anonymous
  • anonymous
expand function f(z)= (z^2)-1 in Taylor series at z=1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
f(z) = z^2 -1 f'(z) = 2z f''(z) = 2 f'''(z) = 0
amistre64
  • amistre64
is it to the (z-1)^n or (z+1)^n ??
anonymous
  • anonymous
Thanks. It is (z^2)-1 . as you did the function is not differentiable after f''' (z)

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amistre64
  • amistre64
it is ...but its pointless to add up alot of zeros ;)
anonymous
  • anonymous
Thanks . still think about it :)
amistre64
  • amistre64
0 2 2 --- + --- (x+1) + --- (x+1)^2 ...this it? 0! 1! 2!
amistre64
  • amistre64
...(x-1)s right?
amistre64
  • amistre64
2x -2 +x^2 -2x +1 x^2 -1...if im seeing it right lol
anonymous
  • anonymous
Is it \[z^2-1?\] OR \[{(z^2)}^{-1}?\]
anonymous
  • anonymous
z ^2 −1
anonymous
  • anonymous
at z=1
amistre64
  • amistre64
....i take it that formula subs z for x then lol
amistre64
  • amistre64
so after we expand it; we apply z=1 and get 0 right?
amistre64
  • amistre64
i just started readig taylors the other day :)
anonymous
  • anonymous
Yeah, I think what you did is right.
anonymous
  • anonymous
It just should be (x-1) rather than (x+1). Right?
amistre64
  • amistre64
yes...(x-1) :) the general is (x-a) so i assumed a=1
anonymous
  • anonymous
Exactly. It's z, not x anyways :)
anonymous
  • anonymous
Thanks guys :)

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