How to prove this inequality using Taylor approximation?

Question

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<x \leq1 $ 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?

Those pairs of terms are all positive for $0 < x < \sqrt{42}.$ — David K, Nov 09 '20 at 13:27
Here is a Taylor series approach: https://math.stackexchange.com/a/803163/42969. — Martin R, Nov 09 '20 at 13:30

score 3 · Accepted Answer · answered Nov 09 '20 at 13:26

3

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 $x-\frac{x^3}{3!}< -1$.

answered Nov 09 '20 at 13:26

Surb

55,662

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ohh now I recognize the usefulness of Lagrange remainder expression :P – Tortar Nov 09 '20 at 13:35

How to prove this inequality using Taylor approximation?

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