Error Bound on Tangent Maclaurin Series

[Juanriq]

Salutations! Just checking if my logic is correct.

Homework Statement

I need to bound the error for $$\tan x$$ on $$[0, \frac{\pi}{2}]$$

Homework Equations

$$R_n(x) = \displaystyle \frac{\tan^{n+1}(\zeta)}{(n+1)!}x^{n+1}$$

The Attempt at a Solution

So...I thought that the error should go to 0 since the factorial will eventually overtake the polynomial. Then, I thought that this logic might break down since $$\tan x$$ goes to infinity on the interval. Am I overthinking this?

Thanks!

 

[lippyka]

No, you are not overthinking this. The error will indeed go to 0, since the factorial will eventually overtake the polynomial. However, you need to be careful because the error will be larger near the boundary values (i.e. 0 and \frac{\pi}{2}) due to the fact that \tan(\zeta) can become very large near these values.