Question

Consider the Maclaurin series: 

Part C: The series:  has a partial sum  when x = 1. What is an interval, |S − S5| ≤ |R5| for which the actual sum exists? Provide an exact answer and justify your conclusion. (10 points)

 

             

 

Answer 1

Remainder for Maclaurin series is R₅(x)=sin(x)-S₅(x). When x=1:

R₅(1)=sin(1)-305353/362880,

R₅(1)=∑(-1)ⁿ/(2n+1)!-(1-1/3!+1/5!-1/7!+1/9!) where summation limits are 0≤n<∞.

R₅(1)=∑(-1)ⁿ/(2n+1)! where summation limits are 5≤n<∞. We can take S₅ to be equivalent to S₅(1) and R₅ to be equivalent to R₅(1). The first term of this reduced series is -1/11! and continues +1/13!-1/15!+1/17!-... Because of alternating signs, the Maclaurin series for sin(1) oscillates between being a little below sin(1) and being a little above sin(1). The discrepancy (absolute distance between true sin(1) and the series summation up to a given n), represented by |S-S_n|, decreases with progressive terms. R₅<0 because the lead term is negative and the next term (positive 1/13!) has a lower magnitude so the overall sum is negative. The same applies to the next pair of terms, implying that the overall sum is negative: -1/11!+1/13!=-1/11!+1/(13×12×11!)=(1/11!)(-1+1/156)=-155/13!. We need to compare magnitudes using |S-S₅| and |R₅|. S as an approximation to sin(1) to which it converges, will always be slightly above or below the true value. Similarly R₅ will always be slightly above or below the true difference. -R₅≤S-S₅≤R₅ is another way to express the relationship.

I hope this helps.