New answers tagged

0

The comments to this question point to this question which shows that if a function is Riemann integrable over an interval and vanishes on a dense subset then its integral is $0$. This happens because in this case given any partition of the interval we can choose the tags so that the Riemann sum becomes $0$. Now the answer to the current question is ...


0

As pointed out by PhoemueX, the proof is correct. However, we do not have $\lVert \varphi_n\rVert\leqslant 1/2$. For example, if we take $x$ such that $x_k=\sqrt k$ for $1\leqslant k\leqslant n$ and $x_k=0$ for the other $k$'s, then $$\frac{\left|\varphi_n(x)\right|}{\lVert x\rVert}=\frac 1n\sum_{k=1}^n k\cdot \frac 1{\sqrt{\sum_{k=1}^n k}}=\frac ...


2

The bijection can be realized by the negative of the exponential function: $$ f:\mathbb R\to (-\infty, 0)\quad x\mapsto - e^{x} $$


2

No, $f$ need not be uniformly continuous on $(0,1)$. Let $f_n=\left\{\begin{array}{ll}n+1&\text{if }x<\frac1{n+1}\\\frac1x &\text{if }x\geq \frac1{n+1}\end{array}\right.$. Then $f_n\xrightarrow{\text{unif} }f$ on every compact subset of $(0,1)$ where $f(x)=\frac1x$. But $f$ is not uniformly continuous on $(0,1)$.


0

For the first question, consider the following: Let $$ g(x)=\begin{cases}1&x=0\\-1&x\not=0\end{cases}. $$ Let $q=0$; then $\lim_{x\rightarrow q}g(x)=\lim_{x\rightarrow 0}g(x)=-1$. Therefore, $r=-1$. Let $$ f(x)=0. $$ Suppose that $p=0$, then $\lim_{x\rightarrow p}f(x)=\lim_{x\rightarrow 0}f(x)=0=q$. Now, $g(f(x))=g(0)=1$. Therefore, ...


0

Let (1) $a_n = \frac{2^n}{3^{n+1}}$ Then it's easy to see that: (2) $a_{n+1} = \frac{2}{3} \cdot a_n$ Let's say we know that the sequence has a limit $L$. Then by letting $n$ go to infinity in (2) we get: (3) $L = \frac{2}{3} \cdot L$ So $L=0$. Now we just have to prove that the sequence has a limit. To do this it suffices to note that ...


1

A floor function is rather annoying to integrate (in my opinion), but you have a very easy area over which you integrate: the triangle with vertices $(-2,0), (0,0)$ and $(0,2)$. You can work out what the value of the floor function $\lfloor x+y\rfloor$ will be on this triangle: $\lfloor x+y\rfloor = 0 \iff 0 \leq x+y < 1 \iff -x \leq y < -x+1$, in ...


1

Draw the Region of the plane $$\{(x,y)\mid y\in [0,2],x\in[0,2-y]\}$$ Then note that \begin{align*} & [\dots]\\\text{if }\ & 0\ge x+y<1\quad \lfloor x+y\rfloor=0\\ \text{if }\ & 1\ge x+y<2\quad \lfloor x+y\rfloor=1\\ & [\dots] \end{align*} and so on for the various integer values. Can you identify these regions in the picture? What is ...


0

If $x = \dfrac{2k + 1}{2^{2m + 1}}$ then the binary representation of $x$ consists of binary representation of $k$ till $2m$ binary digits and then followed by $1$. This means that if $$k = b_{1}b_{2}\ldots b_{2m}$$ in binary notation then $$x = 0.b_{1}b_{2}\ldots b_{2m}1$$ in the binary notation. It follows that $\beta_{i}(x) = b_{i}$ for $i = 1, 2, \ldots ...


4

$$\frac{2^n}{3^{n+1}}=\frac13\left(\frac23\right)^n$$ Now, what is $\;\lim\limits_{n\to\infty}x^n\;$ for $\;|x|<1;$ ?


2

Every subset of finite-dimensional Euclidian space is totally bounded if and only if it is bounded. Which means that every subset of $\left[0;1\right]$ is totally bounded.


0

This is not accurate, but it may help. If $f(0)=0$ or $f(1)=1$ everything is proved, so let's assume the contrary. Draw a graph $y_1:g(x)=x$. If there exists $f:\left[0;1\right]\rightarrow\left[0;1\right]$, so that $\forall t\in\left[0;1\right]\;f(t)\neq t$, $f$ is continuous and $y_2$ is its plot, means that $y_2$ lies upper or lower than than $y_1$. ...


