Taylor’s TheoremJon LeeMarch 1, 2007Given a function f(x) whose derivatives f0(x), f00(x), . . . all exist at x = 0 (or moregenerally, any point x = a), we can find its Taylor series centered around 0, which is givento bef(x) ≈∞Xn=0f(n)(0)n!xn= f(0) + f0(0) · x + f00(0)/2 · x2+ f(3)(0)/6 · x3+ · · · .In computing a Taylor series using summation notation, the hardest part is coming upwith a closed form expression for f(n)(x), the n-th derivative of f(x). A few steps make thiseasier:1. compute a few actual derivatives f0(x), f00(x), . . .2. after having computed enough derivatives, spot a pattern and guess a general formulafor f(n)(x)3. to be safe, prove your formula is correct via inductionWhat happens if we try this on the function f(x) = log(1 − x)? Following the stepsabove, we find that1. Differentiating madly, we find thatf(x) = log(1 − x)f0(x) = −1/(1 − x)f00(x) = −1/(1 − x)2f(3)(x) = −2/(1 − x)3f(4)(x) = −6/(1 − x)4f(5)(x) = −24/(1 − x)5f(6)(x) = −120/(1 − x)6. . .12. It’s not too hard to guess a formula for f(n)(x) — after differentiating enough times,you get a goo d feel of the pattern. Here, we hope thatf(n)(x) = −(n − 1)!/(1 − x)nfor n ≥ 1 .3. Supposing we want to check that our formula is correct, we can prove it by induction.Indeed, assuming it holds for some value of n ≥ 1, we havef(n+1)(x) =ddxf(n)(x)=ddx−(n − 1)!/(1 − x)n= −n!/(1 − x)n+1by the chain rule ,which is what our formula gives.Since f(x) = log(1 − x) and f(n)(x) = −(n − 1)!/(1 − x)nfor n ≥ 1, we see that f(0) = 0and f(n)(0) = −(n − 1)! for n ≥ 1. The formula for a Taylor series then shows that theTaylor series for log (1 − x), centered around 0, islog(1 − x) ≈∞Xn=1−(n − 1)!n!xn=∞Xn=1−xnn= −x −x22−x33− · · · .We can also find error estimates by Taylor’s theorem, which states that the error betweenthe n-degree Taylor polynomial Tn(f), and a function f itself on the interval [−d, d] isbounded byM(n + 1)!|x|n+1,where M is a bound on f(n+1)on the interval [−d, d].In this example, choose n = 3 and the interval [−0.3, 0.3]. Then,Tn(f) = T3(f) = −x −x22−x33.For a value of M, we bound f(n+1)(x) = f(4)(x) on the interval [−0.3, 0.3]. It can be checkedthat the least value that works isM =0.344.Hence, on the interval [−0.3, 0.3], our error T3(x) − f(x) is bounded byM(3 + 1)!|x|3+1=0.344 · 4!|x|4which is less than0.344 · 4!· 0.34on [−0.3, 0.3]