Question

Expand \(\sinh z\) in a Taylor series about the point \(z=i \pi\).

Short answer

The Taylor series expansion of \(\sinh(z)\) about the point \(z=i \pi\) is \(-1*(z-i \pi) + 1*(z-i \pi)^3 - 1*(z-i \pi)^5 + 1*(z-i \pi)^7 - ...\)

Step-by-step solution

Evaluate function at the point of interest

Determine the value of \(\sinh(z)\) at \(z=i \pi\). The hyperbolic sine function is given by \(\sinh(z) = \frac{e^z - e^{-z}}{2}\). Hence, \(\sinh(i \pi) = \frac{e^{i \pi} - e^{-i \pi}}{2} = 0\).

Evaluate first derivative at the point of interest

Determine the first derivative of \(\sinh(z)\) at \(z=i \pi\). The derivative of \(\sinh(z)\) is \(\cosh(z)\), which is \(\frac{e^z + e^{-z}}{2}\). Hence, \(\cosh(i \pi) = \frac{e^{i \pi} + e^{-i \pi}}{2} = -1\).

Evaluate second derivative at the point of interest

Determine the second derivative of \(\sinh(z)\) at \(z=i \pi\). The derivative of \(\cosh(z)\) is again \(\sinh(z)\). Hence, \(\sinh''(i \pi) = \sinh(i \pi) = 0\).

Compute and summarize the Taylor series

Given the results from Steps 1 to 3, and knowing that the pattern of derivatives of \(\sinh(z)\) at \(z=i \pi\) will continue in the series, the Taylor series expansion of \(\sinh(z)\) around \(z=i \pi\) is: \(0 + -1*(z-i \pi) + 0*(z-i \pi)^2 + 1*(z-i \pi)^3 - ...\).

Key concepts

Hyperbolic Functions

Hyperbolic functions, similar to trigonometric functions, are a set of functions that have important applications in various branches of mathematics, including algebra, geometry, calculus, and particularly in solving problems involving hyperbolic geometry and complex analysis. Two fundamental hyperbolic functions are the hyperbolic sine ( ( ( ( ( ( ( … … … … …              …                               )