1

We can use the inequalities between quadratic mean (a.k.a. square root mean), arithmetic mean, geometric mean and harmonic mean $$\sqrt{\frac{x^2+y^2}2} \ge \frac{x+y}2 \ge \sqrt{xy} \ge \frac2{\frac1x+\frac1y}.$$ These inequalities are true for any $x,y>0$. The equality holds if and only if $x=y$. They can be generalized for more than two variables (also ...


1

Let $\varepsilon>0$. By uniform continuity on $(a,b]$ and $[b,c)$, there is $\delta_1>0$ s.t. $$\forall x,y\in (a,b],\ |x-y|<\delta_1\implies |g(x)-g(y)|<\frac{\varepsilon}{2}$$ and there is $\delta_2>0$ s.t. $$\forall x,y\in [b,c),\ |x-y|<\delta_2\implies |g(x)-g(y)|<\frac{\varepsilon}{2}.$$ Let $\delta=\min\{\delta_1,\delta_2\}$ and ...


2

Let $m> n$ be such that $l=m-n$ satisfies $$\left|\frac{\sum_{j=n}^{m-1} a_j^2}{l}-\rho\right|<\epsilon$$ for all $m>n$, where $\epsilon>0$ is chosen such that $\rho+\epsilon<1$. Then, an application of AM-GM inequality gives $$\prod_{j=n}^{m-1} a_j^2\le \left(\frac{\sum_{j=m}^n a_j^2}{l}\right)^l\le (\rho+\epsilon)^l$$ Since this is true for ...


1

The differential $df(p,\delta p)$ of a function $f:U\to \mathbb{R}^m$ is a function of both a point $p\in U$ and of a tangent vector $\delta p\in T_pU$ at $p$. As you note, the differential is always linear in the tangent vector; this is a defining property of the differential. Generically $df$ is an arbitrary nonlinear function of $p$, though. Some special ...


0

Consider a function $f$ which is the sum of triangles, $\Gamma_n$ , where $\Gamma_n$ has base centered at $n$, having vertices at $(n+1/n^3,0)$, $(n-1/n^3,0)$ and $(n,n)$ i.e. function takes value $f(n)=n$ and outside the triangle $f$ takes value $0$, then, this function satisfies all hypothesis, and shows that all options are false.


0

Let $A = \mathbb{Q} \cap (a,b)$. It should be clear that $a,b$ are lower & upper bounds respectively. There are sequences $a_n \downarrow a, b_n \uparrow b$ with $a_n,b_n \in A$, hence $a \le \inf A \le a_n$ and $b \ge \sup A \ge b_n$ for all $n$, hence we have the desired result.


0

Let $y = \cos \alpha$ and $x = \cos(\alpha/3)$ then we know that $$y = x(4x^{2} - 3)\tag{1}$$ Your approximation says that $$x \approx \frac{1}{2}\sqrt{\frac{2y}{y + 3} + 3}\tag{2}$$ or using $(1)$ we get $$\frac{y}{x} = 4x^{2} - 3 \approx \frac{2y}{\sqrt{y + 3}}\tag{3}$$ From $(1)$ we see that this implies $$\frac{1}{x}\approx \frac{2}{\sqrt{y + ...


2

The quantity $u(x,t)$ can describe: The height of a wave at point $x$ at time $t$ The temperature at point $x$ at time $t$ The concentration of some substance at point $x$ at time $t$ The number of cars on a 50-meter stretch of a road... PDEs are not tied to physical quantities; they describe physical processes. The PDE $u_t+cu_x$ describes the ...


0

Suppose $|f_n(x)|$ is bounded for all $n$, although not necessarily uniformly with respect to $x$, on the interval $I = (x_0 - \delta_1,x_0 + \delta_1).$ In other words, suppose for each $x \in I$ there exists $M(x)$ such that $|f_n(x)| \leqslant M(x)$ for all $n \in \mathbb{N}$. If, in addition, each $f_n$ is continuous on $I$, then there exists a ...


1

Suppose $0<\mu (A)<\infty$ and $0<\mu (B)<\infty,$ and $A\cap B=\phi.$ (1).Let $f(x)=1/\mu (A)$ when $x\in A$ and $f(x)=0$ when $x\not \in A.$ Let $g(x)=1/\mu (B)$ when $x \in B$ and $g(x)=0$ when $x\not \in B.$ Then $f, g$ are linearly independent, and $\|(f+g)/2\|_1=\|f\|_1=\|g\|_1=1.$ (2). Let $f(x)=1$ when $x\in A$ and $f(x)=0$ when $x ...


0

Just use the definition of supremum. We need to show that (1) $b$ is an upper bound of $\mathbb Q\cap(a,b)$, and (2) $\forall b'<b$ is not an upper bound of $\mathbb Q\cap(a,b)$. We only need to prove (2). Without loss of generality, we can suppose $b'>a$. Since $b'<b$ and $\mathbb Q$ is dense in $\mathbb R$, i.e. we can find $q\in\mathbb Q$ such ...


