# Remainder Estimation Theorem & Maclaurin Polynomials :[

## Homework Statement

Use the Remainder Estimation Theorem to find an interval containing x=0 over which f(x) can be approximated by p(x) to three decimal-place accuracy throughout the interval. Check your answer by graphing |f(x) - p(x)| over the interval you obtained.

f(x)= sinx
p(x)= x - (x^(3)/3!)

## Homework Equations

Remainder Estimation Theorem:
|Rn(x)| is less than or equal to (M/(n+1)!)|x-xo|^(n+1)

## The Attempt at a Solution

i honestly have NO idea how to do this problem.. am i supposed to find the Maclaurin polynomial of sinx?

any help would be sooo appreciated!

## The Attempt at a Solution

Dick
Homework Helper
p(x) IS the Maclaurin series for sin(x) up to n=4. Carefully write down the remainder term R and figure out how small x should be to guarantee R<0.001. Right?

Hmm well that makes sense, but how do I find the remainder term? I dont understand that..

Dick
Homework Helper
You wrote down a somewhat garbled form of it in part 2). Look it up and figure out what the parts are in this particular case (sin(x) expanded around x=0).

Oh.. is it Rn(x) = f(x) - pn(x) = f(x) - [sigma] (f^(k)*(xo))/k! * (x-xo)^k ?

That's the only other equation I can find..

Dick
Homework Helper
Noooo. No quite. It's actually more like your form in 2. Let's use that if you can tell me what M and x0 are.

I guess I would say xo is 0? And M is |F^(n+1)(x)| ?

Dick
Homework Helper
That's a pretty good guess. Except that M is the maximum of the expression you wrote over all of the x in the interval where we are going to use R. Is that what you meant to say? Now can you think of a good ESTIMATE (upper bound) for M?

Right.. M is the upper bound.. And I'm trying to find an interval with x=0 in it, so could the upper bound be pi? or 1? Ok, I'm not sure.

Dick
Homework Helper
What is f^(n+1)(x) in this case? While we are at it what is n? You are getting there.

Isnt it the nth+1 derivative of ...sinx?
Hahah I dont feel like I'm getting it at all :[

Dick
Homework Helper
Why so glum? You are exactly right. What is n (there are two right answers!).

Hmm.. I dont know isnt just infinity? Or does it get a value?

Dick
Homework Helper
It definitely gets a value! The series expansion may be infinite, but the remainder term estimates the error in truncating the infinite series to a finite one. What is the highest power in our FINITE series? (It's on your question page.)

OH! Hmm..all that's on my question page is.. 0?

Dick
Homework Helper
f(x)= sinx
p(x)= x - (x^(3)/3!)

One of these is a truncated series. Exercise: DO work out the Maclaurin expansion of sin(x). Maybe this is not what you are getting.

Dick
Homework Helper
Sorry. Meant to say "This is maybe what you are NOT getting."

So: let xo = 0?
f(x)= sinx, f(0)=0 po(x)=0
f'(x)= cosx, f'(0)=1 p1(x)=x
f''(x)= -sinx, f''(0)=0 p2(x)= x
f'''(x)= -cosx, f'''(0)=-1 p3(x)=x - (1/3!)x^3

Dick
Homework Helper
Bravo! So we want the remainder term for which value of n?

Hahah ummmmm... the p3?

Dick
Homework Helper
Right again! So you want R3(x). You could also use R4(x) since the x^4 term in the series vanishes, right? Now give me R3 and R4 - and tell me how to estimate the max(f^(n+1)(x)) part. Then you are practically done.

Ohh ok awesome!
So R3(x) would be f(x) - p3(x) since its the remainder?
maybe... .008?

Dick
Homework Helper
More or less, yes. But give me the remainder term in the form you quoted in in part 2 of your question. It will indeed estimate the difference between sin(x) and p3(x).

" |Rn(x)| is less than or equal to (M/(n+1)!)|x-xo|^(n+1) "

so am I using 3 as n?
Sorry for completely not getting this!

Dick