1

Approximation of a function by a polynomial function 1. step: Assuming that f(x) is differentiable at x = a, from the picture we see: 𝑓(𝑥) = 𝑓(𝑎) + ∆𝑓 ≈ 𝑓(𝑎) + 𝑓′(𝑎)∆𝑥

Linear approximation of f(x) at point x around x = a 𝑓(𝑥) = 𝑓(𝑎) + 𝑓′(𝑎)(𝑥 − 𝑎) 𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑹 = 𝑓(𝑥) − 𝑓(𝑎) − 𝑓′(𝑎)(𝑥 − 𝑎) will determine magnitude of error Let’s try to get better approximation of f(x). Let’s assume that f(x) has all derivatives at x = a. Let’s assume there is a possible power series expansion of f(x) around a. ∞ 2

3

𝑓(𝑥) = 𝑐0 + 𝑐1 (𝑥 − 𝑎) + 𝑐2 (𝑥 − 𝑎) + 𝑐3 (𝑥 − 𝑎) + ⋯ = ∑ 𝑐𝑛 (𝑥 − 𝑎)𝑛

∀ 𝑥 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎

𝑛=0

n 0 1 2 3 ⋮ n ⋮

𝑓 (𝑛) (𝑥) 𝑓 (𝑥) = 𝑐0 + 𝑐1 (𝑥 − 𝑎) + 𝑐2 (𝑥 − + 𝑐3 (𝑥 − 𝑎)3 + ⋯ 𝑓′(𝑥) = 𝑐1 + 2𝑐2 (𝑥 − 𝑎) + 3𝑐3 (𝑥 − 𝑎)2 + ⋯ 𝑓′′(𝑥) = 2𝑐2 + 3 ∙ 2𝑐3 (𝑥 − 𝑎) + 4 ∙ 3(𝑥 − 𝑎)2 ∙∙∙ 𝑓′′′(𝑥) = 3 ∙ 2𝑐3 + 4 ∙ 3 ∙ 2(𝑥 − 𝑎) + 5 ∙ 4 ∙ 3(𝑥 − 𝑎)2 ∙∙∙

𝑓 (𝑛) (𝑎) 𝑓(𝑎) = 𝑐0 𝑓′(𝑎) = 𝑐1 𝑓′′(𝑎) = 2𝑐2 𝑓′′′(𝑎) = 3 ∙ 2𝑐3

𝑓 (𝑛) (𝑥) = 𝑛(𝑛 − 1) ∙∙∙ 3 ∙ 2 ∙ 1𝑐𝑛 + (𝑛 + 1)𝑛(𝑛 − 1) ⋯ 2𝑐𝑛+1 (𝑥 − 𝑎)

𝑓 (𝑛) (𝑎) = 𝑛! 𝑐𝑛

𝑎)2

Taylor’s theorem If f (x) has derivatives of all orders in an open interval I containing a, then for each x in I, ∞

𝑓(𝑥) = ∑ 𝑛=0 𝑛

𝑓 (𝑛) (𝑎) 𝑓 (𝑛) (𝑎) 𝑓 (𝑛+1) (𝑎) (𝑥 − 𝑎)𝑛 = 𝑓(𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) + ⋯ + (𝑥 − 𝑎)𝑛 + (𝑥 − 𝑎)𝑛+1 + ⋯ (𝑛 + 1)! 𝑛! 𝑛! ∞

= ∑ 𝑘=0

𝑓 (𝑘) (𝑎) 𝑓 (𝑘) (𝑎) (𝑥 − 𝑎)𝑘 + ∑ (𝑥 − 𝑎)𝑘 𝑘! 𝑘! 𝑘=𝑛+1

convention for the sake of algebra: 0! = 1 = 𝑇𝑛 + 𝑅𝑛 𝑛

𝑇𝑛 = ∑ 𝑘=0

𝑅𝑛 (𝑥) =

𝑓 (𝑘) (𝑎) (𝑥 − 𝑎)𝑘 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑜𝑓 𝑛𝑡ℎ 𝑜𝑟𝑑𝑒𝑟 𝑘! 𝑓 (𝑛+1) (𝑧) (𝑥 − 𝑎)𝑛+1 𝑖𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑅𝑛 (𝑛 + 1)!

Taylor’s theorem says that there exists some value 𝑧 between 𝑎 and 𝑥 for which: ∞

∑ 𝑘=𝑛+1

𝑓 (𝑘) (𝑎) 𝑓 (𝑛+1) (𝑧) (𝑥 − 𝑎)𝑘 can be replaced by 𝑅𝑛 (𝑥) = (𝑥 − 𝑎)𝑛+1 𝑘! (𝑛 + 1)!

