0$. ( taylor series)

Asked May 07 '21 at 10:15

Active May 16 '21 at 11:44

Viewed 82 times

the question in a random study problem says show that x - $\frac{x^3}{6} < sinx< x $ for $x>0$. ( taylor series). I am assuming the question means about x=0

I have tried finding the maximum error which you end up as lagrange $R (2) < \frac{x^3}{6}$ but that only lets me prove $sinx<x$

edited May 16 '21 at 11:44

user577215664

40,625

asked May 07 '21 at 10:15

candyflossmh

Integrate $1<\cos x$ three times from $0$ to $x$. – May 07 '21 at 10:20
@YvesDaoust What is that inequality? Whenever is that true? – May 07 '21 at 10:25
2

@fundamentalform: maybe $1>\cos x$ is more appropriate ;-) – May 07 '21 at 10:27
5

Does this answer your question? Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$ – May 07 '21 at 10:28
@YvesDaoust Indeed! :-) – May 07 '21 at 10:37
I you integrate , where do the constants go? – candyflossmh May 07 '21 at 11:07

show that $x-x^{3}/6 < \sin x< x\text{ for }x>0$. ( taylor series)

0 Answers0