Tag Info

New answers tagged

1

If you take the equation which defines convexity, and multiply f by -1, it is straightforward to see that this is equivalent to the equation defining concavity. The logic works both ways so it truly is "if and only if".


1

We are dealing with the wave equation $u_{xx}+u_{tt}-u_{tt}=0$. Notice that it has a mixed term. So we cannot use d'Alembert's formula here. (The formula assumes a difference of squares when "factoring" the derivatives.) Here, we have to "factor" the derivatives manually. Note that we can rewrite our wave equation as $$\left( \frac{\partial^2}{\partial x^2} ...


1

I would try a solution of type $F(z)=F(x-ct)$, or $G(w)=G(x+ct)$. For $F$ this gives $$ u_{xx}=F''(z)\\ u_{xt}=-cF''(z)\\ u_{tt}=c^2F''(z) $$ so $$ (1-c-c^2)F''(z)=0\implies c_{1,2}=\frac{1\pm\sqrt{5}}{2}. $$ So $F(x-c_it)$ is a solution for every choice of $F$. Same reasoning for $G$.


0

I have use proof by contradiction; Assmue there exsit $x_{0}\in [a,b]$, $$\dfrac{f'^2(x_{0})}{f(x_{0})}>2\max_{x\in[a,b]}|f''(x)|\Longrightarrow 1-\dfrac{2f(x_{0})f''(\theta)}{f'(x_{0})}<0,\forall \theta\in[a,b]$$ Use Taylor's Formula,we have $$f(x)=f(x_{0})+f'(x_{0})(x-x_{0})+\dfrac{f''(\theta)}{2}(x-x_{0})^2,\theta\in [a,b]$$ Let ...


3

Hint: note that your equation can be expressed as follows: $$(D^2_x + D_x D_t - D^2_t)\, u = (D_x- \varphi^- D_t) (D_x - \varphi^+ D_t) u = 0, $$ where $\varphi^{\pm} = (-1 \pm \sqrt{5})/2$ are the two solutions of $s^2+s-1=0$. What if we now define $v := (D_x - (-1+\sqrt{5})D_t/2) u$? Can you solve 1st order PDEs? Cheers! Of course, you will have ...


0

First, choose $\alpha:= h(x,y)\beta$ for some real number $\beta$ in order to get rid of the real part. Assuming $h(x,y)\neq 0$ (otherwise there is nothing to prove), we have $$\frac{h(x,y)}{|h(x,y)|^2 }+\beta^2h(y,y)\geqslant 2\beta.$$ Completing the squares, we find the choice $\beta=h(y,y)^{-1/2}$, which yields $$\frac{h(x,x)}{|h(x,y)|^2}-\frac ...


0

Suppose we have that $\|S_R f\|_{p}\le C_p\|f\|_{p}$ for some constant $C_p >0$, $p\in (0,\infty)$. Since Schwartz functions are dense in $L^p$, for every $\varepsilon >0$, there exists $g\in\mathcal{S}$ such that $\|f-g\|_{p}<\varepsilon$. \begin{align} \|S_R f-f\|_{p} & \le\|S_R f-S_R g\|_{p}+\|S_R g-g\|_{p}+\|g-f\|_{p}\\ & \le ...


1

Perhaps you will be interested in this: http://arxiv.org/pdf/math/0701039.pdf It uses means and methods that you will likely don't have difficulty in understanding-with a bit of further studying perhaps-while being one of the best approaches to the problem I have seen..


1

This is a proof I stumbled upon while taking Fourier analysis, it uses Fourier series to show that $$\sin(z)=z\prod_{n=1}^{\infty}\left ( 1-\frac{z^2}{n^2 \pi^2}\right )$$ As one would expect the derivation was already known. This derivation doesn't use the Weierstrass Factorization Theorem and if you have some experience with integrals and series then it ...


0

What is doubtless meant by $f(x)$ being Laplace transformable is that $\int_0^\infty e^{-sx} f(x) \, dx$ converges, at least for certain values of $s \in \mathbb{R}$ or $\mathbb{C}$. See also Chapter 5 of Transforms and Applications Handbook, 3rd ed., Poularikas.


