Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the Maclaurin polynomial of order 4 for \(f(x)\) and use it to approximate \(f(0.12) .\) $$ f(x)=\sinh x $$

Step-by-Step Solution

Answer

The approximation for \( f(0.12) \) using the 4th-order Maclaurin polynomial is approximately 0.120288.

Step 1: Understand the Maclaurin Series

The Maclaurin series for a function is a Taylor series expanded at zero. It is given by: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \]. To construct the Maclaurin polynomial of order 4 for \( f(x) = \sinh x \), we only need terms up to \( x^4 \).

Step 2: Find the Derivatives of \( \sinh x \)

We need the derivatives of \( \sinh x \) evaluated at \( x = 0 \).- \( f(x) = \sinh x \), so \( f(0) = 0 \).- First derivative: \( f'(x) = \cosh x \), so \( f'(0) = 1 \).- Second derivative: \( f''(x) = \sinh x \), so \( f''(0) = 0 \).- Third derivative: \( f'''(x) = \cosh x \), so \( f'''(0) = 1 \).- Fourth derivative: \( f^{(4)}(x) = \sinh x \), so \( f^{(4)}(0) = 0 \).

Step 3: Write the Maclaurin Polynomial

Using the derivatives, we can write the Maclaurin polynomial of order 4 as:\[P_4(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4\]Substituting the values: \[ P_4(x) = 0 + x + 0 + \frac{1}{6}x^3 + 0 \].

Step 4: Simplify the Polynomial

After simplification, the polynomial becomes: \[ P_4(x) = x + \frac{1}{6}x^3 \].

Step 5: Approximate \( f(0.12) \)

We use the polynomial to approximate \( f(0.12) \):\[P_4(0.12) = 0.12 + \frac{1}{6}(0.12)^3\]Calculate \( (0.12)^3 = 0.001728 \). Then, \[\frac{1}{6}(0.001728) = 0.000288\].Therefore, \[ P_4(0.12) \approx 0.12 + 0.000288 = 0.120288 \].

Key Concepts

Taylor Series

The Taylor series is a powerful tool in calculus that allows us to represent complex functions using infinite sums of polynomial terms. In simple terms, it breaks down complicated functions into simpler, more manageable forms. When you expand a function's Taylor series at zero, it becomes a Maclaurin series. This particular form is useful because it starts at zero, making calculations straightforward. In the Maclaurin series, every term includes derivatives of the function evaluated at zero. This series becomes incredibly useful for approximations and analytical studies, especially around the point of expansion.

• The formula for the Maclaurin series is similar to the Taylor series but centered at zero.
• Simplifies calculations for functions like exponential, trigonometric, and hyperbolic functions.
• Provides approximations when using a finite number of terms.

Derivatives

Derivatives are the backbone of calculus providing rates of change of functions. In the context of Taylor and Maclaurin series, we use them to compute each term of the series. The derivatives tell us how a function behaves and changes, which is crucial when approximating using polynomials.To construct a Maclaurin polynomial, you calculate the derivatives of the function at zero up to the necessary order. For hyperbolic functions, like the \( \sinh x \), you alternate between the function itself and its companion hyperbolic cosine \( \cosh x \).

• For \( f(x) = \sinh x \):
  • First derivative: \( \cosh x \), evaluated at \( x = 0 \), gives 1.
  • Second derivative: \( \sinh x \), evaluated at \( x = 0 \), gives 0.
  • Continuing with these, we calculate higher order derivatives as required.
• Including derivatives in the series allows capture of function's behavior around a point.

Approximation

Approximations are essential in mathematics when exact solutions are difficult or impossible to find. Maclaurin polynomials provide approximations by using a finite number of terms from their series representation.By choosing a Maclaurin polynomial of a specific order, like order 4 in this exercise, you get a polynomial that mimics the behavior of the original function around zero. This makes it easier to compute values for complex functions such as hyperbolic functions or even trigonometric functions.The example in the exercise showed that:

• With derivatives calculated and using \( P_4(x) = x + \frac{1}{6}x^3 \) , we can approximate \( f(0.12) \).
• Higher polynomial order increases accuracy but requires more computing power.

Hyperbolic Functions

Hyperbolic functions, such as \( \sinh x \) and \( \cosh x \), are analogs of the common trigonometric functions and arise often in mathematical physics. They have properties similar to sine and cosine but relate to hyperbolas instead of circles.These functions are uniquely defined:

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

The hyperbolic sine function \( \sinh x \) grows exponentially and is odd, meaning its derivative is hyperbolic cosine \( \cosh x \), which is even.In the context of Taylor and Maclaurin series, hyperbolic functions transform series into simple polynomials, allowing approximations and deeper analysis.

• Useful in areas like calculus, physics, and complex analysis.
• Acts as building blocks similar to trigonometric functions, but for different kinds of equations.