# How to prove this inequality using Taylor approximation?

I'd like to prove the following via Taylor approximation :

$$x > 0 \implies \sin x > x-\frac{x^3}{3!}$$

I tried to estimate the error but I found $$E_3(x) \leq |\frac{x^4}{4!}|$$, which doesn't give information about the sign of the error. Then, I tried to compare the sign of each couple of terms in the error but it gives the right result just when $$0 because in this case

$$(\frac{x^5}{5!} - \frac{x^7}{7!} )+(\frac{x^9}{9!} -\frac{x^{11}}{11!} ) + \dots$$

all the couples are positive. But from here I don't know how to continue to include the case $$x>1$$.

How can I find a way using Taylor approximation ? (I know other methods like derivatives can be used, but I want just to use Taylor if possible).

Besides, Is there a general method to find the intervals where the sign of the error is positive/negative?

## 1 Answer

Hint

If $$x\in [0,\pi]$$, there is $$b\in [0,x]$$ s.t. $$\sin(x)=x-\frac{x^3}{3!}+\frac{\sin(b)}{4!}x^4.$$

If $$x\geq \pi$$ then $$x-\frac{x^3}{3!}< -1$$.

• ohh now I recognize the usefulness of Lagrange remainder expression :P – Tortar Nov 9 '20 at 13:35