Using the definition of cn(u) in the previous problem and defining another elliptic function dn(

Step 1: Prove part (a) $frac{d}{d u}(operatorname{cn}(u))=-operatorname{sn}(u) operatorname{dn}(

Question

Using the definition of $\mathrm{cn}(u)$ in the previous problem and defining another elliptic function $\operatorname{dn}(u)=\sqrt{1-k^2 \sin ^2 \phi}=\sqrt{1-k^2 \operatorname{sn}^2(u)}$, prove the following: (a) $\frac{d}{d u}(\operatorname{cn}(u))=-\operatorname{sn}(u) \operatorname{dn}(u)$; (b) $\frac{d^2}{d u^2}(\operatorname{cn}(u))=\left(2 k^2-1\right) \operatorname{cn}(u)-2 k^2 \operatorname{cn}^3(u)$; (c) $\int \operatorname{cn}(u) d u=\frac{1}{k} \arccos (\operatorname{dn}(u))$.

   Using the definition of $\mathrm{cn}(u)$ in the previous problem and defining another elliptic function $\operatorname{dn}(u)=\sqrt{1-k^2 \sin ^2 \phi}=\sqrt{1-k^2 \operatorname{sn}^2(u)}$, prove the following:
(a) $\frac{d}{d u}(\operatorname{cn}(u))=-\operatorname{sn}(u) \operatorname{dn}(u)$;
(b) $\frac{d^2}{d u^2}(\operatorname{cn}(u))=\left(2 k^2-1\right) \operatorname{cn}(u)-2 k^2 \operatorname{cn}^3(u)$;
(c) $\int \operatorname{cn}(u) d u=\frac{1}{k} \arccos (\operatorname{dn}(u))$.

Computer Algebra Recipes

Richard H Enns,… 1st Edition

Chapter 4, Problem 25

Instant Answer

Step 1

- Recall that $\operatorname{cn}(u) = \cos(\phi)$ where $\phi$ is such that $u = \int_0^\phi \frac{d\theta}{\sqrt{1-k^2 \sin^2 \theta}}$. - Differentiate $\operatorname{cn}(u) = \cos(\phi)$ with respect to $u$ using the chain rule: \[ \frac{d}{du} Show more…

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Using the definition of $\mathrm{cn}(u)$ in the previous problem and defining another elliptic function $\operatorname{dn}(u)=\sqrt{1-k^2 \sin ^2 \phi}=\sqrt{1-k^2 \operatorname{sn}^2(u)}$, prove the following: (a) $\frac{d}{d u}(\operatorname{cn}(u))=-\operatorname{sn}(u) \operatorname{dn}(u)$; (b) $\frac{d^2}{d u^2}(\operatorname{cn}(u))=\left(2 k^2-1\right) \operatorname{cn}(u)-2 k^2 \operatorname{cn}^3(u)$; (c) $\int \operatorname{cn}(u) d u=\frac{1}{k} \arccos (\operatorname{dn}(u))$.

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