Question

Find the Taylor polynomial of degree n=4 for x near the point a=pi/2 for the function cos(2x)

P4(x)=

Answer 1

i dont understand what to do with the a....

Answer 2

@agent0smith

Answer 3

http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries_files/eq0021M.gif

a is that a. the other stuff is just the value of the function at a, and it's various derivatives

It's much more complex than it looks, though.

Answer 4

so what do i do? :/

Answer 5

Find f(a) which is f(pi/2)
find the first derivative, then find f'(pi/2)
find the second... etc

down to the fourth derivative

Answer 6

eg f(a) = cos(2*pi/2)

f' = -2sin(2x)
f'(a) = -2sin(2pi/2)

Answer 7

Oh you only need the third derivative, for n=4.

Answer 8

cos(2a) = cos(2*pi/2) = cos(pi) = -1.
sin(2a) = sin(pi) = 0.

The taylor series of cos(2x) at a=2pi is:

\[\cos(\pi)-\frac{ 2\sin(\pi) }{ 1!}(x-2\pi)+\frac{ 4\cos(\pi) }{ 2! }(x-2\pi)^2-\frac{ 8\sin(\pi) }{ 3! }(x-2\pi)^3+\frac{ 16\cos(\pi) }{ 4! }(x-2\pi)^4\]

Answer 9

a=pi/2

Answer 10

\[= -1 -2(x-2\pi)^2 -\frac{ 2 }{ 3 }(x-2\pi)^4\]

Answer 11

it says its wrong:/

Answer 12

Woops, It's pi/2 instead of 2pi and the symbols are wrong. Sorry. D:

Answer 13

no its okay!

Answer 14

Give me a second, though.

Answer 15

Okay. The Taylor polynomial of degree four near point a is:
\[p(x) = f(a) + \frac{ f'(a)}{ 1! }(x-a)+\frac{ f''(a) }{ 2! }(x-a)^2+\frac{ f'''(a) }{ 3! }(x-a)^3 +\frac{ f''''(a) }{ 4! }(x-a)^4\]
So, be f(x)=cos(2x) and a =pi/2
\[f(a) = \cos(\frac{ 2\pi }{ 2 }) = \cos(\pi) = -1\]
\[f'(a) = -2\sin(2a) = -2\sin(\frac{ 2\pi }{ 2 }) = -2 \sin(\pi) = 0\]
\[f''(a)=-4\cos(2a) = -4\cos(\pi) = 4\]
\[f'''(a) = 8\sin(2a)=8\sin(\pi)=0\]
\[f''''(a)=16\cos(2a) = 16\cos(\pi)=-16\]

Answer 16

its not right

Answer 17

Substituting this values in the taylor polynomial we get:
\[p(x) = -1 + \frac{ 0 }{ 1! }(x-a)+\frac{ 4 }{ 2! }(x-a)^2+\frac{ 0 }{ 3! }(x-a)^3-\frac{ 16 }{ 4! }(x-a)^4\]
\[= -1 +2(x-a)^2-\frac{ 2 }{ 3 }(x-a)^4\]
Remembering that a = pi/2 ....
\[P(x) = -1+2(x-\frac{ \pi }{ 2 })^2-\frac{ 2 }{ 3 }(x-\frac{ \pi }{ 2 })^4\]

Answer 18

ooooh okay yeah its right! thank you:)

Answer 19

You just have to calculate f(a), f'(a), f''(a), f"'(a) and f""(a), plug those in the taylor polynomial P(x) and substitute where necessary. It's a real chore. D:
Just remember to get the positive/negative sign correctly when dividing/subs.