Question

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the function. (Enter your answers as a comma-separated list.)
y = 6 sec(3x)

Answer 1

Answer:

So, the first three nonzero terms in the Maclaurin series are:

$6, \, 27x^2, \, \frac{405x^4}{4}.$

Step-by-step explanation:

From exercise we have the next function y = 6 sec(3x).

We know that Maclaurin series for the function sec x, have the following form:

$\sec x=1+\frac{x^2}{2}+\frac{5x^4}{24}+...$

So, for the given function y = 6 sec(3x), we get:

$y = 6 \sec(3x)=6+27x^2+\frac{405x^4}{4}+...$

So, the first three nonzero terms in the Maclaurin series are:

$6, \, 27x^2, \, \frac{405x^4}{4}.$