Maclaurin remainder interval estimate

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Homework Help Overview

The discussion revolves around estimating the remainder of the Maclaurin series for the function f(x) = sinh(x) on the interval |x| ≤ 1. Participants are exploring how to derive an upper bound for the remainder term R3(x) and its implications.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of the polynomial approximation P3(x) and the expression for the remainder R3(x). There are inquiries about how to estimate the remainder and the rationale behind choosing specific bounds for sinh(c).

Discussion Status

Some participants have provided guidance on how to approach the estimation of the remainder, suggesting the need to maximize |x^4 sinh(c)| under given constraints. There is an ongoing exploration of the relationship between sinh(1) and the number 2, with various interpretations being considered.

Contextual Notes

Participants are working within the constraints of the problem, specifically the interval |x| ≤ 1 and the requirement to estimate the remainder in a way that is both rational and informative. There is mention of a specific estimate provided in the text, which some participants are trying to reconcile with their own calculations.

chief10
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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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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.
 
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.
 
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?
 
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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No ideas anyone?
 
i'm going to beg lol
 
chief10 said:
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]
 
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
chief10 said:
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
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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