1

The key observation here is that the level sets of the function are compact (as you observed the set of all feasible points is not compact). In fact the set of all $P$ satisfying $P(0)=1$ $$ \int_0^1 |P(t)|dt\le 1 $$ is bounded in $L^1$. Now the underlying space is finite-dimensional, hence this set is compact (is it closed as the intersection of the $L^1$ ...


0

In other words, what you have to show is that the norm $\|\cdot\|_1$ induced by the inner product $\langle \cdot, \cdot \rangle_1$ on $H'$ is equal to the usual dual norm on $H'$. If you unwind this, it comes down to showing something like the following: for every $z \in H$, $$\|z\| = \sup_{v \in H, \|v\|=1} |\langle v,z \rangle|.$$ Prove this by proving ...


1

Since $\sum\limits_{n=1}^{\infty}\dfrac1{n^2}$ converges, by Cauchy Criterion $$ \forall \epsilon>0, \exists N>0, \forall m,n>N, m>n, \sum\limits_{k=n}^{m}\dfrac1{k^2}<\epsilon $$ By telescopic nature of series, for all $m,n>N, m>n$, there is $$ x_m-x_n<\sum\limits_{k=n}^{m-1}\dfrac1{k^2}<\epsilon \hspace{5 mm} \text{or} \hspace{5 ...


1

We have $a_{n+1} = 1 + \frac{1}{1+a_{n}} = \frac{2+a_n}{1+a_{n}} $. Therefore, $a_n > 1$ for all $n$. Also, $\begin{array}\\ a_{n+1}-\sqrt{2} &= \dfrac{2+a_n}{1+a_{n}}-\sqrt{2}\\ &= \dfrac{2+a_n-\sqrt{2}(1+a_n)}{1+a_{n}}\\ &= \dfrac{2-\sqrt{2}-a_n(\sqrt{2}-1)}{1+a_{n}}\\ &= \dfrac{(\sqrt{2}-1)(\sqrt{2}-a_n)}{1+a_{n}}\\ \end{array} $ so ...


0

Hint: Applying the recurrence relation twice, we find that $$ a_{n+2} = 1 + \frac{1}{1+a_{n+1}} = 1 + \frac{1}{1+\left(1 + \frac{1}{1+a_{n}}\right)} = \\ 1 + \frac{1}{2 + \frac{1}{1+a_{n}}}=\\ 1 + \frac{1+a_n}{2 + 2a_n + 1} =\\ 1 + \frac{1+a_n}{3 + 2a_n} = \\ \frac{3 + 2a_n + 1+a_n}{3 + 2a_n} = \\ \frac{4 + 3a_n}{3 + 2a_n} $$ This recurrence gives us both ...


0

Hint: $\sqrt{2}$ is a piece of cheese, $a_{2n}$ is a slice of bread, $a_{2n+1}$ is another slice of bread. What would you do next?


1

Check here or here for the definition of a Radon measure. A measure $\nu: \mathcal{F} \to \mathbb{R}$ is concentrated on a set $A \in \mathcal{F}$ if $\nu(A^c) = 0$. Thus if $\nu$ is concentrated at a point (say $\omega$), then for any $A \in \mathcal{F}$ we have $$ \nu(A) = \begin{cases} \nu(\Omega) &\text{ if $\omega \in A$} \\ 0 &\text{ if ...


3

If $f(x)$ is not identically $0$, some $a_n$ is nonzero. Suppose $a_m$ is the first nonzero coefficient. Then $g(x) = f(x)/x^m = a_m + a_{m+1} x + \ldots$ is a convergent power series, therefore continuous, and $g(0) = a_m \ne 0$. By continuity, there is $\epsilon > 0$ such that $g(x) \ne 0$ for $|x| < \epsilon$, and thus $f(x) \ne 0$ for $0 < |x| ...


3

By continuity of $f$ we have that $\,f(0)=\lim_{n\to\infty}f(x_n)=0$. As $f\in C^\infty(-R,R)$, there exist $y_n\in(x_{n+1},x_n)$, such that $$ 0=\frac{f(x_{n+1})-f({x_n})}{x_{n+1}-x_n}=f'(y_n), $$ due to Mean Value Theorem. Clearly $y_n\to 0$, and due to continuiuty of $f'$, we have $f'(0)=\lim_{n\to\infty}f'(y_n)=0$. Similarly, we may found a ...


