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Chapter 16: Problem 5

\(\sqrt{4+x}\) can be written as the series a) \(2-\frac{x}{4}+\frac{x^{2}}{64}+\cdots\) c) \(2-\frac{x^{2}}{4}-\frac{x^{4}}{64}+\cdots\) b) \(2+\frac{x}{8}-\frac{x^{2}}{128}+\cdots\) d) \(2+\frac{x}{4}-\frac{x^{2}}{64}+\cdots\)

Short Answer

Expert verified
In summary, the power series expansion of the function \(\sqrt{4+x}\) is given by: \(\sqrt{4+x}= 2+\frac{x}{4}-\frac{3x^{2}}{64}+\cdots\), which corresponds to option d. We found this series by first identifying a known function, \((1+x)^{k}\), and then adjusting it to fit the given function. We then used the Taylor series formula to calculate the series expansion and compared it to the provided options.

Step by step solution

01

Identify the function to expand

To expand the given function \(\sqrt{4+x}\) as a power series, we will use the Taylor series formula.
02

Identify a known function whose Taylor series we can manipulate

Our given function is \(\sqrt{4+x}\). Notice that it closely resembles the general function \(\sqrt{1+x}\), which can be written as \((1+x)^{\frac{1}{2}}\). In this case, we have a similar function \((1+\frac{x}{4})^{\frac{1}{2}}\).
03

Use the known Taylor series for \((1+x)^{k}\)

We know that the Taylor series for the function \((1+x)^{k}\), with \(k\neq-1,0\), is given by \((1+x)^{k} = 1+ kx + \frac{k(k-1)}{2!}x^{2}+\frac{k(k-1)(k-2)}{3!}x^{3}+\cdots\) Now, substitute \(k=\frac{1}{2}\) and replace \(x\) with \(\frac{x}{4}\) in this general formula.
04

Find the Taylor series of the given function

Using the formula from step 3, we have the Taylor series for \((1+\frac{x}{4})^{\frac{1}{2}}\) as: \((1+\frac{x}{4})^{\frac{1}{2}} = 1+\frac{1}{2}\cdot\frac{x}{4}-\frac{1\cdot3}{2\cdot4\cdot4}x^{2}+\cdots\) Simplify the series: \((1+\frac{x}{4})^{\frac{1}{2}} = 1+\frac{x}{8}-\frac{3x^{2}}{128}+\cdots\) Now, add a factor of 2 outside the series to match the given function \(\sqrt{4+x}\): \(\sqrt{4+x}=2\left(1+\frac{x}{8}-\frac{3x^{2}}{128}+\cdots\right)\) Expanding the series and combining the terms, we get: \(\sqrt{4+x}= 2+\frac{x}{4}-\frac{3x^{2}}{64}+\cdots\)
05

Compare the calculated series to the given options

Comparing the calculated series to the options given in the exercise, we can see that the correct power series for \(\sqrt{4+x}\) is: \(\sqrt{4+x}= 2+\frac{x}{4}-\frac{3x^{2}}{64}+\cdots\) This matches option d.

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