A-level Mathematics/MEI/NM/Approximation/Errors

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Approximations

Even when we are trying to be as precise as possible, it is often very hard to give the exact answer to something. Like if you were to measure your height, because of different factors the value you get is likely not your actual height; but it is very close to your actual height. for example, the nearest centimetre. This value, would be an approximation of your actual height.

If we take $e$, as an example, you could never write out $e$ exactly, as it is irrational; it goes on forever. The first 100 decimal places of $e$ are:

2. 7182818284 5904523536 0287471352 6624977572 4709369995 9574966967 6277240766 3035354759 4571382178 5251664274

However precise this number is to the actual value of $e$, it is still an approximation.

Approximations occur because of a few key reasons:

• Problems with accurate measurements
• Rounding
• Simplifying a model
• Computers being used can only use a finite number of decimal places
• It isn't worth getting any more accurate

Errors

At the time of writing this, the 'population clock' run by the USA states that there are 316,356,429 people living in the USA. However, many sources will say that the population is 300,000,000. This means that these sources are approximating the value.

To find the error in this approximation, we do:

$300000000 - 316356429 = -16356429$

Mei define an error

'When an exact value x is approximated by X, the error ε is given by: ε = X - x

Notice that if the approximation is too big the error is positive, and if the approximation is too small the error is negative.'

To put it simply, the error is a measure of the distance between and estimate of the value, and the actual value.

As well as this, we have what is called the absolute error, which is effectively the magnitude of the error.

Mei define it as this:

'When an exact value x is approximated by X, the absolute error is defined as the modulus of the error.

absolute error = |ε| = |X - x|'

The modulus of the errors means its magnitude; how big it is. The positive value of that number.

Although knowing the errors and absolute errors of estimates is very good, it is hard to compare them. Say I'm comparing two measurement techniques.One looks at the sizes of atoms, and the other looks at the sizes of galaxies. The error is automatically going to be greater with the measurements of the galaxies as it is just so much larger than an atom, which isn't a very fair comparison. This is when we use something called the relative error.

Mei define the relative error as:

'A useful measure of error is the relative error. This is defined by:

$\text {relative error} = \frac{\text {error}}{\text {exact value}} = \frac{\epsilon}{x}$'

The relative error is measuring the ratio of the error to the exact value, making it a better representation of the error's you're dealing with.

There like before, we have a different version of the relative error;Absolute relative error

Mei define it as:

'When an exact value x is approximated by X, the absolute relative error is the modulus of the relative error, and is defined as:

$\text {absolute relative error} = \left| \frac{\text{error}}{\text{exact value}} \right| = \left| \frac{\epsilon}{x} \right| = \left| \frac{X - x}{x} \right|$'

Again, it is the ratio between error and exact value, although this time the value is always positive.

Rounding numbers and Interval estimates

Rounding

Rounding numbers is a way we can simplify a number where its full value is not necessary.

Usually numbers are rounded to a certain number of either decimal places, or significant figures. The new rounded value is now an approximation of the original.

If x is the number of places we are to round to,then when we round to x number of decimal places, it means we round x many places after the decimal point. If we round to x number of significant figures, we round x places after the first non-zero term.

example

1. 'round 48.7564 to 2 decimal places'

solution
look at the 3rd decimal place, it is greater than 5. This means, we will be rounding up.
⇒ =48.76

2. 'round 690354.23 to 4 significant figures'

solution
Look at the 5 digit. It is less than 5, so we round down
⇒ =690300

It is usual to write the number of significant figures (s.f) the number has been rounded to after it, as with our last example, 690300 could be either the to 4s.fs, 5s.fs, or an exact value. Our one would be =690300(to 4 s.f)

Interval Estimates

Think about the following statement: 'the value of y, correct to 1 decimal place, is 2.5' This means that if you were to round y to 1 decimal place, you would get an answer of 2.5. This means that y could be any value that round to 2.5 to 1 decimal place.

To help get over this, we must think about the lowest number, and the highest number y can take for it to round to 2.5. These numbers would be 2.45, and 2.55. So our value y lies between 2.45, and 2.55. We can write this like this:

2.45≤y<2.55

This means that y is more than or equal to 2.45, but less than 2.55. It can not be equal to 2.55 as this would round to 2.6, however, 2.549999999999999... would round to 2.5, and this is effectively 2.55.

This is an example of an interval. It could also be written like so: [2.45,2.55), where the square bracket include the number being talked about, and the round bracket does not include the number. In this case, 2.45 is said to be the lower bound of the number, and 2.55 is said to be the upper bound.

Any statement where the value of a number,x, is between two numbers, is called an interval estimate for x.

This is what MEI says:

'When an exact value, x, is known to be between two values a and b, there is an interval estimate to x.

This could take one of the following forms:
a < x < b, a < x ≤ b, a ≤ x < b, a ≤ x ≤ b.

"The value a is said to be a lower bound for x and the value b is said to be an upper bound for x.'

It is also possible to obtain an approximation of a value from its interval estimate; for example:

In the interval 1.13≤x<1.141, we want to find x correct to the most decimal places we can. we can see that both of these numbers round to a value of 1.1 to 1 decimal place. This is the nearest we can get to its actual value, as to 2 decimal places, it could either be 1.3, or 1.141.

Now, in the interval estimate 7.8≤x<8, we want to find the best approximation we can of the value x. This would mean that the absolute error of this value would be as small as possible. If we were to take the midpoint of 7.8 and 8, we would find that 7.9 is the midpoint. It is exactly 0.1 away from each of these numbers, so this means that our best estimate of x is 7.9, with a maximum absolute error of 0.1

 To do: Content practically done, however needs to be rewitten in the future. More examples and questions in the future. ~ollz272