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

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

Mathematics
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f(z) = z^2 -1 f'(z) = 2z f''(z) = 2 f'''(z) = 0
is it to the (z-1)^n or (z+1)^n ??
Thanks. It is (z^2)-1 . as you did the function is not differentiable after f''' (z)

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Other answers:

it is ...but its pointless to add up alot of zeros ;)
Thanks . still think about it :)
0 2 2 --- + --- (x+1) + --- (x+1)^2 ...this it? 0! 1! 2!
...(x-1)s right?
2x -2 +x^2 -2x +1 x^2 -1...if im seeing it right lol
Is it \[z^2-1?\] OR \[{(z^2)}^{-1}?\]
z ^2 −1
at z=1
....i take it that formula subs z for x then lol
so after we expand it; we apply z=1 and get 0 right?
i just started readig taylors the other day :)
Yeah, I think what you did is right.
It just should be (x-1) rather than (x+1). Right?
yes...(x-1) :) the general is (x-a) so i assumed a=1
Exactly. It's z, not x anyways :)
Thanks guys :)

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