Need help with part b)! How do we show that this is true?

Question

I need help with part b). By the Lagrange error bound, I know that the error is going to be less than the absolute value of the next term.

Indeed, #2^5/5! = 4/15# but how do I show this rigorously on a math exam?

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Answer 1 · 2018-05-11T07:43:38

OK, I'll assume for part a, you got #x-x^3/6+x^5/120#

And we have #abs(sinx-x+x^3/6)<=4/15#

By substituting the Maclaurin series, we get:

#abs(x-x^3/6+x^5/120-x+x^3/6)<=4/15#

#abs(x^5)/120<=4/15# (since 120 is positive we can just take it out of the #abs()#)

#abs(x^5)<=32#

#abs(x)^5<=32#

#abs(x)<=32^(1/5)#

#abs(x)<=2#