Problem 36 - Calculus

Question

The hyperbolic sine and cosine are differentiable and satisfy the conditions \(\cosh 0=1\) and \(\sinh 0=0,\) and $$ \frac{d}{d x}(\cosh x)=\sinh x \quad \frac{d}{d x}(\sinh x)=\cosh x $$ (a) Using only this information, find the Taylor approximation of degree \(n=8\) about \(x=0\) for \(f(x)=\) \(\cosh x\) (b) Estimate the value of \(\cosh 1\) (c) Use the result from part (a) to find a Taylor polynomial approximation of degree \(n=7\) about \(x=0\) for \(g(x)=\sinh x\)

Step-by-Step Solution

Answer

(a) \(\cosh x \approx 1 + \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^6}{720} + \frac{x^8}{40320}\). (b) \(\cosh 1 \approx 1.54308\). (c) \(\sinh x \approx x + \frac{x^3}{6} + \frac{x^5}{120} + \frac{x^7}{5040}\).

Step 1: Determine the Taylor Series Formula

To find the Taylor approximation, we use the formula for the Taylor series centered at \(x = 0\):\[ f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} x^k \]where \(f^{(k)}(0)\) represents the \(k^{th}\) derivative of \(f\) evaluated at \(x=0\).

Step 2: Compute Derivatives of cosh and sinh

Given the derivative formulas:\(\frac{d}{dx}(\cosh x) = \sinh x\) and \(\frac{d}{dx}(\sinh x) = \cosh x\),we can compute the derivatives:- \(\cosh x\): 1, 0, 1, 0, 1,...- \(\sinh x\): 0, 1, 0, 1, 0,...

Step 3: Evaluate Derivatives at x=0 for cosh

From Step 2, substituting \(x=0\), we have:- \(\cosh(0) = 1\) - \(f^{(1)}(0) = \sinh(0) = 0\)- \(f^{(2)}(0) = \cosh(0) = 1\) - Continue this pattern up to \(n=8\).

Step 4: Construct Taylor Series for cosh

Substituting into the Taylor series formula gives:\[ \cosh x \approx 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \frac{x^8}{8!} \],using only even-numbered derivatives because odd derivatives are zero at 0.

Step 5: Estimate cosh(1) Using Taylor Polynomial

Substitute \(x=1\) into the 8th-degree Taylor polynomial:\[ \cosh 1 \approx 1 + \frac{1^2}{2} + \frac{1^4}{24} + \frac{1^6}{720} + \frac{1^8}{40320} \],Calculating these:\[= 1 + 0.5 + 0.0417 + 0.0014 + 0.0000248 \approx 1.54308\]

Step 6: Evaluate Derivatives at x=0 for sinh

Using the derivatives pattern from Step 2, for \(\sinh x\), we have:- \(\sinh(0) = 0\)- \(g^{(1)}(0) = \cosh(0) = 1\)- \(g^{(2)}(0) = \sinh(0) = 0\)- Continue this pattern up to \(n=7\).

Step 7: Construct Taylor Series for sinh

Substituting into the Taylor series formula gives:\[ \sinh x \approx x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} \],since odd derivatives are non-zero at 0.

Key Concepts

Hyperbolic Functions

Hyperbolic functions, such as hyperbolic sine \(\sinh x\) and hyperbolic cosine \(\cosh x\), are analogous to the trigonometric functions sine and cosine but are defined using exponentials. Specifically, they are defined as follows:

• \(\sinh x = \frac{e^x - e^{-x}}{2}\)
• \(\cosh x = \frac{e^x + e^{-x}}{2}\)

These functions are critical in many areas of mathematics and engineering, particularly because they are solutions to certain differential equations. For example, they appear in problems related to hyperbolic geometry and the shape of a hanging cable, known as a 'catenary.'

One key characteristic of hyperbolic functions is their relationship through differentiation. The derivatives are:

• The derivative of \(\cosh x\) is \(\sinh x\).
• The derivative of \(\sinh x\) is \(\cosh x\).

Differentiation

Differentiation is a fundamental concept in calculus, which involves finding the rate at which a function changes at any given point. This rate of change can be understood as the function's slope or gradient.

For the hyperbolic functions discussed above:

• The derivative of \(\cosh x\) is \(\sinh x\), which implies that the slope of \(\cosh x\) at any given point is equal to the value of \(\sinh x\) at that point.
• Similarly, the derivative of \(\sinh x\) is \(\cosh x\), signifying that the slope of \(\sinh x\) is the value of \(\cosh x\) at that specific point.

This interrelationship between hyperbolic sine and cosine through differentiation helps simplify their analysis and study in mathematical contexts, especially when using Taylor or Maclaurin series to approximate these functions.

Differentiation facilitates the calculation of Taylor series because it provides the necessary derivatives evaluated at a point (often zero), which are used to construct polynomial approximations of functions.

Maclaurin Series

The Maclaurin series is a specific type of Taylor series centered at zero. It is used extensively to approximate functions as a power series. The formula for a Maclaurin series is:

• \(f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} x^k \)

This series provides a way to express a function as an infinite sum of its derivatives evaluated at zero, each multiplied by powers of \(x\) divided by the factorial of their order.

In the provided exercise, the goal is to develop Maclaurin series for the hyperbolic functions \(\cosh x\) and \(\sinh x\).

For \(\cosh x\), only even derivatives are non-zero at \(x=0\), leading to:

• \(\cosh x \approx 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \frac{x^8}{8!}\)

Similarly, for \(\sinh x\), only odd derivatives are non-zero, resulting in:

• \(\sinh x \approx x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}\)

Thus, Maclaurin series provide a powerful way to approximate complex functions using polynomials, which makes them easier to work with analytically and computationally.