anonymous
  • anonymous
Estimate the value of cos(pi) with error less than 10^-11. (Taylor Series Problem)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
So, we know that the Taylor series of cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! ... and we also know that it is equal to -1. I think it's basically asking for you to use the Taylor series to approximate it such that cos(pi) = -0.999999999999 (twelve decimal places, to be safe). Is that basically what the question is asking for?
anonymous
  • anonymous
I think I need to choose an estimation point maybe 0 (bad approximation) or 3 or maybe I'm just over thinking it. I get what you are saying..but I think that is too easy
anonymous
  • anonymous
I see what you mean. I'm not exactly sure of this wording; haven't seen it asked that way before. :P

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anonymous
  • anonymous
It weird because I could do something of cos(.31) because I could set my c=0 but is not the case for cos(pi).
anonymous
  • anonymous
By "estimation point", do you basically mean the point about which the Taylor series is approximated? (the "a" in (x-a)?)
anonymous
  • anonymous
yes.
anonymous
  • anonymous
did you try to use remainder term formula R(x)? you can set up amount of error
anonymous
  • anonymous
like f(x)-Pn(x) = blah?
anonymous
  • anonymous
im trying to get it from net... to complicated to copy from the book
anonymous
  • anonymous
maybe cos(pi) = \[\sum_{n=0}^{\infty} (-1)^{n}* (\Pi ^{2n}/(2n)!\]
anonymous
  • anonymous
inik I see in my book by what you mean with the Rn thing..but we've never used that in class so idk about it
anonymous
  • anonymous
does that make sense^
anonymous
  • anonymous
have to go... check this site: http://scidiv.bellevuecollege.edu/dh/ccal/CC10_11.pdf
anonymous
  • anonymous
It's from Taylor's original theorem...basically says that if you can represent a function f(x) by a taylor series T(x) to the nth order, the remainder function is defined as \[R(x) = f(x) - T(x) = 1/n! * \int\limits\limits_{a}^{b} \ f^{(n+1)} (t) * (x-t)^n dt\] to account for anything extra that the Taylor polynomial may have missed. Not sure how you can apply it here though..
anonymous
  • anonymous
I see. But yeah, still stuck

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