Uncertainty Calculations (Multiplication) - Wilfrid Laurier...

Calculations with UncertaintiesRecap

Uncertainty Calculations - MultiplicationWilfrid Laurier University

Terry Sturtevant

Wilfrid Laurier University

May 9, 2013

Terry Sturtevant Uncertainty Calculations - Multiplication Wilfrid Laurier University

Multiplication by a constantMultiplication with Multiple UncertaintiesMultiplication with Multiple Uncertainties

Calculations with uncertainties

When quantities with uncertainties are combined, the resultshave uncertainties as well.

Following is a discussion of multiplication.

For the following examples, the values of x = 2± 1 andy = 32.0± 0.2 will be used.

Multiplication by a constant

Multiplication by a constant with uncertainties

Multiplication by a constant

Multiplication by a constant with uncertainties

Multiplication by a constant - Example

Suppose we have a number with an uncertainty, and wemultiply it by a constant.

(A constant is a number with no uncertainty.)

What happens to the uncertainty?

x = 2± 1

→ 4x can be as small as 4× 1 = 4

since 2− 1 = 1

→ 4x can be as big as 4× 3 = 12

since 2 + 1 = 3

so 4x = 8± 4 = (4× 2)± (4× 1)

x = 2± 1

since 2− 1 = 1

since 2 + 1 = 3

so 4x = 8± 4 = (4× 2)± (4× 1)

x = 2± 1

since 2− 1 = 1

since 2 + 1 = 3

so 4x = 8± 4 = (4× 2)± (4× 1)

x = 2± 1

since 2− 1 = 1

since 2 + 1 = 3

so 4x = 8± 4 = (4× 2)± (4× 1)

x = 2± 1

since 2− 1 = 1

since 2 + 1 = 3

so 4x = 8± 4 = (4× 2)± (4× 1)

x = 2± 1

since 2− 1 = 1

since 2 + 1 = 3

so 4x = 8± 4 = (4× 2)± (4× 1)

x = 2± 1

since 2− 1 = 1

since 2 + 1 = 3

so 4x = 8± 4 = (4× 2)± (4× 1)

x = 2± 1

since 2− 1 = 1

since 2 + 1 = 3

so 4x = 8± 4 = (4× 2)± (4× 1)

x = 2± 1

since 2− 1 = 1

since 2 + 1 = 3

so 4x = 8± 4 = (4× 2)± (4× 1)

Graphically,

The nominal value of x is here. (i.e. the value without consid-ering uncertainties)

Graphically,

If we multiply by 1/2, both x and ∆x get smaller.

To summarize,

So in general, ∆ (Cx) = C∆x

When multiplying by a constant, we multiply theuncertainty by the constant as well.

To summarize,

Multiplication with Multiple Uncertainties

What if both numbers have uncertainties?

Multiplication with Multiple Uncertainties

What if both numbers have uncertainties?

Multiplication with Multiple Uncertainties - Example

If we multiply these numbers,

z = (x = 2± 1)× (y = 32.0± 0.2)

→ z can be as small as 1× 31.8 = 31.8

since x can be as small as 1 and y can be as small as 31.8

→ z can be as big as 3× 32.2 = 96.6

since x can be as big as 3 and y can be as big as 32.2

z = (x = 2± 1)× (y = 32.0± 0.2)

→ z can be as big as 3× 32.2 = 96.6

z = (x = 2± 1)× (y = 32.0± 0.2)

→ z can be as big as 3× 32.2 = 96.6

z = (x = 2± 1)× (y = 32.0± 0.2)

→ z can be as big as 3× 32.2 = 96.6

z = (x = 2± 1)× (y = 32.0± 0.2)

→ z can be as big as 3× 32.2 = 96.6

z = (x = 2± 1)× (y = 32.0± 0.2)

→ z can be as big as 3× 32.2 = 96.6

z = (x = 2± 1)× (y = 32.0± 0.2)

→ z can be as big as 3× 32.2 = 96.6

since x can be as big as 3

and y can be as big as 32.2

z = (x = 2± 1)× (y = 32.0± 0.2)

→ z can be as big as 3× 32.2 = 96.6

z = (x = 2± 1)× (y = 32.0± 0.2)

→ z can be as big as 3× 32.2 = 96.6

To summarize,

z can be as small as 1× 31.8 = 31.8

z can be as big as 3× 32.2 = 96.6

The nominal value of z is

z = 2× 32.0 = 64.0

So we can say z ≈ 64.0± 32.4

and we see that ∆z ≈ 32.4 =(

12 + 0.2

)64.0 =

(∆xx + ∆y

So in general, ∆ (xy) = xy(

∆xx + ∆y

)When multiplying numbers, we add proportionaluncertainties.

