How Accurate is Lagrange Interpolation for Approximating Cos(0.75)?

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Discussion Overview

The discussion revolves around the accuracy of the Lagrange interpolation method for approximating the cosine function at 0.75, specifically using a degree three polynomial and four-digit chopping arithmetic. Participants explore the calculation of the approximation, the actual value of cos(0.75), and the error bound associated with the approximation.

Discussion Character

  • Technical explanation
  • Homework-related
  • Meta-discussion

Main Points Raised

  • The problem involves using Lagrange interpolation to estimate cos(0.75) based on given cosine values at nearby points.
  • One participant calculated the approximation as 0.7313, with an actual error of 0.0004 and an error bound of 2.7 × 10-8.
  • The discrepancy between the actual error and the error bound is attributed to the precision of the provided data, which is limited to four decimal places.
  • Another participant expressed disappointment over the lack of responses to their initial query.
  • Several participants engaged in a meta-discussion about the norms of posting solutions and the expectations of sharing work on the forum.

Areas of Agreement / Disagreement

Participants generally agree on the calculations related to the approximation and error analysis. However, there is no consensus on the appropriateness of sharing solutions before course completion, leading to a mix of perspectives on posting practices.

Contextual Notes

The discussion reflects limitations in the precision of the cosine values provided and the implications of four-digit chopping arithmetic on the accuracy of the approximation. The exact nature of the error bound calculation remains unresolved.

Hero1
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Problem:
Use the Lagrange interpolating polynomial of degree three or less and four digit chopping arithmetic to approximate cos(.750) using the following values. Find an error bound for the approximation.

cos(.6980) = 0.7661
cos(.7330) = 0.7432
cos(.7680) = 0.7193
cos(.8030) = 0.6946

The actual value of cos(.7500) = 0.7317 (to four decimal places). Explain the discrepancy between the actual error and the error bound.

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Solution:
The approximation of cos(.7500) 0.7313. The actual error is 0.0004, and an error bound is 2.7 × 10-8. The discrepancy is due to the fact that the data are given only to four decimal places.

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Can anyone help me figure out the intermediary steps from the problem to solution?
 
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I managed to figure out the solution to the problem myself, however, I am disappointed that no one responded to my thread.
 
Hero said:
I managed to figure out the solution to the problem myself, however, I am disappointed that no one responded to my thread.


Hi Hero,

I'm sorry no one responded to your thread. Over 95% of our threads get responded to, so believe me it's not a normal occurrence. Can you post how you solved the problem for us? Other people in the future might find it useful :)

Jameson
 
Jameson said:
Hi Hero,

I'm sorry no one responded to your thread. Over 95% of our threads get responded to, so believe me it's not a normal occurrence. Can you post how you solved the problem for us? Other people in the future might find it useful :)

Jameson

I can't post it until after the course is over. I don't want my instructor thinking that I stole online content.
 
Hero said:
I can't post it until after the course is over. I don't want my instructor thinking that I stole online content.

You posted the question here hoping someone else would answer it. How is your posting your own work different than someone else posting their work?

I guess if you just won't post your solution, then ok but I don't quite get it. As a rule of thumb I would suggest that you be comfortable with anything you post on this site to be read by anyone.

Again, sorry you didn't get any help but give it one more try and when another question comes up and I'm sure you'll find some help. I'll make sure our staff knows that you have a question if you post next time.
 
Jameson said:
You posted the question here hoping someone else would answer it. How is your posting your own work different than someone else posting their work?

I guess if you just won't post your solution, then ok but I don't quite get it. As a rule of thumb I would suggest that you be comfortable with anything you post on this site to be read by anyone.

Again, sorry you didn't get any help but give it one more try and when another question comes up and I'm sure you'll find some help. I'll make sure our staff knows that you have a question if you post next time.

I will post it
 

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