if f(x) = e^x and Pn(x) is the n-th order tylor polynomial for f at x = 0?

Question

then how large must n be to be certain that Pn(0.2) approximates e^0.2 accurate to 6 decimal places.
Thanks
Dave

Answer 1

Per the Lagrange form of the remainder term, there exists ξ∈(0, 0.2) such that e^(0.2) - P_n(x) = e^(ξ) (0.2)^(n+1)/(n+1)!. So we need to find n such that e^(ξ) (0.2)^(n+1)/(n+1)! < 1/1,000,000. Since e^(0.2) is definitely less than 2 (which follows from the fact that 2^5 = 32>e), it suffices to find n such that (0.2)^(n+1)/(n+1)! < 1/2,000,000. Checking, we see that (0.2)^6/6! = (1/5)^6/6! = (1/15,625)/720 = 1/11,250,000, whereas (0.2)^5/5! = (1/5)^5/5! = (1/3,125)/120 = 375,000, so n+1=6 is the minimum value that will suffice. Therefore, n must be greater than or equal to 5.