Question

Taylor series of Cos(6x) in summation notation?

Answer 1

I keep getting alternating zeroes

Answer 2

When plugging in
\[f^n(0)=\cos(6x)\]

\[f(0)=1 |_{n=0}\]
\[f(0)=0|_{n=1}\]
\[f(0)=-6^2|_{n=2}\]
\[f(0)=0|_{n=3}\]
\[f(0)=6^4|_{n=4}\]

Sorry if my notation is weird.
Im just not sure how to write this in summation notation

Answer 3

@TheSmartOne

Answer 4

Im assuming the pattern would continue like this.

Answer 5

One of my answers has \[\sum_{k=0}^{\frac{ n }{ 2 }}\] as Sigma.
I have a feeling this might take care of the alternating zeroes, but I cant tell for sure

Answer 6

\(\Large f^{(n)}(x)\) is the nth derivative of f(x), ie, how many derivatives you apply to f(x).


\[\Large f^{(0)}(x) = f(x) = \cos(6x)\]
\[\Large f^{(1)}(x) = \frac{d}{dx}\left[f(x)\right] = -6\sin(6x)\]
\[\Large f^{(2)}(x) = \frac{d}{dx}\left[f^{(1)}(x)\right] = -36\cos(6x)\]
\[\Large f^{(3)}(x) = \frac{d}{dx}\left[f^{(2)}(x)\right] = 216\sin(6x)\]
\[\Large f^{(4)}(x) = \frac{d}{dx}\left[f^{(3)}(x)\right] =1296\cos(6x)\]


Hopefully you see a pattern emerging?

Answer 7

But dont I need to plug 0 in for x?

Answer 8

yes, that will generate the coefficients (well part of them anyway)

Answer 9

so the pattern is 6^n

Answer 10

that's part of it

Answer 11

just do Cos(x) and then replace x by 6x