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Not exactly, since $x^Tx$ results in a scalar, not a vector. You could do
$\nabla x^Tx = \nabla \sum_k x_i^2 = [2x_1, 2x_2, ...]^T$ and so on.
Not exactly, since $x^Tx$ results in a scalar, not a vector. You could do
$$\nabla x^Tx = \nabla \sum_k x_i^2 = [2x_1, 2x_2, ...]^T$$ and so on.
Including: Uncaught Error: Token type math_inline_double not supported by Markdown parser
The text was updated successfully, but these errors were encountered:
inducer
changed the title
ProseMirro Markdown parse issues
ProseMirror Markdown parse issues
Oct 24, 2024
Including:
Uncaught Error: Token type
math_inline_doublenot supported by Markdown parser
The text was updated successfully, but these errors were encountered: