Errors - Approximated Number

Errors 1. Approximated number: "a" is an approximated value of "A" if "a" is slightly different from "A" and may replace it in the calculation. We write a A Example a= √2 →1.41 rounding down 1.42 rounding up 2. Errors a) Absolute error: /Aa|=|A-al 1A-a/SA/A is the upper bound of Aal -ΔΣ Α-ας > - Δ+as As Ata b) Relative error: Sa= |Da~ Aal A lal 3. Origin of the errors a) Error related to the position of the problem. Ex: An irregular plate has different -1- temperature across its parts. If We Want to calculate temperature of each regular plate in Imm, we consider a order to simplify calculation. b) Error related to infinite process. Sinxxx 3 31 + truncation error c) Error related to the numerical system: 0.333 rounding off. = 3 Example: 3.141/5 3.142 3.144/5 3.144 3.141/6 3.142 3.142/6 3.143 d) Error related to the implementation of a number in the Computer: Normalized representation of a decimal Exponent + sign of the number mantissa 2- o.aadadda load mantissa In order to avoid a sign for the exponent, We take 50 as origin. Ex:.6.02 3 x 10 23 0.60 23 x lo 24 + 746023000 .. 6.023 × 10-23 0.60 23x10 -22 + 286023000 Remark: If the mantissa has more than 7 digits, we will have truncation error. 4. Majoration of the error in function evaluation a) Error on one variable function (y = f(x)) We propose the evaluation of f(x) for x= a where f is differentiable in neighberhood of x=a. We replace "a" by an approximated value "a'". By applying MVT (mean value theorem), -3- we get: f'(x) = fla)-f(a') a-a Afla) = (f'(x)) Δα Af(x)/5 M, 14x) where M₁ = sup/ f'(x)) [a', a] b) Multiple variable function: f(x,y,3) By applying mean value theorem, there B, ४) ∈ Ja, a'[, ]b, b'[, ]c, c'[ Such that f(a', b', c') = f(a,b,c) + [fx (2,3,r)] (a-a) + [fy(d, B, 8)] (b-b) + [fo (a, B, 8)] (c-c) => f(a, b, c')-f(a,b,c) = fxAa + fy Ab + foc y lofs M,/AX + M₂/Ay + M3/A31 2 where M₁ = Sup | f'(x); M₂= Sup|fy|; XE [α, α] M3= sup/fl ZE [C] [b, b] -4- Ex: We would like to calculate f(x, y) = sqrt(4 + xy) for x = 10.25 plus/minus 0.02 and y =9.75 pm0. 5 Estimate f(10.25, 9.75). Then, determine the associated error to this approximation. f(io * 0.25, 9.75) = sqrt(4 + (io * 0.25)(9.75)) = |Delta*f| <= M_{1}|Delta*x| + M_{2}|Delta*y| |Delta*x| = 0.02 ; |Delta*y| =0. 5 M prime = sup[f_{x}'] 10.195 10.25-0.02; 10.25 +0.02 rfloor M_{2} = sup|f_{y}'| M_{i} = 0.479 y*epsilon [9.75-0. 5 ; 9.75 +0.05] f_{y}' = x/(2sqrt(4 + xy)) = (io * 0.25)/(2sqrt(4 + io * 0.25y)) f_{x}' = y/(2sqrt(4 + xy)) = 9.75/(2sqrt(4 + 9.75x)) |f x ^ prime | longrightarrow fhen*M_{i} = 9.75 2 sqrt 4+9.75(i0.25\ 0.02) -5- 2/4+10.25y then M2= 10.25 = 0.5039 24+10.25 (9.75-0.05) 1Afls (0.479) (0.02) + (0.5039) (0.05) 50.0347 5. Significant digits: Significant digit of an approximated number is a digit i) ≠ 0 ii) The zeros between 2 significant digits ≠0. iii) The zeros following the last significant digit ≠0. Example: 0.00 78 0 0 807000 ii LLL ii -6- 6. Exact significant digits Let "A" be the exact value and "a" is the approximated value of "A" The first n significants digits of "a" are exact if | | Delta a|=|A - a|<= i 2 | e ^ (m - n + 1) with α then am) (m-1), am_n+, are exacts. m7 a= alpha m | 0 ^ m + alpha m-1 |o^ m-1 +...+ alpha lo^ m-n+1 +... Ex: A = 35.9 a = 36 Determine the number of exact significant digits that "a" approaches "A" |Delta*a| = |A - a| =|36. -3.97|=0. 3 + 6x10+0x101/40x10² < a' approaches "A" by 3 exact Significant digits. 2 0. 3 <= 1/2 * 10 ^ - 2 * X 3 <= 1/2 * 10 ^ - 1 3 <= 1/2 -7-