Public
Edited
Dec 24
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sum = (f) => (n) => {
let sum = 0;
for (let k = 0; k <= n; k++) sum += f(k);
return sum;
}
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A = sum((k) => (4 * (k % 2 ? -1 : 1)) / (2 * k + 1))
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//S = (A) => (n) =>
// (A(n + 1) * A(n - 1) - A(n) ** 2) / (A(n + 1) - 2 * A(n) + A(n - 1))

// An exactly equivalent and more numerically stable form:
S = (A) => (n) =>
A(n + 1) - (A(n + 1) - A(n)) ** 2 / (A(n + 1) - 2 * A(n) + A(n - 1))
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Shanks = (A, n = 1) => [...Array(n).keys()].reduce(S, A)
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B = sum((k) => k * Math.pow(-1, k - 1))
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B(10000)
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Shanks(B, 4)(4)
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C = sum((k) => Math.pow(-2, k))
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C(100)
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Shanks(C)(1)
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