arithemetic errors - comprehensive study approach

Propagation of errors is when error is accounted for in arithmetic error operations with the measured values. The measured values are being used for addition, subtraction, multiplication and division along with their errors.

What do we do with errors when we have found them?

propagation of errors in Addition and subtraction

Addition and subtraction is when we account for errors while summing up values that have errors. we discuss this with an example:

consider the rectangle with lengths 14.75 cm, 8.96, cm, 14.75 cm, 8.96 cm as shown.

A rectangle measured with propagation of errors

suppose we need to find it’s perimeter. To achieve that, we are to add the lengths of each side. we have the following measurements:

The errors in measurement of length = (1/2) x 0.01 = 0.005 cm

The errors in measurement of width= (1/2) x 0.01 = 0.005 cm

maximum possible length = 14.755 cm

minimum possible length = 14.745 cm

maximum possible width = 8.965 cm

minimum possible width = 8.955 cm

because we have upper and lower limits of lengths then we have 3 possible sums that gives the perimeter:

when we use the upper limits in measured lengths to find perimeter
when we use the lower limits in measured lengths to find perimeter
perimeter when we use the actual measured values without errors.

maximum possible perimeter =14.755 cm + 14.755 cm + 8.965 cm +8.965 cm = 47.44 cm

minimum possible perimeter = 14. 745 cm + 14. 745 cm + 8.955 cm + 8.955 cm = 47.40 cm

working perimeter = 14.75 cm + 14.75 cm + 8.96 cm + 8.96 cm =47.42 cm

absolute error in sum = maximum sum – working sum

That is : | 47.44 – 47.42 | = 0.02

similarly; absolute error = | working sum – minimum sum | =| 47.42 – 47.40 | = 0.02

$$\text{relative error in perimeter} = \frac{0.02}{47.42} = 0.000422 $$

Example in propagation of errors

Calculate error in 1.23 m – 0.67 m

solution

$$ \text{The measurements lie within } 1.23 \pm 0.005 \text{ and } 0.67 \pm 0.005 $$

The maximum possible difference will be obtained when we subtract the minimum value of subtrahend from maximum of the minuend.

that is, maximum difference = 1.235 – 0.665 = 0.57

The minimum possible difference is when we subtract the the maximum value of subtrahend from minimum of the minuend.

that is; minimum =1.225 – 0.675 = 0.55

the working difference = 1.23 – 0.67 = 0.56

absolute error = | 0.56 – 0.55| or | 0.57 – 0.56 | = 0.01

propagation of errors in Multiplication: product errors

To calculate errors involving multiplication, we need to find working area, maximum possible area and the minimum possible area.

consider our earlier rectangle which we redraw here:

suppose we need to find it’s area.

$$ \text{The measurements lie within } 14.75 \pm 0.005 \text{ and } 8.96 \pm 0.005 $$

$$ \text{maximum possible error} = 14.755 \times 8.965 = 132.278575 $$

$$ \text{minimum possible error} = 14.745 \times 8.955 = 132.041475 $$

$$ \text{working area} = 14.75 \times 8.96 = 132.1600 $$

The working area is NOT midway between the lower and the upper limits

$$ \text{maximum area} – \text{working area} = 132.278575 – 132.1600 = 0.118575$$

$$ \text{working area} – \text{minimum area}= 132.1600 – 132. 041475= 0.118525$$

The absolute error in products is the average of the two differences above expressed as:

$$\text{absolute error in area} = \frac{0.118575 + 0.118525}{2} = 0.11855$$

the absolute error in area can also be calculated from average difference between the maximum area and minimum area

that is :

$$\text{absolute error in area} = \frac{1}{2}{\text{maximum area}-\text{minimum area}}$$

in our example: we have

$$\text{absolute error in area} = \frac{1}{2}({132.278575}-{132.041475}) =\frac{1}{2} \times 0.2371 = 0.11855$$

calculating relative error in the area

$$\text{relative error in area} = \frac{\text{absolute error in area}}{\text{working area}} $$

in our case:

$$\text{relative error in area} = \frac{0.11855}{132.1600} = 0.000897 $$

Alternative procedure of calculating relative error in area

relative error in product can also be calculated from relative errors of the individual measurements. Relative error can also be expressed as the sum of the relative errors in individual measurements.

for example considering our triangle.

$$\text{relative error in length} = \frac{0.005}{14.75} = 0.000339$$

$$\text{relative error in width} = \frac{0.005}{8.96} =0.000558$$

$$\text{relative error in area } = ( 0.000339 + 0.000558) = 0.000897$$

Error in quotient (Division error)

While working on relative error in division, we calculate largest possible quotient, smallest possible quotient and the working quotient.

consider finding density of substance given it’s mass as 5.79 g and it’s volume as 4.63 cm³ .

$$Density (\rho) = \frac{mass}{volume} $$

$$Density (\rho) = \frac{5.79}{4.63} = 1.25054$$

The maximum possible quotient will be given by the maximum possible value of the numerator and the smallest possible value of the denominator.

$$ \text{maximum possible quotient} = \frac{\text{maximum numerator}}{\text{miminum numerator} }$$

in our example:

$$ \text{maximum possible quotient} = \frac{5.795}{4.625 } = 1.25297$$

$$ \text{minimum possible quotient} = \frac{5.785}{4.635 } = 1.24811$$

$$\text{absolute error in quotient} = \frac{1}{2} \times (\text{maximum possible quotient} – \text{minimum possible quotient})$$

$$\text{absolute error in quotient} = \frac{1}{2} \times (1.25297 – 1.24811)=\frac{1}{2}\times 0.00486=0.00243$$

Relative error in quotient

$$\text{Relative error in quotient} = \frac{\text{absolute error in quotient}}{\text{working quotient}} $$

in our example:

$$\text{Relative error in quotient} = \frac{0.00243}{1.25054} = 0.00194 $$

Tag: arithemetic errors

propagation of errors: Arithmetic errors