Question

Find the Taylor series for f(x) = sin 3x centered at a = pi/2 you must use the definition of Taylor series.

Answer 1

any idea?

Answer 2

So by definition they just mean use this, right?\[\Large\rm \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\]

Answer 3

So if I'm interpreting this correctly, they just want us to generate a bunch of terms?

Answer 4

Oh I guess we need to condense it down to a summation after we generate some terms and see the pattern.

For n=0 we have,\[\Large\rm f^{(0)}\left(\frac{\pi}{2}\right)=\sin\left(\frac{3\pi}{2}\right)=-1\]
So our first term is,\[\Large\rm \frac{-1}{0!}\left(x-\frac{\pi}{2}\right)^2=-\left(x-\frac{\pi}{2}\right)^2\]

Answer 5

You'll need to calculate some derivatives for the other terms.
Understand the process?

Answer 6

do you mean
f(x)=sin 3x
f '(x)=3 cos 3x
f''(x)=-9 sin 3x
f'''(x)=-27 cos 3x

Answer 7

Yes.

Answer 8

So using the definition of the Taylor Series the terms generated are,\[\large\rm \frac{f\left(\frac{\pi}{2}\right)}{0!}\left(x-\frac{\pi}{2}\right)^{0}+\frac{f^{'}\left(\frac{\pi}{2}\right)}{1!}\left(x-\frac{\pi}{2}\right)^{1}+\frac{f^{''}\left(\frac{\pi}{2}\right)}{2!}\left(x-\frac{\pi}{2}\right)^{2}+...\]

Answer 9

Blah I put a square on the first term I made before.. my bad..

Answer 10

no problem thanks a lot