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

??

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# Homework Help: Taylor innequality prove

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