Question

Derive the Maclaurin series for f(x)=sinx.

Answer 1

you want answer or method?

Answer 2

or easy way of remembering?

Answer 3

Maclaurian series is just taylor series when a = 0.

f(o) + f'(0)*x + (f''(0)*x^2)/2 + (f'''(0)*x^3)/6

Answer 4

derive is in come up with? or take the derivative of?

Answer 5

How many terms of the series do you need?

Answer 6

...my eyes!!!

Answer 7

oh no that is wrong!!!

Answer 8

The directions does not specify but I would imagine three terms
How did you go from sin(x) to the answer

Answer 9

\begin{eqnarray*}f(0) &=& \sin(0) = 0 \\ f'(0) &=& \cos(0) = 1\\ f''(0) &=& -\sin(0) = 0 \\ f^{(3)}(0) &=& - \cos(0) = -1 \\ f^{(4)}(0) &=& \sin(0) = 0\end{eqnarray*}and then it cycles over those values. So, \begin{eqnarray*}f^{(2n)}(0) &=& 0 \\ f^{(2n+1)}(0) &=& (-1)^n\end{eqnarray*}and finally\[f(x) = \sum_{n = 0}^\infty \frac{f^{(n)}(0)x^n}{n!} = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}.\]

Answer 10

looked so nice too. let me try again
\[\sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!}\]

Answer 11

Ok I see.

Answer 12

Thanks guys

Answer 13

here is how you can remember.
sine is odd, so all powers are odd
it alternates
sin(0)=0 so start with x
done

Answer 14

I will make a note of that