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Maclaurin formula: finding a delta for a given error

  1. Aug 13, 2010 #1
    Approximate the function [tex]f(x)=\sin(x)[/tex] using the corresponding Maclaurin polynomial: [tex]P_5(x)[/tex], in a bound [tex]\epsilon(0,\delta)[/tex]. Determine a value of [tex]\delta>0[/tex], so that the rest [tex]R_5(x)[/tex] verifies [tex]|R_5(x)|<0.0005[/tex] for all [tex]x\in{\epsilon(0,\delta)}[/tex]

    Well, the first thing that puzzles me a bit is that the order for [tex]P_5[/tex] and [tex]R_5[/tex] is the same. I assume that it is a mistake, and that the polynomial must be [tex]P_4[/tex], or the rest [tex]R_6[/tex]. I have chosen to leave the degree of the polynomial as it was and turn up one grade to the rest.

    So I have to find a delta for which any value within the environment [tex](0,\delta)[/tex] is less than the error 0.0005

    So I did:

    [tex]R_6(x)=|\displaystyle\frac{\sin(\alpha)x^6}{6!}|<0.0005[/tex] [tex]0<\alpha<x[/tex]

    So I must find x:

    [tex]R_6(x)=|\sin(\alpha)x^6|<0.36[/tex] [tex]0<\alpha<x[/tex]

    I must find x that satisfies the equation: [tex]\sin(x)x^6<0.36[/tex]

    I don't know how to go ahead.

    Bye there.
     
    Last edited: Aug 13, 2010
  2. jcsd
  3. Aug 13, 2010 #2

    vela

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    I think you're just getting confused by the notation. The usual notation is

    [tex]R_n(x) = f(x)-P_n(x)[/tex]

    The remainder term turns out to depend on xn+1 and f(n+1)(x).
    Hint: Use the fact that x ≥ sin x when x ≥ 0.
     
  4. Aug 13, 2010 #3
    I don't get it. I know that 1≥sin(x) how should I use the fact you mentioned? I wouldn't noted it if you weren't.

    Thanks vela.
     
  5. Aug 13, 2010 #4

    vela

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    The fact that x ≥ sin x for x ≥ 0 is one of those things you pick up along the way, typically when you do a problem like this one.

    You're trying to solve for x, right? And the difficulty is that you can't solve an equation of the form xn sin x = c. So you use the usual approach. Instead of solving the equation exactly, you look for an upper bound that you can solve that gives you a good estimate. In this case, you know that x6 sin x ≤ x7.
     
  6. Aug 13, 2010 #5
    Thanks man!

    One doubt. I was thinking that as the sine is always less than one I could use x⁶ as an upper bound too. Right?
     
  7. Aug 13, 2010 #6

    Mark44

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    What does this say - x⁶ ?
     
  8. Aug 13, 2010 #7

    vela

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    Yes, you can. So what is the difference between the two choices? Why would you choose one over the other?
     
  9. Aug 13, 2010 #8

    Mark44

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    sin(x) = 1 for x = pi/2 + 2k*pi, k an integer.

    What is x⁶? I can't read the exponent.
     
  10. Aug 13, 2010 #9

    vela

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    Telemachus's idea is to replace sin x by 1, so he'd end up with the inequality x^6 ≤ 0.36.

    It's weird. His exponent is in a smaller font than mine.
     
  11. Aug 13, 2010 #10
    Sorry, I was studying :P

    x⁶ is [tex]x^6[/tex]

    I think I would choose [tex]x^6[/tex] cause its a better approximation than [tex]x^7[/tex], so I got [tex]|x<\sqrt[6]{\frac{9}{25}}[/tex]

    Edit: Ok, I've seen that I was wrong, and that [tex]x^7[/tex] is a better approximation. But now I have more doubts, damn. For this: x ≥ sin x, or even using x^6 ≥ x^6 sin x, couldn't it happen that using that approximation I would get outside the boundary I was looking for? In this case I've noted that it works, but in general I would like to know what could happens. I mean, while using the substitution it could happens that the new value could be bigger than the error I was looking for, right? cause, I know that [tex]x^7\geq{x^6 sin x}[/tex] and so [tex]x^8\geq{x^6 sin x}[/tex] it is, and even for any n>6 [tex]x^n\geq{x^6}[/tex] right?

    Mmm ok, I am wrong for values of x, such that [tex]x\leq{1}[/tex]. I've tried with x^8 and it seems to work too, and its a better approximation for the boundary I am looking for.

    http://www.wolframalpha.com/input/?i=x^6+sin(x)+=+9/25

    From here I get x ~~ 0.8807530939284029...

    And http://www.wolframalpha.com/input/?i=(9/25)^(1/8)

    0.88011173679339339727
     
    Last edited: Aug 13, 2010
  12. Aug 13, 2010 #11

    vela

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    Nope. As long as your steps are reversible, which they usually are when you're doing these algebraic manipulations, you can start with the condition you found for x and deduce that the error will be within the given bound.
     
  13. Aug 13, 2010 #12
    Thanks vela, this is really helpful and very useful.
     
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