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Maclaurin remainder interval estimate

  1. Dec 7, 2012 #1
    1. The problem statement, all variables and given/known data

    The question asks to estimate the remainder on the interval |x|≤ 1.
    f(x) is given as sinh(x).

    I solved the polynomial P3(x) = x + (1/6)(x3)

    I then went ahead and solved R3(x) up to the point shown below.

    R3(x) = (sinh(c)*x4)(1/24)


    I then don't know how to go about getting the estimate. It clearly lies between 0 and x but it escapes me.. Any help would be great. Thanks a lot.
     
    Last edited: Dec 7, 2012
  2. jcsd
  3. Dec 7, 2012 #2

    pasmith

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    Re-read whatever is in your notes or textbook on estimating the remainder, or just look it up on wikipedia. That should give you an idea of what to do next.
     
  4. Dec 7, 2012 #3
    I did do that.. however I can't seem to procure the answer given in the text of

    |R3(x)| < (1/12) as sinh(1) < 2

    I can't make sense of that.
     
  5. Dec 7, 2012 #4

    pasmith

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    The question is basically asking you for an upper bound (not necessarily a least upper bound) for [itex]|x^4\sinh(c)|/24[/itex] if [itex]|x| \leq 1[/itex] and [itex]|c| \leq |x|[/itex].

    So how you maximize [itex]|x^4\sinh(c)|[/itex] subject to those constraints?
     
  6. Dec 7, 2012 #5
    sub in 1 i guess since it's inclusive?

    i still don't get however how you ascertain that R3(x) is less than 1/12th
     
    Last edited: Dec 7, 2012
  7. Dec 7, 2012 #6
    No ideas anyone?
     
  8. Dec 8, 2012 #7
    i'm gonna beg lol
     
  9. Dec 8, 2012 #8

    pasmith

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    Correct. That gives you [itex]|R_3(x)| \leq \sinh(1)/24[/itex].

    Now you need to show, from the definition of sinh, that [itex]\sinh(1) < 2[/itex]. Then you can conclude that
    [tex]|R_3(x)| \leq \sinh(1)/24 < 2/24 = 1/12.[/tex]
     
  10. Dec 8, 2012 #9
    why would you need to show that it's less than 2 though? i mean of all the numbers to pick, if you didn't know the answer, why would you pick 2?
     
  11. Dec 8, 2012 #10

    pasmith

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    The basic answer is that simply leaving the estimate as sinh(1)/24 feels incomplete. For many reasons it is preferable to give a rational estimate, which had better be bigger than sinh(1)/24 (although not so much bigger as to be useless).

    The crudest way is to note that [itex]\sinh(1) = (e - e^{-1})/2 < e/2 < 2[/itex], since [itex]e < 4[/itex].

    But that's really the largest useful estimate: an immediate improvement is to note that actually [itex]e < 3[/itex], so [itex]e/2 < 3/2[/itex], which gives an estimate of [itex]|R_3(x)| < 3/(2 \times 24) = 1/16.[/itex]
     
  12. Dec 8, 2012 #11
    okay i'll make note of that

    thanks a lot for your help pasmith, it's been much appreciated.. do you mind if i shoot a few more maclaurin's your way in the future if i have an issue? possibly via PM?
     
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