You compute e by summing the Taylor series of e^x at x = 1. How many terms do you need so the truncation error is below 10^{-6}, and what bound justifies the count?

You compute e by summing the Taylor series of e^x at x = 1. How many terms do you need so the truncation error is below 10^{-6}, and what bound justifies the count?

Approach: Use the Lagrange form of the Taylor remainder, bound the derivative on the interval by a crude constant, then find the smallest factorial that clears the target.

10. Truncating the Taylor series of e^x at x = 1 leaves the Lagrange remainder after the term in x^n, which is R_n = e^c / (n+1)! for some c in (0, 1), and e^c < 3 on that interval, so R_n < 3/(n+1)!. At n = 8 the bound is 3/9! = 3/362880 = 8.3e-6, which misses the target, and at n = 9 it is 3/10! = 3/3628800 = 8.3e-7, which clears it. So the terms from x^0 through x^9 are needed, that is 10 terms, and the true error there is 1/10! + 1/11! + ... = 2.7e-7, comfortably inside the bound. The factorial growth in the denominator means each extra term buys roughly another order of magnitude at this point, so the count is insensitive to how crudely the derivative was bounded.

Follow-up: How many terms would you need at x = 10, and why is summing that series directly a poor way to compute e^{-10}?

Key concepts: lagrange remainder, taylor series, factorial growth, truncation error.