Section 3C

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Generalities

Many numbers come up as a result of a measurement

The measurements, however, are never exact

Sometimes the numbers we see are already rounded

Sometimes the numbers may be only rough estimates and/or projections

Today we learn how to operate with these uncertainties.

Significant digits

These are the digits that represent actual measurement.

There is a substantial difference between

"My weight is 154 pounds" and

"My weight is 154.00 pounds"

In the case of 154 pounds it may well be in fact 154.27, while

that cannot be the case when the number is 154.00

These two zeros are significant because they tell something.

General idea on whether a digit is significant or not:

While a non-zero digit is always significant,

a zero may be significant, and may be not.

That depends on what this zero wants to tell you.

A zero is $$not$$ significant if it stays there only to keep the position for a correct order of magnitude. To check up: this non-significant zero disappears in scientific notation.

Otherwise a zero is significant.

Significant digits: summary (p.152)

Always significant:

nonzero digits;

zeros right from decimal point $$and$$ right from a nonzero digit
example: 4.20, 3.00, but not 60, and not 0.01

zeros between non-zero digits (as in 302), or other significant zeros (as in 10.0).

Not necessarily significant digits: summary (p.152)

Always non-significant:

Zeros to the left of the first non-zero digit. Example: 0.00036

Not significant unless stated otherwise:

zeros to the right of the last non-zero digit but before the decimal point. Examples: 500,000 or 340

Rounding ( Brief review on p.152)

Step 1 Decide about the last place to be kept.

Step 2 If the $$next$$ digit is below 5, round down, while if the next digit is 5 and above, round up.

Example: 398.3 rounded to the nearest one is 398

Example: 398.3 rounded to the nearest ten is 400

Measurement errors

These errors may be random and systematic.
Random errors are always with us. They are not predictable, but we typically can bound them.
Systematic errors are embedded in our measurement system. They are typically related to a problem with the system.

Size of errors

Is a two pound error of a scale big or small?

That may depend on what is weighted and why that thing is weighted

There are two ways to measure an error:

$$\text{absolute error} = \text{measured value} - \text{true value}$$

$$\text{relative error} = \frac{\text{measured value} - \text{true value}}{\text{true value}} \times 100\% = \frac{\text{absolute error}}{\text{true value} }\times 100\%$$

Both absolute and relative errors may be important.

Example: 2 pounds of absolute error is a small and acceptable error if you weight a truck because the relative error is small.

Example: 2 pounds of absolute error is too big and not acceptable if you weight a puppy because the relative error is big.

Example: 2 pounds of absolute error is too big and not acceptable if you weight an artificial satellite although the relative error may be small. It is the absolute error which may affect the trajectory badly.

Accuracy and Precision

Precision is about the amount of digits, while the accuracy is about being close to the true value.

Example. My true weight is 154 pounds.
A digital scale shows 151.23 pounds while a mechanical scale shows 155 pounds.
The digital scale is more precise, while the mechanical is more accurate.

Operations with numbers which you know only approximately

Rounding $$after$$ the operations are performed.

Rules to follow

round to the precision of the least precise number used. Indeed, if you know one weight up to a pound, and another one up to a milligram, you still know the sum no better than up to a pound.

Multiplication/division:
round such that the amount of significant digits in your answer is the same as the fewest in your operands.

Operations with numbers which you know only approximately ... examples

Exercise 64, p. 161

At a hardware store, you buy a 55-kilogram bag of sand. You also buy a 1.25 kilograms of nails. What is the total weight of your purchases?

Solution. While clearly
$$55 + 1.25 = 56.25$$,
this is not yet a good answer. One has to round it.

Exercise 64, p. 161 ... continued

While you know the weight of the nails up to a hundredth of kilogram, you know the weight of sand only up to a kilogram.
The least precise operand is the weight of sand, and we have to round our sum $$55 + 1.25 = 56.25$$ up to a kilogram.
60,000= $$6 \times 10^4$$ has only one significant digits. Thus we have to round the product
$$60000 \times 445.79 = 26474000$$
Answer: $$\ 30000000 = 3 \times 10^7 =30$$ million dollars.