New answers tagged

0

I have proved it as follows. Taking the expression for $f'(x)$, multiplying through by $(1+x^{2} \sin^{4} \frac{1}{x})$, we need to prove the following inequality, call it (a): $$\left(1+x^{2} \sin^{4} \frac{1}{x}\right) \arctan(x \sin^{2} \frac{1}{x}) + x\sin^{2}\frac{1}{x} > 2\sin\frac{1}{x} \cos\frac{1}{x} \qquad \mbox{(a)}$$ Observe that since ...


1

Since $$ \left\|\frac{f(x+hy)-f(x)}{h}\right\|\ge c\|y\|, $$ for $y\ne0$ you get $$ \|d_xfy\|=\left\|\lim_{h\to0}\frac{f(x+hy)-f(x)}{h}\right\|\ge c\|y\|\ne0. $$ So $d_xf$ is invertible.


4

In my view it is not really the definition of the radius of convergence. Rather the definition of the radius of convergence of $$\sum_{n = 0}^{\infty} a_n(z - c)^n $$ is $R=\sup\{|z_0 - c | \colon \sum_{n = 0}^{\infty} a_n(z_0 - c)^n \text{ converges}\}$. With this definition, it is clear that this is the largest disk where the convergence could ...


0

Sorry to take so long. I've been busy. On the interval in question, $$\lfloor x \rfloor = \begin{cases}1, & 1 \le x < 2\\2, & x = 2\end{cases}$$ Given $n$, the intervals are $\left[1 + \frac {k-1}n,1 + \frac{k}n\right], k = 1, \ldots, n$. So the sums are $$L_n = \sum_{k=1}^{n}\left\lfloor 1 + \frac {k-1}n \right\rfloor\frac1n = ...


1

Getting a finite open cover $K$ consisting of metric balls seems to be a red herring. Your assumptions imply there is an open $U$ containing $K$ over which $f$ is injective. Let $\epsilon=d(K,\mathbb R ^n \setminus U)$. All you have to show is $\epsilon>0$, because clearly $\{x\in \mathbb R^n : d(x,K) < \epsilon \}\subseteq U$. Suppose ...


1

This is good. Alternatively, for a direct proof, note $A$ is non-empty and bounded below. Every lower bound $c$ of $A$ must satisfy $c \notin \mathbb{R_+}$ since for all $\epsilon \in \mathbb{R_+}$ there is some $n$ such that $\epsilon > \frac{1}{n}$ and $\frac{1}{n} \in A$. Since $\max \mathbb{R} \setminus \mathbb{R_+} = 0$, and $0$ is a lower bound of ...


0

I'd like to point out that $$\frac b2<b\leq\text{inf}(A).$$ This being a contradiction implies that $b>\inf(A)$.


1

Your answer is fine. Just take some care with the assumption that "since $b \in A$, b can not be a lower bound of A". For example: $A = [0, 1]$. The infimum of this set is also $0$ and $0 \in A$.


1

Yes, but you have to verify this. Suppose $x=\sum f(v)v\in\operatorname{span}\beta$, where $v\in\beta$ and the sum is finite. Show that $\Vert Tx\Vert^2=\Vert x\Vert^2= \sum|f(v)|^2$ (use orthonormality of $\beta$). Thus, if $Tx=0$, we must have $x=0$, which means that $T$ is injective.


2

Yes. This is even true for a real analytic section of a real analytic bundle over a real analytic manifold. To prove this, suppose $f=g$ on a set $U$ with nonempty interior (which we will henceforth also call $U$); if $U$ is not everything, then there is a point $x \in \partial U$. Work instead with $f-g$; we wish to show it vanishes in a neighborhood of ...


3

Actually, it depends on what you call a continuous curve. If you mean an object defined in a parametric way $t \mapsto (x(t),y(t))$ where $x$ and $y$ are two continuous functions on an interval $I$ then $x+y$ is also a continuous function on $I$. But by the intermediate value theorem, if you require $x+y$ to take values in $\mathbf{Q}$ then it forces this ...


0

If you know a bit about connected components: The set $P=\left\{(x,y):x+y\in\mathbb{Q}\right\}$, which is the union of all lines $y=-x+q$ for $q$ rational, has precisely these lines as connected components, so these are in fact all "curves" which satisfy the equation $x+y\in\mathbb{Q}$.


1

For each $x,y\in\mathbb R$ define $f_x$ and $f^y$ on $\mathbb [0,1]$ by $y\mapsto f(x,y)$ and $x\mapsto f(x,y)$, respectively. Since $\mathbb Q$ has zero Lebesgue measure, $f_x=0$ a.e. for all $x$ so $$0=\int_0^1 f_x(y)\ \mathsf dy=\int_0^1\int_0^1f(x,y)\ \mathsf dy\ \mathsf dx.$$ However, $f$ is not Riemann integrable on $[0,1]^2$. Let $(x,y)\in[0,1]^2$ ...


