Proving an Interval for Arctan Function: A Convincing Approach

[tomboi03]

Identify an interval [-a,a] so that
|arctan(x)-x+x^3/3| < or = 10^-15
for all x E [-a,a]. Your result needs to be supported by a convincing proof

I'm not sure how to go about this... can someone help me out?
all i could probably do is... since.. 10^-15 is close to zero, i can find out what the x's are equal to... but, i don't know what else i can do.

How would i be able to find the x's if there's an arctan? Is there any trig identities that i use?

Thanks

 

[CompuChip]

Hint: as you say that 10^{-15} is so close to zero, so probably will the boundary x's have to be. So try Taylor-expanding the arctan around x = 0. You might want to go up to 5th order

 

[tomboi03]

I'm not sure, how I'm suppose to find the boundaries of x... i got down to
|x^5/5| \leq 10^-15

Is -a and a equal to.. -10^-15 and 10^-15 respectively?

 

[CompuChip]

If x^5/5 has to be smaller than 10^(-15), i.e. \frac{x^5}{5} \in [-10^{-15}, 10^{-15}], then which interval can x itself lie in?

 

[HallsofIvy]

I'm not sure, how I'm suppose to find the boundaries of x... i got down to
|x^5/5| \leq 10^-15

Is -a and a equal to.. -10^-15 and 10^-15 respectively?

So you want to solve the inequality -10^{-15}< x^5/5< 10^{-15}.