Answer:
So, the first three nonzero terms in the Maclaurin series are:
[tex]6, \, 27x^2, \, \frac{405x^4}{4}.[/tex]
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:
[tex]\sec x=1+\frac{x^2}{2}+\frac{5x^4}{24}+...[/tex]
So, for the given function y = 6 sec(3x), we get:
[tex]y = 6 \sec(3x)=6+27x^2+\frac{405x^4}{4}+...[/tex]
So, the first three nonzero terms in the Maclaurin series are:
[tex]6, \, 27x^2, \, \frac{405x^4}{4}.[/tex]
The Maclaurin series for the function is shown below,
[tex]6sec(3x)=6+27x^{2} +\frac{405}{4}x^{4}[/tex]
The Maclaurin series for the function y = sec x is given as,
[tex]sec(x)=1+\frac{x^{2} }{2} +\frac{5}{24}x^{4}[/tex]
To find series for sec(3x), just replace x by 3x in above series.
[tex]sec(3x)=1+\frac{(3x)^{2} }{2}+\frac{5}{24}(3x)^{4} \\\\sec(3x)=1+\frac{9x^{2} }{2}+\frac{405}{24}x^{4}[/tex]
To find series for 6sec(3x), multiply 6 in above series.
[tex]6sec(3x)=6+27x^{2} +\frac{405}{4}x^{4}[/tex]
Learn more about the Maclaurin series here:
https://brainly.com/question/7846182