1

Use a "diagonal": Given $n$, divide the interval $[-n,n]$ in $2n^2$ intervals of length $1/n^2$, namely $I^n_i=[n+(i-1)/n^2,n+i/n^2]$, $1\leq i\leq 2n^2$, and define "rectangular functions" $f^n_i=n1_{I^n_i}$. Then $\int f^n_i=1/n$. ($1_A$ denotes the characteristic function of $A$). The sequence ...


1

If you have such a sequence $(f_n)_{n\geq0}$ on $[0,1]$, then $g_n(x)=1/(\lfloor x\rfloor^2)f_n(\{x\})$ works on $\mathbb{R}$.


0

For each $x,y\in\mathbb R$ define $f_x$ and $f^y$ on $\mathbb [0,1]$ by $y\mapsto f(x,y)$ and $x\mapsto f(x,y)$, respectively. Since $\mathbb Q$ has zero Lebesgue measure, $f_x=0$ a.e. for all $x$ so $$0=\int_0^1 f_x(y)\ \mathsf dy=\int_0^1\int_0^1f(x,y)\ \mathsf dy\ \mathsf dx.$$ A similar argument may be used to show that the other iterated integral is ...


0

Simple, use derivative, not approximations. You get: $(x\arctan(x\sin^2(\frac{1}{x})))'=\arctan(x\sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(-2x\sin\frac{1}{x}\cos\frac{1}{x}(-\frac{1}{x^2}))$ so all are positive for $x \geq 1$.


3

Check for $m = 0$ and $n = 0$. Then ($n \neq 0 \wedge m \neq 0$) divide numerator and denominator by $mn \neq 0$. $$ A = \left\lbrace \frac{1}{4 \frac{m}{n} + \frac{n}{m}} \left| m \in \mathbb{Z}\smallsetminus \lbrace 0 \rbrace \wedge n \in \mathbb{N}\smallsetminus \lbrace 0 \rbrace\right.\right\rbrace \cup \lbrace 0 \rbrace $$ Now let $q = \frac{m}{n}$. $q ...


1

You made a mistake. $$|f|=\sqrt{(\langle f|f\rangle)}.$$ Angle depends on the definition of inner/dot product. You have to define the inner product first (i.e. including fixing your $a$ and $b$, different $a$ and $b$ will give you different angle) before computing its angle.It has nothing to do with intersection point.


0

The domain of $f$ is broken into two parts: the triangle $A$ above the diagonal in $[0,1]^2$ and the triangle $B$ below the diagonal. The parts of $f$ are continuous (being a composition of continuous functions) on the interiors of $A$ and $B$, respectively. What you have to check is that $f$ is still continuous on the boundaries $\partial A$ and $\partial ...


0

I think you are confused with the definition of cover and compactness. Let's deal only with subsets of $\mathbb{R}$ to make things clearer: Definition: A cover of a subset $A$ of $\mathbb{R}$ is a collection $\mathcal{C}$ of subsets of $\mathbb{R}$ such that $A\subseteq \bigcup\left\{B:B\in\mathcal{C}\right\}$, that is, the union of all elements of ...


-1

It's also worth noting that this representative of is a strong bias towards analysis. An algebraist generally cares minimally about $\mathbb{R}$, generally field extensions of $\mathbb{Q}$ and $\mathbb{F}_p$ are far more interesting. The badly named so-called Real Numbers are very much unimportant in Combinatorics and Number Theory as well, but ...


1

First $(\epsilon,1)_{\epsilon>0}$ is an open cover of $(0,1)$ because: for any $\epsilon>0$, $(\epsilon,1)$ is open $(0,1)\subset \bigcup_{\epsilon>0}(\epsilon,1)$ (in fact it is equal) Now assume that this open cover has a finite subcover. This means that there exists $\epsilon_1,\dots,\epsilon_n$ such that $(0,1)\subset ...


3

You write some $\epsilon$ as close as possible to zero But there is no such $\epsilon$! For any positive $\epsilon$, there is a strictly smaller positive $\epsilon$ - for instance, $\epsilon/2$. And the interval $({\epsilon/2}, 1)$ will contain some points in $(0, 1)$ that $(\epsilon, 1)$ does not (for instance, $\epsilon$ itself). While no specific ...


1

Denote $f_k(x)=f(kx)$ then we have $$\lim_{k\to\infty}f_k(x)=r\;\text{ a.e.}$$ so by the dominated convergence theorem we have $$\lim_{k\to\infty}\int_0^a f_k(x)dx=\int_0^a rdx=ra$$


0

If not, then you immediately have a disconnection of $X$ as $$f^{-1}((-\infty,y))\cup f^{-1}((y,\infty))$$ (can you see why these two sets are disjoint, nonempty, open, and exhaust $X$?).


