Proof that |arctan⁡(x)−π4−(x−1)2|≤(x−1)22I'm trying to prove that, for every x≥1 :|arctan⁡(x)−π4−(x−1)2|≤(x−1)22I could do it graphically...

mairie0708zl

mairie0708zl

Answered

2022-01-26

Proof that |arctan(x)π4(x1)2|(x1)22
I'm trying to prove that, for every x1 :
|arctan(x)π4(x1)2|(x1)22
I could do it graphically on R, but how to make a formal algebraic proof?

Answer & Explanation

chaloideq1

chaloideq1

Expert

2022-01-27Added 11 answers

We have that for x1
arctanxπ4(x1)20|arctanxπ4(x1)2|=π4+(x1)2arctanx
then consider
f(x)=(x1)22+arctanxπ4(x1)2f(x)=1x2+1+x320
and since f(1)=0 we have that f(x)0 and the inequality is proved.
euromillionsna

euromillionsna

Expert

2022-01-28Added 16 answers

Hint: For each x>1, Taylor gives
arctan(x)=π4+(x1)2+arctan(cx)2(x1)2
where 1<cx<x Thus all you need to show is that |arctan(c)|1 for all c1

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