Taylor polynomial approximation (HELP ME)

  • Thread starter frasifrasi
  • Start date
  • #1
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Ok, we are asked to determined the degree of the the taylor polynomial about c =1 that should be used to approximate ln (1.2) so the error is less than .001


the book goes throught the steps and arrives at:

|Rn(1.2)| = (.02)^(n+1)/(z^(n+1)*(n+1)


but then, it states that

(.02)^(n+1)/(n+1) < .001

and there is an error pointing this expression to

1000 < (n+1)(5^(n+1))

--> I have no idea how the book arrived at this second expression, could anyone please explain?

- Also, if anyone knows, why was the z term left out in the second step?


Thank you.
 

Answers and Replies

  • #2
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why was the z term dropped?

does anyone know?
 
  • #3
HallsofIvy
Science Advisor
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I just don't believe your book has what you wrote. For one thing, |Rn(1.2)| = (.02)^(n+1)/(z^(n+1)*(n+1) is missing a ")". For another, the remainder formula has z in the numerator not the denominator- You can't make the error smaller by choosing z larger!
 
  • #4
276
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OK, in general how does this process work? I am having a little trouble following the steps...

*Yeah, I forgot the last parenthesis
 

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