Question

find the taylor series for f centered at 4 if...

Answer 1

\[f ^{n}(4) = (-1)^{n}n!/3^{n}(n+1)\]

Answer 2

\[\sum_{n=0}^{\infty}\frac{f^{(n)}(4)(x-4)^n}{n!}\]

replace \(f^{(n)}(4)\)

Answer 3

replace it with what?

Answer 4

with the expression you wrote above

Answer 5

okay but how did he get that equation?

Answer 6

The definition of a taylor series is

Answer 7

okay how did he get (x-4)^n

Answer 8

ohhhh nvm got it

Answer 9

j\[\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\]

Answer 10

lol ok

Answer 11

so the answer should be: \[\sum_{0}^{\infty} (-1)^{n}(x-4)^{n}\] ?

Answer 12

actually no i messed up

Answer 13

\[\sum_{0}^{\infty} (-1)^{n}(x-4)^{n}/3 ^{n}(n+1)\]

Answer 14

assuming:
\[f ^{n}(4) = (-1)^{n}\frac{n!}{3^{n}(n+1)}\]

and given the generic from Zarkon:
\[\sum_{n=0}^{\infty}\frac{f^{(n)}(4)}{n!}(x-4)^n\]
lets combine the two of them

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

Answer 15

looks good to me