Question

using taylor's theorem prove >x-x^3/6 < sin x< x-x^3/6+x^5/120, for x>0

Answer 1

Answer:

$3y4522xy)7()752 - 2 + - \times \div 123yx4 - 45xy + 11p()7862$

Answer 2

Answer:

Proof given below

Step-by-step explanation:

Taylor's series expansion of any function $f(x)$ about any point $x=X$ is given by $f(x)= \sum_{n=0}^{n=\infty} \frac{f^{(n)}(X)}{n!} (x-X)^n$ where $f^{(n)}(X)$ denotes nth derivative of f w.r.t x.

Expansion of sin(x) is

$sinx=x-x^3/3!+x^5/5!-$......

so $sinx-x+x^3/6 =F_1 =x^5/5!-x^7/7!+......$ since for convergence we must have absolute values of negative terms less than previous positive one's, this says that $F_1$ $> 0$, which implies that $x-\frac{ x^3}{6} < sinx.$

Using simillar argument take $sinx-x+x^3/6-x^5/120 =F_2 =-(x^7/7!-x^9/9!+....)$ ,this time it is easy to see that  $F_2 < 0$, which implies that $sinx < x-x^3/6+x^5/120 .$ Hence Proved