In this problem you show that a Taylor Series for a function actually converges

Question

In this problem you show that a Taylor Series for a function actually converges to the function. Show that the Taylor Series for f(x) = sin x converges to sin x for all x. This background information will be useful: lim_n rightarrow infinity x6n/n! = 0 for all x. Outline of strategy: Get an upper bound M for |f^(n = 1)(x)|on the interval from a to x. Write down the nth degree error bound for R_n(x). Take the limit of this bound for R_n(x) as n rightarrow infinity, show it is 0, for all x. State the conclusion:

Explanation / Answer

Maclaurin 's expansion for f(x) = sinx is

f(x) = sin x= x - x3/ 3! +x5/ 5! - x7 /7! + ------

this is an alternating series and Leibnitz test can be applied to show that he series is convergent

1 . Lim n-> inf an  = Lim n-> inf xn/ n! =0 as x is very very small compare to xn

2. an+1 = xn+1 /n+1 ! < xn /n! as xn+1 <xn and also 1/ (n+1) ! < 1/n!

ie an+1 < an   for all n

from 1 and 2 we conclude the given alternaing series is convergent when |x|<1

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