Recent News

Lasfierbintitoatesezoaneledownload16 🔗

By January 16, 2023No Comments

Lasfierbintitoatesezoaneledownload16 🔗




The string you posted definitely looks like binary data. The reason why it’s showing up as text is probably because you downloaded the text version of the website.
Check under help->reading options -> check mark all radiobutton
I think it is the flag – | – in the end. Make sure that is unchecked.

|x-x_0|^{ -\alpha-1}.$$ Therefore, the parameter regime where the approximation works is $$\label{eq:singapprox2}
|x-x_0| \ll \tfrac12 \min\{|x-x_0|^{\alpha},|x-x_0|^{ -\alpha-1}\}.$$ We have $|x-x_0|^{\alpha-1} \lesssim {\lVert\rho(x)\rVert}_\infty \leq \tfrac12 {\lVert\rho(x)\rVert}_2 \leq \tfrac12 {\lVert\rho(x)\rVert}_\infty$. This proves that the approximation error decays as $x \to x_0$ at fixed ${\lVert\rho(x)\rVert}_\infty$.

Now we are ready to prove the theorem. Let us assume that ${\lVert\rho(x)\rVert}_\infty \leq \gamma$ for $x \in {\mathbb{R}}$ and prove the lower bound in. Set $\beta := \alpha + 1$ and $\gamma := \tfrac12 {\lVert\rho(x)\rVert}_\infty$. If $\gamma > 1$ and $\beta = 2 \alpha$, then $|x-x_0|^\beta \gtrsim {\lVert\rho(x)\rVert}_\infty^{\alpha}$ and we are done. Otherwise, we use the approximation inequality and conclude that for $|x-x_0| \leq \tfrac12 \min\{|x-x_0|^{\alpha},|x-x_0|^{ -\alpha-1}\}$ we have $$\begin{aligned}
{\lVert\rho(x)\rVert}_2 &\gtrsim {\lVert\rho(



Author ulbhon

More posts by ulbhon