Remainder Estimation Theorem & Maclaurin Polynomials :[

[raincheck]

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!

 

[Dick]

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?

 

[raincheck]

Hmm well that makes sense, but how do I find the remainder term? I don't understand that..

 

[Dick]

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).

 

[raincheck]

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]

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.

 

[raincheck]

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

 

[Dick]

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?

 

[raincheck]

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]

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

 

[raincheck]

Isnt it the nth+1 derivative of ...sinx?
Hahah I don't feel like I'm getting it at all :[

 

[Dick]

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

 

[raincheck]

Hmm.. I don't know isn't just infinity? Or does it get a value?

 

[Dick]

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.)

 

[raincheck]

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

 

[Dick]

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]

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

 

[raincheck]

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]

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

 

[raincheck]

Hahah ummmmm... the p3?

 

[Dick]

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.

 

[raincheck]

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

 

[Dick]

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).

 

[raincheck]

" |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]

Yesssss. Use n=3 (or 4). ESTIMATE M. It's a sin or a cos of something right? How big can it possibly be?

 

[raincheck]

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

so (1/4!)|-xo|^(4) ?
I already know xo right? I still don't know what it is though..

 

[Dick]

Let's call that good enuf. So R3(x)<=|x^4|/4!. (x0=0 right?). R3(x) is the size of the error you make when you when you evaluate p3(x) as an approximation to sin(x). You can see that as x gets big, the error can get big. The question you are being asked to answer is how big can x get and still keep R3(x)<0.001 (the three decimal places).
Sorry, I've got to run. I'll check in tomorrow... Good luck!

 

[raincheck]

so .001 = (x^4)/4! ? Would I solve for x?
Thanks so much for helping me! :]