a. Find the first four nonzero terms of the Maclaurin series for the given function.
b. Write the power series using summation notation.
c. Determine the interval of convergence of the series.
\[
f(x)=e^{-2 x}
\]
Answer
The interval of convergence for the series is \(-\infty < x < \infty\), because the series for \(e^x\) converges for all real numbers \(x\), and the substitution \(-2x\) for \(x\) does not change the interval of convergence.
Step 1 ：The Maclaurin series for the function \(e^x\) is given by \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\).
Step 2 ：Substitute \(-2x\) for \(x\) in the Maclaurin series for \(e^x\) to get the Maclaurin series for \(e^{-2x}\), which is \(\sum_{n=0}^{\infty} \frac{(-2x)^n}{n!}\).
Step 3 ：The first four nonzero terms of the Maclaurin series for \(e^{-2x}\) are \(1, -2x, 2x^2, -\frac{4}{3}x^3\).
Step 4 ：The power series in summation notation is \(\sum_{n=0}^{\infty} \frac{(-2x)^n}{n!}\).
Step 5 ：The interval of convergence for the series is \(-\infty < x < \infty\), because the series for \(e^x\) converges for all real numbers \(x\), and the substitution \(-2x\) for \(x\) does not change the interval of convergence.
