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Chapter 9: Problem 6

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, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=(\sin x) \sqrt{1+x}\)

Short Answer

Expert verified
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 by step solution

01

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:\[\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \ldots \]
02

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 \(x^5\).
03

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.
04

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.
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( x \).

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Most popular questions from this chapter

Recall that the Second Derivative Test for Local Extrema (Section 4.3) does not apply when \(f^{\prime \prime}(c)=0\). Prove the following generalization, which may help determine a maximum or a minimum when \(f^{\prime \prime}(c)=0 .\) Suppose that $$ f^{\prime}(c)=f^{\prime \prime}(c)=f^{\prime \prime \prime}(c)=\cdots=f^{(n)}(c)=0 $$ where \(n\) is odd and \(f^{(n+1)}(x)\) is continuous near \(c\). 1\. If \(f^{(n+1)}(c)<0\), then \(f(c)\) is a local maximum value. 2\. If \(f^{(n+1)}(c)>0\), then \(f(c)\) is a local minimum value. Test this result on \(f(x)=x^{4}\).

Suppose that Peter and Paul alternate tossing a coin for which the probability of a head is \(\frac{1}{3}\) and the probability of a tail is \(\frac{2}{3}\). If they toss until someone gets a head, and Peter goes first, what is the probability that Peter wins?

Find the Taylor series in \(x-a\) through the term \((x-a)^{3}\). \(\tan x, a=\frac{\pi}{4}\)

Find the Taylor polynomial of order 3 based at a for the given function. $$ \sec x ; a=\frac{\pi}{4} $$

. Did you ever wonder how people find the decimal expansion of \(\pi\) to a large number of places? One method depends on the following identity (see Problem 34 of Section 1.9). $$ \pi=16 \tan ^{-1}\left(\frac{1}{5}\right)-4 \tan ^{-1}\left(\frac{1}{239}\right) $$ Find the first 6 digits of \(\pi\) using this identity and the series for \(\tan ^{-1} x\). (You will need terms through \(x^{9} / 9\) for \(\tan ^{-1}\left(\frac{1}{5}\right)\), but only the first term for \(\left.\tan ^{-1}(1 / 239) .\right)\) In 1706, John Machin used this method to calculate the first 100 digits of \(\pi\), while in 1973 , Jean Guilloud and Martine Bouyer found the first 1 million digits using the related identity $$ \pi=48 \tan ^{-1}\left(\frac{1}{18}\right)+32 \tan ^{-1}\left(\frac{1}{57}\right)-20 \tan ^{-1}\left(\frac{1}{239}\right) $$ In \(1983, \pi\) was calculated to over 16 million digits by a somewha different method. Of course, computers were used in these recen calculations.
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