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Then $z_N$ would approximate $z$ as $N \to \infty$.
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The below example solves the case where $d = 1$, $\Omega = [0, 1]$, $p$ (`random_samples`) is the uniform distribution on $[0, 1]$, $k(x, y) = \|x - y\|^{-0.4}$ (`kernel`), and $f(x) = 1$ (`rhs`), and evaluates the solution at $x = (0.5,)$.
...return np.linalg.vector_norm(x - y, axis=-1) **-0.4
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>>>defrhs(x):
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... x0 = x[..., 0]
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...return np.ones_like(x0)
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>>> z_N = montecarlo_nystrom(
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...random_samples=random_samples,
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...kernel=kernel,
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...rhs=rhs,
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...n=100,
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...n_mean=10,
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... )
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>>> np.round(z_N(np.asarray((0.5,))), 6) # Evaluate at x=0.5
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np.float64(0.272957)
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```
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## References
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- Feppon, F., & Ammari, H. (2022). Analysis of a Monte-Carlo Nystrom Method. SIAM J. Numer. Anal. Retrieved from https://epubs.siam.org/doi/10.1137/21M1432338
description="A flake providing a dev shell for Numba with CUDA without installing Numba via nix. Also supports PyTorch yet being minimal for Numba with CUDA.";
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