Upper Bound Error for Maclaurin Polynomial of Sin(x) on the Interval [0,2]

[Batmaniac]

Homework Statement

Find the 3rd-order Maclaurin Polynomial (i.e. P3,o(u)) for the function f(u) = sin u, together with an upper bound on the magnitude of the associated error (as a function of u), if this is to be used as an approximation to f on the interval [0,2].

I did the question fine except for the upper bound error.

Homework Equations

|Rn,o(u)| <= (k|u|^4)/(4!)

Where 0 <= sin(u) <= 2, and |sin(u)| <= k

The Attempt at a Solution

Okay, since 0 <= sin(u) <= 2 and |sin(u)| <= k, shouldn't k = sin(2)? The actual answer is k = 1, but if sin(u) is bounded between 0 and 2, it can't have a value greater than sin(2) so saying that it's max is 1 (at sin(pi/2) is an overestimation for this particular problem by my thought process.

Any enlightening comments as to why it's 1 and not sin(2)? Thanks!

 

[Gib Z]

The interval is [0,2] for values of u. So 0\leq u \leq 2. What is the maximum value of sin u on that interval?

 

[Batmaniac]

Right, pi/2 = 1.57 which is less than 2.

Okay then the max value is 1. But if u was bounded between zero and 1 then the max value would be sin1 right?

 

[Gib Z]

Yes, because sin u is a strictly increasing function over that interval.