can i get some help with finding
the first 5 terms of taylor series of sin(2x) centered @ 0

Question

can i get some help with finding
 the first 5 terms of taylor series of sin(2x) centered @ 0

accidentalaichan · Answer

Won't be giving the answers, but hopefully this will help you find them. http://www.math.ufl.edu/~vatter/teaching/m8w10/m8l12.pdf

anonymous · Answer

$$f(x)=2^0\sin2x\
f'(x)=2^1\cos 2x\
f''(x)=-2^2\sin2x\
f'''(x)=-2^3\cos2x\
~~~~~~~~~~~\vdots$$
For $x=0$, all the even order derivatives will be zero, so let's just consider the odd ($n=2k+1$) powered ones.
$$f^{(2k+1)}(x)=(-1)^{k}2^{2k+1}\cos2x~~~~~~~~~~\text{where }k=0,1,2,...$$
To see why this formula works, you can plug in values of $k$ and see if it matches up:
$$k=0~\Rightarrow~f'(x)=2\cos2x\
k=1~\Rightarrow~f'''(x)=-2^3\cos2x\
\text{and so on...}$$
So, the formula for a Taylor series centered at 0 is given by
$$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$
Or, in terms of odd values of $n$, like above:
$$f(x)=\sum_{k=0}^\infty \frac{f^{(2k+1)}(0)}{(2k+1)!}x^{2k+1}$$
Noting that $f^{(2k+1)}(0)=(-1)^{k}2^{2k+1}$, you have
$$f(x)=\sum_{k=0}^\infty \frac{(-1)^{k}2^{2k+1}}{(2k+1)!}x^{2k+1}$$
Now you have a formula you can use to find the first five terms.

anonymous · Answer

thank you!

anonymous · Answer

You're welcome