Question

The hyperbolic functions cosh \(z\) and \(\sinh z\) are defined as follows: $$\cosh z=\frac{e^{z}+e^{-z}}{2} \quad \text { and } \quad \sinh z=\frac{e^{z}-e^{-z}}{2}$$ for any \(z,\) real or complex. (a) Sketch the behavior of both functions over a suitable range of real values of \(z\). (b) Show that \(\cosh z=\cos (i z)\). What is the corresponding relation for \(\sinh z ?\) (c) What are the derivatives of cosh \(z\) and \(\sinh z ?\) What about their integrals? ( \(\mathbf{d}\) ) Show that \(\cosh ^{2} z-\sinh ^{2} z=1\) (e) Show that \(\int d x / \sqrt{1+x^{2}}=\operatorname{arcsinh} x\). [Hint: One way to do this is to make the substitution \(x=\sinh z .]\)

Short answer

(b) \( \sinh z \) corresponds to \( i \sin z \); (c) Derivatives: \( \cosh' = \sinh \), \( \sinh' = \cosh \); Integrals: \( \int \cosh = \sinh + C \), \( \int \sinh = \cosh + C \); (d) \( \cosh^2 - \sinh^2 = 1 \); (e) \( \int dx/\sqrt{1+x^2} = \mathrm{arcsinh } x \).

Step-by-step solution

Sketch the Functions

To sketch the functions \( \cosh z \) and \( \sinh z \), plot them over a real range, for example, \( z \) from \(-3\) to \(3\):- \( \cosh z \) is an even function, symmetric around the y-axis, resembling a "U" shape starting from 1 when \( z = 0 \). Endpoints rise exponentially as \( z \) increases or decreases. \( \sinh z \) is an odd function, passing through the origin, it resembles a reflected exponential graph that increases positively in positive z, and negatively in negative z.

Show Identity for \( \cosh z \)

Given the identity \( \cos(i z) = \cosh z \), substitute \( i z \) in \( \cos x = \frac{e^{ix} + e^{-ix}}{2} \),\( \cos(i z) = \frac{e^{-z} + e^{z}}{2} \). This matches \( \cosh z = \frac{e^{z} + e^{-z}}{2} \). Both expressions are equal, proving the identity.

Identity for \( \sinh z \)

To find \( \sinh(i z) \), substitute \( x = i z \) in \( \sin x = \frac{e^{ix} - e^{-ix}}{2i} \), yielding \( \sin(i z) = \frac{e^{-z} - e^{z}}{2i} = -i \cdot \sinh z \). Thus, \( \sinh(i z) = i \sin z \).

Derivatives and Integrals

The derivative of \( \, \cosh z \) is \( \sinh z \), because differentiating \( \frac{e^{z} + e^{-z}}{2} \) results in \( \frac{e^{z} - e^{-z}}{2} = \sinh z \). The derivative of \( \sinh z \) is \( \cosh z \) since differentiating \( \frac{e^{z} - e^{-z}}{2} \) results in \( \frac{e^{z} + e^{-z}}{2} = \cosh z \). For integrals, \( \int \cosh z \, dz = \sinh z + C \) and \( \int \sinh z \, dz = \cosh z + C \).

Prove Hyperbolic Identity

Use the expressions: \( \cosh^2 z - \sinh^2 z = \left(\frac{e^{z} + e^{-z}}{2}\right)^2 - \left(\frac{e^{z} - e^{-z}}{2}\right)^2 \). Simplify to \( \frac{(e^{2z} + 2 + e^{-2z}) - (e^{2z} - 2 + e^{-2z})}{4} = \frac{4}{4} = 1 \). This proves the identity.

Evaluate Integral

Use the substitution \( x = \sinh z \), hence, \( dx = \cosh z \, dz \). The integral becomes \( \int d x / \sqrt{1+x^{2}} = \int d(\sinh z) / \sqrt{\cosh^2 z} \) which simplifies to \( \int dz = z + C \). Since \( \sinh^{-1} x = z \), the integral equals \( \operatorname{arcsinh} x + C \).