5

First of all, since $f$ is continuous, $f(0) = 0$, so $a_0 = 0$. Now, think about $f(x)/x = \sum_{n=0}^\infty a_{n+1}x^n$, which is again continuous. For any $x_n$, we have $f(x_n)/x_n = 0$, so $f(x)/x \rightarrow 0$, as $x \rightarrow 0$. Hence, $a_1 = 0$. Continuing this way, we show that $a_n = 0$ for all $n$. Note that we need to assume that all $x_n$ ...


2

$$(x-y) \log \left(1-\frac{y}{x}\right) = (x-y) \log \left(\frac{x-y}{x}\right) = \log \left(\left(\frac{x-y}{x}\right)^{x-y}\right)$$ $$x \log\left(1-\frac{y}{x+y}\right) = x \log\left(\frac{x}{x+y}\right) = \log\left(\left(\frac{x}{x+y}\right)^{x}\right)$$ Now, since log is strictly increasing, we might as well compare $(\frac{x}{x+y})^{x}$ and ...


3


1

Yes, you need to prove compactness. This is an instance of Rellich–Kondrachov theorem, but presumably the point of the problem is to show compactness directly. To get uniform convergence, it suffices to have Fourier coefficients converging in $\ell^1(\mathbb{Z})$, so let's look at those. Let $g_n^{(k)}$ be the $k$th coefficient of $g_n$. Then $g_n''$ has ...


0

$\lim\limits_{x\rightarrow 0} -(-x)^n=-\lim\limits_{x\rightarrow 0} (-x)^n=\lim\limits_{x\rightarrow 0} (-1)^nx^n=(-1)^n\lim\limits_{x\rightarrow 0} x^n$ And the latter limit is $0$ for $n>0$.


2

$\int\dfrac{\sin x}{1+x^2}dx$ $=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!(x^2+1)}dx$ $=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{2(2n+1)!(x^2+1)}d(x^2+1)$ $=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!(x^2+1)}d(x^2+1)$ ...


1

To expand @NotAloner's answer above, $$\cos^n x = \left( { e^{ix} + e^{-ix} \over 2} \right)^n = {1 \over 2^n} \sum_{j=0}^n {n \choose j} e^{(n-2j)ix}$$ Hence $$D^k \cos^n x = {1 \over 2^n} \sum_{j=0}^n {n \choose j} i^k(n-2j)^k e^{(n-2j)ix}$$ and evaluated at $x = 0$ $$c(n,k) = {1 \over 2^n} \sum_{j=0}^n {n \choose j} i^k(n-2j)^k$$ As commented in the ...


0

\begin{align} f(x)&=\cos^n{x}\\ f'(x)&=n\cos^{n-1}{x}\cdot(-\sin{x})\\ f''(x)&=n(n-1)\cos^{n-2}{x}\cdot(-\sin{x})^2-n\cos^n{x}\\ f'''(x)&=n(n-1)(n-2)\cos^{n-3}{x}\cdot(-\sin{x})^3+2n(n-1)\sin{x}\cdot\cos^{n-1}(x)-nf'(x)\\ \ldots\\ ...


0

i don't have a general expression but a sum. we have $$\begin{align} 2^{n-1}\cos^n x &= {n \choose 0} \cos nx + {n \choose 1} \cos(n-2)x + {n \choose 1} \cos(n-2)x + \cdots\end{align} $$ taking derivatives $k$ times and use the fact that $\frac{d^k}{dx^k}\cos(x) = \cos(x + k\pi/2),$ we have $$\begin{align} 2^{n-1}\frac{d^k}{dx^k}\left( \cos^n ...


0

$(\cos x)^n = \left(\dfrac{e^{ix}+e^{-ix}}{2}\right)^n$. I think some pattern should emerge when taking derivatives of higher order like the $k$th one. You edited your question so my answer would not be useful anymore...fine. As for the edited question, you have $f(x) = \cos x$, taking derivatives: $f'(x) = -\sin x, f''(x) = -\cos x, f'''(x) = \sin x, ...


0

Assuming you mean $a_n \le a_{n-1}^2$ you can still not say the growth is no more than polynomial. Let $a_n=2^{n+1}$. This grows exponentially but satisfies your constraint.


