Use this list of basic taylor series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the taylor series for f(x) = sin2(3x) based at 0. give your answer using summation notation, write out the first three non-zero terms, and give the interval on which the series converges. (if you need to enter [infinity], use the [infinity] button in calcpad or type "infinity" in all lower-case.)

The Taylor series for , the first three non-zero terms are and the interval of convergence is

Step-by-step explanation:

These are the steps to find the Taylor series for the function

   2. The Taylor series of

Substituting y=6x we have:

   3. Find the Taylor series for

(1)

(2)

Substituting (2) in (1) we have:

Bring the factor inside the sum

Extract the term for n=0 from the sum:

To find the first three non-zero terms you need to replace n=3 into the sum

To find the interval on which the series converges you need to use the Ratio Test that says

For the power series centered at x=a

suppose that . Then

So we need to evaluate this limit:

Simplifying we have:

Next we need to evaluate the limit

-(n+1)(2n+1) is negative when n -> ∞. Therefore

You can use this infinity property when a>0 and n is even. So

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is .

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8

23

step-by-step explanation: