So, if we want to guarantee that our error is within a tolerance of +0.001, we can choose n so that efc+1 (n+1)! < 0.001

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So If We Want To Guarantee That Our Error Is Within A Tolerance Of 0 001 We Can Choose N So That Efc 1 N 1 0 001 1 (54.66 KiB) Viewed 69 times

So, if we want to guarantee that our error is within a tolerance of +0.001, we can choose n so that efc+1 (n+1)! < 0.001 Here is a table of values in the case c= 1.5. (1.5) (n+1)! 06.722533605507097 15.0419002041303225 2 2.5209501020651612 30.9453562882744354 40.2836068864823307 50.07090172162058266 6 0.015193226061553426 70.0028487298865412674 80.0004747883144235446 n Notice that when n=8, we can guarantee that the remainder |R() is less than 0.001. (In fact, it's less than 0.0005.) This should agree pretty well with what you saw in Exercise 2. Notice that the theory doesn't say that the choice n= 7 isn't good enough, it just guarantees that n = 8 is. n = Exercise 5. Make a similar table for the interval (-1,1). Stop when n is large enough to guarantee that R, () <0.001 on this interval