Taylor Series with Remainder

  • Thread starter Batmaniac
  • Start date
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

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

3. 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

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

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

Want to reply to this thread?

"Taylor Series with Remainder" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads