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anonymous
 5 years ago
expand function f(z)= (z^2)1 in Taylor series at z=1
anonymous
 5 years ago
expand function f(z)= (z^2)1 in Taylor series at z=1

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1f(z) = z^2 1 f'(z) = 2z f''(z) = 2 f'''(z) = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1is it to the (z1)^n or (z+1)^n ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks. It is (z^2)1 . as you did the function is not differentiable after f''' (z)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1it is ...but its pointless to add up alot of zeros ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks . still think about it :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.10 2 2  +  (x+1) +  (x+1)^2 ...this it? 0! 1! 2!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.12x 2 +x^2 2x +1 x^2 1...if im seeing it right lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is it \[z^21?\] OR \[{(z^2)}^{1}?\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1....i take it that formula subs z for x then lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1so after we expand it; we apply z=1 and get 0 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i just started readig taylors the other day :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, I think what you did is right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It just should be (x1) rather than (x+1). Right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1yes...(x1) :) the general is (xa) so i assumed a=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly. It's z, not x anyways :)
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