Key concepts

Cosh and Sinh Identities

Hyperbolic functions, such as cosh and sinh, are similar to trigonometric functions like cosine and sine, but they differ in significant ways due to their foundation in hyperbolas rather than circles. Hyperbolic cosine, or \( \cosh z \), is defined as \( \cosh z = \frac{e^z + e^{-z}}{2} \), and hyperbolic sine, or \( \sinh z \), is defined as \( \sinh z = \frac{e^z - e^{-z}}{2} \). These identities are integral in showing how hyperbolic functions can be used to mimic complex trigonometric identities. By understanding how these functions are constructed through exponential expressions, you can better grasp their behavior. For instance, these expressions reveal that \( \cosh z \) is inherently an even function, meaning it behaves the same for \( z \) and \( -z \). Conversely, \( \sinh z \) is an odd function, illustrating symmetry about the origin.

Moreover, an interesting identity can be drawn where \( \cosh z \) relates to trigonometric functions: notably, \( \cosh z \) can be shown to equal to \( \cos(iz) \). This comes from substituting \( iz \) into the general trigonometric cosine function. Likewise, the function \( \sinh z \) has a complex relationship with the trigonometric sine function. Through these identities, hyperbolic functions demonstrate how closely linked they are with circular trigonometric functions, yet offer unique benefits, particularly in solving differential equations and modeling real-world problems such as catenary curves in physics.

Differentiation and Integration of Hyperbolic Functions

Both differentiation and integration in calculus are essential tools when working with hyperbolic functions. The derivatives and integrals of hyperbolic functions mimic those of their trigonometric counterparts, but they have their own elegant rules and patterns.

For derivatives:

• The derivative of \( \cosh z \) is \( \sinh z \). Differentiating the definition of \( \cosh z = \frac{e^z + e^{-z}}{2} \) gives us \( \frac{e^z - e^{-z}}{2} \), which is \( \sinh z \).
• Similarly, the derivative of \( \sinh z \) is \( \cosh z \). By differentiating \( \sinh z = \frac{e^z - e^{-z}}{2} \), we obtain \( \frac{e^z + e^{-z}}{2} \), the definition of \( \cosh z \).

For integrals:

• The integral of \( \cosh z \) with respect to \( z \) is \( \sinh z + C \).
• The integral of \( \sinh z \) with respect to \( z \) is \( \cosh z + C \).

These differentiations and integrations demonstrate their simplicity and beauty. They also highlight a surprising symmetry: whereas integration and differentiation often involve more complex functions, hyperbolic functions maintain an almost effortless balance. Learning these derivatives and integrals can aid you significantly, especially when encountering more complex integrals in calculus and higher-level applied math.

Hyperbolic Function Relationships

Hyperbolic functions have relationships that are crucial to both their theoretical development and application in solving real-world problems. One of the foundation stones of hyperbolic function relationships is their parallels to classic trigonometric identities, but with key differences.

One of the most vital identities is \( \cosh^2 z - \sinh^2 z = 1 \). This equation resembles the Pythagorean identity from trigonometry, \( \cos^2 x + \sin^2 x = 1 \). This identity can be derived from the definitions of \( \cosh z \) and \( \sinh z \) by using algebraic manipulation:

• Expand each square: \( (\cosh z)^2 = \left(\frac{e^z + e^{-z}}{2}\right)^2 \) and \( (\sinh z)^2 = \left(\frac{e^z - e^{-z}}{2}\right)^2 \).
• By subtracting these squared terms, the resulting simplification shows that the exponential terms cancel out, ultimately equating this identity to one.

This identity is not only mathematically beautiful but also practically beneficial, often used in fields such as physics and engineering, especially when modeling hyperbolic structures or processes.

Moreover, these relationships enable substitutions that simplify integration, such as the integral \( \int \frac{dx}{\sqrt{1 + x^2}} = \operatorname{arcsinh} x \). By substituting \( x = \sinh z \), the calculation benefits from the identity above, leading to a straightforward evaluation. Understanding these relationships enhances comprehension and solves complex integrals and differential equations effectively.