Oppenbhaimer

Testing

Usual new blog test.

the five boxing wizards jump quickly.

THE FIVE BOXING WIZARDS JUMP QUICKLY.

The five boxing wizards jump quickly.

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float Q_rsqrt( float number )
{
  long i;
  float x2, y;
  const float threehalfs = 1.5F;
 
  x2 = number * 0.5F;
  y  = number;
  i  = * ( long * ) &y;                       // evil floating point bit level hacking
  i  = 0x5f3759df - ( i >> 1 );               // what the fuck?
  y  = * ( float * ) &i;
  y  = y * ( threehalfs - ( x2 * y * y ) );   // 1st iteration
  // y  = y * ( threehalfs - ( x2 * y * y ) );   // 2nd iteration, this can be removed
 
  return y;
}

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How to integrate $\int_C \frac{\log z}{z} dz$, $C = [i, 1]$? (Taken from This MSE thread)

$$ \newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert} \begin{align} & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\, -\ \overbrace{\int_{1}^{\epsilon}{\ln\pars{x} \over x}\,\dd x} ^{\ds{\text{over}\,\,\, \pars{1,\epsilon}}}\ -\ \overbrace{\int_{0}^{\pi/2} {\ln\pars{\epsilon} + \ic\theta \over \epsilon\expo{\ic\theta}}\, \epsilon\expo{\ic\theta}\ic\,\dd\theta} ^{\ds{\text{over}\,\,\, \epsilon\expo{\ic\pars{0,\pi/2}}}}\ -\ \overbrace{\int_{\epsilon}^{1}{\ln\pars{y} + \ic\pi/2 \over \ic y}\,\ic\,\dd y} ^{\ds{\text{over}\,\,\,\pars{\ic\epsilon,\ic}}} \\[5mm] = &\ -\,{1 \over 2}\,\ln^{2}\pars{\epsilon} - \bracks{\ic\ln\pars{\epsilon}\,{\pi \over 2} - \color{#f00}{\pi^{2} \over 8}} - \bracks{-\,{1 \over 2}\,\ln^{2}\pars{\epsilon} -\ic\ln\pars{\epsilon}\,{\pi \over 2}} \,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\Large \to}\,\,\, \boxed{\pi^{2} \over 8} \end{align} $$

Nice. How do you decompose the ELBO? (Taken from Understanding Diffusion Models: A Unified Perspective)

$$ \begin{align} \log p(\boldsymbol{x}) &\geq \mathbb{E}_{q(\boldsymbol{x}_{1:T}\mid\boldsymbol{x}_0)}\left[\log \frac{p(\boldsymbol{x}_{0:T})}{q(\boldsymbol{x}_{1:T}\mid\boldsymbol{x}_0)}\right]\\ &= \begin{aligned} \underbrace{\mathbb{E}_{q(\boldsymbol{x}_{1}\mid\boldsymbol{x}_0)}\left[\log p_{\boldsymbol{\theta}}(\boldsymbol{x}_0\mid\boldsymbol{x}_1)\right]}_\text{reconstruction term} &- \underbrace{\mathcal{D}_{\text{KL}}(q(\boldsymbol{x}_T\mid\boldsymbol{x}_0) \mid\mid p(\boldsymbol{x}_T))}_\text{prior matching term} \\ &- \sum_{t=2}^{T} \underbrace{\mathbb{E}_{q(\boldsymbol{x}_{t}\mid\boldsymbol{x}_0)}\left[\mathcal{D}_{\text{KL}}(q(\boldsymbol{x}_{t-1}\mid\boldsymbol{x}_t, \boldsymbol{x}_0) \mid\mid p_{\boldsymbol{\theta}}(\boldsymbol{x}_{t-1}\mid\boldsymbol{x}_t))\right]}_\text{denoising matching term} \end{aligned} \end{align} $$

I think that’s it. Maybe a hint tag, but I’ll add that on my own. Embedding HTML works here, so iframes/embeds can be added as pleased.

Look how well that aged, huh