Truncation errors: using Taylor series to approximation functions - PowerPoint PPT Presentation

Truncation errors: using Taylor series to approximation functions

Approximating functions using polynomials: Let’s say we want to approximate a function 𝑔(𝑦) with a polynomial 𝑔 𝑦 = 𝑏 ! + 𝑏 " 𝑦 + 𝑏 # 𝑦 # + 𝑏 $ 𝑦 $ + 𝑏 % 𝑦 % + ⋯ For simplicity, assume we know the function value and its derivatives at 𝑦 ! = 0 (we will later generalize this for any point). Hence, 𝑔 " 𝑦 = 𝑏 # + 2 𝑏 $ 𝑦 + 3 𝑏 % 𝑦 $ + 4 𝑏 & 𝑦 % + ⋯ 3×2 𝑏 % 𝑦 + (4×3)𝑏 & 𝑦 $ + ⋯ 𝑔 "" (𝑦) = 2 𝑏 $ + 𝑔 """ (𝑦) = 3×2 𝑏 % + (4×3×2)𝑏 & 𝑦 + ⋯ 𝑔 "' (𝑦) = (4×3×2)𝑏 & + ⋯ 𝑔 "# 0 = (4×3×2) 𝑏 $ 𝑔 0 = 𝑏 ! 𝑔′′ 0 = 2 𝑏 # 𝑔 ()) 0 = 𝑗! 𝑏 ) 𝑔′ 0 = 𝑏 " 𝑔′′′ 0 = (3×2) 𝑏 !

Taylor Series Taylor Series approximation about point 𝑦 ! = 0 𝑔 𝑦 = 𝑏 ! + 𝑏 " 𝑦 + 𝑏 # 𝑦 # + 𝑏 $ 𝑦 $ + 𝑏 % 𝑦 % + ⋯ 𝑔 𝑦 = 𝑔 0 + 𝑔 " 0 𝑦 + 𝑔 "" 0 𝑦 $ + 𝑔 """ 0 𝑦 % + ⋯ 2! 3! - 𝑔 ) (0) 𝑦 ) 𝑔 𝑦 = 0 𝑗! )+, Demo “Polynomial Approximation with Derivatives” – Part 1

Taylor Series In a more general form, the Taylor Series approximation about point 𝑦 ! is given by: 𝑔 𝑦 = 𝑔 𝑦 ! + 𝑔 $ 𝑦 ! (𝑦 − 𝑦 ! ) + 𝑔 $$ 𝑦 ! (𝑦 − 𝑦 ! ) # + 𝑔 $$$ 𝑦 ! (𝑦 − 𝑦 ! ) % + ⋯ 2! 3! - 𝑔 ) (𝑦 ! ) (𝑦 − 𝑦 ! ) ) 𝑔 𝑦 = 0 𝑗! )+,

Iclicker question Assume a finite Taylor series approximation that converges everywhere for a given function 𝑔(𝑦) and you are given the following information: 𝑔 1 = 2; 𝑔 & (1) = −3; 𝑔 && (1) = 4; 𝑔 ' 1 = 0 ∀ 𝑜 ≥ 3 Evaluate 𝑔 4 A) 29 B) 11 C) -25 D) -7 E) None of the above

Demo “Polynomial Approximation with Derivatives” – Part 2 Taylor Series We cannot sum infinite number of terms, and therefore we have to truncate . How big is the error caused by truncation? Let’s write ℎ = 𝑦 − 𝑦 ! ) 𝑔 & 𝑦 % , 𝑔 & (𝑦 % ) ℎ & = 0 (ℎ) & 𝑔 𝑦 % + ℎ − 0 𝑗! 𝑗! &'( &')*+ ) 𝑔 & 𝑦 % And as ℎ → 0 we write: ℎ & ≤ C 5 ℎ )*+ 𝑔 𝑦 % + ℎ − 0 𝑗! &'( ) 𝑔 & 𝑦 % Error due to Taylor ℎ & = 𝑃(ℎ )*+ ) approximation of 𝑔 𝑦 % + ℎ − 0 𝑗! degree n &'(

