prove this inequality for x>0(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

x-\frac{x^3}{6}+\frac{x^5}{120}>\sin x

[/tex]

this is a tailor series for sin x

[tex]

sinx=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5

[/tex]

for this innequality to be correct the remainder must be negative

but i cant prove it because there are values for c when the -sin c expression will be possitive

[tex]

R_5=\frac{-sin(c)}{6!}x^6

[/tex]

??

i got another approach

[tex]

f(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-sin

[/tex]

[tex]

f'(x)=1-\frac{x^2}{2}x+\frac{x^4}{24}-\cos x

[/tex]

[tex]

f''(x)=-x+\frac{x^3}{6}+\sin x

[/tex]

[tex]

f^{(3)}(x)=-1+\frac{x^2}{2}+\cos x

[/tex]

[tex]

f^{(4)}(x)=x-\sin x

[/tex]

in this point how to know that the 4th derivative is always positive for x>0

??

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Taylor innequality prove

**Physics Forums | Science Articles, Homework Help, Discussion**