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- anonymous

Estimate the value of cos(pi) with error less than 10^-11. (Taylor Series Problem)

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- anonymous

Estimate the value of cos(pi) with error less than 10^-11. (Taylor Series Problem)

- jamiebookeater

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- 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

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

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

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

By "estimation point", do you basically mean the point about which the Taylor series is approximated? (the "a" in (x-a)?)

- anonymous

yes.

- anonymous

did you try to use remainder term formula R(x)? you can set up amount of error

- anonymous

like f(x)-Pn(x) = blah?

- anonymous

im trying to get it from net... to complicated to copy from the book

- anonymous

maybe cos(pi) = \[\sum_{n=0}^{\infty} (-1)^{n}* (\Pi ^{2n}/(2n)!\]

- 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

does that make sense^

- anonymous

have to go... check this site:
http://scidiv.bellevuecollege.edu/dh/ccal/CC10_11.pdf

- 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

I see. But yeah, still stuck

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