Question

Expand the following function using Taylor Series about a = pi / 4
\[\cos 2x\]

Answer 1

cos 2x comes where?

Answer 2

f(x) = cos 2x. expand using taylor series about a = pi / 4

Answer 3

cos(x) = 1/√2 - 1/√2(x-π/4) - 1/√2(x-π/4)^2/2 + 1/√2(x-π/4)^3/6 + 1/√2(x-π/4)^4/24 + .

Answer 4

can you use the equation thing. i cant make up the answer.

Answer 5

i know a cos(x) but im not sure where the 2x comes into play

Answer 6

cos 1
- sin 0
- cos 1

would suggest

1 -(x-pi/4)^2 ... even powers to me alternating plus minuses divided by the exponents factorial

but then again there is that "2" to place

Answer 7

hint: evaluate the derivatives

Answer 8

hmm, so use 2x in the derivatives ...

Answer 9

cos (2x)
-2sin (2x)
-4cos(2x)
-6 cos(2x) like that

Answer 10

2(pi/4) = pi/2 so all these are the 90 degrees

Answer 11

which makes sin the dominant one if im reading it right

Answer 12

yes youre getting there

Answer 13

if I were to take a gander, and chk it with the wolf:

\[cos(2x)~-\frac{2}{1!}(x-\frac{pi}{4})+\frac{6}{3!}(x-\frac{pi}{4})^3-\frac{10}{5!}(x-\frac{pi}{4})^5+\frac{14}{7!}(x-\frac{pi}{4})^7 ...\]

Answer 14

it ate the = sign :)

Answer 15

Theres something wrong but its nearly done

Answer 16

http://www.wolframalpha.com/input/?i=y%3Dcos%282x%29+and+y%3D-2%28x-pi%2F4%29%5E1%2B6%28x-pi%2F4%29%5E3%2F3%21+-10%28x-pi%2F4%29%5E5%2F5%21+%2B14%28x-pi%2F4%29%5E7%2F7%21-18%28x-pi%2F4%29%5E9%2F9%21+%2B22%28x-pi%2F4%29%5E11%2F11%21+-26%28x-pi%2F4%29%5E13%2F13%21

its taking shape i think :)

Answer 17

\[\sum_{n=1}^{\infty} (-1)^{n}\frac{4n-2}{(2n-1)!}(x-\frac{pi}{4})^{2n-1}\]
maybe :) but thats all i got

Answer 18

Its actually:\[\sum_{n = 0}^{\infty}\frac{(-1)^{n}2^{2n + 1}(x - \frac{\pi}{4})^{2n + 1}}{(2n + 1)!}\]

Answer 19

or\[\cos 2x = -\frac{2}{1!}(x - \frac{\pi}{4}) + \frac{2^{3}}{3!}(x - \frac{\pi}{4})^{3} - ... \]

Answer 20

2*2*2*2 ..... yeah, sooo close :)

Answer 21

you might want to try the puzzle i posted. Not related to calculus though.

Answer 22

http://openstudy.com/study#/updates/4f390ddae4b0fc0c1a0df2df

Answer 23

find Taylor expansion of X-1/X+1 , at Xo = 0