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5

Yes, we can always partition an infinite set to countable subsets. But the partition itself is not going to be finite, and certainly not countable. Namely, if $T$ is uncountable, and $T=\bigcup_{i\in I}A_i$ where each $A_i$ is countable, then $I$ is necessarily uncountable. In fact we can prove that $|I|=|T|$. Similarly, you can partition every set into ...


5

All ultrapowers of $\mathbb{C}$ of cardinality the cardinality of the continuum are isomorphic to $\mathbb{C}$ as fields. They are not all isomorphic to $\mathbb{C}$ as fields with additional unary function $\text{Conj}$, the conjugate function. We can consider the full structure on $\mathbb{C}$, by adding function symbols, relation symbols for every ...


5

Your approach for the proof is incorrect. You want to prove an implication $P \to Q$. So if you assume that $Q$ is true, then you have nothing left to prove. You should just begin with an integer $n$ then try to see what can be said about $n^3+n^2$. For example, you can say $$n^3+n^2=n^2(n+1).$$ Now if $n$ is odd then $n+1$ is even and vice versa. So you ...


4

Hint $$n^3+n^2=n(n(n+1))$$ $n(n+1)$ is the product of two consecutive integers.


4

Alternatively, by Fermat's Little Theorem, $$\begin{align}n^2 + n^3 &= n^2 + n\cdot n^2\\ &\equiv n + n\cdot n \pmod 2\\ &= n + n^2 \pmod 2\\ &\equiv n + n \pmod 2\\ &= 2n \pmod 2\\ &\equiv 0 \pmod 2\end{align}$$


4

Indeed, in your setting, $f''\geqslant g''$ implies $f\geqslant g$. To show this, note that, since $f(0)=g(0)=0$, for every nonnegative $x$, $$f(x)=\int_0^xf'(t)\,\mathrm dt,\qquad g(x)=\int_0^xg'(t)\,\mathrm dt.$$ Using the hypothesis that $f'(0)=g'(0)=0$, an integration by parts of these identities yields $$f(x)=\int_0^x(x-t)f''(t)\,\mathrm ...


4

The integrand is less than $1/\sqrt{x}$, so that $0<\int_\varepsilon^1 \frac{e^{-nx}dx}{\sqrt{x}}<\int_0^1 \frac{dx}{\sqrt{x}}=2$, and $\int_0^1 \frac{e^{-nx}dx}{\sqrt{x}}$ exists. As to $\int_1^\infty \frac{e^{-nx}dx}{\sqrt{x}}$, the integrand rapidly decreases to $0$ as $x\rightarrow\infty$, in particular, $0<\int_1^N ...


3

Compute the generating function of the harmonic numbers: $$ \begin{align} \sum_{n=1}^\infty H_nx^n &=\sum_{n=1}^\infty\sum_{k=1}^n\frac{x^n}{k}\\ &=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{x^n}{k}\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{x^{n+k}}{k}\\ &=-\frac{\log(1-x)}{1-x}\tag{1} \end{align} $$ Integrating $(1)$ yields $$ ...


3

One option is to check directly that the definition of a convex function is satisfied. It's useful to know that any norm on $\mathbb R^n$ is a convex function. Proof: If $x,y \in \mathbb R^n$ and $0 \leq \theta \leq 1$, then \begin{align*} \| \theta x + (1 - \theta) y \| & \leq \| \theta x \| + \| (1 - \theta) y \| \\ &= \theta \| x \| + (1 - ...


3

Consider the rectangular path with vertices at $-R,\, R,\, R+it,\, -R+it$. When $R\to \infty$, the integrals over the vertical sides tend to $0$ by the standard estimate, and you are left with $$\int_{-\infty}^\infty \exp \left\{-\frac{1}{2}(x-it)^2\right\}\,dx = \int_{-\infty +it}^{\infty +it} \exp \left\{-\frac{1}{2}(z-it)^2\right\}\,dz,$$ where the ...


3

A man walks into a restaurant, orders food, eats, then leaves. Did the man have to cook his food? Did he pay? Did he sit down? Did he eat the food he ordered or just eat something else entirely. Your problem is known in Psychology as the interpretation and application of one's schema. http://en.wikipedia.org/wiki/Schema_(psychology) Because people are ...


3

There are many possible conventions, but you can't get around the fact that a $2\pi$ will appear somewhere. In order to have a way of comparing different conventions easily, I wrote up restrictions on the various coefficients in the definitions and comparison between different choices.


3

Here's a shorter proof: Select any $n \in \mathbb{Z}$. There are two cases: $n \equiv 0, 1 \pmod{2}$. In either case, use modular arithmetic to compute $n^2 + n^3 \pmod{2}$. What can we conclude?


3

Outline: We want to construct an arithmetic sequence of length $n$ that is a subsequence of the sequence $(\lfloor i\alpha\rfloor)$. For $i=1,2,3, \dots, 10n+1$, let $b_i$ be the fractional part of $i\alpha$, that is, $i\alpha-\lfloor i\alpha\rfloor$. These fractional parts are all distinct, since $\alpha$ is irrational. By the Pigeonhole Principle, there ...


2

1. Metrics that satisfy $d(x+r,y+r)=d(x,y)$: There are many metrics of this kind, as remarked by other users. Also the sum of any such metrics still has the property. 2. Metrics that satisfy $d(xr,yr)=d(x,y)$: There is such a metric on $(0,\infty)$, namely $d(x,y)=|\log(x)-\log(y)|$. There are also such metrics on $\mathbb R$, and something can be said ...


2

The statement is false. Consider $f(x)=\sin \frac{\pi}{2} x$. The LHS goes to $0$, while the RHS does not converge.