Taylor series with remainder Let 𝑔 be (𝑜 + 1) - times differentiable on the interval (𝑦 ! , 𝑦) with 𝑔 (') continuous on [𝑦 ! , 𝑦] , and ℎ = 𝑦 − 𝑦 ! ) 𝑔 & 𝑦 % , 𝑔 & (𝑦 % ) ℎ & = 0 (ℎ) & 𝑔 𝑦 % + ℎ − 0 𝑗! 𝑗! &'( &')*+ Then there exists a 𝜊 ∈ (𝑦 ! , 𝑦) so that ) 𝑔 & 𝑦 % ℎ & = 𝑔 )*+ (𝜊) (𝑜 + 1)! (𝜊 − 𝑦 % ) )*+ 𝑔 𝑦 % + ℎ − 0 𝑔 𝑦 − 𝑈 𝑦 = 𝑆(𝑦) 𝑗! &'( And since 𝜊 − 𝑦 ! ≤ ℎ Taylor remainder ) 𝑔 & 𝑦 % ℎ & ≤ 𝑔 )*+ (𝜊) (𝑜 + 1)! (ℎ) )*+ 𝑔 𝑦 % + ℎ − 0 𝑗! &'(

Demo: Polynomial Approximation with Derivatives

Demo: Polynomial Approximation with Derivatives

Iclicker question A) B) C) D) E) Demo “Taylor of exp(x) about 2”

Demo “Polynomial Approximation with Derivatives” – Part 3 Making error predictions

Using Taylor approximations to obtain derivatives Let’s say a function has the following Taylor series expansion about 𝑦 = 2 . 𝑔 𝑦 = 5 2 − 5 2 𝑦 − 2 ! + 15 𝑦 − 2 " − 5 4 𝑦 − 2 # + 25 32 𝑦 − 2 $ + O((𝑦 − 2) % ) 8 Therefore the Taylor polynomial of order 4 is given by 𝑢 𝑦 = 5 2 − 5 2 𝑦 − 2 ! + 15 𝑦 − 2 " 𝑢 𝑦 8 8 where the first derivative is 6 4 𝑢 " (𝑦) = −5 𝑦 − 2 + 15 𝑦 − 2 ! 2 2 𝑔 𝑦 1 2 3 4

Using Taylor approximations to obtain derivatives We can get the approximation for the derivative of the function 𝑔 𝑦 using the derivative of the Taylor approximation: 𝑢 " (𝑦) = −5 𝑦 − 2 + 15 𝑦 − 2 ! 2 For example, the approximation for 𝑔′ 2.3 is 𝑔 " 2.3 ≈ 𝑢 " 2.3 = −1.2975 𝑢 𝑦 8 6 (note that the exact value is 𝑔 " 2.3 = −1.31444 4 2 What happens if we want to use the same method to 𝑔 𝑦 approximate 𝑔′ 3 ? 1 2 3 4

Iclicker question The function 𝑔 𝑦 = cos 𝑦 𝑦 # + )*+(#,) ,-#, & ' is approximated by the following Taylor polynomial of degree 𝑜 = 2 about 𝑦 = 2π 𝑢 - 𝑦 = 39.4784 + 12.5664 𝑦 − 2π − 18.73922 𝑦 − 2𝜌 - Determine an approximation for the first derivative of 𝑔 𝑦 at 𝑦 = 6.1 A) 18.7741 B) 12.6856 C) 19.4319 D) 15.6840

Computing integrals using Taylor Series A function 𝑔 𝑦 is approximated by a Taylor polynomial of order 𝑜 around 𝑦 = 0 . ) 𝑔 & 0 𝑦 & 𝑢 ) = 0 𝑗! &'( / 𝑔 𝑦 𝑒𝑦 by integrating the We can find an approximation for the integral ∫ . polynomial: / 𝑦 0 𝑒𝑦 = / ()* . ()* 0-" − Where we can use ∫ . 0-" Demo “Computing PI with Taylor”

