Question

Find the Taylor series for \[f(x)= \frac 1x; a=-3\]
\[f(x)=\sum_{n=0}^{\infty} \frac{f^n (-3)}{n!} x^n= \frac{1}{-3}-\frac{\frac 19}{1!}x-\frac{\frac{1}{27}}{2!}x^2-\frac{\frac{1}{81}}{3!}x^-\frac{\frac{1}{243}}{4!}x^4-.....\]

how did I do?

Answer 1

No.

Answer 2

Whaaaaaatt?

Answer 3

:(

Answer 4

<tilting head>

Answer 5

\[
f(x)=\sum_{n=0}^{\infty} \frac{f^n (-3)}{n!} (x+3)^n= \frac{1}{-3}-\frac{\frac 19}{1!}(x+3)-\frac{\frac{1}{27}}{2!}(x+3)^2\cdots
\]

Answer 6

where did you find the (x-3)?

Answer 7

This the definition of the Taylor Series as a =-3
\[
f(x) =\sum_{n=0}^\infty\frac {f^{n}(a)}{n!}(x-a)^n
\]

Answer 8

was I doing the MacLaurin series at (-3) ?

Answer 9

sorry I'm new to Maclaurin, Taylor, and Co.

Answer 10

Ok so if it is not centered at around zero I use the Taylor series?

Answer 11

Well the question asked for Taylor series...I'm soo silly

Answer 12

\[
-\frac{1}{3}-\frac{x+3}{9}-\frac{1}{27}
(x+3)^2-\frac{1}{81} (x+3)^3-\frac{1}{243}
(x+3)^4-\frac{1}{729}
(x+3)^5-\frac{(x+3)^6}{2187}-\\\frac{(x+3)^7
}{6561}-\frac{(x+3)^8}{19683}-\frac{(x+3)^
9}{59049}-\frac{(x+3)^{10}}{177147}+O\left
((x+3)^{11}\right)
\]

Answer 13

the denominator goes to infinity

Answer 14

If you notice that
\[
f^n(x)=(-1)^n \frac { n!}{x^{n+1}}
\]

Answer 15

then

\[
\frac {f^n(a)}{n!}(x-a)^n =(-1)^n \frac {(x-a)^n}{a^{n+1}}
\]
Then the series will be
\[

\sum_{n=0}^\infty (-1)^n \frac {(x-a)^n}{a^{n+1}}

\]

Answer 16

give me a min to take that in...

Answer 17

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

Answer 18

makes sense. Remind me what the first equation is
\[f^n(x)=(-1)^n \frac { n!}{x^{n+1}}\]

Answer 19

Power series?

Answer 20

I remember doing a ratio test with
\[f^n(x)=(-1)^n \frac { n!}{x^{n+1}}\] to test for convergence....alternating series?