff$$ ((A+iI)f,g) = 0,\;\;\; f \in \mathcal{D}(A). $$Conjugating the above gives$$ ((A-iI)\overline{f},\overline{g})=0,\;\;\; f \in \mathcal{D}(A). $$... 2 A bounded function is a function whose range is contained in a finite interval. A sequence of bounded functions has the property that each of its individual functions f_n has such a limited range (but the overall range of the collection may still be infinitely large). The sequence is pointwise bounded if for every x separately the sequence f_n(x) is ... 2 The function f: R^2 \to R where f(x, y) = |x|+xy is an example See that for the following function$$f_{yx} = 1$$for every (x, y) \ \epsilon \ R^2 But f_x doesn't exist at the following set of points \{ (0, y) \ \epsilon \ R^2 \ | \ y \ \epsilon \ R \}  Thus for the set \{ (0, y) \ \epsilon \ R^2 \ | \ y \ \epsilon \ R \} \subset R^2, ... 2 As Daniel fischer noted this argument only works for the special case of non negative terms b_i . Hint: if the set \{n: b_n \geq 1\}  is infinite then you should know what to do. However if that is finite,then try to make use of the following fact: for every n  such that b_n<1 we have$$\frac {b_n}{1+b_n} > \frac {b_n}{2} $$1 First of all, using 2xy \le x^2 + y^2,$$|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 + ...

1

This sometimes called the Popcorn Function, since if you look at a picture of the graph, it looks like kernels of popcorn popping. (Also called the Dirichlet function.) To prove it is discontinuous at any rational point, you could argue in the following way: Since the irrationals that are in $(0,1)$ are dense in $(0,1)$, given any rational number $p/q$ in ...

1

Let $x$ be an integral curve through $p$: $$x'(t) = V\bigl(x(t)\bigr),\quad x(0) = p.$$ Since $V$ is differentiable at $p$ and $V(p) = 0$, there exists an $M > 0$ such that $\|V(x)\| \leq M \|x - p\|$ for all $x$ in some neighborhood of $p$. By the flow equation and the triangle inequality, \begin{align*} \|x(t) - p\| ...

1

Let's assume for contradiction that $\lim_{n \to \infty}(-x)^n$ converges to $l$ for $x>1$, then $\lim_{n \to \infty}|(-x)^n| = |l|$ (I hope you can use this). Finally, $|(-x)^n| = |x^n|$ $\implies$ $\lim_{n \to \infty}x^n = |l|$, which is a contradiction, because $\lim_{n \to \infty}x^n$ does not converge when $x>1$ (as you has proved before). I hope ...

1

The notion of a metric is supposed to capture the main features of the everyday idea of distance between two points. The most important parts of this idea are that a point is at zero distance from itself; distinct points are a positive distance apart; the distance from a point $x$ to a point $y$ is the same as the distance from $y$ to $x$; and the ...

1

Instead of the limiting process you can choose $M$ as the least upper bound: $$A := \{ | f(x) | : x \in \mathbb R \} \, , \quad M := \sup A \, .$$ The supremum exists because every non-empty bounded set of real numbers has a supremum (https://en.wikipedia.org/wiki/Least-upper-bound_property). Then for all $x \in \mathbb R$ $$|f(x)| = \left| ... 1 First off, if E is dense in V, then E=V since E is closed. (Not clear from your post if you noticed that, just pointing out.) Now (in general) for any x do: (a) if x>\sup E then let f(x)=f(\sup E) (just as you already did, (b) if x<\inf E then let f(x)=f(\inf E) (note also that instead of \inf or \sup we could write \min or ... 1 One reason why I quite like the problem is that it starts with a nice magic trick (a misdirection if you like). Since it is posed for continuous functions one immediately starts by thinking about that, when in fact you should be simply investigating the nature of the set of local extrema. As others have pointed out it is just a countable vs. uncountable ... 1 Hint: For all we know, f is continuous everywhere but at c and if \alpha(x) (where \alpha(x) is the monotonically increasing function that we integrate f with respect to) is continuous at every discontinuity of f (for f with finitely many discontinuities) then f\in\mathcal{R}(\alpha). To avoid this, we must pick an \alpha that is ... 1 It is not sufficient to know that the sine stays between the boundaries, you need to use the fact that the boundaries are actually hit an infinite number of times as x comes arbitrarily close to 1. 1 Function, how it is defined at the moment, is discontinuous, because it is not defined at x=1. The right question: is it possible to define the fuction for x=1, that it becomes continuous. Limit at point zero does not exist, neither from right nor from left. 1 The sets A_j are disjoint, and so \nu_j\perp\nu_k for j\ne k. Then by countable subadditivity$$\sum_j\nu_j(E)=\sum_j\nu(E\cap A_j).$$Again, the sets E\cap A_j are disjoint, so$$=\nu\left(E\cap(\cup_jA_j)\right)=\nu(E\cap X)=\nu(E).$$Similar results apply to \lambda_j and \rho_j=f_jd\mu_j. You can use this to show that \lambda, \rho are ... 1 since gof is Surjective for every y \in A \exists x \in A such that$$gof(x)=y$$\implies$$g(f(x))=y \tag{1}$$But since we know that g is surjective, for every y \in A \exists z \in B such that$$g(z)=y \tag{2}$$Using (2) in (1) we get$$g(f(x))=g(z)$$But since g is Injective$$f(x)=z which proves $f$ is Surjective.

1

Use the following facts to do your heavy lifting: Proposition A. Given functions $g : Z \leftarrow Y$ and $f : Y \leftarrow X$... If $g \circ f$ is surjective, then so too is $g$. If $g \circ f$ is injective, then so too is $f$. The proof is left as an exercise for the reader. Proposition B. Given function a function \$f : Y \leftarrow ...

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