We want to find the Maclaurin series
for f of x equals five, cosine three x squared,
which can be written in this form here,
and then find the first few coefficients
c sub zero, c sub two, through c sub eight.
Because we're looking for a Maclaurin series
we know the series is centered at c equals zero.
One way to find the Maclaurin series
would be to use this formula here,
where we have the summation from n equals zero to infinity
of the nth derivative of f,
evaluated at zero divided by n factorial
times x to the nth.
In expanded form, it would look like this.
But notice how to find the first derivative
we'd have apply the chain rule,
and then to find a higher order derivatives
we'd have to apply the product rule and the chain rule.
So instead of doing all this extra work,
we'll build a Maclaurin series
using the known Maclaurin series for cosine x given here.
Most texts do provide
a list of power series
for elementary functions similar to the one given here.
So again, in our example we using this power series
to build a Maclaurin series.
Looking at the expansion of the power series for cosine x,
notice how we could write this as a summation
of n equals zero to infinity of negative one to the nth
since the terms alternate sine
times x to the power of two n,
because the exponents are even,
divided by two n factorial.
So to find the Maclaurin series for our function,
we can substitute three x squared for x
then multiply by five.
So five
cosine three x squared
would be equal to five
times, substituting three x squared for x,
we would have one minus, instead of x squared,
we'd have three x squared squared
divided by two factorial
plus, instead of x to the fourth, we'd have three x squared
to the fourth divided by four factorial
minus, instead of x to the sixth, we'd have three x squared
to the sixth, divided by six factorial and so on.
Now notice here, if we distribute five,
we'd actually find c sub zero through c sub eight,
where c sub eight would be the coefficient
of the [mumbles] term.
And notice how this term here would have to [mumbles].
So let's go ahead and find c sub zero through c sub eight.
And then we'll come back and write this Maclaurin series
in a form similar to this form to this form here
using summation notation.
If we distribute five, we would have five minus
five times three x squared squared
over two factorial, which is two
plus five times three x squared
to the fourth divided by four factorial, which is 24,
and so on.
We can stop here again
because this term has to [mumbles] eight.`
So we'd have five minus, this would be five times
nine x to the fourth divided by two.
It'd be 45 halves x to the fourth
plus, here we have five times three x squared to the fourth.
That'd be five times 81, that's 405,
divided by 24.
Well 405 divided by 24,
common factor of three here.
That simplifies to 135 divided by eight.
So we have plus 135
eighths x to the eighth, and so on,
but now we can find the coefficients that we need.
C sub zero is the coefficient of the degree zero term
and the constant term, which is five.
C sub two is the coefficient of the degree two term.
There is no degree two term, so it's zero.
C sub four would be negative 45 halves.
C sub six would be zero, since there's no degree six term.
And c sub eight would be 135
divided by eight.
So now we have found the coefficients,
but let's take the time to write the Maclaurin series
using summation notation similar to this form here.
So this Maclaurin series would be equal to
five times the summation
from n equals zero to infinity of
negative one to the nth,
and then instead of x to the power of two n
we would have 3 x squared to the power of two n
divided by two n factorial.
Now let's go ahead and make this fit this form here.
So we'll bring the factor of five in.
So we would have f of x, which equals five cosine
three x squared equals the summation from n equals
zero to infinity of
negative one to the nth times five
times three x squared to the power of two n
divided by two n factorial.
I hope you found this helpful.