0

Let $A=\Bbb{Q}\cap (a,b)$. Of course, $b\geq x$ for all $x\in A$. Suppose that the supremum of $A$ is $m<b$. By density, there exists $r\in\Bbb{Q}\cap(m,b)$. Hence, $r\in A$ and $r>m$, which contradicts the maximality of $m$.


4

Norms are a measure of distance. One has different ways to define what is the distance between points in multiple dimensions, which collapse to the usual notion of the absolute value in 1D. in particular, Euclidean distance is defined by the 2-norm, $$ \left\| \begin{pmatrix} x_1 \\ \vdots \\ x_k \end{pmatrix} \right\|_2 = \sqrt{\sum_{i=1}^k x_i^2} $$ ...


0

Hint: If $\sum f_n$ converges uniformly on a set $E,$ then we must have $\sup_E|f_n| \to 0.$ Is that true here?


0

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} ...


1

Hint for $(1)$: What happens if you take the preimage of $[0,1]$? Hint for $(2)$: How many points does the set $\{x\in[0,1]:f(x)=a\}$ have?


0

Hint: $f$ is not measurable since $f^{-1}((0,1))=N-\{0\}$.


0

As a generalization, if we subtract the first $m$ terms of the series for $e^x$ and divide by $x^m$, we get this: $f_m(x) =e^x-\sum_{k=0}^{m-1} \dfrac{x^k}{k!} = $ so $\begin{array}\\ \int_0^1 \dfrac{f_m(x)dx}{x^m} &=\int_0^1 \dfrac{\sum_{k=m}^{\infty} \dfrac{x^k}{k!}dx}{x^m}\\ &=\sum_{k=m}^{\infty} \dfrac1{k!}\int_0^1 \dfrac{x^kdx}{x^m}\\ ...


2

To me it seems Jeremy Gray put here some words of warning that interpreting the work of the old masters in terms of one or two modern concepts only is too often superficial and not appropriate to fully grasp what was really going on. If we look at the last sentence of OPs cited paragraph we can see that the historical development is more than that. ...


2

You proof looks basically correct to me. $\lim_{n\to\infty}\frac{e^x-1}{x}=1$ should be $\lim_{x\to 0+}\frac{e^x-1}{x}=1$. I would write (1) as $$(1)\int_0^1\frac{e^x-1}{x}dx=\int_0^1g(x)dx.$$


3

A real function $f(x)$ may be analytic at $x=0$ with a finite radius of convergence, due to the existence of singularities in the complex plane: take, for instance, $f(x)=\frac{1}{x^2+1}$ or its primitive $\arctan x$. In such a case the Taylor series actually converges to $f(x)$ only if $x$ is sufficiently close to the origin, so, even if $$\forall ...


1

Setting $x=\cos 2t $ we get $1-x =2\sin^2 t ,$ $1+x =\cos^2 t ,$ $1 -x -\sqrt{1-x^2 } =2\sin^2 t -\sin 2t =2\sin t (\sin t -\cos t ),$ $1 +x -\sqrt{1-x^2 } =2\cos^2 t -\sin 2t =2\cos t (\cos t -\sin t )$ So we get a equation $$\sin^{2n} t + 2n (\cos^2 t -\sin^2 t ) -\cos^{2n} t =0$$ Hence $$(\sin t -\cos t )\cdot [(\cos t + \sin t) \sum_{k=0}^{n-1} ...


1

Several comments: It is not enough to say that $f(x_n,y_n)$ is bounded and hence it has a convergent subsequence because then you would have to write $f(x_{n_k},y_{n_k}) \rightarrow L$ and you wouldn't a priori know that $L$ has the form $L = f(x',y')$ for some $(x',y') \in \mathbb{R}^2$ like you assume in the proof. In the way your argument is written, ...


5

HINT: For (a) let $x$ and $y$ be distinct points in $[0,1]$, let $U$ be an open nbhd of $x$, and let $V$ be an open nbhd of $y$. What do you know about the sets $[0,1]\setminus U$ and $[0,1]\setminus V$? How big is the union of two countable sets? Is $\Bbb R$ countable? For (b), start by showing that a sequence $\langle x_n:n\in\Bbb N\rangle$ in ...


0

Hint: $D$ ia the diagonal matrix that has as diagonal elements the eigenvalues of $A$ and $M$ is a matrix that has as columns the corresponding eigenvectors. Do you know how to find these? ( see here)


1

Let us consider the unit sphere under $|| \cdot ||_1$: all polynomials $p(x)$ such that $\int_0^1|p(x)|dx \le 1$. We have $$\int_0^1 |p(x)| dx \le \int_0^1 \sum |a_i||x^i| dx = \int_0^1 \sum |a_i|x^i dx = |a_0|+\frac{1}{2}|a_1| + \dots + \frac{1}{n+1}|a_n| $$ by the triangle inequality and similar standard inequalities. EDIT: Compare the upper bound with ...