1

As you remarked in a comment: $f:[0,\infty)\rightarrow\{0,1\}$ can be presented as $1_{[s,\infty)}$. This function is right-continuous everywhere and if $s>0$ then it is left-continuous everywhere except at $s$. So if $s>0$ then $f$ is continuous everywhere except at $s$. If $s=0$ then $f$ is constant hence continuous everywhere.


4

$$\int_0^1 \frac{1}{2} \ dx = \int_0^1 x \ dx$$


1

$\newcommand{\Cpx}{\mathbf{C}}$Not sure I've seen a formal definition of "Wick rotation", but I'd probably call your example "complexification" rather than "Wick rotation". "Wick rotation", in my experience, is more along the lines of replacing the Lorentz metric $dx^{2} - dt^{2}$ on a $1$-dimensional spacetime with the Euclidean metric $dx^{2} - (i\, ...


1

Yes, f is continuous everywhere except at $t_0 =s $. If $ s > t_0 $, then for all sequences $t_n$ with $\lim_{n \to \infty} t_n = t_0$ there is an $N \in \mathbb{N}$ such that $ s > t_n $ (otherwise the sequence cannot converge to $t_0$). But then $f(t_n) = 0 $ for all $ n > N$, so $\lim_{n\to\infty} f(t_n) = 0 = f(t_0)$. The same argument works for ...


0

An equation like $$a^x-b^x=c$$ usually doesn't have closed-form solutions. Anyway, using the substitution $t^n=b^x$, it becomes $$t^{n\ln a/\ln b}-t^n=t^m-t^n=c.$$ For $n,m\in\{1,2,3,4\}$, you get a polynomial and analytical solutions are possible, though sometimes quite complicated. For instance, $$8^x-4^x=9$$yields by the Cardano formula ...


2

Note that $\displaystyle x_{n+k}\leq x_n+\frac{1}{n^2}+\cdots+\frac{1}{(n+k-1)^2}\leq x_n+\frac{1}{n^2}+\cdots$. We have \begin{eqnarray} \varlimsup_{k\to\infty}x_{n+k}\leq x_n+\frac{1}{n^2}+\cdots \end{eqnarray} Let $\displaystyle\varepsilon(n)=\frac{1}{n^2}+\cdots$. We have $\displaystyle\lim_{n\to\infty}\varepsilon(n) =0$ as the series ...


0

Hint: $$x_1\le x_0+1$$ $$x_2\le x_1+\frac14\le x_0+1+\frac14$$ $$x_3\le x_2+\frac19\le x_0+1+\frac14+\frac19$$ $$\cdots$$ $$x_n\le x_0+\sum_{k=1}^n\frac1{k^2}.$$


0

André Nicolas showed that $C(q)$ is finite and $\ge q$. Claude Leibovici showed that $C(q)\ge \frac12(2^{q}-1)$ and conjectured that $C(q)\approx 2^{q-1}(1+\epsilon)$. We are still missing an explicit bound on $C(q)$ from above; here's a very simple one: By the extended mean value theorem with $f(x)=(1+x)^q$ and $g(x)=x+x^q$ we find that $$ ...


1

There is a slick solution to (i) using compactness: simply observe that if $X,Y$ are compact then so is $X\times Y$. Then $f(x,y)=x+y$ is continuous, so $f(X\times Y)=X+Y$ is also compact (as the continuous image of a compact set). You can get (ii) by using subsequential convergence of compact sets. (iii) is false, as is seen by the example $X=\mathbb Z$ ...


0

Let $$ f_a(x)=\frac{1}{\sqrt{|x|}} \chi_{[-a,a]} $$ Observe that $f_a \in L^1$ but not in $L^2$. Now just compute the fourier transform and play with $a$. \begin{align} \hat{f_a}(y) &= \int_{-a}^a \frac{1}{\sqrt{|x|}}e^{-iyx} \, dx \\ &=\int_0^a\frac{1}{\sqrt{x}}e^{-iyx}+\frac{1}{\sqrt{x}}e^{iyx} \,dx \\ & =2 \int_0^{\sqrt{a}}e^{-iys^2} + ...


