Maclaurin remainder interval estimate

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Homework Statement



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.
 
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Answers and Replies

  • #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.
 
  • #3
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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.
 
  • #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?
 
  • #5
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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
 
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  • #6
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No ideas anyone?
 
  • #7
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i'm gonna beg lol
 
  • #8
pasmith
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sub in 1 i guess since it's inclusive?

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]
 
  • #9
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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?
 
  • #10
pasmith
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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?

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]
 
  • #11
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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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