Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=(\sin x) \sqrt{1+x}\)

Step-by-Step Solution

Answer

The Maclaurin series for \(f(x)\) up to \(x^5\) is \(x + \frac{x^2}{2} - \frac{5x^3}{24} + \frac{x^4}{48} + \frac{x^5}{120}\).

Step 1: Recall Known Maclaurin Series

First, recall the Maclaurin series for \( \sin x \) and \( \sqrt{1+x} \). The Maclaurin series for \( \sin x \) is given by:\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots \]The Maclaurin series for \( \sqrt{1+x} \) is given by:\[\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \ldots \]

Step 2: Calculate the Product of Two Series

To find the Maclaurin series for \( f(x) = (\sin x) \sqrt{1+x} \), you must multiply the series of \( \sin x \) and \( \sqrt{1+x} \). Expand and combine the terms as follows:\[(x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots)(1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \ldots)\]Perform the multiplication and collect like terms up to \(x^5\).

Step 3: Perform Termwise Multiplication

Focus on the terms that will give contributions up to \(x^5\):- Multiply \(x\) by \(1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16}\).- Multiply \(-\frac{x^3}{6}\) by \(1 + \frac{x}{2}\).- Multiply \(\frac{x^5}{120}\) by \(1\).Each of these should be expanded and simplified.

Step 4: Sum Like Terms

Combine terms up to \(x^5\):- From \(x(1 + \frac{x}{2} - \frac{x^2}{8})\), you get \(x + \frac{x^2}{2} - \frac{x^3}{8} + \frac{x^4}{16}\).- From \(-\frac{x^3}{6}(1 + \frac{x}{2})\), you get \(-\frac{x^3}{6} - \frac{x^4}{12}\).- From \(\frac{x^5}{120}\times1\), you get \(\frac{x^5}{120}\).Combine these results.

Step 5: Write Down the Resulting Series

Summing the contributions, we obtain:\[f(x) = x + \frac{x^2}{2} - \frac{5x^3}{24} + \frac{x^4}{48} + \frac{x^5}{120}\]So the series for \(f(x)\) up to \(x^5\) is complete.

Key Concepts

Taylor series

The Taylor series is a powerful mathematical tool used to approximate smooth functions with polynomial expressions. It is centered around a specific point, and each term of the series involves higher-power derivatives of the function.
A Taylor series takes the form:

• If centered at the point 0, the series is particularly known as the Maclaurin series.
• For a function \( f(x) \), we express the Taylor series as:\[ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots\]
• We calculate these terms using the function's derivatives at the point of expansion.
• This polynomial form allows us to approximate complex functions, keeping terms only to the desired degree for precision.

In this Maclaurin series exercise, we have expanded at 0, to provide a polynomial representation for specific positive terms.

polynomial expansion

Polynomial expansion involves expressing a function or expression as a sum of powers, usually using known series.

• Given functions like \(\sin x\) and \(\sqrt{1 + x}\), their expansions can simplify multiplication and division operations.
• The expansion allows for the exact multiplication of two series by carefully combining like terms.
• When multiplying the series \(\sin x\) with \(\sqrt{1 + x}\), each term in the multiplication contributes to the final result up to the desired degree.
• Ensuring combinations of like terms up through \(x^5\) allows for accuracy in approximations up to that degree.

For example, the function \( f(x) = (\sin x) \sqrt{1+x} \) required explicit polynomial expansion to accurately combine and sum terms.

function approximation

Function approximation is a vital concept in mathematics and engineering that involves finding simpler functions that closely resemble more complex ones. The Taylor and Maclaurin series are examples of such approximations.

• These series provide polynomial approximations of functions, expanding them into sums with increasingly higher degree terms.
• By choosing the appropriate number of terms (e.g., up to \(x^5\)), we ensure the approximation reaches a satisfactory level of precision.
• Approximating functions is useful for calculations where exact function values are hard to compute, such as in calculus or computational methods.
• The Maclaurin series helps simplify complex expressions into familiar and manageable polynomials that are easier to differentiate or integrate.

In this problem, we've encapsulated \( f(x) \) up to a polynomial of degree 5, ensuring a close approximation for small values of \( x \).