1

Hint: Suppose it is not true and $\lim\inf a_{n}=\delta>1$. Then, for large enough $n$, $a_{n}\geq \frac{1+\delta}{2}>1$ for all $n$ (since the inf of the remaining terms converges up to $\delta$)... once all your terms are strictly bigger than one, the rest of the sum must be growing faster than $x$.


1

The function's amplitude on $(a,b)$ is bounded by $m:= f(b)-f(a) \lt \infty$. Look at the right-side limit at a point of discontinuity $c$ (existence of the left-side limit follows in the same way). Fix $m \gt \epsilon \gt 0$. Pick a point $x_1$ in the interval to the right of $c$. If $y_1 := f(x_1) - f(c) \lt \epsilon$, you are done (monotonicity). If ...


2

Note that $f(x) \le f(c)$ for $a < x < c$. Therefore the following exists $$ \sup_{x < c} f(x) $$ and you can show that the following limit exists, too: $$ \lim_{x\uparrow c}f(x) = \sup_{x < c} f(x). $$


3

Suppose that $T$ is bounded. Then $$ \|x\|^{2}+\|Tx\|^{2}=((I+T^{\star}T)x,x) \le \|(I+T^{\star}T)x\|\|x\| \\ \|x\|^{2} \le \|(I+T^{\star}T)x\|\|x\| \\ \|x\| \le \|(I+T^{\star}T)x\|. $$ The last inequality can be use to show that the range of $I+T^{\star}T$ is closed, and $I+T^{\star}T$ has a bounded inverse on ...


0

Let $(M,g)$ be a Riemannian manifold and $p \in M$. Recall that normal coordinates about $p$ map a neighbourhood of the identity in $T_{p}M$ to a neighbourhood of $p$ via the exponential map $exp|_{p}$. The metric in normal coordinates is defined (e.g. in the context of the Gauss lemma) by taking polar coordinates on $T_{p}M$ and noting that for $\epsilon$ ...


1

Since $T^*T$ is Hermitian, the eigenvalues $\lambda_i$ are real. Suppose $v_i$ is an eigenvector corresponding to $\lambda_i$. Then \begin{eqnarray} \langle \lambda_i v_i, v_i\rangle =\langle T^*Tv_i, v_i\rangle= \|Tv_i\|^2\geq 0 \end{eqnarray} The eigenvalues $\lambda_i$ of $T^*T$ are real and nonnegative, and the eigenvalues of $I+T^*T$ are $1+\lambda_i$, ...


0

I will write $n'$ instead of $n++$ for brevity. Let $P(n)$ be the statement $n' \neq n$. We have the base case $P(0)$. Assume $P(n)$ is true, that is, $n \neq n'$. By an axiom, different natural numbers have different successors (i.e. $'$ is injective), hence $n' \neq (n')'$, which is precisely $P(n')$.


1

No. Let $$ f(x)=\begin{cases}x^3\sin\dfrac{1}{x} & \text{if }x\ne0,\\0 & \text{if } x=0.\end{cases} $$ Then $$ f'(x)=\begin{cases}3\,x^2\sin\dfrac{1}{x} -x\cos\dfrac1x& \text{if }x\ne0,\\0 & \text{if } x=0.\end{cases} $$ $f'$ is continuous and $f(x)=o(x^2)$, but $f'$ is not differentiable at $x=0$.


3

You have already shown that $$S = \{ z : \lvert f(z)\rvert < \delta\} \subset C = \{ z : \lvert z\rvert \leqslant r\}.$$ From that, it follows directly that $\lvert f(z)\rvert \geqslant \delta$ for all $z$ with $\lvert z\rvert > r$, and that shows that $f$ cannot have an essential singularity at $\infty$, so it is a polynomial. As a non-constant ...


1

This is not an answer but it is too long for a comment. For the function $$F(x,q) = \frac{(1 + x)^q - 1}{x^q + x}$$ André Nicolas showed that $$F(0,q)=q$$ $$F(\infty,q)=1^{+}$$ What I looked at is the problem of the derivative and here, there are a few things which are interesting as soon as $q\gt 2$. The derivative $$F'(0,q)=\frac 12 q(q-1)\gt 1$$ means ...



Top 50 recent answers are included