How do I prove x - \frac{x^3}3 < \arctan x

iocasq4

iocasq4

Answered question

2022-01-30

How do I prove xx33<arctanx<x?
for every x>0?
I tried taking the limit of xx33 and arctan(x) as x approaches 0, but I get 0 which makes sense since theyre

Answer & Explanation

Roman Stevens

Roman Stevens

Beginner2022-01-31Added 10 answers

Note that 
1t211+t21 
for a real t (t=0 is the only value where the inequalities hold with equality). Therefore,
0x(1t2)  dt 0x11+t2  dt 0x1  dt  
for x0. That is, 
xx33arctan(x)x 
Only when x=0 do the inequalities hold with equality.

When x>0, as stated in the comments, we have
x2x(1t2)  dt <x2x11+t2  dt <x2x1  dt  
Hence, 
0x(1t2)  dt <0x11+t2  dt <0x1  dt  
for x>0

tsjutten20

tsjutten20

Beginner2022-02-01Added 13 answers

Hint:
ddxarctanx=(1+x2)1
(1+xk)n=r=0n(n1)(nr+1)r!xkr

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?