Possible webpage title: Proving the Taylor Inequality for Positive Values of x

[transgalactic]

prove this inequality for x>0
$$x-\frac{x^3}{6}+\frac{x^5}{120}>\sin x$$

this is a tailor series for sin x
$$sinx=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5$$
for this innequality to be correct the remainder must be negative
but i can't prove it because there are values for c when the -sin c expression will be possitive

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

??


i got another approach
$$f(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-sin$$
$$f'(x)=1-\frac{x^2}{2}x+\frac{x^4}{24}-\cos x$$
$$f''(x)=-x+\frac{x^3}{6}+\sin x$$
$$f^{(3)}(x)=-1+\frac{x^2}{2}+\cos x$$
$$f^{(4)}(x)=x-\sin x$$
in this point how to know that the 4th derivative is always positive for x>0

??

 

[Mark44]

prove this inequality for x>0
$$x-\frac{x^3}{6}+\frac{x^5}{120}>\sin x$$

this is a tailor series for sin x

Yes, this is a Taylor's series, but more precisely it is a Maclaurin series since it is in powers of x or x - 0.

$$sinx=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5$$
for this innequality to be correct the remainder must be negative
but i can't prove it because there are values for c when the -sin c expression will be possitive

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

??

I seem to remember that you asked this same question before. How did it turn out that time?
The remainder formula gives bounds on c. What are they for your problem? Your problem also states what values of x to consider. Understanding the bounds on c and the possible values for x will help you with evaluating -sin(c).

i got another approach
$$f(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-sin$$
$$f'(x)=1-\frac{x^2}{2}x+\frac{x^4}{24}-\cos x$$
$$f''(x)=-x+\frac{x^3}{6}+\sin x$$
$$f^{(3)}(x)=-1+\frac{x^2}{2}+\cos x$$
$$f^{(4)}(x)=x-\sin x$$
in this point how to know that the 4th derivative is always positive for x>0

??

 

[transgalactic]

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

??

 

[Mark44]

Did you look at what I wrote in my previous post?

 

[transgalactic]

Did you look at what I wrote in my previous post?

you say that i should bound c
so the -sinc will give a ngative value

but i cent do it
i have x>0

and in this area it can be both possitive and negative .

i prefer to shoe that the 4th derivative is always positive for x>0

how to show that?

 

[Mark44]

I don't understand what you're doing, but here's some help at what you're trying to do.
g(x) = x - sin(x) is 0 at x = 0, and g'(x) > 0 except at x = k*pi. That should tell you something about the sign of x - sin(x).

 

[transgalactic]

its not proving that the 4th derivative is positive

 

[Gib Z]

Really, at the level of answering Taylor Series questions I don't think anyone would think twice about you simply stating that the inequality x > sin x, x>0 holds.

Once you agree on that, one could integrate both sides of that inequality between 0 and some number, with gives us an inequality for cos in terms of t. Then integrate that one from 0 to x, you'll see where this is getting at.

 

[transgalactic]

what do you man that level
i can't just say that x>sinx
x>0
i need to base this fact on some math or whatever
??