Question

Taylor series

picture #12

Answer 1

i came up with \[\sum_{0}^{\infty} -(x+2)^n\]

Answer 2

first one you want to expand about 1 so it should look like
\[f(1)+f'(1)(x-1)+\frac{f''(1)}{2}(x-1)^2+\frac{f'''(1)}{3!}(x-1)^3+...\]

Answer 3

\[f(1)=\frac{1}{2}\]
\[f'(x)=-\frac{1}{(1+x)^2}\]
\[f'(1)=-\frac{1}{4}\] etc a nice patter will emerge

Answer 4

ok i'll give it a try

Answer 5

why is it (x-1)?

Answer 6

its centered at c=-2 so i thought it would be (x+2)

Answer 7

we can take the derivatives for the first one easily
\[f'(x)=-(x+1)^{-2}\]
\[f''(x)=2(x+1)^{-3}\]
\[f'''(x)=-6(x+1)^{-4}\] etc

Answer 8

oh i was doing the wrong one sorry

Answer 9

yes the second one should be in powers of (x+2)

Answer 10

that one is much easier, and now that i look i see you have the answer, so ignore what i wrote

Answer 11

how do i find the radius of convergence?

Answer 12

i did limit as n-> infinity abs(a_n+1/a_n)