Edwin Franks
• Member for 1 year, 1 month
• Last seen more than a month ago
• Sydney NSW, Australia

Let $f(x,y)=2(x+1)^2+((x+y)+1)^2$ and $g(x,y)=(x-1)^2+((x+y)-1)^2$. Since the second partial derivatives of $f$ and $g$ are positive and both $H_f=\begin{bmatrix} 6 & 2 \\ 2 & 2 \end{bmatrix}$ ...
Let $\{v_1,v_2,\dots\}$ be a complete orthonormal basis for $\mathcal{X}$. Let $w_1=v_1$, $w_2=\frac{1}{\sqrt 2}(v_1+v_2)$ and more generally for $k=3,4,\dots$ let w_k=\frac{1}{\sqrt k}(v_1+v_2+\... View answer Accepted answer 3 votes $$\frac1{\Delta x}\big[(A+(x+\Delta x)B)^{-1}-(A+xB)^{-1}\big]=$$ $$\frac1{\Delta x}(A+xB)^{-1}(A+xB)\big[(A+(x+\Delta x)B)^{-1}-(A+xB)^{-1}\big] (A+(x+\Delta x)B) (A+(x+\Delta x)B)^{-1}$$ =-(A+xB)^{... View answer 2 votes \begin{align*} \int_0^1\int_0^1 & (x-y)f^2(x)f(y)\mathrm{d}x\mathrm{d}y = \int_0^1\int_0^y(x-y)f^2(x)f(y)\mathrm{d}x\mathrm{d}y\ + \\ & \int_0^1\int_0^x(x-y)f^2(x)f(y)\mathrm{d}y\mathrm{d}... View answer Accepted answer 2 votes Fix s>1. For k=2,3,\dots, one has 2^k-1=3,7,15,\dots and \begin{align*} \sum_{n=1}^{2^k-1}\frac{1}{n^s} & = 1 + %% \Big(\frac{1}{2^s}+ \frac{1}{3^s}\Big) + %% \Big(\frac{1}{4^s}+... View answer Accepted answer 2 votes Let \Phi\in\mathcal{S}(\mathcal{H}) and suppose that for any unitary conjugation operator \mathcal{U}\in\mathcal{S}(\mathcal{H}), \mathcal{U}\Phi=\Phi\mathcal{U}. Let \{e_1,e_2,\dots,e_d\} be ... View answer 2 votes Let \displaystyle{F(x)=\sum_{n=1}^\infty x_n}. |F(x)|\leq \|x\|_1<\|x\|_1+\|x\|_p so that \|F\|^*\leq 1. Let u_k be defined by u_{k,n}=1/k if n\leq k and u_{k,n}=0 if n> k. Since ... View answer 1 votes Let [x],[y]\in X_{>1}/\langle3\rangle with x,y\in X_{>1} the least element of their respective equivalence classes. I claim that \label{eq:pseudometric} d([x],[y])=|3^{-i}... View answer Accepted answer 1 votes The answer I believe is no. Let V=(0,1)\times(0,1), let F be the square with corners (-1,-1),\ (2,-1),\ (2,2), and (-1,2). The minimum distance between \partial V and F is a=1. However, ... View answer 1 votes Let A and B be defined by \begin{align*} A & =\big\{a\in\mathbb{R}: a^2=\sum_{i=1}^\infty a_i10^{-2i}, a_i\in\{1,2\,\dots,9\}, a^2\not\in \mathbb{Q}\big\} \\ B & =\big\{b\in\mathbb{R}: b^... View answer 1 votes \begin{align*} %% \Big\|\sum_{k=0}^{n} \binom{n}{k}& \frac{1}{n^k} B^k-e^B\Big\| = %% \Big\|\sum_{k=0}^{n} (1-\frac1n)(1-\frac2n)\cdot(1-\frac{k-1}{n}) %% \frac{B^k}{k!} -e^B\Big\|\\ %... View answer 1 votesf'(t)=H(t-a)\int_{0}^{t-a} g(s) f(t-s)\mathrm{d}s =\int_{0}^\infty g(s)H\big((t-a)-s\big) f(t-s)\mathrm{d}s.$$Note that for t<a, f'(t)=0 and hence f(t)=f(0) for 0\leq t\leq a. The ... View answer 1 votes If f(y) = c_0 + c_2 y^2 + \int f(z) e^{f(y-z)}\mathrm{d}z differentiating three times gives f'''(y)=\int f'''(z)e^{f(y-z)}\mathrm{d}z. Thus, f'''=f'''*e^f= f'''*\underbrace{e^f \ast \cdots \ast e^... View answer Accepted answer 0 votes To show that s<\infty implies S<\infty for p>1 set a_n:=\mu\big(\{x:2^n<f(x)\leq2^{n+1}\}\big), b_n:=\mu\big(\{x:2^n<f(x)\}\big), q=p/(1-p) and M_n=\sup\{a_n\bigm| m\geq n\}... View answer 0 votes With positive consumption, the marginal utility of x_2 is negative for the first consumer, and positive for the second consumer,$$\frac{\partial u_{1}}{\partial x_2}<0 \text{ and } \frac{\... View answer 0 votes See equation (3) page 7,\eta(a)=a\nu'(a)$. Thus if$\nu$is concave,$\nu''<0$so, $$\eta'(a)=\nu'(a)+a\nu''(a)<\nu'(a)=\frac1a(a\nu'(a))=\frac{\eta(a)}{a}.$$ View answer 0 votes If$\lim_{x \to \infty}f(x) = \lim_{x \to \infty}g(x) = 0\$ and \lim_{x \to \infty} \frac{f(x)}{g(x)}=L_1 \text{ and } \lim_{y \to 0+} \frac{f^{-1}(y)}{g^{-1}(y)}=L_2\tag{1} \end{...