Question

Okay, this one is also giving me a hard time, can anyone help me with this one as well? Find the first three nonzero terms of the Maclaurin series expansion of the function sec x by finding the reciprocal of the series for cos x.

Answer 1

The Maclaurin series is a special case of a Taylor series expansion where a=0:
http://en.wikipedia.org/wiki/Taylor_series_expansion#Definition

You don't even have to do the math, wikipedia also gives the first four terms of the Taylor/Maclaurin series expansion for trig functions:
http://en.wikipedia.org/wiki/Trigonometric_functions#Series_definitions

Lets say we did want to do the math (cause its fun!), then the solution would look like:
\[\cos(x) =1+x \cos \prime (0) /2! + x^2 \cos \prime \prime (0) / 3! + ...\]
\[\cos(x) =1- x \sin\ (0) /2 - x^2 \cos \ (0) / 6 + ...\]
\[\cos(x) =1- x^2 \cos \ (0) / 6 + ...\]

The inverse would be one over that. I'll leave you to find the third non-zero term : )

Answer 2

oh wow, Thanks a lot! you're awesome. ^_^

Answer 3

Woops, I made a mistake in the math, one sec.

Answer 4

\[\cos (x) = \cos (0) + x \cos \prime (0) / 1! + x^2 \cos \prime \prime (0) / 2! + x^3 \cos \prime \prime \prime (0) / 3!\]
\[\cos (x) = 1 - x \sin\ (0) / 1 - x^2 \cos (0) / 2 + x^3 \sin\ (0) / 3!\]
\[\cos (x) = 1 - x^2 / 2 \]

There. You know what the third term is (it's on Wikipedia) you just have to solve for it. Good luck!