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

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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?

1 Answer
May 11, 2018

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