Question

expand function f(z)= (z^2)-1 in Taylor series at z=1

Answer 1

f(z) = z^2 -1
f'(z) = 2z
f''(z) = 2
f'''(z) = 0

Answer 2

is it to the (z-1)^n or (z+1)^n ??

Answer 3

Thanks. It is (z^2)-1 . as you did the function is not differentiable after f''' (z)

Answer 4

it is ...but its pointless to add up alot of zeros ;)

Answer 5

Thanks . still think about it :)

Answer 6

0 2 2
--- + --- (x+1) + --- (x+1)^2 ...this it?
0! 1! 2!

Answer 7

...(x-1)s right?

Answer 8

2x -2 +x^2 -2x +1

x^2 -1...if im seeing it right lol

Answer 9

Is it \[z^2-1?\]
OR
\[{(z^2)}^{-1}?\]

Answer 10

z ^2 −1

Answer 11

at z=1

Answer 12

....i take it that formula subs z for x then lol

Answer 13

so after we expand it; we apply z=1 and get 0 right?

Answer 14

i just started readig taylors the other day :)

Answer 15

Yeah, I think what you did is right.

Answer 16

It just should be (x-1) rather than (x+1). Right?

Answer 17

yes...(x-1) :) the general is (x-a) so i assumed a=1

Answer 18

Exactly. It's z, not x anyways :)

Answer 19

Thanks guys :)