If-x-0-0-5pi-show-that-sinx-x-1-6-x-3-Hence-prove-that-1-3000-0-1-10-x-2-1-x-sinx-2-dx-2-5999-

Question Number 2089 by Yozzi last updated on 01/Nov/15

I f x \in [0, 0.5 π], s h o w t h a t s i n x ⩾ x - \frac{1}{6} x^{3} .

H e n c e p r o v e t h a t

\frac{1}{3000} ⩽ \int_{0}^{1 / 10} \frac{x^{2}}{{(1 - x + s i n x)}^{2}} d x ⩽ \frac{2}{5999} .

Commented by prakash jain last updated on 02/Nov/15

\sin x ⩾ x - \frac{x^{3}}{6}

0 ⩽ {(\frac{x}{1 - x + \sin x})}^{2} ⩽ {(\frac{x}{1 - x + x - \frac{x^{3}}{6}})}^{2}

\int \frac{36 x^{2}}{{(6 - x^{3})}^{2}} d x

6 - x^{3} = u \Rightarrow - 3 x^{2} d x = d u

\int \frac{- 12 d u}{u^{2}} = \frac{12}{u} = \frac{12}{6 - x^{3}} = \frac{12000}{5999} - 2 = \frac{2}{5999}

\sin x ⩽ x

x^{2} ⩽ {(\frac{x}{1 - x + \sin x})}^{2}

\int x^{2} d x = {[\frac{x^{3}}{3}]}_{0}^{1 / 10} = \frac{1}{3000}

Not a complete proof but an outline of approach .

Commented by 123456 last updated on 01/Nov/15

x - \frac{x^{3}}{6} = T_{3} (x)

T_{n} (x) = \underset{i = 0}{\sum^{n}} \frac{d^{i} f}{{d x}^{i}} (a) \frac{{(x - a)}^{i}}{i!}, f (x) = \sin x

Commented by prakash jain last updated on 02/Nov/15

\sin x ⩾ x - \frac{x^{3}}{3!}, can be seen from \sin x series

expansion .

Commented by Yozzi last updated on 02/Nov/15

T h a n k s!

Facebook Tweet Pin