To summarize,

z can be as big as 3× 32.2 = 96.6

z = 2× 32.0 = 64.0

So we can say z ≈ 64.0± 32.4

12 + 0.2

)64.0 =

(∆xx + ∆y

∆xx + ∆y

To summarize,

z can be as big as 3× 32.2 = 96.6

z = 2× 32.0 = 64.0

So we can say z ≈ 64.0± 32.4

12 + 0.2

)64.0 =

(∆xx + ∆y

∆xx + ∆y

To summarize,

z can be as big as 3× 32.2 = 96.6

z = 2× 32.0 = 64.0

So we can say z ≈ 64.0± 32.4

12 + 0.2

)64.0 =

(∆xx + ∆y

∆xx + ∆y

To summarize,

z can be as big as 3× 32.2 = 96.6

z = 2× 32.0 = 64.0

So we can say z ≈ 64.0± 32.4

12 + 0.2

)64.0 =

(∆xx + ∆y

∆xx + ∆y

To summarize,

z can be as big as 3× 32.2 = 96.6

z = 2× 32.0 = 64.0

So we can say z ≈ 64.0± 32.4

12 + 0.2

)64.0 =

(∆xx + ∆y

∆xx + ∆y

To summarize,

z can be as big as 3× 32.2 = 96.6

z = 2× 32.0 = 64.0

So we can say z ≈ 64.0± 32.4

12 + 0.2

)64.0 =

(∆xx + ∆y

∆xx + ∆y

To summarize,

z can be as big as 3× 32.2 = 96.6

z = 2× 32.0 = 64.0

So we can say z ≈ 64.0± 32.4

12 + 0.2

)64.0 =

(∆xx + ∆y

∆xx + ∆y

When multiplying numbers, we add proportionaluncertainties.

To summarize,

z can be as big as 3× 32.2 = 96.6

z = 2× 32.0 = 64.0

So we can say z ≈ 64.0± 32.4

12 + 0.2

)64.0 =

(∆xx + ∆y

∆xx + ∆y

To show this graphically, remember that the product of twonumbers is the area of a rectangle with sides equal to the twolengths.

For illustration purposes, we’ll only show uncertainties in onedirection.

Just remember that uncertainties can be in either direction.

Graphically,

This is the nominal value of x

Graphically,

x ∆x

The maximum value of x includes ∆x .

Graphically,

This is the nominal value of y .

Graphically,

The maximum value of y includes ∆y .

Graphically,

This is the nominal value of the area; i.e. xy .

Graphically,

max(xy)

This is the maximum value of the area; i.e. (x + ∆x) ×(y + ∆y).

Graphically,

This is the difference between the nominal value of the areaand the maximum value of the area.

Graphically,

This part of the difference has a size of x∆y

Graphically,

This part of the difference has a size of y∆x

Graphically,

∆x∆y

This part of the difference has a size of ∆x∆y

Graphically,

∆x∆y

Because this is relatively small, we’ll ignore it.

Graphically,

This is approximately the difference, and has a size of y∆x +x∆y

So finally,

If the uncertainty in xy is approximately equal to y∆x + x∆y ,or

∆ (xy) ≈ y∆x + x∆y

We can multiply both the top and bottom of the right side byxy so we get

∆ (xy) ≈ xyxy (y∆x + x∆y)

which becomes

∆ (xy) ≈ (xy)(

∆xx + ∆y

)Remember that if x or y can be negative, we’ll needabsolute value signs around the appropriate terms, sinceuncertainty contributions should always be given aspositive numbers.

So finally,

∆ (xy) ≈ y∆x + x∆y

which becomes

∆ (xy) ≈ (xy)(

∆xx + ∆y

So finally,

∆ (xy) ≈ y∆x + x∆y

which becomes

∆ (xy) ≈ (xy)(

∆xx + ∆y

So finally,

∆ (xy) ≈ y∆x + x∆y

which becomes

∆ (xy) ≈ (xy)(

∆xx + ∆y

So finally,

∆ (xy) ≈ y∆x + x∆y

which becomes

∆ (xy) ≈ (xy)(

∆xx + ∆y

So finally,

∆ (xy) ≈ y∆x + x∆y

which becomes

∆ (xy) ≈ (xy)(

∆xx + ∆y

So finally,

∆ (xy) ≈ y∆x + x∆y

which becomes

∆ (xy) ≈ (xy)(

∆xx + ∆y

Remember that if x or y can be negative, we’ll needabsolute value signs around the appropriate terms, sinceuncertainty contributions should always be given aspositive numbers.

So finally,

∆ (xy) ≈ y∆x + x∆y

which becomes

∆ (xy) ≈ (xy)(

∆xx + ∆y

1. When multiplying by a constant, the uncertainty getsmultiplied by the constant as well.

4× (2± 1) = (4× 2)± (4× 1) = 8± 4

2. When multiplying numbers, we add the proportionaluncertainties.

(2± 1)× (32.0± 0.2) = (2× 32.0)± (2× 32.0)

)= 64.0± 64.0 (0.5 + 0.00625)

= 64.0± 32.4

4× (2± 1) = (4× 2)± (4× 1) = 8± 4

(2± 1)× (32.0± 0.2) = (2× 32.0)± (2× 32.0)

)= 64.0± 64.0 (0.5 + 0.00625)

= 64.0± 32.4

4× (2± 1) = (4× 2)± (4× 1) = 8± 4

(2± 1)× (32.0± 0.2) = (2× 32.0)± (2× 32.0)

)= 64.0± 64.0 (0.5 + 0.00625)

= 64.0± 32.4

4× (2± 1) = (4× 2)± (4× 1) = 8± 4

(2± 1)× (32.0± 0.2) = (2× 32.0)± (2× 32.0)

)= 64.0± 64.0 (0.5 + 0.00625)

= 64.0± 32.4

4× (2± 1) = (4× 2)± (4× 1) = 8± 4

(2± 1)× (32.0± 0.2) = (2× 32.0)± (2× 32.0)

)= 64.0± 64.0 (0.5 + 0.00625)

= 64.0± 32.4

Recap - continued

3. Uncertainties in final results are usually expressed to onesignificant figure, so the above result becomes

64.0± 32.4 = 60± 30

Recap - continued

3. Uncertainties in final results are usually expressed to onesignificant figure, so the above result becomes

64.0± 32.4 = 60± 30

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