Question

Use series (7a) to determine the first four nonvanishing terms of the Maclaurin series for (a) \(\sinh t=\frac{e^{t}-e^{-t}}{2}\) (b) \(\cosh t=\frac{e^{t}+e^{-t}}{2}\)

Short answer

Question: Determine the first four nonvanishing terms of the Maclaurin series for sinh(t) and cosh(t). Answer: For sinh(t), the first four nonvanishing terms are \(sinh(t) \approx t + \frac{t^3}{3!}\). For cosh(t), the first four nonvanishing terms are \(\cosh(t) \approx 1 + \frac{t^2}{2!}\).

Step-by-step solution

Find the first four derivatives of sinh(t)

To determine the first four derivatives of sinh(t), we will differentiate the function with respect to t repeatedly. 1. First derivative: \(\frac{d}{dt} \sinh(t) = \frac{d}{dt} (\frac{e^t - e^{-t}}{2}) = \frac{e^t + e^{-t}}{2} = \cosh(t)\) 2. Second derivative: \(\frac{d^2}{dt^2} \sinh(t) = \frac{d}{dt} \cosh(t) = \frac{d}{dt} (\frac{e^t + e^{-t}}{2}) = \frac{e^t - e^{-t}}{2} = \sinh(t)\) 3. Third derivative: \(\frac{d^3}{dt^3} \sinh(t) = \frac{d}{dt} \sinh(t) = \cosh(t)\) 4. Fourth derivative: \(\frac{d^4}{dt^4} \sinh(t) = \frac{d}{dt} \cosh(t) = \sinh(t)\)

Evaluate the derivatives at 0

Now, we will evaluate these derivatives at t=0. 1. \(\sinh(0) = \frac{e^0 - e^{-0}}{2} = \frac{1 - 1}{2} = 0\) 2. \(\cosh(0) = \frac{e^0 + e^{-0}}{2} = \frac{1 + 1}{2} = 1\) 3. \(\cosh(0) = 1\) 4. \(\sinh(0) = 0\)

Write the Maclaurin series for sinh(t)

Using the evaluated derivatives and the following formula for the Maclaurin series: \(f(t) \approx f(0) + f'(0) * t + \frac{f''(0)}{2!} * t^2 + \frac{f'''(0)}{3!} * t^3 + ...\) The Maclaurin series for sinh(t) is: \(\sinh(t) \approx 0 + 1 * t + \frac{0}{2!} * t^2 + \frac{1}{3!} * t^3 + ...\) First four nonvanishing terms: \(\sinh(t) \approx t + \frac{t^3}{3!}\) #Phase2: Cosh(t)#

Find the first four derivatives of cosh(t)

We have already found the first four derivatives of cosh(t) while finding derivatives for sinh(t). We can use those here: 1. First derivative: \(\cosh(t)\) 2. Second derivative: \(\sinh(t)\) 3. Third derivative: \(\cosh(t)\) 4. Fourth derivative: \(\sinh(t)\)

Evaluate the derivatives at 0

We have already evaluated these derivatives at t=0 in Phase1: 1. \(\cosh(0) = 1\) 2. \(\sinh(0) = 0\) 3. \(\cosh(0) = 1\) 4. \(\sinh(0) = 0\)

Write the Maclaurin series for cosh(t)

Using the evaluated derivatives and the formula for the Maclaurin series: \(f(t) \approx f(0) + f'(0) * t + \frac{f''(0)}{2!} * t^2 + \frac{f'''(0)}{3!} * t^3 + ...\) The Maclaurin series for cosh(t) is: \(\cosh(t) \approx 1 + 0 * t + \frac{1}{2!} * t^2 + \frac{0}{3!} * t^3 + ...\) First four nonvanishing terms: \(\cosh(t) \approx 1 + \frac{t^2}{2!}\)

Key concepts

Differential Equations

Differential equations play a crucial role in mathematics and various scientific fields, representing relationships between functions and their derivatives. Think of them as mathematical sentences that describe how quantities change.

For example, in the context of Maclaurin series and the exercise provided, we see differential equations in action when finding the derivatives of hyperbolic functions like \(\text{sinh}(t)\) and \(\text{cosh}(t)\). The derivatives of these functions yield new functions that themselves can be differentiated, indicating a relationship that differential equations are designed to represent and resolve. By solving these equations, we're able to discover patterns that lead to the formulation of power series expansions.

Hyperbolic Functions

Hyperbolic functions, including \(\text{sinh}(t)\) and \(\text{cosh}(t)\), are analogs of the sine and cosine functions but for a hyperbolic geometry. They are defined using exponential functions, as seen in the formulas \(\text{sinh}(t) = \frac{e^t - e^{-t}}{2}\) and \(\text{cosh}(t) = \frac{e^t + e^{-t}}{2}\).

These functions share many properties with their circular counterparts, like periodicity and specific identities. Understanding hyperbolic functions is important for the study of linear differential equations and the modeling of real-world phenomena such as wave propagation, electric circuits, and fluid dynamics.

Power Series Expansion

Power series expansion is a technique in calculus for expressing a function as an infinite sum of terms. Each term is a constant multiplied by a power of the variable. For instance, the Maclaurin series, a type of power series, represents a function as a sum of its derivatives at zero.

The form of the Maclaurin series is given by: \[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots\]Power series expansion is an invaluable tool for simplifying complex functions, approximating non-polynomial functions, and solving differential equations.

Calculus

Calculus is the branch of mathematics dealing with continuous change. It's divided primarily into two fields: differential and integral calculus. Differential calculus concerns calculating instantaneous rates of change (derivatives), while integral calculus focuses on finding the total size or value, such as areas under curves.

In the example exercise, differential calculus is used to find the derivatives of hyperbolic functions to build the Maclaurin series, showcasing the interplay between different areas of calculus. Calculus concepts are essential for solving real-life problems in engineering, physics, economics, and beyond.