transgalactic
- 1,386
- 0
prove that..
<br /> x-\frac{x^3}{6}+\frac{x^5}{120}>\sin x\\<br />
<br /> R_5=\frac{f^{5}(c)x^5}{5!}<br />
i need to prove that the remainder is negative .
<br /> \sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5<br />
<br /> R_5=\frac{cos(c)x^5}{5!}<br />
<br /> x-\frac{x^3}{6}+\frac{x^5}{120}>\sin x\\<br />
<br /> R_5=\frac{f^{5}(c)x^5}{5!}<br />
i need to prove that the remainder is negative .
<br /> \sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5<br />
<br /> R_5=\frac{cos(c)x^5}{5!}<br />