2

Approximating function by a polynomial function So we are good to go. We can find value of 𝒇(𝒙) around 𝒂 by calculating 𝑇𝑛 , and then adding 𝑅𝑛 . Instead of adding infinite number of terms (how in the world???), we have finite number of terms. Beautiful. A little problem: We know z exists, BUT we don’t know how to find z . That’s why we use approximation: By approximating 𝒇(𝒙) with 𝑇𝑛 we neglect 𝑅𝑛 . We can do it only if 𝑅𝑛 is very small compared to 𝑇𝑛 . So if we find maximum possible value for 𝑅𝑛 , and that value is small compared to 𝑇𝑛 , we found good approximation: 𝑛 𝑓 (𝑘) (𝑎) (𝑥 − 𝑎)𝑘 𝑓(𝑥) ≈ ∑ 𝑘! 𝑘=0

error is remainder 𝑅𝑛 (𝑥) =

𝑓 (𝑛+1) (𝑧) (𝑥 − 𝑎)𝑛+1 (𝑛 + 1)!

𝑧 is the value between 𝑎 and 𝑥 , that yields maximum 𝑅𝑛

A Maclaurin series is a Taylor series expansion of a function about 0

for n = 0 𝑓′(𝑧) =

𝑓(𝑥) = 𝑓(𝑎) + 𝑓′(𝑧)(𝑥 − 𝑎)

𝑓(𝑥) − 𝑓(𝑎) → 𝑀𝐸𝐴𝑁 𝑣𝑎𝑙𝑢𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑥−𝑎

Problems: • For what values of x can we expect a Taylor series to represent f (x)?→ 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 • How accurately do Taylor polynomials approximate the f (x)? → magnitude of the error 𝑅𝑛

example 1: Find Taylor’s series for sin x centered at a = 0 (McLaurin series). 𝑓(𝑥) = 𝑠𝑖𝑛 𝑥 𝑓 ′ (𝑥) = 𝑐𝑜𝑠 𝑥 𝑓′′(𝑥) = −𝑠𝑖𝑛 𝑥 𝑓 ′′′ (𝑥) = − cos 𝑥 𝑓 (4) (𝑥) = sin 𝑥 𝑡ℎ𝑖𝑠 𝑝𝑎𝑡𝑒𝑟𝑟𝑛 𝑤𝑖𝑙𝑙 𝑟𝑒𝑝𝑒𝑎𝑡 ∞

𝑓(𝑥) = ∑ 𝑛=0

𝑓(0) = 0 𝑓′(1) = 1 𝑓 ′′ (1) = 0 𝑓 ′′′ (1) = −1 𝑓 (4) (1) = 0

𝑓 (𝑛) (𝑎) −1 3 (𝑥 − 𝑎)𝑛 = 0 + 1𝑥 + 0𝑥 2 + 𝑥 + 0𝑥 4 + ⋯ 𝑛! 3! ∞

=𝑥−

𝑥3 𝑥5 𝑥7 (−1)𝑛 𝑥 2𝑛+1 + − +⋯= ∑ 3! 5! 7! (2𝑛 + 1)! 𝑛=0

Signs alternate and the denominators get very big; factorials grow very fast. Ratio test: 𝑥 2𝑛+3 (2𝑛 + 1)! 1 lim | ∙ | = |𝑥 2 | lim | | = |𝑥 2 | ∙ 0 2𝑛+1 𝑛→∞ (2𝑛 + 3)! 𝑛→∞ (2𝑛 + 3)(2𝑛 + 2) 𝑥 This converges for any value of x. The radius of convergence is infinity.

3 example 2: Find fifth order Taylor’s approximation for f(x) = ln x centered at a = 1. 𝑓(𝑥) = ln 𝑥 𝑓 ′ (𝑥) = 1/ 𝑥 𝑓′′(𝑥) = −1/𝑥 2 𝑓 ′′′ (𝑥) = 2!⁄𝑥 3 𝑓 (4) (𝑥) = − 3!⁄𝑥 4 𝑓 (5) (𝑥) = 4!⁄𝑥 5 𝑓 (𝑛) (𝑥) =

𝑓(1) = 0 𝑓′(1) = 1 𝑓 ′′ (1) = −1 𝑓 ′′′ (1) = 2! 𝑓 (4) (1) = − 3! 𝑓 (5) (1) = 4!

(−1)𝑛+1 (𝑛 − 1)! 𝑥𝑛

→

𝑓 (𝑛) (1) = (−1)𝑛+1 (𝑛 − 1)! 𝑐𝑛 =

∞

𝑓 (𝑛) (1) (−1)𝑛+1 = 𝑛! 𝑛

∞

ln(𝑥) = ∑ 𝑛=0

(−1)𝑛+1 𝑓 (𝑛) (1) 1 1 1 (𝑥 − 1)𝑛 = ∑ (𝑥 − 1)𝑛 = (𝑥 − 1) − (𝑥 − 1)2 + (𝑥 − 1)3 − (𝑥 − 1)4 + ⋯ 𝒇𝒐𝒓 |𝒙 − 𝟏| < 𝟏 𝑛! 𝑛 2 3 4 𝑛=1

𝟎