Thread: Inequality from a Taylor series

Hey everyone.

I am working through a Taylor expansion of sin(x). Can anyone explain to me how we can draw this inequality from the Taylor series?

The Taylor series of sin(x), around c=0, is sin(x) = x - x^3/3! + x^5/5! +...

So, for x >= 0 : x - x^3/3! <= sin(x) <= x.

Thanks!

Originally Posted by cgiulz

Hey everyone.

I am working through a Taylor expansion of sin(x). Can anyone explain to me how we can draw this inequality from the Taylor series?

The Taylor series of sin(x), around c=0, is sin(x) = x - x^3/3! + x^5/5! +...

So, for x >= 0 : x - x^3/3! <= sin(x) <= x.

Thanks!

Taylor series cannot prove this inequality.The right way to solve this problem is to contruct a function like and use the derivation.