User:Murad Semed

Error in Addition
Let's consider that X and Y have an approximation value of x and y respectively.

If we let Z=X+Y, then X=x+∆x, Y=y+∆y, and Z=z+∆z with z=x+y. Thus Z=X+Y⟹z+∆z=x+∆x+y+∆y⟹∆z=x+∆x+y+∆y-(x+y). ∆z=∆x+∆y ⟹|∆z|=|∆x+∆y|≤|∆x|+|∆y|⟹|∆z|≤|∆x|+|∆y| which is the absolute error. The relative error is R_er=|∆z|/Z≤(|∆x|+|∆y|)/(X+Y).

Error in subtraction
If we let Z=X-Y, then X=x+∆x, Y=y+∆y, and Z=z+∆z with z=x-y. Thus Z=X-Y⟹z+∆z=x+∆x-(y+∆y),

⟹∆z=x+∆x-y-∆y-(x-y).

∆z=∆x-∆y

⟹|∆z|=|∆x+(-∆y)|≤|∆x|+|∆y|

⟹|∆z|≤|∆x|+|∆y| which is the absolute error.

The relative error is R_er=|∆z|/Z≤(|∆x|+|∆y|)/(X-Y).

Example 1. If one is calculating X+Y where X = 1.5 ± 0.05, and Y = 3.4 ± 0.04. Find the bounds for the propagation error in adding two numbers.

Solution. By looking at the numbers, the maximum possible values of X and Y are X = 1.55 and Y = 3.44. Hence X +Y = 1.55 + 3.44 = 4.99  is the maximum value of X + Y. The minimum possible values of X and Y are X = 1.45 and Y = 3.36. Hence X + Y = 1.45 + 3.36 = 4.81 is the minimum value of X +Y.

Hence 4.81 ≤ X+Y≤4.99.

2. Given A=1.652 and B=9.78. Then find the maximum error in A - B.

Solution:- Let maximum error in A-B be denoted by E_(A-B), then E_(A-B)=|∆a|+|∆b| where Then E_(A-B)=|∆a|+|∆b|=0.0005+0.005=0.0055. Which is the same with E_(A+B).
 * ∆a|≤1/2×〖10〗^(-3)=0.0005 and  |∆b|≤1/2×〖10〗^(-2)=0.005.

3. Find the absolute error of the value of 3A-6B+9C, if a,b, and c are used to approximate the value of A, B, and C respectively, and let A=3.456, B=2.67 and C=1.6784.

Solution. 3A-6B+9C=3A+6(-B)+9C=3|∆a|+6|∆b|+9|∆c|=3(0.0005)+6(0.005)+9(0.00005)=0.03195.

Propagated error in Multiplication
If we let Z=XY, then X=x+∆x, Y=y+∆y, and Z=z+∆z with z=xy. Thus Z=XY⟹z+∆z=(x+∆x)(y+∆y), ∆z=xy+y∆x+x∆y+∆x∆y-z⟹∆z=y∆x+x∆y+∆x∆y. Now by neglecting ∆x∆y for its very small, we get ∆z=y∆x+x∆y

And its absolute error is |∆z|=|y∆x+x∆y|≤|y∆x|+|x∆y|.

Relative error is R_er=|∆z|/z=(|y∆x+x∆y| )/xy≤(|∆x|  )/x+(|∆y|  )/y.

Propagated error in division
If we let Z=X/Y, then X=x+∆x, Y=y+∆y, and Z=z+∆z with z=x/y. Thus, Z=X/Y⟹z+∆z=((x+∆x))/((y+∆y) )=(x+∆x)(y-∆y)/(y+∆y)(y-∆y) =(xy+y∆x-x∆y-∆x∆y)/(y^2-(∆y)^2 ). Now neglecting ∆x∆y and (∆y)^2, we get ∆z=(xy+y∆x-x∆y)/y^2 -x/y=(xy+y∆x-x∆y-xy)/y^2 =(y∆x-x∆y)/y^2

Then the absolute error is


 * ∆z|=|(y∆x-x∆y)/y^2  |≤(y|∆x|+x|∆y|)/y^2 =x/y [(|∆x|  )/x+(|∆y|  )/y]

Relative error is

R_er=|∆z|/z=|(y∆x-x∆y)/y^2  |/(x/y)≤(x/y  [(|∆x|  )/x+(|∆y|  )/y])/(x/y)=(|∆x|  )/x+(|∆y|  )/y

Remark:- If f(x) is a function of several variables x_1,x_2,…,x_(n-1),x_n, then the maximum possible value of the error in f is

∆f≈|∂f/(∂x_1 ) ∆x_1 |+|∂f/(∂x_2 ) ∆x_2 |+⋯+|∂f/(∂x_(n-1) ) ∆x_(n-1) |+|∂f/(∂x_n ) ∆x_n |.