Remainder Estimation Theorem & Maclaurin Polynomials :[

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  • #1
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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!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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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?
 
  • #3
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Hmm well that makes sense, but how do I find the remainder term? I dont understand that..
 
  • #4
Dick
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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).
 
  • #5
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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..
 
  • #6
Dick
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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.
 
  • #7
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I guess I would say xo is 0? And M is |F^(n+1)(x)| ?
 
  • #8
Dick
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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?
 
  • #9
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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.
 
  • #10
Dick
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What is f^(n+1)(x) in this case? While we are at it what is n? You are getting there.
 
  • #11
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Isnt it the nth+1 derivative of ...sinx?
Hahah I dont feel like I'm getting it at all :[
 
  • #12
Dick
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Why so glum? You are exactly right. What is n (there are two right answers!).
 
  • #13
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Hmm.. I dont know isnt just infinity? Or does it get a value?
 
  • #14
Dick
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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.)
 
  • #15
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OH! Hmm..all that's on my question page is.. 0?
 
  • #16
Dick
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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.
 
  • #17
Dick
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Sorry. Meant to say "This is maybe what you are NOT getting."
 
  • #18
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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
 
  • #19
Dick
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Bravo! So we want the remainder term for which value of n?
 
  • #20
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Hahah ummmmm... the p3?
 
  • #21
Dick
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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.
 
  • #22
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Ohh ok awesome!
So R3(x) would be f(x) - p3(x) since its the remainder?
maybe... .008?
 
  • #23
Dick
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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).
 
  • #24
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" |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!
 
  • #25
Dick
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Yesssss. Use n=3 (or 4). ESTIMATE M. It's a sin or a cos of something right? How big can it possibly be?
 

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