Question

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that R(x) → 0.]

4cos(x) at a= 3pi

Answer 1

you should start by creating a list of the derivatives of f(x)

Answer 2

i got the series as this:

\[-4 + 0 (x-3 pi) + \frac{4}{2!}(x-3pi)^2 + 0(x-3pi)^3 - \frac{4}{2!}(x-3pi)^4+....\]

Answer 3

derivatives:

-4sinx
-4cosx
4sinx
4cosx

Answer 4

lets see if that works :) recall that 3pi is the same as pi

cos(pi) = -1 ; sin(pi) = 0

cos
-sin
-cos
sin
cos
-sin
-cos
...

we can ignore all the sin bits and the cos parts swap signs

-4, 4. -4. 4. -4. 4 are the coeefs we are looking for

Answer 5

lets plug those into the standard set up for the taylor:

\[T_s=c_0\frac{(x-a)^0}{0!}+c_1\frac{(x-a)^1}{1!}+c_2\frac{(x-a)^2}{2!}+c_3\frac{(x-a)^3}{3!}+...\]

\[T_s=-4\frac{(x-3\pi)^0}{0!}+4\frac{(x-3\pi)^1}{1!}-4\frac{(x-3\pi)^2}{2!}+4\frac{(x-3\pi)^3}{3!}+...\]

Answer 6

we can simply that with summation notation

Answer 7

\[T_s=\sum_{0}^{inf}\frac{(-4)^{(n+1)}(x-3pi)^n}{n!}\]

Answer 8

ahh ok. Thank you!

Answer 9

i see an error

Answer 10

i forgot to modify my exponents when I deleted the sins

Answer 11

I was going to say, you want even powers of n...

Answer 12

make the the bottom 2n! and that x^(2n)

Answer 13

:) good catch phi

Answer 14

tyler looks good except for a typo on the last term 4! (not 2!)

Answer 15

Ohh ok, Thank you both!!

Answer 16

Getting the correct sign seems tricky

Answer 17

Yeah in my first answer i had (-1)^n and not n+1. Thanks for the help

Answer 18

tweak amistre's answer, the 4 does not belong inside the exponent (n+1)