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Question Number 212177 by MrGaster last updated on 05/Oct/24
f(x)=arctan(((1−x)/(1+x)))=? ask:f^(2023) (0)
f(x)=arctan(1−x1+x)=?ask:f2023(0)
Answered by a.lgnaoui last updated on 05/Oct/24
f(0)=(π/4)⇒ f^(2023) (0)=(π^(2023) /4^(2023) )
f(0)=π4⇒f2023(0)=π202342023
Answered by Ghisom last updated on 05/Oct/24
if you mean (d^(2023) f/df^(2023) ) the answer is 2022! ((d^(2n) f(0))/df^(2n) )=9 ((d^(2n+1) f(0))/df^(2n+1) )= { ((−(2n)!; n=2k)),(((2n)!; n=2k+1)) :} [2023=2n+1=2(2k+1)+1=4k+3 ⇒ k=505; n=1011]
ifyoumeand2023fdf2023theansweris2022!d2nf(0)df2n=9d2n+1f(0)df2n+1={−(2n)!;n=2k(2n)!;n=2k+1[2023=2n+1=2(2k+1)+1=4k+3⇒k=505;n=1011]

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