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Formula

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nd(u, k)
1.0294659946
dimensionless
dn(u, k) 0.9713773988
sn(u, k) 0.475082936

What is the Jacobi nd function?

The Jacobi elliptic functions sn, cn and dn generalize the ordinary trigonometric functions and arise from inverting the incomplete elliptic integral of the first kind. The function nd(u, k) is simply the reciprocal of the delta-amplitude function: \(\operatorname{nd}(u,k) = \frac{1}{\operatorname{dn}(u,k)}\). It depends on a real argument u and the elliptic modulus k (note this is the modulus, not the parameter \(m = k^{2}\) and not the modular angle).

Graph of the Jacobi nd function oscillating above a baseline of 1
The nd(u,k) function oscillates periodically and stays at or above 1.

How to use this calculator

Enter the argument u (any real number) and the modulus k (typically \(0 \le k \le 1\)). The calculator returns nd(u, k) to about ten significant figures, along with the intermediate values dn(u, k) and sn(u, k).

The formula explained

Setting \(m = k^{2}\), the amplitude is \(\phi = \operatorname{am}(u, k)\) and $$\operatorname{dn}(u,k) = \sqrt{1 - m\cdot\operatorname{sn}^{2}(u, k)}.$$ We evaluate sn using the arithmetic-geometric mean (AGM) and a descending Landen transformation: build sequences a, b, c with \(a_{0}=1\), \(b_{0}=\sqrt{1-m}\), \(c_{0}=k\), iterate the AGM until c is negligible, then descend the angle \(\phi = 2^{N}\cdot a_{N}\cdot u\) back down. Finally \(\operatorname{nd} = \frac{1}{\operatorname{dn}}\).

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Diagram showing nd as the reciprocal of the dn elliptic function
nd(u,k) is the reciprocal of dn(u,k); the two curves meet at the value 1.

Worked example

For \(u = 0.5\) and \(k = 0.5\) (\(m = 0.25\)): \(\operatorname{sn} \approx 0.479262\), so $$\operatorname{dn} = \sqrt{1 - 0.25\cdot 0.479262^{2}} \approx 0.970864$$ and $$\operatorname{nd} = \frac{1}{0.970864} \approx 1.0300.$$

FAQ

What happens at k = 0? \(\operatorname{dn}(u, 0) = 1\) for every u, so \(\operatorname{nd}(u, 0) = 1\) exactly.

What about k = 1? \(\operatorname{dn}(u, 1) = \operatorname{sech}(u) = \frac{1}{\cosh(u)}\), so \(\operatorname{nd}(u, 1) = \cosh(u)\).

Is nd ever undefined? For \(0 \le k < 1\), dn is bounded below by \(\sqrt{1 - k^{2}} > 0\), so nd is always finite. Only at \(k = 1\) does dn approach 0 (as \(u \to \pm\infty\)), where nd grows without bound.

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