How do you find MacLaurin's Formula for #f(x)=sinxcosx# and use it to approximate #f(1/2)# within 0.01?

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Answer 1 · 2017-01-14T06:16:17

0.42

Taylor series about x = 0, with error term, for f(x) is

#f(x)= f(0) + x/f'(0) = ...+x^(n-1)f^(n-1)/((n-1)!)+x^n/(n!)f^n(theta x)#,

#0 < theta < 1#

#=1/2[2x-(2x)^3/(3!)+(2x)^5/(5!)+,,]. So,

#f(1/2)= 1/2[1-1/6+1/120]+E#, where

#|E|=1/2(1/5040 |sin (theta/2) |)<=1/10080=O(0.0001)#

And so,

f(1/2)=1/2(0.8417)=0.421, correct to three decimals, after rounding.