0

(1). The set of congruences of squares, modulo $10$, is $A=\{0,1,4,5,6,9\}$ And the set of congrences, modulo $10,$ of $\{2 x: x\in A\}$ is $B=\{0,2,8\}.$ And $A\cap B=\{0\}$. So, modulo $10,$ we have $$(2). \quad n^2-2 m^2\equiv 0\iff n^2\equiv 2 m^2\equiv 0\iff ((n\equiv 0 \land (m\equiv 0\lor m\equiv 5)).$$ (3). Observe that $n\equiv 0\pmod {10}\iff ...


2

The continuous image of a connected space is connected, and the only connected spaces in $\;\Bbb R\;$ are intervals, so since $\;f(X)\subset\Bbb R\;$ is connected it is an interval. Try now to take it from here.


0

the signs depend on the solutions of $k^x=x^k$, which cannot be solved by elementary means. Rewrite $$x=\left(\sqrt[k]k\right)^x=e^{x\ln(k)/k}$$ and set $u=-x\ln(k)/k$. Then $$e^{-u}=-\frac k{\ln(k)}u,\\ ue^u=-\frac{\ln(k)}k,\\ u=W\left(-\frac{\ln(k)} k\right),\\ x=-\frac k{\ln(k)}W\left(-\frac{\ln(k)} k\right).$$ where $W$ denotes the Lambert function. ...


0

By definition the product measure is the outer measure defined using covers by countable unions of "rectangles". If $$\Delta\subset\bigcup_jA_j\times B_j$$then it's not hard to show that $$\sum\mu(A_j)c(B_j)=\infty;$$hence $(\mu\times c)(\Delta)=\infty$.


0

You are dealing with $$\frac{\exp((\ln n)^2)}{n!}$$ when $n$ is large. (If I'm not misinterpreting it, you mean $(\ln n)^{\ln n}$ rather than $\ln(n^{\ln n})$.) For an elementary approach, two facts, which every sensible calculus textbook should include, might be helpful here: 1). For any $a>0$ (however terribly small), and any $\epsilon>0$, there ...


1

It's easy to see that $\ln n\lt n/2$ and $n!\gt(n/2)^{n/2}$ for $n\gt1$. Thus $${(\ln n)^{\ln n}\over n!}\lt{(n/2)^{\ln n}\over(n/2)^{n/2}}={1\over(n/2)^{(n/2)-\ln n}}$$ and the latter tends to $0$ for any number of reasons.


1

For $0\le m\le n$, let $S_m:=\sum_{k=1}^m a_k-\sum_{k=m+1}^na_k$. If $S_m=0$ for some $m$, we are done. Hence assume $S_m\ne 0$ for all $m$. As $S_n=-S_0$, there are $m$ such that $S_m$ is positive and others such that $S_m$ is negative. Hence there exists $m>0$ such that $S_{m-1}$ and $S_{m}$ have opposite sign. But $S_{m}-S_{m-1}=2a_m$ (or equivalently ...


0

It is easy to show that $\ln (n!) > \frac{1}{3} n \ln n$.$^{(\dagger)}$ Using this, rewrite $$ 0 < \frac{(\ln n)^{\ln n}}{n!} < \frac{(\ln n)^{\ln n}}{e^{ \frac{1}{3} n \ln n}} = e^{(\ln n)\ln \ln n - \frac{1}{3} n \ln n} = e^{(\ln n)( \ln \ln n - \frac{1}{3} n )} \xrightarrow[n\to \infty]{} 0 $$ where the limit follows from observing that the ...


1

$$a_{e^n}=\frac{n^n}{\Gamma(1+e^n)}=\frac{n}{e^n}\frac{n}{e^n-1}...\frac{n}{e^n-n+1}\frac{1}{\Gamma(1+e^n-n)}\le\left(\frac{n}{e^n-n+1}\right)^n\frac{1}{\Gamma(1+e^n-n)}$$ Each factor converges to $0$.


0

Let $$a_n=\frac{\left(\ln n\right)^{\ln n}}{n}$$ Then $$\ln a_n=\ln\frac{\left(\ln n\right)^{\ln n}}{n}=\ln\left(\left(\ln n\right)^{\ln n}\right)-\ln n=\ln n\cdot\ln\ln n-\ln n=\ln n\cdot\left(\ln\ln n-1\right)$$ It is easy to see that $\ln a_n\to\infty$, so also $a_n\to\infty$ EDIT: I just noticed that $$a_n=\frac{\left(\ln n\right)^{\ln n}}{n!}$$ I ...