2

This is not a rigorous answer. I just need the space to easily type the equations. Let us consider the ODE holds near $x\to\infty$: $$(f'(x))^2+f^3(x)=0 \tag{1}$$ The solution to (1) is: $$f(x)=-\frac{4}{(x+c)^2} \implies f'(x)=\frac{8}{(x+c)^3}$$ So $$\lim_{x\to\infty}f(x)=0,\lim_{x\to\infty}f'(x)=0 \tag{2}$$ Now let us consider the next ODE holds ...


2

For $x,y\geq 0$, $\left(x^{\frac{1}{3}} \right)^2+\left(y^{\frac{1}{3}} \right)^2=1$, then there exists $\theta\in \Bbb R$ such that $x^{\frac{1}{3}}=\cos(\theta)$ and $y^{\frac{1}{3}}=\sin(\theta)$, that is $x=\cos^3(\theta)$ and $y=\sin^3(\theta)$ the original form!


2

Regarding $\int_{0}^{\infty}e^{-nx}/\sqrt{x} dx$, the tail is not a problem because for $x > 1$ we have $$0 < \frac{e^{-nx}}{\sqrt{x}} < e^{-nx}$$ and the right hand side is integrable. The only other cause for concern is the singularity at $x=0$. But this is also not a problem, because for all $x > 0$, we have $e^{-nx} \leq 1$, and so $$0 < ...


2

I assume you mean $C[0,1]$ with sup-norm. Hint: Try to show that for each $n$ and arbitrary $\varepsilon>0$ there exists $g\notin E_n$ such that $\|g\|_\infty<\varepsilon$. Try to show that $f\in E_n$ and $g\notin E_{2n}$ implies $f+g\notin E_n$. Using these two facts you should be able that if $f\in E_n$ then in any ball $B(f,r)$ there is a ...


2

It can't be a vector space because scalar multiplication doesn't behave as required. For instance, given your algebraic operations, we'd have $$\frac{1}{2} \star (2 \star \pi) = \frac{1}{2} \star 0 = 0$$ but $$\left(\frac{1}{2} \cdot 2\right) \star \pi = 1 \star \pi = \pi \ne 0$$ so scalar multiplication doesn't satisfy the necessary conditions. To ...


2

Yes, this is true. The supremum part of $C^1$ was already understood, so let's look at the derivative. With convolutions we can choose where the derivative falls: $$\nabla (f*\rho_\epsilon) = f*(\nabla \rho_\epsilon)$$ Let $\eta=\nabla \rho$ and $\eta_\epsilon = \epsilon \nabla \rho_\epsilon$. By the chain rule, $\eta_\epsilon(x) = ...


1

Outside a sufficiently large ball centered at $0$, $\log f$ is holomorphic Not really, unless your definition of "holomorphic" allows multi-valued functions, which would be unusual. For example $\log z$ does not have a holomorphic branch in $|z|>1$. I think what you should do is: Observe that $f$ has finitely many zeros. Divide $f$ by a ...


1

Let $g(x)=\lvert x\rvert$ in $[-1,1]$, and extend $g$ to be periodic in $\mathbb R$, with period $2$. Set $$ g_{k,\ell}(x)=\frac{1}{k} g(\ell x). $$ It is not hard to see that $$ g_{k,\ell}\in E_n \quad\text{iff}\quad n\ge \frac{\ell}{k}. $$ Fix now $n\in\mathbb N$. We shall show that $E_n$, which is a closed subset of $C[0,1]$, has empty interior. Let ...


1

You can certainly define a 'scalar multiplication' as you have. But that doesn't make it a vector space. Why? Because 1) if it were a vector space, it would obviously be 1-dimensional because any element in $(0, 2\pi)$ is a generator under your scalar multiplication but 2) the additive group of the angles is not isomorphic to $\mathbb{R}$, because it ...


1

Without giving too much away, there is a way to simply evaluate the integral. It is related to the Gamma function $\Gamma(\frac{1}{2})$ but there is a subtlety involing the variable $x$ in place of $nx$. By performing a variable substitution, you should get a Gaussian integral which can be directly evaluated. For starters, let $u=\sqrt x$ and go from there. ...


1

Adding a few things to the good answers you already received, using the change of variable suggested by TeeJay, you obtain $$\int{\frac{e^{-nx}}{\sqrt{x}}}dx=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{n} \sqrt{x}\right)}{\sqrt{n}}$$ from which $$\int_{0}^{\infty}{\frac{e^{-nx}}{\sqrt{x}}}dx=\frac{\sqrt{\pi }}{\sqrt{n}}$$ provided $n \gt 0$.


1

It suffices to write $f$ as the pointwise supremum of some family of affine functions, here $f=\sup\{g,h\}$ with $g:x\mapsto x$ and $h:x\mapsto-x$, since every such supremum defines a convex function and every convex function can be written as such a supremum.


1

At least the second equality is not true for $n=1$ : $$\left(e(\sqrt e-1)=\right)e^{1/1}\left(e^{1/(1+1)}-1\right)\not =e^{1/1}-\log(1)*\frac{e^{1/(1+1)}-1}{1/(1+1)}*\frac{1}{1+1}\left(=e\right).$$


1

You could use Simpson's rule to discretize the integral, which is indepedent from $x$ (I mean none of the limits are $x$) and for every $x$ sweeps the domain $(a,b)$. For your problem, this reads: $${ \color{blue}{\phi_i = f_i + \sum^{n-1}_{j=0} \frac{t_{j+1}-t_j}{6} \left[ K(x_i,t_j) \phi_j + 4 K \left( x_i,\frac{t_{j+1}+t_j}{2} \right) ...



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