tex2svg-page: global svg cache not rendering in safari #44
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I'm not able to reproduce the issue. Running produces a file that displays properly for me in Safari 14.1.2 on Mac OS X 10.14.6. Can you provide a (minimal) sample the doesn't work for you, but the source and the MathJax output? |
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I think I was able to create a minimal reproduction - I think that if the page is large enough (I copy pasted a fragment many times), safari starts rendering the SVG before the global cache is parsed (it is at the bottom of the HTML), which makes a different number of statements rendered based on how fast the parser/svg renderer are in relation to each other (different amount of rendering on each refresh). Here is the reproduction code, which was then converted with <!DOCTYPE html>
<html lang="en">
<head>
<title>Test</title>
</head>
<body>
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:Recall that the covariance between \(X\) and \(Y\) is defined \(\sigma_{xy} = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\). To simplify the math (without loss of generality), let's assume that \(X\) and \(Y\) are normalized, meaning that the data has mean 0 and variance 1. This makes \(\sigma_{xy} = \mathbb{E}[XY]\). With this, we can derive the solution to \(\beta\) with a bit of calculus, which gives:
</body>
</html>And here is a screenshot of the resulting rendered page on my iPad: I think a possible fix would be to alter the output of the renderer to put the SVG cache at the top of the page. Is this doable with the litedom adapter? |
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When running
tex2svg-pageon a set of HTML files, (both litedom and jsdom), math is rendered correctly when cache isnoneorlocal, but when the cache isglobal, latest safari on macos and on iPad (as well as chrome on iPad, which I think uses the safari webview) fail to render the SVG, leaving blank spaces.Both chrome and firefox render correctly with the global cache, it is only safari which seems to have issues.
When using the mathjax javascript with global cache on the site, instead of pre-rendering math, everything is rendered correctly, so there seems to be some difference between pre-rendered by
tex2svg-pageand rendered live on Safari.The text was updated successfully, but these errors were encountered: