Question

Find the first three nonzero terms of the Maclaurin expansion of the function; \[\large f(x)=8*\cos(x)\]

Answer 1

have you tried it?

Answer 2

sure have and I seem to fail almost immediately :(
I just don't understand where to go with it

Answer 3

so, x = 0
f(x)= 8cos(x) => f(0) = 8cos(0) = 8

f'(x) = -8sin(x) => f'(0) = -8(0 = 0

f''(x) = -8cos(x) => f''(0) = -8(1) = -8

Answer 4

have you done something like that?

Answer 5

oh shoot, is macaluarin expansion and series the same lol?

Answer 6

Ah Maclaurin is a different form of test is what I am aware of. My notes say that where a function and all its derivatives are computed where x=0, we then are dealing with a Maclaurin series :S

Answer 7

ok, so im on the right track :D

Answer 8

yaaaay :D good to hear

Answer 9

continuing with the above
f'''(x) = 8sin(x) => f'''(0) = 0

\[ f(x) = f(x) + \frac{f'(x)}{1!}x + \frac{f''(x)}{2!}x^2 + \frac{f'''(x)}{3!}x^3+...\]

Answer 10

so being nonzero terms we must use each derivative where x=0?

Answer 11

so subbing it in
\[ f(x) = 8 + \frac{0}{1!}x + \frac{-8}{2!}x^2 +\frac{0}{3!}x^3 + \frac{8}{4!}x^4+..\]
\[ f(x) = 8 + -\frac{8}{2!}x^2+ \frac{8}{4!}x^4+..\]

Answer 12

non-zero simply means to not include the terms that have a \(0\), if that make sense.

Answer 13

Well, it would be better, if you have the answer for it. Don't always trust me.

Answer 14

I understand that now :). Yes but you helped me out in understanding it! I wish I had then answers for it now, at least the solutions to see. But because I didn't know it I thought I would ask :)

Answer 15

alrighty, good luck :)

i confirmed it with wolf, and my solution is right, so dw :)
http://www.wolframalpha.com/input/?i=maclaurin+exansion+8cos%28x%29

Answer 16

I had a look, but where does it show the answer for it? :)

Answer 17

http://gyazo.com/70c331395db6c26ccd6e2d85be403d58

Answer 18

exact same answer, of what i got, but i was too lazy to simplify it, lol

Answer 19

oh haha well thank you @Mimi_x3 !! :D

Answer 20

No worries :)