1

Yes. Def'n : A linear order $<_S$ on a set $S$ is order-dense iff $\forall x,y\in S\;(x<_S y\implies$ $ \exists z\;(x<_Sz<_Sy)).$ Theorem.(Cantor). If $S$ is countably infinite and $<_S$ is an order-dense linear order on $S$ with no end-points (no $<_S$ max or min) then there is an order-isomorphism from $S$ to $Q .$ So let $S$ be the ...


7

If $f_1$ and $f_2$ differ in values just at let us say in the middle of the interval, thus at a point, this would not change the integral value, it would vanish, but $f_1$ and $f_2$ would be considered different.


6

There's no way to do justice to "Why is mathematics about real numbers?" within the length constraints of a Math.SE post, but here are some relatively philosophical observations and opinions (meant to be a bit provocative, in the spirit of answering a soft question). First, as multiple people have commented, the real numbers are not universally regarded as ...


1

You are pretty confused. They are used as such in the definition. When you substitute $x = F^{n}(y)$, it changes to $y - F^{-n}y$. Since $f$ and $g$ commutes, their lifts too. Only that property is being exploited here.


2

Suppose that there exist $(a,b,c,d,u,v)$ such that $$ad-bc=u^2+v^2$$ $$a+d=2u$$ where $a,b,c,d\ge 0$ with $v\not=0$. Since $u=(a+d)/2$, we have $$ad-bc=\left(\frac{a+d}{2}\right)^2+v^2,$$ Multiplying the both sides by $4$ gives $$4ad-4bc=a^2+2ad+d^2+4v^2$$ i.e. $$-4bc=(a-d)^2+4v^2$$ The LHS is non-positive and the RHS is positive. This is a contradiction.


1

Hint: Multiply the second equation by $a$ (or $d$, it doesn't matter) and substitute into the first equation.


0

What have you tried so far? And, as for being one-to-one, can you think of examples where a function maps, say, every open interval to $R$? what would such a function look like? is it one-one? onto?


3

Writing $x_n = \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}$, we see: $$x_n>x_{n-1}\iff \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}>\frac{1-p^{n}}{1-p^{n-1}}\frac{n-1}{n}$$ $$\iff(1-p^{n-1})(1-p^{n+1})n^2>(1-p^n)^2(n^2-1)$$ $$\iff (1-p^n)^2 > n^2[(1-p^n)^2-(1-p^{n-1})(1-p^{n+1})]=n^2[p^{n-1}+p^{n+1}-2p^n]$$ ...


1

Since $f$ is $C^1$, existence and uniqueness of a local solution is guaranteed by the Existence and Uniqueness Theorem (a.k.a. Picard's theorem,...) What you are asked to prove is that the solution is defined on $[0,\infty)$. For this it is enough to prove that the solution does not blow-up, that is, it is bounded on any finite interval $[0,T]$ (with a bound ...


0

The idea behind the proof is quite simple once you get read of the useless verbosity. You start by using the fact that $\{n: |u_n-t|<1\}$ is infinite, take $n_1$ in it. Then you can use the fact that $\{n: |u_n-t|<\dfrac12\}$ is infinite, take $n_2$ in it strictly bigger than $n_1$. Proceeding as this by induction, for each $k$ you can choose $n_k$ ...


0

@ChristopherCarlHeckman : it is a Laplace Transform (see remark at the end) Precisely, it is the Laplace Transform (LT) of a very important function, the cardinal sine (sinc). This LT can be found in most LT tables as $$\int_0^\infty {\sin(k)\over k}\exp(−sk)\,dk=\frac\pi2-\arctan(s)$$ from which it is easy to deduce by an elementary change of ...


0

In every metric space $(X,d)$ a sequence $x_n$ converges to $x$ if and only if every subsequence $x_{n_k}$ has a further subsequence converging to $x$ (easy proof by contradiction). All you need to now is thus that convergence in measure is convergence in a metric space (e.g. $d(f,g)=\int \min\lbrace 1,|f(x)-g(x)|\rbrace \, d\mu(x)$ is a suitable metric on ...


2

Your equation can be changed to $$5(5^{x-1}-1)=24(3^{x-1}-2^{x-1})$$ At least $x=0$, $x=1$ and $x=3$ are solutions. If your question is a Diophantine problem: With $n_1,n_2\in\mathbb{N}$ you have the conditions $$3^{x-1}-2^{x-1}=5n_1$$ $$5^{x-1}-1=24n_2$$ EDIT: $x\in\mathbb{R}$ o.k. But with the values 0,1,3 you can check ...


3

We assume $x>0$ and $s>0$. Then by differentiating the following identity with respect to $s$, $$ f(s)=\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk $$ one may write $$ f'(s)=-\int_0^\infty \exp(−sk)\sin(kx)\,dk=-\frac{x}{x^2+ s^2} $$ giving $$ f(s)=-\arctan \left( \frac{s}x\right)+C. $$ Observing that, as $s \to \infty$, $f(s) \to 0$, we then obtain ...



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