Question

Comparing Maclaurin Polynomials (a) Compare the Maclaurin polynomials of degree 4 and degree \(5,\) respectively, for the functions \(f(x)=e^{x}\) and \(g(x)=x e^{x} .\) What is the relationship between them? (b) Use the result in part (a) and the Maclaurin polynomial of degree 5 for \(f(x)=\sin x\) to find a Maclaurin polynomial of degree 6 for the function \(g(x)=x \sin x\). (c) Use the result in part (a) and the Maclaurin polynomial of degree 5 for \(f(x)=\sin x\) to find a Maclaurin polynomial of degree 4 for the function \(g(x)=(\sin x) / x\).

Short answer

The relationship between the Maclaurin polynomials of \(e^x\) and \(xe^x\) is that each term in the series for \(g(x),\) is the derivative of the corresponding term in \(f(x),\) multiplied by \(x.\) Based on this, the Maclaurin series for \(x \sin x\) to the 6th degree is \(x^2 - x^4/4! + x^6/6!\), and the series for \((\sin x) / x\) to the 4th degree is \(1 - x^2/3! + x^4/5!\).

Step-by-step solution

Find the Maclaurin polynomials

Start with expressing the given functions as their Maclaurin polynomials. Recall the series expansions for \(e^x\) and \(x e^x\). For \(f(x)=e^x\), the Maclaurin series up to the 5th degree is given by: `\(f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}\)`For \(g(x) = x e^x\), the Maclaurin series expansion can be found by taking the derivative of the expansion for \(e^x\) and then multiplying by \(x\). So we get: `\(g(x) = x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + \frac{x^5}{4!} + \frac{x^6}{5!}\)`

Identify the relationship

We can see from the formulas in Step 1 that the relation between the two series is that the Maclaurin series for \(g(x)\) is simply the derivative of the series for \(f(x)\), with each term multiplied by \(x\).

Apply the relationship to find new Maclaurin series

Given the expansion for \(f(x) = \sin(x)\), which is `\(f(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}\)` apply the derived relationship to find the series for `\(g(x) = x\sin(x)\), which gives us `\(g(x) = x^2 - \frac{x^4}{4!} + \frac{x^6}{6!}\)`.

Find the Maclaurin series for \((\sin x) / x\)

Likewise for \(g(x) = (\sin x) / x\), note that \(xg(x)\) is simply \(f(x) = \sin(x)\). So, \(g(x)\) can be gotten from \(f(x)\) by dividing every term by \(x\), giving `\(g(x) = 1 - \frac{x^2}{3!} + \frac{x^4}{5!}.\)`