1

It's not quite true, as David C. Ullrich points out in the comments. If one of the two factors is constant, the other has no reason to be bounded. But for $g$ of the given specific form, that is the only thing that allows unboundedness of one of the factors. If $p$ is not constant, we choose $x_1,x_2$ with $p(x_1) \neq p(x_2)$, and then find $$\lvert ...


4

This is Tauber's Theorem, the original tauberian theorem. Hence there must be answers all over the web, presumably including MSE. I'm going to justify the existence of this answer by proving a stronger statement, and also by giving what I consider to be the "right" way to organize the proof (consistent with the general theme that convergence results should ...


2

Let $\{p_n\}$ be a sequence of all rationals in $[0,1]$. For each $n$, put $f_n(p_m) = 1$ for $m\leq n$, and $f_n(x)=0$ everywhere else on $[0,1]$. Each $f_n$ has a finite number of discontinuities, hence is Riemann-integrable. Their limit is not.


1

The second interpretation doesn't make any sense. In algebra, one way to view a formal power series is as an infinite ordered sequence of numbers. In this case, the powers of the variable are used only to indicate the order of the coefficients. The coefficient of $x^5$ is just the fifth term in the sequence, while the coefficient of $x^0$ is the zeroth ...


1

A more general theorem in Folland's Real Analysis can answer this question:


0

Another method $$\sum_{n=0}^{\infty}x^{n+1}=x\sum_{n=0}^{\infty}x^n=\frac{x}{1-x}\quad \textrm{for}\quad |x|< 1$$ Now, integrating term-wise we have $$\int\left(\sum_{n=0}^{\infty}x^{n+1}\right)\text{d}x=\sum_{n=0}^{\infty}\int x^{n+1}\text{d}x=\sum_{n=0}^{\infty}\frac{x^{n+2}}{n+2}=x^2\sum_{n=0}^{\infty}\frac{1}{n+2}x^n=\int \frac{x}{1-x}\text{d}x$$Hence ...


1

Hint: $$f(x)=\sum_{n=0}^{\infty}{\frac{1}{n+2}x^n}=\frac{1}{x^2}\sum_{n=0}^{\infty}{\frac{1}{n+2}x^{n+2}}$$ Now differentiate w.r.t. $x$ and obtain $f'(x)$ in explicit form. Then integrate the result. Don't forget to check the convergence interval. You will also need a particular value of the original series when integrating later. Use $f(0)=\frac{1}{2}$ ...


1

No you have several errors. First Since $[0,1]\cap\mathbb{Q}$ is dense in $[0,1]$, then for any $x\in[0,1]$, there exists an open neighbourhood $U_x=\{y:|x−y|<\epsilon\}$ of $x$ so that $U_x\cap \mathbb{Q}Q\not=\not 0$. doesn't really make sense. What density implies is that for all $x\in [0,1]$ and $\epsilon>0$ $B_d(x,\epsilon)\cap ...


3

First note that for $v_1,v_2\in S^{n-1}$ we have that $$ |f(x+v_1t_1+v_2t_2)- f(x)| \\= |f(x+v_1t_1+v_2t_2)-f(x+v_1 t_1)+f(x+v_1 t_1) -f(x)| \\\leq |t_2|L(v_2) + |t_2|L(v_1) $$ Now take an orthonormal basis $(e_i)$ of $\mathbb{R}^n$ and define $$ L_i = L(e_i) $$ to get $$ |f(x+ \sum_i t_i e_i)-f(x)| \leq \sum_i |t_i| L_i \leq \sqrt{\sum_i ...


3

HINT: Use $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ and $$2I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b\left(f(x)+f(a+b-x)\right)dx$$ $$2I=\int_{-1}^1\dfrac{dx}{x^2+1}$$


2

Probably the MVT helps: $$ f_n(x) = n \left( f'(\xi(x,n)) \frac{1}{n} \right) $$ for some $$ x < \xi(x,n) < x+ \frac{1}{n}. $$ Now use the fact that $\xi(x,n) \to x$ as $n \to +\infty$ and that $f'$ is uniformly continuous on $[a,b]$.


7

Hint. One may write $$ \begin{align} \int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)} &=\int_{-1}^0 \frac{dx}{(e^x+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)} \\\\&=\int_0^1 \frac{dx}{(e^{-x}+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)} \\\\&=\int_0^1 \frac{e^x\:dx}{(e^{x}+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)} ...



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