Question

Show that for $k=0$: $$u=F(\varphi ,0)=\varphi ,\mathrm{sn}u=\sin u,\mathrm{cn}u=\cos u,\mathrm{dn}u=1$$ and for $k=1$: $$\begin{array}{c}u=F(\varphi ,1)=\ln (\sec \varphi +\tan \varphi )\text{ or }\varphi =\mathrm{gd}u\text{ (Problem }19)\\ \mathrm{sn}u=\tanh u,\mathrm{cn}u=\mathrm{dn}u=\mathrm{sech}u\end{array}$$

Short answer

For k=0: u = \phi, sn(u) = \sin(u), cn(u) = \cos(u), dn(u) = 1. For k=1: u = \ln(\sec\phi + \tan\phi), sn(u) = \tanh(u), cn(u) = sech(u), dn(u) = sech(u).

Step-by-step solution

Verify for k=0

For the case where k=0, the given expressions are: 1. $u=F(\varphi ,0)=\varphi$2. $\mathrm{sn}u=\sin u$3. $\mathrm{cn}u=\cos u$4. $\mathrm{dn}u=1$Since for k=0, the Jacobi elliptic functions simplify as shown. The integral representation is trivial and shows $u=\varphi$ is valid. Additionally, the trigonometric identities align perfectly with the elliptic function definitions at k=0.

Verify for k=1 - Expression for u

When k=1, the problem asks us to confirm the given expressions:1. $u=F(\varphi ,1)=\ln (\sec \varphi +\tan \varphi )$2. $\varphi =\mathrm{gd}u$By looking up the relation between elliptic integrals for k=1, it is known that the integral corresponds to hyperbolic functions. Thus, $u=F(\varphi ,1)=\ln (\sec \varphi +\tan \varphi )$ is correct.

Verify for k=1 - Jacobi elliptic functions

Further, for k=1, the Jacobi elliptic functions have the forms:1. $\mathrm{sn}u=\tanh u$2. $\mathrm{cn}u=\mathrm{sech}u$3. $\mathrm{dn}u=\mathrm{sech}u$This can be confirmed by the definitions of Jacobi elliptic functions when k=1, where they reduce to hyperbolic functions as shown. Hence the expressions are verified to be correct.

Key concepts

Elliptic Functions

Elliptic functions are a class of complex functions that generalize the trigonometric functions. They are defined using elliptic integrals and have important applications in various fields like physics, engineering, and mathematics.
In the context of the given exercise, the Jacobi elliptic functions such as $\text{sn}u,\text{cn}u,$ and $\text{dn}u$ are of particular interest. These are functions that depend on a parameter $k$ and simplify to well-known trigonometric and hyperbolic functions for specific values of $k$.
For $\text{k=0}$, the Jacobi elliptic functions simplify as follows:

• \text{sn} u = $\text{sin}u$


• \text{cn} u = $\text{cos}u$


• \text{dn} u = 1

These results show a direct analogy with common trigonometric functions.

For $\text{k=1}$, the Jacobi elliptic functions transform into hyperbolic functions:

• \text{sn} u = $\text{tanh}u$


• \text{cn} u = $\text{sech}u$


• \text{dn} u = $\text{sech}u$

Understanding these connections helps you see how different mathematical concepts are interconnected.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for hyperbolic geometry. They are defined using exponential functions and have wide applications in many areas of mathematics and science.
The primary hyperbolic functions include $\text{sinh}u$ and $\text{cosh}u$, similar to how $\text{sin}u$ and $\text{cos}u$ are the primary trigonometric functions.
In the given problem, for $\text{k=1}$, we see the Jacobi elliptic functions relate directly to hyperbolic functions:

• \text{sn} u = $\text{tanh}u$
• \text{cn} u = $\text{sech}u$
• \text{dn} u = $\text{sech}u$

These relationships are quite insightful as they illustrate how Jacobi elliptic functions interpolate between trigonometric functions and hyperbolic functions, depending on the parameter $k$. When $\text{k=0}$, the functions reduce to trigonometric identities, while for $\text{k=1}$, they reduce to hyperbolic functions.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the included variables. These identities are useful in simplifying expressions and solving equations.

• The basic trigonometric identities are the Pythagorean identities like ${\text{sin}}^2\theta +{\text{cos}}^2\theta =1$

In the given exercise, these identities play an important role in connecting elliptic functions to simpler functions.
For instance, when $\text{k=0}$, we have:

• \text{sn} u = $\text{sin}u$
• \text{cn} u = $\text{cos}u$
• \text{dn} u = 1

These are directly related to the basic trigonometric functions and are derived using fundamental trigonometric identities. Learning these identities helps to more easily understand and solve problems involving Jacobi elliptic functions.

Integral Representation

Integral representations provide a way to express functions in terms of integrals. This approach can often simplify the evaluation and understanding of complex functions.
In the context of the given problem, the integral representation is crucial for defining the variable $u$.
For $\text{k=0}$, we have $\text{F}(\theta ,0)=\theta$, which corresponds to the integral representation:
$$u=F(\theta ,0)=\theta$$ This simplifies the problem significantly.

For $\text{k=1}$, the integral representation becomes more involved. The exercise shows that:

• \text{F}(\theta, 1) = $\text{ln}(\text{sec}\theta +\text{tan}\theta )$

This form is derived by analyzing the integral representation of elliptic functions at $\text{k=1}$, linking them to hyperbolic function forms. Understanding these integral representations is key to grasping the behavior of elliptic functions under different parameters.