Iclicker question A function 𝑔 𝑦 is approximated by the following Taylor polynomial: 𝑢 . 𝑦 = 10 + 𝑦 − 5 𝑦 - − 𝑦 ! 2 + 5𝑦 $ 12 + 𝑦 . 24 − 𝑦 / 72 " 𝑔 𝑦 𝑒𝑦 Determine an approximated value for ∫ 1% A) -10.27 B) -11.77 C) 11.77 D) 10.27

Finite difference approximation For a given smooth function 𝑔 𝑦 , we want to calculate the derivative 𝑔′ 𝑦 at 𝑦 = 1. Suppose we don’t know how to compute the analytical expression for 𝑔′ 𝑦 , but we have available a code that evaluates the function value: We know that: 𝑔 𝑦 + ℎ − 𝑔(𝑦) 𝑔′ 𝑦 = lim ℎ 2→4 Can we just use 𝑔′ 𝑦 ≈ > ?@A B> ? ? How do we choose ℎ ? Can we get A estimate the error of our approximation?

For a differentiable function 𝑔: ℛ → ℛ , the derivative is defined as: 𝑔 𝑦 + ℎ − 𝑔(𝑦) 𝑔′ 𝑦 = lim ℎ +→- Let’s consider the finite difference approximation to the first derivative as 𝑔′ 𝑦 ≈ 𝑔 𝑦 + ℎ − 𝑔 𝑦 ℎ Where ℎ is often called a “perturbation”, i.e. a “small” change to the variable 𝑦 . By the Taylor’s theorem we can write: 𝑔 𝑦 + ℎ = 𝑔 𝑦 + 𝑔 . 𝑦 ℎ + 𝑔′′(𝜊) ℎ ! 2 For some 𝜊 ∈ [𝑦, 𝑦 + ℎ] . Rearranging the above we get: 𝑔 . 𝑦 = 𝑔 𝑦 + ℎ − 𝑔(𝑦) − 𝑔′′(𝜊) ℎ ℎ 2 + Therefore, the truncation error of the finite difference approximation is bounded by M ! , where M is a bound on 𝑔 .. 𝜊 for 𝜊 near 𝑦 .

Demo: Finite Difference 𝑓𝑠𝑠𝑝𝑠 ℎ 𝑔 𝑦 = 𝑓 0 − 2 We want to obtain an approximation for 𝑔′ 1 𝑒𝑔𝑓𝑦𝑏𝑑𝑢 = 𝑓 0 𝑒𝑔𝑏𝑞𝑞𝑠𝑝𝑦 = 𝑓 0*1 − 2 − (𝑓 0 −2) ℎ 𝑓𝑠𝑠𝑝𝑠(ℎ) = 𝑏𝑐𝑡(𝑒𝑔𝑓𝑦𝑏𝑑𝑢 − 𝑒𝑔𝑏𝑞𝑞𝑠𝑝𝑦) 𝑓𝑠𝑠𝑝𝑠 < 𝑔 "" 𝜊 ℎ 2 truncation error

Demo: Finite Difference Should we just keep decreasing the perturbation ℎ , 𝑔 𝑦 + ℎ − 𝑔(𝑦) 𝑔′ 𝑦 = lim in order to approach the limit ℎ → 0 and obtain a ℎ +→- better approximation for the derivative?

What happened here? Uh-Oh ! 𝑔 𝑦 = 𝑓 0 − 2 𝑔′ 𝑦 = 𝑓 0 → 𝑔′ 1 ≈ 2.7 𝑔 1 + ℎ − 𝑔(1) 𝑔′ 1 = lim ℎ +→- Rounding error! 1) for a “very small” ℎ ( ℎ < 𝜗 ) → 𝑔 1 + ℎ = 𝑔(1) → 𝑔′ 1 = 0 2) for other still “small” ℎ ( ℎ > 𝜗 ) → 𝑔 1 + ℎ − 𝑔 1 gives results with fewer significant digits (We will later define the meaning of the quantity 𝜗 )

𝑓𝑠𝑠𝑝𝑠~𝑁 ℎ Truncation error: Minimize the error 2 2𝜗 ℎ + 𝑁 ℎ 2 𝑓𝑠𝑠𝑝𝑠~ 2𝜗 Gives Rounding error: ℎ = 2 𝜗/𝑁 ℎ