Question

Find the Taylor series for \[f(x)=sinx\] centered at \[a=\frac{\pi}{2}\]
\[\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}\]

Answer 1

I've figured out what the sum is

Answer 2

are u from socrates's class?

Answer 3

who's that?

Answer 4

nvm

Answer 5

i have exam 5 on this tmr, and final on the day after that =]

Answer 6

oh I see.

Answer 7

Oh I think what I wrote is a maclaurin series

Answer 8

i don get this section of the chapter either

Answer 9

I have to write it from scratch every time...

Answer 10

oh I messed up, the factorials should be 1 less, I skipped n=1 factorial...\[f(x)=\sum_{n=0}^\infty{f^{(n)}(a)\over n!}(x-a)^n\]\[=\frac1{1!}(x-\frac\pi2)-\frac1{3!}(x-\frac\pi2)^3+\frac1{5!}(x-\frac\pi2)^5-...\]

Answer 11

so where does the pi go ?
(no pun intended :P)

Answer 12

give me a second to look it this...I'm kinda new and slow at this.

Answer 13

oh no I've messed up terribly, this is all wrong...
I should probably have just slept :P
lol

Answer 14

No it's not that. I guess the x and a's and n confuse me.

Answer 15

The formula for the Taylor expansion around \(x=a\) is\[f(x)=\sum_{n=0}^\infty{f^{(n)}(a)\over n!}(x-a)^n\]in your case \(a=\frac\pi2\)
let's go check each term:\[\{a_0\}={f^{(0)}(\frac\pi2)\over0
!}(x-\frac\pi2)^0=\sin\frac\pi2=1\]\[\{a_1\}=\frac1{1!}\cancel{\cos\frac\pi2}^{\huge0}(x-\frac\pi2)^1=0\]\[\{a_2\}=-\frac1{2!}\sin\frac\pi2(x-\frac\pi2)^2=-\frac1{2!}(x-\frac\pi2)^2\]\[\{a_3\}=-\frac1{3!}\cancel{\cos\frac\pi2}^{\huge0}(x-\frac\pi2)^3=0\]so the pattern is that all odd n terms stay

Answer 16

Oh darn! I get it now. For a second I forgot what \[f^{(0)}\] and \[f^{(1)}\] meant. Durrr
so the zero derivative is sin(pi/2) and \[(x-\frac{\pi}{2})^0 =1\]
makes sense

Answer 17

I get the first line....now on to the second line

Answer 18

one sec, I have to deal with a potentially problematic user...sorry brb

Answer 19

all odd n terms are 0 though

Answer 20

right I said it backwards :/
sorry I'm pretty tired I guess

Answer 21

no worries :P

Answer 22

\[\{a_1\}={f^{(1)}(\frac\pi2)\over1!}(x-\frac\pi2)^1\]and\[f(x)=\sin x\implies f'(x)=\cos x\implies f'(\frac\pi2)=0\]so\[\{a_1\}={1\over1!}(0)(x-\frac\pi2)^1=0\]

Answer 23

yep that makes sense

Answer 24

\[\{a_2\}={f^{(2)}(\frac\pi2)\over2!}(x-\frac\pi2)^2\]and\[f'(x)=\cos x\implies f''(x)=-\sin x\implies f''(\frac\pi2)=-1\]so\[\{a_2\}={1\over2!}(-1)(x-\frac\pi2)^1=-{1\over2!}(x-\frac\pi2)\]

Answer 25

next one will be zero...

Answer 26

*another typo, the last one should be\[\{a_2\}=-\frac1{2!}(x-\frac\pi2)^2\](I forgot the n=2 in the exponent on the parentheses)

Answer 27

\[\{a_4\}={f^{(4)}(\frac\pi2)\over4!}(x-\frac\pi2)^4\]and\[f(x)=\sin x\implies f^{(4)}(x)=\sin x\implies f^{(4)}(\frac\pi2)=1\]so\[\{a_4\}={1\over4!}(1)(x-\frac\pi2)^4={1\over4!}(x-\frac\pi2)^4\]by now a pattern should be emerging

Answer 28

yep I see the pattern now.
Thanks @TuringTest

Answer 29

welcome :)

Answer 30

my suggestion ... keep it easy as much as possible
http://www.wolframalpha.com/input/?i=expand+sin%28x%29+at+pi%2F2&dataset=&equal=Submit
http://www.wolframalpha.com/input/?i=expand+cos+x+at+0

Answer 31

let x = u + pi/2, u=0 ... it is equivalent to expansion of cos u at u=0, change back u = x - pi/2

Answer 32

I won't be able to use wolfram on my final exam though

Answer 33

yeah I realized that @experimentX but I though it good practice to do it manually first time around
thanks for reminding me to point that out though, I had almost forgotten

Answer 34

yep ... math hacks!!