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anonymous
 5 years ago
Estimate the value of cos(pi) with error less than 10^11. (Taylor Series Problem)
anonymous
 5 years ago
Estimate the value of cos(pi) with error less than 10^11. (Taylor Series Problem)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0I see what you mean. I'm not exactly sure of this wording; haven't seen it asked that way before. :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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
 5 years ago
Best ResponseYou've already chosen the best response.0By "estimation point", do you basically mean the point about which the Taylor series is approximated? (the "a" in (xa)?)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you try to use remainder term formula R(x)? you can set up amount of error

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like f(x)Pn(x) = blah?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im trying to get it from net... to complicated to copy from the book

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maybe cos(pi) = \[\sum_{n=0}^{\infty} (1)^{n}* (\Pi ^{2n}/(2n)!\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0inik 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
 5 years ago
Best ResponseYou've already chosen the best response.0does that make sense^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0have to go... check this site: http://scidiv.bellevuecollege.edu/dh/ccal/CC10_11.pdf

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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 \[R(x) = f(x)  T(x) = 1/n! * \int\limits\limits_{a}^{b} \ f^{(n+1)} (t) * (xt)^n dt\] to account for anything extra that the Taylor polynomial may have missed. Not sure how you can apply it here though..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I see. But yeah, still stuck
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